Air Conditioning System Fundamentalsfor Architecture Students

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Authors Chung, Chenwu

Publisher The University of Arizona.

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AIR CONDITIONING SYSTEM FUNDAMENTALS FOR ARCHITECTURE STUDENTS

BY CHENWU CHUNG

A MASTER REPORT SUBMITTED TO THE FACULTY OF THE COLLEGE OF ARCHITECTURE IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF ARCHITECTURE UNIVERSITY OF ARIZONA 1989

COMMIITEE lledli

ACKNOWLEDGEMENT

I would like to thank the College of Architecture and all the

faculty and staff, especially Professor Matter and Professor Clark,

for their continous help and support throughout this study.

I would also like to thank my family for their support and

encouragement throughout the course of this study.

A special appreciation goes to my wife Mei who spent many

hours helping me in typing this report.

J

I

PREFACE

This report is not only submitted in partial fulfillment of

requirements for a Master of Architecture but also is designed to

be used as guide to help architectural professionals and students

in the selection and analysis of HVAC systems.

This report divided into four sections. Part I is an

introduction of basic concepts and terminologies, such as how

energy is transfered, sun -earth relations, thermal resistance, and

what is entropy. Part II is a review of basic HVAC systems and

principles, such as heating and cooling loads, variable volume

system, and Constant Volume system. In Part III, duct design will

be introduced, for example, low- velocity duct design. Part IV will

be applications of above principles and methods.

* Note: The number in [ ] refers to the number of the reference

book

z

PART I FUNDANMENTALS:

Chapter 1: Basic Thermodynamics and Heat Transfer

1 -0 Introduction

1 -1 Conservation of Energy

1 -2 Entropy and the Second Law of Thermodynamics

1 -3 Characteristics of Thermodynamic systems

1 -4 Conduction

1 -5 Convection

1 -6 Radiation

Chapter 2: Solar Radiation

2 -0 Introduction

2 -1 Sun -Earth relations

2 -2 Time

2 -3 Solar Radiations

Chapter 3: Comfort Zone and Heat Transmission

3 -0 Introduction

3 -1 Comfort Zone

3 -2 Heat Transmission in Structures

3 -3 Tabulated Overall Heat -Transfer Coefficients

Chapter 4: Moist Air Properties and The Conditioning Process

4 -0 Introduction

4 -1 Psychrometric Chart

4 -2 Classic Moist Air Processes

3

4 -3 Design Conditions

PART II HVAC SYSTEMS AND LOADS

Chapter 5: Air Conditioning System

5 -0 Introduction

5 -1 The Basic Central System

5 -2 All -Air Systems

5 -3 Air -and -Water Systems

5 -4 All -Water Systems

5 -5 Heat Pump Systems

5 -6 Heat Recovery Systems

5 -7 Economizer Systems

Chapter 6: Heating and Cooling Loads

6 -0 Introduction

6 -1 Heat Loads

6 -2 Cooling Loads and CLTD Method

PART III DUCTS SYSTEMS

Chapter 7 : Ducts Systems

7 -0 Introduction

7 -1 Fluid Flow Basics

7 -2 Air Flow In Ducts

4

7 -3 Air Flow In Fittings

7 -4 Turing Vanes and Dampers

7 -5 Duct Design Fundamentals and

Pressure Gradient Diagrams

7 -6 Equal- Friction Method

PART IV APPLICATIONS

Chapter 8 : Applications

8 -0 Introduction

8 -1 Plan and Elevations

8 -2 Analysis and Calcultions

APPENDIX

Appendix A: Glossary

Appendix B: Water Properties of Liquid and Vapor

Appendix C: References

5

PART I FUNDAMENTALS

Chapter 1: Basic Thermodynamics and Heat Transfer

Chapter 2: Solar Radiation

Chapter 3: Comfort Zone and Heat Transmission

Chapter 4: Moist Air Properties and Conditioning Process

e2

Chapter 1: Basic Thermodynamics and Heat Transfer

1 -0 Introduction

1 -1 Conservation of Energy

1 -2 Entropy and The Second Law

1 -3 Charcteristics of Thermodynamics System

1 -4 Conduction

1 -5 Convection

1 -6 Radiation

7

1 -0 Introduction

In this chapter the author will introduce the reader to one

of the most important areas of engineering science

thermodynamics.

From the study of thermodynamics, the reader can appreciate

the interactions of a system with its surroundings. However,

thermodynamics deals with the end states of the process during

which an interaction occurs and provides no information concerning

the nature of the interaction at which it occurs. Heat transfer

is used to explain how heat is transfered between the inside and

outside of a structure, therefore, heat transfer can show the

nature of the interaction which thermodynamics can not tell.

Thermodynamics is concerned about the notion of energy; the

idea that energy is conserved is the first law of thermodynamics.

It is the starting point for the science of thermodynamics and for

engineering analysis. A second concept in thermodynamics is

entropy; entropy provides a means for determining if a process is

possible, the definition of entropy will be introduced latter.

Processes which produce entropy are possible; those which destroy

entropy are impossible. This idea is the basis for the second law

of thermodynamics.

Heat transfer is energy in transit due to a temperature

difference. Whenever there exists a temperature difference in a

medium or between media, heat transfer must occur. When a

temperature gradient exists in a stationary medium, which may be

solid or fluid, we use the term conduction to refer to the heat

transfer that will occur across the medium. In contrast, the term

8

convection refers to heat transfer that will occur between a

surface and a moving fluid when they are at different temperatures.

The third mode of heat transfer is termed thermal radiation. All

surfaces of finite temperature emit energy in the form of

electromagnetic waves. Hence, in the absence of an intervening

medium, there is net heat transfer by radiation between two

surfaces at different temperatures.

1 -1 Conservation of Energy

A fundamental aspect of the energy concept is that energy is

conserved, that is, the energy of an insolated system is constant.

The term system mentioned above needs to be explained. We shall

use the term system in a very broad sense to identify the subject

of discussion or analysis. The system is something defined by the

analyst for the particular problem at hand. The system is whatever

we wish to discuss, and we must be very careful each time to

describle precisely just what it is that we are talking about.

Having carefully defined the system, everything else is

automatically its environment. The interactions between a system

and its environment are the main

interest in thermodynamics.

"Since the system and its environment form an isolated system,

if the energy of the system increases the energy of the environment

must decrease a corresponding amount in order to maintain overall

conservation of energy. We can therefore view the interaction as

a process of energy transfer, and work is one of the mechanisms for

such energy transfer.

The amount of energy transfer to a system as work (the "work

done on the system ") associated with some infinitesimal change in

the position of matter inside is

1/1,2= dx J (1-1)

Here F is a force exerted by the environment on matter within the

system, and dx is the infinitesimal motion of that matter in the

direction of F which occurs during the period of observation.

10

The piston -cylinder system provides an easy way to appreciate

and easy to compute energy transfer as work associated with a

change in the fluid volume. The force (F) per unit area exerted

by the fluid on the piston is the fluid pressure P. Denoting the

piston area by A, the force on the piston is PA, and the amount of

energy transfer as work from the fluid to the piston. "[1]

1^-)11: 1-12F de CoNTtzaL SVOF -ArcE - f, )2s Ay.dx

_S2 v

-Dote : 4' ofx = AY.e4kx 1:xs-b5)4,ce \ D(u»ie = d 1/

Fig 1 -1

"We have discussed how observable work can be done on a

system, causing its energy to change. However, it is possible to

transfer energy to a system in ways which are not observable as

work. Consider the system shown in Fig 1 -2

)- r,uuUI I

IlM.uW"

-------

: out - s tale VbeYat' hlofecuIe-S

o : - be'vathil yYtvleck Í-eS

Fig 1 -2

We might picture the atoms in the walls surrounding the system as

little vibrating masses. Some will be heading toward the boundary

when some are heading away and, as a result, there might be no

macroscopically observable motion. However, the interaction of

these atoms with the molecules in the system can result in changes

in the energy of individual particles, hence a change in the

internal energy of the system. The macroscopically observable work

is zero, so we must find some other way to account for the energy

change. This second mechanism of energy transfer is called energy

transfer as heat. "[1] There is no atom transfer through the

boundary but the system temperature is still changed.

"Heat, like work, is energy in the process of being

transferred. Heat and work are not stored within matter; they are

"done on" or "done by" matter. Energy is what is stored, and work

and heat are two ways of tranferring energy across the bundaries

of a system. Once energy has entered the system, it is impossible

to tell whether the energy was transferred in as heat or work. "[1]

12

1 -2 Entropy and the Seconnd Law of Thermodynamics

"The central idea in engineering science is that nature

behaves in a manner which is predictable. We have seen how energy -

balance analysis is used to predict the change in state of a system

due to transfers of energy as heat and work or to spontaneous

internal changes. But we also know from experience that, while

certain spantaneous changes in state can occur in isolated systems,

the reverse changes are never observed. For example; oxygen and

hydrogen readily react to form water; but who has ever seen water

spontaneously separate into two basic elements? By itself, first -

law analysis can't reveal the possibility or impossibility of a

process; it cannot point the direction of time.

A falling object will warm up when it is brought to rest by

striking the ground; but no one has ever seen an object cool down

and leap up? The ability to rule out impossible processes is

clearly esential to any complete predictive theory of nature. "[1]

The second law of thermodynamics provides the necessary

structure for this second type of analysis.

To explore this idea further, let us consider the system of

Fig 1 -3 -1.-Jole_K v

1

P1Ï1 -Fl S-t at (VII (( i,

1

I

1 j Ss-E -- _____,

w#riv, e5

Fig 1 -3

"It consists of a flywheel surrounded by a gas, in a rigid

I3

adiabatic container; it is an isolated system. Suppose we start

out with the system in state A, with the flywheel spinning and the

gas and wheel fairly cool. As time passes we expect that

collisions between the gas molecules and the wheel will eventually

cause a transfer of energy from the flywheel to the gas molecules,

so the wheel will slow down and the gas will start to spin around

in the box. The random motions of the gas molecules will tend to

randomize the energy of the organized swirl of the gas, and

eventually both the wheel and gas will come to rest

(macroscopically), and all the energy will reside in the random

motions of the gas and wheel molucules, i.e., in the form of

internal energy, so the gas and wheel will be warmer. We denote

this by state B. The energy balance for this process is

(U + KE)A. = Ug initial final energy energy

(1 -3)

Where U is the total internal energy of the system, and KE is

the rotational kinetic energy of the flywheel. All this is very

rational. "[1]

"Now, let us propose a process with less credence. Suppose

that we start out with the system in state B, i.e., with a warm,

motionless system, and allow the system to change to state A, where

enveything is cooler and the wheel is spinning. The energy balance

for this process is

Uß = (U +KE)A initial final energy energy

(1 -4)

1 4

Equations (1 -3) and (1 -4) are identical, so if Eq. (1 -3) is

satisfied, so is Eq. (1 -4). Certainly this second process will

never happen, but the conservation -of- energy principle, i.e., the

first law of thermodynamics, does not tell us that the process

cannot happen. The first law is insensitive to the direction of

the process.

Let us consider this example further. In the initial state

A most of the energy is in a highly organized form. All the

molecules of the flywheel are rotating around the axis together,

and this organization makes it possible to extract that energy

quite easily as useful work. We can simply attach a generator

to the flywheel, generate electricity, and use this energy to raise

weights, run trains, etc. But once the system has reached state

B, where all the energy is microscopically disorganized, it is much

more difficult to extract the energy as useful work; we certainly

cannot do it just with a generator.

Something has been lost by the process that resulted in a

randomization of organized energy; we have lost some ability to do

useful work, something has also been produced: a higher state of

molucular chaos. This loss and gain always go hand in hand;

whenever molecular chaos is produced, the ability to do useful work

is reduced. "[1]

Entropy is a property of matter that measures the degree of

randomization or disorder at the microscopic level. The natural

state of affairs is for entropy to produced by all processes.

Associated with entropy production is a loss of ability to do

useful work. Energy is degraded to a loss useful form, and

l5

it is sometimes said that there is a decrease in the availability

of energy. The notion that entropy can be produced, but never

destroyed, is the second law of thermodynamics.

It is also useful to think about entropy as a measure of our

uncertainty about the microscopic state. As the molecules move

about, collide, and change directions, the microscopic state is

continually changing.

At any instant a system could be in any one of billions of

billions of microscopic states, and we never can be very sure

exactly which state exists at a particular time. The magnitude

of the entropy reflects uncertainty about the microscopic state.

So, the fact that we are uncertain about microscopic details

makes it impossible for us to convert all the molecular energy to

useful work. With increased uncertainty, in other words, with

increased entropy, our ability to do useful work with a given

amount of energy is reduced.

"Let us apply the second law to the isolated system of

Fig 1 -3. Entropy is usually denoted by the symbol "S ", and we will

denote the production of entropy as "Ps ". Since this is an

isolated system, no entropy can flow in or out, hence any entropy

change inside must be due to entropy production inside. So, the

entropy bookkeeping is

Ps = S - S (1 -5) final initial

entropy increase in entropy storage production

l&

The second law requires that the entropy production be zero

or greater, so

S - S final initial

>= o

once we learn how to evaluate the entropy of matter as a aunction

of its state, we can calculate the entropies of states A and B,

and we will find that Sg > S, Therefore, it is impossible for

the system to go from state B to state A; the second law of

thermodynamics will not allow it. "(l)

Processes that do not violate the second law can be classed

as reversible or irreversible. The concept of a reversible process

is very important in thermodynamics, and the ability to recognize,

evalute, and reduce irreversibilities in a process is essential to

a competent engineering thermodynamic specialist.

Suppose the system of interest is an isolated system. The

second law says that any process that would reduce the entropy of

the isolated system is impossible. Suppose a process takes place

within the isolated system in what we shall call the forword

direction. If the change in state of the system is such that

entropy increses for the forward process, then for the backward

process (that is, for the reverse change in state) the entropy

would decrease. The backward process is therefore impossible,

hence we say that the forward process is irreversible. If the

entropy is unchanged by the forward process, such a process is

called reversible. The key idea of a reversible process is that

it does not produce any entropy.

Since a reversible process does not produce any entropy, the

17

total molecular disorgnization within the isolated system remains

constant. The reversible process is an idealization, like

frictionless pulleys and resistanceless wires.

An irreversible process is one that is not reversible, that

is, one that produces entropy. All real processes (with the

possible exception of conducting current flows) are in some measure

irreversible, though many processes can be analyzed quite

adequately by assuming that they are reversible.

"Recognition of the irreversibilities in a real process is

especially important in engineering. Irreversibility, or departure

from the ideal condition of reversibility, reflects an increase in

the amount of disorganized energy at the expense of organized

energy. The organized energy (such as that of a raised weight) is

easily put to practical use; disorganized energy (such as the

random motions of the molecules in a gas) requires "straightening

out" before it can be used effectively. "[1]

Process that are usually idealized as reversible include:

Frictionless movement

Restrained compression or expansion

Mixing of two samples of the same substance at same state

etc.

Process that are irreversible include

Movement with friction

Electric current flow through a nonzero resistance

Mixing of matter of different composition or state

etc.

I8

1 -3 The Characteristics of Thermodynamic Systems

Before introducing the characteristics of themodynamic

systems, there is a thermodynamic property which we shall find to

be of particular importance. It is enthalpy. (i defined by i =

U + PV).

The product of pressure (P) and specific volume (V) has the

unit of energy per unit of mass, as does U. Enthalpy is a term

created just for convenience, because total energy is equal to

internal energy (U) plus flow work (PV), kinetic energy, potential

energy, chemical energy etc.. During the analysis procedure

internal energy (U) and flow work (PV) are heavily used, kinetic

energy, potential energy, and chemical energy are often negligible,

under this circumstance, enthalpy was created.

We can use T -S (temperature vs. entropy) diagram and P -i (pressure

vs. enthalpy) diagram to analyze the thermodynamic systems.

The theoretical single -stage cycle is a perfect thermodynamic

system for introduction.

Refrigerant entering the compressor is assumed to be dry

saturaded vapor at the evaporator pressure. This is a convenient

place to begin an analysis because we can easily determine all

fluid properties here. The compression process 3 -4 is assumed to

be reversible and adiabatic and, therefore, isentropic,and is

continued until the condenser pressure is reached.

Point 4 is obviously in the superheated region. The process

4 -1 is carried out at constant pressure with the temperature of

the vapor decreasing until the saturated vapor condition is reached

at 4'; then the process is both at constant temperature and

T Sat,ya.t44( ..E.

Su,Cooled izG {0 pN

4

CO 141 Ssok-

Eva Po 0.t0 j > 3 Ca.) 1

t23 Grì-E.'C poin-f. ,c./

Sk-ttiva-fcd a.4..14+.e 4

i"t r`x- r lt 1Ze.3 I`O

((1)2ribva .er.c

J ¡`

SatAdra, rzA ).ú.ot U

SA.t.,,-u-Ea 54.4

4 kezzfect

C )

CC

2

constant pressure during tha condensation from 4' to 1.

At point 1 the refrigerant leaves the condenser as a saturated

liçuid. It is then expanded through a throttling value.

rarLial evacraticn occurs as the rressure drops accross the

value. ____ throttling- process 1 -2 is irreversible with an increase

in entropy occuring. For this reason the process is shown as a

das`ed line in Fig 1 -4. For a throttling process the enthalpy at

exit and at inlet are equal.

_ a_ _. _ a -i: :_: e cycle, factors cause the

complete cycle to deviate from the theoretical. These are the

pressure losses in all connecting tubing that increase the power

requirement for the cycle and the heat transfer to and from the

various components. Heat transfer will generally be away from the

system on the high -pressure side of the cycle and will improve

Exposed surfaces cn the low-pressure side cf the cycle will

aenerally degrade cycle performance. As shown in Fig 1 -5, q23 will

increase compressor load and q4 decreases compressor load.

- - should be r._.. __oned that the condenser and evaporator heat -

_'aces have ~ressure losses that also contribute to the

-F I G 2I

1 -4 Conduction

At mention of the word "conduction ", we should immediately

conjure up concepts of atomic and molecular activity, for it is

processes at these levels that sustain this mode of heat

transfer. Conduction may be viewed as the transfer of energy

from the more energetic to the less energetic particles of a

substance due to interactions between the particles.

The physical mechanism of conduction is most easily

explained by considering a gas and using ideas familiar from a

background in thermodynamics. Consider a gas in which there

exists a temperature gradient and assume that there is no bulk

motion. The gas may occupy the space between two surfaces that

are maintained at different temperatures as shown in Fig 1 -6.

T

ó O

I

,

N

cb..x -,--7 , /7 I O ,o ,

1

1 i D \ D I

I

Fig 1 -6

I.

/ V

----> X

Tz

g

We associate the temperature at any point with the energy of

the gas molecules in the vicinity of the point. This energy is

related to the random translational motion, as well as to the

internal rotational and vibrational motions, of the molecules.

"Moreover, higher temperatures are associated with higher molecular

energies, and when neighboring molecules collide, as they are

constantly doing, a transfer of energy from the more energetic to

the less energetic molecules must occur. In the presence of a

temperature gradient, energy transfer by conduction must occur in

the direction of decreasing temperature. This transfer is evident

from Fig 1 -6. The hypothetical plane at X is constantly being

crossed by molecules from above and below due to their random

motion. However, molecules from the left are associated with a

larger temperature than those from the right, in which case there

must be a net transfer of energy in the positive X direction. We

may speak of the net transfer of energy by random molecular motion

as a diffusion of energy. "[2]

This situation is much the same in liquids, although the

molecules are more closely spaced and the molecular interactions

are stronger and more frequent. Similarly, in a solid, conduction

may be attributed to atom activity in the form of lattice

vibrations. The modern view is to ascribe the energy

transfer to lattice waves induced by atomic motion. In a

nonconductor, the energy transfer is exclusively via these lattice

waves; in a conductor it is also due to the translational motion

of the free electrons.

23

Examples of conduction heat transfer are legion. The exposed

end of a metal suddenly immersed in a cup of hot coffee will

eventually be warmed due to the conduction of energy through the

spoon. On a winter day there is significant energy loss from a

heated room to the outside air. This loss is principally due to

conduction heat transfer through the wall that seperates the room

air from the outside air.

4It is possible to quantify heat transfer processes in terms

of approprite rate equations. These equations may be used to

compute the amount of energy being transferred per unit of time.

For heat conduction, the rate equation is known as Fourier's law.

For the one -dimensional plane wall shown in Fig 1 -7, having a ,

T

2). It

X

T2

Fig 1 -7

temperature distribution T(x), the rate equation is expressed as

r

the heat flux X per unit area perpendicular to the direction of transfer, and it

is propitional to the temperature gradient, dT /dX, in this

direction. The propotionality constant K is a transport property

know as the thermal conductivity (W /m *k) and is a characteristic

arr

(1 -7)

CI: Os the heat transfer rate in the X direction

24

of the wall material.t2]Where W is Watt, m is Meter and K is Kelvin.

The minus sign is a consequence of the fact that heat is

transferred in the direction of decreasing temperature. Under the

conditions shown in Fig 1 -7, where the temperature distribution is

linear, the temperature gradient may be expressed as

T -i" -T. (1 -8)

d X l_--

and the heat flux is then

T2 --5 T -i;

Note that this equation provides a heat flux, that is, the rate of

heat transfer per unit area. The heat rate by conduction, qx(W),

through a plane wall of area A is then the product of the flux and

the area, 43-X -^X A

25

1 -5 Convection

The convection heat transfer mode is comprised of two

mechanisms. In addition to energy transfer due to random molecular

motion (diffusion), there is also energy being transferred by the

bulk, or macroscopic, motion of the fluid./YC ZJ

This fluid motion is associated with the fact that, at any instant,

large numbers of molecules are moving collectively or as

aggregaties. Such motion, in the presence of a temperature

gradient, will give rise to heat transfer. Because the molecules

in the aggregate retain their random motion, the total heat

transfer is then due to a superposition of energy transport by the

random motion of the molecules and by the bulk motion of

the fluid. It is constomary to use the term convection when

referring to this cummulative transport and the term advection

when referring to transport due to bulk fluid motion.

\ \We are especially interested in convection heat transfer,

which occurs between a fluid in motion and a bounding surface when

the two are at different temperatures. Consider fluid flow over

the heated surface of Fig 1 -8. A consquence of the fluid -

surface interaction is the development of a region in the fluid

through which the velocity varies from zero at the surface to a

finite value V(oo) associated with the flow. This region of the

fluid is known as the hydrodynamic, or velocity, boundary layer.

Moreover, if the surface and flow temperatures differ, there will

be a region of the fluid through which the temperature varies from

Ts at Y = 0 to T(en) in the outer flow. This region, called the

thermal boundary layer, may be smaller, larger, or the same

2(9

size as that through which the velocity varies. In any case, if

Ts > Tom, convection heat transfer will occur between the surface

and the outer flow. C2]

Fig 1 -8 C 2]

-r

The convection heat transfer mode is sustained both by

random molecular motion and by the bulk motion of the fluid

within the boundary layer. The contribution due to random

molecular motion (diffusion) generally dominates near the surface

where the fluid velocity is low. In fact, at the interface

between the surface and the fluid (Y = 0), heat is transferred by

this mechanism only. The contribution due to bulk fluid motion

originates from the fact that the boundary layers grow as the

flow progresses in the X direction. Appreciation of boundary

layer phenomena is essential to understanding convection heat

transfer. It is for this reason that the discipline of fluid

mechanics plays a vital role in our analysis of convection.

17

AConvection heat transfer may be categorized according to the

nature of the flow. We speak of forced convection when the flow

is caused by some external means, such as by a fan, a pump, or

atmospheric winds. In constrast, for free (or natural) convection

the flow is induced by buoyancy forces in the fluid.

These forces arise from density variatins caused by temperature

variations in the fluid. An example is the free convection heat

transfer that occurs from a hot pavement to the atmosphere on a

still day. Air that is in contact with the hot pavement has a

lower density than that of the cooler air above the pavement.

Hence, a circulation pattern exists in which the warm air moves up

from the pavement and the cooler air moves downward. However, in

the presence of atmospheric winds, heat transfer from the pavement

to the air is likely to be dominated by forced convection, even

though the free convection mode still exists.+' C2J

We have described the convection heat transfer mode as energy

transfer occurring within a fluid due to the combined effects of

conduction and bulk fluid motion. In general, the energy that is

being transferred is the sensible, or internal thermal, energy of

the fluid. However, there are convection processes for which there

is, in addition, latent heat exchange.

This latent heat exchange is generally associated with a phase

change between the liquid and vapor states of the fluid.

Regardless of the particular nature of the convection heat

transfer mode, the appropriate rate equation is of the form

q" = h (Ts - Tx)) (1 -10)

where q ", the convective heat flux ( 1..J/1112 ), is propotional to

the surface and fluid temperatures, Ts and Too, respectively. This

expression is known as Newton's law of cooling, and the

proportionality constant h (IJ / rr IG ) is referred to as the

convection heat transfer coefficient. It emcompasses all the

effects that influence the convection mode. It depends on

conditions in the boundary layer, which are influenced by surface

geometry, if the roughness of the surface is bigger then the

boundary layer will be higher, the nature of the fluid motion, and

a number of the fluid thermodynamic and transport properties,

for example, enthalpy and viscosity. Moreover, any study of

convection ultimately reduces to a study of the means by which h

may be determined.

21

1 -6 Radiation

Thermal radiation is energy emitted by matter that is at a

finite temperature. Although we focus primarily on radiation from

solid surfaces, emission may also occur from liquids and gases.

Regardless of the form of matter, the emission may be attributed

to changes in the electron configurations of the constituent atoms

or molecules. The energy of the radiation field is transported by

electromagnetic waves (or alternatively, photons). While the

transfer of energy by conduction or convection requires the

presence of a material medium, radiation does not. In fact,

radiation transfer occurs most efficiently in a vaccum.

l The maximum flux ((e-1 /v ) at which radiation may be emitted

from a surface is given by the Stefan -Boltzmann law,

W STS (1 -lia)

where Ts is the absolute temperature (K) of the surface and a-

is the Stefan -Boltzmann constant (G= ., (:-7x I c7- 5 wi' WZ i : ) .

Such a surface is called an ideal radiator or black body. The heat

flux emitted by a real surface is less than that of the ideal

radiator and is given by

_ a -Tsé.

where is a radiative property of the surface called the

emissivity. This property indicates how efficiently the surface

emits compared to an ideal radiator. "t 2J

The equation q":-. Ts4 which determines the rate at

which radiation is exchanged between surfaces is generally a good

deal more complicated. However, a special case, for example, a

house to the universe, that occurs very frequently in practice

involves the net exchange between a small surface and a much large

surface that completely surrounds the smaller one. The surface and

the surroundings are seperated by a gas that has no effect on the

radiation transfer. The net rate of radiation heat exchange

between the surface and its surroundings, expressed per unit area

of the surface, is _`_'r (T54 -Tu4) (1 -12)

In this expression, A is the surface area and E is emissivity,

while T,,,r is the temperature of the surroundings. For this

special case, the area and emissivity of the surroundings do not

influence the net heat exchange rate.

There are many applications for which it is convenient to

express the net radiation heat exchange in the form

CTS -- Ts,Y (1 -13)

where the radiation heat transfer coefficient hr is

hr = e C T y) T2+ TT.(A -2)

Here we have modeled the radiation mode in a manner similar to

convection. In this sense we have linearized the radiation rate

equation, making the heat rate propotional to a temperature

difference rather than to the difference between two temperatures

to the fourth power. Note, however, that hr depends strongly

on temperature, while the temperature dependence of the convection

heat transfer coefficient h is generally weak .t70

The surface within the surroundings may also simulatneously

transfer heat by convection to the adjoining gas (Fig 1 -9).

/ he-t Ya4pEA-.4`1.`

V ?7cofr, V.

Cohvec-tA4`" 17t.2ad- fiiVZt+S6)^"

Fig 1 -9 C2J

The total rate of heat transfer from the surface is then the sum

of the heat rates due to the two modes. That is

or

q=

=

se.couv,-r irctoC, (1 -14)

11AC l5- lro) fef"Tq- t5 r5u,+'> (1-15)

32

CHAPTER 2

Solar Radiation

2 -0 Introduciton

2 -1 Sun -Earth Relations

2 -3 Time

2 -4 Solar Position Charts

35

2 -0 Introduction

Throughout the history of humankind, the sun has been the

source of all of the energy used by humans. Solar radiation has

important effects on both the heat gain and heat loss of a

building. This effect depends to a great extent on both the

location of the sun in the sky and the clearness of the atmosphere

as well as on the nature and orientation of the building.

It is useful at this point to discuss ways of predicting the

variation of the sun's location in the sky during the day and with

the seasons for various locations on the earth's surface.

It is also useful to know how to predict for specified weather

condtions, the solar irradiation of a surface at any given time

and location on the earth, as well as the total radiation striking

a surface over a specified period of time.

2 -1 SUN -EARTH RELATIONS

The earth moves in a slightly elliptical orbis about the

sun. The plane in which the earth rotates around the sun

(approximately once every 365 1/4 days) is called the ecliptic

plane or orbital plane. The mean distance from the center of the 67

earth to the center of the sun is approximately 92.9 x 10 miles

or 1.5 x 10 km. The perihelion distance when the earth is

closest to the sun, is 98.3 percent of the mean distance and

occurs on January 4. The aphelion distance, when the earth is

farthest from the sun, is 101.7 percent of the mean distance and

occurs on July 5. Because of this the earth receives about 7

percent more total radiatin in January than in July.

Mawc1t 21 Attie 1tt )T e Vevcia,Q

tom= ef4u`1no

44 la h

e 2 l dY 22

Su4.041P^'

Sa iStt.R.

'2-22t5 c7

,4t-turi 04t

gefX 2 0- 2-3

Tvyi'c. af-&. + 12

W ctfW Sos4 to-

Pee. , 21 vY Z

eY7. ̀Cz--

z3 5°

Trorc. 06 eAPv,co,,

Fig 2 -1

TroPcc. Ca,4 ca-k

TrE 14 6aPr40->,

35

Fig 2 -1 shows the movement of the earth around the sun. As the

earth moves it also spins about its own axis at the rate of one

revolution each 24 hours. There is an additional motion because

of a slow wobble or gyroscopic precession of the earth.

The earth's axis of rotation is tilted 23.5 degrees with

respect to the orbital plane. As a result of this dual motion and

tilt, the position of the sun in the sky, as seen by an observer

on earth, varies with the observer's location on the earth's

surface and with the time of day and the time of year.

For practical purposes the sun is so small as seen by an observer

on earth that it may be treated as a point source of radiation.

Fig 2 -1 can be used to explain the effect of the earth's tilt and

radiation about the sun. At the time of the vernal equinox (Mar.

21) and of the autumnal equinox (Sep. 22 or 23) the sun appears to

be directly overhead at the equator and the earth's poles are

equidistant from the sun. Equinox means "equal nights" and during

the time of the two equinoxes all points on the earth (except the

poles) have exactly 12 hours of darkness and 12 hours of daylight.

II During the summer solstice (June 21 or 22) the north pole is

inclined 23.5 degrees toward the sun. All points on the earath's

surface north of 66.5 degrees north latitude (the Arctic Circle)

are in continuous daylight, whereas all points south of 66.5

degrees south latitude (the Antarctic Circle) are in continuous

darkness. Relatively warm weather occurs in the northern

hemisphere and relatively cold weather occurs in the southern

hemisphere. The word solstice means sun standing still. ' E 3 J i

During the summer solstice the sun appears to be directly

30

overhead along the Tropic of Cancer, whereas during the winter

solstice it is overhead along the Tropic of Capricorn. The torrid

zone is the region between, where the sun is at the zenith,

directly overhead, at least once during the year. In the temperate

zones (between 23.5 and 66.5 degrees latitude in each hemisphere)

the sun is never directly overhead but always appears above the

horizon each day. The frigid zones are those zones with latitude

greater than 66.5 degrees where the sun is below the horizon for

at least on full day (24 hours) each year. In these two zones the

sun is also above the horizon for at least one full day each year. u

[33.

37

2 -2 Solar Angles

From the previous section, we know, the earth rotates about

the sun, it spins about an axis which points to the North Star and

is inclined at 23.45 degrees to the orbital plane. Therefore, the

angle between the earth's equatorial plane and the earth -sun line

varies between ±23.45 degree throughout the year.

This angle is called the declination, o& . Declinations north of

the equator are positive (summer in the northern hemisphere) those

south are negative. The declination is tabulated in Table 2 -1 and

represented graphically in Fig 2 -2. It can be approximated by

(284+ & -f-:- 23,4-&.S.'11 C3Co 3365

3 (2 -1)

where n is the day of the year (Table 2 -1).

The value of ($ calculated from Eq(2-1) will be correct within +0.37

degree (with maximum positive deviation on May 1) and -1.70 degree

(with maximum negative deviation on October 9).

The sun's location in the sky relative to a point on the

surface of the earth can be defined with two angles, the solar

altitude o< , and the solar azimuth k's as illustrated in

Fig 2 -3.

The solar altitude at a point on the earth is the angle

between the line passing through the point and the sun and the line

passing through the point tangent to the earth and passing below

the sun. The solar azimuth is the angle between the line under the

sun and the local meridian pointing to the equator, or due sourth

in the northern hemisphere. It is positive measured to the east

38

:x- "

f ° `

" /»'

-^/ @

um

(1

| . |i.

_1 .1

| |

! |

J _ l ' i |

| _L|

| `

. _ J

! |

|

L|

i | ]

| |

/ |

| }

/ /| |

/ |

|

.

.

.

/

.

Date

/ :.J ` ,. t-;,,., c¢. 7-1` "1 e [3 D cf

Day n

Declination 0, deg+

ET. minutes of time §. Date Day n

Declination 3, deer

:.

minutes ol"

time ;;

.'.1:.. I

10

l

10

- 2.7 k,

-22.0 -_.l -7.6

July 9

19

190 200

22.9 20.9

-5.2 -6.2

20 20 - 20.2 -11.1 29 210 18.8 -6.4 30 _,, -17.., - 13.3

A::_. Q 220 16.2 -5.6 1- : h. a 41; - 14 S - 14 3 18 230 13.2 -3.3

1) _ ,_) - H.4 - ::1 .) 28 240 9_8 -1.2

\!.=r. 1 60 -?7 -12.4 Sept. 7 270 6.2 _1_1.0

i . -..) - !(;!) r - A r ' _ ... k,' - .1 - -.2 2 '-)) - 1.5 i 0

31 +IJ / ) -4.2 0ct.7 280 -5.4 12.6

-Npr. 10 1C.0 7.S -1.3 17 290 -9.1 14.6

_'!! 110 11.4 +1.1 27 30( 1 - 12.7 ;6.1 30 120 14.7 2.5

Nov. 6 310 -15.9 16.3

May 10 13O 17.5 3.6 16 320 -18.7 15.2

20 146 19.a 3.5 26 330 -20.9 12.' .. _:.- 2:

1-;-:..-:. :, . -,) -22..5 .9 Jt,:r.e ) ,rr 2=. G.:i 16 _: _ì -23.3 4.4

.:. .".t _. -!.3 26 3:-.) -23.-: -06 2

,, _ - - .31 -` -2: : -3.0

4c

and negative to the west (in both hemispheres).

The solar zenith angle os , is the angle between a solar ray

and local vertical direction, is the complement of c, .

kiest

(lor t!,

e

Fig 2 -3

The sun's location in the sky is a function of the location

on the earth, the time of year, and the time of day. The location

on the earth is specified by the latitude (g . At the equator,

0 = 0. North of the equator, latitude are positive; south of it

they are negative. The time of year is specified by the solar

delination o> , previsouly defined. The time of day is

specified by the hour angle to. The hour angle is defined as zero

at local solar noon (d-S = 0) , and increases by 15 degrees for each

hour before local solar noon [ e.g. for 8 AM (solar tince), 00= 60

degree) and decreases by 15 degrees for each hour after solar noon

[ e.g. at 3 PM (solar time), W = -45 degrees] in both hemispheres.

With the help of spherical geometry, expressions for the solar

altitude and the solar azimuth can be developed in terms of 9 , and 0). Thus,

4- (

Sìh OKs _ ca -fi Cos cb Co5 ,;$ Cos (.0 (2-2)

Cos 6 S hi (-J

C9s °<s (2 -3)

For example, we want to determine the location of the sun at 9 AM

solar time at Tuscon, AZ 415= 32 degree on Sep. 7.

From Equation 2 -1

6 _ Sri L 36o . 650) el Cor -S-rc l Taile 2 -!, Ç=

From Equation 2 -2 for Tucson = 32 degree and CO= +45 degree

(9AM) .

50$1046 = 5itn C 2 ) 5,1;-) C6.4) -+ Cos C3 2.) CosC5.4) 6613. C4 O t i s = 4c7. 3 °

From Equaiton 2 -3

Cos £3)

Therefore,

Tucson, AZ, Sep. 7, 9 AM

- S,4', =3'Z ° / w -er°l o(s = 4a, 3°, 1'6=614.

The surrise time, sunset time, and the day length can be

determined using Eq. (2 -2). At sunrise or sunset

42

oC 5=0 = -1- Cos et Cos is, Cos 5

where 6J1,5 is the sunset (or sunrise) hour angle, which may then

be expressed as

(45 = Cos -1 (- "t ah ) (2 -4)

The day length is twice the sunset hour angle. From Eq. (2 -4)

the day length can be expressed in hours as

t.1 -611 s (2 -5)

when (,)AS is expressed in degrees. The actural time during which

the sun is visible is somewhat longer than that indicated by Eq.

2 -5. First, the refractive effect of the atmosphere allows one to

see "around" the earth's horizon. Second, the sun is not a point

source of light as assumed in the development of Eq. (2 -4) but it

has a finite size so that a "limb" of it is visible before its

center appears.

The path of the sun on any given day is approximately in a

plane tilted at an angle equal to the local latitude from the

vertical. Relative to a given point on the earth's surface, the

plane moves north (summer in the northern hemisphere) and south

(winter in the northern hemisphere) throughout the year.

'At equinox, the plane contains the point. It is only at equinox

that every point on the earth's surface moves essentially in the

relative orbit plane of the sun. At equinox the sun rises due east

and sets due west for every point on the earth. These statements

can be better understood by examining Fig 2 -4.

43

5

_- - ust_azes

. ----f t'

\ r; .i

.,..:

.

ca )

Fig 2-4 [33

s ,n-.,a t-- d'ars for Houston, IX, at 20

N latitude at the equinoxes and at solstices. The vie:. from

_ :-1b) clears_ the sun's relative orbit plane

cc,.s`an 7_1 i_ .4/[33

Example (2 -1):

For 30 degree N latitude, determine the sunrise time, the day

length, the sumrise solar azimuth angle, and the solar noon

zenith angle for the equinox and check the solution with

Fig 2 -4.

Solution: For equinox

Wss = cos ( -tan 30 tan 0) = 90° = 6h or sunset and

sunrise at 6 P.M. and 6 A.M. solar time (as expected).

The daylength (from Eq. 2 -5) is

td = 2/15 cos (-tan 30 tan 0) = 2/15(90) = 12h

as expected. For the sunrise solar azimuth (from Eq. 2 -3)

Cos o° 5 C Qo`) n Cos o°

so, --5,s= -t qc°

That is, the sun rises 90 degree east of south (due east) and

sets 90 degree west of south (due west). The solar noon zenith

1can be determined from

Eq. 2-2:

Cos s , = S h oc sn = S ols C d"c.= D) S -- Cos cl) Cos C6° Cos( - )

which implies that

o5h=el7- ` since oS = 0 at equinox.

This procedure can be repeated for the other cases, with

the results being summerized in the following:

4kinoX

. wss r5s ZSh

30° O' 6 hrs /VP's Roc. 30°

-5

2 -3 TIME

A worldwide system of 24 time zones has been established by

international agreement, so that all "days" can start at

approximately the same local time. The earth revolves 360 degrees

in 24 hours or 15 degrees per hour. Therefore, each of the 24 time

zones represents a different hour and physically consists of the

land contained between two meridians 15 degree apart, although the

latter is often approximate since various political and natural

boundaries cause some modificatin of the time zone boundaries.

The time zones are measured westward (positive) from the prime

meridian through Greenwich, England. In each time zone the

zone mean time (ZMT) or clock time is Greenwich time less the

number of hours equal to the time zone meridian divided by 15.

In the United States, the time zone meridians associated with

the various time zone are Eastern, 75 degree W; Central,

90 degree W; Mountain, 105 degree W; and Pacific, 120 degree W;

Alaska and Hawaii are in the time zone associated with the 150

degree W meridian. When it is noon (local time) in Greenwich, the

zone mean time (local time) in Tucson is 5 A.M. (of the same day).

A solar day at a given point on the earth is defined as the

length of time between successive solar noons. However, because

of the eccentricity of the earth's orbit and the inclination of

the earth to the earth -sun orbit plane, the length of a solar day

is not constant and in fact varies by up to about 30 seconds

between successive days at certain times of the year. The

cumulative effect of the phenomenon is a shift in the time scale

of up to 15 minutes from the mean time. This shift (the difference

between true solar time and mean solar time), called the equatin

of time (ET), is shown in Fig 2 -2 and tabulated in Table 2 -1.

The relatinship between true solar time and zone (clock) time

can now be developed. True solar time is the zonal time plus the

equation of time correction plus an ajustment for the difference

between the zonal meridian and the local meridian.

Since the earth rotates 15 degrees per hour, this adjustment is 4

minutes for each degree of longtitude. Solar time can therefore

be determined as

solar time = zonal time (or clock time)

+ 4 (zonal meridian - local meridian)

+ equation of time

or solar time = ZMT + 4 (s.t -% ) + ET (2 -6)

where Ast is standard meridian and k is local meridian in

degrees and the two correctin terms are in minutes of time. "C3]

Example(2 -2):

Determine the sunrise and sunset time (local time) on OUnter

30 at Phoenix (33 degree N , 12 degree W).

Solution:

For October 30 and from Fig 2 -2 or Table 2 -1

- 13. 4-` / ET = +16.2 min.

From Eq. (2 -4)

GJss Cos -(C-t-ah (33 °.) -tav, (-t 3,4 °)] = SI, I °

Sunrise occurs at 81.1/15 = 5.41 hrs. before solar noon or

6:35 A.M. solar time.

Sunset occurs at 5:25 P.M. solar time.

47

From Eq. (2 -6), sunrise occurs at

local time = 6:35 A.M. - 4 (105 -112) min - 16 min

= 6:47 A.M. (Mountain Standard Time).

Similarly, sunset is at 5:25 + 26 -16 = 5:37 P.M.

Note that most states in the United States switch to

daylight saving time the last Sunday in April and change

back to standard time the last Sunday in October.

Arizona, however, does not. In a state that does change to

daylight saving time, the June 1 local time would be

advanced 1 hour, e.q. 7 A.M. MST becomes 8 A.M. MDT.

The local time on Oct. 30 may be daylight or standard time.

depending upon the year.

MST - Mountain Standard Time

MDT - Mountain Daylight Time

2 -4 SOLAR CHART AND SHADING DIAGRAM

Although the previous section can be used to define the sun's

position in the sky, the formulas are too cumbersome for certain

types cf estirates cr calculations. This is particularly true in

determining the effect of obstructions that cast shadows upon a

certain site. This problem is easier solved by plotting the

equations on a solar chart, we will see an example in

Fig 2 -6.

Solar charts have various formats, one useful format is a plot

of the solar azimuth d-5 and zenith angle sis for selected days

of the year and times of the day at a given latitude, as presented

by ::azria. A solar chart of this format for 40 degree N latitude

is presented in Fig 2 -5.

ñ N

r /

r / r

;i 1 141

-r-" \LA

3 Pm

&o Esp - --

Fig 2 -5

is the solar azimuth, while the vertical axis

is the solar altits cr zenith the solid lines represent the sun's

path on the 21st day of each month. Dotted lines give various

41

hours of the day. A definite point in time is then represented by

the intersection of a solid and dotted line. The sun's position

et that time can then be found by reading the chart. The chart is

convenient for determining the tines when a given point on a site

falls within the shadows of obstructions such as neighbouring

buildings, trees, and hills. They may also be used to design

shadowing obstructions to block the solar energy during selected

period.

An example -.:ith an obstructing tree and building is shown in

Fig 2 -6. For example, the sun shine is blocked by the tree from

9 A.M. to 11 A.M. in February.

;cß!1

Fig 2 -6

One person stands at the pcsition shown on the site map in

_ __g 2-7. A bu__di. _ 7T' tall is located soutn and west of

the pronoc=d location. This site is 44 degree N latitude.

s. chart to find the times when the person will eb

in the .-__ado . of this building

Fig 2 -7

Solution:

From Fig. 2 -3, we can know that azimuth ( -5) and altitude

(DU should be found for corners (1), (2), (3).

(1) _ --a C 3 ) =6 3 , lti%e,J)rt

] =1Q6 ° z3oo (2)

(t-z= - - ¡ ° Waal l

01 = (

-v 5 Z,( -d

6 t 5 +s3op ) (3)

3 - (4; 2 6, 6 °

= IY.Ge (l vo f2oo )] This building then masks the sun from the site as shown in

Fig. 2 -8. Obstruction may be included into a building

design to control the solar influx. For example, it may be

desirable to have the sun shining through windows during the

winter, but not during the summer. This would allow a

passive solar -heat gain when heat is needed and prevent it

when heat is not needed. E 33

51

X11. 2. -m 2v 4o by wQ. ó 12'°

One such device is a vertical shade placed along one or both

sides of the window as illustrated in Fig 2 -9. Depending upon

the orientation of the window, orientation of the shades,

relative sizes and locations, these shades will block the sun's

rays at selected times.

Fig 2 -9

Plotting of the azimuths of the window and of the shade's

vertical rays on the solar chart may be used to determine the

times when the window is shaded, partially shaded, and completely

unshaded. The procedure is illustrated in the following example.

52

E }:AMPLE (2 -4) :

A window that faces 15 degree east of south and is located

at 40 degree N latitude uses vertical sun shades shown in

Fig 2 -10. Determine the morning and afternoon solar

azimuths at which the window is fully shaded. The shades

are tall comparaed to the window height.

Solution:

Fig 2 -10

h $' V

- I \ Fig 2 -11

For a completely shaded window (Fig 2 -11).

Om = tan(1 /4 +2) = 9.5

= tan (1/ 4) = 14.0

V soil = 15 + (90 - 9.5) = 95.5 east

= 15 - (90 -14) = -61 west

(9m= & of morning

(9-a = of afternoon

Ì-C,, wt = azimuth of morning

4101"n .'cc SGtn.% - /

rep, = azimuth of afternoon

Plotting these azimuths on the solar chart of Fig 2 -12

reveals that the shades permit most of the morning sun to

53

shine upon the window, while partially blocking the

afternoon sun.

I

. .`

-

/// .

;-

. .

r

/ .,

.

: ,

.

i

¡

.

;

/ i , / //, / / ' ``\ ̀

\

. ` :

-

,

.

.

_ , . , > .

, , ,'/,í,rf » / -- , `

f , ' / f .

% \ ... r j ' '

i : : ' - ,

/ / /

/-`

No / \/2 \ 2. ! \ í

' r ! ir . // i. '1

' i

1 / /

-

\ \ , ` . ̂y, Y ,i i % r /i

, /, í \C/ ' n

:+i: .

east -=

,,;.

The solar profile angle is very important when we want to use

We-St -c. !;t',

s a?na c, a.., for an overhang. In the Fig 2 -13, when the

..... .r . - / ;r l l `/. /".. i t"

r

.

.

i

, / -rt L. L4; i".'--.7"- :-

i 7/ ' I ' '

Lr .., /

s;hßz (r) 't r

1.7 t( .

. .

3 - - =

j _ _.._a _ ._ .. _..\w-:

its ...ag;îl}..:.:L:e becomes si?^e/ ' CO5C Y$- I J ; while the `r e_ `1`.Gi

sla. __.a_. _ cc: n 7 between

s.a~ _s the solar r_cfile angle:

ep= th1Lth&'Cos(s.-r)J (2-7)

reach its ..._n_:"' .. iv

and its maximum at either sunrise or sunset, when 6p is at its

54

maximum. When the window faces due south, this result simplifies

to C917. (+a44 ßE Cos à-s )

(2 -8)

The minimum will now occur at solar noon. The maximum occurs at

sunrise and sunset. Equation 2 -7 and the solar chart may also be

used to determine when an overhang shades a window.

For example, when the sun is at the solar profile angle c, of

Fig 2 -14 the vertical window is entirely in the shade. When the

sun is at 8p2 , the entire window is out of the shade. These

two angles are determined by the relative position and sizing of

the window and overhang, which are known once the design is set.

Rearranging Eq. 2 -7 gives -1 lGi1n 6:7 - J

Co5 C rs - ?-

(2 -9)

Fig 2 -14

Notice that ( r5 - - ) is the solar azimuth measured with respect

to the window azimuth rather than with respect to due south. For

a fixed value of 8r , this result may be plotted on the same

coordinates as the solar chart. This plot is known as shading

55

chart (Fig 2 -15). The use of this chart in conjuction with the

solar chart is demonstrated in the following example.

/ ^' f

/``

¡" ._..; r i

Gt C'i r L' .L ^ . . . . ' i , 1 :.-L V' `

Fig 2-15

o

Example:

Tht ,:indo:.- overhang combination shown in Fig 2 -16 is

errected at 40 degree N latitude. This window faces 15

cf degree

scut-h. Use the shading and solar charts to determine

the times when the window is completely shaded and unshaded.

,_....

r

_

Fig 2 -16

5C

Solution:

For a shaded window:

r,_ For a unshaded window:

617- -env1-1 (2 / 1 )

These lines are then traced from the shading chart onto the

solar chart. The azimuth r = 15 degree of the window is

accounted for by aligning the 0 degree relative -azimuth line

of the shading chart with the window azimuth on the solar

chart. The result is shown in Fig. 2 -17.

Ne s t Fig 2 -17

5

Chapter 3

Comfort Zone and Heat Transmission

3 -0 Introduction

3 -1 The Comfort Zone

3 -2 Heat Transmission in Structures

3 -3 Tabulated Overall Heat -Transfer

Coefficients

55

3 -0 Introduction

In this chapter we will discuss about two different topics,

the first one is Comfort Zone which is in section 3 -1 and the

second is about Heat Transmission in 3 -2 and 3 -3.

A comfort zone is one in which there is freedom from annoyance

and distraction, so that working or pleasure tasks can be carried

out unhindered physically or mentally. Everyday experience tells

us that there are a host of factors which are relevant to this

concept. Not only do air and surface temperatures, humidity, air

movement and air purity play a part, but psycho -sociological

factors also have an important role. The attitudes of people

around us, the organization of space, color schemes and many other

factors all can have an influence on our mood and work output.

Since there is an interaction between all these factors, the

problem is complicated further. A deficiency in one of the

physical factors can spoil the balance of the environment; equally

to, surroundings which contain disturbing social or psychological

aspects can be uncomfortable.

Heat transmission is the fundamental of Heating, Ventilating,

and Air Conditioning (HVAC). The design of HVAC systems, including

building insulation selection, sizing ducts and evaluating thermal

performance of these systems is based on principles of heat

transfer given in Chapter 1. Information presented in this chapter

and in the more technical discussion of heat transfer found in

Chapter 1 is based on steady -state or equilibrium conditions. Heat

transfer under dynamic or changing conditions is discussed

elsewhere. Because most of the calculations require a great deal

E9

of repetitive work, tables that list coefficients and other data

for typical situations are used. Thermal resistance is a very

useful concept and will be used extensively in this and following

chapters. Generally all three modes of heat transfer - conduction,

convection, and radiation - are important in building heat gain and

loss.

GG

3 -1 The Comfort Zone

It is very hard to find a unanimous comfort zone to satisfy

everyone. The Institude for Environmental Research at Kansas State

University under ASHRAE contracts has conducted extensive research

on thermal comfort of clothed sedentary subjects, and most of the

following is based on that work.

The perception of comfort, temperature, and thermal

acceptability is related to one's rate of metabolic heat

production, its rate of transfer to the environment, and the

resulting physiological adjustments and body temperatures. The

heat transfer rate is influenced by the environmental factors of

air temperature, thermal radiation, air movement, and humidity,

and by the personal factors of activity and clothing. Thermal

sensations can be described by feelings of hot, warm, slightly

warm, neutral, slightly cool, cool, and cold. Judgements as to

whether the environment is thermally acceptable is related to

environmental parameters and to thermal sensation.

In a uniform thermal environment the 80 percent thermal

acceptability limits occur at conditions that produce thermal

sensations near slightly cool and slightly warm. Clothing, through

its insulation properties, is an important modifier of body heat

loss and comfort. Clothing insulation can be described in terms

of its clo value [1 clo = 0.88 (ft2 -hr -F) /BTU]. A heavy two -piece

business suit and accessories has an insulation value of about 1

clo, whereas a pair of shorts is about o.05 clo.

The operative or adjusted db temperatures and clo values

corresponding to the optimum sensation of neutral, and the 80%

thermal acceptability limits of ASHRAE are given in Fig 3 -1.

Sedentary 50% :

50% of the total

time is seated.

'V =< 30 fpm :

average wind

velocity equal to

or smaller than

30 feet per minute.

5'e de.tr-vv ;-c%

1-73v P

t

64 6) 3" '72 7 6

- Fig 3 -1

The Insulation value of clothing worn indoors is influenced

by the season and outside weather conditions. During the summer

months, typical clothing in commercial establishments consists of

lightweight dresses, lightweight slacks, short -sleeved shirts or

blouses, and accessories. These ensembles have insulation values

ranging from 0.35 to 0.6 clo. At other times, between seasons,

the clothing may have an insulation value in the range of 0.6 to

0.8 clo. Because of the seasonal clothing habits of building

occupants, the temperature range for comfort in summer is higher

than for winter. The acceptable range of operative temperatures

and humidities for the winter and summer is defined on the

psychrometric chart of Fig 3 -2. The zones overlap in the 73 -75 F

range. In this region people in summer dress would tend to be

C'o2

slightly cool, whereas those in winter clothing would be near t:e

slightly warm sensation.

Due to individual, clothing, and activity differences, the

boundaries of each comfort zone are not actually as sharp as

shown in Fig 3 -2.

--/rj

1..18 -1-e4.,,.-Per.:

Fig 3-2

The air temperature in a room generally increases from floor

to ceiling. If this increment is sufficiently large, local

discomfort can occur. To prevent this, the vertical air

temperature difference between head and ankles should not exceed

5 F.

To minimize foot discomfort, the surface temperature of the

floor for people wearing appropriate indoor footwear should be

between about 65 F and 84 F.

C,3

Comfort conditions for clothing levels different from those

just described can be determined approximately by lowering the

temperature ranges of Fig 3 -2 by 1 F for each 0.1 clo of increased

clothing.

In general, children below 12 years of age require a reduction

in temperature of 1 F, whereas adults over 60 may require an

increase of 1 F.

Humidity is described in terms of dew point temperature. In

the zone occupied by the sedentary or near sedentary people the

dew point temperature should not be less than 35 F or greater than

62 F. The thermal effect of humidity on the comfort of sedentary

persons is small (Fig 3 -2). The upper and lower humidity limits

are based on considerations of comfort, respiratory health, mold

growth, and other moisture -related phenomena.

It should be noted that humidification in winter may need to

be limited to prevent condersation on windows and other building

surfaces. To minimize respiratory distress in winter, relative

humidity should be as high as the structure will allow without

condensation, but not more than 50 percent. Within the thermally

acceptable temperature ranges of Fig 3 -2, there is no minimum air

movement that is necessary for thermal comfort. The maximum

average air movement allowed in the occupied zone is lower in

winter than in summer. In winter, the average air movement in the

occupied zone should not exceed 30 fpm. If the temperature is less

than the optimum, the maintenance of low air movement is important

to prevent local draft discomfort.

In summer, the average air movement in the occupied zone

should not exceed 50 fpm. The comfofrt zone can be extended above

79 F, however, if the average air movement is increased

30 fpm for each degree F of increased temperature to a maximum

temperature of 82.5 F.

The mean radiant temperature (MRT) is the unifor surface

temperature of an imaginary black enclosure with which a person,

also assumed to be a black body, exchanges the same heat by

radiation as in the actural environment. This parameter is

important when surrounding surfaces are at a temperature different

from the body.

In effecting heat loss and comfort, the mean radiant

temperature can be as important as air temperature. The mean

radiant temperature is given by

tmr = tbg + C * V - (tbg - ta) (3 -1)

where

tbg = black globe temperature, F

C = constant, 0.157

V = air velocity, f t /Th h

ta = air dry bulb temperature, F

For an indoor environment, where air movement is low, the operative

temperature is approximately the average of air temperature and

mean radiant temperature. The operative temperature is the uniform

temperature of an imaginary enclosure with which an individual

exchanges the same heat by radiation and convection as in the

actural environment. This index attemps to include the effect of

convection as well as radiation.

For usual practical applications the operative temperature is the

mean of the dry bulb and mean radiant temperatures at a given

location in the space and is refferred to as the adjusted dry bulb

temperature. When the mean radiant temperature in an occupied zone

differs from the air temperature, adjustments should be made to

keep the operative temperature within the appropriate comfort zone.

With the increasing need to conserve energy, adjustments may be

mandated in the allowable winter and summer temperatures. The

extent to which comfort can be achieved by clothing adjustments is

indicated in Fig 3 -1-

Example:

Determine the operative temperature for a work station in a

room near a large window where the dry bulb and black globe

temperatures are measured to be 75 F and 81 F, respectively.

The air velocity is 30 fpm at the station.

Solution:

The operative temperature depends on the mean radiant

temperature, which is given by Eq. 3 -1

tmr = 81 + 0.157 (30)/2 (81 -75) = 86 F

then the operative temperature is:

to = (ta +tmr) /2 = (75 +86)/2 = 81 F

A person working at this location would probally be

uncomfortable. u C53

Note:

The black globe temperature is the equilibrium temperature fo

a 6 -in diameter black globe and is used as a single temperature

index describing the combined physical effects of dry bulb

temperature, air movement, and the radiant energy received from

G&

various surrounding areas.

The comfort conditions discussed so far relate to sedentary

activity, which accounts for most applications. However, in

research laboratories, machine shops, assembly lines, and general

manufacturing areas, for example, the occupants may be engaged in

rather active work. The comfort zone temperatures should be

decreased when the average steady -state activity level of the

occupants is higher than sedentary or slightly active, reader can

refer to ASHRAE Fundamentals, 1985, Chapter 8 for more details.

6 7

3 -2 Heat Transmission in Structures

In Chapter one, we have learned some basic principles and

formulas about heat transfer. In this chapter, we are going to

forcus on heat transmission especially on building structures

based on those fundamentals we have already learned.

Consider the flat wall of Fig 3 -3

-t,

X Z X

( aì

Fig 3 -3

where uniform temperatures tl and t2 are assumed to exist on each

surface. If the thermal conductivity, the heat transfer rate,

and the area are constant, Eq. 1 -7 may be integrated to obtain:

q = - KA *(t2- t1) /(x2 -xl) (3 -2)

q = heat transfer rate, BTU /hr

K = thermal conductivity, BTU /(hr -ft -F)

A = area normal to heat flow f *t squre

4= temperature difference (F)

ßX = distance difference (ft)

Another very useful form of Eq. 3 -2 is

q = -(t2-tl)/R' (3-3)

where R' is the thermal resistance defined by

68

R' = (x2- xl) /KA = o x /KA (3 -4)

The thermal resistance for a unit area of material is very

commonly used in handbooks. The term, R' sometimes called the

"R- factor ", is referred to as the unit thermal resistance, or

simply the unit resistance, R. for a plane wall the unit resistance

is R =dX /K (3 -5)

Thermal resistance R' is analogous to electrical resistance,

and q and (t2-t1) are analogous to current and potential difference

in Ohm's law. This analogy provides a very convenient method

ofanalyzing a wall or slab made up of two or more layers of

dissimilar material. Fig 3 -3(b) shows a wall constructed of three

different materials.

The heat transferred by conduction is given by Eq. 3 -4 and

Fig 3 -4, where

R' = R1' + R2' + R3' = -t (3-6) k, A 2,4 k 3

z1X1 X2 00(3 -,__ -*-------- - .,

oX2 oX. KzÁ KaA

Fig 3 -4

61'

Vo (e 3-la C 6] Thermal Properties of Typical Building and Insulating Materials -Design Values'

Description Densitq Conduc- Conduc- Resistance e(R) Specific lb/ft3 tivit' tance Heat

A (C) Per inch For thick- Btu /Ib Btu In./ Btu /h thickness ness listed deg F h 11 =F 10F (1/A) (1 /Cl)

h11í hft F /Btu F /Btu

BUILDING BOARD

Boards, Panels, Subflooring, Sheathing Woodboard Panel Products

Asbestos -cement board 120 4.0 - 0.23 - 0.24 Asbestos- cement board 0 125 in. 120 - 33.00 - 0.03 Asbestos- cement board 0 25 in. 120 - 16.50 - 0.06 Gypsum or plaster board 0 375 in. 50 - 3.10 - 0.32 0.26 Gypsum or plaster board 0 5 in. 50 - 2.22 - 0.45 Gypsum or plaster board 0 625 in. 50 - 1.78 - 0.56 Plywood (Douglas Fir)° 34 0.80 - 1.25 - 0.29 Plywood (Douglas Fir) 0 25 in. 34 - 3.20 - 0.31 Plywood (Douglas Fir) 0 375 in. 34 - 2.13 - 0.47 Plywood (Douglas Fir) 0 5 in. 34 - 1.60 - 0.62 Plywood (Douglas Fir) 0 625 in. 34 - 1.29 - 0.77 Plywood or wood panels 0 75 in. 34 - 1.07 - 0.93 0.29 Vegetable Fiber Board

Sheathing, regular density 0 S in. 18 - 0.76 - 1.32 0.31 0 78125 in. 18 - 0.49 - 2.06

Sheathing intermediate density 0 5 in. 22 - 0.82 - 1.22 0.31 Nail -base sheathing 0 S in. 25 - 0.88 - 1.14 0.31 Shingle backer 0 375 in. 18 - 1.06 - 0.94 0.31 Shingle backer 0 3125 in. 18 - 1.28 - 0.78 Sound deadening board 0 5 in. 15 - 0.74 - 1.35 0.30 Tile and lay -in panels, plain or

acoustic 18 0.40 - 2.50 - 0.14 0 5 in. 18 - 0.80 - 1.25

075 in. 18 - 0.53 - 1.89 Laminated paperboard 30 0.50 - 2.00 - 0.33 Homogeneous board from

repulped paper 30 0.50 - 2.00 - 0.28 Hardboard

Medium density 50 0.73 - 1.37 - 0.31

High density, service temp. service underlay 55 0.82 - 1.22 - 0.32

High density, std. tempered 63 1.00 - 1.00 - 0.32 Particleboard

Low density 37 0.54 - 1.85 - 0.31

Medium density 50 0.94 - 1.06 - 0.31 High density 62.5 1.18 - 0.85 - 0.31 Underlayment 0 625 in. 40 - 1.22 - 0.82 0.29

Wood subfloor 0 75 in. - 1.06 - 0.94 0.33

BUILDING MEMBRANE

Vapor -permeable felt - - 16.70 - 0.06 Vapor -seal, 2 layers of mopped

IS -lb felt - - 8.35 - 0.12 Vapor -seal, plastic film - - - - Ncgl.

FINISH FLOORING MATERIALS

Carpet and fibrous pad Carpet and rubber pad

- - - - 0.48 0.81

- - 2.08 1.23

0.34 0.33

Cork tile 0 125 in. - - 3.60 - 0.28 0.48

Terrazzo 1 in. - - 12.50 - 0.08 0.19

Tile -asphalt, linoleum. vinyl, rubber - - 20.00 - 0.05 0.30 vinyl asbestos 0.24 ceramic 0.19

Wood, hardwood finish 0 75 in. 1.47 - 0.68

INSULATING MATERIALS

Blanket sad Batt° Mineral Fiber, fibrous form processed

from rock, slag, or glass approx.e 3-4 in. 0.3 -2.0 - 0.091 - /pi approx.e 3.5 in approx.e 5.5 -6.5 in. approx.e 6-7.5 in

0.3 -2.0 0.3 -2.0 0.3 -2.0

- - - 0.077 0.053 0.045

- - - 13d 19d 21d

approx.e 9-10 in approx.e 12 -13 in.

0.3 -2.0 0.3 -2.0

- - 0.033 0.026

- - 30d 38d

Tahiv 3-I 6 [63_ Thermal Properties of Typical Building and Insulating Materials- Design Values'

Description Density Conduc- Conduc- Resistance e(R) Specific Ib /ft3 lisilys tance Heal,

A (C) Per inch For thick- Btu /lb Btu in./ Btu /h thickness ness listed deg F h'fi2F ft2F (I /A) (1 /Cf)

hft2 heft F /Btu F /Btu

Board and Slabs

Cellular glass 8.5 0.35 - 2.86 - 0.18 Glass fiber, organic bonded 4 -9 0.25 - 4.00 - 0.23 Expanded perlite, organic bonded 1.0 0.36 - 2.78 - 0.30 Expanded rubber (rigid) 4.5 0.22 - 4.55 - 0.40 Expanded polystyrene extruded

Cut cell surface 1.8 0.25 - 4.00 - 0.29 Smooth skin surface 1.8 -3.5 0.20 - 5.00 - 0.29

Expanded polystyrene, molded beads 1.0 0.26 - 3.85 - - 1.25 0.25 - 4.00 - - 1.5 0.24 - 4.17 - - 1.75 0.24 - 4.17 - - 2.0 0.23 - 4.35 - -

Cellular polyurethanes (R -11 exp.)(unfaced) 1.5 0.16 - 6.25 - 0.38 Cellular polyisocyanurate "(R -11 exp.) (foil

glass fiber reinforced core) faced,

2.0 0.14 - 7.20 - 0.22 Nominal 0.5 in - 0.278 - 3.6 Nominal 1.0 in - 0.139 - 7.2 Nominal 2.0 in - 0.069 - 14.4

Mineral fiber with resin binder 15.0 0.29 - 3.45 - 0.17

Mineral fiberboard, wet felted Core or roof insulation 16-17 0.34 - 2.94 - Acoustical tile 18.0 0.35 - 2.86 - 0.19 Acoustical tile 21.0 0.37 - 2.70 -

Mineral fiberboard, wet molded Acoustical tiles, 23.0 0.42 - 2.38 - 0.14

Wood or cane fiberboard Acoustical tiles 0 5 in. - - 0.80 - 1.25 0.31 Acoustical tile' 0 75 in. - - 0.53 - 1.89

Interior finish (plank, tile) 15.0 0.35 - 2.86 - 0.32 Cement fiber slabs (shredded wood

with Portland cement hinder 25 -27.0 0.50 -0.53 - 2.0 -1.89 - - Cement fiber slabs (shredded wood

with magnesia oxysulfide binder) 22.0 0.57 - 1.75 - 0.31

LOOSE FILL Cellulosic insulation (milled paper or

wood pulp) 2.3 -3.2 0.27 -0.32 - 3.70 -3.13 0.33 Sawdust or shavings 8.0 -15.0 0.45 - 2.22 0.33 Wood fiber, softwoods 2.0 -3.5 0.30 - 3.33 0.33 Perlite, expanded 2.0 -4.1 0.27 -0.31 - 3.7 -3.3 - 0.26

4.1 -7.4 0.31 -0.36 - 3.3 -2.8 - 7.4 -11.0 0.36 -0.42 - 2.8 -2.4 -

Mineral fiber (rock, slag or glass) approx.( 3.75 -5 in 0.6 -2.0 - - 11.0 0.17 approx.' 6.5 -8.75 in. 0.6 -2.0 - - 19.0 approx.( 7.5 -10 in 0.6 -2.0 - - 22.0 approx.' 10.25 -13.75 in 0.6 -2.0 - - 30.0

Mineral fiber (rock, slag or glass) approx.t3.S in. (closed sidewall application) 2.0 -3.5 - - - 12.0 -14.0

Vermiculite, exfoliated 7.0 -8.2 0.47 - 2.13 - 0.32 4.0 -6.0 0.44 - 2.27 -

FIELD APPLIEDcl Polyurethane foam 1.5 -2.5 0.16-0.18 6.25 -5.26 Ureaformaldehyde foam 0.7 -I.6 0.22 -028 3.57 -4.55 Spray cellulosic fiber base 2.0.6.0 0.24 -0.30 3.33 -4.17

PLASTERING MATERIALS Cement plaster, sand aggregate 116 5.0 0.20 0.20

Sand aggregate 0 375 in. - 13.3 0.08 0.20 Sand aggregate 0 75 in. - 6.66 0.15 0.20

Gypsum plaster: Lightweight aggregate 0 5 in. 45 3.12 0.32 Lightweight aggregate 0 625 in. 45 2.67 0.39 Lightweight agg. on metal lath O 75 in. - 2.13 0.47 Perlite aggregate 45 1.5 0.67 0.32

Table 3-1 G C6] Thermal Properties of Typical Building and Insulating Materials -Design Values'

Description Density Conduc- Conduc- Resistance t(R) Specific lb/ft3 tivlob Lance Hew.

A (C) Per inch For thick- Btu/lb Btuin./ Btu /h thickness cress listed deg F hft2F ft2F (1/x) (1/(1

hft2 h IV F /Btu F /Btv

PLASTERING MATERIALS Sand aggregate Sand aggregate Sand aggregate Sand aggregate on metal lath Vermiculite aggregate

0 5 in. 0 625 in. 0 75 in.

105 105 105 - 45

5.6

1.7

11.10 9.10 7.70

0.18

0.59

0.09 0.11 0.13

0.20

MASONRY MATERIALS Concretes

Cement mortar 116 5.0 - 0.20 Gypsum -fiber concrete 87.5% gypsum,

12.5% wood chips 51 1.66 - 0.60 0.21 Lightweight aggregates including ex- 120 5.2 - 0.19

panded shale, clay or slate; expanded 100 3.6 - 0.28 slags; cinders; pumice; vermiculite; 80 2.5 - 0.40 also cellular concretes 60 1.7 - 0.59

40 1.15 - 0.86 30 0.90 - 1.11 20 0.70 - 1.43

Perlite, expanded 40 0.93 - 1.08 30 0.71 - 1.41 20 0.50 - 2.00 0.32

Sand and gravel or stone aggregate (oven dried) 140 9.0 - 0.11 0.22

Sand and gravel or stone aggregate (not dried) 140 12.0 - 0.08

Stucco 116 5.0 - 0.20

MASONRY UNITS

Brick, common' 120 5.0 - 0.20 - 0.19 Brick, face' 130 9.0 - 0.11 - Clay tile, hollow:

1 cell deep 3 in. - - 1.25 - 0.80 0.21 1 cell deep 4 in. - - 0.90 - 1.11 2 cells deep 6 in. - - 0.66 - 1.52 2 cells deep 8 in. - - 0.54 - 1.85 2 cells deep 10 in. - - 0.45 - 2.22 3 cells deep 12 in. - - 0.40 - 2.50

Concrete blocks, three oval core: Sand and gravel aggregate 4 in. - - 1.40 - 0.71 0.22

Bin. - - 0.90 - 1.11 -

12in. - - 0.78 - 1.28 Cinder aggregate 3 in. - - 1.16 - 0.86 0.21

4 in. - - 0.90 - 1.11 8 in. - - 0.58 - 1.72

12 in. - - 0.53 - 1.89 Lightweight aggregate 3 in. - - 0.79 - 1.27 0.21

(expanded shale, clay, slate 4 in. - - 0.67 - 1.50 or slag; pumice)- 8 in. - - 0.50 - 2.00

12 in. - - 0.44 - 2.27 Concrete blocks, rectangular core.i.k

Sand and gravel aggregate 2 core, 8 in. 36 lb. 0.96 1.04 - 0.22 Same with filled cores' 0.52 1.93 0.22

Lightweight aggregate (expanded shale, clay, slate or slag, pumice): 3 core, 6 in. 191b. 0.61 1.65 0.21 Same with filled cores' 0.33 2.99 2 core, 8 in. 24 lb. 0.46 2.18 Same with filled cores' 0.20 5.03 3 core, 12 M. 38 lb. 0.40 2.48 Same with filled cores' 0.17 5.82

Stone, lime or sand 12.50 0.08 0.19

Gypsum partition tile: 3 12 30 in. solid 0.79 1.26 0.19 3 12 30 in.4 -cell 0.74 1.35 4 12 30 in. 3 -cell 0.60 1.67

METALS (See Chapter 39. Table 3)

D

To, He 3-( ci - Thermal Properties of Typical Building and Insulating Materials- Design Values'

Description Density Conduc- Conduc- Resistance e (R) Specific lb/ft3 thit)b lance Heat,

A (C) Per inch For thick- Btu /Ib Btuin./ Btu /h thickness ness listed deg F hft2F ft2 I: (1 /A) (1/Ç)

hfi2 hft F /Btu F /Btu

ROOFINGS

Asbestos- cement shingles Asphalt roll roof¡ng Asphalt shingles Built -up roofing 0 375 in. Slate 0 5 in. Wood shingles, plain and plastic film faced

120 70 70 70 - -

- - - - - -

4.76 6.50 2.27 3.00

20.00 1.06

- - - - - -

0.21 0.15 0.44 0.33 0.05 0.94

0.24 0.36 0.30 0.35 0.30 0.31

SIDING MATERIALS (on flat surface)

Shingles Asbestos -cement 120 - 4.75 - 0.21 Wood, 16 in., 7.5 exposure - - 1.15 - 0.87 0.31 Wood. double, 16 -in., 12 -in. exposure - - 0.84 - 1.19 0.28 Wood, plus instil. backer board, 0.3125 in - - 0.71 - 1.40 0.31

Siding Asbestos- cement, 0.25 in., lapped - - 4.76 - 0.21 0.24 Asphalt roll siding - - 6.50 - 0.15 0.35 Asphalt insulating siding (0.5 in. bed.) - - 0.69 - 1.46 0.35 Hardboard siding, 0.4375 in. 40 1.49 - 0.67 0.28 Wood, drop, 1 8 in - - 1.27 - 0.79 0.28 Wood, bevel, 0.5 8 in., lapped - - 1.23 - 0.81 0.28 Wood, bevel, 0.75 10 in., lapped - - 0.95 - 1.05 0.28 Wood, plywood, 0.375 in., lapped - - 1.59 - 0.59 0.29

Aluminum or Steelm, over sheathing Hollow- backed - - 1.61 - 0.61 0.29 Insulating -board backed nominal

0.375 in. - - 0.55 - 1.82 0.32 Insulating -board backed nominal

0.375 in., foil backed 0.34 2.96 Architectural glass -- - 10.00 - 0.10 0.20

WOODS (12% Moisture Content) °P

Hardwoods 0.39 Oak 41.246.8 1.12 -1.25 - 0.89 -0.80 - Birch 42.6 -45.4 1.16 -1.22 - 0.87 -0.82 - Maple 39.8.44.0 1.09 -1.19 - 0.94 -0.88 - Ash 38.4-41.9 1.06 -1.14 - 0.94 -0.88

Softwoods 0.39 Southern Pine 35.6 -41.2 1.00 -1.12 - 1.00 -0.89 - Douglas Fir -Larch 33.5 -36.3 0.95 -1.01 - 1.06 -0.99 - Southern Cypress 31.4 -32.1 0.90-0.92 - 1.11 -1.09 - Hem -Fir, Spruce- Pine -Fir 24.5 -31.4 0.74 -0.90 ' - 1.35 -1.11 - West Coast Woods, Cedars 21.7 -31.4 0.68 -0.90 - 1.48 -1.11 - California Redwood 24.5 -28.0 0.74 -0.82 - 1.35 -1.22 -

Notes for Table 3 - 'Except where otherwise noted, all values are for a mean temperature of 75 F. Representative values for dry materials. selected by ASH RAE TC 4.4, are intended

as design (not specification) values for materials in normal use. Insulation materials in actual service may have thermal values that vary from design values depending on their in -situ properties (e.g., density and moisture content). For properties of a particular product, use the value supplied by the manufacturer or by unbiased tests.

"To obtain thermal conductivities in But /h ft2 F. divide the A value by 12 in. /ft. e Resistance values are the reciprocals of C before rounding off C to two decimal places. 'Does not include paper backing and facing. if any. Where insulation forms a boundary (reflective or otherwise) of an air space, see Tables 2A and 2B for the

insulating value of an air space with the appropriate effective emittance and temperature conditions of the space. °Conductivity varies with fiber diameter. (See Chapter 20. Thermal Conductivity section.) Insulation is produced in different densities, therefore, there is a wide

variation in thickness for the same Rvalue among manufacturers. No effort should be made to relate any specific R -value to any specific density or thickness. tValues are for aged, unfaced, board stock. For change in conductivity with age of expanded urethane, see Chapter 20, Factors Affecting Thermal Conductivity. SInsulating values of acoustical tile vary, depending on density of the board and on type. size and depth of perforations. bASTM C 855 -77 recognizes the specification of roof insulation on the basis of the C- values shown. Roof insulation is made in thickness to meet these values. ¡Face brick and common brick do not always have these specific densities. When density differs from that shown, there will be a change in thermal conductivity. iAt 45 F mean temperature. Data on rectangular core concrete blocks differ from the above data on oval core blocks. due to core configuration, different mean

temperatures, and possibly differences in unit weights. Weight data on the oval core blocks tested are not available. "Weights of units approximately 7.625 in. high and 15.75 in. long. These weights are given as a means of describing the blocks tested, but conductance values are

all for I ft2 of area. I Vermiculite, perlite, or mineral wool insulation. Where insulation is used, vapor barriers or other precautions must be considered to keep insulation dry. O1 Values for metal siding applied over flat surfaces vary widely, depending on amount of ventilation of air space beneath the siding; whether air space is reflective

or nonreflective; and on thickness, type, and application of insulating backing -board used. Values given are averages for use as design guides, and were obtained -om several guarded hotbox tests (ASTM C236) or calibrated hotbox (ASTM C 976) on hollow- backed types and types made using backing -boards of wood fiber,

.named plastic, and glass fiber. Departures of±50% or more from the values given may occur. "Time -aged values for board stock with gas -barrier quality (0.001 in. thickness or greater) aluminum foil facers on tow major surfaces. °See Ref. 5. PSee Ref. 6. 7, 8 and 9. The conductivity values listed are for heat transfer across the grain. The thermal conductivity of wood varies linearly with the density and

the density ranges listed are those normally found for the wood species given. If the density of the wood species is not known. use the mean conductivity value.

Table 3 -1 gives the thermal conductivity K for a wide variety

of building and insulating materials.

Other useful data are also given in Table 3 -1; for example,

the reciprocal of the unit thermal resistance and the unit thermal

conductance, C. Note that K has the units of

(BTU- in) /(ft2 hr -F). With a x given in inches, the unit thermal

conductance C is given by

C = _ Ctstu /11Y'- f t2-ß) (3 -7) z)(

We established some basic ideas about convection in the

Chapter One. The convection heat transfer coefficient or the film

coefficient h appearing in Eq. (1 -10) depends on the fluid, the

fluid velocity ect.. Many correlations exist for predicting the

covection heat transfer coefficeint under various conditions.

More details about correlations for forced convection are given in

Chapter 3 of ASHRAE Handbook, 1985 Fundamentals.

Most building structures have forced convection along outer

walls or roofs, and natural convection occurs inside narrow air

spaces and on the inner walls. There is considerable variation in

surface conditions, and both the direction and magnitude of the air

motion on outdoor surfaces are very unpredictable.

Eq. 1 -10 may also be expressed in terms of thermal resistance:

tT"'_ -t5 -too

where

or

(3 -8)

RI _ (,%r F)/13-N (3-9)

hA

R = e

FJ/ßtK(3-10) h

74

The thermal resistance given by Eq. 3 -9 may be summed with

the thermal resistance arising from pure conduction given by Eq.3-

4.

For building structures, the convective heat transfer

coefficient usually ranges from about 1.0 BTU /(hr- F -ft2)

for free convection up to about 6 BTU /(hr -F -ft 2) for force

convection with an air velocity of about 15 miles per hour.

Because of the low convective heat transfer coefficient, especially

with free convection, the amount of heat transferred by thermal

radiation may be equal to or larger than that transferred by

convection."[ 53 'Thermal radiation is the transfer of thermal energy by

electromagnetic waves and is an entirely different phenomenon from

conduction and convection. In fact thermal radiation can occur in

a perfect vaccum. "C3

Fig 3 -5 shows situations where radiation may be a significant

factor. For the wall

and for the air space

a .ì ' c The resistances can be combined to obtain an equivalent

overall resistance R' with which the heat transfer rate can be

computed using Eq. 3 -11

`C-6o - ) tz'

(3 -11)

75

Fig 3 -5

AThe thermal resistance for radiation is not easily computed,

however, because of the fourth power temperature relationship of

Eq. -1 -12. For this reason and because of the inherent

uncertainty in describing the physical situation, theory and

experiment have been combined to develop,.combined or effective

unit thermal resistances and thermal conductances for many

typical surfaces and air spaces. "[E J

Table 3 -2 gives effective film coefficients and unit thermal

resistances as a function of wall position, direction of heat

flow, air velocity, and surface emittance for exposed surfaces

such as outside walls." [5J

-7&

Surface Conductances (131u /(h ft' F)j and Resistances 1(h ft= F) /fltu] fur Air'

Surface Enrirtance Position of Direction Non- Reflective Reflective

Surface of Heat' reflective c = 0.20 c = 0.05 Flow c = 0.90

STII.L AIR Horizontal Sloping-45 deg Vertical Sloping --45 deg Horizontal

CNIOVING AIR (Any Position) 15 -mph Wind

(for winter) 7.5 -mph Wind (for summer)

hi R ht R h R

Upward 1.63 0.61 0.91 1.10 0.76 1.32 Upward 1.60 0.62 0.88 1.14 0.73 1.37 Horizontal 1.46 0.68 0.74 1.35 0.59 1.70 Downward 1.32 0.76 0.60 1.67 0.45 2.22 Downward 1.08 0.92 0.37 2.70 0.22 4.55

hp R ho R ho R

Any 6.00 0.17

Any 4.00 0.25

allo surface has both an air space resistance value and a surface resistance value. No air space value exists for any surface facing an air space of less than 0.5 in.

bFor ventilated attics or spaces above ceilings under summer conditions (heat flow dos. n) see Table 4.

cConductances are for surfaces of the stated emittance facing virtual blackbody surroundings at the same temperature as the ambient air. Values are based on a surface -air temperature difference of 10 deg F and for surface temperature of 70 F.

dSce Fig. 2 for additional data.

Table 3 -2 [6]

Table 3 -3 gives representative valuse of emittance E for

some building and insulating materials

Reflectivity and Emittance Values of Various Surfaces and Effective Emittances of Air Spaces

Effective Emittance E of Air Space

Surface Reflectivity Average in Percent Emittance e

One surface emit-

tance t; the other

0.90

Both surfaces

emit- tances e

Aluminum foil, bright 92 to 97 0.05 0.05 0.03 Aluminum sheet 80 to 95 0.12 0.12 0.06 Aluminum coated paper,

polished 75 to 84 0.20 0.20 0.11 Steel. galvanized. bright 70 to 80 0.25 0.24 0.15 Aluminum paint 30 to 70 0.50 0.47 0.35 Building materials: wood,

paper, masonry, nonmetallic paints 5 to 15 0.90 0.82 0.82

Regular glass 5 to 15 0.84 0.77 0.72

Table 3 -3 [6]

77

For example, a typical vertical brick wall exposed to the

outdoors has an effective emittance E of about 0.8 to 0.9.

In still air the average film coefficient from table 3 -2, is about

1.46 BTU /(hr- F -ftZ) and the unit thermal resistance is

0.68 (hr- F -ft2) /BTU.

If the surface were highly effective, E = 0.05, the film

coefficent would be 0.59 BTU /(hr -F -ft 2) and the unit thermal

resistance would be 1.7 (hr- F -ftZ) /BTU. It is evident that thermal

radiation is a large factor when natural convection occurs. If the

wind velocity were to increrse to 15 mph, the convective heat

transfer coefficient would increase to about

6 BTU /(hr- ft2 -F). With higher air velocities the relative effect

of radiations dimishes.

Table 3 -4 and 3 -5 give conductances and resistances for air

spaces as a function of position, direction of heat flow, air

temperature, and the effective emittance of the space. The

effective emittance E is given by

l I (3 -12)

where e, and E2 are for each surface of the air space.

The effect of radiation is quite apparent in tables 3 -4 and

3 -5, where the thermal resistance may be observed to decrease by

a factor of two or three as E varies from 0.05 to 0.82.

The preceding paragraphs cover thermal resistances arising

from conduction, convection and radiation. Eq.3 -6 may be

generalized to give the equivalent resistance of n resistors in

series.

8

`Th6 le3-4 C6] Îf:CrCi:a Tì(ti1F Ilil' of li:lIICa Air S1;:tict"c 1.lI-F,lllu l'uihnn

of Air

5;a<c

Picc.;i+t Air 1i:cni 01 NJcan

Meat Tcmp.° 1 lu. (FI

'1 emir Di! l,`' (dc; F)

0.5-in. A;r Sparc` 0.75-1r.. Air?pa. t`

0.03 l':iluc+,f/d.e

0.05 0.2 (15 0.82 0.03 LalucOfLd.e

O.b_ 0.2 0.5 11.63 9:1 10 2.13 2.03 1.5; 0.99 0,73 2.34 2.2: 1.61 1.04 o -< 50 30 I.62. 1.57 1.29 0.96 0 75 1.71 1.5.-5 1.35 0.9'J o -- 50 10 2.13 2.05 l.(.1 Il 0.54 '_.30 2.2; 1.70 1 IG 0>; Hori. 1:;. 0 2l1 1.'3 1.70 1.45 12 091 1.?3 I. 9 f.5_' 1.16 0 3 (1 10 2. 10 2.04 1.?f1 .27 1.00 2.23 2.15 1.7); 1.31 I`'3 -50 0 l'9 I.a6 1.49 .23 I A4 l. î7 1. 4 1.55 l." l' - -50 IU 2.ú4 2.00 I.75 40 1.16 2.16 2.11 1.84 1.46 1.2_,

90 Ir) 2 44 2.31 I.65 .05 0.76 2.96 2.75 1.58 1.15 0 51 50 30 2 C6 1.98 156 .IO 0 33 1.99 1.92 1.52 I 05 0.3- 45° / 50 10 2.55 2.44 I 83 22 0.90 2.90 2.75 2.00 1.29 0.94 Slope L';. 0 20 2.20 2.14 1.76 .30 1.02 2.13 2.07 1.72 1.28 1t ri /

0 IO 2.63 2.54 2 03 .44 1 10 2.72 2.62 2.08 1.47 1.13 -50 20 2 (+5 2.04 1.73 .42 1.17 2.05 2.01 1.76 1.41 I.14 -50 10 2.62 2.56 2.17 .66 1.33 2.53 2.47 2.10 1.62 1.30

90 10 2 4' 2.34 1.67 .06 0.77 3.50 3.24 2.05 1.22 0 S4 50 30 2.57 2.46 1.34 .23 0.90 2.91 2.7" 2.01 1.30 0 94 50 0

10 2.56 2.54 1.58 .24 0.91 3.70 3.46 2.35 1.43 1.01 Vertical 20 2 S2 2.72 2.14 .50 1.13 3.14 3.02 2.32 1.58 1.18 Horiz. -t

0 10 2.93 2.82 2.20 .53 1.15 3.77 3.59 2.64 1.73 1.26 -50 21 2.90 2.82 2.35 .76 1.39 2.90 2.53 2.36 1.77 1.39 -50 10 3.20 3.10 2.54 .87 1.46 3.72 3.60 2.87 2.04 1.56

90 10 2.45 2.34 1.67 .06 0.77 3.53 3.2" 2.10 1.22 0 5: 50 30 2.64 2.52 1.87 .24 0.91 3.43 3.23 2.24 1 .39 0 99 45° 50 10 2.67 2.55 1.89 .25 0.92 3.81 3.5" 2.40 1.45 1.02 Slope Do n 0 20 2.91 2.80 2.19 .52 1.15 3.75 3.5" 2.63 1.72 1.76 ' 0 10 2.94 2 83 2.21 .53 1.15 4.12 3.91 2.61 1.80 1.30 -50 20 3.16 3.07 2.52 .86 1.45 3.78 3.65 2.90 2.05 1.5" -50 10 3.26 3.16 2.55 .89 1.47 4.35 4.1' 3.22 2.21 1.66

90 10 2.43 2.34 1 67 .06 0.77 3.55 3.29 2.10 1.22 0 S5 50 30 2 66 2.54 1.83 .24 0.91 3.77 3.52 2.33 1.44 I 02 50 10 2.67 2.55 1.89 .25 0.92 3.54 3w 2.41 1.45 1.02 Horiz Do» c 0 20 2.94 2.83 2.20 .53 1.15 4.18 3.96 2 83 1.81 1.1,1 0 I0 2.96 2.85 2.2' .53 1.16 4.25 4.02 2.87 1.52 1.31 -5(1 20 3.25 3.15 2.58 .89 1.47 4.60 4.41 3.36 '-.'-S 1 69 -crl 10 3.25 3.IS 2.60 .90 1.47 4.71 4.51 3.42 2.30 1.-I

ai, te :5-S C6 Thermal Resistances of Plane' Air Spaces (Concluded) 1F.CLK)V A

Position of

Air Space

Direction of

Heat Flow

Air Space Mean Temp

Temp,d Diff.a (FI (del! H

1.5 -in. Air Spacee 3.5 -in. Air Spacec

0.03 Value ofEde

0.05 0.2 0.5 0.82 0.03 %aloe of Ed.'

0.05 0.2 0.5 0.82 90 10 2.55 2.41 1.71 1.08 0.77 2.84 2.66 1.83 1.13 0.80 SO 30 1 87 1.81 1.45 1.04 0.80 2(19 2.01 I.58 1.10 0.84 50 It) 2 50 2.40 I 81 1.21 0.89 2.80 2.66 1.95 1.28 0.93 Ilunz !!p 0 2u 2 01 195 1 63 1.23 097 2.25 2.IS 1.79 1.32 1.03

t1 10 2.43 2.35 1.90 1.38 1.06 2.71 2.62 2.07 1.47 1.12 -50 20 194 1.91 1.68 1.36 1.13 2.19 2.14 1.86 1.47 1.20 -50 Itl 2.37 2.31 1.99 1.55 1.26 2.65 2.58 2.18 1.67 1.33

90 I() 2.92 2.73 1.86 1.14 0.80 3.18 2.96 1.97 1.18 0.82 50 30 2.14 2.06 1.61 1.12 0.84 2.26 2.17 1.67 1.15 0.86 45°

Slope 50 /1 0 Up

10 20

2 SS 2.30

2.74 2.23

1.99 1482

1.29 1.34

0.94 1.04

3.12 2.42

2.95 2.35

2.10 1.90

1.34 1.38

0.96 1.06 0 10 2.79 2.69 2.12 1.49 1.13 2.9S 2.87 2.23 1.54 1.16 -50 2() 2 22 2.17 1.68 1.49 1.21 2.34 2.29 1.97 1.54 1.25 -50 10 2.71 2.64 2.23 1.69 1.35 2.S7 2.79 2.33 1.75 1.39

90 10 3.99 3.66 2.25 1:27 42.87. 3.69 3.40 2.15 1.24 'Q.á5- 50 30 2.58 2.46 1.84 1.23 0.941 2.67 2.55 1.89 1.25 t,.91 50 10 3.79 3.55 2.39 1.45 1.02 3.63 3.40 2.32 1.42 1.01 Vertical 211 2.76 2.66 2.10 1.48 1.12 2.58 2.78 2.17 1.51 1.14 Horiz. ----la.- 0

0 10 3.51 3.35 2.51 1.67 1.23 3.49 3.33 2.50 1.67 1.23 -50 2t) 2.64 2.58 2.18 1.66 1.33 2.82 2.75 2.30 1.73 1.37 -50 10 3.31 3.21 2.62 1.91 1.48 3.40 3.30 2.67 1.94 1.50

90 10 5.07 4.55 2.86 1.36 0.91 4.81 4.33 2.39 1.34 0.90 50 30 3.5d 3.36 2.31 1.42 1.00 3.51 3.30 2.28 1.40 1.00 45' 50 10 5.10 4.66 2.85 1.60 1.09 4.74 4.36 2.73 1.57 1.08 Slope Down 0 2() 3.55 3.66 2.68 1.74 1.27 3.81 3.63 2.66 1.74 1.27 0 10 4.92 4.62 3.16 1.94 1.37 4.59 4.32 3.02 1.88 1.34 -50 20 3.62 3.50 2.60 2.01 1.54 3.77 3.63 2.90 2.05 1.57 -50 10 4.67 4.47 3.40 2.29 1.70 1.50 4.32 3.31 2.25 1.68

90 IO 6.09 5.35 2.79 1.43 0.94 0.07 8.19 3.41 1.57 1.00 50 30 6 27 5.63 3.18 1.70 1.14 9.60 8.17 3.56 1.88 1.22 5() 10 6.61 5.90 3.27 1.73 1.15 1.15 9.27 4 09 1.93 1.24 Horiz. Down 0 20 7.03 6.43 3.91 2.19 1.49 0.90 9.52 4.S7 2.47 1.62

1 o

-St) 10 20

7.31 7.73

6.66 7.20

4.00 4.77

2.22 2.85

1.51 1.99

1.97 1.64

10.32 10.49

5.08 5.02

2.52 3.25

1.64 2.18 -50 I0 8 09 7.52 4.91 2.89 2.01 2.98 11.56 ,.36 3.34 2.22

7q

Re' = Rl' + R2'+ R3' + R4' .... +Rn' (3-13)

Fig 3 -6 is an example of a wall being heated or cooled by a

combination of convection and radiation on each surface and

having five different resistances through which the heat must be

conducted.

The equivalent thermal resistance Re' for the wall is given

by Eq. 3 -13 as

Re' = Ri' + Rl' + R2' + R3' + Ro' (3 -14)

Each of the resistances may be expressed in terms of

fundamentals variables using Equation 3 -4 and 3 -9.

11....11- li -6C,

a4C---' .... j ---0 X L

g

cKi

f

Fig 3-6

c - 1

-f- öX -t-

ßXz -t 6 t --1- C3 - 1ft A4 kl 4 1

K 2-14 z 1-3/4 3 ! 1 ct' a

The convective heat transfer coefficient may be read from

Table 3 -2 and the thermal conductivities may be obtained from

Table 3 -1. For Fig. 3 -6, a plane wall, the areas in Eq. 3 -15

are all equal and cancelled.

Thermal resistances may also occur in parallel. In theory

the parallel resistance can be combined into an equivalent

thermal resistance in the same way as electrical resistances,

Eq. 3 -16 gives a reasonable approximation of the equivalent

thermal resistance.

I

Fe! + f - --- - -t ,

h (3 .--I G)

80

For large variations in the thermal resistance of parallel

conduction paths, numerical techniques using the computer

suggested.

k The concept of thrermal resistance is very useful and

convenient in the analysis of complex arrangements of building

materials. After the equivalent thermal resistance has been

determined for a configuration, however, the overall unit thermal

conductance, usually called the overall heat transfer coefficient

U, is frequently used.

U.= _ 1 (63TV /*-ft2-F) A 1R. (3 -17)

The heat transfer rate is then given by

gf= U A of where

(3 -18)

UA = conductance, BTU /(hr -F)

A = surface area, ft 2.

.6t = overall temperature difference, F

For a plane wall the area A is the same throughout the wall.

In dealing with a curved wall, we select the area for convenience

of calculation. For example, in the problem of heat transfer

through the ceiling- attic -roof combination, it is usually most

convenient to use the ceiling area.

The area selected is then used to determine the appropriate

value of U from 3 -17. "E]

o

3 -3 Tabulated Overall Heat -Transfer Coefficients

Since repetitive calculations have to be done for

calculating the coefficients and other data, tables have been

constructed that give overall coefficients for many common

building sections including walls and floors, doors, windows and

skylights.

The tables used in the ASHRAE Handbook have a great deal of

flexibility and are summarized in the following pages.

Walls and roofs vary considerably in the materials from

which they are constructed. Therefore, the thermal resistance or

the overall heat transfer coefficient is usually computed on an

individual basis using Eqs. 3 -15 and 3 -17.

NExample(3 -1):

A frame wall is modified to have 3 1/2 in. of mineral fiber

insulation between the studs. Compute the overall heat -

transfer coefficient U if the unit thermal resistance

without the insulation is 4.44 (hr -F -ft ) /BTU (2* 4Itstuds

i 2

on 16 in. centers) N

';'.."4511114SPaniti

3

Fig 3 -7

Solution A:

Total unit resistance given 4.44

Deduct the air space unit resistance -1.01

Add insulation unit resistance 11.00

Total R (hr -F -ft Z) /BTU 14.43

2`;,ca.'

(Table 3 -5)

(Table 3 -1)

82

then based on one square foot, we see that

u = --- = f,,--7 T v/ c t-- f 2 - F / - },43 `

Solution B:

Eq. 3 -16 may be used to correct for framing

1 /Rc' = 1 /R' + 1 /Rf or UcAt = UAb + UfAf

where

At = total area

Ab = area between studs

Af = area occupied by the studs

The unit thermal resistance of a section through the 2 * 4

stud is equal to the total resistance less the resistance of the

stud from Table 3 -1. A 2 * 4 stud is only 3 1/2 inch deep.

Rf = 1 /Uf = 4.4 -0.97 + (1.25) (3.5) = 7.81

or

Uf = 0.128 (BTU)/(hr-F-ft )

Then using Eq. 3 -16 we get

Uc = (7,7)04, )- ( ,o29-)(lir)

z c0-1 gT ti- z) 17 / (6

The following example illustrates the calculation of an

overall heat transfer coefficient for an unvented roof -ceiling

system. U(5.J

Example(3 -2):

Computed the overall heat -transfer coefficient for the

roof -ceiling combination shown in Fig 3 -8. The wall has an

overall heat- transfer coefficient of 0.16 BTU /(hr -F -ft 2 .

0-3

The roof without ceiling has a ccn:luctance of 0.13

BTU /(hr- F -ft2). The ceiling has a conductance of 0.2

BTU /(hr- F -ft ). The air space is 2.0 ft in the vertical

direction. The ceiling has an area of 15,000 ft `-and a

perimeter of 500 ft.

Solution:

Fig 3-8

It is customary to base the overall heat -transfer

coefficient on the ceiling area horizontal to the floor. Note

that heat can enter or leave the air space through the roof or

the wall that closes the space around the perimeter.

The thermal resistances of the roof and the wall are

parallel and in series with the resistance of the ceiling. Then

for roof and wall.

Crw = CwAw + CrAr

and

R'rw = 1/Crw = 1/(CwAw + CrAr)

Further

R'o = R'rw + R'c

and (

Rio= rt '

CcJrgw 4 Cv-A-Y Cc c

PARA L_

S C-tz. ( S

where

C = unit conductance; Crw: for roof and wall;

Cw: for wall; Cr: for roof; Cc: for ceiling.

R' = thermal resistance; R'rw: for roof and wall;

R'c: for ceiling; R'o: for overall

A = area; Aw: for wall: Ar: for roof;

Ac: for ceiling.

substitution yields

Then

12-' = ° (ar3 62 C >r,00a

Uo A c

ULP _ - d 0 83 t3rvC r-f-- F (a oo478-0 7) 0 oc>>

_

bufelooY

dar %/

I

MINloor koo-f-

1/41t l kJoolr of-tf cfooe,

Wa.e,Q. S Ch oi,,4.-60[S 60-v- et.

i o (/-2 -ex-444ra- l9 3 -

OS

()reran Coefficients of Heat Transmission (U-Factor) of Windols, Sliding Patio Doors. and Skylights for Use in Peak Load Determination and Mechanical Equipment Sizing Only and Nut in An)

of Ar.ii.:1! Energ:. Us.gc, Bi h12-

Part A. Exterior` Vertical Panels

No Sturm Sash

No

i-uer Sha,:r Indoor Sh.ole

S.Jrnmer Suri=er N1, inter

(

1 IO 0 S3

C

e = t. =

1.02 0.9!

1.(,.-; 0.76 0 4) 0 6S

0O 0 70

0.79 6.75 0.59 0.55 6.1ts.

3 I, 0.62 0.65 0.52 0 5? I 4-in. a.r 0.5S 0.45 1 2. r. 0.49 0 56 0.42 0 52

, 04 0 53 0 35 0 4

4 1

4 :71 zir 0 39

(1,

0 44 0 31 O.40 I, 2-In. au space' 0.31 0.39 0.26 0.36

Ird..6r. Storrn Sash 1in. Air Spaceb Added

to Described Product

s'3]4e Indoor Shade

inier Surnrner W inter Surnrror

(ilitss Duititior Sporn Siish 1-in. Air Spat eh /1,1tlitt to Described flr duet

s.0 lud-or Shale

inter Stuntner' inter Summer"

il 0 50 (1.44 0.49

0.47' 0() 0.39 0 55

0 44 0 60 0.37 0.55 0.40 0.50 0.33 0,45

0.37 04(1 0 35 0.39 0.3: 0.39

0 29 0 28 0.25

If,

0.37

0 11 0 30 F1 34 0 I ' 2 - 0 IQ

0 24 1. I ì rl 2.)

0 27 o 32 .

, - U 2 1 0.31 0.19 0.29

..Cr' lit Indoor Stur in s.th 1-in. Air Space Added

to Described Product

No .'hide Indoor 'harle

S.ng.e Glass, Loss Erniitance Coatined . . .

6. 5..

r; 4- t.

ose

o5:: e 45

0.4 t

03 P 31

04)

U 45 ,r 40

C

ti -V o li

e - t. 37 035 032 0 30 0 36 ::-..........r.:: Cl',. J ...r z:

3 16-tn...,:r. spate' 0.37 0 40 02 0.3r 0 35

I 4-in. air sracci 0 35 0 39 02S 0 36 (1 34

I 2 tn. al:. SrasS.g 0 31 0.13 02 0 35 0 "l0

! 2. -., ..7 sr:',...,

c = ! .) 0 ...') 0 I" 0 24 () II 0 2S

c = t... -1; 02 o 23 0 :2 030 0 :6 0.25 0 :s 0 :0 0.26 0 2 z

: ..! " t. 21

0 I l tt

F.v.r:or.' Horizontal Panels

I .4 1!. r

1 23

suir-ur-

II '4 .

1 1,

.1

Pia.s tic I;r)m, 3 i

Single Walled 1.15 0.80 Double Walled 0.70 0.46

Sointiltl V.inter Suromrr

0.42 0 47

f 0 ,;( 0 45 0 35 0 35 O 31

0 39 0 2? 0 15 0 27 ti 3- 0 :1

o It 0 23

o 3: (1 22 ri 2s 0 20

r i 0 2:

0 45

-10

0 30

0 35

0 34

o 35

o 7.;

0.29 0 25

Table =-"D-& c.3

T7,11(2: 3-7 C Coefficients of Transmission (U) for Wood Doors*, B1u/hftiF

/I( ir

i;l.d 14'st:1'4)0011

NVinterb Summerc

No Storm Door

Wood Storm Door'

Metal Storm 1)oorf

No Storm Door

-. s

_ __

I lor.,),. -:,,,,.. II.: Ii ,! Nor 0.47 0.30 0.32 0.45 .-, .- -o!:.1.:0!c !11,:., Jr 0.39 0.26 0.28 0.38 ! s. 1'.:..i...i ...t.).1:. \ i: it 7 I(-in. r.,:lel, 0.57 0.13 0.37 0.54

-3 4 I lor,,,..\ ,:oi ..: fi.,,h J,,,,..- 0.46 0.29 0.32 0.44 %%II h ,Itick 1.1.n.):.;.:, 0.56 0.33 0.36 0.54

-3 4 ',oh,' core roisli door 0.33 0.28 0.25 0.32

\\ oh ,in,...1-...,:laziol...z 0.46 0.29 0.32 0.44 Wiili ;;I-o:i.itin ,,..1.1-i: 0.37 0.25 0.27 0.36

3 4 pand dor Nith 7 16111. l'allehh 0.54 0.32 0.36 0.52 \\ vsh ,.;i1,.1e1.1.1..111-,2' 0.67 0.36 0.41 0.63 \1:11 1:! wi.tioi,2......1.1-.... 0.50 0.31 0.34 0.48

-1 4 l'.v.,.c1 .10,)- ... i,1'. I 1 .,. ;:i. 1-,inels.' 0.39 0.26 0.28 0.3S

\\-10: ,ir..!: H....,:i..:' 0.61 0.34 0.38 0.58 V...1., :::- ,.:.:::,..,.... .!.. -,, 0.44 0.28 0.31 0.42

i 4 `.,::.: ...c:c :1::]: do,-.. 0.27 0.20 0.21 0.26 V. !;!: i:i.21::»...iiiv...: 0.41 0.27 0.29 0.40

0.33 0.23 0.25 0.32

b C63 Coefficients of Transmission (U) for Steel Doors', Btu!h ft2. F

Door I hickne,,,,

1)escripti,in

1-3 4

0.h I c.:1,

-3 4

Wintert' Summere

No Wood Metal No Storm Storm Storm Storm Door Door' Doorr Door

0.19 0.16 0.17 0.18

0.40 0.39

-111ji1-tnient Factors for Various Window, Sliding Patio Door, and Skylight Ty pes

i act in Parts -1, and B hy These Fac:orsl

'.t I,

;

Storm Sash Storm Sash Applied Over

D,uhle Triple Applied Double or Triple 1r,u1..ning insulating Over Single Insulating

tdass Glass Glass Glass

i t ,) I 00 1.00 1.00 , -.., . I no 0.95 - 1.00 0.90- 1.00 0.95 - 1.00 - 1.2o 1.50 - 1.30 1.40 - 1.20 1.50 - 1.30

. - 1.15 1 ( ) - ..2.5. 0.90 -1.2rí 0 95 - !.25

87

Table 3 -6 contains overall heat transfer coefficients for

windows, skylights, and other light- transmitting panels.

Notice that there are values for summer and winter coefficients.

This difference arises because coefficients for use in the winter

assume a different velocity than in the summer.

Table 3 -6 applies only for air -to -air heat transfer and does not

account for solar radiation.

Table 3 -7 gives overall heat transfer coefficients for common

doors. Again note the difference between summer and winter.

Solar radiation has not been included.

Table 3 -8 gives the adjustment factors for coefficients U of

various window, sliding pation door, and skylight types.

Sb

Chapter 4

Moisture Air Properties and Air Conditioning Process

4 -0 Introduction

4 -1 Psychrometric Chart

4 -2 Classic Moist Air Processes

4 -3 Design Conditions

Sei

4 -0 Introduction

In 1911 Willis H. Carrier made a significant contribution to

the air -conditioning field when he published relations for moist

air properties together with a psychrometric chart. These formulas

became fundamental to the industry.

In about 1945 Goff and Gratch published themodynamic

properties for moist air that were for many years the most accurate

available. New formulations have recently been developed at the

National Bureau of Standards. The properties based on these

formulations are the basis for the thermodynamic properties of

moist air given in the 1985 ASHRAE Handbook, Fundamentals Volume

and Appendix A of this book. Bedford and Warner have shown that

error in caculation of the major

percent when perfect gas

relations are used. This chapter

gas relations.

Material in this chapter emphasizes the thermodynamic

analysis. That is, only the states at the beginning and end of a

process are considered. In the final analysis the nature of the

physical process, the path of the process, will not be included in

this introductive book.

properties

emphasizes

will be less than 0.7

the use of the perfect

9 D

4 -1 The Psychrometric Chart

Some fundamental parameters have to be introduced before

introducing the psychrometric chart.

Atmospheric air is a mixture of many gases plus water vapor

and countless pollutants.

Dry air is the atmospheric air which has no water vapor and

pollutants.

Dry air behaves as a perfect gas. Moist air is a mixture of

air and water vapor. The amount of water vapor may vary from zero

to a maximum determined by the temperature and pressure of the

mixture. The latter case is called saturated air.

Humidity Raatio(W) is the ratio of the mass of the water vapor

my to mass of the dry air ma in mixture.

Temperature is a degree of hotness or coldness of a substance.

Relative humidity is the ratio, in percent, of the amount of

moisture in a volume of air to the total amount which that volume

can hold at the given temperature and atmospheric pressure.

Dew point temperature: Saturation is usually reached by the

air being cooled until its saturation vapor pressure equals the

actual vapor pressure. The temperature of the air at that point

is called the dew -point temperature.

Dry bulb temperature is the air temperature measured with the

familiar mercury or alcohol thermometer. The liquid from a small

reservoir expands into a long column with a very small inside

diameter. If the sunlight strikes the bulb of the thermometer the

reading will be higher than the air temperature because of the

g

direct radiation.

Wet bulb temperature is the temperature of evaporating water

contained in a piece of tissue surrounding the thermometer bulb.

The amount that the evaporating surface will cool is determined

by the difference between the vapor pressure (dry bulb) and the

saturation vapor pressure (wet bulb), if the air is saturated then

the wet bulb temperature is the same as dry bulb temperature. A

graphical representation of the properties of moist air can now be

introduced in the form of psychrometric chart.

In Fig 4 -1 dry bulb temperature is plotted along the horizontal

axis in degrees Fahrenheit or Celsius. The dry bulb temperature

lines are straight but not exactly parallel and incline slightly

to the left. Humidity ratio is plotted along the vertical axis on

the right -hand side of the chart in lbmv /lbma or kgv /kga. The

scale is uniform with horizontal lines. The saturation curve with

values of the wet bulb temperature curves upward from left to

right. Dry bulb, wet bulb, and dew point temperatures all coincide

on the saturation curve.

Relative humidity lines with a shape similar to the saturation

curve appear at regular intervals. The enthalpy scale is drawn

obliquely on the left of the chart with parallel enthalpy lines

inclined downward to the right. Although the wet bulb temperature

lines appear to coincide with the enthalpy lines, they diverge

gradually in the body of the chart and are not parallel to one

another.

The spacing of the wet bulb lines is not uniform. Specific

volume lines appear inclined from the upper left to the lower right

g2

. . . 1...

. . -

1.

Y

' :

IYAIN11 M11\/ iw1 ó111iÒá MOISTURE PER routio ow AIR

-_-_I----I-----_- '. _-.1111 al .Y1I IiG. - 11116011111Preir" __y

i _ _ Í' SS _ . ENE/ .' . g'. . 'lt5 I Pm

.4"

-- --}--: -"..: -..;. .- )

-J. - .

\ i ;?':'

e-t

' '/

r ''1 . `.

0Q L 4 c t ,-

\

i

120'. .r' : I ., . , .,,.

O

1

/. .,!%` ,

/. 1 .J . ;I.

Ì.. +:. : / . / I' HUMIOITY RTIO (W) -POUNDS MO: :T RE PER UND 1

OR` /

AIR -. .

l: V ::.: i' , . ° i °

O P

N Ì1 Willi/ Tï/iI/It1 //IiL/J

il; be117/

,r /flip 7;77/ íi/ü O

A

and are not parallel. A protractor with two scales appears at the

upper left of ASHRAE Chart No.l. One scale gives the sensible heat

ratio and the other ratio of enthalpy difference to humidity ratio

difference. Notice that the enthalpy, specific volume, and

humidity ratio scales are all based on a unit mass of dry air and

not a unit mass of the moist air.

Example:

Read the properties of moist air at 75 F db, 60 F wb, and

standard sea level pressure from ASHRAE psychromatric chart N.1.

Solution:

Humidity Ration W = 0.0077 lbmv /lbma

Relative Humidity 0= 41 %

Enthalpy i = 26.4 Btu /lbma

Specific volume ¿7= 13.65 -f-t37 1bmA.

Dew point temperature, td = 50 F

94

4 -2 Classic Moist Air Processes

The most powerful analytical tools of the air -conditioning design

engineer are the first law of thermodynamics or energy balance, and

the conservation of mass or mass balance. These conservation laws

are the basis for the analysis of moist air processes. It is

customary to analyze these processes by using the bulk average

properties at the inlet and outlet of the device. In actual

practice the properties may not be uniform across the flow area

especially at the outlet, and a considerable

length may be necessary for complete mixing.

In this section we will consider the basic processes that are

a part of the analysis of most systems.

95

Heating or Cooling of Moist Air

When air is heated or cooled without the loss or gain of

moisture, the process yields a straight horizontal line on the

psychrometric chart because the humidity ratio is constant. Such

processes can occur when moist air flows through a heat exchanger.

In cooling, if the surface temperature is below the dew point

temperature of the moist air, dehumidification will occur. This

process will be considered later. Fig 4 -2 shows a schematic of a

device used to heat or cool air. Under steady- state -steady -flow

conditions the energy balance becomes:

l'Yl a 2 2 -{- - Vl Gt i (4-1)

m : mass divided by time, for example lbm /hr, in here means Sm

rate of the dry air.

i : enthalpy Btu /lbma

g : energy divided by time, for example BTU /hr, heat

transfer rate.

The energy balance technique yields a positive number for q for

both cooling and heating, and the direction of heat trnsfer is

implied by the terms heating and cooling. It is to be emphasized

that

and

= Z GL 1/11 Z(l

z- tia 2 4. 1^121,cI'z

(4 -2)

(4 -3)

where it and i2 may be obtained from the psychrometric chart. The

convenience of the chart is evident.

Fig 4 -2

Fig 4 -3 shows heating and cooling processes. Because the moist

air has been assumed to be a perfect gas, Eq. 4 -1 may be rearranged

and written

q = ma (il - i2) (cooling or heating) (4- 4)4[G]

t2

Fig 4 -3 C6]

i

Example:

Find the heat transfer rate required to warm 1500 cfm (4-1*;1/1

of air at 60 F and 90 % relative humidity to 120 F without the

addition of moisture.

Solution:

Equation 4 -1 or 4 -4 may be used to find the required heat

transfer rate. First it is necessary to find the mass flow

rate of the dry air.(m).

YY1e k (4 -5)

Q : volume flow rate

The specific volume (U') is read from chart 1 at tl = 60 F and L

Cb= 90 % as 13.31 7 tXbota:

11)-1 13 6'D) 2 r

Also from chart 1, it = 25.3 BTU /lbma and i2 = 40 BTU /lbma.

Then by using Eq. 4 -4, we get

= 6762 (40.0 - 25.3) = 99,400 BTU /hr

The above example shows that the relative humidity decreases when

the moist air is heated. The reverse process of cooling results

in an increase in relative humidity.([ S]

8

"Cooling and Dehumidifying of Moist Air

When moist air is cooled to a temperature below its dew point,

some of the water vapor will condense and leave the air stream.

Fig 4 -4 shows a schematic of a cooling and dehumidifying device,

Fig 4 -4

and Fig 4 -5 shows the process on the psychrometric chart. Although

the actual process will vary considerably depending on the type of

surface, surface temperature, and flow conditions, the heat and

mass transfer can be expressed in terms of the initial and final

states.

FL1 . 4 -S

By referring Fig 4 -4, we see that the energy balance becomes

g°1

111 a e = 1g- f kr'1 a -( 2 -f- 111 W 14) (4 -6)

and the steady flow mass balance for the water in the air is

*IV W( V) W --i' 1410.(42Z (4 -7)

combining Eq.(4 -6) and (4 -7) we get

= 1 (' (- ,` ,y) 111 c. ( fit), (%)ßc`1,,,, (4-8)

Equation 4 -8 represents the total amount of heat transfer from

the moist air. The last term on the right -hand side of

Eq.(4 -8) is usually small compared to the others and is often

neglected. The following example illustrated this point. "[ G]

Example:

Moist air at 80 F db and 67 wb is cooled to 58 db and 80 %

relative humidity. The volume flow rate is 2,000 cfm and the

condensate leaves at 60 F. Find the heat transfer rate.

Solution:

Eq. (4 -8) applies to this process, which is similar to

Fig 4 -5. The following properties are read from chart 1:

V = 13.85 -ft3/16WlC& it = 31.6 BTU /lbma,

W1 = 0.0112 lbmv /lbma, i2 = 22.9 BTU /lbma,

W2 = 0.0082 lbmv /lbma.

The enthalpy of the condensate is obtained from Appendix B,

iw = 28.08 BTU /lbmw. The mass flow rate ma is obtained from

Eq.(4-5).

Ma = 200o (6o) 8664 lb1g/hr

I3,ß & then from Eq. (4-8)

q = 8664 [(31.6 - 22.9) - (0.0112 - 0.0082) * 28.08]

q = 8664 [(8.7) - (0.084)]

I Oo

The last term, which represents the energy of the

condensate, is quite insignificant in this case. For most

cooling and dehumidifying processes this will be true.

Finally, neglecting the condensate term, q = 75,000 BTU /hr. A

ton of refrigeration is 12,000 BTU /hr

then q = 6.25 tons." C 5] The cooling and dehumidifying process involves both sensible

and latent heat transfer, where sensible heat transfer is

associated with the decrease in dry bulb temperature and the

latent heat transfer is associated with the decrease in

humidity ratio. These quantities may be expressed as , s=Ñt-a-ti1)

and

ejr 1/1 a Ci,- Lo-)

The energy of the condensate has been neglected. Obviously

(4 -11)

The sensible heat factor SHF is defined as qs /q. This

parameter is shown on the semicircular scale of Fig 4 -5. The

use of this feature of the chart is shown later.

101

Heating and Humidifying Moist Air

4A device to heat and humidify moist air is shown schematically

in Fig 4 -6. This process is generally required during the cold

months of the year. An energy balance on the device yields

mat , 11- 4- hi t.J ti4tJ = Wick -LL (4-12)

and a mass balance on the water gives

h Lt/ -f- _ l')'la ("02_ (4-13)

r4.6,:.6;h1 fried ,utyI

{

Fig 4 -6

Equations (4 -12) and (4 -13) may be combined to obtain

or

22¡/e

) 41'1) % VY1 ck 6s- wl /

(4 -14)

(4 -15)

Equation (4 -14) or (4 -15) gives the direction of a straight line

¡D2

that connects the initial and final states on the psychrometric

chart.

Fig 4 -7 shows a typical combined heating and humidifying process.

Fig 4 -7

A graphical procedure makes use of the semicircular scale on

chart 1 to locate the process line. The ratio of the change in

enthalpy to the change in humidity ratio is 4i 1,- -4- Z(J (4 -16)

Fig 4 -7 shows the procedure where a straight line is laid out

parallel to the line one the protractor through state 1. Although

the process may be represented by one line from state 1 to state

2, it is not practical to do this. The heating and humidification

processes are usually carried out sperately, as shown in Fig 4 -6

and 4 -7 as processes 1 -a and a -2.11C S]

Let us then consider the humidification of air without the

addition of heat.

03

Adiabatic Humidifiation of Moist Air

' When moisture is added to moist air without the addition of heat,

Eq. (4 -15) becomes N 1 I 2

- 7 Z w ¿ C A ) (&),_- I.v i

(4 -17)

The direction of the process on the psychrometric chart can vary

considerably. If the injected water is saturated vapor at the dry

bulb temperature, the process will proceed at a constant dry bulb

temperature (1 --t 2). If the water enthalpy is greater than the

enthalpy of saturated vapor at the dry bulb temperature, the air

will be heated and humidified (1-42a). If the water enthalpy is

less than the enthalpy of saturated vapor at the dry bulb

temperature, the air will be cooled and humidified (1--.4 2b).

Fig 4 -8

t b4

Fig 4 -8 shows these processes. One other situation is worthy of

mention. When liquid water at the wet bulb temperature is

injected, the process follows a line of constant wet bulb

temperature, as shown by (1 to 2u) in Fig 4 -8.y['

\'Example:

Moist air at 60 F db and 20 % relative humidity enters a

heater and humidifier at the rate of 1,600 cfm. It is necessary

to heat the air followed by adiabatic humidification so that

it leaves at a temperature of 115 F and a relative humidity of 30

%.

Saturated water vapor at 212 F is injected. Determine the

required heat transfer rate and mass flow rate of the

steam.

Solution:

Fig 4 -6 is a schematic of the apparatus. It is first necessary

to locate the states as shown in Fig 4 -7 from

the given information and Eq. 4 -17 using the protractor

feature of the psychrometric chart. Process 1 -a is sensible

heating; therefore, a horizontal line to the right of state 1 E

constructed.

Process a -2 is determined from Eq.4 -17 and the protractor. elA ow =z =lt 'I BtiA/!bh1

where iw is read from Appendix A. A parallel line is drawn fran

state 2 as shown in Fig 4 -7. State a is determined by

the intersectin on lines 1 -a and a -2. the heat transfer me is then given by

(7 _ MON. C i,a'tI)

1 os

where Vlik _ a(00) ((oo 60 3a lion

and il and is are read from chart 1 as 16.8 and 28.6 B'IRJ /lbma.

Then ç - 7.34::> o (20.6--(0,8) = gof G1-0 8t-( /),) r The mass flowrate of the steam is given by

,,, = Y1 a C t/tl -- i,J, )

where W2 and W1 are read from chart 1 as 0.01275 and 0.0022

lbmu /lbma.

Then mw = 7300 * (0.01275 - 0.0022) = 77.016mv /hr. «C J

Adiabatic Mixing of Two Streams of Moist Air

IThe mixing of air streams is quite common in air -conditioning

systems. The mixing process usually occurs under adiabatic

conditions and with steady flow. Fig 4 -9 illustrates the mixing

of two air streams.

An energy balance gives

Mat -f WIA2Z2= Ni1a3".3

The mass balance on the dry air is

Mal 111 ct z. = tir) Gt 3

And the mass balance on the water vapor is

111ai vi -f }) (12,w2. = 1)\-) Gt3 w3 (4 -20)

By combining Eqs. 4 -18, 4 -19, and 4 -20 and eliminating

we obtain the following result:

(t22 i/t)3

3 _, tA-21

Vti`1cc -- -- l) 1(;cZ

(4 -21)

--_-----

--}- O 1, ¡ fr'Ú2,, i2i/2 Fig 4 -9

I b-7

The state of the mixed streams must lie on a straight line between

states 1 and 2. This is shown on Fig 4 -10. It may be further

inferred from Eq. 4 -21 that the length of the various line segments

are propotional to the masses of dry air mixed.

32 )1k1 ,/ yrl Gc i - _ _-_- -

l 3 / = 2

2

13 tti'

- - 1G(3 12

. kJ,

rsT I 1

' I

f

i

-6, f3 tz

Fig 4 -10

The truth of these statements is most easily shown by solving

Eq. 4-21 for i3 and W3

YVI \

I1,s1aZ -

1

(4 -23)

YY1Gt) z ,t

(A)3 _ yl aZ (4-24)

Iog

Clearly for given states 1 and 2, a straight line will be

generated when various values of h1 ek1 /{4'7a Z are used and the result

plotted on the psychrometric chart. It is also clear

that the location of state 3 on the line is propotional to ' V in3

Consider the case when 1 Gì i/ 1'1Gì 2 = 1, for example. This fact

provides a very convenient graphical procedure for solving mixing

problems in contrast to the use of Eqs. 4 -18, 4 -19 and 4 -20.

It should be noted that the mass flow rate is used when the

graphical procedure is employed; however, the volume flow rates

may be used to obtain approximate results. "CÇJ

`Example:

2000 cfm of air at 100 F db and 75 F wb are mixed with 1000 cfm

of air at 60 F db and 50 wb. The process is adiabatic, a t

steady flow rate and at standard sea level pressure.

Find the condition of the mixed streams.

Solution:

A combination graphical and analytical solution is first

obtained. The initial states are first located on Chart 1

as illustrated on Fig 4 -10 and connected with a stright lie.

Eqs. 4 -19 and 4-20 a re combined to obtain

(A)3=-- + VYla2

( (AL.-1,2 (4 -25) 3

By using the property values from Chart 1, we obtain

h1a1 (VC 2? 4 42 (1 bh1 /hr ) 13

Ir}) G- 2p,90C6Ói 14,4

33 W3 = p, o b 53 t ( O, 2/2- p,00 r3) (.454 2t83.2)

_ o. 0(03 ( IbPrt`/Ibrna)

lo9

The intersection of W3 with the line connecting state 1 and 2

gives the mixture state 3. The resulting dry bulb temperature

is 86 F and the wet bulb temperature is 68 F. The complete

graphical procedure could also be used where

3 Z'1 c 2 X33 2 ßa3 (324-4561--2-) 065

and 1.3= 0,&''(12)

The length of line segments (2 and l3 depends on the scale o f

the psychrometric chart used. However, when the length 7 is laid out along !2 from state 1, state 3 is accurately

determined." C 5J

1 LO

4 -3 Design Conditions

The complete air -conditioning system may involve two or more of

the processes just considered. In this section we are going to

discuss how to combine more than one process. Various systems that

carry out these conditioning processes will be described in Chapter

5.

Sensible Heat Factor

The sensible heat factor (SHF) was defined in section 4 -2 as the

ratio of the sensible heat transfer to total heat transfer for a

process: -s s

SH-F t (4 -26)

If we recall Eq. 4 -9 and 4 -10 and refer to Fig. 4 -5, it is evident

that the SHF is related to the parameter 1/4"(A) .

The SHF is plotted on the inside scale of the protractor on

Chart 1. The following examples will demostrate the usefullness

of SHF. //CD

Example;

Conditioned air is supplied to a space at 15 °C db abd 14 °C w b

at the rate of 0.5 W\Vs. The SHF for the space is 0.7 and

the space is to be maintained at 24 C db. Determine the

sensible and latent cooling coils for the space.

Solution:

Chart la can be used to solve this problem conveniently.

A line is drawn on the protractor through a value of o.7

on the SHF scale, a parallel line is then drawn from the

initial state (15 C db and 14 C wb) to the intersection of 2 4

C db line, which defines the final state.

Fig 4 -11 illustrates the procedure. The total heat transfer rE

for the process is given by

%a Ctiz- L, )

and the sensible heat transfer is given by

8-5 CsH F ) 112.

AS

Fift

kl-

PSY

(, ;

.,1

. 1 .1

01i 1

N

o.

1 ..'

N

(_1I

;? I

nt..

I 1

VI

I t

.i II

AR

OM

C

t r'

Am

CR

ICA

II

;;OC

IET

Y

(r

11

1"

111

1110

.1

11

, i.

.1

1

1,1

1'I.

, n

1111

4I

11

ill .1

1

111i

1111

I

pl..

1111

'

I

Y

2.',

7h

_ y.

lo

.... 1

, I.

IMT

l11

TO

A A

I,

1.,W

11A

i, i1

141\

,1

in

r1..

DR

Y B

UIE

TE

MP

ER

AT

UR

E °

C

40

¡AR

T

1a

- 1 n

and o. _ 6? //(71 = '

O 2-1 - e7, 6 O

where vl is read from Chart la. Also from Chart la,

il = 39.3 kj /kg dry air and i2 = 52.6 kj /kg dry air, then

eri = 0, 60& C 52, 6 -- 39 _)- ß,o4 K0/5

=EsH )E = Ft c4- 6

and 7/1'

The process 1 -2 of this example and its extension to the left s

called the condition line for the space. Assuming that

state 2, the space condition, is fixed, air supplied at any dAn

on the condition line will statisfy the load requirement.

However, as the point is moved, different quantities of air

must be supplied to the space. In general, the closer point 1

is to point 2, the more air is required; the converse is also

true.

Fig 4 -11

"4

We will now consider several examples of single -path, constant -

flow systems. Heat losses and gains to the ducts and fan power

will not be included in this example.

Example:

A given space is to be maintained at 78 db and 65 wb. The trtal

heat gain to the space has been determined to be 60,000 BTU /hr

of which 42,000 BTU /hr is sensible heat transfer.

The outdoor air requirement of the occupants is 500 cfm.

The outdoor air has a temperature and RH of 90 F and 55 %,

repectively. Determine the quantity and the state of the air

supplied to the space and the required capacity of the

cooling and dehumidifying equipment.

Solution:

A simplified schematic is shown in Fig. 4 -12. The given

quantities are shown and stations are numbered for reference.

Losses in connecting ducts will be neglected.

Let us first consider the steady flow process for the

conditioned space. By Eq. 4 -26 the sensible heat factor is

SHF = 42,000/60,000 = 0.7

The state of the air entering the space lies on the line

defined by the SHF on psychrometric Chart 1. Therefore,

state 3 is located as shown on Fig 4 -13. and a line drawn

through the point parallel to the SHF = 0.7 line on the

protractor.

State 2 may be any point on the line and is determined by

the operating characteristics of the equipment, desired

indoor air quality, and by what will be comfortable for

115

the occupants.

(tc = b;7 Qc = G-GG4V, D

O

Retc1Y'L Fa PI

COD('ny Gc.a(

Unì,

5(.4.12PI7 Fa,,

Fig 4 -12

{ >

ex l2ccit st

T ?5"=6(7, o F3t0, r

If75 = 42, c'° C7. f3t `A 1,

Fig 4 -13

I1

To the knowledge we have, we assume that the dry bulb

temperature t2 cannot be more than 20 F less than t3. Then t2 =

58 F and state 2 is determined. The air quantity required may

now be found from an energy balance on the space

h1CIa = Ì11Cí3 z = h%1 a2 C -C 3- 2 )

6-r142= (Z. 3- -C2)

From Chart 1, i3 = 30 BTU /lbma, i2 = 23 BTU /lbma,

or

and

and Gco YYla2= YlA3 - o = g G-7c lbk1a/h !

also from Chart 1, V2 = 13.21 and

2 - m ck - 6c = (g o f Attention is now directed to the cooling and dehumidifying

unit. However, state 1 must be determined before

continuing. A mass balance on the mixing section yields

then

i_\ _ , ,,.,1 4 = Mai = 0-in 2_

o )

(jo = ( 4, 23 ft37lbn1

0(7)( 60/ 14,23 = 2L( o I1:7b1'vtA, r i1% lac=

a4 = YYl a 2._ vy) a o - rv = 646o 1 ioY.

By using the graphical technique discussed in Fig 4 -10,

17

we see that iî1 Gt z ( ( c

4c c') state 1 is located at 81 F db and 68 F wb. A line

constructed from state 1 to state 2 on Chart 1 then

represents the process taking place in the conditioning

equipment.

An energy balance gives

}III A( C. c or

,

Gc C4c= h'lt,,-z From Chart 1, il = 32.4 BTU /lbma 7 D j24-2 &oi S("0(8 Y)=6 , tD/1J

The sensible heat factor (SHF) for the cooling unit is found

to be 0.6 using the protractor of Chart 1 (Fig. 4 -13).

Then 2-05 = c 6

and _ 2a ßc,/h The sum of ..cc_ and is known as the coil refrigeration load

in contrast to the space cooling load. "E SJ

t iÚ

CHAPTER 5

Air -Conditioning Systems

5 -0 Introduction

5 -1 The Basic Central System

5 -2 All -Air Systems

5 -3 Air -and -Water Systems

5 -4 All -Water Systems

5 -5 Heat Pump Systems

5 -6 Heat Recovery Systems

5 -7 Economizer Systems

WI

5 -0 Introduction

All air -conditioning systems generally have the same basic

elements; however, the physical appearance and arrangement of the

various components may vary dramatically.

Although the same elements are present, the manner in which

the systems are controlled and operated to satisfy a given

environmental requirement may be quite different. Therefore,

various systems that meet the requirements of different building

types and users, load variations, and economic condiserations are

discussed in the following sections.

The original basic system of air conditioning was the forced

warm air heating and ventilating system with centrally located

equipment that distributed tempered air through ducts. It was

found that cooling and dehumidification equipment added to these

systems produced satisfactory air conditioning in spaces where heat

gains were relatively uniform throughout the conditioned area. The

use of this system was made complicated by the presence of variable

heat sources within the space served.

When this occurred, the area had to be divided into sections or

zones and the central system supplemented by additional equipment

and controls.

As the science of air conditioning progressed, variations in

the basic designs were required to meet the functional and economic

demands of individual buildings.

120

5 -1 The Basic Central System

The basic central system is an all -air, single -zone system

that is used as part of most systems. It can be design for low -

,medium-, and high -pressure air distribution. Normally the

equipment is located outside the conditioned area in a basement,on

the roof, or in a service area at the core of a commercial

building. It can be installed in any convenient area of a factory,

particularly in the roof truss area or on the roof. The basic

central system can be located adjacent to the heating and

refrigeration equipment, for example, boiler and condensor, or at

a considerable distance from it by using a circulating refrigerant,

chilled water, hot water, or steam for energy transfer sources.

It is important that the temperature within the area conditioned

by a central system be uniform if a single -zone constant -air -volume

duct system is to be used because air temperature is sensed at only

one place in the building for control.

j2

(1) Space with Uniform Loads

In spaces with relatively large open areas and small external

loads such as theaters, auditoriums, department stores, and the

public spaces of most buildings; air -conditioning loads are fairly

uniform throughout. Adjustments for minor variations can be made

by supplying more or less air, by changing the supply air

temperatures in the original design, and by balance of the system.

In commercial buildings the interior areas generally meet these

criteria when local heat sources such as computers are treated

seperately. These interior areas usually require year -round

cooling and any isolated spaces of limited occupancy may require

special attention.

(2) Space Requiring Precision Control

Spaces with stringent requirements for cleanliness, humidity,

and temperature control, and /or air distribution are usually

isolated within the larger building and require precision control.

The components of a central system can be selected and assembled

to meet the exact requirements of the area. Locating the equipment

outside the conditioned space permits routine inspection and

maintenance without interference.

(3) Multiple Systems for Large Areas

In spaces such as large office buildings, factories, and large

department stores, practical considerations require multiple

installation of control systems.

The size of the individual system is usually limited only by the

112

physical limitations of the structure. In multiple systems the

equipment is often located in the truss space, against the outside

wall, or on the roof where it least interferes with the operations

within the conditioned space and outdoor air is readily available.

(4) Primary Source of Conditioned Air for Other Systems

There are various systems for controlling conditions in

individual zones in which a constant supply of conditioned outdoor

air is used for ventilation and for some of the

air -conditioning load. This reduces the amount of conditioned air

handled by the central system and, consequently, the space required

for ductwork. By utilizing high velocities and designing for the

resultant higher pressure and sound levels, the ductwork can be

further reduced in size.

(5) System for Environmental Control

For applications requiring close aseptic or contamination

control of the environment, all -air type systems generally are used

to provide the necessary air supply to sustain adequate dilution

of the controlled space. These applications are usually

combinations of supply systems and scavenging exhaust systems to

circulate the diluting air through the controlled environment

space.

12

5 -2 All -Air Systems

An all -air system provides complete sensible heating and

sensible and latent cooling by supplying only air to the

conditioned space. In such systems there may piping connecting

the refrigerating -and heat -producing devices to the air -handling

device. No additional cooling is required at the zone. The term

"zone" implies a provision or the need for seperate thermostatic

control, whereas the term "room" implies a partitioned area that

may or may not require seperate control.

All -air systems may be briefly classified in two basic

categories; (1) single -path systems and (2) dual -path systems.

Single -path systems contain the main heating and cooling coils in

a series air flow path using a common duct distribution system at

a common air temperature to feed all terminal apparatus.

Dual -path systems contain the main heating and cooling coils in a

parallel flow or series parallel air flow path using either (1)

a seperate cold and warm air duct distribution system that is

blended at the terminal apparatus (dual -duct system) or (2) a

single supply duct to each zone with a blending of warm and cold

air at the main supply fan (multi -zone system).

The all -air system may be adapted to all types of air -

conditioning systems for comfort or process work. It is applied

in buildings requiring individual control of conditions and having

a multiplicity of zones such as office buildings, schools and

universities, laboratories, hospitals, stores, hotels and ships.

Air systems are also used for many special applications where a

need exists for close control of temperature and humidity,

(24

including clean rooms, computer rooms, hospital operating rooms,

textile and tobacco factories.

(1) Heating Considerations

Heating may be accomplished by the same duct system used for

cooling, by a separate perimeter air system, or by a separate

perimeter radiation system using hot water or steam. A perimeter

heating system is usually used in conjunction with the all -air

systems. However, its greatest application is with variable -air-

volume systems used for cooling only. During the times when heat

is required, the cooling system is used for ventilation only.

(2) Single -Zone System

The simplest all -air system is a supply unit serving a single

zone. The unit can be installed either within a zone or remote

from the space it serves and may operate with or without ductwork.

q i7PTON.LI

Fig 5 -1 C 7 J

125

A single -zone system responds to only one set of space

conditions. Thus, it is limited in application to where variations

in load are uniform throughout the zone. Fig 5 -1 shows a schematic

of a single zone constant -volume all -air system.

(3) Reheat Systems

The reheat system is a modification of the single -zone

constant -volume system. Its purpose is to permit zone or space

control for areas of unequal loading or to provide heating or

cooling of perimeter areas with different exposures, or for process

or comfort applications where close control of space conditions is

desired.

As the word "reheat" implies, the application of heat is a

secondary process, being applied to either preconditioned primary

air or recirculated room air. A single low- pressure reheat system

is produced when a heating coil is inserted in the duct system.

The more sophisticated systems utilitze higher pressure duct

designs and pressure- reduction devices to permit system balancing

at the reheat zone. The medium for heating may be hot water,

steam, or electricity.

Conditioned air is supplied from a central unit at a fixed

cold air temperature designed to offset the maximum cooling load

in the space. The control thermostat activates the reheat unit

when the temperature falls below the limit of the controlling

instrument's setting. A schematic arrangement of the components

for a typical reheat system is shown in Fig 5 -2. To conserve

energy reheat should not be used unless absolutely necessary. At

(2

oTSoe .0"

vìg S -2 C

- ¡;p A C N.A,

ì9 5-3 C7]

the very least, reset control should be provided to maintain the

cold air at the highest possible temperature to satisfy the space

cooling requirement.

(4) Variable -Volume System

The variable- volume system compensates for varying loads by

regulating the volume of air supplied through a single duct.

Special zoning is not required because each space is supplied by

a controlled outlet is a separate zone. Fig 5 -3 is a schematic of

a true variable- air -volume (VAV) system.

Significant advantages of the variable -volume system are low

initial cost and low operating costs. The first cost of the system

is far lower in comparison with other systems that provide

individual space control because it requires only single runs of

ducts and a simple control at the air terminal. Where diversity

of loading occurs, smaller equipment can be used and operating

costs are generally the lowest among all the air systems.

Because the volume of air is reduced with a reduction in load, the

refrigeration and fan horsepower follow closely the actual air -

conditioning load of the building.

Although some heating may be done with a variable volume

system, it is primarily a cooling system and should be applied only

where cooling is required the major part of the year. Buildings

with internal spaces with large internal loads are the best

candidates. A secondary heating system should be provided for

boundary surfaces during the heating season. Baseboard perimeter

heat is often used. During the heating season, the VAV system

12E7

simply provides tempered ventilation air to the exterior spaces.

An important aspect of VAV system design is fan control.

There are significant fan power savings where fan speed is reduced

in relation to the volume of air being circulated.

Single -duct variable -volume systems should be considered in

applications where full advantage can be taken of their low cost

of installation and operation. Applications exist for office

buildings, hotels, hospitals, apartments, and schools.

(5) Dual -Duct System

In the dual -duct system the central station equipment supplies

warm air through one duct run and cold air through the other. The

temperature in an individual space is controlled by mixing the warm

and cold air in proper proportions. Variations

of the dual -duct system are possible, with one form shown in

Fig 5 -4.

M x.NG BOX

SuVIve AIR

R[TyRN a R

Fig 5 -4 [ -r

For the best performance some form of constant volume

regulation should be incorprorated into the system to maintain a

constant flow of air. Without this the system is difficult to

control because of the wide variations in system pressure drop that

occur from the normal demand from loading changes.

Many dual -duct systems are installed in office buildings,

hotels, hospitals, schools, and large laboratories. A common

characteristic of these multiroom buildings is their highly

variable sensible heat load. This system provides great

flexibility in satisfying multiple loads and in providing prompt

and opposite temperature response as required.

Space or zone thermostats may be set once to control year -

round temperature conditions. All outdoor air can be used when

the outdoor temperature is low enough to handle the cooling load.

A dual -duct system should be provided with control that will

antomatically reset the cold air supply to the highest temperature

acceptable and the hot air supply to the lowest temperature

acceptable.

A variable -volume system may be incorporated into the dual -

duct system, with various arrangements possible. Two supply fans

are usually used in this case, one for the hot deck and one for

the cold deck.

(6) Multizone System

The multizone central station units provide a single supply

duct for each zone and obtain zone control by mixing hot and cold

air at the control unit in response to room or zone thermostats.

(3p

For a comparable number of zones this system provides greater

flexibility than the single duct and involves lower cost than the

dual -duct system, but it is physically limited by the number of

zones that may be provided at each center unit.

Typical multizone equipment is similar in some respects to

the dual -duct system, but the two air streams are propotioned

within the equipment instead of being mixed at each space served,

and the proper temperature air is provided as it leaves the

equipment. Fig 5 -5 shows a sketch of a multizone system.

OU'S DE iR

.v7 20,E i

I -c.rsT .- E ] . ) St-Lv .A - 1 ^. ' ZOE 2 ER.JST

I , - r : < REL.iEr

R = ZCNE ! GPTIONr: "ERy`SrT

-----.1 - - ,

.+ r

TO ZOE 2

1 R

IRETuRN R

.1-R

Fig 5 -5 C -I

The system conditions groups of rooms or zones by means of a

supply fan having heating and cooling coils in parallel downstream

from the fan.

The use of many duct runs and control systems can make the

initial cost of this system high compared to other all -air systems.

15 i

Also to obtain very fine control this system might require larger

refrigeration and air -handling equipment, which should be

considered in estimating both initial cost and operating cost.

132

5 -3 Air and Water Systems

In the all -air systems discussed in the previous section the

spaces within the building are cooled solely by air supplied to

them from the central air -conditioning equipment. In contrast, in

an air -and -water system both air and water are distributed to each

space to perform the cooling function. In virtually all air -water

systems both cooling and heating functions are carried out by

changing the air or water temperatures (or both) to permit control

of space temperature during all seasons of the year.

There are several basic reasons for the use of this type of

system. Because of the greater specific heat and much greater

density of water compared to air, the cross -sectional area required

for the distribution pipes is markedly less than that required for

ductwork to accomplish the same cooling task. Consequently, the

quantity of air supplied can be low compared to an all -air system,

and less building space need be allocated for the cooling

distribution system.

The reduced quantity of air is usually combined with a high -

velocity method of air distribution to minimize the space required.

If the system is designed so that the air supply is equal to the

air needed to meet outside air requirements or that required to

balance exhaust or both, the return air system can be eliminated

for the area conditioned in this manner.

The pumping horsepower necessary to circulate the water

throughout the building is usually significantly less than the fan

horsepower to required deliver and return the air. Thus not only

space but also operating cost savings can be realized.

Systems of this type have been commonly applied to office

buildings, hospitals, hotels, schools, appartments, laboratories,

and other buildings. Space saving has made these systems

benificial in high -rise structures.

The air side of air -and -water systems is comprised of control

air -conditioning equipment, a duct distribution system, and a room

terminal. The air is supplied at constant volume and is often

referred to as primary air to distinguish it from room air that is

recirculated over the room coil.

The water side in its basic form consists of a pump and piping

to convey water to the heat transfer surface within each

conditioned space. The heat exchange surface may be a coil that

is an integral part of the air terminal (as with an induction

unit), a completely separature component within the conditioned

space (radiant panel), or both (as is true of fan -coil units).

Individual room temperature control is obtained by varying

the capacity of the coils within the room by regulation of either

the water flow through it or the air flow over it. The coil may

be converted to heating service during the winter, or a second coil

or a heating device within the space may provide heating capacity

depending on the type of system.

(1) Air -Water Induction System

The basic arrangement for air -water induction units is shown

in Fig. 5 -6. Centrally conditioned primary air is supplied to the

unit plenum at high pressure. The plenum is acoustically treated

to attenuate part of the noise generated in the duct system and in

l

the unit. A balancing damper is used to adjust the primary air

quantity within limits.

Nitta 1-1,

t y

4,

,--7.--- %:. ̂-- ..,-. ,

C____ , .. ; Í

.{ w

T /, ¡/i

/i/ '1i (;(_,`i , / / v,,-

, , ';/ / '11. -----------cer,..ick,14,s.4r.---d-

I

ti -t ¡/ / / ( !

Fig 5 -6

The high- pressure air flows through the induction nozzles and

induces secondary air from the room and over the secondary coil.

This secondary air is either heated or cooled at the coil depending

on the season, the room requirement or both. Ordinarily no latent

cooling is accomplished at the room coil, but a drain pan is

provided to collect condensed noisture resulting from unusual

latent loads of short duration. The primary and secondary air are

mixed and discharged to the room.

Induction units are usually installed at a perimeter wall

under a window, but units design for overhead installation are

available. During the heating season the floor -mounted inductiL..

unit can function as a convector during off hours with hot .a`_- to the coil and wihtout a primary air supply.

135

(2) Fan -Coil Conditioner System

The fan -coil conditioner unit is a versatile room terminal

that is appled to both air -water and water only systems.

" The basic elements of fan -coil units are a finned -tube coil

and a fan seciton, Fig 5 -7. "[ J

fr

(

1 5

c

Fig 5 -7 Es] 1. Finned tube coil

2. Fan scrolls

3. Filter

4. Fan motor

5. Auxiliary condensate pan

6. Coil connections

7. Return air opening

8. Discharge air opening

9. Water control value

The fan section recirculates air continuously from within the

perimeter space through the coil, which is supplied with either

hot or chilled water.

In addition, the unit may contain an auxiliary heating coil,

which is usually of the electric resistance type but which can be

of the steam or hot water type. Thus the recirculated room air is

either heated or cooled. Primary air made up of outdoor air

sufficient to maintain air quality is supplied by a separate

central system usually discharged at ceiling level. The primary

air is normally tempered to room temperature during the heating

season, but is cooled and dehumidified in the cooling season. The

primary air may be shut down during unoccupied periods to conserve

energy.

137

5 -4 ALL -WATER SYSTEMS

All water systems are those with fan -coil, unit ventilator,

or valance -type room terminals, with unconditioned ventilation air

supplied by an opening through the wall or by infiltration.

Cooling and humidification are provided by circulating chilled

water or brine through a finned coil in the unit. Heating is

provided by supplying hot water through the same or a separate coil

using two, three, or four -pipe water distribution from

centrol equipment. Electric heating or a separate steam coil may

also be used.

Humidification is not practical in all -water systems unless

a separate package humidifier is provided in each room.

The greatest advantage of the all -water system is its

flexibility for adaption to many building module requirements.

A fan -coil system applied without provision for positive

ventilation or one taking ventilation air through an aparture is

one of the lowest first -cost central station type perimeter systems

in use today. It requires no ventilation air ducts, is

comparatively easy to install in existing structures, and as with

any central station perimeter system utilizing water in pipes

instead of air ducts, its use results in considerable space savings

throughout the building.

All -water systems have individual room controls with quick

response to thermostat settings and freedom from recirculation of

air from other conditioned space. When fan -coil units are used

with three -or four -pipe water arrangements, each is its own zone

with a choice of heating or cooling at all times and no seasonal

135

changeover is required. All -water systems can be installed in

existing buildings with a minimum of interference in the use of

occupied space.

There is no positive ventilation unless openings to the

outside are used, and then ventialtion can be affected by wind

pressures and stack action on the building.

Special precautions are required at each unit with an outside

air opening to prevent freezing of coil and potential water damage

from rain. Because of these problems and energy conservation

considerations; it is becoming standard practice to eliminate the

outdoor air feature of these systems and rely on other means to

provide outdoor air. This type of system is not recommended for

applications requiring high indoor air quality.

Seasonal changeover is required in most climates with two -pipe

system, and zoning and piping are required to reduce operating

difficulties during intermediate seasons when a sun -exposed zone

may need cooling while other zones need heat.

If a two -pipe system has only one pump, the same quantity of

hot and cold water is circulated even though the requirements for

each may be different. With three- and four -pipe systems hot and

cold water may be furnished throughout the year.

Maintenance and service work has to be done in occupied areas;

as the units become older, the fan noise can become objectionable.

Each unit requires a condensate drainline. it is difficult to

limit bacterial growth in the unit. In extremely cold weather it

is often necessary to close the outside air dampers to prevent

freezing of coils, reducing ventilation air to that obtained by

13.=(

infiltration. Filters are small and inefficient and require

Frequent chr.ging to maintain air volume.

AFig 5 -8 illustrates a typical unit ventilator, used in all-

;pater systers, with two seperate coils, one used for heating and

the other for cooling with a four -pipe system. In some cases the

unit ventilator may have only one coil, such as the fan coil of

Fig 5-7.4/C6]

i.. JL'

r//// ///,/%--- /

I ,41J

U !'` /1J l o

- 'r^.--t, Cti-.

. ¡''//' tYLt (cl1L

J

Fig 5 -8 C5,3

1 4-0

5 -5 Heat Pump Systems

The term heat pump as applied to HVAC systems is a system in

which refrigeration equipment is used such that heat is taken from

a heat source and given up to the conditioned space when heating

service is wanted and is removed from the space and dis charged to

a heat sink whencooling and dehumidification are desired. The

thermal cycle is identical with that of ordinary refrigeration, but

the application is equally concerned with the cooling effect

produced at the evaporator and the heating effect produced at the

condenser. In some applications both the heating and cooling

effects obtained in the cycle are utilized.

Unitary heat pumps are shipped from the factory as a complete

preassembled unit including internal wiring, controls, and piping.

Only the ductwork, external power wiring, and condensate piping are

required to complete the installation.

For the split unit it is also necessary to connect the refrigerant

piping between the indoor and outdoor sections.

In appearance and dimensions, casings of unitary heat pumps closely

resemble those of conventional air -conditioning units having equal

capacity.

Capacities of unitary heat pumps range from about 1 1/2 to 25

tons, although there is no specific limitation. This equipment is

almost univerally used in residential and the smaller commercial

and industrial installations. The multiunit type of installation

with a number of individual units of 2 to 20 tons of cooling

capacity is particularly advantageous to obtain zoning and to

provide simultaneous heating and cooling.

141

(1) Heat Pump Types

The air -to -air heat pump is the most common type. It is

particularly suitable for factory -built unitary heat pumps and has

been widely used for residential and commercial applications.

Outdoor air offers a universal heat -source, heat -sink medium for

the heat pump. Extended -surface, forced -convecton heat transfer

coils are normally employed to transfer the heat between the air

and the refrigerant.

The capacity of an air -to -air heat pump is highly dependent

on the outdoor temperature. Therefore, it is usually necessary to

provide supplemental heat. The most common type of supplemental

heat for heat pumps in the United States is electrical -resistance

heat. This is usually installed in the air -handler unit and is

designed to turn on automatically, some times in stages, as the

indoor temperature drops. Heat pumps that have fossil fuel -fired

supplemental heat are referred to as hybrid or bivalent heat pumps.

These are more common in Europe than in United States.

Air -to -water heat pumps are sometimes used in large buildings

where zone control is necessary and are also sometimes employed for

the production of hot or cold water in industrial applications as

well as heat reclaiming. Heat pumps for water heating are

commercially available in residential size.

A water -to -water heat pump uses water as the heat source and

sink for both cooling and heating operation. Heating -cooling

changeover may be accomplished in the switching in the water

circuits.

A water -to -air heat pump uses water as a heat source and sink

(42

and uses air to transmit heat to or from the conditioned space.

Water is a satisfactory and in many cases an ideal heat source.

Well water is particularly attractive because of its relatively

high and nearly constant temperature, generally about 50 F in

northern areas and 60 F and higher in the south.

However, abundant sources of suitable water are not always

available and the application of this type of system is limited.

Frequently sufficient water may be available from wells, but the

conditon of water may cause corrosion in heat exchangers or

it may induce scale formation. ß'C5]

D4

*7c,E.Dì l

T.re lnÌater HeateY

Pu""P

$Y Pass

----3

+.19. `---+

H

=,d e vrd t4a ! kIate --to --A r

1-ka t Pt-tMP iil Each Zvoe

Fig 5-9 [FJ]

(2) Closed -Loop Systems

stIn many cases a building may require cooling in interior zones

while needing heat in exterior zones. The needs of the north zones

of a building may also be different from those of the south. In

some cases a closed -loop heat pump system is a good choice. Closed -

loop systems may be solar assisted. A closed -loop system is shown

in Fig 5 -9.1'[ Fj]

Individual water -to -air heat pumps in each room or zone accept

energy from or eject energy to a common water loop, depending on

whether that area has a call for heating or for cooling. In the

ideal case the loads will balance and there will be no surplus or

deficiency of energy in the loop. If cooling demand is such that

more energy is rejected to the loop than is required for heating,

the surplus is rejected to the atmosphere by a cooling tower.

The ground has been used successfully as a source -sink for

heat pumps, using either a vertical or a horizontal run of plastic

pipe as the ground -coupling device. Water from the heat pump.

tests and analysis have shown rapid recovery in earth temperature

around the well after the heat pump cycles off.

144

5 -6 Heat Recovery Systems

In large commercial applications considerable heat energy is

generated internally and may require removal even during the

coldest weather. This condition usually occurs within the

central spaces, which do not have exterior walls. It is necesary

to exhaust considerable quantities of air from large commercial

structures because of the introduction of outdoor ventilation

air. Considerable savings in energy can be realized if the heat

energy from the interior spaces and the exhaust air can be

recovered and used in heating the exterior parts of the

structure. Heat energy may also be recovered from waste water.

Redistribution of heat energy within a structure can be

accomplished through the use of heat pumps of the air -to -air or

water -to -water type.

Recovery of heat energy from exhaust air is accomplished

through the use of air -to -air heat exchangers, rotating heat

exchangers, and air -to -water heat exchangers connected by a

circulating water loop. Sometimes spray systems are used.

Fig 5 -10 and 5 -11 illustrate the air -to -water and rotating heat

recovery systems.

Chw i

l"-f 4 Lk_ e v

F16- 5- D

14-7

and uses air to transmit heat to or from the conditioned space.

Water is a satisfactory and in many cases an ideal heat source.

Well water is particularly attractive because of its relatively

high and nearly constant temperature, generally about 50 F in

northern areas and 60 F and higher in the south.

However, abundant sources of suitable water are not always

available and the application of this type of system is limited.

Frequently sufficient water may be available from wells, but the

conditon of water may cause corrosion in heat exchangers or

it may induce scale formation. "CL.]

Ceo(e/r

c.'.--* ì i

T-ire * tA)ate-

H.ea -te Y

13y pass Ni___

1---i 1-1, f.

4-y 1-t.P. Ir..), ,-.,

H.

H, p,---->,

`--4 H P, '----

Tr-Kiev/Lot l^la'f.e--to -A ov-

Heat RMP i Eat i, 717o .e.

Fig 5-9 [5]

Ka

While Fig 5 -12 shows how an air -to -air system might be

arranged. The air -to -air and rotating type systems are the most

effective in recovering energy but require that the intake and

exhaust to the building be at the same location, whereas the air -

to -water system may have the exhaust and intake at widely

separated locations; however, the air -to -water system is not as

effective because of the added thermal resistance of the water.

To prevent freezing, glycol must be introduced, which further

reduces the effectiveness of the air -to -water system.

All of previous described systems may also be effective during

the cooling season when they function to cool and dehumidify

outdoor ventilation air.

i4"7

5 -7 Economizer Systems

It is often possible to cool a space either totally or

partially using outdoor air. This will be true during the spring

and fall of the year and at night in northern regions and at high

altitudes. Outdoor air can also be used to cool interior spaces,

which may otherwise require operation of the cooling system. This

results in economics of energy and money, the general term

economizer has been adopted to describe these systems.

The design of economizers are quite varied but generally have

provisions for introducing outdoor air and exhausting indoor air

(Fig 5 -13) . "L5J3

(i e, ta to p e\--

\ \ \ \ \ I I

PctiArh

\ \\ Cc7htrc

''M Mot--

td

/ j E--- Di ta kc / j A i` Y 1

Fig 5 -13 C5] There is a great variety of control arrangements, and the

economizer may be combined with the heat recovery system. It is

impoetant to sense both outdoor temperature and humidity so that

hot humid air is not introduced to the space. This would defeat

the purpose of the economizer. Adequate precautions must be taken

to prevent contamination of the space by dirty or odorous outdoor

air.

t4e,

CHAPTER 6

Heating and Cooling Loads

6 -0 Introduction

6 -1 Heating Loads

6 -2 Cooling Loads and CLTD Method

141

6 -0 Introduction

This chapter is going to cover two of the most important

categaries in HVAC calculations. The first part is Heating loads,

the secondpart is cooling loads. The CLTD (Cooling Load

Temperature Difference) method is one of the methods used in

calculating cooling loads because it is easy to learn. We are going

to have one section particularly for the CLTD method.

Prior to the design of the heating system an estimate must be

made of the maximum probable heat loss of each room or space to be

heated. There are two kinds of heat losses: (1) the heat

transmitted through the walls, ceiling, floor, glass, or other

surfaces, and (2) the heat required to warm outdoor air entering

the space. The actual heat loss problem is transient because the

outdoor temperature, wind velocity, and sunlight are constantly

changing.

A larger number of variables are considered in making cooling-

load calculations than in heating -load calculations. In both

situations the actual heat loss or gain is a transient one.

However, for design purposes heat loss is usually based on steady

state heat transfer, and the results obtained are usually quite

adequate. In design for cooling, however, transient analysis must

be used if satisfactory results are to be obtained. This is

because the instantanious heat gain into a conditioned space is

quite variable with time, primarily because of the strong transient

effect created by the hourly variation in solar radiation. There

may be an appreciable difference between the heat gain of the

structure and the heat removed by the cooling equipment at a

t SO

particular time. This difference is caused by the storage and

subsequent transfer of energy from the structure and it's contents

to the circulated air. If this is not taken into account, the

cooling and dehumidifying equipment will usually be grossly

oversized and estimates of energy requirements meaningless.

Is'

6 -1 Heating Loads

Heating load calculations, comparatively, are more simplified

than cooling load calculations. The main reason is for design

purposes the heat loss is usually for steady -state heat transfer

for some reasonable design temperature.

The actual heat loss problem is transient as we know from the

introduction. The transfer function method discussed in the next

section in connection with the cooling load, may be used under

winter conditions to account for changing solar radiation, outdoor

temperature, and the energy storage capacity of the structure.

During the coldest months, however, sustained periods of very cold,

cloudy, and stormy weather with relatively small variation in

outdoor temperature may occur. In this situation heat loss from

the space will be relatively constant and in the absence of

internal heat gains will peak during the early morning hours.

Transient analyses are often used to study the actual energy

requirements of a structure in simulation studies. In such cases

solar effects and internal heat gains are taken into account.

ti In the ASHRAE Handbook 1985 Fundamentals Inch -Pound Edition

chapter on The Heating Loads has more details about estimating the

maximum probable heat losses and procedure. Here is the general

procedure for calculation of design heat losses of a structure:

1. Select the our door design conditions: temperature,

humidity, and wind direction and speed.

2. Select the indoor design condtions to be maintained.

3. Estimate the temperature in any adjacent unheated spaces.

4. Select the transmission coefficients and compute the heat

152

losses for walls, floors, ceilings, windows, doors, and

floor slabs.

5. Compute the heat load due to infiltration.

6. Compute the heat load due to outdoor ventilation air.

This may be done as part of the air quantity calculation.

7. Sum the losses due to transmission and infiltration. I /C5j Following equations are for calculating design heating loads:

Heating loads for roofs, ceilings, walls, and glass:

q = U * A * TD (6 -1)

U = coefficient of transmission, BTU /hr * ft 2 * F

(From ASHRAE chapter 23, tables 3 and 4)

A = area calculated from plans

TD = temperature difference between inside and outside

design dry bulb.

Heating loads for walls below grade:

q = U * A * TD (6-2)

2 U = coefficient of transmission BTU /hr *ft *F

(From ASHRAE chapter 25, table 3)

A = area calculated from plan

TD = temperature difference between inside and outside,

outside temperature can be obtained from ASHRAE

Handbook, 1985 Fundamentals (IP), P25,6 Fig 4. by

using formula to -A . ta. is the mean winter

air temperature dry bulb, A is from the Fig 4.

Heating loads for floors above grade, on grade, and below

grade:

Above grade:

153

q = U * A * TD (6 -3)

U = coeffecient of transmission (BTU /hr *ft *F)

A = area calculated from plan

TD = temperature difference between inside and outside.

On grade:

q = F2 * P (6-4)

F2 = Heat loss coefficient of slab floor construction

(BTU /h *ft *F) from ASHRAE Handbook, 1985

Fundamentals (IP) P.25.9 table 5.

P = perimeter of slab (ft)

Below grade:

q = U * A * TD (6 -5)

U = heat transfer coefficient (BTU /hr *ftZ *F)

from ASHRAE Handbook 1985 Fundamentals P25.6

table 4.

A = area calculated from plan

TD = temperature difference (same as equation 6 -2)

Heating loads for infiltration and ventilation air sensible

heat and latent heat.

Sensible heat:

-5= .1 2do.U TD (6 -6)

v = volume flow rate of outdoor air entering building

1D= temperature difference

Latent heat: I.1

F__e -

p 2 , S y. o(

(6 -7)

v = volume flow rate of outdoor air entering building

154

(ev)= Humidity ratio difference from psychrometric chart

The rest of this section is composed of some examples so that

t:ze reader may have some chances to practice the above equations.

First cf all,t-,:o terms should be carefully distinguished, C and U.

C is conductance and the unit is BTU /hr *ft *F ; U is overall heat

transfer coefficient and the unit is also BTU /(hr * ft * F).

C is for materials; U is for overall spaces and materials.

U' is heat -loss coefficient and the unit is BTU /(hr- ft -F).

It

--- ' c,'

E !

(r??. /i

. ' ,- ¡ S ... v

LcciZ ao l S , C:ì -.. 2 ;; " G i C a' ILI i"7! C :, --- i¡=

(6-,)

Fig 6 -1 [.3

Esa.i...a*-e the te-.perature in the crawl space of Fig 6 -1.

:h` Cc:.., ._:_....._ floor is 0.2 ETU/hr-ft -F

including the. air film on each side. The conductance for

S5

the foundation wall including the insulation and inside and

outside air film resistances is 0.12 BTU /(hr -ft '-F)

Assume an indoor temperature of 70 F and an out door

temperature of 10 F at a location where the degree days

are about 3000. The building dimensions are 50 x 75 (ft Z)."[,]

Solution:

The first step is to make an energy balance for the crawl

space. Heat will be gained through the floor and heat will

be lost through the foundation wall and through the ground

around the perimeter, much like a floor slab. Infiltration

of outdoor air will also represent a heat loss but that

will be neglected in this example.

or

Cf CA-f ,) Cf4'- =cfbAfcC-tc,- ctc- _6c= (CA)+, + u'r Cc,4) f,

( CA -t (c-4) fo

where

= Heat gain through the floor

= Heat lost through the foundation wall

hi-oknci = Heat lost through the ground

Cff = Conductance of the floor

p = Area of the floor

= Indoor temperature

-tc = Crawl space temperature

-o = Outdoor temperature

5

U = Heat loss coefficient

f = Perimeter

Now the the area of the floor is 50 x 75 = 3750 ft and

assuming the foundation wall averages a height of 2 ft, the

area of the foundation wall is 2[2x50 +(2x75)] = 500 ft

The perimeter, P, of the building is (2x50) +(2x75) = 250 ft.

The heat -loss coefficient is 0.8 BTU /(hr -ft -F)

then

CG,1(.2) 1;ov +(tv Cob xcc)+-1c (0,zX7c) 1

(C., Z x 3 76 o) f Cc, ( 2 xi--c-)c)--f Cc; /( 2- Çj c J _rte- &F

Example (6 -2):

Estimate the temperature in the unheated room shown in Fig -6

2. The structure is built on a slab. The exterior walls

of the unheated room have an overall heat transfer

coefficient of 0.2 BTU /(hr -ft 2

F) , the ceiling- attic -roof

combination has an overall coefficient of 0.07 BTU /(hr -ft2 -

F),and the interior walls have an overall coefficient of

0.06 BTU /(hr -ft 2 F). Inside and outdoor design temperatures

are 72 F and -5 F, respectively.

Solution:

R" is thermal resistance (hr -F) /BTU

R is overall thermal resistance (hr -ft2 F) /BTU

Vh hefd

k e ate cl

Fig 6 -2

The heat transfer from the heated to the unheated room has

a single path, neglecting the door, and may be represented

as

q = Ui * Ai * (ti-tu) (6-8)

Ui = U value between heated and unheated

Ai = Area of wall between heated and unheated

ti = heated room temperature

tu = unheated room temperature

The heat transfered from the unheated room to the outdoor

air has parallel paths through the ceiling and walls

(neglecting the door and floor)

q = Uc * Ac * (tu -to) + Uo * Ao * (tu -to) (6 -9)

Uc = U value of ceiling

Uo = U value of exterior wall

The heat loss to the floor can be neglected because of the

anticipated low temperature in the room. Eq. 6 -9 may be

written

l,D ) -t - )

R'Ceift1K l.I vu-Es de 1` (6 -10)

158

where

tzI ll

-f - L/c Ac U f-1 0 (6-11) G ì l tG FiktS ,,,k

By combining Eqs. (6 -8) and (6 -11) , we obtain

and solving for tu

-E, _ -t.t` i C

(6 -12)

Ì--E- C ) (6-13)

F4' I _ Ù,' A,1 Cp,c- C 8> (z)

_1; U fi c o _ C°,677 (012) Zz L) 43,

011--F)/ettA

Then from equation (6 -13)

-72-I- Cr,c>q/c.r?.2)-S Ifi C ,o`tS

(o

6 -2 Cooling Loads and CLTD Method

In the beginning of this section, it is important to

differentiate between heat gain, cooling load, and heat extraction

rate. Heat gain is the rate at which energy is transferred to or

generated within a space. It has two components, sensible heat and

latent heat, which must be computed and tabulated saperately.

Heat gains usually occur in the following forms:

1. Solar radiatin through openings

2. Heat conduction through boundaries with convection and

radiation from the inner surface into the space.

3. Sensible heat convection and radiation from internal

objects.

4. Ventilation (outside) and infiltration air.

5. Latent heat gains generated within the space.

The coling load is the rate at which energy must be removed

from a space to maintain the temperature and humidity at the design

values. The cooling load will generally differ from the heat gain

at any instant of time. This is because the radiation from the

inside surface of walls and interior objects as well as the solar

radiation coming directly into the space through openings does not

heat the air within the space directly.

This radiant energy is mostly abosrbed by floors, interior walls,

and furniture, which are then cooled primarily by convection as

they attain temperatures higher than that of the room air.

Only when the room air receives the energy by convection does this

energy became part of the cooling load.

%The heat storage characterstics of the structure and interior

1 &O

objects determine the thermal lag and therefore the

relationship between heat gain and cooling load. For this reason

the ther-.;ül mass (product of mass and specific heat) of the

structure and its contents must be considered in such cases.

Fig 6 -3 illustrates the phenomenon.r[]

LT o-E) 44' -J ec.-tx.(

Fig 6 -3 E&J

The reduction in peak cooling load because of the thermal lag can

te Ç :yre im_ orrar_t in siz=ng the cooling equipment.

The heat extraction rate is the rate at which energy is

r _d from th` space by the cooling and dehumidifying equipment.

This rare is equal tc the cooling load when the space conditions

are constant and the equipment is operating. However, this is

fluctuation in room temperature is

necessary for the control system to operate. Because the cooling

load is also below the peak or design value most of the time,

intermittent or variable operation of the cooling equipment is

required.

°Fig 6 -4 shows the relation between heat gain and cooling load

and the effect of the mass of the structure. The attenuation and

delay of the peak heat gain is very evident especially for heavy

construction. 4[ J3] InStahfd S t-ÍFat

- Leiht StrHCf"' Medici 1 Stt suc`f.i,ti-e

7"IEáVGJ tY[tCt

Heaf

Fig 6 -4 C623

> -(' c' Vt 1 E_

off Fig 6 -5 C 6 7

1/2

Fig 6 -5 shows the cooling load for flourescent lights that

are used only part of the time. The sensible heat component from

people acts in a similar way. The part of the energy produced by

the lights or people that is radiant energy is momentarily stored

in the surroundings. The energy convected directly to the air by

the lights and people, and later by the surrounding objects, goes

into the cooling load. The area under the curves of Fig 6 -4 are

all approximately equal. This means that about the same total

amount of energy must be removed from the structure during the day;

however, a larger portion is removed during the evening hours for

heavier construction .11C 5]

(1) The CLTD Method

ttThe methodology discussed here is a basic load calculation

procedure that produces cooling loads on an hourly basis if

desired. The purpose of the cooling load is to provide a basis

for system design. As discussed in chapter 5, there are many

different system types and controls. Therefore, it follows that

the results of a load calculation may have to be manipulated in

accordance with the building and system to be used5JThis will not

be included in our discussion.

The CLTD method makes use of a temperature difference in the

case of walls and roofs and Cooling Load Factors (CLF) in the case

of solar gain through windows and internal heat sources. The CLTD

and CLF vary with time and are a function of environmental

conditions and building parameters. They have been derived from

computer solutions using the transfer function procedure. A great

1 &3

deal of care has been taken to sample a wide variety of conditions

in order to obtain reasonable accuracy.

These factors have been derived for a fixed set of surface and

environmental conditions; therefore, correction factors must often

be applied.4 ]

In general, calculations proceed as follows:

For walls and roofs:

& = UA(CLTI7)& (6 -14)

U = overall heat transfer coefficient, BTU /(hr -ft -F)

A = area, ftt

CLTD = temperature difference which gives the cooling load

at time &, F

`The CLTD accounts for the thermal response (lag) in the heat

transfer through the wall or roof, as well as the response (lag)

due to radiation of part of the energy from the interior surface

of the wall to objects within the space."

For solar gain through glass:

?6= A(SC) (SHGF) (CLF)&. (6 -15)

where

A = area, ft2

SC = shading coefficient (internal shade)

SHGF = solar heat gain factor, (BTU /hr -ft )

CLF = cooling load factor for time

The SHGF is the maximum for a particular month, orientation,

and latitude. The CLF accounts for the variation of the SHGF with

time, the massiveness of the structure, and internal shade. Again

the CLF accounts for the thermal response (lag) of the radiant part

104

of the solar input.

For internal heat sources

QQ l . CC L FJc, (6 -16)

where

= instantaneous heat gain from lights, people, and

equipment, BTU /hr.

(CLF)6, = cooling load factor for time G.

The CLF accounts for the thermal response of the space to the

various internal heat gains and is slightly diferent for each.

If the CLTD /CLF procedure has been adapted to a computer or

when the transfer function method is used directly, the cooling

load for each hour of the day is computed and the hours of interest

may be selected. However, when the calculations are made by hand,

calculations for 24 hours in the day are too time consuming. In

such a case, the time of day when the peak cooling loads will occur

may be estimated. In fact, two different types of peaks may need

to be determined.

ItFisrt, the time of the peak load for each room is needed in

order to compute the air quantity for that room.' /C53 q Second, the time of the peak load for a zone served by a

central unit is required to size the unit. It is at these peak

times that cooling load calculations should be made. The estimated

times when the peak load will occur are determined from the tables

of CLTD and CLF values together with the orientation and physical

characteristics of the room or space [LJ These tables will be

introduced later. The times of the peak cooling load for walls,

roofs, and windows are obvious in the tables, and the most dominant

165

cooling load components will then determine the peak time for the

entire room or zone. For example, rooms facing west with no

exposed roof will experience a peak load in the late afternoon or

early evening. East -facing rooms tend to peak during the morning

hours. High internal loads may dominate the cooling load in some

cases and cause an almost uniform load throughout the day. As

mentined earlier,further discussion will be presented later

regarding the use of the cooling load results.

The details of computing the various cooling load components

will be discussed in the following articles. ASHRAE Handbook

Fundamentals should be consulted for complete data and an original

source of information.

(2) Cooling Load -External Surfaces

The calculation procedure is similar for walls, roofs, and

conduction through glass. A different procedure is used for the

glass solar gain.

(1) WALLS AND ROOFS

Table 6 -1 and 6 -2 give the CLTD values in degrees F that were

computed for the following conditions: SI

1. Dark surface for solar radiation absorption.

2. Inside temperature of 78 F.

3. Outdoor maximum temperature of 95 F and an outdoor daily

range of 21 F. This corresponds to a mean outdoor

temperature of 85 F.

4. Solar radiation for 40 degree north latitude on July 21.

5. Outside convective film coefficient of 3.0 BTU /(hr -ft -

F) .

6. Inside convective film coefficient of 1.46 BTU /(hr -ft -

F) .

7. No forced ventilation or air ducts in the ceiling space.' ",

To convert CLTD values from degree F to C simply multiply by 5/9.

When conditions differ from the above, the CLTD should be adjusted

according to the following relation:

CLTDcor = (CLTD +LM)K + (78 - ti)+ (tom -85) (6 -17)

where

cor = corrected

LM = is a correction for latitude and month from Table 6 -3,F

K = color adjustment factor

ti = room design temperature, F

tom = outdoor mean temperature, tom = to - DR /2, F

llThe color correction factor K should be used with cautin

especially for roofs. It has a value of 1.0 for dark surfaces and

0.5 for permanently lighted -colored surfaces. The designer must

be confident that a light -colored surface will remain in that G

condition before using a value of K less than 1.0.[S]

Table 6 -4, 6 -5, and 6 -6 describe the roofs and walls given in

Table 6 -1 and 6 -2 in detail. Note, for example, in Table 6 -1

each roof is further described by use of code numbers for each

layer. The details of each layer are given in Table 6 -6.

This same general procedure is used in Table 6 -5 for walls, except

a capital letter is used to denote construction groups that are

thermally simillar. Again the layers are described in Table 6 -6.

7

L61 Cooling Load Temperature Differences for Calculating Cooling Load from Flat Roofs

Roof Ik.crpiiun ur !`o ( nstr cuon

Hour of ( - slue Maul. Mlnl- Maul. Differ.

)% eight Iltu ,h Solar Time,k mum mum mum once h ft) ft'F) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24CLTDCLTDCLTDCLTD

ithout Suspended Ceiling Stv:I sheet with I.in. 0 :13 I -3 -3 -5 -3 6 19 34 49 61 71 78 79 77 70 59 45 30 18 12 8 5 3 14 -5 79 84 r ^ 2-in linsu(a: on IS) (0.1:41

2 I -in .ocd with I in. 8 0 1 "0 6 3 0 -1 -3 -3 -2 4 14 27 39 32 62 70 74 74 70 62 51 38 28 20 14 9 16 -3 74 77

3 4..- ! w _cr.cte:e 1S 0 ::? > 5 2 0 -2 -3 -3 I 9 20 32 44 55 64 70 73 71 66 57 45 34 25 18 13 16 -3 73 76 t 2-in. !'..0 ,: avete

.i:h 1., :9 0:_h 12 8 5 3 0 -I -I 3 11 20 30 41 SI 59 65 66 66 62 54 45 36 2^ 22 17 16 -I 67 68

5 .

:as.a':Ca 9 0 IN 3 0 -3 -4 -5 -7 -6 -3 5 16 27 39 49 57 63 64 62 57 48 37 26 18 II 7 16 -7 64 71

6 5-in I.w.cencrete 24 0.158 22 17 13 9 6 3 I 1 3 7 15 23 33 43 51 58 62 64 62 57 50 42 35 28 18 1 64 63

7 2.5-in. s.00d s.ith I-in. insulation 13 0 1)0 29 24 20 16 13 10 7 6 6 9 13 20 27 34 42 48 53 55 56 54 49 41 39 34 19 6 56 50

8 8-in I 4 ;orcree ?! 0 1:5 ?5 10 :6 22 18 14 11 9 7 7 9 13 19 25 33 39 46 50 53 54 53 49 45 40 20 7 54 Al

9 _. . ..s. .:,...ttv ? : ' :. +_ If 15 12 9 8 8 10 14 20 26 33 40 46 50 53 53 52 48 43 38 34 30 18 8 53 45

(^2.-in :insulation 152) (O.GO) 1,3 _ r.:a

:in :-_:a:.,n ;s r ,KJt 31 :5 :3 ;9 16 13 10 9 8 9 13 17 23 29 36 41 46 49 51 50 47 43 39 35 19 8 SI 43

II Rce(:er.2;e 34 31 28 25 22 19 16 14 13 13 15 18 22 26 31 36 40 44 45 46 45 43 40 37 20 13 46 33 . :on:r.c

.... . 0 IS. 31 28 25 _. 20 17 15 14 14 16 18 n :5 31 36 40 43 45 45 44 42 40 37 34 19 14 45 31

ICI trsula:icn 75 (0.117)

:7 , ... . O 76 33 30 24 25 .2 :0 IS 17 16 17 18 2: 24 28 32 36 39 41 43 43 42 40 22 16 43 27 -.. .. ..a.. . .°)

N itll Suspended Ccili ,1 : S:e-'_Fee4ith 1 9 0124 2 J -2 -3 -4 -4 -1 9 23 37 50 62 71 77 71 74 67 56 42 28 18 12 8 5 15 -4 78 12

(Cr ^9a 2 .eh I-in. IO 0.115 20 15 II 8 S 3 2 3 7 13 21 30 40 48 35 60 62 61 58 51 44 37 30 25 17 2 62 60

i-.u:z::n I : ) 0 :7= ' :4 10 7 4 2 0 0 4 10 19 29 39 44 55 62 65 64 61 54 46 38 30 24 17 0 65 65

:.. :.;, ^ 17I .. .. .3 .. 17 15 13 13 U 16 20 25 30 33 39 43 46 47 46 44 41 38 35 32 18 13 47 34

n+ C w b:;n. :0 0_: :! :3 16 U IO 5 5 7 12 I8 25 33 41 48 53 57 37 56 52 46 40 34 29 18 5 57 52 -

' - n ' . _ - - . . . : 5 ! : S 23 19 16 13 10 8 7 8 11 16 :2 :9 35 42 sf 52 e4 54 51 47 42 37 20 7 54 47

. . - -. _ ... .. :? :5 23 21 I! 16 15 15 16 14 :1 25 30 34 38 41 43 4a 44 42 40 37 21 IS 44 29 I. .. ,as,; _.,n sn . .;=r- .. . ' 6 13 :9 :5 :3 20 If 15 14 14 15 17 20 25 :9 14 38 42 45 46 45 44 42 21 14 46 32

_ 4 __ _, _. :0 21 :4 _ :9 ?: 74 16 13 :4 :? r ?5 34 33 19 20 38 I8

s. n . 15 0:'2 -- 73 30 28 26 24 .... 2A 18 18 18 20 22 25 28 72 35 73 40 41 41 40 39 37 21 18 41 23

_ 7.._.1:.0:1

1I ñx::era:en::er.^ ' OLs: 70 :9 2S :" 26 25 :4 23 22 22 22 :3 23 25 26 28 29 31 32 33 33 33 33 32 22 22 33 I1 ,2 t- :.. . . CJ-:'[:.

- ._. :7 :5 : 26 25 24 23 22 21 21 22 23 :5 :6 :3 30 32 33 34 34 34 33 32 31 20 21 17

.7 .7 .

- ' ...

15 34 21 32 31 29 . 26 24 23 ... .. .. ... 24 25 27 30 32 34 35 36 37 36 23 21 3" 16

..:.a.1.= " ''.:11

165

GJ Cooling Load Temperature Differences for Calculating Cooling Load from Sunlit Wzlls

0100 0200 0301 0400 05(0 0aYt 0 "C4) 0800 0900 1000 se., Time, h

1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400

Hr of Meal- mum

CLT D

Mini- Maxi. mum mum

CLTD CLTD

Dlffer- tm,ee

CLTD

Is -:nn C.a'nude H a l l F at i n t Group A W alb

N 14 14 14 13 13 13 12 12 II 11 10 10 10 IO 10 IO 11 II 12 12 13 13 14 14 2 10 14 4

NE 19 19 19 II l' 17 16 15 15 15 15 15 16 16 17 18 18 18 19 19 20 20 20 20 22 15 20 5

E 24 24 23 23 2. ! 20 i9 19 IB 19 19 20 21 2.2 23 24 24 25 25 25 25 25 25 22 18 25 7

SE 24 23 23 22 21 20 :0 19 18 IS 18 18 18 19 20 21 22 23 23 24 24 24 24 24 22 18 24 6

S 20 20 19 19 18 IF 1' 16 16 15 14 14 14 It II IS 16 17 IB 19 19 20 20 20 I3 14 20 6

S'.\' :5 25 25 24 24 n 22 21 20 19 19 18 17 17 17 17 18 19 20 22 23 24 25 25 24 17 25 8

27 26 :6 25 24 :4 :J 22 21 20 19 19 18 18 18 IS Iq 20 22 23 25 26 26 I 18 27 9

NA 2' 21 21 20 20 19 t9 19 17 16 15 15 15 14 14 14 15 15 16 17 18 19 20 21 1 14 21 1

Group B U as \ .5 14 . IS 12 II I: I^ 9 9 9 8 9 9 9 10 II 12 13 14 14 aS 15 IS 24 8 15 7

`. !9 1+ . 16 .. i4 12 1: 13 14 15 16 17 IS 19 19 20 :0 21 21 21 20 20 21 12 21 9

E 23 _ :1 20 I8 l' ;6 15 I! IS 17 19 21 22 21 25 26 26 27 :7 26 26 25 24 20 15 27 12

SE 23 22 21 20 18 17 16 15 14 14 IS 16 18 20 21 23 24 25 26 26 26 26 25 24 21 14 26 12

S 21 20 19 IB 17 IS li 13 12 II 11 II 11 12 14 IS 17 19 :0 21 22 22 22 21 23 11 11

SW 27 26 25 24 22 21 19 18 16 IS 14 14 13 13 14 IS 17 20 2.2 25 27 28 28 28 24 13 28 15

W 29 28 27 26 24 23 21 19 18 17 16 15 14 14 14 IS 17 19 22 25 27 29 29 30 24 14 30 16

NW 23 22 21 20 19 18 17 15 14 13 12 12 12 11 12 12 13 IS 17 19 21 22 23 23 24 II 23 9

Group C H alb N IS 14 13 12 11 10 9 8 8 7 7 8 8 9 10 12 13 14 15 16 17 17 17 16 22 7 17 10

NE 19 17 16 14 13 11 10 10 II 13 15 17 19 20 21 22 22 23 23 23 23 22 21 20 20 10 23 13

E 22 21 19 I' 15 14 12 12 14 16 19 22 25 27 29 29 30 30 30 29 28 27 26 24 18 12 30 18

'E 22 21 19 1" 15 14 12 12 I: 13 16 19 2 24 26 28 29 29 :9 29 20 27 26 24 19 12 29 17

S 21 19 IS 16 13 13 I: 10 9 9 9 10 II 14 11 20 2: 24 25 :6 25 25 24 22 20 9 26 17

S\t 29 27 2! 22 20 18 16 IS 13 12 II II II IS 15 IB 2.2 26 :9 32 33 33 32 31 22 II 33 22

W. 31 29 2 25 22 20 18 16 14 13 12 12 12 13 14 16 20 24 29 32 35 35 35 33 22 12 35 23

\ __ 23 20 IS 16 14 II II IO 10 10 10 II 12 13 15 18 2.2 25 27 27 27 26 22 10 27 17

Grasp D w alb N :5 13 . .

!3 10 9 7 6 6 6 6 6 7 8 10 12 13 15 17 18 19 19 19 18 16 21 6 19 13

s E 17 15 II iC 8 7 8 10 14 17 20 22 23 23 24 24 25 25 24 23 22 20 18 19 7 25 18

19 17 15 i3 1I 9 9 9 17 30 32 33 :3 32 31 .)0 28 26 2. 22 16 8 33 23

SE :0 17 ., 13 II 10 S 8 IJ 13 17 :6 t9 31 32 3: 32 :1 30 28 26 24 22 17 8 32 24

S 19 17 15 13 11 9 8 7 6 6 7 9 12 16 :a 24 2' 29 29 29 27 26 24 22 19 6 29 23

SW 28 IS 22 19 16 14 12 10 9 8 8 8 10 12 16 21 27 32 36 38 38 37 34 31 21 S 38 30

H' 31 r :4 21 IS aS 13 11 10 9 9 9 10 11 14 18 21 30 36 40 41 40 38 31 21 9 41 32

7.N 2. 22 19 r 14 ,. 10 9 8 7 7 8 9 10 12 14 18 22 27 31 32 32 30 27 22 7 32 25

Group EA ans

N 12 10 5 ` 4 3 4 . 6 7 9 II 13 15 17 19 20 21 23 20 18 16 14 20 3 22 19

`E 13 11 9 ' 6 4 5 9 15 20 :4 25 25 26 26 26 26 :6 :5 24 22 19 17 15 16 4 26 22

E. 14 1: I^ 8 6 5 6 II 18 26 33 36 38 37 36 34 33 32 30 28 25 22 20 17 13 5 38 33

<<: 15 1: 10 S 7 5 3 8 12 19 25 31 35 37 37 36 34 33 31 28 26 23 20 17 15 5 37 32

c 15 1: 1G 9 7 5 4 3 4 , 9 13 19 24 29 32 34 33 31 29 26 23 20 17 17 3 34 31

"t __ l. I. I2 F O ! 5 6 9 12 18 24 32 35 43 45 44 40 35 30 26 19 5 45 40

r 2.3 2 . 9 6 6 9 11 14 20 36 43 49 49 15 40 34 29 20 6 49 43

N'A =:i 11 14 I' + 6 1 . ! 6 9 IO 13 16 :0 26 32 37 39 36 32 28 24 :A 5 33 33

Groot) Fl:a:b 6: S 6 . 2 ! . 4 5 . 9 II 14 17 19 21 22 23 24 23 20 16 13 II 19 1 23 23

4 ' . '4 2' :e :3 29 27 27 :7 r :5 :4 .. 19 16 13 II II I 30 :9

:3 .

, .

_

_ .

6 17 :8 7?

.. 44 4' 3 39 36 31 3: 3) _" 24 21 l' IS 1: 12 2 45 43

SE. 10 S 4 3 10 19 25 26 41 43 42 39 36 34 21 :9 23 21 IB IS 12 13 2 43 41

S 10 5 6 i 3 2 . I 3 7 13 20 27 34 38 39 38 35 31 26 22 18 15 12 16 1 39 38

15 II 9 6 5 3 2 2 4 5 S 11 17 26 35 44 SO 53 52 45 3' 28 23 18 18 2 53 48

\ 1" 13 10 7 5 4 3 J 4 6 8 11 14 20 28 39 49 5' 60 51 43 34 2' 21 19 3 60 57

N.\ 14 10 9 6 4 3 _ _ 3 5 F IO 13 IS 21 27 35 42 46 43 35 28 2.2 18 19 2 16 44

GropG A.Os 1 0 -I 2 - 9 3 ._ 15 I9 21 23 24 :4 _ :6 22 15 11 9 7 S 18 -I 26

s.E 3 2 -I 9 26 39 'a 30 26 26 27 27 25 25 22 18 14 11 9 7 5 9 -I 39 40

4 , . -I .. ' 17 . $0 10 33 31 30 29 27 24 19 IS 12 IO 8 6 10 -I 55 56

_ __ 4) ., 4,, 42 16 j2 30 2' :4 ;9 15 12 10 8 6 11 -I 51 52

- . -! - . _ ._ .. ' Je 41 3' 31 25 23 .< 12 70 9 5 :4 -1 46 4' J . . .. 5 :5 36 53 19 63 ::1 3' :4 l' 13 10 9 16 0 63 63

. _ . . . 3 :! .. 19 . 41 !S 6' 72 6' 43 29 2J IS 11 8 17 t 72 71

3 11 IS '+ 21 _ 37 4" SS 35 4. .. r 13 10 7 19 0 55 35

lt7Ü_ í E V CLTD Correction For Latitude and Month Applied to Walls a.td Roofs, North Latitudes

Lat. ó:onth N

NNE NNW'

NE NW

ENE WNW

L W

FS WSW

SE SW

SSE SSW S HOR

G Dec -3 -5 -5 -5 -2 0 3 6 9 -1 Ja^ Nov -3 -5 -4 -4 -1 0 2 4 7 -1 Fe Oct -_ -2 -2 -2 -1 -1 0 -1 ' 0 0

Ntar/ Seht -3 0 i -1 -I -3 -3 -5 -8 0 Apr Aug 5 4 3 0 -2 -5 -6 -8 -8 -2 M1ta /Jul 10 7 5 0 -3 -7 -8 -9 -8 -4

Jun 12 9 5 0 -3 -7 -9 -10 -8 -5

8 Dec -4 -6 -6 -6 -3 0 4 8 12 -5 Jan Nov -3 -5 -6 -5 -2 O 3 6 10 -4 Feb.Oct -3 -4 -3 -3 -1 -1 1 2 4 -1 Nt:ir'Scpt -3 -2 -I -1 -1 -2 -2 -3 -4 0

Arr - :\ue 2 2 2 0 -1 -4 -5 -7 -7 -I Mav%Jul ? 5 4 0 -2 -5 -7 -9 -7 -2

Jun . 6 4 0 -2 -6 -8 -9 -7 -2 16 i.%:_ -6 -8 -8 -4 -1 4 9 13 -9

Jan: No -4 -6 -7 -7 -4 -1 4 8 12 -7 Feb/Oct -3 -5 -5 -4 -2 0 2 5 7 -4

Mar/Sept -3 -3 -2 -2 -1 -1 0 0 . 0 -1 Apr/Aug -1 0 -1 -1 -1 -3 -3 -5 -6 0 \ta ,Jul 4 3 3 0 -1 -4 -5 -7 -7 0

Jun (, 1 4 1 -1 -4 -6 -8 -7 0

24 17e.: -5 -7 -9 -10 -7 -3 3 9 13 -13 Jan 'Nov -4 -6 -8 -9 -6 -3 3 9 13 -11 Feb 'Oct -4 -5 -6 -6 -3 -I 3 7 l0 -7 Mar 'Sept -3 -4 -3 -3 -1 -1 1 2 4 -3 Arr Aug -2 -I 0 -I -1 -2 -1 -? -3 0

MavJul 1 2 2 0 0 -3 -3 -5 -6 1

Jun 3 3 3 1 0 -3 -4 -6 -6 1

32 -5 - -10 -11 -s -5 2 9 12 -17 'ar. N"v -5 -7 -9 -1I -8 -4 2 9 12 -15 FeO Oct -4 -6 -7 -8 -4 -2 4 8 11 -10

Mar Sert -I -4 -4 -4 -2 -1 3 5 7 -5 Arr. Aug -2 -2 -I -2 0 -1 0 1 I -1 Ma'. 'Jul 1 I I 0 0 -1 -1 -3 -3 1

Jun 1 2 2 I 0 -2 -2 -4 -4 2

_:t I'-,: -6 -4 -IO -13 -10 -7 0 7 10 -21 kill Nov -5 -7 -10 -12 -9 -6 1 8 11 -19 Fc- Oct -5 -7 -8 -9 -6 -3 3 8 12 -14 Mar Scot -4 -5 -5 -6 -3 -1 4 7 10 -8 Ar- A.:c -- -3 -2 -2 0 0 2 3 4 -3

`

..., Jul C 0 0 0 o o o o i 1

.1::n 1 I 1 0 1 0 0 -I -1 2

t':; -^ -9 -I I -14 -I3 -IO -3 2 6 -25 ,.. .. - -_ -ll -13 -I1 -S -1 5 8 -24 1---:t. 0...t -_ -- -10 -II -3 -_ 1 8 11 -18 Mar: :':pt -4 -6 -6 -7 -4 -I 4 8 11 -11 :,^- At.; -_ -3 -3 -3 -1 0 4 6 7 -5 \fav Jul r. -1 0 0 I I 3 3 4 0

Jun 1 1 2 1 2 I 2 2 3 2

1)r: -- -9 -12 -16 -15 -14 -9 -5 -3 -28 J:;n ti..,. -5 -11 -15 -14 -12 -6 -1 2 -21 Fci C?:: -. -S -10 -12 -!0 -7 0 6 9 -22 \ M7 5. . -. , - -? -5 -- 4 s 12 -15

r :\ _ _ -- -4 -4 -1 I 5 7 9 -8 `.. t. n O n ' 2 5 6 7 -2

2 1 3 i 4 5 6 I

,:.,; -- -9 -!2 -16 -17 -IS -16 -14 -12 -30 Jan, Nov -7 -9 -12 -16 -16 -16 -13 -IU -8 -29 Feb/Oct -6 -8 -11 -14 -13 IO -4 I 4 -26 Mar/Sept -5 -7 -9 -10 -7 -4 2 7 11 -20 ,1rr 'Aug -3 -4 -4 -4 -1 1 5 9 11 -ll May/Jul I 0 1 0 3 4 6 8 10 -3

Jun 2 2 2 2 4 4 6 7 9 0 1

Roof Construction Code Roof No.

Description

I Sled Sheet o.th Idn. insulation 2 1- in...ocd with In insulation 3 4 -in. I. w. concrete 4 2 -in. h.w. concrete with I -in. insulation 5 I -in. wood s.tth 2 -in. insulation 6 6-in. 1. w. concrete 7 2.5-in. wood with . in. insulation 8 8.1n. I. w concrete 9 .1 -in. h w. co:- fete w ith I.in. insulation

10 2.3.rn. wood with 2 -in rrwlaucn 11 Roof retrace system 12 6-in. h. w. concrete with 1 -in. insulation 13 4.in. wood »it 1.in. i :nutar.on

Code Numbers of l.ayus.43,

AO, E2, F3, BS, M, EO AO, El, £3, BS, B7, EO AO, E2, E), Cla, EO A0, E2, E3, BS, C12, Ell A0, E2, E3, B6. 137. EO A0, E2, E3, CIS, EC,

AO, E2, E3, BS, B8, EO AO, E2, E3, C16, EO

A0, E2, E3, B5, CS, LO A0, E2. E3, 86. B8. EO A0, C12, BI, B6, E2, E3, CS, Ell AO, E2, E3, B5, C13, E0 A0, E2, E3, B5, 89, EO

Thermal Properties and Code Numbers of Layers Used in Calculation of Coefficients for Roof and Wall

Ecscription Code Number L

Thickness and Thermal Properties' k D SH R Mass

Outside surface resistance AO 0.333 1 -in. Stucco (asbestos cement or wood siding plaster, etc) Al 0.0833 0.4 116 0.20 0.208 9.66 4-in. facebrick (dense concrete) A2 0.3333 0.75 130 0.22 0.444 43.3 Steel siding (aluminum or other lightweight cladding) A3 0.0050 26 0 480 0.10 0.000 2.40 Outside surface resistance 0.333 0.5 -in. slag, membrane A4 0.0417 0.83 55 0.40 0.375in. felt 0.0313 0.11 70 0.40 Finish A6 0.0417 0.24 78 0.26 0.174 3.25 4-in. facchrick A7 0.3333 0.77 125 0.22 0.433 41.6 Air Space Resistance 131 0.91 1 -in, insulation B2 0.0833 0.025 2.0 0.2 332 0.17 2 -in. insulation B3 0.1667 0.025 2.0 0.2 6.68 0.33 3 -in. insulation B4 0.2500 0.025 2.0 0.2 10.03 0.50 1 -in. insulation 35 0.0833 0.025 5.7 0.2 3.33 0.47

B6 0.1667 0.025 5.7 0.2 6.68 0.95 1 -in. v.) b7 0.0S33 0.067 37.0 0.6 1.19 3.u8 2.5 -in.wood B8 0.2063 0.067 37.0 0.6 2.98 7.71 4 -in. wccd B9 0.3333 0.067 37.0 0.6 4.76 2.3 2 -in. wood BIO 0.1667 0.067 37.0 0.6 2.39 6.18 3-in. wcod BI I 0.2500 0.067 37.i.) 0.6 3.58 9.25 3 -in. insulation BI2 0.2500 0.025 5.7 0.2 10.00 1.42

t :'.idLI:. n B13 0.3333 0.025 5.7 0.2 13.33 1.90 S -i i :,suta;c.fr. B14 0.4167 0.025 5.7 0.2 16.67 2.38 6-1n. insulation B15 0.5000 0.025 5.7 0.2 20.00 2.85 4 -in. cl.;. ule CI 0.3333 0.33 70.0 0.2 1.01 23.3 4-0: L,.. concrete block C2 0.3333 0 22 38.0 0.2 1.51 12.7 4-00100 ..0 ,.crete C3 0.3333 0 47 61.0 0.2 0.71 20.3 4 -in. common brick C4 0.3333 0.42 120.0 0.2 0.79 40.0 4-in. I.... c.mcrete CS 0.3333 1.00 140 0 0.2 0.333 46.6

C6 0.6667 0.33 70.0 0.2 2.02 46.7 C7 0.6667 0.33 38.0 0.2 2.02 25.4

. 00-. c :c C8 0 666' 0.6 61.0 0.2 1.11 46.7 3- :1...07.mc1 brick 0.9 0.6667 0.42 120 0 0.2 1.59 80.0 .000 ..... ,Oft.erele C:0 0 6667 1.00 140.0 0.2 0.667 93.4 120r.. 1.w. concrete CI 1 1.0000 1.00 140.0 0.2 1.08 140.0 20n. i.... concrete C12 1.6667 1.00 140.0 0.2 1.16 7 23.4 6-in. I... ccrcrete C13 0.5000 I.CO 143.0 0.500 0.2 70 ,

C14 0.3333 0.10 40.0 0.2 3.333 13.3 6-.n. C15 0.5000 0.10 40.0 0.2 5.000 20 0

C16 0 6667 0.10 40.0 C 2 6.667 26 7 :t i .....,,re:_ bleek :..,ed .s.,iat:cn) C17 0 bC67 t) O3 10.0 t; 2 9.00 12.0 n.: N conc,_'ebl.ck I :Ilo:1i C18 0.6667 0.34 53 0 0.2 1.98 35.4

C19 1.0000 0.08 19.0 0.2 13.5 19.0 C23 1.06.0 u 39 :5.0 C.2 2.59 56.0

tta.e rest 0,01,c E0 .665 0.75-in. plaster; 0.75 -in. gypsum or other similar fint..hing layer EI 0 0025 0.42 100 0.2 0.149 ó.25 0.5 -in. slag or stone E2 0.0417 0.83 55 0.40 0.050 2.29 0.375 -in. felt & membrane E3 0.0313 0.11 70 0.40 0.285 2.19 Ceiling air space E4 1.0 Acoustic tile E5 0.0625 0.035 30 0.20 1.786 1.8S

17(

Rl le Wall Construction Group Description

Group Nu. Description of Construction

Weight (Ib /fi2)

U-Value (Btu/h ft=F)

Code Numbers of Layers / see Tabs

4 -i Face 3ri_k + (Brick) C Air Space + 4 -in. Face Brick 83 0.358 A0, A2, BI, A2, E0 D 4 -in. Common Brick 90 0.415 A0, A2, C4, El , E0 C 1 -in. Insulation or Air Space + 90 0.174 -0.301 AO, A2, C4, BI/B2, El, E0

4 -in. Common Brick B 2 -in. Insulation + 4 -in. Common Brick 88 0.111 AO, A2, B3, C4, El, EO B 8 -in. Common Brick 130 0.302 AO, A2, C9, El, EO A Insulation or Air Space + 8 -in. Common brick 130 0.154 -0.243 AO, A2, C9, BI/B2, El, EO

4-in. Face Brick + (H. W. Concrete) C Air Space + 2 -in. Concrete 94 0.350 A0, A2, B1, C5, El, E0 B 2 -in. Insulation + 4 -in. Concrete 97 0.116 AO, A2, B3, CS, El, E0 A Air Space or Insulation + 8 -in. or more Concrete 143 -190 0.110 -0.112 AO, A2, BI, C10/11, El, E0

4-in. Face Brick + (L. W. or H. W. Concrete Block) E 4 -in. Block 62 0.319 AO A2, C2, El, E0 D Air Space or Insulation + 4 -in. Block 62 0.153 -0.246 AO, A2, C2, B1/B2, El, E0 D 8 -in. Block 70 0.274 AO, A2, C7, A6, E0 C Air Space or 1 -in. Insulation + 73 -89 0.221 -0.275 AO, A2, BI, C7/C8, El, E0

6 -in. or 5 -in. Block B 2 -in. Insulation + 8 -in. Block 89 0.096 -0.107 AO, A2, B3, C7/C8, E1, E0

4-in. Face Brick + (Clay Tile) D 4 -in. Tile 71 0.381 AO, A2, Cl, El, EO D Air Space + 4 -in. Tile 71 0.281 AO, A2, CI, BI, El, EO C Insulation + 4 -in. Tile 71 0.169 AO, A2, Cl, B2, E1, EO C 8 -in. Tile 96 0.275 AO, A2, C6, E 1, E0 B Air Space or 1 -in. Insulation + 8 -in. Tile 96 0.142 -0.221 AO, A2, C6, BI/B2, El, E0 A 2 -in. Insulation + 8 -in. Tile 97 0.097 AO, A2, B3, C6, El, E0

H.W. Cenerete Wall + (Finish) E 4 -in. Concrete 63 0.585 AO, Al, C5, El, EO D 4 -in. Concrete + 1 -in.

or 2 -in. Insulation 63 0.119 -0.200 AO, Al, CS, B2/B3, El, E0

C 2 -in. lasulation + 4 -in. Concrete 63 0.119 AO, Al, B6, CS. El, E0 C 8 -in. Concrete 109 0.490 AO, AI, C10, El, E0 B 8 -in. Concrete + 1 -in.

or 2 -in. Insulation 110 0.115 -0.187 AO, Al, C10, B5/B6, El, E0

A 2 -in. Insulation + 8 -in. Concrete 110 0.115 AO, Al, B3, C10, El, E0 B 12 -in. Concrete 156 0.421 AO, Al, C11, El, EU A 12 -in. Concrete + Insulation 156 0.113 A0, C11, B6, A6, £0

L. W. and H.W. Concrete Block + (Finish) F 4 -in. Block + Air Space /Insulation 29 0.161 -0.263 AO, Al, C2, BI/B2, El, E0 E 2 -in. insulation + 4 -in. Block 29 -37 0.105 -0.114 AO, Al. B3, C2/C3, Ei, E0 E 8-in. Block 47 -51 0.294 -0.402 AO, Al, C7/C8, El, E0 D 8 -in. Block + .Air Space/Insulation 41 -57 0.149 -0.173 AO, Al, C7/C8, BI/B2, El, EU

Clay Til,. + (Finish) F 4 -in. Tile 39 0.419 AO, Al, Cl, El, EO F 4 -in. Tile + Air Space 39 0.303 AO, Al, Cl, BI, E1, E0 E 4-in. Tit + 1 -in. Insulation 39 0.175 AO, Al, Cl, B2, El, EO D 2-in. Insulation + 4 -in. Lle 40 0.110 AO, AI, B3, Cl, El, EO D 8 -in. Tile 63 0.296 AO, Al, C6, BI/B2, El, E0

- Air S7ace'l -in. Insulation r

63 0.151 -0.231 AO, Al, C6, BI/B2, El, E0 . B _- 1n.Irt; a::on 8 -in. I:le 63 0.099 AO, Al, B3, C6, EI, EO

Metai Carta: i .....OUI ; .0 SpoLc 1- in.' 5-6 0.091 -0.2 30 AO, A3, B5/B6/B12, A3, E0

3 -,n. Frame Wall

G 1 -in. to 3 -in. Insulation 16 0.081-0.1 78 AO, Al, BI, B2/B3/B4, EI, E0

112.

The overall heat transfer coefficients given in Table 6 -1 for roofs

and Table 6 -5 for walls are to be used as guides.

The actual value of U calculated as described in Chapter 5

should be used in calculations. When the actual roof or wall

cannot be found in Table 6 -1 or 6 -5, a thermally similar one should

be selected. Thermally similar walls or roofs have similar mass

and heat capacity.

ttA wall or roof may have more insulation than those given in

Tables 6 -1 or 6 -5 if this is not possible use CLTD equals to 29 F.

For each R -7 increase in insulation, choose a roof of similar mass

and heat capacity with a peak CLTD that occurs 2 hours later. For

walls, simply move upward one letter in Table 6 -2 to the next group

for each R -7 increase in insulation. To reiterate, the actual

overall heat transfer coefficient for the wall or roof should be

used. It is important in selecting a wall or roof group that the

mass and insulation be oriented properly. Walls with indentical

thermal resistance and mass will respond differently if the

individual component layers are reordered. More simply stated, a

wall or roof responds differently if the insulation is on the

inside rather than the outside of the massive component.4/ C EJ Example(6 -3):

A building located in Tallahassee, FL., has a roof

described as number 1 in Table 6 -1 with a suspended ceiling.

There are 2.5 in. of fibrous glass batts laid on top of the

ceiling. Compute the cooling load per ft at 10:00 AM and

5:00 PM solar time using standard design conditions for

August.

173

Solution:

The cooling load is calculated using Eq. 6 -14 where U

is computed for the actual roof. Data can be obtained from

Table 6 -6. The overall heat -transfer coefficient including

the ceiling air space and acoustic tile ceiling is 0.134

BTU /(hr -ft'2 F) as given in Table 6 -1. This addition of the

2.5 in. batts, which have a thermal resistance of 7 (hr -ft 2

-F) /BTU, will change U to 0.07 BTU /(hr -ft 2 -F). The CLTD

should then be obtained from Table 6 -1 for a thermally

similar roof with a peak CLTD occuring 2 hours later than roof

number 1.

It appears that roof number 2 is the best choice. The

uncorrected CLTD values for 10:00 AM and 5:00 PM are 13 F

and 62 F, respectively. Eq. 6 -17 is used to correct these

values as follows:

LM = -0.5 F; Table 6 -3 at 30 °N.Lat.

K = 1.0; dark surface assumed

ti = 78 F; ASHRAE standard

tom = 92 - (19/2) = 82.5

For 10:00 AM

CLTD 10 = (13- 0.5)(1.0) +(78 -78) +(82.5 -85) = 10 F

For 5:00 PM

CLTD 17 = (62- 0.5)(1.0) +(78 -78) +(82.5 -85) = 59 F

Then the cooling load at 10:00 AM is

qc /A = U (CLTD10) = 0.07 (10) = 0.7 BTU /(hr -ft )

at 5:00 PM

qc /A = U (CLTD17) = 0.07 (59) = 4.13 BTU/(hr-ft ) 41E6,)

(14

(2) FENESTRATION

Heat admission or loss through fenestration areas is affected

by many factors of which the following are the most significant.

tI1. Solar radiation intensity and incident angle.

2. Difference between outdoor and indoor temperature.

3. Velocity and direction of air flow across the exterior

and interior suraces.

4. Low temperature radiation exchange between the surfaces

of the glass and the surroundings.

5. Exterior or interior shading. p L-0

When solar radiation strikes an unshaded window, about 8

percent of the radiant energy is typically reflected back outdoors,

from 5 to 50 percent is absorbed within the glass, depending on the

composition and thickness of the glass, and the remainder is

transmitted directly indoors, to become part of the cooling load.

The solar gain is the sum of the transmitted radiation and the

portion of the absorbed radiation that flows inward. Because heat

is also conducted through the glass whenever there is an outdoor -

indoor temperature difference, the total rate of heat admission is

Total heat admission through glass

= Radiation transmitted through glass

+ Inward flow of absorbed solar radiatioon

+ Conduction heat gain (6 -18)

The first two quantities are related to the amount of solar

radiation falling on the glass and the third quantity occurs

whether or not the sun is shining. In winter the conduction heat

flow may will be outward rather than inward. We can rewrite Eq.

175

6 -18 to read:

Total heat gain = Solar heat gain + conduction heat gain (6 -19)

The conduction heat gain per unit area is the product of the

overall coefficient of heat transfer U for the existing

fenestration and the outdoor - indoor temperature difference (to -ti).

Value of U for a number of widely used glazing systems are given

in Table 3 -6. The cooling load is computed using a CLTD much the

same as a wall or roof. The glass represents a much more simple

situation, however, with small thermal capacity.

Table 6 -7 gives CLTD values for glass. The Table is for 78 F

indoor temperature, 95 F maximum outdoor temperature, and 21 F

daily range. Corrections may be made as shown for walls and roof.

The cooling load due to conduction heat gain through window glass

is then given by

4c = UA (CLTD) (6 -20)

The solar heat gain is much more complex because the sun's

apparent motion across the sky causes the irradiation of a surface

to change minute by minute. It was mentioned earlier that the

cooling load due to solar heat gain is expressed in terms of a

shading coefficient, a solar heat gain factor, and a cooling load

factor (Eq. 6 -15). We will first consider the shading coefficient.

The ASHRAE procedure for estimating solar heat gain assumes

that a constant ratio exists between the solar heat gain through

any given type of fenestration system and the solar heat gain

(under exactly the same solar conditions) through unshaded clear

sheet glass (i.e., the reference glass). The reference glass is

double- strength glass with a transmittance of 0.86, a reflectance

I7,

of 0.08, and an 0.06 absorptance. This ratio, called the shading

coefficient, is unique for each type of fenestration or each

combination of glazing and internal shading device.

Solar heat gain of fenestration SC = (6 -21)

Solar heat gain of double- strength glass

The shading coefficients for several types and combinations

of glass are given in Table 6 -8. The shading coefficient for any

fenestration will rise above the tabulated values when the inner

surface coefficient is increased and when the outer surface

coefficient is decreased. The converse is also true. Table 6 -8

shows the effect for outer film coefficient of 3 and 4

BTU /(hr- ft2 -F) where the larger value is for a 7.5 mph (miles per

hour) and the smaller value is for a lower wind velocity. The

effectof the film coefficient on the shading coefficient is related

to energy absorbed by the glass and then transferred away by

convection.

Blinds, shades, and drapes or curtains that are often

installed on the inside next to windows decrease the solar heat

gain. The shading coefficient is used to express this effect.

Tables 6 -9 and 6 -10 give representative data for single sheet and

insulating glass with indoor shading by blinds and shades.

Note that the shading coefficient applies to the combination of

glass and the shading device.

Shading coefficients for draperies are a complex function of

color and the weave of the facric. Although other variables also

have an effect, reasonable correlation has been obtained using only

color and openness of the weave. Fig 6 -6 and Table 6 -11 are a

Table i -7 C 6J _ Cooling Load 7 emperature Differences for Conduction through Glass

Solar Time, h 0100 0200 0300 0400 0500 0600 0700 0800 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 CLTD

F I 0 -I -2 -2 -2 -2 0 2 4 7 9 12 I) 14 14 13 12 10 8 6 4 3 2

Corrections; 7 he clues in the table were .aLulatrd for an Inside temrirrarure of 78 F and an outdoo, maximum temperature of 95 1 %Ith an outdoor dady range nt 21 r. The table temam, Ir. proximate]) correct for other outdoommaurnums 93102 f and other outdoor dally ranges I$.31 F. provided the outdoor daily a.eragr temperature remains approximately 65 F. If the room air temperature is different from 78 F and/or the outdoor daily aseragc temperature is different from 85 F. the following rules apply: (a) I or room air temperature less than 78 F. add the difference between 78 F and room air temperature: if greater than 78 F, subtract the difference. (b) For outdoor daily average temperature Ins than 85 F. subtract the difference between 85 F and the daily average temperature; if greater than 85 F. add the difference.

Shading Coefficients for Single Glass C6] Ta b le ar- -&.., and Insulating Glass'

A. Single Glass

Type of No mi nad Solar Shading Coefficient Glass Thicknessb Trans. he.4.0 hp 3.0 Clear 1/8 in. 0.86 1.00 LOO

1/4 in. 0.78 0.94 0.95 3/8 in. 0.72 0.90 0.92 1/2 in. 0.67 0.87 0.88

Heat Absorbing 1/8 in. 0.64 0.83 0.85

1/4 in. 0.46 0.69 0.73 3/8 in. 0.33 0.60 0.64 1/2 in. 0.24 0.53 0.58

B. Insulating Glass

Clear Out, Clear in 1/8 in.' 0.71e 0.88 0.88

Clear Out, Clear in 1/4 in. 0.61 0.81 ' 0.82

Heat Absorbing° Out, Clear In 1/4 in. 0.36 0.55 0.58 aRcfers to factory-fabricated units with 3/16, 1/4 or 1 /2 -in. air space or to

prsbmc windows plus storm sash. Refer to manufacturer's literature for values.

(Thickness of each pane of glass, not thickness of assembled unit. Refers so gray. bronze and green tinted hat -absorbing float glass.

(Combined transmittance for assembled unit.

/ .1 -r!1: _., ;C6 J Shading Coefficients for Single Class with Indoor Shading by Venetian Blinds or Roller Shades

Nominal Thickness'

in.

Solar Trans.b

Type of Shading Venetian Blinds Roller Shade

Opaque Translucent Medium Light Dark White Light

Clear Clear Clear Pattern Heat - Absorbing Pat :.rn Tinted

3/32 w 1/4 1/4 to 1/2 1/8 to 1/2

1/8 3/16, 7/32

0.87 to 0.80 0.80 to 0.71 0.87 to0.79 - 0.74, 0.71

0.64 0.55 0.59 0.25 0.39

Heat -Absorbing° 3/16, 1/4 0.46 H.at- Absorbing Pattern 3. 16, li4 - 0.57 0.53 0.45 0.30 0.36 Tined 1/8, 7/32 0.59. 0.45

Heat-Absorbing or Pattern 0.44 to 0.30 0.54 0.52 0.40 0.28 0.32

Heat-Absorbingd 3/8 0.34

Heat -Absorbing 029ío0.15 or Pattern 0.24 0.42 0.40 0.36 0.28 0.31

Re(lecti%e Coated Glass

S.C.`= 0.30 0.25 0.23 0.40 0.33 0.29 0.50 0.42 0.38 0.60 0.50 0.44

!Refer to manufacturer's literature for values. t F or Bern. t:.ad, c; ague r, t.:te and beige louvers in the tightly closed positsoa, SC is 0.25 and 0.29 when used .tin glass of 0.71 too BO transmittance.

c St: for :., >ssv::hno, g desace d R -ors to a., : rccu, :.d peen ::teal heatat...o'b.ag glns.

o -ía[63 ..

Shading Coefficients for Insulating Glass' with Indoor Shading by Venetian Blinds or Roller Shades T) pe of Shading

Nominal

Solar Trans.° Venetian Blinds` Roller Shade

Opaque Translucent

Type of Glass Thickness. Outer Inner Each l.i2ht Paie Pane Medium Light Dark White Light

C';e_r Out 32, 1.-6 in. 0.87 0.87 Cacar Ice 0.57 0.51 0.60 0.25 0.37 Clc rO.c

I'4in. 0.F0 0.80

H__, -it".: r,.mgd Out 1. 4 tr;. 0.46 0.80 3.39 0.35 0.40 0.22 0.30 Clear In

P.;:tlecute Coated Glass SC`= 0.20 0.19 0.1P

0.30 0.27 0.26 0.34 0.33

Rcers :o a_:_ry ;ab:tca:eo ar.us a nh ?, 10. 1-'4. or I 2-in. air space, or to prune wusdoas plus storm wsndov.s. r. r;e: :,..r.ar..:a.r.,:er's:nc,:_:e: cla:;

-.. . ,. ... - ..., c; :A : :e iosca. SC ;s a;.; os::nainy the same as for opagae rAnr ro!;r. sudes .:

. . . -. .....Bt.rce......._....., e ., sr.ao,ng.:-.:.e

brief summary of information given in the ASHRAE Handbook.

Many manufacturers of drapery materials furnish data on the

radiation properties of their products. This is usually in the

form of the reflectance and transmittance, the abscissa and

ordinate for Fig 6 -6. The shading coefficient index letter is read

from Fig 6 -6 (A to J) and used in Table 6 -11 to determine the

shading coefficient for the combination glass and drapery system.

The reflectance and transmittance of the drapery are often not

known. In this case the index letter may be estimated from the

openness of the weave and color of the material. Openness is

classified as: Open,I; semiopen, II; and closed, III. Color is

classified as: dark, D; medium, M; and light, L. A light -colored,

close -weave material would then be classified IIIL. From Fig 6 -6

the index letter varies from G to J for this classification, and

judgement is required in making a final selection.

Heat gain through sunlit double- strength sheet glass, the

reference glass, is designated as the Solar Heat GAin Factor

(SHGF). Values of this quantity have been calculated for the

twenty -first day of each month, for daylight hours, for various

directions and various latitudes.

The solar heat gain factor of Eq. 6 -15 is the maximum value

for the month for a given orientation and latitude. The cooling

load factor then represents the ratio of the actural solar heat

gain, which becomes cooling load, to the maximum solar heat gain.

This approach reduces the amount of tabulated data. Table 6 -12

gives maximum solar heat gain factors for 32, 40, and 48 degree

north latitude. Table 6 -13 gives maximum solar heat gain factors

(00

e ¡ l C éJ S :.at ;i :tä Coeffieicnts for Single and Insulating Glass with Draperies

G II I J

SC for Index Letters in Fig. 23'

D E F Glazin;

Glass Trans.

Glatia SCb

A E3 C

G:ass I: 4 in. Clear 0.80 0.95 0.80 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.40 0.35 1/2 in. Cte.tr 0.71 0.88 0.74 0.70 0.66 0.61 0.56 0.52 0.48 0.43 0.39 0.35 1/4 in. Heat Abs. 0.46 0.67 0.57 0.54 0.52 0.49 0.46 0.44 0.41 0.38 0.36 0.33 1/2 in. Heat Abs. 0.24 0.50 0.43 0.42 0.40 0.39 0.38 0.36 0.34 0.33 0.32 0.30

Reflective Coated 0.60 0.57 0.54 0.51 0.49 0.46 0.43 0.41 0.38 0.36 0.33 (see manufa .turers' literature 0.50 0.46 0.4-4 0.42 0.41 0.39 0.38 0.36 0.34 0.33 0.31 for exact values) 0.40 0.36 0.35 0.34 0.33 0.32 0.30 0.29 0.28 0.27 0.26

0.30 0.25 0.24 0.24 0.23 0.23 0.23 0.22 0.21 0.21 0.20

Ins atin3Gla_s 1 /2 -in. Air Space Clear Out and Clear In 0.64 0.83 0.66 0.62 0.58 0.56 0.52 0.48 0.45 0.42 0.37 0.35

Heat Abs. Out and Clear In 0.37 0.55 0.49 0.47 0.45 0.43 0.41 0.39 0.37 0.35 0.33 0.32

Ref cctice Coated 0.40 0.33 0.37 0.37 0.36 0.34 0.32 0.31 0.29 0.28 0.28 (see manufacturers' literature 0.30 0.29 0.28 0.27 0.27 0.26 0.26 0.25 0.25 0.24 0.24 for exact s alues) 0.20 0.19 0.19 0.18 0.18 0.17 0.17 0.16 0.16 0.15 0.15

tint glass duct wtr no drrr: cr). 1 1 I

Coe'r..:co; .an: C] for t ".e SC:tnrs ir. Fit 23 for rcpresenunse slar:c ¡s Substitut for SC index Inters in Fis. 23 .alun on the line of the slums selected.

SN aDING COEFFICIENT INDEX LETTER GLAZING INDICATED IN TABLE 39

DRAPERIES ARE 100°'a FULLNESS (Iatr.. ::ncsdra,^cJN: I `.:cs: I. ShaoirF Cxfficients are for draped

fat rt;s. .. (::her ,^roperr,es arc for fabr.rs in flat

?. !',e Fa^r.c Rc(le: arce zr,d Tra.:- -.t:tan:c co ocu:n S:-.aa;r.g

4. l.se Pe :'.t s and Yarn Re.^,e::ance or r--nnr:i ar.,. Far.:. Rc;._.._...e :o , ar:.o;a En,.. °-r..l

Cr to ot:at.^. Ap- r:os:.::ate .`.,..7:ng CaerFtarts.

CL \SSii It.-ATiON OF FAI3I' iCS I - ();-,en \1'ease

11 - Se-.:-open \L'esse Ill ° C:_..S \Ve_se

D - Dar. . !cr" 1 - \!c_._... 'Col"

- i... .. -Co., r'

RI 6--6 C6J

0.70

/ (A) (6)

REYARN' FLECTANCE

IOr20 25 3C40 r56 p,0 / 0.60 è astiara+Emo, Ì. ,. - W

050 v z d

040 ll

C 30' H U

0 2C

01`'

(G)

O 1 i.1/::'IIL 1 aj (H) WNW; °a ,,

l W d )

O KVSi 9F`

)

//,,Eas.,.ur f

PAN. I.ri a L Ii /ece ' MEMO ('"PF,r,titi.

S ra S r i V

D

û O33 c40 050 FABRIC REFLECTANCE

0.60 073

`SI

;,z k_ i2 C7 Maximum Solar Heat Gain Factor, Btu /h ft2 for Sunlit Glass, North Latitudes

0 °N Lar 20°N 1.21

N NNE/ NNW

NE/ NW

ENE/ WNW

E/ W

ESE/ WSW

SE/ SW

SSE/ SSW S HOR -N

NNE/ NNW

NE/ NW

ENE / WNW

El W

ESE/ WSW

SE/ SW

SSE/ SSW S HOR

Jan. Fch. Slat. Apr. '.3 J_-

_. .

G. 1-

.

D-:

34 36

. 38 71

: '

I:' 1:1

75

_7

35

24

34 39

87

:34 163 173

:::3 134

84 43

35

34

F8

132

170

193

203 206 :.J! 187

1u3

1:9 5S

71

177

205 223 224 2

2 i ".. 216 213

1,4 175 164

234 245 242 2:1 :01 191

195

212 231

235 2:) .28

254 241

223 I94 151

140 149

175

213

_:S 250 :53

235 210 170

116 SO

66 77

11: 163

202 2:0 :40

182

141

87 38

37 37

38

39 64

135

179 196

118

67

38 37

37 37

38 39 40

66 117 138

296 306 303 284 265 255 260 276 293 299 293 288

Jan. Fcb. "14r. Apr. May June July Aug. Scp. 0:1. No. .

Dec.

29 31

34

38

47 59

48 40

36 32

29 27

29 31 49 32

123

135 124

91

46

32 29 27

48

88 132

166 184

189

182

162

127

87 48

35

138

173

200 213 217 216 213 206 191

167 136

122

201 226 237 228 217 210 2:2 220 225

217 197

187

243 244 236 208 184

173

179

200 225 236 239 238

253 238 206 158

124

108

119

152

199

231 249 254

233 201

152 91

54 45

53

88

148 196

229 241

214 174

115

56 42 42 43

57 114 170 211

226

232 263 284 287 283 279 278 280 275 258 230 217

4 N 1 11 24' N. Lat

N NNE/ NNW

NE.' NW

ENE/ WNW

E' K

ESE' WSK

SE.' SW

SSE/ SSW S HOR N

NNE/ NNW

NE/ NW

ENE/ WNW

E/ W

ESE/ WSW

SE/ SW

SSE/ SSW S HOR

Jan. Feb Mar. A,r. 613.

A_¡ Sc:. C,.: No.. Dc;.

33

35

3e °5 93

1 :2 96

_

36

34 33

33

35 77

125

154

164

154

1:4 75

26

34 33

19 12.3

163

IS9 2

1,- 14 156

120 79

62

170

199

219 223 2:3

215

215 2: :9 193

168

157

2 :9 242 242 225 2:6 I --

214 231 234 226 221

248

227 190

IN 4

1`.6

Iil 216 234 243 250

237 215 177 126

S9

"3

35

I:) 110 207 232 243

193

152 96 43

38 38

39 42

93

148 190

206

141

88 43 38

38

38

36

40 44

86 139

160

286 301

302 287 272 263

267 279 293 294 284 277

Jan. Fcb. Mar. Apr. Slay J.tne

Jc1y Aug. Sep. Qt. No.. Dec.

27 30 34 37

43

55

45

38

35 31

27 26

27

30 45 68

117 1 37

116 67 42

31

27 26

,41 BO

124 159 176 164

176

156

119 79

42

2'.

128

165 195

209 214 214

210 203 183

159

1:6 112

190

220 234 228 218 2.2

213 220 222 211

167

180

240 244 237 212 190 179

185

293 225 237 236 234

253 243 214 169

132 117

129

162

2-i6 235 249 247

241

213 168 107

67 55

65

103

163

207 237 247

227 192

137 75

46 43

46 72

134 187

224 237

214 249 275 263 282 279

273 277 266 244 213 199

8\ I,e 28 N. Lat

N NNE/ \\W

NE' \W

ENE W\K

it K

ESE.' W6K

SE' SK'

SSE/ SSK s HOR

N (Shade)

\\E./ \ \W NE/ NW

ENE/ %1 NW

E/ K

ESE/ WSW

SE/ SW

SSE/ SSK S HOR

Jan.

Fc7.

Mar. A;r. f=.

1.,7:e

1.!y

A.:

.:.

: Je;

32

34

37 44

74

90 77

4"

3S

15

33

31

32 -. 67

11'

149

153

145

II' 66

35

33

31

71

114

156

IS4

196

200 195

1'9 149

;I: '1

5

163

193

215 _21

::J

217

215

214

SA 107

161

149

2:4

239 241

27.5

_

200

204

:;6 230

231

._ 2!5

:f0

24S

2)0 145

1:7'

141

162

leb 219

::0 _.. 246

242

219 164

134

97

62

93

1:9 176

21: 233

:4'

201 165

110

53

39

39 40

51

10'

160

20) 2 15

162

110

55

39

36

39 39 41

56

tu8 160

179

275

294

300 289

277

269

272

282 290

:39 273 265

Jan. Fcb. Mar. Apr S1a) June July Aag. St 7. O:I. Nos. Ilea.

25

29 33

36 40

SI 41

38

14

30

26

24

25

:9 4) 84

113 125

114

83

33

30

26

24

35 72

116

151

172 It 170 149

III 71

35

24

117

157

169

:05 211

211

203 199

179

151

115

99

183

213 231

228

219 213

215 2:0 2 :9 2'v4

161

172

235 244 237 216 195 184 190

207

226 236 232 227

251

246 221

178

144 128

140

172

213

238 247

243

247 224 182

124

83 68 80

120

177

217

243 251

236 2.07

157

94

58 49 57 91

154

202 235

246

196

234 265 278

280 278 276 272 256 229 195

179

\

Ie k:

K

32' \. 1 at N\F. \\K \F..

\W ENE, K\K 84 ..84

SE, SW

SNE SSW S HOR N

(Shade) NNE' NW

NE/ NW

ENE, WNW

E, K

ESE/ KSK

SE' SW

SSE/ SSW S HOR

52i F?o `.1sr

_' \r. ...

-._: -..

.

...

31 34

36

--

,

"5

._ :" :4 ._

31

34

59

1.:5

139

144

1 37

!;rì

-

34

32

_?

63

103

146 l"6 14 ;';S 191

'"4

- .. .

cì a'

155

156

::8 :19 2:J 217

215 212

'

.. 53

..i

217 :35 :4.)

_. 2'4 _.. _.i --a

_

2:4

246 24,5

273 :.' 173

.61 1 x5

. .

--- ..... 241

247 _6 I)) I4:

I,i6 v)

10:

135

.._ 2:4 243

212 177

1:4 64

40 40 41

42 I:1 172

_.,9 223

182

133 73 40

43 40 41

142

73

133 179

197

262 :86 :97 290

260 274 275 282 287 :60 260 250

Jan Fcb. Afar. Apr. Ma) Jure J0: Aug Sept 0.r No.. Dew

24

27

32 36

38 44 40 37 33

2.3

24

22

24 2" 3" 60

111

1:2 I11

35

:5 24 22

29 65

107

146

170 176 167

141

103

63

29 22

105

149

197

200 203 2ì8 :04 195

1 "3 143

103

84

175

`05

227 2:0 214 215 219 215 195

173

162

229 242 :37 219 199 139 194 :10 2:1 234 225 218

249 248 227 147

155

139

150

151

:is 239 245

:46

250 232 195

141

99 83 96

136 169

225 246 252

246 221

176 115 74 60 72

III 171

215 243

252

176 217 252 271 277 276 273 265 244 213 175 158

:6' \ I ae 36'\ 1.1

\\F 1.>1 ek NE' ESE ' E ESE, SE SSE/ \ \\K NW NSA K K\4 SK !INN S HOR (Shade) \\K \K W\W W KSH SK SSW S HOR .-.

. .. . _... _.. ... 2:3 1-.9 :43 14n 22 . ;ob :1L :4' 212 25: 155 . . , . . _, _-- _ . ,,. '4 -5 1r. _ _. . + . 39 2.46 234 _,_ 199

. . .:i :i .. t3i 93 :91 1114r :J :ì 'N 1"c 2'3 :33 .._ :06 192 238

.r r) -_ .. _- :'4 -7 45 2:9 Alt 35 "6 144 198 ... :21 '-6 l56 135 262 .. _ ._ ..r . . . .. 45 41 . N1.. 3. . - :69 . ._. : ... to 93 272 :I' .

.. +I 41 2-7 J....c 47 i ._ !"5 .. _. :;4 .'.: 9`I 77 273

. ._ . ... .. . - ... 44 42 2'7 J.. 39 . I65 .iI _16 !v9 .,I !i3 43 :`8

.. . . . . : ? _. , .45 . -6 282 A-¡ 36 'S 13S 90 ;s . i9 1:1 13; 237 5J 134 lvb 2 :24 Iv) 114 93 262 Scp 31 31 95 167 210 228 223 200 187 230 O:r. 33 33 95 174 223 237 225 183 ISO 270 Qt. 27 27 56 133 187 230 239 231 225 195

Nov. 30 30 55 145 206 241 247 2I0 196 246 Nov. 22 22 24 87 163 215 243 248 246 154 Dec 29 29 41 132 198 241 254 233 212 234 Dec 20 20 20 69 151 204 241 253 254 136

1 g2

l4Y; %; -' i L J Maximum Solar Heat Gain Factor, Btu /h ft2 for Sunlit Glass, North Latitudes (concluded)

40'N.Lai 60'N.Eat N

(Shade) NNE; NNW

! ,'_W

EN:J NNW

E/ W

LSE1 NSW

SF' SW

SSFJ SSW 5 HOR

:tn. _ 20 2C 74 151 205 241 25. 254 133

I -b. 50 129 IF6 234 216 244 241 I8O Mar. 29 :9 93 163 219 239 2_'6 216 206 223

\pr. 34 71 110 190 224 222 203 170 154 i52 May !02 _ ._. '?S 17 1 :'3 113 265 J c . ̂ . e - 1!3 . _ 2.75 2:6 IN 16: 116 95 :67 Jul. _ 1r1 v.'s 2 1',) 1:9 109 262 Aug. :5 ifs :16 214 1`4 165 149 247 S:;7. 33 _ 5" i') :'3 __ 2 :6 209 2Y) 215 0:t. 2$ 25 49 123 180 225 233 236 234 177 Nn.. 20 20 :0 151 20! 248 250 132 Dec. 19 IF 18 60 135 138 222 249 253 113

N (Slade)

NN E./ NNW

NE/ NW

ENE/ WNW

E/ W

ESE/ WSW

SF./ SW

SSE/ SSW S HOR

Jan. 7 7 7 7 46 88 130 152 16.4 21

Feb. 13 13 13 58 118 168 204 225 231 69 Mar. 20 20 56 125 173 215 234 241 212 128 Apr. 27 59 118 169 206 222 225 220 218 178 May 43 99 149 192 212 2:0 211 198 194 203 June 58 110 162 197 213 215 202 186 181 211 Ely 44 97 147 189 208 215 206 193 190 20' Aug. 28 57 114 161 199 214 217 213 211 176 Sep. 21 21 50 115 160 202 222 229 231 123 Oct. 14 14 14 56 III 159 193 215 221 67 Nov. 7 7 7 7 45 86 127 148 160 22 Dec. 4 4 4 4 16 51 76 100 107 9

44' \. Lat 64.N. Lat N

::n. Fe.

31:31:.! t'- Ayr Stay

June July Aug. Sep.

C \'.

\r ..

:x;.

'Shade! SNE \\N

\E \t4

F.\E: N\3t'

I. N

FSF.i 14 61%*

St: SN

SSE.' 5.1:W S HOR

i7 17 64 111.1 1S9 213 2 +2 109

__ __ :' '9 __ _46 249 24' 160 _ _ _ 2:1 2:4 2 3 2_1 :!9 :(K ly N, 136 135 : 21 224 2;0 IF) 171 240 36 66 16' 201 219 211 153 119 132 257 47 VI 169 ICS 215 203 I'1 132 115 261 37 96 159 198 215 206 179 144 128 254 34 66 132 180 214 215 102 177 165 236 29 28 FO 152 199 226 __ 216 211 199

21 :3 _ III 1 "I 21' : -) 239 157

!s IS át !15 :F6 __ 249 109

IS 15 47 115 1 "5 :10 246 89

48' \. l at

- \ \\F \F F\F. F

'*n'. \\\l \ll NNV, N

;an. IS .. . .. 119

Feb. 10 :0 36 103 168

\14r. _ :6 1,- 1'4 :^.t A- t: _.

... . ' . ' '11

... -. ...

2 . 2I5 3uI+ '" ^e :16 19s :14 Aug. _3 , Ca 1"4 _: Sep. 27 . . 144 191

Oct. 21 21 35 a1 I61

cv 15 .. ._ t'. i?cc. '6 31

52' N. Lai

\\F. \F F\F F

.,... \\1.t

Fen

a a 'S

7 1.3

_ !ti

t--

.

`6

1,11

\ 1 at

. ' '

.

aune

.

13

.. 111

.

13' .

1-,9

_., :13

July 31 98 147 :92 211

Aug. 30 56 119 165 203 Sc n 21 23 58 126 171

Co :6 !F, _ '° 112

Na.. 10 Ij 10 21 7: De_. 7 7 4'

ESF NS

SF:-

SN SjF_.

SSA S 110 1"5 2.16 .39 245 85

216 242 249 250 138

:7.4 239 _._ 223 196

__. .. 191 136 2:6 :1 _ :63 150 247

^ ,z) 14: 131 2.52

`9 I,- 159 146 244

_. _ 6 1E9 l0.7 223

2:3 228 123 220 182

20" 131 241 24: 136

1-2 .._ :31 243 85 .. . . :_5 213 65

EcE 4: F_

'.t.N s HOR

... ___ _ 1 62 _._ . __. 2.`) 115 t 236 169 :24 2:4 t 211

!t" 215 : 142

21 2 161 :13 2.14 (9' 191 :08

__9 __ 163

:33 :1.' 114

:1' __. F2

I177 42

c :Ft 7

. . :4 - < +.1

. ' -- '. 4 -"' rl

. ::I 24: ':4 _'. _.

u . A , 37 131 22

213 1r6 174 168 231

214 201 183 177 221

216 215 206 203 193

211 2. 210 211 144

1"6 _13 2:9 214 91

122 165 190 200 40 92 135 159 171 23

N

(Shade) N Er NNW

NFJ NW

FNE/ WNW

El W

ESE/ WSW

SE SW

SSE/ SSN S HOR

Jan. 3 3 3 3 IS 45 6' 89 96 8

Feb. 11 11 11 43 89 111 177 202 :10 3j Mar. IS 18 47 113 159 203 2:6 236 239 111.5

Arr. 25 59 113 163 201 219 225 225 224 ;C4

May 48 97 150 189 211 220 ?15 207 204 123

June 62 114 162 193 213 216 208 196 193 203 July 49 96 148 186 207 215 211 202 200 192 Aug. 27 58 109 157 193 211 217 217 217 159 Sept. 19 19 43 103 149 lag 213 224 227 101

Oct 11 Il it 40 83 _. I57 191 199 46

Nov. 4 4 4 4 IS 44 66 87 93 8

Dec. 0 0 0 0 1 5 II 14 15 1

E 6 Maximum Solar Beat Gain Factor For

Externally Saaded Glas, Btu/ h ft=

(Based on Ground Reflectance of 0.2)

lise for latitudes 0.24 deg. For I7titudcs greater than 24. use north orientation. Table I IA. Fcr horizontal glass in shade, use the la ulctet1 salues for all

N

\\F. NNW

\t: sot

1 N -

'1.\N E `t

irr '

''-t`04. SF./ SSE.

141 L I.A1.1 HUR

Jan. 31 31 31 32 :4 35 37 3- 35 16

+4 11 34 21 31 1' 33 7x 39 !6

'',' n 35 3' 19 19 _ 46 1; .4 I9

A ^r. 40 40 41 _ 1: 41 ? 4.7 :4

May 43 44 45 46 45 41 41 ¿' 4` 18

Jere 45 46 47 47 46 41 41 s) 4.) 31

July 45 45 46 47 47 43 42 41 41 31

Aug 42 42 43 95 46 45 43 42 42 28

Sept. 37 37 38 40 41 42 42 41 41 23

Oct. 34 34 34 36 38 39 40 40 40 19

Nor. 32 32 32 32 34 36 38 33 19 17

Dec. 30 30 30 31 32 34 36 37 37 IS

I g3

¡4 C67_] Cooling Load F2ctors for Glass without Interior Shading. North Latitudes Fend- (ration Facing

koom Con -

struclion Solar Time, I 1 2 3 4 S 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

L 0 17 0.14 0.11 0 09 0.08 0.33 0.42 0.48 0.56 0 63 0.71 0.76 0.60 0.82 0.82 0.79 0.75 0.84 0.61 0.48 0.38 0.31 0.25 0.20 N Sl 0 23 0 20 0.18 0 16 0.14 0.34 0 41 0 46 0.53 0 59 0 6S 0.70 0.73 0.75 0.76 0.74 0.75 0.79 0.61 0.50 0.42 0.36 0.31 0.27

LS"adc±) H 0.25 0.21 0.21 0.20 0.19 0 38 0.45 0.49 0.55 0.60 0.65 0.69 0.72 0.72 0.72 0.70 0.70 0.75 0.57 0,46 0.39 0.34 0.31 0.28

L 0 06 0.03 0.04 0 03 0 03 0.26 0.43 0 47 0 44 0.41 0.40 0.39 0 39 0.38 0.36 0.33 0.30 0.26 0.20 0.16 0.13 0.10 0.08 0.07 NNE M 0 09 0 08 0 07 0 06 0.06 0.24 0 38 0 42 0 39 0.37 0.37 0.36 0.36 0 36 0 34 0.33 0.30 0.27 0.22 0.18 0.16 0.14 0.12 0.10

H 0 11 0.10 0 09 0.09 0.08 0.26 0.39 0 42 0 39 0.36 0.35 0.34 0 34 0.33 0.32 0 31 0.28 0.25 0.21 0.18 0.16 0.14 0.13 0.12

L 0.04 0 04 0.03 0.02 0.02 0.23 0.41 0 51 0 51 0 45 0.39 0.36 0 33 0.31 0.28 0.26 0.23 0.19 0.15 0.12 0.10 0.08 0.06 0.05 NE M 007 006 0.06 005 0.04 021 036 0.44 045 040 0.36 0.33 031 030 0.28 026 0.23 0.21 0.17 0.15 0.13 0.11 009 0.08

H 0.09 0 08 0 08 0 07 0.07 0.23 0 37 0 44 0 4-4 0.39 0.34 0.31 0.29 0.27 0 26 0 24 0.22 0 20 0.17 0.14 0.13 0.12 0.11 0.10

L 0 04 0 03 0 03 0.02 0.02 0.21 0 40 0 52 0 57 0.53 0.45 0.39 0.34 0.31 0 28 0.25 0.22 0 18 0.14 0.12 0.09 0.08 0.06 0.05 ENE M O0' O C6 0J5 005 004 0:0 035 045 049 047 041 0.36 033 0.30 0:8 0.26 0.23 0.20 0.17 0.14 0.12 0.11 009 0.08

fi 009 0 .9 008 0 0 0.07 0:2 036 046 049 045 0.38 0.33 030 017 025 023 0.21 0.19 0.16 0.14 0.13 012 0.11 010

L. 0 04 0 03 0.03 0 02 0 02 0 19 0 37 0.51 0 57 0.57 0.50 0.42 0 37 0.32 0 29 0.25 0.22 0.19 0. i S 0 12 0.10 0.08 0 06 0.05 E M 007 006 006 0.05 0.05 018 0.33 0.44 050 05I 0.46 0.39 035 031 029 026 0.23 021 0.17 015 0.13 0.11 010 0.08

H 0 09 0.09 0 08 0 08 0.07 0.20 0 34 0.45 0 49 0.49 0.43 0.36 0.32 0.29 0.26 0.24 0.22 0.19 0.17 0. I S 0 13 0.12 0. I 1 0 10

L 0.05 004 003 003 0.02 017 034 0.49 058 061 0.57 0.48 041 0.36 032 028 024 020 0.16 0.13 0.10 009 007 0.06 ESE t1 0 CB 0J" 006 005 005 016 031 043 OSI 054 0.51 044 039 035 032 0:9 0'.6 022 019 016 014 012 011 0.09

H 0.10 005 009 008 0.u8 019 032 047 050 052 0.49 0.41 036 032 029 026 024 021 0.18 016 0.14 0.13 0.12 0.11

L. 005 004 004 003 003 013 028 0.43 055 062 063 0.57 048 042 037 033 028 0.24 019 0.15 0.12 010 0.08 007 SE 3.1 0C9 008 007 006 005 014 0:6 038 049 0.54 0.56 0.31 045 040 036 033 029 0.25 021 018 016 0.14 0.12 0.10

H O.II 010 OIO 009 008 017 028 040 049 0.53 0.53 0.48 041 036 033 0.30 0.27 0.24 020 0.18 0.16 0.14 0.13 0.12

L 0.07 0.05 0.04 0.04 0 03 0.06 0 15 0.29 0.43 0.55 0.63 0.64 0 60 0.52 0.45 0 40 0 35 0.29 0.23 0.18 0.15 0.12 0.10 0.08 SSF. m 0 11 0 .9 0 C8 0A7 0 06 0 OS 0 16 0 :6 0.38 0 48 0.55 0.57 0 54 0 48 0 43 0 39 0.35 0.30 0.25 0.21 0.18 0.16 0.14 0.12

11 0.12 0.11 O II 010 009 0.12 019 029 0.40 0.49 0.54 0.55 0.51 044 039 035 031 0.27 0.23 0.20 0.18 0.16 0.15 0.13

L 0.08 0.07 0.05 0.04 0.04 0.06 0.09 0.14 0 22 0 34 0.48 0.59 0 65 0.65 0.59 0.50 0.43 0.36 0.28 0.22 0.18 0.15 0.12 0.10 S M 0 12 0.11 0.;9 0.;8 0 07 O O3 O I I 0. 14 0 21 0 31 0 42 0 52 0 57 0 58 0 53 0 47 0 41 0 36 0.29 0 25 0.21 0.18 0 16 0.14

H 0;3 C12 OC 0.11 010 011 014 017 C24 033 C 43 0S1 056 0 SS 053 043 037 032 026 022 0.20 0.18 0.16 0.15

L 0.10 0 08 0 0' 0 06 0.05 0.06 0 09 0 11 0.15 0.19 0.27 0.39 0.52 0.62 0.67 0.63 0.58 0.46 0 36 0 28 0.23 0 19 0.15 0.12 SSw M 0 1 4 0.12 0. 1 1 0.09 0 08 0 0 9 0 I 1 0 13 0. I S 0.18 0 25 0 35 0 46 0 55 0 59 0 59 0 53 0 44 0.35 0.30 0.25 0.22 0.19 0.16

li 0.15 0.14 0.11 0.12 0.11 0.12 014 016 0.18 0.21 0.27 0.37 046 0.51 057 055 049 0.40 032 0.26 0.23 0.20 0.18 0.16

L 0 1 2 0 . 1 0 3 8 006 005 C.06 0CS 0 1 0 r 1 2 0 1 4 0.16 0.24 036 049 060 066 066 G 5 8 043 0 3 3 027 022 0IE 0.14 SI 015 0.14 012 010 009 003 010 012 013 OIS 0.17 023 033 044 053 058 059 0.53 0.41 033 0 2 0.V 0.21 0.18 It 0 : 5 0 1 4 0 I 3 0 12 0.: 1 0 12 0 13 0 14 0 16 0 :7 0.19 0 25 0 34 0 44 0 52 0 56 0:6 0.49 0.37 0 30 02.5 0 2.; 0 19 0.17

L 0.12 0.10 008 007 005 006 007 009 010 0.12 0.13 0.17 026 040 052 062 066 061 044 034 027 022 0.18 0.15 w Sw NI 0 1 5 0.13 0 1 2 0.10 0 09 0 09 0 1 0 O I I 0 12 0 13 0 14 017 0 24 3.35 0 46 0 54 0 58 0 55 0 42 0 34 0 28 0.24 0 21 0.18

H 0.1! 014 0.13 0.12 0 11 011 0.12 013 0.14 015 0.16 019 026 036 046 053 056 051 0.3E 030 025 021 0.19 0.17

L 0.12 0 10 0 CS 0.06 0 05 0.36 0 07 0 08 0.10 0.11 0.12 0.14 0.20 0.32 0.45 0.57 0.64 0 61 0.44 0.34 0.27 0.22 0.18 0.14 NI 3 :5 0.i3 011 9:0 c09 N 0,:9 0,0 0.11 0.12 0.13 0.14 0 i9 0.29 0.40 0.50 0.56 0.55 031 0.33 0.27 0.23 020 0.17 H 014 013 012 0.11 010 011 0 C 013 0.14 0.14 0.15 016 021 0.30 040 049 0.54 0.52 0.38 0.30 0.24 021 0.18 0.16

L 0i2 0.10 008 006 005 006 00'. 003 0.10 0.12 0.13 0.15 0.17 0.26 040 0.53 063 062 044 0.34 0.27 022 0.18 0.14 'A \'.l .1 0.15 0'3 0 1 010 0 0 005 0:0 0 1 012 0.13 0.14 0 1 017 0:4 035 047 055 0.55 041 033 0.27 0.:3 0:0 017

1i 0;4 013 0:: 0:1 0:0 C!1 0;2 013 014 0.15 0.16 0.17 0.18 025 036 046 0.53 052 0.38 030 0.24 0.20 0.18 0.16

L 0.11 0.09 0.08 0 06 0 05 0 06 0 08 0.10 0.12 0.14 0.16 0.17 0.19 0.23 0.33 0.47 0.59 0.60 0.42 0.33 0.26 0.21 0.17 0.14 5.1 0.14 C 12 011 1 0,79 0 38 0 TO 0.10 0 I 1 0 :3 0.14 0 16 0.17 0.18 0 21 0 30 0 42 0.51 0.54 0.39 0.32 0.26 0.22 0.19 0.16 H 0.14 0 12 0.11 0.10 0.IJ 0 10 0.i2 0 13 0.13 0.16 0.18 0.11 0 19 0.22 0 30 0.41 0.50 0.51 0.36 0.29 0.23 010 0 17 0.15

L 0.12 0 09 C 08 0 06 0 05 0 07 0 11 0 14 0 18 0 22 0.25 0.27 0.29 0.30 0.33 0.44 0.57 0.62 0.44 0.33 226 0.21 0.17 0.14 'I 0 If ^:3 .3 :0 9J9 C;0 012 3;5 018 C:I 023 0:6 ú27 0.29 031 039 0.51 056 041 033 0.27 C23 0.:0 0'7 ;3 014 0:3 0:: O.1 7;O 01: 0;: 3 17 0:0 0:3 0.25 026 0:8 0:! 031 038 945 053 038 0.30 C25 021 Old 016

L C I I 0 39 0.07 0.06 0 OS 0 07 0 14 0 24 0 36 0 48 0.58 0 66 0.72 0.74 0 73 0.67 0.59 0.47 0.37 0.29 0.24 0.19 0.16 0.13 ' 3 . :-+ . . ,. ., :1 i:4 C 3 3 0 43 0.57 0 59 0 64 0 67 0 M O t: 0 56 0.47 0.38 0 32 018 0 24 C 21 0.18

.. . . .., . , .... .. .._ ^45 052 059 0 t 064 012 0 5 OSI 042 035 029 0.25 0:3 0:1 0.19

. . a.' 2.r,. :-..;: ete ^.;-.r ::_^. S-; r; a:e.uri/ 3010 er r.utenal/f11 of f ax area .... .... r. . .:a'e _r.ei S.in ..:_:c:<C..rs a:c1.703bofbudJ-n¡rra:enal It1 (loot area .:: c.' .. _.';:e ca.^:cr +a... rin t.4: ..; ;. a.matn> 13010 of bu1:0e4 r:a:e:.aia. ti' JI Iloor atea.

i s4

t, C63 1. ,

Cooling Load Factors for Glass with Interior Shading, North Latitudes (All Room Constructions)

Fctcs- (ration Facing

0100 0'00 0300 0400

N 0.080.0' 0.06 0.06

NNE 0.03 0.03 0.02 0.02

NE 0.03 0.02 0.02 0.02

ENE 0.03 0.02 0.02 0.02

E 0.03 0.0' 0.02 0.02

ESE 0.03 0.03 0.02 0.02

SE 0.03 0.03 0.02 0.02

SSE 0.04 0.03 0.03 0.03

S 0.04 0.04 0.03 0.03

SSW 0.05 0.C4 0.04 0.03

SW 0.U5 0.05 0.04 0.04

WS«' 0.05 0.05 0.04 0.04

O:5 0.05 0.04 0.04

N«' C,..5 C.35 0.04 0.03

MV 0.05 0.04 0.04 0.03

NNW 0. aQ 0.05 0.04 0.03

NOR. O.t, ,;. 5 0.04 0.04

Solar Time, h 0500 0600 0700 0800 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400

0.07 0.73 0.66 0.65 0.73 0.80 0.86 0.89 0.89 0.86 0.82 0.75 0.78 0.91 0.24 0.18 0.15 0.13 0.11 0.10

0.03 0.64 0.77 0.62 0.42 0.37 0.37 0.37 0.36 0.35 0.32 0.28 0.23 0.17 0.08 0.07 0.06 0.05 0.04 0.04

0.02 0.56 0.76 0.74 0.58 0.37 0.29 0.27 0.26 0.24 0.22 0.20 0.16 0.12 0.06 0.05 0.04 0.04 0.03 0.03

0.02 0.52 0.76 0.80 0.71 0.52 0.3; 0.26 0.24 0.22 0.20 0.18 0.15 0.11 0.06 0.05 0.04 0.04 0.03 0.03

0.02 0.47 0.72 0.80 0.76 0.62 0.41 0.27 0.24 0.22 0.20 0.17 0.14 0.11 0.06 0.05 0.05 0.04 0.03 0.03

0.02 0.41 0.67 0.79 0.80 0.72 0.54 0.34 0.27 0.24 0.21 0.19 0.15 0.12 0.07 0.06 0.05 0.04 0.04 0.03

0.02 0.30 0.57 0.74 0.81 0.79 0.68 0.49 0.33 0.28 0.25 0.22 0.18 0.13 0.08 0.07 0.06 0.05 0.04 0.04

0.02 0.12 0.31 0.54 0.72 0.81 0.81 0.71 0.54 0.38 0.32 0.27 0.22 0.16 0.09 0.08 0.07 0.06 0.05 0.04

(3.03 0.09 0.16 0.23 0.3S 0.58 0.75 0.83 0.80 0.68 0.50 0.35 0.27 0.19 0.11 0.09 0.08 0.07 0.06 0.05

0.03 0.09 0.14 0.18 0.22 0.27 0.43 0.63 0.78 0.84 0.80 0.66 0.46 0.25 0.13 0.11 0.09 0.08 0.07 0.06

0.03 0.07 0.11 0.14 0.16 0.19 0.22 0.38 0.59 0.75 0.83 0.81 0.69 0.45 0.16 0.12 0.10 0.09 0.07 0.06

0.03 0.07 0.10 0.12 0.14 0.16 0.17 0.23 0.44 0.64 0.78 0.84 0.78 0.55 0.16 0.12 0.10 0.09 0.07 0.06

0.03 0.06 0.09 0.11 0.13 0.15 0.16 0.17 0.31 0.53 0.72 0.82 0.81 0.61 0.16 0.12 0.10 0.08 0.07 0.06

0.03 0.G' 0.10 0.12 0.14 (+.16 0.17 0.18 1..22 0.43 C A5 0.80 0.84 0.66 0.16 0.12 0.10 0.08 0.07 0.06

0.03 0.07 0.11 0.14 0.17 0.19 0.20 0.21 0.22 0.30 0.52 0.73 0.82 0.69 0.16 0.12 0.10 0.08 0.07 0.06

0.03 0.11 0.17 0.22 0.26 0.30 0.32 0.33 0.34 0.34 0.39 0.61 0.82 0.76 0.17 0.12 0.10 0.08 0.07 0.06

0.03 0.12 0.27 0.44 0.59 0.72 0.81 0.85 0.85 0.81 0.71 0.58 0.42 0.25 0.14 0.12 0.10 0.08 0.07 0.06

lQ

for a special case of external shade.

The cooling load factors depend on the actual solar heat gain

for a particular time, the internal shade, and the building

construction when there is no internal shade. The interior shading

has two effects. First, it refects some of the solar radiation so

that it never enters the space. Second, the shading device

prevents the solar radiation from being absorbed by the floor,

interior walls, and furnishings. Instead the shading device absorbs

the radiation, which is subsequently convected to the room air.

Table 6 -14 and 6 -15 give CLF for glass without and with interior

shadings.

Example(6 -4):

The building of Example 6 -1 has a 4x5 ft single glass window

in the south wall. The window has light -colored venetian

blinds. Compute the cooling load due to the window at 10:00

A.M. and 5:00 P.M. solar time for August using standard

design conditions.

Solution:

There are two components of cooling load for the window.

One is due to heat conduction, the other to solar radiation.

Eq. 6 -20 gives the cooling load due to conduction where the

overall coefficient is 0.81 BTU /(hr- ft2 -F). The CLTD values

are given in Table 6 -7 as 4 F and 13 F for 10:00 A.M., and

5:00 P.M.; respectively. Because the daily average

temperature is about 83 F, the table values should be isdned

by 2 F. The cooling load due to conduction at 10:00

AM is:

I8&

qc = 20 (0.81) (4 -2) = 32 BTU /hr

and for 5:00 PM

qc = 20 (0.81) (13 -2) = 178 BTU /hr

Considering the solar radiation and using Eq. 6 -15 with a

shading coefficient of 0.55 from Table 6 -9; a maximum solar

heat gain factor of 111 BTU /(hr -ft2) is obtained from Table

6 -12 for 32 degree north latitude. The cooling load factors

from Table 6 -15 are 0.58 and 0.27. The cooling load for

10:00 A.M. is:

qc = (20) (0.55) (111) (0.58) = 708 BTU /hr

and for 5:00 P.M. is:

qc = (20) (0.55) (111) (0.27) = 330 BTU /hr

The total cooling load for the window at 10:00 A.M. is then

qc = 32 + 78 = 740 BTU /hr

and for 5:00 P.M. is:

9c = 178 + 330 = 508 BTU /hr

The shaded portion of a window is easily estimated using the

methods and data presented in Chapter 2. Seperate calculations

are then made for the sunlit and shaded portions of the glass to

obtain the cooling load; however, the cooling load due to

conduction is the same for both parts. Common situations are

overhangs, side projections, or setbacks.

At latitudes greater than 24 degree, the shaded portion of

the glass is treated as a north -facing window with the SHGF and

CLF read from Tables 6 -12, 6 -14, or 6 -15 for the north orientation.

The shading coefficient is the same for the sunlit and shaded

parts.

ls7

For the special case of shaded horizontal glass and shaded

glass for latitudes less than 24 degre, the SHGF is given in Table

6 -13, CLF values are read from Tables 6 -14 and 6 -15 for the north

orientation.

(3) Cooling Load -Internal Sources

Internal sources of heat energy may contribute significantly

to the total cooling load of a structure, and poor judgement in

the estimation of their magmitude can lead to unsatisfactory

operation and /or high costs when part of the capacity is unneeded.

These internal sources fall into the general categories of people,

lights, miscellaneous equipment, infiltration and ventilation.

Infiltration and ventilation were discussed in Heating Loads using

the same calculation procedure.

Fans and duct heat gain will be discussed in Chapter 7 and Chapter

8.

(A) People

In ASHRAE Handbook 1985 Fundamentals Chapter 8 contains

detailed information containing the rates at which heat and

moisture are given up by occupants engaged in different levels of

activity and Table 6 -16 summarizes the data needed for heat gain

calculations. Although the data of Table 6 -16 are quite accurate,

large errors are often made in the computation of heat gain from

occupants because of poor estimates of the periods of occupancy or

the number of occupants. Great care should be taken to be

realistic about the allowance for the number of people in a

structure. It should be kept in mind that rarely will a complete

Rases of Heat Gain from Occupants o; Conditioned Spaces'

Degree of Actisit Typical Application

Total Heal Adults, Male

Total Hest Adjustedb Sensible Heat

Btu'h

Latent Heat

Bluth Btu /h Btu;h

Seated at rest Theater, ntosle 400 350 210 140

Seated, very light work writing Offices, hotels, apis 480 420 230 190

Seated, eating Restauranic 520 580 e 255 325

Seated, light work, typing Offices, hotels, apes 640 510 255 255

Standing, licl: ..a: i, or wail;:tg ,... Retail Store, bank 800 640 315 325

Light bench work Factory 880 780 345 435

Walling, 3 mph, light ma :::ir.e work Factory 1040 1040 345 695

Bowling Bowling alley 12C0 960 345 615

Moderate dancing Dance hall 1360 1280 405 875

He3sy w, rk, h-.1sy nta:hlr:e wurl., r.tbng Factory 1600 1600 565 1035

Hcasc worl.. arl.ic:i:, Gymnasium 2000 1800 635 1165

Ta b1 e - l 7 CO Sensible Heat Cooling Load Factors for People

Hwrrler Ia 6I nir) InioNpa,e

5 o 7 8 9 IO II 12 13 14 15 It, l' I4 19 20 II _. 2 24

. u49 _ u 1' J t3 0I0 0.:8 IIJ7 0 0 005 0M u04 003 003 002 002 002 UO_ 0,)I OOI 001 ODI act out vol a d 44 0 59 J u " 1 0_" 0.21 0 1 6 0 1 4 0 1 1 0 10 0.08 0 07 0 06 006 0 05 0 04 0 04 0 03 0 03 0.0) 0 02 O 92 0 0 O ut n 05u .. (. n'2 0'6 0'9 034 0:5 :21 018 0.15 013 0 I 010 0 6 00' 0 0 OJn D05 D04 00-1 003 0-3 003 i 01 ,. ^" L1 22 0'6 Ce.) 032 014 038 0.3J 02S 021 018 015 013 012 010 0,N 0L,8 OU' 006 0 0 ^.? 0 0

11 ' - s i 6 S3 0 t. 0 57 ús9 0 42 0 34 0 :8 0 23 0:0 0 17 0 15 _ . 1 ;1 0 10 0 cN t- ._ 0:" 7 rh _ '1' .. ..,.'9 0 .,t 0 l4 Oro 0 "!:8 0..9 09. Uv'_ 0 45 0 36 r-3 J 025 o:t ,.. , t 014 01: 0;, U9 .. _

. ., .c -_ 077 0 J53 oe5 05' 059 c)v 091 0 9 093 014 047 O38 O JI _ ,:3 0'0 01' 0;5 Oi3 0.11 .-J r-` "9 Oa2 ot< r5; 0 041 092 093 094 095 045 096 049 ,?d o23 0:r 024 O:d 0.15 0.16

. - . .4 0 . , i ,.. , . _ . 4,1 0 P 3 ? ' 0 21

office staff be present or a classroom be full. On the other hand,

a theater may often be completely occupied and sometimes may

contain more occupants than it is designed for.

Each design problem must be judged on its own merits. With the

exception of theaters and other high- occupancy space, most spaces

are designed with too large an allowance for their occupants. One

should not allow for more than the equivalent full -time occupants.

The heat gain from people have two components: sensible and

latent. Latent heat gain goes directly into the air in the space;

therefore, this component immediately becomes a cooling load with

no delay. However, the sensible component from a person is delayed

due to storage of a part of this energy in the room and

furnishings. The cooling load factor is used to express this

delayed effect. The CLF depends on the total hours the occupants

are in the space and varies from the time of entry.

Table 6 -17 gives CLF values for people. The data are for

continuously operating cooling equipment. When the cooling

equipment is turned on and off when the occupants arrive and leave,

the CLF is 1.0. This is because any energy stored in the building

at closing time is still essentially there the next morning. When

the cooling equipment is operated for several hours after

occupancy, however, the CLF values are about the same as for

continuous equipment operation. When the density of people is

large such as in a theater, a CLF of 1.0 should be used.

Example (6 -5) :

An office suite is designed with 10 private offices, a

secretarial area with space for four secretaries, a

reception area and waiting room and an excutive office with

a connecting office for a secretary. Estimate the cooling

load from the occupants at 3:00 P.M. solar time.

Solution:

In the absence of additional data the following approach

seems reasonable. For the private offices assume that an

average of 7 out of the 10 will be occupied between 8:00

A.M. to 5:00 P.M. Assume that 3 of the 4 secretaries are

always present and a receptionist is always there. The

waiting room will have a transient occupancy, assume 2

people. The executive office will probably experience

variable occupance. However, an average of one continuous

occupant is about right.

Assume the secretary is always present. The total number

of people for which the heat gain is to be based is then 15.

We will assume sedentary, very light work and use data from

Tables 6 -16 and 6 -17. Assuming an adjusted group of males

and females and very light work,the sensible and latent heat

gains per person are 230 and 190 BTU /hr, respectively.

The latent cooling load due to people is:

ql = 15 (190) = 2850 BTU /hr

There is a question about where the occupants go for lunch.

Assume they stay in the space, then the total hours in the

space are 9 and 3:00 P.M. represents the seventh hour after

entry. From Table 6 -17, CLF is between 0.82 and 0.83. The

cooling load due to sensible heat is:

qs = 15 (230) 0.825 = 2860 BTU /hr

ia l

and the total cooling load at 3:00 P.M. is:

q = ql + qs = 2850 + 2850 = 5700 BTU /hr

Note that the occupants could enter and leave the space on

different schedules. In such a case, make a seperate

caculation for each group. Cj]

(B) Lights

The cooling load due to lighting is often the major component

of the space load and an accurate estimate is essential. A number

of factors need to be considered because the heat gain to the air

may differ significantly from the power supplied to the lights.

Some of the energy emitted by the lights is in the form of

radiation that is absorbed in the sapce. The absorbed energy is

later transferred to the air by convection. The manner in which

the lights are installed, the type of air distribution system, and

the mass of the structure are important. Obviously a recessed light

fixture will tend to transfer heat to the surrounding structure,

whereas a hanging fixture will transfer more heat directly to the

air. some light fixtures are designed so that air returns through

them, absorbing heat that would otherwise go into the space.

Lights are often turned off and on to save energy, which makes

accurate computations difficult. Lights left on 24 hours a day

approach an equilibrium condition where the cooling load equals the

power input.

The designer should becareful to use the correct wattage in

making calculations. It is not good practice to assume nominal

wattage per unit area. Additionally, all of the installed lights

may not be used all the time. The instantaneous heat gain for

l2

Cooling Load Factors When Lights Are on for 8 Hours "a" Coef-

nclenu "b" Class- IflcaUon 0 1 2 3 4 5 6 7 8

Nimber of boues ante ItBhts are lorned on 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

A 0.02 0.46 0.57 0 65 0.72 0.77 0.82 0.85 0 88 0.46 0.37 0.30 0.24 0.19 0.13 0.12 0.10 0.08 0.06 0.05 0.04 0.03 0.03 0.02 B 007 0.51 0.S6 061 065 068 071 0.74 0.77 0.34 0.31 028 0.25 0.22 020 0.18 0.16 0.15 0.13 0.12 0.11 0.10 0.09 0.08

0.45 C O. I I 0.55 0.58 0 60 0.63 0 65 0.67 0.69 0.71 0.28 0.26 0 25 0-23 0.22 0 20 0.19 0.18 0.17 0.16 0.15 0.14 0.13 0.12 0.12 D 0 14 0.58 0.60 0 61 0.62 0.63 0.64 0.65 0.66 0.22 0.22 0.21 0.20 0.20 0.19 0.19 0.18 0.18 0.17 0.16 0.16 0.16 0.15 0.15

A 0 01 0.56 0 65 0 72 0.77 0 82 0.85 0.88 0.90 0.37 0 30 0.24 0.19 0.16 0.13 0.10 0.08 0.07 0.05 0.04 0.03 0.03 0.02 0.02 B 0 06 0 60 0 64 0 68 0.71 0 74 0.76 0.79 0.81 0.28 0.25 0.23 0.20 0.18 0 16 0.15 0.13 0.12 0.11 0.10 0.09 0.08 0.07 0.06

0.55 C 0 09 0.63 0.66 0.68 0.70 *0 71 0.73 0.75 0.76 0.23 0.21 0.20 0.19 0.18 0.17 0.16 0.15 0.14 0.13 0.12 0.11 0.1 I 0.10 0.10 D 0. I 1 0.66 0.67 0.68 0 69 0.70 0.71 0.72 0.72 0.18 0.18 0.17 0.17 0.16 0.16 0.15 0.15 0.14 0.14 0.13 0.13 0.13 0.12 0.12

A 0 71 0 66 0.73 0 78 0.82 0.56 0.88 0 91 C 93 0 29 0.23 0.19 0.15 0.12 0.10 0.08 0.06 0.05 0.04 0.03 0.03 0.02 0.02 0 01

B 0 Ca 0 69 0.72 0.75 0 77 0.80 0 82 0 84 0 85 0.22 0 19 0.18 0.16 0.14 0 13 0.12 0.10 0.09 0.08 0.08 0.07 0.06 0 06 0.05 0.65 C 0 0' 0 72 0.73 0 75 0.76 0 78 0.79 0.80 0 82 0.18 0 17 0.16 0.15 0.14 0.13 0.12 0.11 0.11 0.10 0.10 0.09 0.08 0.08 0.07

D 0 09 0.73 0.74 0 75 0.76 0.77 0.77 0.78 0.79 0.14 0.14 Cl) . 0.13 0.13 0.12 0.12 0.11 0.11 011 0.10 0.10 0.10 0.10 0.09

A 0 01 0.76 0.80 0.84 0 87 0 90 0.92 0.93 0 95 0.21 0.17 0.13 0.11 0.09 0.07 0.06 0.05 0.04 0.03 0.02 0.02 0.02 0.01 0.01

B 0.03 0.78 0 80 0.92 0 84 0 85 0.87 0 88 0 89 0.15 0.14 0.13 0.11 0.10 0 09 0.08 0.07 0 07 0.06 0.05 0.05 0.04 0 04 0.04

0.75 C 0.05 0.80 0 81 0 82 0.83 0 84 0 85 0.86 0.87 0.13 0.12 0.1 I 0.10 0.10 0 09 0.09 0.08 0.08 0.07 0.07 0.06 0.06 0.06 0.05

D 0.06 0.81 0.82 0.82 0.83 0.83 0.84 0.84 0.85 0.10 0.10 0.10 0.09 0.09 0.09 0.08 0.08 0.08 0.08 0.07 0.07 0.07 0.07 0.07

Cooling Load Factors When Lights Are on for 10 Hours "a"Coef- "b" Clase- Nombre of boon alter IlBhts are rtarned on ficlenu Iticatton 0 1 2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 II 19 20 21 22 23

A e ,."" 0 4' 0 58 0 66 0.73 0'9 0 82 0.86 0.88 0.91 0.93 0.49 0.39 0.32 0.26 0.21 0.17 0.13 0.11 0.09 0.07 0.06 0.05 0.04 R . .. O54 059 C`63 0(:i O'0 073 0'6 078 080 0.82 039 035 0.32 0.29 026 0.23 0.21 0.19 0.17 0.15 0.14 0.12 0.11

0.45 C ':5 029 0 6 4a Dvù 0.9 070 072 073 075 076 033 031 029 027 0:6 0.24 0.23 0.21 0.20 0.19 0.18 017 016 D 0:: 062 063 064 066 06' 065 069 069 070 071 0.27 '0.26 0.26 025 0.24 0.23 0.27 0.22 0.21 0.21 0.20 019 0.19

A 0^: 057 065 0'2 078 092 085 088 091 092 0.94 0.40 032 026 021 0.17 0.14 0.11 0.09 0.07 006 0.05 004 003 B 0 Cs 0 62 0.66 0.69 0 73 0 75 0.78 0.80 0.82 0.84 0.85 0.32 0.29 0.26 0.23 0.21 0.19 0.17 0.15 0.14 0.12 0.11 0.10 0.09

0.55 C 0 1: 0 66 0.69 0.70 0.72 0 73 0.75 0.77 0.78 019 0.81 0.27 0 25 0.24 0 22 0.21 0.20 0.19 0.17 0.16 0.15 0 14 0.14 0.13

D C 15 0 69 0.70 0' I 0 72 0 73 0 73 014 0.75 0.76 0.76 0.22 0 22 0 21 0.:0 0.20 0.19 0.18 0.18 0.17 0.17 0.16 0.16 0.15

A C:: 0.66 O'3 09 053 055 059 0.91 093 094 095 031 0.25 0.20 0.16 0.13 0.11 008 007 0.05 0.04 004 0.03 0.02

B 0 0 0'1 074 0.76 0"9 091 083 081 0.66 087 089 025 0.22 020 0.18 0.16 0.15 0.13 0.12 0.11 010 009 006 007 0.65 C 0C^7 074 075 0'7 078 080 081 082 0.83 094 085 0.21 0.20 0.18 0.17 0.16 0.15 0.14 0.14 0.13 01: on 0.11 0.:0

D 0.11 0 76 0 77 0 77 0.78 0.79 0.79 0.80 0.81 0.81 0.82 0.17 0.17 0.16 0.16 0.15 0.15 0.14 0.14 0.14 0.13 0 13 0.12 0.12

A ON 0'6 0?1 044 055 040 092 093 095 0% 0.97 022 0.18 014 0.12 0.09 0.08 006 005 004 0.03 0.03 002 0.02

H 0.7-4 0-9 0F1 O93 095 OH 088 OS9 090 09! 092 0.18 0.16 014 0.13 0.12 0.10 009 0.08 0.08 0.07 0.06 006 0.05

0.75 C 00' 081 OF: 0F3 054 095 086 097 099 099 089 015 014 0.13 012 0.12 0.11 0.10 0.10 0.09 009 008 0.08 0.07 D 005 0.93 0.93 084 0.81 0.85 085 0.86 086 0.87 0.87 0.12 0.12 0.12 0.11 0.11 0.11 0.10 0.10 0.10 0.09 0.09 0.09 0.09

Cooling Load Factors When Lights Are on for 12 Hours ,';.uf -1,-C1,11. N.oher of oan aller BBhu arr tvretd on

f;cirnu 1 1: - a t l o n 0 1 2 3 4 5 6 7 8 9 10 1 1 1 2 13 14 IS 16 17 18 I9 20 2I 22 23

A 025 049 009 067 073 0"8 O83 096 089 091 093 094 095 0.51 0.41 013 0.27 0.22 0.17 014 0.11 009 0.07 0.06 R 013 057 061 065 00 072 075 077 079 082 0S) 0.85 087 043 039 0.35 C.31 028 025 an 021 0.18 0.17 00

0.45 C 0!9 063 065 067 0 R9 071 073 074 076 077 079 080 O81 037 an 033 031 0:9 0.27 026 0:4 023 011 0.20

D ,22 C61. OC 059 0; n.0 :?71 0-2 073 C-4 074 07t 0.76 04: 031 030 0:9 0:9 0.27 0.26 o:6 025 0_L 023

\ "a C59 0^^ 071 0-5 O?: 055 OF9 091 093 094 095 0% 042 0.34 or 0:: 019 014 0.11 09 007 006 0.05

B ..I 065 ,T,rE 072 074 077 079 051 063 085 086 088 089 035 on 025 026 0.:3 021 0.19 017 015 014 0.12

0.55 C 0.5 069 071 an 0'2 an on on 0.80 ou 0.83 ou 0.85 on on or an 0:4 an an on 0.19 0.17 0.16

D 0 i 5 0.72 073 074 0-5 076 0 76 or an 0.78 an 0.80 0.80 an on ON 0.24 0.23 0.22 0.22 0.21 0.20 0.20 0.19

A 0J1 O6' 0'4 0-9 043 0F6 099 091 013 094 095 09 097 033 0.26 on 0.17 0.14 0.11 0.09 0.07 006 0.05 0.01 B 0' 0-1 0-5 079 OF) 091 094 095 087 058 099 090 092 027 on 022 on 018 0.16 013 0.13 an 0.11 0.10

062 C .. 0", on 7'9 0 91 0.' 1 on O Q4 0 95 0 86 0 96 0 87 0.88 024 0.22 021 010 019 0.17 0.16 00 0.14 014 0.13 D _ '9 r1-4 ' 9.3 0,1 ,^51 0 F2 082 C 53 013 014 084 0.85 0.:0 0.20 019 0.18 0.18 0.17 0.17 0.16 016 0.15 0.15.

A , . .7-7 O S I 0 9 5 0,4 0 90 092 094 095 0% 0.91 or 0 98 on an 015 0.12 0.10 0.08 0.06 005 ON 0.03 OU R 0. ^91 O': 054 096 0S' 098 090 091 092 092 093 094 0.19 018 0.16 014 013 0.12 0.10 0.09 0.08 008 0.07

0-5 C 0F1 0 1 CF5 C', 0 09 09S 089 090 090 091 0.91 01' 016 015 014 013 012 0.12 0 I 0.10 0.10 0.09 .) n F< o ,. , c6 C <6 0 c6 0 R7 0 57 0.88 0.88 0.88 0.89 0.89 0.14 0.14 0.14 0.13 013 0.12 0.12 0 12 0.11 0.11 0.11

Cooling Load Factors When Lights Are on for 14 Hours ..a..Coef. Namb rr of boon alter Itghts are turned ou C. :n'A ifinucn O 1 2 3 4 5 6 7 8 9 IO II 12 13 14 15 16 il 18 19 20 21 22 23

011 01 7.55 074 ^? 093 097 00 091 093 094 0.99 0.95 or an on 0.34 or 0:: an on on 009 ? ` , , n^' .ìRQ .. . 0'4 C"7 079 v91 08) 085 LQ6 088 059 OW 045 041 or 0.34 030 0:7 024 0.22 0C

4 .. . G 071 r:11 '' '1 07;, 077 079 052 on L R2 0.83 0.84 0 85 0.41 030 u.36 0.34 on an 0.:9 0.:7 0:3

.. '7!1 '': O.": 0"7 ^ J'S 06 or an 0'a 0-9 0.80 050 0.40 0.36 075 0.34 an 0.32 0.31 an 0.:9 ON ,74 . , .;a ,:? n 0. o;1 7 04 095 091, 097 O 9R 00 on 0:8 0.22 0.18 015 0.12 0 09 01

. ! . ,t ':5 1Y, a 'S e;9 040 071 0 /1 018 014 031 0:. 0 25 0 22 0.:0 011 n: o a + 1-9 .,. 1 0 3 064 095 096 056 09" 059 o34 on o30 an o:6 o:5 o:3 0:2 c:l

, - .. ..-.-9 ,1c7 092 092 0173 051 0R4 029 on on or on on 0:4 0:4 0.:3

- `. -.'.' _-"1 7;5 0'u 0's6 607 n-Q 0c9 034 0:- 02'_ 0l' 014 0i1 00J 00' 0 9 1 ..< .., C.c¡ O?) 091 092 O41 n<4 0.:7 C:5 0:1 0:1 019 017 Oft 014 20

., . . .

o . _ ''0 .:4 CS . C6 047 0ES 089 n99 '>.1 09! 0 :6 0 :: 0.27 _ 0:? tti9 0 IS C'17 016 ll 017 051 C/2 an 043 on UF4 O64 Qè5 0.85 0.86 0.86 0.87 OF' U97 023 on on on 0:0 on 019 0.1E 0.3

A 0 03 0' 9 0 92 0 86 0 SS 0 91 0 92 0 94 0 95 0.96 0 97 0 97 0.9E 0 98 0.99 0.24 0.19 0.16 0.12 0 10 0 08 0.07 0.05 0.04

B 0N on O84 096 097 095 0'0 091 0.92 0.92 013 094 0.94 095 0.96 011 0.19 0.17 0.15 0.14 012 0.11 0.10 0.09

075 oz; 0L. Oa' O:c Ci; 059 790 096 091 091 092 092 097 0.93 0.19 018 0.17 OM 015 n14 0.0 0.12 011

D C : : Ci' 6,7 O " , . 0 S5 0"r9 0.69 0 49 0.90 OW 0.90 0.90 091 091 0.16 Oft 0.15 0.15 0.14 0.14 014 00 00

lq3

6-161 C6 Design Values of "a" Coefficient

Features of Room Furnishings, Light Fixtures, and Ventilation Arrangements

a Furnishings Air Supply and Return

Type of Light Fixture

0.45 Heavyweight, simple furnish- ings, no carpet

0.55 Ordinary furni- ture, no carpet

0.65 Ordinary furni- ture, with or without carpet

0.75 or Any type of greater furniture

Low rate; supply and return below ceiling (V40.5)1 Medium to high ven- tilation rate; supply and return below ceiling or through ceiling grill and space (V0.5)a Medium to high ven- tilation rate or fan coil or induction type air -conditioning terminal unit; supply through ceiling or wall diffuser; return around light fixtures and through ceiling space. (V O.5)a Ducted returns through light fixtures

Recessed, not vented

Recessed, not vented

Vented

Vented or free - hanging in air stream with ducted returns

al, a room air supply rate rn cfm,(t z of floor area.

The "b" Classification Values Calculated for Different Ern elope Constructions

and Room Air Circulation Rates

Room F :n. elope Construction,

Room Air Circulation and Tspe of Supply and Return°

(mass of floor area, Its ftz) Los Medium High Ven High

2 -in. Wood Floor (l0) B A A A 3 -in. Co .ocie floor 140) B B B A 6-in. Con :rete. i loor (75) C C C B

8in. Concrete Floc r (I 20) D D C C I2 -in. Concrete Floor (160) D D D D

aFlorr cosered +nh arct anJ r.hher pad; for a floor co,ered only ruh floor rile rake ne.t ;!a>. :fi :atzen co the rbhl .n the ,ame :or.

bLc +. f or seniiiacon rate -minimum required to cope ruh cooling load from light, :-te::o: zone. Surrt) :h.rough !loot, shall or ceiling di:fuser. Crang spare

riet ,er_.cd an.: h - O 4 Bra n (:'F w here . inside surface :on,ecuon coci(i :cent used to

of :Lu:cr. ci 13,111-...,10.,1 \Ica_ -. 1ca,_r ,rr. :.ia :..n rare. ,rar ̂ h through floor. ritt or cnlmg diffuser Ceiling

if..: . . . ... _.._ r umt rr S tan h- uàb:.. a :'f-

fie. H : )1 ro::n i. >ed n-on :mize teapezature gradients to a room. Rc_n ,, -.: ana h I _ B::. h :: r

l q4

lights may be expressed as

qi = 3.412 WFuFs (6-22)

where

W = summation of all installed light wattage, watts

Fu = use factor - ratio of wattage in use to that installed

Fs = special allowance factor for lights requiring more

power than their rated wattage. For typical 40 W

fluorescent lamps, Fs = 1.20, for tungsten, Fs = 1.0

The cooling load is then given by

4 = qi (CLF) (6 -23)

The cooling load factor is a function of the building mass,

air circulation rate, type of fixture, and time. Table 6 -18 gives

cooling load factors as a function of time for lights that are on

for 8, 10, 12, and 14 hours. The "a" and "b" classifications are

given in Tables 6 -19 and 6 -20. The "a" classification depends on

the nature of the light fixture, the return -air system, and the

types of furnishings, whereas the "b" classification depends on the

construction of the building and the type of supply and return -air

system. The CLF values of Table 6 -18 are for the use of

continuously operating cooling equipment. If the cooling equipment

is turned on and off on the same schedule as the lights, the CLF

is equal to 1.0 because this condition is similar to the case when

lights and cooling equipment are on continuously. When the cooling

equipment is operated for a few hours after the lights are used,

however, the system behaves as if the cooling equipment were

operating continuously.

In the case of light fixtures that are cooled by the return

air (vented fixtures or recessed fixtures in a ceiling return

plenum) , the cooling load given by Eq. 6 -23 using the specified

"a" coefficient is not all imposed on the space. The return air

carries part of the load directly back to the coil. The amount of

load absorbed by the return air is a function of many variables but

can be up to 50% of the light power.

Example (6 -6):

The office site of Example(6 -5) has total installed light

wattage of 8,400W. The fluorescent light fixtures are

recessed with 40 -W lamps. Supply air is through the ceiling

with the air returning through the ceiling plenum. The

lights are turned on at 8:00 A.M. and turned off at 6:00

P.M. Estimate the cooling load at 10:00 A.M. and 4:00 P.M.

The floor is 6 -in concrete.

Solution:

Assuming that about 15% of the lights are off due to

unoccupied offices, the use factor Fu is 0.85. The

special allowance factor for lights (Fs) is 1.2. Then from

Eq. 6 -22

qi = 3.412 (8400) (0.85) (1.2) = 29,234 BTU /hr

The "a" and "b" classifications are 0.55 and c,

respectively, then from Table 6 -18, CLF10 is 0.68 and CLF16

is 0.78. Then

410 = 29,234 (0.68) = 19,880 BTU /hr

416 = 29,234 (0.78) = 22,800 BTU /hr

Since the return air is flowing through the ceiling space

over the light fixtures, some of the cooling load due to

the lights never reaches the space.

Let us assume a 20% reduction in the load to the space, Then

q10 (1 -0.2) 19,880 = 15,900 BTU /hr

q16 (1 -0.2) 22,800 = 18,240 BTU /hr

One must remember, however, that the 20% reduction in space

load must be added to the coil load.

(C) Miscellaneous Equipment

There is an infinite variety of equipment and appliances that

will provide a heat gain when installed in the conditioned space.

If the equipment is completely enclosed, the heat gain equals the

input to the equipment. However, this does not always happen. An

electric motor maybe within the space, but the device powered by

the motor may be outside it. In this case the heat gain results

from the inefficiency of the motor and will be only a small

percentage of the rated power of the motor. Kitchen appliances

usually have a hood through which air is exhausted, thus reducing

the heat gain appreciably. Remember, however, that the exhausted

air must be replaced by outdoor air. The ASHRAE Handook gives

extensive data for commercial cooking equipment.

Just as with people and lights, care must be taken to properly

estimate the actual periods of use, so that an unreasonably large

heat gain is not obtained.

After the heat gain is determined, the cooling load is

computed using a cooling load factor in a fashion identical to that

for people. Table 6 -21 gives CLF values for the use of unhooded

equipment. Typical office equipment, small computers and motors,

and small electric appliances fall in this category.

Similar data are given in ASHRAE Handbook Fundamentals for hooded

appliances that are mainly commercial kitchen equipment.

-t-e.. E63 Sensible Heat Cooling Load Factors for Appliances -Unhooded Toul Operational

Muni Hours after ap9Uaaces are out

1 2 3 4 S 6 7 II 9 10 II 12 13 14 15 16 17 IS 19 20 21 22 23 24

2 0.56 0.64 0.15 MU I 0.08 0.07 0.06 0.05 MU 004 on an an 0 02 0.02 0.02 0.01 0 01 0.01 0.01 0.01 0.01 0.01 0.01

4 0 57 0.65 0.71 0. 75 0.23 0.18 0.14 0 12 0.10 0.08 0.07 0.06 OM 0.03 0.04 0.07 0.03 0.03 0.02 0.02 0.02 0.02 0.01 0.01

6 0.57 0.65 0.71 0.76 0.79 an an an OIS OIS 0.13 0.11 O.IO 0.08 0.07 0.06 0.06 005 0.04 004 0.03 0.03 0.03 0.02

8 M58 0.66 0.72 0.76 0.80 082 0 85 0.87 0 33 0.26 0.21 0.I8 0.15 0.13 0.11 0.10 0.09 0.08 0.07 0.06 0 .OS 0.04 0.07 0.03

10 0.60 0.68 0.73 0.77 MU an OM an an 0 90 0.36 0.29 0.24 0.20 0.17 0.15 0.13 0.11 0.10 0.08 0.07 0.07 0.06 0 05

12 0.62 0.69 075 0.79 0 82 084 0 86 0 88 an 0.91 0 92 0.93 0 38 an an 0.21 0.18 0.16 0.14 0.12 0.11 0 09 0.08 0.07

14 OM 0.71 0.76 010 0 83 0.85 0.87 an an an an an 0.94 0 95 0.40 0 32 0.27 an 0.19 MP au S 0.13 0.1 I 0.10

16 0 67 0 74 0.79 0.82 0 85 0.87 0.89 0.90 0.91 0.92 0.93 0 94 0.95 0.% 096 0.97 0.42 0.34 0.28 0.24 0 20 0.18 0.13

IB 071 0.78 082 0 85 0 87 089 0.90 0 92 0 91 094 094 0.95 0.96 0.96 0 97 0.97 097 0 98 0.43 0.35 an 0.24 0.21 0.IB

Example (6 -7):

Suppose the office suite previously described has a

collection of typewriters,duplicating machines, electric

coffee pot, and the like, with a total nameplate rating of

3.7 KW. It is estimated that only about one -half of the

equipment is in use on a continuous basis. Estimate the

cooling load at 5:00 P.M.

Solution:

Assume the equipment is turned on at 8:00 A.M. and operates

until 4:00 P.N. From Table 6 -21 the CLF is 0.33 because

5:00 P.N. is 9 hours after startup time. The cooling load

is:

.. f: .

q= (3.7 x 10 ) x (3.412) x (0.33) / 2

.....,. ,

tqf

= 2,080 BTU /hr

It should be noted that all the equipment in a space does

not have to be lumped together to make a calculation.

Equipment can be started and stopped at different times and

separate calculations made for each.

(D) Infiltration and Ventilation

For both infiltration and ventilation have sensible and latent

heat gain. In here, only equations are provided, theoriotical

discussion refer to Chapter 22 of ASHRAE handbook 1985 Fundamentals

for sensible heat

qs = 1.1 (cfm) .At (6 -24)

(cfm) = volume flow rate

a t = temperature difference between indoor and outdoor

for latent heat

ql = 4840 (cfm) A w (6 -25)

o w = humidity ratio difference between indoor and outdoor both equations 6 -24 and 6 -25 apply to infiltration and ventilation.

CHAPTER 7

Ducts System

7 -0 Introduction

7 -1 Fluid Flow Basics

7 -2 Air Flow in Ducts

7 -3 Air Flow in Fittings

7 -4 Turning Vanes and Dampers

7 -5 Duct Design Fundamentals and

Pressure Gradiant Diagrams

7 -6 Equal- Friction Method

2.19

7 -0 Introduction

This chapter discusses the details of distributing the air

through the ducts to the various spaces in the structure. Proper

design of the duct system and accessories are essential. A poorly

designed system is inefficient. Correction of faulty design is

expensive and sometimes practically impossible.

Sections 1 to 4 are about how air streams go within the duct

system, what kind of fittings are usually used, and why dampers and

vanes should be used. Section 5 is about fundamental ideas of

duct design and pressure gradiant. Section 6 is about one of the

low- velocity system design methods, high -velocity system design

methods will not be involved in this book because of its

complicated calculations and concepts.

20!

7 -1 Fluid Flow Basics

The distribution of fluids by pipes, ducsts, and conduits is

essential to all heating and cooling systems. The fluids

encountered are gases, vapors, liquids, and mixtures of liquid and

vapor (two -phase flow). From the standpoint of overall design of

the buidling system, water and air are of greatest importance. The

fundamental principles in this section are not only for air but

also for water.

\\Bernoulli equation is the most fundamental equation has to be

introduced for 1 fluid flow basics. When the fluid flow has

(1) steady flow, (2) incompressible flow, (3) frictionless flow,

and (4) flow along a streamline, conditions then

-? + F -f- 2 -Ce t.

The Bernoulli equation is a powerful and useful equation

because it relates pressure changes to velocity andelevation

changes along a streamline.

The Bernoulli equation can be applied between any two points

on a streamline provided that the other three restrictions are

satisfied. The result is

p 2 l

f + fi _ 2 2v27-.+ 2

(7 -2)

where subscripts 1 and 2 represent any two points on a streamline.

Neglecting elevation difference because in most cases

then we get

VI 2 2 f I V2

P r (7 -3)

202

The static pressure is that pressure which could be measured

by and instrument moving with the flow. However, such a

measurement is rather difficult to make in a prectical situation.

Since no pressure variation normal to flow streamlines when those

streamlines were straight, then we can measure the static

pressure in a flowing fluid using a wall pressure "tap ", placed

in a regionwhere the flow streamlines are straight, as shown in

Fig 7 -1. The pressure tap is a small hole, drilled carefully in

the wall, with its axis perpendicular to the duct wall and free

from burrs, accurate measurements of static pressure can be made

by connecting the tap to a suitable measuring instrument. "efj t Total pressure is the sum of static pressure and velocity

pressure. We have seen that static presure at a point can be

measured with a static pressure tap. If we knew the total

pressure at the same point, then the flow speed could be i

computed." [4J o /27

77(

P

P2-

(6)

z-{iPz Z= F -p

= CPi-P) (9

-f-

(7 -4)

then we can compute the velocity pressure. The experimental

setup is shown in Fig 7 -2.

203

-Flo(A)

S Veak4bhYç

e71'/%"_;/

7

PYe.J4(4.1,-.e -FIG 7-I C4-

To tA.( D}'.e..1.Lt-t Y-e s

Low -- ;-

Ír7c5 i 7i

P T-(G 7 2 [4,]

In Fig 7 -2 the static pressure corresponding to point A is read

from the wall static pressure tap. The total pressure is

measured directly at A by the total head tube, as shown.

The adiabatic, steady flow of a fluid in a duct or pipe is

governed by the first law of thermodynamics and Bernoulli

equation which may be written

where

z i =`

e, 29

z

Z rtbz q -r2t{ e?. ; 9 Ye

p = static pressure, lbf /ft

p = mass density at a cross section, lbm /ft

= average velocity at a cross section, ft /sec

g = local acceleration of gravity, ft /sec 2

gc = constant, 32.17(lbm- ft) /(lbf -sect)

z = elevation, ft

(7 -5)

2o4

w = work of fan or pump (ft- lbf) /lbm

= lost head, ft

Each term of Eq. 7 -5 has the units of energy per unit mass

or specific energy. The last term on the right in eq. 7 -5 is the

internal conversion of energy due to friction. The first three

terms on each side of the equality are the pressure energy,

kinetic energy, and potential energy, respectively. A sign

convection has been selected such that work done on the fluid is

negative.

Another governing relation for steady flow in a conduit is

the conservation of mass. For flow along a single conduit the

mass rate of flow at any two cross sections 1 and 2 is given by

C(71 - 2 / - 2 H,

where

(7 -6)

m = mass flow rate, lbm /sec

A = cross -sectional area normal to the flow, ft-2-

when the fluid is incompressible, Eq. 7 -2 becomes

where

(7 -7)

Q = volume flow rate, ft3 /sec

Equation 7 -5 has other useful forms. If is multiplied by the

mass density, an equation is obtained where each term has the

units of pressure.

Fi )3o (49t1) (7-8)

205

In this form the first three terms on each side of the equality

are the static pressure, the velocity pressure, and elevation

pressure, respectively. The work term now has units of pressure

and the last term on the right is the pressure lost due to

friction.

Finally, if Eq. 7 -5 is multiplied by 9% , an equation

results where each term has the units of length, commonly

referred to as head.

9p 2 3c r /Z2 G i + / _ --L -' -f -1-_ C w (7-9) Pf 9 Py z Z -, f

J

The first three terms on each side of the quality are the static

head, velocity head, and elevation head, respectively. The work

is now in terms of head and the last term is the lost head due to

friction.

Equations 7 -5 and 7 -6 are complementary because they have

the common variables of velocity and density. When Eq. 7 -5

multiplied by the mass flow rate m, another useful form of the

energy equation results, assuming

2= constant

[F1_F 71 z 41 /- gC-

'

9c. J(7-10)

where

* = power, work per unit time, lbf -ft /sec

All terms on the right -hand side of the equality may be positive

or negative except the lost energy, which must always be

positive.

Some of the terms in Eqs. 7 -5 and 7 -10 may be zero or

negligibly small. When the fluid flowing is a liquid, such as

water, the velocity terms are usually rather small and can be

neglected. In the case of flowing gases, such as air, the

potential energy terms are usually very small and can be

neglected; however, the kinetic energy terms may be quite

important. Obviously the work term will be zero when no pump,

turbine, or fan is present.

The total pressure, a very important concept, is the sum of

the static pressure and the velocity pressure.

In terms of head Eq. 7 -7 is written

9c r, gc P

Z

5 P 5*/0 -+ 23 or

Ho = H + Hv

where

(7 -11)

Po = total pressure

Ho = total head

H = static head

Hv = velocity head

Equation 7 -5 may be written in terms of total head and with

rearrangement of terms become

. (poi-P0-,) 9 P

where

(7-14)

207

Poi = total pressure at point 1

Po2 = total pressure at point 2

This form of the equation is much simpler to use with gases

because the term (Z1 -Z2) is negligible, and when no fan is in the

system, the lost head equals the loss in total pressure head.

20b

7 -2 Air Flow in Ducts

The general subject of fluid flow in ducts was discussed in

the last section. The special topic of air flow is treated in

this section. Although the basic theory is the same for fluid

flow in ducts and pipes, certain simplifications and

computational procedures will be adopted to aid in the design of

air ducts.

Equation 7 -9 applies to the flow of air in a duct.

Neglecting the elevation head terms for the gravity does not

effect the air mass significantly, assuming that the flow is

adiabatic since we do not consider any heat transfer through the

ducts so far, and no fan is present, Eq. 7 -5 becomes: y [ 3

[772- r Je - ¡ -fr

l3j G f (7 -15) J l

and in terms of the total head with 7 constant c 17, f {2

e where

P = static pressure, lbf /ft

(7 -16)

P01 = total pressure at point 1. lbf /ft2

(2= mass density at a cross section, lbm /ft 3

V'= average velocity at a cross section, ft /sec

g = local acceleration of gravity, ft /sec Z

gc = constant, 32.17 (lbm- ft) /(lbf -sect)

if = lost head, ft

Equation 7 -15 provides a great deal of insight into the duct flow

problem. The only important terms remaining in the energy

2ot

equation are the static head (

T7Z velocity head ( Ìk2

The static and velocity hea

9c Pt ) , the

9 P, , 5 ) , and the lost head ( ) .

ds are interchangeable and may

increase or decrease in the directin of flow depending on the

duct cross -sectional area. Because the lost head if must be

positive, the total pressure always decreases in the direction of

flow. Figure 7 -3 illustrates these prinicples.

For duct flow the units of each term in Eq. 7 -15 are usually

in.H20 because of their small size.

following form for constant density(e ):

¡q Pi -{ / D.3 9

Eq. 7 -15 then takes the

(7 -17)

To simplify the notation, the equation may be multiplied by

for each term then we can get

and

and

where

5c Z

P9 . Hsl + Hvl = Hs2 + Hv2 + lf

HO1 = H02 + lf

Hsl = rl q static head at point 1

Z

Hvl = q velocity head at point 2

Hs2 = Ç static head at point 1

(7-18)

21,9

sfat.c. p res4.4.44e

FLoW`C" 1 -

e(i ` °; !

I

t

At i« oSPIt I

Prz,se.t4+^e

( 1

. . , ,

Tct.c-c F ,c

( ve(oúl P44 e

PS

{'U. : Ve.` f,y pr. aio,wr e

Po : TotwQ. rt,o.d{..,r .e

Ps .s S-eett`c.

p,'j -7-sC2)CJ

2 ((

'7;4 Hv2 =

9 velocity head at point 2

HO1 = Hsl + Hvl total head at point 1

H02 = Hs2 + Hv2 total head at point 2

Because the conputational procedure of lost head becomes

tedious when designing ducts, special charts have been prepared.

Figures 7 -4, 7 -5, 7 -6 and 7 -7 are examples of such charts for

air flowing in galvanized steel ducts with approximately 40

joints per 100 ft or 30m. The charts are based on standard air

which has oxygen, 0.2095; Nitrogen, 0.7809; Argon, 0.00934;

Carbon dioxide, 0.00031, of volume fraction. For the temperature

range of 50 F or 10 C to about 100 F or 38 C there is no

correction for viscosity and density changes. Above 100 F or 38

C however, a correction should be made, the procedure and the

formula of the correction will not be introduced in this book.

11The effect of roughness is a more important consideration.

A common problem to designers is determination of the roughness

of fibrous glass duct liners and fibrous ducts. This material is

manufactured in several grades with various degrees of roughness.

Futher, the joints and fasteners necessary to install the

material affects the overall pressure loss. Smooth galvanized

ducts tyupically have a friction factor of about 0.02, whereas

fibrous liners and duct materials will have friction factors

varying from about 0.03 to 0.06 depending on the quality of the

material and joints, and the duct diameter.) /C6'

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7 -3 Air Flow in Fittings

Whenever a change in area or direction occurs in a duct or

when the flow is divided and diverted into a branch, substantial

losses in total pressure may occur. These losses are usually of

greater magnitude than the losses in the straight duct and are

referred to as dynamic losses. Furthermore, fittings are

classified as either constant flow, such as an elbow or transition,

or as divided flow, such as a wye or tee.

'Elbows are generally efficient fittings in that their losses

are small when the turn is gradual. When an abrupt turn is used

without turning vanes, the lost pressure will be four or five times

larger. '/ CiJ 1When considering the lost pressure in divided flow fittings,

the loss in the straight- through section oulet must be considered.

The angle of the branch takeoff has a great influence on the

pressure loss. The velocity may increase, decrease, or remain

constant through the fitting. In every case there will be some

loss in total pressure.' [ J

It is often convenient to express the effect of fittings in

terms of equivalent length. The equivalent length is a function

of the air velocity and the size (diameter) of the fitting.

Figures 7 -8 and 7 -9 give some equivalent lengths commonly used for

residential system design. This approach simplifies calculations

and is helpful in the design of simple low- velocity systems.

(Example (7 -1) :

Compute the lost pressure for each branch of the simple duct

system shown in Figure 7 -10 using the equivalent length

217

35

35

10 in. (25 cm) minimum

30 (91

55 (17)

5 (1.51

,-I , : .5 15 15!

1

70 (21)

-7 -- cc]

15 (51

10 (31

5 11.5)

28

system shown in Figure 7 -10 using the equivalent length

approach with data from Figures 7 -8 and 7 -9 and SI units. 2goc -fvn ((-'13 % coc o,, /s> sv f--f. Cir-D/n)

of -t- (t ) glncitci 03 , )

t o cc. o (-2. c ) O 1'2'1\ 1 t- O

Fig 7 -10

2C c1-14,1 ( lc f -( ( 12c1, )

)

Solution:

The lost pressure will first be computed from i to a; then

sections a to 2 and a to 3 will be handled separately.

The equivalent length of section 1 to a consists of the

actual length of the 25 cm duct plus the equivalent for the

entrance from the plenum of 11 m given in Fig 7 -8. Then

La = 15 + 11 = 26 cm

From Fig. 8 -4 at 0.19 m /s and for a pipe diameter of 25 cm,

the lost pressure is 0.85 Pa /m of pipe. Then

d po 1 0. = (0.85) (26) = 22.1 Pa

Section 1 to 3 has an equivalent length equal to the sum of

the actual length, and the equivalent for the 45 degree

branch takeoff, one 45 degree elbow, and one 90 degree elbow

l_q3 = 12 + 11 + 1.5 + 3 = 27.5 m

2(.1

From Fig. 7 -6, at 0.057 m /s and for a 15 -cm diameter duct,

the lost pressure is 1.0 Pa /m of pipe. Then , Fc a3 = (1.0) (27.5) = 27.5 Pa

Section a to 2 has an equivalent length equal to the sum of

the actural length and the equivalent length for the

straight- through section of the branch fitting. A length

of 2 in is assumed for this case

La2 = 15 + 2 = 17 m

From Fig. 7 -6, the loss per meter of length is 0.6 Pa and

Poa2 = (0.6) (17) = 10.2 Pa

Then the lost pressure for section 1 to 2 is

4 Pol2 =6 Pola +O Poa2 = 22.1 + 10.2 = 32.3/Pa

and for section 1 to 3

QPo13 =óPola +a Poa3 = 22.1 + 27.5 = 49.6 Pa i/ f f )J

7 -4 Turning Vanes and Dampers

The two main accessory devices used in duct systems are vanes

and dampers. It is the responsibility of the designer to specify

the location and use of the devices.

Turning vanes have the purpose of preventing turbulence and

consequent high loss in total pressure where turns are necessary

in rectangular ducts. Although large- radius turns may be used for

the same purpose, this requires more space. When turning vanes are

used, an abrupt 90 degree turn is made by the duct, but the air is

turned smoothly by the vanes, as shown in Fig. 7-11.1/CG]

Fig 7 -11 no

221

Turning vanes are of two basic designs. (Fig 7 -12). The vanes

shown in Figs 7 -12a and 7 -12b are used in ducts having a maximum

width of 36 in.; the vanes of Fig. 7 -12c are used in larger ducts.

Dampers are necessary to balance a system and to control

makeup and exhaust air. The dampers may be hand operated and

locked in position after adjustment or may be motor operated and

controlled by temperature sensors or by other remote signals.

The damper may be a single blade on a shaft or a multiblade

arrangement as shown in Fig. 7 -13. The blades may also be

connected to operate in parallel.

A combination damper and turning assembly, called an extractor

is adjustable from outside the duct and may be extended into the

air stream to regulate flow to a branch as shown in Fig 7 -14.

222

Duct

Air extractor

F( e,-- 14 rsJ

Shaft eNtension

Single vane elbow

9D oca.

-

4 ,', in . R/

90 deg

Small double vane

2á in.

(c)

4 in. R

2 in. R-J / \ y0 deg. \/, /

Large double vane

7---1Z C5.3

-1- 1. C 5 223

7 -5 Duct Design Fundamentals and Pressure Gradiant Diagrams

The purpose of the duct system is to deliver a specified

amount of air to each diffuser in the conditioned space at a

specified total pressure. This is to ensure that the space load

which is from insolation of the sun, people, lights, appliance etc.

will be absorbed and the proper air motion within the space will

be realized. The method used to lay out and size the duct system

must result in a reasonably quiet system and must not require

unusual adjustments to achieve the proper distribution of air to

each space. A low noise level is achieved by limiting the air

velocity, by using sound -absorbing duct materials or liners, and

avoiding drastic restrictions in the duct such as nearly closed

dampers. Figure 7 -15 gives recommended velocities and pressure

losses for duct systems. A low velocity duct system will generally

have a pressure loss of less than 0.15 in.H2O per 100 ft (5.7

Pa /m) .

The use of fibrous glass duct materials has gained wide

acceptance in recent times because they are very effective for

noise control. These ducts are also attractive from the

fabrication point of veiw because the duct, insulation, and

reflective vapor barrier are all the same piece of material. Metal

ducts are usually lined with fibrous glass material in the vincity

of the air distribution equipment and for some distance away from

the equipment. The remainder of the metal duct is then wrapped or

covered with insulation and a vapor barrier. Insulation on the

outside of the duct also reduces noise.

The duct system should be relatively free of leaks, especially

224

Low -velocity systems High -velocity systems

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when the ducts are outside the conditioned space.

Air leaks from the system to the outdoors results in a direct loss

that is proportional to the amount of leakage and the difference

in enthalpy between the outdoor air and the air leaving the

conditioner. Presently 1 percent of the total volume flow of the

duct system is an accepted maximum leakage rate in high -velocity

systems. No generally accepted level has been set up for low -

velocity systems. However, care should be taken to tape or

otherwise seal all joints to minimize leakage in all duct systems

and the sealing material should have a projected life of 20 to 30

years.

The layout of the duct system is very important to the final

design of the system. Generally the location of the air diffusers

and air -moving equipment is first selected with some attention

given to how a duct system may be installed. The ducts are then

laid out with attention given to spaceand ease of construction.

It is very important to design a duct system that can be

constructed and installed in the allocated space. If this is not

done, the installer may make changes in the field that lead to

unsatisfactory operation.

The total pressure requirements of a duct system are an

important consideration. From the standpoint of first cost, the

ducts should be small; however, small ducts tend to give high air

velocities, high noise levels, and large losses in total pressure.

Therefore, a reasonable compromise between first cost, operating

cost, and practice must be reached. The cost of owning and

operating an air -distribution system can be expressed in terms of

22 <v

system parameters, energy cost, life of the system and interest

rates such that and optimum velocity or friction rate can be

established. A number of computer programs are available for this

purpose.

The total pressure requirements of a duct system are

determined in two main ways. For residential and light commercial

applications, all of the heating, cooling, and air -moving equipment

is determined by the heating and /or cooling load. Therefore, the

fan characteristics are known before the duct design is begun.

Furthermore, the pressure losses in all other elements of the

system except the supply and return ducts are known. The total

pressure available for the ducts is then the difference between the

total pressure characteristic of the fan and the sum of the total

pressure losses of all of the other elements in the system

excluding the ducts. Figure 7 -16 shows a typical total pressure

profile for a system. In this case the fan is capable of

developing 0.6 in.H2O at the rated capacity. The return grille,

filter, coils, and diffusers have a combined loss in totalpressure

of 0.38 in.H2O. Therefore, the available total pressure for which

the duct must be designed is 0.22 in.H2O

This is usually divided for low- velocity systems so that the

supply -duct system has about twice the total pressure loss of the

return ducts.

Large commercial and industrial duct systems are usually

designed using velocity as a limiting criterion and the fan

requirements are determined after the design is complete. For these

larger systems the fan characteristics are specified and the

227

correct fan is installed in the air handler at the factory or on

the job.

The total pressure in a duct system at any location is the

sum of the static pressure and the velocity pressure. As

frictional and dynamic effects occur in the airstream in the course

of flow, an energy loss occurs that appears as a reduction in total

pressure. Therefore, in any real duct system, except where energy

is added with a fan, the total pressure will decrease in the

direction of flow.

Figure 7 -17 is a pressure gradient diagram for a simple fan

system, where

Po = total pressure

Ps = static pressure

Pv = velocity pressure

Pb = barometric pressure

The line connecting all of the respective total pressure

points along the duct system is called the energy grade line (EGL) .

The line connecting all of the static pressure points along the

duct is called the hydraulic grade line (HGL). The vertical

distance between these lines at any section is the velocity head.

The reference pressure may be any value, although for simplicity

local atmospheric pressure is nomally used. Several points should

be noted regarding the relationships depicted on the diagram.

First, the only location at which there is a total pressure

increase is a the point of external energy input, namely the fan.

At all other locations, total pressure will decrease, in straight

equal -area ducts at a rate equal to pressure drop per unit lenght,

228

O

ó

0.4-

Element Return grille Return duct Filter Heat and cool coils Supply ducts Diffusers Fan total pressure-in. H2O

Total pressure loss, in. H,0 0.04 0.08 0.08 0.23 0.14 0.03 0.60

-0.12

-0.2 -

Distance ---.-

Return Heat gr ille Return Filter element Cool

I duct Fan coil

= v,., r 1'

_Y L `J l

á- -- P= P,í7.4

'

Supply ducts Diffusers

riff r

1

U : Vel-6-C; ú,

; Totu.0 f> e Ps S+-cdt,ti, -14 -t-,c.

FlG _f

22ct

and in fittings at a rate determined by the applicable dynamic

losses. Second, the total pressure always equals or exceeds the

static pressure by a quantity equal to the velocity pressure.

Third, though the total pressure may not increase except at the

fan, the static pressure may increase or decrease in relation to

the arrangement of the system, at times in unusual manners.

The pressure in any building served by an air system is

dictated by the location of the fan and the duct system

arrangement. The problem of determining, understanding, or

controlling the relationship of the pressure in a building to the

ambient or surrounding pressure is best understood through the use

of the pressure gradient diagram. Figure 8 -16 will begin to

develop this concept clearly. In this system, one additional

element has been added: an outdoor air intake. This system would

be operable in this manner only if an exhaust system were operating

to exhaust the outdoor air. Otherwise, the space pressure would

have to be positive to permit the outdoor air to be exfiltrated

from the building. There is no guarantee that this system will

operate satisfactorily, because the volume flow rate of

exfiltration is very hard to be measured, consequently, we will not

be able to know how much air is exhausted.

23C7

7 -6 The Equal Friction Method

The method described in this section is for a low- velocity

system. This method can be used for high -velocity system design,

but the results will not be satisfactory in some cases, for

example, the high -velocity has smaller duct size which will make

the roughness of the duct becomes more significant, in this case,

the pressure loss per foot will not be the same throughout the

whole duct system.

The principle of this method is to make the pressure loss per

foot of duct length the same for the entire system. If the layout

is symmetrical with all runs from fan to diffuser about the same

length, this method will produced a good balanced design. However,

most duct systems have a variety of duct runs ranging from long to

short. The short run will have to be dampered, which can cause

considerable noise.

The usuali procedure is to select the velocity in the main

duct adjacent to the fan in accordance with Fig. 7 -15. The known

flow rate then establishes the duct size and the lost pressure per

unit of length using Fig. 7 -4 or 7 -5. This same pressure loss per

unit length is then used throughout the system.

A desirable feature of this method is the gradual reduction of air

velocity from fan to outlet, thereby reducing noise problems.

After sizing the system, the designer must compute the total

pressure loss of the longest run (largest flow resistance), taking

care to include all fittings and transitions. When the total

'pressure available for the system is known in advance, the design

loss value may be established by estimating the equivalent length

2 3(

of the longest run and computing the loss in pressure per unit

length.

Example(7 -2):

Select duct size for the simple duct system of Fig. 7 -19

using the equal friction method and SI units. The total

pressure available for the duct system is 0.12 in.H20 or

30 Pa and the loss in total pressure for each diffuser at

the specified flow rate is 0.02 in.H20 or 5 Pa. Because

the system is small, the velocity in the main supply duct

should not exceed about 1000 ft /min or 5 m /s, and the branch

duct velocities should not exceed about 600 ft /min or 3 m /s.

(Note : Pa = Pascal, Pressure unit, 1 atm = 101,325 Pa)

Solution:

The total pressure available for the ducts, excluding the

diffusers, is 30 Pa - 5 Pa = 25 Pa

and the longest run is 1 -2 -3. The equivalent- length method

will be used to account for losses in the fittings. Then

for simplicity if we use Figs. 7 -8 and 7 -9

L123 = (L1 + Lent) + (L2 + Lst) + (L3 + Lwye + Lel + Lboot)

L123 = (11 + 6) + (1.5 + 4.6) + (3 + 11 + 3 + 9) = 49.1 m

and Po' = Po /L123 = 25/49.1 = 0.51 Pa /m

P = total pressure from plenum to the end of

L123 = Length of Q This value will be used to size the complete system using

Fig. 7 -6 or 7 -7. Table 7 -1 summarizes the results showing

the duct size, and velocity in each section. The

rectangular sizes selected are rather arbitary in this case.

232

1--- .,,,1--c>,,,,)

O loft C31-^) 1-1)&f-14-1

d---* C°, t>71 k+47>

© 1.-f -(.1,1-14, )

(ra°Gfwx kft 9

o, o l ((''`S)u (1,8-01)

ir ft (.4,600 -\ Ci.-t-3-1)

I

(I i -t- w1) .. CPC) G7ki °' °g- %s f

T...j 17-1c1

2 33

It is interested to check the actual loss in total pressure

from the plenum to each outlet.

(,Po)123 = 8.67 + 3.11 + 13.26 = 25.04 Pa

(pPo)124 = 8.67 + 3.11 + 9.18 = 20.96 Pa

(6Po)15 = 8.67 + 15.30 = 23.97 Pa

S.ec -C,'c-. )

h Lt-im 6 e %-

Q (m S )

t7

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Table 7 -1

The loss in total pressure for the three different runs are

unequal when it is assumed that the proper amount of air is

flowing in each. However, the actual physical situation is

such that the loss in total pressure from the plenum to the

conditioned space is equal for all runs of duct. That is,

if the actual energy grade lines for each duct run were

constructed, they would all begin at the plenum pressure and

end at the room pressure. Therefore, the total flow rate

from the plenum will divide itself among the three branches

in order to satisfy the lost pressure requirement. If no

adjustments are made to increase the lost pressure in

34

section 4 and 5, the flow rates in these sections will

increase relative to section 3 and the total flow rate from

the plenum will increase slightly because of the decreased

system resistance. However, dampers in section 4 and 5

could be adjusted to balance the system. Nevertheless this

duct sizing method is used extensively. It is not always

necessary that the system be designed to balance without

adjustments.

2aS

Chapter 8: Application

8 -0 Introduction

8 -1 Plan and Elevations

8 -2 Analysis and Calculations

8 -0 Introduction

A one -story office building is located in Tucson, Arizona.

The adjoining storage space is not conditioned and its inside air

temperaturate is basically equal to the outdoor air temperature at

any time of the day.

* Roof construction : 0.375 -inch built -up roofing; concrete

slab, lightweight aggregate, density 30 lb /ft , 2 -in.;

10 -in. fiberglass insulation; nonreflective air space 2.5

ft (90 F mean; 10 F temperature difference, E = 0.82);

metal ceiling suspension system with metal hanger rods; a

coustic tile.

* Wall construction : 1 -in. stucco; 8 -in. lightweight

concrete block; 3.5 -in. fibrous glass; 0.75 -in. gypsum

board.

* Floor construction : 3 -in. concrete on ground, no carpet.

* Fenestration : 3 -ft * 5 -ft window of regular plate glass,

1/4" with light -colored venetian blinds, not openable.

* Front door : one. 6 -ft * 7 -ft.

* Side door : two. 3 -ft * 7 -ft.

* Rear door : two. 3 -ft * 7 -ft.

* Door construction : solid core flush door, 2.25 -in.

* For outside walls assuming a wind speed of 7.5 mph. For

party and inside walls, still air was assumed.

* Indoor design conditions : dry -bulb temperature, 75 F;

wet bulb temperature, 65 F.

* Outdoor design conditions : dry -bulb temperature, 104 F;

wet bulb temperature, 66 F.

237

* Occupancy : 55 office workers from 0800 to 1700 with one

hour lunch break.

* Lights : 17,500 watts fluorescent, from 0800 to 1800;

4000 watts tungsten, continuous. The fixtures are not

vented.

* Ventilation : 15 cfm per person.

* Infiltration : 100 ft of outdoor air per person per door

passage, assume the main entrance is used at a rate of 30

persons per hour, for each side door is at a rate of 5

persons per hour.

* Outside ground reflectance is 0.2.

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240

8 -2 Analysis and Calculation

The first step is to find out we have to know when the maximum

cooling load occur.

From table 6 -3 lat. 32 degree, June has the biggest CLTD

correction for the roof. The roof and west facing wall and windows

dominate the cooling load for Tucson, Arizona. The next step is

to find out the hour of maximum cooling load for June. We are

going to choose 1400, 1500, 1600, and 1700, hours for comparison.

We also need to compare the cooling loads of 1400, 1500, 1600,

and 1700, after adding external shades. This is to determine

whether the peak hour of the cooling load without external shades

is the same as with external shades.

After we have found the peak hour of the day, than we

calculate the cooling load of the rest of the windows, doors, walls

and all the internal cooling loads such as people and lights.

<A> External Cooling Loads

<1> q for roof

Roof Construciton :

* 7.5 mph, outside surface (Table 3 -2) 0.25

* Built -up roofing 0.375 -in 70 lb /ft (Table 3 -1d) 0.33

* Concrete slab l.w.(density 30 lb /ft , 2 -in)

(Table 3 -1c) 1.11 x 2 = 2.22

* 10 -in fiber glass (Table 3 -la) 30

* Nonreflective air space 2.5 -ft (90 F mean, 10F

temperature difference, E = 0.82) (Table 3 -5)

* Acoustic tile (Table 6 -6)

* Inside surface air (Table 3 -2)

0.8

1.786

0.61

R = 35.99

then U = 0.0278

The next step is calculation of the mass, so that the roof

number of Table 6 -1 can be found:

70 lb /ft * 0.375 in. * ft /12 in. = 2.1875 lb /ft .

30 lb /ft * 2 in. * ft /12 in. = 5 lb /ft .

30 lb /ft * 0.0625 ft = 1.875 lb /ft .

total = 9.0625 lb /ft . matched with Table 6 -1. The total is

close to number 1 of the roof number with suspension.

Since we have R -30, for each R -7 increase, the peak CLTD will

be 2 hours later, so 30/7 = 4, 4 * 2 = 8, Therefor the peak will

be 8 hours late. We can not find the peak CLTD with similar mass

in Table 6 -1, according to ASHRAE we use 29 as the peak CLTD.

CLTDcor = (CLTD + LM) * K + (78 - Ti) + (Tom - 85)

= (29 + 2) * 0.65 + (78 - 75) + (91 - 95)

= 29.15

where

LM = 2 for June, Tucson (32 degree)

K = 0.65 for light colored

Ti = 75 F

Tom = 104 - 26/2 = 91 (Equation 6 -17)

q = U A CLTDcor = (0.0278) * (80 * 50) * (29.15)

242

= 3241.48 (Btu /h)

<2> q for west -window

U = 0.81 (Table 3 -6)

(1) q cond. = U A CLTDcor

from Table 6 -7 for 1400, 1700 CLTD = 13

for 1500, 1600 CLTD = 14

a. 1400, 1700:

CLTDcor = 13 * 0.65 + (78 - 75) + (91 - 85) = 17.45

b. 1500, 1600:

CLTDcor = 14 * 0.65 + (78 - 75) + (91 - 85) = 18.1

a. 1400, 1700:

q cond. = 0.81 * (3 * 5 * 8) * 17.45 = 1696.14

b. 1500, 1600:

q cond. = 0.81 * (3 * 5 * 8) * 18.1 = 1759.32

(2) q rad. = A SC SHGF CLF

A = 3 * 5 * 8 = 120

SC = 0.55 (Table 6 -9)

using maximum SC in equation for window with external

shading for safety reason.

SHGF = 214 without external shading (Table 6 -12)

SHGF = 46 with external shading ground reflectance is

0.2 (Table 6 -13)

CLF = 0.53 (1400) (Table 6 -15)

= 0.72 (1500)

= 0.82 (1600)

= 0.81 (1700)

2 *3

q rad. w/o ext. = 120 * 0.55 * 214 * 0.53 = 7485.7 (1400)

= 120 * 0.55 * 214 * 0.72 = 10169.3 (1500)

= 120 * 0.55 * 214 * 0.82 = 11581.7 (1600)

= 120 * 0.55 * 214 * 0.81 = 11440.4 (1700)

q rad. w/ ext. = 120 * 0.55 * 46 * 0.53 = 1609.1 (1400)

= 120 * 0.55 * 46 * 0.72 = 2185.9 (1500)

= 120 * 0.55 * 46 * 0.82 = 2489.5 (1600)

= 120 * 0.55 * 46 * 0.81 = 2459.2 (1700)

(3) Sum (without external shading)

1400 1500 1600 1700

q cond. 1696.14 1759.32 1759.32 1696.14

q rad. 7485.7 10169.3 11581.7 11440.4

total 9181.84 11928.62 13341.02 13136.54

Sum (with external shading)

1400 1500 1600 1700

q cond. 1696.14 1759.32 1759.32 1696.14

q rad. 1609.1 2185.9 2489.5 2459.2

total 3305.24 3945.22 4248.82 4155.36

use solar chart and shading diagram for profile angle

17 degree at 1 pm

72 degree at 5 pm for Aug. 21

tan 72 = x /1.5, x = 4.62 ft

tan 17 = y /5, y = 1.53 ft

2-44

4 ' D/ :jj,í1?i'ttiTj

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SECT! ot-I

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Sum

245

Find the solar azimuth at Aug. 21, 5 pm

g= 23.45 * sin 360(284 +233)/365

= 11.75

from equation 2 -2

sin Go = sinq( sins + cos 4'cos S cos (.?

= sin(32) * sin(11.75) + cos(32) * cos(11.75) * cos(75)

= 0.53 * 0.2 + 0.848 * 0.979 * 0.259

= 0.321

= 18.7

from equatioin (2 -3)

á.= sins os () * sin (W) /cos c çj

= sin Icos(11.75) * sin(- 75)/cos(18.72)

= -87.0

west 87 degree is the answer

tan z/3 = 87, z = 57.2 ft

so, the shading is like

-E >

O!////!!/!!//i

4, 1,1

i <S >

'2 4(2

<3> q for west -wall

A = 80 * 14 - 3 * 5 * 8 - 3 * 7 * 2 = 958 ft

wall construction group D : (Table 6 -5) R

outside surface resistance 0.333

1 -in. stucco 0.208

8 -in 1.w. concrete block 2.02

3.5 -in. fibrous glass 13

0.75 -in. gypsum board 0.149

inside surface resistance 0.685

R = 16.395

then U = 0.061

In Table 6 -2, since we have R -13, we move from group D to

group B (move up 1 group for each R -7)

Table 6 -2 group B

CLTD 1400 1500 1600 1700

west 14 14 15 17

south 12 14 15 17

east 22 24 25 26

north 9 9 10 11

CLTD correctin for wall (Table 6 -3)

June N E S W

1 0 -4 0

2.47

For west wall

q = U A CLTDcor

(1) 1400 :

CLTDcor = (14 + 0)

= 18.1

q = 0.061 *

(2) 1500 :

CLTDcor = (14 + 0)

= 18.1

.

q = 0.061 *

(3) 1600 :

CLTDcor = (15 + 0)

= 18.75

q = 0.061 *

(4) 1700 :

CLTDcor = (17 + 0)

= 20.05

q = 0.061 *

*

958

0.65 + (78

* 18.1 =

-75) + (91

1057.73

- 85)

* 0.65 + (78 - 75) + (91 - 85)

958 * 18.1 = 1057.73

* 0.65 + (78 - 75) + (91 - 85)

958 * 18.75 = 1095.71

* 0.65 + (78 - 75) + (91 - 85)

958 * 20.05 = 1171.68

<4> for door Table 3 -7a

solid core flush door 2 1/4 ". U = 0.26

4 = U A T = 0.26 * (3 * 7 * 2) * (104 - 75)

= 316.68

<5> Total of roof + west -wall + west- window + doors

(1) without external shading

248

1400 1500 1600 1700

roof 3241.48 3241.48 3241.48 3241.48

wall 1057.73 1057.73 1095.71 1171.68

windows 9181.84 11928.62 13341.02 13136.54

doors 316.68 316.68 316.68 316.68

total 13797.73 16544.51 17994.89 17866.38

(2) with external shading

1400 1500 1600 1700

roof 3241.48 3241.48 6241.48 3241.48

wall 1057.73 1057.73 1095.71 1171.68

windows 3305.24 3945.22 4248.82 4155.36

doors 316.68 316.68 316.68 316.68

total 7921.13 8561.11 8902.69 8885.2

r From above, 1600 (4 pm) has the peak cooling load, the maximum

external cooling load for the building at 1600 of June is the total

N roof + walls + doors + windows (with external shading) , we have

calculated the roof, west -doors, west -windows, and west -wall. Now,

we need q for doors, windows, and windows on east, south, and north

sides.

<6> East

(1) east -windows

U = 0.81 (Table 3 -6)

CLTD = 14 (1600) (Table 6 -7)

CLTDcor = (14 + 0) * 0.65 + (78 -75) + (91 - 85)

= 18.1

q cond. = 0.81 * (3 * 5 * 8) * 18.1 = 1759.32

q rad. = A * SC (Table 6-9) * SHGF (Table 6-13) * CLF

(Table 6-15)

= 3 * 5 * 8 * (0.55) * 46 * (0.17) = 516.12

(2) east -wall

A = 958

U = 0.061

CLTD = 25 (Table 6 -2)

CLTDcor = (25 + 0) * 0.65 + (78 - 75) + (91 - 85)

= 25.25

q = 0.061 * 958 * 25.25 = 1475.56

(3) east -doors

q = 316.68

(4) total

q = 2275.44 + 316.68 + 1475.56 = 4067.68

<7> north

(1) this wall like a partition wall, we do not have to

calculate CLTD

q = U A T

= 0.061 * (50 * 14 - 3 * 7 * 2) * (104 - 75)

= 1164.1

(2) doors

q = 316.68

(3) total

q = 1164.1 + 316.68 = 1480.78

<8> south

(1) windows

U = 0.81

CLTD = 14

CLTDcor =

q cond. =

=

q rad. =

(Table 3 -6)

(Table 6 -7)

14 * 0.65 + (78 - 75)

0.81 * (3 * 5 * 4) *

879.66

(3 * 5 * 4) * (0.55) *

+ (91

18.1

40 *

- 85)

(0.35)

= 18.1

=462

q window = 879.66 + 462 = 1341.66

(2) doors

q = 316.68

(3) wall

U = 0.061, LM = -4 (Table 6 -3)

A = 598, CLTD = 15 (Table 6 -2)

CLTDcor = (15 - 4) * 0.65 + (78 - 75) + (91 - 85)

= 16.15

= 0.061 * 598 * 16.15 = 589.1

(4) total

q = 1341.66 + 316.68 + 589.1 = 2247.44

<9> Grand External Loads

q total = q east + q north + q south + (q west + q roof)

= 4067.68 + 1480.78 + 2247.44 + 8902.69

= 16698.59

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Z6a 253

<B> Internal Cooling Loads

<1> lighting

q tungsteng = 3.412 * w * Fu * Fs

= 3.412 * 4000 * 1 * 1.0

= 13650

q flouresent = 3.412 * w * Fu * Fs

= 3.412 * 17500 * 1 * 1.2

= 71650

"a" = 0.55 (Table 6 -19)

"b" = B (Table 6 -20)

Lights are on from 0800 to 1800 then we should use Table 6-

18b, calculation of cooling load is for 1600, it is 8 hours after

0800, so from Table 6 -18b "a" = 0.55, "b" = B. We get CLF = 0.82,

then the cooling load at 1600 from the flouresent lights is (Eq.

6 -23)

q = 71650 * 0.82 = 58753

total for lights: q = 13650 + 58753 = 72403

<2> people

latent heat: ql = 55 * 190 (Table 6 -16)

= 10450

CLF for people from Table 6 -17 (one hour lunch break),

CLF = 0.84

sensible heat: qs = 55 * 230 (Table 6 -16) * 0.84 (Table 6-

17) = 10626

<3> infiltration

from psychrometric chart: 75 F db, 65 F wb, Wi

104 F db, 66 F wb, Wo

100 ft per person per door passage

100 * (30 + 5 * 4) = 5000 (ft /h)

5000/60 = 83.3 cfm

Eq. 6 -24 is = 1.1 * 83.3 * (104 - 75) = 2657.3

Eq. 6 -25 ql = 4840 * 83.3 * (0.015 - 0.005) =

= 0.015;

= 0.005.

4031.7

<4> ventilation

55 * 15 = 825 cfm

qs = 1.1 * 825 * (104 - 75) = 26317.5

ql = 4840 * 825 * (0.015 - 0.005) = 39930

<c> Grand Total for Cooling Loads

q external = 16698.59

qs internal = q lights + q people + q infiltration +

q ventilation

= 72403 + 10626 + 2657.3 + 26317.5

= 112003.8

ql internal = q people + q infiltration + q ventilation

= 10450 + 4031.7 + 39930

= 54411.7

qs total = 16698.59 + 112003.8 = 128702.39

ql total = 54411.7

q grand total = 183114.09 (Btu /h)

25S

SHF = 128702.39/183114.09 = 0.7

Assume dry bulb temperature is 15 F less than state 3.

Energy Balance:

ma2 i2 + q = ma3 i3

or q = ma2 (i3 - i2)

ma2 = q / (i3 -i2)

from psychrometric chart i3 = 30, i2 = 25 (Btu /lbm)

ma2 = ma3 = 183114.09 / (30 - 25) = 36622.8 (lbm /h)

Also, from psychrometric chart v2 = 14.2 ft / lbma

62 ma2 * v2 = (36622.8 * 14.2) / 60 = 8667.4 cfm

State 1 must be determined before continuing, a mass balance

on the section yields

ma0 + ma4 = mal = ma2; ma0 = 6 / v0; v0 = 14.3 ft / lbma

ma0 = (825 * 60)/ 14.3 = 3461.5 (lbma /h)

then ma4 = ma2 - ma0 = 36622.8 - 3461.5 = 33161.3 (lbma /h)

31 / 30 = ma0 / mal = 3461.5/36622.8 = 0.0945

4///i/1/11//1,,//,/,4,,,, / :.-1 - ti L. /OI I I o /

a1 , . ll..

hti S ,,/,11//r¡ '//, t1G ///ui ur//////('/'r/ru/ ¡rr/tr/rru/i/u ..,Ñ/ i N ,..9. O O O ' :i I. .:, N. _.i

T r

tl AYO ONn N3d 3tl 1510M SONnod-(M1 OI va ALOIYInN

I!A , ,

P , ' My

" 11111 \j TAM 1"PS 47:14r/

tlnl01331 Bino ANO

Py

9 h

//

/ . '.-^, 1,--

. :. };.!

.:illIMPIS

= -' ®BE

.E 1/iirter 0s..Ne.MME 'r rie / on_w - moil-------------

_,'§ ...NI Ylt AYO wino. YU 1YUSroW town/

DA m,.. ,canr

"2-r,

31 = 0.1 30

State 1 is located: 78 F db; 50 %

An energy balance gives

mal * it = qc + ma2 * i2

qc = mal * (il - i2)

From psychrometric chart il = 30 Btu /lbma

qc = 36622.8 * (30 - 25) = 183114 (Btu /h) = 15 (ton)

From psychrometric chart, we can find SHF for cooling unit

ie. line from State 2 to State 1, SHF = 0.87

159309.2 (Btu /h)

= 23804.82 (Btu /h)

Use KRUEGER Model RS -1 Round Ceiling diffuser: 10 ",

total pressure 0.145 in. H2O, 325 cfm, 9 ft throw is 75 fpm

8667.4 / 325 = 27.28 (28 diffusers)

Assume main duct velocity is 1600 fpm, from Fig 8 -13, we can

use 8667.4 cfm and 1600 fpm to get the sizes

52c1 I16fZCfm 1F4 -2cfil

qcs = 0.87 * (183114) =

qcl = 183114 - 159309.2

Total cfm is 8667.4 cfm

2Ì7 ,H31 EE7cf# I ¡F 17cí i

217 4 2.167 cf

217 l 433,C-fhi

o-- 217 325 32f 3ZS 32C-, 326

`-1-wl crw, c f- r, c-f- w, c f-t-n

áS26 7 f ç- ¡; t

8667.4 cfm 34"

4333.5 cfm 24"

2167 cfm 16"

1842 cfm 16"

1517 cfm 16"

1192 cfm 14"

867 cfm 14"

542 cfm 10"

217 cfm 7"

( hj v-t To SCc, I-e )

l0 CIr ----- ---------

Use Figs. 8 -6 and 8 -7:

(11' + 6') + (5' + 5') + (15' + 5') + (3') + 70' + 30' + (5'

* 6) = 180'

(180' / 100') * 0.1 = 0.18

0.18 + 0.145 = 0.325 (in. H2O) Total Pressure

26

A rre i1 d t`X A .

Glossary:

Heat Exchanger:

This term is usually applied to a device in which two fluid steams separated by a solid surface exchange heat energy. Ten

devices may take many forms. However, ordinary metal tubes are

the main components of many types. The heat exchanger is the

most widely used device in HVAC applications.

Schematic:

i

,,...., -63 -t*3 k -t4

Co n-be rf l c c,w

hea-t, -2xc%.31e

Steady- State -Steady -Flow:

It means that the velocities and thermodynamic

properties at each point in space are unchanging in time.

26a

/1p?errci

Water (Refrigerant 713) Properties of Liquid and Vapor'

Abee4ete Premise Lb/i.2

Specific Volume es It per* Eeehelpy, ate per lb Eetropy. ate/lb F

Febr. Sat. Fehr.

Tulip. Set. Liquid Su. Vapor Su. Liquid E.g. Set. Vapor Su. Liquid Evap. Vapor Temp.

KFl P. .J v8 bJ bh 148 at' a18 i tlF)

33 0.092227 0.01602 3180.5 1.01 1074.59 1057.60 0.00205 2.1811 2.1831 ` 33

40 0.12164 0.01602 2445.4 8.04 1070.64 1078.68 0.01623 2.1426 2.1588 40

30 0.17799 0.01602 1704.3 18.07 1064.99 1083.06 0.03610 2.0895 2.1256 50

60 0.23618 0.01603 1207.1 28.08 1059.34 1087.42 0.05553 2.0385 2.0940 60

70 0.36304 0.01605 867.97 38.07 1033.71 1091.78 0.07458 1.9893 2.0639 70

80 0.50701 0.01607 633.03 48.05 1048.07 1096.12 0.09325 1.9419 2.0352 80

90 0.69538 0.01610 467.90 58.04 1042.40 1100.44 0.11158 11963 2.0079 90

100 0.94959 0.01613 350.22 68.02 I036.72 1104.74 0.12957 1.8523 1.9819 100

110 1.2754 0.01617 265.26 78.00 1031.01 1109.01 0.14724 11098 1.9570 110

120 1.6933 0.01620 203.18 87.98 1025.28 1113.26 0.16461 1.7687 1.9333 120

130 2.2237 0.01625 157.27 97.97 1019.50 1117.47 0.18170 1.7289 1.9106 130

140 2.8900 0.01629 122.96 107.96 1013.69 1121.65 0.19850 1.6903 1.8888 140

150 3.7194 0.01634 97.038 117.96 1007.83 1125.79 0.21503 1.6530 1.8680 ISO

160 4.7424 0.01639 77.267 127.97 1001.92 1129.89 0.23130 1.6168 1.8481 160

170 5.9936 0.01645 62.045 137.99 995.93 1133.94 0.24733 1.5817 1.8290 170

ISO 7.5119 0.01651 50.220 148.01 989.93 1137.94 0.26312 1 5475 1.8106 180

190 9.3403 0.01657 40.956 158.05 983.84 1141.89 0.27868 1.5143 I.7930 190

200 11.526 0.01663 33.640 168.10 977.68 1145.78 0.29402 1.4820 1.7760 200

212 14.6% 0.01671 26.801 180.18 970.17 1150.35 0.31215 1.4444 1.7565 212

220 17.188 0.016772 23.15 188.22 965.3 1153.5 0.32406 1.4201 1 7441 220

240 24.97 0.016922 16.327 208.44 952.3 1160.7 0.35335 1.3609 1.7143 260

260 35.42 0.017084 11.768 223.76 938.8 1167.6 0.38193 1.3044 1.6864 260

280 49.18 0.017259 8.650 249.18 924.9 1174.1 0.40986 1.2504 1.6602 280

300 66.98 0.017448 6.472 269.73 910.4 1180.2 0.43720 1.1984 1.6356 300

340 117.93 0.017872 3.792 311.30 879.5 1190.8 0.49031 1.0997 1.5901 340

380 195.60 0.018363 2.339 353.62 845.4 1199.0 0.54163 1.0067 1.5483 380

420 308.5 0.018936 1.5024 396.89 807.2 1204.1 0.59152 0.9175 1.5091 420

460 466.3 0.019614 0.9961 441.4 764.1 1205.5 0.6404 0.8308 1.4712 460

500 680.0 0.02043 0.6761 487.7 714.8 1202.5 0.6888 0.7448 I.4335 500

Cotepikd by John A. Goff and S. Gratch.

Temp, F

Viscosity, Ib /ft It Tormal Coadoctivfty. Btu /b ft F Specific Hest.ce, lila /lb., F

Sat. Ligald

Sat. Vapor Gas. P1 It= x10-4$

Sat. Liquid Sat. Vapor Gas. P1 gun x111 t

Sat. Liquid

Sat. Vapor Gas

,)mita Gas

(Co )I ass

Temp, F

32 4.28 0.0195 0.329 0.0100 1.007 0.4438 32 40 3.69 0.0199 0.334 0.0103 1.005 0.4442 40 50 3.12 0.0204 0.339 0.0105 1.003 0.4445 50 60 2.68 0.0210 0.345 0.0107 1.001 0.4446 0.4448 60 70 2.32 0.0215 0.350 0.0109 1.000 0.4454 0.4452 70

80 2.03 0.0221 0.354 0.0112 0.999 0.4464 0.4456 80 90 1.79 0.0226 0.359 0.0114 0.998 0.4475 0.4459 90

100 1.60 0.0232 0.363 0.0116 0.998 0.4488 0.4463 100 120 1.30 0.0243 0.371 0.0121 0.998 0.4522 0.4472 120 140 1.093 0.0253 0.378 0.0125 1.000 0.4567 0.4480 140

160 0.938 0.0264 0.383 0.0130 1.001 0.4626 0.4490 160 180 0.813 0.0275 0.388 0.0135 1.003 0.4699 0.4500 180 200 0.717 0.0286 0.392 0.0140 1.006 0.4791 0.4510 200 212 0.668 0.0293 2.93 0.394 0.0143 14.3 1.008 0.4859 0.4516 0.4892 212 220 0.638 0.0295 2.97 0.395 0.0145 14.4 1.009 0.4902 0.4521 0.4867 220

240 0.574 0.0306 3.08 0.397 0.0151 14.7 1.013 0.5019 0.4532 0.4819 240 260 0.521 0.0316 3.19 0.398 0.0158 15.3 1.017 0.5157 0.4544 0.4780 260 280 0.476 0.0326 3.30 0.398 0.0165 15.8 1.022 0.5319 0.4556 0.4750 280 300 0.439 0.0335 3.41 0.397 0.0172 16.4 1.029 0.5508 0.4569 0.4729 300 320 0.406 0.0345 3.52 0.396 0.0181 17.0 1.040 0.573 0.4582 0.4714 320

340 0.379 0.0354 3.63 0.394 0.0190 17.5 1.053 0.598 0.4595 0.4706 340 360 0.355 0.0363 3.74 0.391 0.0199 18.1 1.066 0.626 0.4609 0.4703 360 400 0.315 0.0382 3.96 0.383 0.0222 19.4 1.085 0.695 0.4638 0.4709 400 500 0.251 0.0432 4.50 0.349 0.0306 22.6 1.180 0.968 0.4715 0.4745 500 600 0.200 0.0510 5.05 0.296 0.0486 26.2 1.528 1.71 0.4798 0.4817 600

700 0.121 0.077 5.60 0.188 0.1203 29.8 0.4885 0.4895 700 706 0.101 0.101 5.63 0.139 0.139 30.0 0.4890 0.4900 706' 800 6.15 33.7 0.4973 0.4977 800

1000 7.24 41.8 0.5150 0.5152 1000

Critical Temperature. Tabulated properties ignore critical region effects. ;Actual value - (Table value) x (Indicated multiplier).

261

APPE\IPx

References:

[1] Engineering Thermodynamics, Reynolds and Perkings, Second

Edition, McGraw -Hill, 1977.

[2] Fundamentals of Heat and Mass Transfer, Incopera and DeWitt,

Second Edition, John Wiley & Sons, 1985.

[3] Solar- Thermal Energy Systems, Howell, Bannerot, and Vliet,

First Edition, McGraw -Hill, 1982.

[4] Introduction to Fluid Mechanics, Fox and McDonald, Third

Edition, John Wiley & Sons, 1985.

[5] Heating, Ventilating, and Air Conditioning, McQuiston and

Park, Third Edition, John Wiley & Sons, 1988.

[6] ASHRAE Handbook Fundamentals, ASHRAE, 1985.

[7] Architecture Graphic Standards, Ramsey and Sleeper, Eighth

Edition, AIA, John Wiley & Sons, 1988.

62