lecture8.pdf

1

Environmental Control

Systems I- Temperature and

Humidity, Arch 353

Lecture # 8

Dr. Hussain Alzoubi

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Total energy in spaces:

1- Solar radiation from windows

2-Heat gain by conduction from walls and windows

3-Appliances and devices

4-Human activities.

-In winter: solar radiation contributes positively

to heating loads in spaces.

-In summer, solar radiation contributes negatively

to cooling loads.

-Heat transfer through walls increases heating loads

in winter, and cooling loads in summer.

3

Example of software for energy calculations and

Thermal comfort:

Ecotect (http://usa.autodesk.com/ecotect-analysis/)

Enegyplus(http://apps1.eere.energy.gov/buildings/energyplus/)

DEROB-LTH(http://apps1.eere.energy.gov/buildings/tools_directory/software.cfm/ID=211/pagename=alpha_list

Energy10(http://apps1.eere.energy.gov/buildings/tools_directory/software.cfm/ID=36/pagename=alpha_list

Trnsys (http://apps1.eere.energy.gov/buildings/tools_directory/software.cfm/ID=58/pagename=alpha_list

Ecotect Tutorialecotect tutorial

http://www.youtube.com/watch?v=JxsOPbjjiHs

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How to convert solar radiation into heat:

1- From tables, find the total incident energy

on the targeted window.

2-Specify the transmittance factor of the window glass.

3-The net solar heat raises the temperature of the air mass as follows

Solar Heat= m c T or:

Solar Heat= m c (T2-T1)

Where :

m : mass of the heat storage (lb or kg)

c : specific heat

T1 : temperature before mass heated by solar heat (c or F

T2 : temperature after mass heated by solar heat

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The largest

6

H= Solar heat,

Wh, or

Calorie, or

Joule,

Btu,…etc

Thermal

storage

Mass, mSpecific heat, c

The temperature of the thermal storage will be

Raised by T;

T = H/mc

For example :

T = Joule / Kg . Joule/Kg. C

T = Joule / Kg . Joule/Kg. C

7

Example :

A 20 ft x 20 ft x 10 ft room has a south window of 20 ft2

with a transmittance factor of 90%,

if the total solar radiation on south

façades in one day is 200 Btu/ft2. How much air temperature

change will be attributed to this solar energy.(assume :

1-No heat loss through walls

2- No infiltration

3-The room is empty and the solar energy will be embodied in

the air mass. (assume air density is 1 kg per 1m3).

Solution:

Solar heat = 20 x 0.90 x 200= 3600 Btu

Air mass = [20 x 20 x10/35.28]*1 kg =113.337kg= 249 lb.

Specific heat from table 3-1 is 0.24Solar Heat= m c T

3600 = 249 x 0.24 x T

T = 3600/ (249 x 0.24) = 60F

Air mass ? Depends on:

-Humidity

- Air temp.

- Altitude

(above sea level)

http://www.denysschen.com/catalogue/density.asp

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• Temperature Degree Celsius,

Centigrade, [ºC]

Kelvin [ºK]

Fahrenheit(English units)

• Energy = Joule, [J]

= Calorie [cal]

1 [cal] = 4.2 [J]

KWh,

Btu (English units)

• Power = Rate of Energy flow = Watts,

[w], [J/s]

Examples of thermal units and energy

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Example : in the following example, you are

given the energy required to heat the cylinder

and you need to find the required rate of energy

Per unit time (power), if you want to heat the

cylinder in 1 hour.

20

Temp

( °C )

Density

pure

water

( g/cm3 )

Density

pure water

( kg/m3 )

Density

pure

water

lb/cu.ft

0 (solid) 0.9150 915.0 -

0 (liquid) 0.9999 999.9 62.42

4 1.0000 1000 62.42

20 0.9982 998.2 62.28

40

0.9922

992.2 61.92

60 0.9832 983.2 61.39

80 0.9718 971.860.65

For heat storage, you need some material densities:

highest

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Heat transfer by conduction

25

1

w/ºC.sq

m

25

26

=UA(T1-T2)

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•In Series

When materials are placed in series, their thermal resistances are added so

that the same area will conduct less energy for a given temperature

difference. An example of this is a cavity-brick wall, with two layers

of brick, an air gap and 12mm of plasterboard - all in series.

Heat transfer by conduction

•In Parallel

When materials are placed in parallel, their thermal conductance are added

and the total energy flow is increased for a given temperature difference.

An example of this would be a cavity-brick wall with a window inserted

within it.

Less conductivity

with multi-layers

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The rate at which heat flows through a homogenous

material under steady-state conditions is given by:

Q =A . T/R OR Q =A C T (C = 1/ R )where:

Q = the resultant heat flow (Watts or Btu/unit time)

A = the surface area through which the heat flows (m² or ft²)

T = the temperature difference between the warm and cold sides

of the material (K , C, or F), and

R = the thermal resistance per unit area of the piece of

material (m²K/W).

C: conductance of material ((W /m²K).

For a composite building element made up of a number of layers

of different materials, its total resistance is given as:

Rt = Rso + SRn + Rsi

where the resistance of the nth layer is:

Rn = tn/kn

where:

Rt = the total overall resistance of the element (m²K/W),

Rn = the resistance of the nth

material within a composite element (m²K/W),

tn = the thickness of the nth

material in a composite element (m), and

kn = is the conductivity of the nth

material in a composite element.

Rso : the resistance of the outside air film

Rsi : the resistance of the inside air film

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Example1: a wall is composed of two common brick layers of Thickness =

15 cm separated by an air gap of 5cm (air at rest). Find the U-value of the

wall (neglect the air films)

Solution :

From table 3-4,

-The K-value of brick is 0.72 W/(m2 ) (C/m)

-The K-value air is 0.024 W/(m2 ) (C/m)

Accordingly,

-The R-value for each brick layer is Rn = tn/kn

= 0.15 / 0.72

= 0.208 m2 C /W

The R-value for the air gap is Rn = tn/kn

= 0.05 / 0.024

= 2.08 m2 C /W

Rt = Rbrick 1 + Rbrick2 + Rair

Rt = 0.208 + 0.208 + 2.08 = 2.5 m2 C /W

U-value = 1/Rt = 1/(2.5) = 0.4 W/ m2 C

Example2: if the wall in example 1 has an area of 10 m2 and separates two

rooms with a temperature difference between them, T=5 C , what is the

rate of heat transfer through this wall in Watt.

Q = AU T = 10 m2 X 0.4 (W/ m2 C) X 5 C = 20 Watt.

b) What is the heat transfer in 10 hours

H = Q (Watt) x T(hour) = 20 X 10 = 200 Watt-hour

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-Concept of conductivity(K), conductance(C), and U-value1- All of them are values for material conductivity

2-Conductivity is the ability of a material to conduct heat from side to

side per unit thickness (inch or meter).

3-Conductance is the ability of a material to conduct heat from side to

side for a specific thickness of a specific material.

4- U-value is a measure of a building elements ability to conduct heat;

the higher the U Value, the greater the conduction of heat ( it measures

the rate of heat transfer through multi-layer walls.)

Accordingly: the heat transfer through walls

could be expressed as follows:

Q = A* T * K /t ( for a wall made of one homogenous material)

Q = A* C* T (for a wall made of one homogenous material with a specific thickness)

Q = A* U* T (for a composite wall made of different materials)

Q = A T/R

Where :

Q: rate of heat transfer through a wall

A : area of wall

K: conductivity

T: Temperature difference between the two sides of the wall

t: thickness of wall

C: conductance

U : U-value of a multi-layer wall.

Notice that :

U-value = 1/(R1+R2+……..Rn)

but NOT EQUAL TO (1/R1) + (1/R2) +…. (1/Rn)

If you have a multi layer wall with specific conductance for each layer,

C1, C2, …Cn, then the U-value = 1/ [(1/C1) + (1/C2) + … (1/Cn)].

but NOT EQUAL TO C1+ C2 + …Cn

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Remember:

Air and water are bad in conduction and good

in convection. Still air can be used as a good insulator.

There are three ways of heat transfer:

1-Radiation (example : solar radiation)

2-Convection (heating by water and air: both of them

are carried from heat source to other spaces.

3-Conduction ( example: heat transfer through walls)

EXAMPLE

The brick conductivity, K, is 0.72 W/(m2)(C/m) and the air conductivity,

K, is 0.024 W/(m2)(C/m). You are required to design a wall with two layers of

brick and air cavity. If each layer of brick has a thickness of 10 cm,

find the minimum width of the air cavity so that you keep

a U-value of 0.5 W/(m2)(C). (Neglect the air films on both sides).

Solution:

Assume air gap is (Ag) m

Rtotal = R(air) + R brick1+R brick2

Rtotal = (Ag/0.024) +(0.1/0.72) +(0.1/0.72)

As Rtotal = 1/U-value, Rtotal = 1/0.5 = 2

2 = Ag/.024 + 0.2/.72

Ag = 0.024(2-1/3.6)

Ag= .0413 meter, or 4.13 cm.

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HEAT,Q = UA(Ti-To)

For varying indoor Temperature:

dQ= U(Ldy)(Ti-To)

Ti is a function of y

In this example:

Ti= 70+(5/h)y

Then dQ = UL(70+(5/h)y-To) dy

HEAT (Q)=UL(70+(5/h)y-To) dy

Assumptions:

1-(To) is constant with

height.

2-Inside temperature

has linear relation

with height y.

L

dy

h

70

75

Outside

wall

section in a wall

T=

T=

Btu or joule

0

h

34

L=6m

H=3m

20c

21.5c

T Outside

=10 c

wall

section in a wall

Ti=

Ti=

Example 1:The internal temperature at a certain wall varies from 20 C at

the skirt of the wall to 21.5 C at the highest point. if the width of the wall is 6 m and the height

is 3 m, find the heat loss through this wall, (outside temperature is 10 C

and U-value of this wall is 1.2 W/ m2 C )

Solution:

Ti= 20+0.5y, y measured in meter.

dQ = UL(20+0.5y-To) dy

HEAT Loss (Q)=UL(20+0.5y-To) dy

HEAT Loss (Q)=1.2*6 (20+0.5y-10) dy

=7.2 (0.5y+10) dy

=(3.6y +72) dy

0

3

0

30

3

= [1.8y2 +72y]

= [1.8(3)2 +72(3)]- [1.8(0)2 +72(0)]

=232.2 Watt OR Joule/second

3

0

35

L=6m

H=3m

20c

21.5c

T Outside

=10 c

wall

section in a wall

Ti=

Ti=

Example 2: (using average temperature instead of integration)

The internal temperature at a certain wall varies from 20 F at

the skirt of the wall to 21.5 F at the highest point. if the width of the wall is 6 m and the height

is 3 m, find the heat loss through this wall. (U-value of this wall is 1.2 W/ m2 C

Alternative Solution:

Average Ti= (20+21.5)/2 = 20.75 C

Q = UA(Ti-To)

HEAT Loss (Q)= 1.2 * (6X3)(20.75-10)

= 1.2 * 18*10.75

= 232.2 Watt OR Joule/second

(the same answer)

Average = 20.75

Area= 18 m2

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Example 2

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Q=312x24(155-154.85)/0.18

=6240 Btu/hour

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Time lag(1)The time delay due to the thermal mass is known as a time lag.The thicker and more

resistive the material, the longer it will take for heat waves to

pass through. The reduction in cyclical temperature on the inside surface compared

to the outside surface is knows and the decrement. Thus, a material with a decrement

value of 0.5 which experiences a 20 degree diurnal (daily) variation in external

surface temperature would experience only a 10 degree variation in internal

surface temperature.

Decrement factor = Ti (max) / To (max)

1- http://new-learn.info/learn/packages/clear/thermal/buildings/building_fabric/properties/time_lag.html

47

Time

51

Door area is 70 ft2

From table No. of casements(length)

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A 4m (W) x 5m (L) x 3m (H) room has an average U-value as follows:

U-value for walls = 0.8 W/(m2)(C)

U-value for roof = 0.5 W/(m2)(C)

Heat loss rate for floor = 6.3 W/(m2) (or 2 Btu/hour/ft2, see lecture 8)

U-value for windows = 1.2 W/(m2)(C)

U-value for door = 0.45 W/(m2)(C)

The area of the door is 2m2 and the area of the window is 1.5 m2.

The indoor temperature is 20 C for all walls and the outdoor temperature is –5 C.

1-Find the rate of heat loss through all surfaces of the room in Watt.

(Neglect heat loss from edges and from opening cracks).

2-Find the total heat loss from 8 am to 5 pm during a workday.

Solution:

-Find area of surfaces: as follows

Area of walls = 2x(4x3)+2x(5x3) – 2 (door) – 1.5(window) = 50.5m2

Area of roof = 4x5 = 20 m2

Area of floor = 4x5 = 20 m2

Area of door = 2 m2

Area of window = 1.5 m2

Heat loss from walls = AUT = AU(Ti – To) = 50.5 x 0.8x(20-(-5)) = 1010 Watt

Heat loss from roof = AUT = AU(Ti – To) = 20 x 0.5x(20-(-5)) = 250 Watt

Heat loss from floor = A X F = 20 x 6.3 = 126 Watt

Heat loss from door = AUT = AU(Ti – To) = 2 x 0.45x(20-(-5)) = 22.5 Watt

Heat loss from window = AUT = AU(Ti – To) = 1.5 x 1.2x(20-(-5)) = 45 Watt

1) The rate of heat loss from all surfaces is = 1010 +250+126+22.5+45 = 1453.5 Watt

OR(Joule/second)

= 1453.5 X 3.412 Btu/hour

2) The heat loss from 8 am to 5 pm = 1453.5 Watt x 9 hours = 13081.5 Watt-hour

= 13.081 Kilo Watt-hour

=13.0815x3412 Btu

5m

ft

4m

Plan

Example:

54

To calculate the heat gain:

1- Calculate all heat gain from windows by radiation

2-Calculate all heat gain from appliances and devices.

3-Calculate all heat from space occupants(human

being)

4-Subtract or add heat gain or loss by conduction

through Windows, walls, ceiling, and floor)

To minimize the heat loss through walls

1-Increase the R-value, decrease the U-value by

material choice or playing with material thickness.

2- For windows, you can increase the R-value by

using multi glass layer ( double pane windows or

triple pane windows, use weather stripped frames)

3-Use special gases between the glass layers.

(typically argon, krypton, or xenon)

55

Scaled R-value

Ti

To

OutsideInside

T1

T2

T3

T4

-Draw temperature profile to know the condensation

point

56

Thickness

Inside

Ti T1

T2

T3

To

T4

Outside

57

T4

T3

T2

T1

Inside Outside

To

Ti

Scaled R-value

To

Outside

Thickness

T1

Inside

Ti

T3

T4

T2

Rt

Ti-To

(Ti-To)/Rt = (Ti-Tx)/Rx

(Ti-To)(Rx/Rt)= Ti-Tx

Tx= Ti- (Ti-To)(Rx/Rt)

Rx

Tx

58

Thickness

To

T2

OutsideInside

Ti T1

glassglass

air

Double pane windows

59

Example:

Draw the temperature profile for the following

wall

C1:

C1=2Btu/(hr)(ft2)(F)


C3=0.3Btu/(hr)(ft2)(F)



Ti=72F

To= 32 F

c1 c2 c3 c4 c5

insulation

Air film

Air film

Solution :

Finding R-values for the wall layers as follows:

R1 = 1/C1 = 1/2 = 0.5

R2 = 1/C2 = 1/3 = 0.34

R3 = 1/C3 = 1/0.3 = 3.34

R4 = 1/C4 = 1/3 = 0.34

R5 = 1/C5 = 1/2 = 0.5

R1 R2 R3 R4 R5

0.50.5 3.34

0.34

0.34

Rt = .5 + .34+3.34+.34+.5 = 5 ft2F/(Btu/hr)

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2- draw R-values

R1 R2 R3 R4 R5

R-value plot

32F

72F

32F

42F

52F

62F

72F T1T2

T3

T4

3-Find the internal temperature by calculations:

Using this equation from the previous slide

Ti at any layer = Tx= Ti- (Ti-To)(Rx/Rt

T1 = 72 – (40)(0.5/5) = 68 F

T2 = 72 – (40)((0.5+0.34)/5) = 65 F

T3 = 72 – (40)(0.5+0.34 +3.34/5) = 39 F

T4 = 72 – (40)(0.5+0.34 +3.34+0.34/5) = 36 F

Example 2: if you have a room with

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Major greenhouse gases[5]

-Greenhouse gases are those gaseous constituents of the atmosphere,

both natural and anthropogenic(caused by human activity), that absorb and

emit radiation at specific wavelengths within the spectrum of infrared radiation

emitted by the Earth's surface

1-Water vapor (H2O)

2-carbon dioxide (CO2),

3-nitrous oxide (N2O),

4-methane (CH4) and

5-ozone (O3)

(Pollution increase it)

62

Wavelength and absorbance

-Some reflected waves cannot go through the gaseous layer

-It depends on wavelength of the reflected waves

The German physicist Wilhelm Wiens proposed the following equation:

63

Example

-as the temperature of the same waves become low

on the surface of the earth, the wavelength of the

reflected heat will be longer and will not go through

the layer of the greenhouse gases.

64

Glass pane acts as the greenhouse gases.

-a pane of glass is wavelength selective in that it

Transmits a great deal of the short-wavelength solar

radiation that strikes its surface , but absorbs most of the

long-wavelength radiations that emanate from low-temperature

surfaces inside buildings. So the glass is ideal material for trapping

solar energy.

Glass

Short waves

long waves

65

-90 percent of the solar energy is accounted between

The near infrared and ultraviolet waves.

66

Effect of heat on building materials

1- Materials expand if heated which causes

structural problems.

2- Expansion depends on

a) Coefficient of linear expansion of material

b) length of building element ( beam, wall, column..etc

c) temperature degrees

67

Effect of solar energy on structure

plan

elevation

68

-Effect of solar heat on structure1-temperature causes expansion and contraction

2- if the materials of the structure are different, the expansion causes cracks

column with walls

beams with slab

foundation with upper walls.

3-if the structure is long, the expansion causes cracks

4-if the wall has two layers with different temperature or different materials,cracks

happen if the wall is not reinforced.

plan

elevation

69

Effect of climate on structure

plan

elevation

70

Linear expansion and temperature in buildings

-depends on temperature

-structure length

-material type, expansion factor

L = L0 T

Where:

= the coefficient of linear expansion

L =the change in length caused by a temperature change T

L0= the length at the initial temperature

At T1

At T2L

71

Material per F per C

Aluminum 13X10-6 24X10-6

Glass 4 to 5 X10-6 7 to 9.5 X10-6

Concrete 6.0X10-6 11X10-6

Steel 6.7X10-6 12 X10-6

Table 1: Coefficient of linear expansion for some

selected Materials () [6]

Example: a 100 m concrete beam is built in

an arid area and has temperature variation between

summer and winter of 40C . Find the expansion of

this beam.

72

for Concrete = 11X10-6

L = L0 T

L = 11X10-6 x100x40

=4.4 cm

Solution:

73

T L

Material per C

Aluminum 24x10-6

Brass 18x10-6

Concrete 11x10-6

Copper 17.1 x10-6

Glass 7 to 9.5 x10-6

Steel 12x10-6

Table 3.2:Coefficient of Linear Expansion for Selected Building Materials

(Reference: Harris, Solar energy system design, page 64)

L L0

Cold weather

Hot weather

expansion

L = L0 T

Example: a 100 ft concrete wall is exposed to sunlight from 7 am to 4 pm

If the wall temperature was raised from 35F to 135F , how much it has

expanded in length (meter)?

Solution:

From table 3.2 below, of concrete is 11x10-6

T = (135 – 35 )x5/9 = 55.6 C

L0 = 100/3.28 = 30.5 meter

L = L0 T = 11x10-6 / C x 30.5mx 55.6 C = 0.01865m = 1.87 cm

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