calculation of thermal loads
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SPACE LOAD CALCULATIONS
Abdullah Nuhait, PhD
King Saud University
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SPACE LOAD CALCULATIONS
Cooling Loads
Heating Loads
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Important Question
Why Does One Need To Compute
Thermal Loads?
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One Answer
In Order to Select Optimum Size of
Cooling and/or Heating Equipment
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Design of Acceptable Air Conditioning
System depends on good estimate of heatgain or heat loss in space to be
conditioned
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Precise Calculation of Heat Transfers
are Difficult Because of
Walls and roofs are complex assemblies ofmaterials
Windows are made of two or more layers ofglass with air space
Windows usually have drapes or curtains
In basement, floors and walls are in contact withthe ground
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Solutions
Experience and experimental data make
reliable estimates possible
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MODES OF HEAT TRANSFER
Modes of heat transfer are conduction, radiation andconvection
The three modes occur simultaneously
All three modes are important in heat gain and heat losscalculations in building structures
They will be considered separately for clarity and ease ofpresentation
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Thermal conduction
Thermal conduction: a mechanism of heat by whichenergy is transported between parts of continuum
q = - k A dt/dx
q = heat transfer rate, Btu/hrk = thermal conductivity, Btu/(hr-ft-F)
A = area normal to heat flow, ft2
dt/dx = temperature gradient, F/ft
note: Equation has negative sign because q flows inpositive direction of x when dt/dx is negative
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Conduction -cont.
Consider flat wall with uniformtemperature t1 and t2 on each surface
q = - k A (t2 - t1)/(x2 - x1)= - (t2 - t1 )/R
Where R is thermal resistance defined by
R = (x2 - x1)/k A = x / k A
Thermal resistance for unit area ofmaterial is very commonly used inHVAC literature
It is called R-factor R = x / k
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R-factor
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Thermal Resistance of Composite Building
Element Thermal resistance R is analogous to electrical resistance
q is analogous to electrical current
t2 - t1 is analogous to electrical potential difference in Ohms law
This analogy provides a very convenient method of analyzing wallmade up of two or more layers of dissimilar material
For wall constructed of three different materials
R= R1 + R2 + R3= x1 / k1A 1 + x2 / k2A 2 + x2 / k2A 2
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Three Different Wall Layers
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Thermal Conductivity Tables
ASHRAE constructed Tables that give
thermal conductivity k for wide variety ofbuilding and insulating materials
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Thermal convection
Thermal convection: transport of energy by mixing inaddition to conduction
q = h A (t t wall)
h = convective heat transfer coefficient, Btu/(hr-ft2
-F)
q = (t t wall)/R
R = 1/h A and R = 1/h
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Convection -cont.
Most building structures have forced
convection along outer walls or roofsand natural convection occurs inside
narrow air spaces and on inner walls
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Thermal radiation
Thermal radiation: transfer of thermal energy by
electromagnetic waves
it is different phenomenon from conduction
Occurs in perfect vacuum
Reduced by intervening medium
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Radiation -cont.
Direct net transfer of thermal energy by radiation betweentwo surfaces that see only each other and that are
separated by non-absorbing medium is given by
q12 = (T1- T2)/ ( (1-1)/A11 + 1/A1F12 + (1-2)/A22 )
= Botzmann constant, 0.1713x10-8 Btu/(hr-ft2-R4)
T = absolute temperature, R
= emittanceA = surface area, ft2
Both surfaces are gray (emittance equals absorptance )
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Two situations where the radiation may
be significant
One situation is found with single surfaceof walls, floors, or roofs
The other is found with two surfaces suchas the false ceiling and the real slab,
called air space
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Single surface
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Two surfaces (air space)
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Combined Resistance
Convective resistance is combined with
fictitious radiative resistance resulting intotal convective resistance that depends
on emittance
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Combined Resistance Tables
Tables give emittance and thermal
resistance
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Overall heat transfer coefficient U
Overall heat transfer coefficient,U, is computed by finding totalresistance, R t , of all
components of wall
U = 1/R A
= 1/R
R = 1/ho + x1/ k1 + x2 / k2
+ x3/ k3 + 1/hi
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Heat transfer rate is given by:
q = U A tU A = conductance, Btu/(hr-F)
A = surface area, ft2
t= overall temperature difference, F
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Tables for U
For convenience of designer, tables have
been constructed that give overall heattransfer coefficients, U, for many common
building sections including walls, roofs,
doors, windows, and skylights
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Heating Load
Estimate of heat loss of each room to be
heated
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Two kinds of heat losses from space
Heat transmitted through walls, ceilings,
floors, roofs, windows, and/or othersurfaces to outside
Heat required to warm outdoor air entering
the space
f
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Transient Heat Transfer
Actual heat loss problem is transient
because of variation of:
Outdoor temperature Wind velocity
Sunlight
O td T t
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Outdoor Temperature
During coldest months, sustained periods
of very cold, cloudy, and stormy weatherwith relatively small variation in outdoor
temperature may occur
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St d St t H t T f
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Steady State Heat Transfer
For design purposes heat loss is
estimated from steady state heat transfer
St f C ti H t l
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Steps for Computing Heat loss
Select outdoor design conditions:
Temperature Humidity
Wind direction and speed
St f C ti H t l
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Steps for Computing Heat loss -cont.
Select indoor design conditions to be
maintained
Estimate temperature in any adjacentunheated spaces
St f C ti H t l
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Steps for Computing Heat loss -cont.
Select the transmission coefficients
Compute heat losses for walls, floors,
roofs, windows etc.
St f C ti H t l t
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Steps for Computing Heat loss -cont.
Compute the heat load due to infiltration
Compute the heat load due to ventilationair
Sum all the losses to obtain the total heatloss by transmission and infiltration
Tables of Recommended Outdoor Air
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Tables of Recommended Outdoor Air
ASHRAE STANDARD 62-1989
gives recommended amount of outdoor air
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Outdoor Design Conditions
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Outdoor Design Conditions
Ideal heating system would provide
enough heat to match heat loss fromstructure
Outdoor Design Conditions cont
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Outdoor Design Conditions -cont.
Weather conditions vary considerably fromyear to year
Heating systems designed for worst
weather conditions on record would have
great excess of capacity most of time
Outdoor Design Conditions cont
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Outdoor Design Conditions -cont.
Failure of system to maintain design
conditions during brief periods of severeweather is usually not critical
Outdoor Design Conditions cont
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Outdoor Design Conditions -cont.
Outdoor air design conditions should be
based on official weather station record forwhich hourly observations were available
for past 12 years
For SUMMER:
Outdoor Design air conditions should be1% value
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Indoor Design Conditions
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Indoor Design Conditions
Indoor air design conditions depends on
activity of occupants and their dresses
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Comfort Zone
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Comfort Zone
Infinite conditions on psychrometric chart
which 80% of people are satisfied
One can select 75 F db and relativehumidity of 20 to 50%
Heat Loss Calculation
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Heat Loss Calculation
Heat transmission: walls, ceilings, roofs,
windows
q s = U A (t i t o)
Outdoor Air Heat Loss
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Outdoo eat oss
Sensible heat is given by
q s = 1.08 Q o (ti to)
Latent heat is given by
q l = 4700 Q o (W i W o)
Calculation of Infiltration
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One method is called crack method
It is based on characteristics of windows
and doors and on pressure difference
between inside and outside
Calculation of Infiltration -cont.
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Calculation of Infiltration cont.
Second method is called air change
method
It is based on assumed number of air
changes per hour for each room
It depends on number of windows and
doors
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Calculation of Infiltration -cont.
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Accuracy of predicting air infiltration is
restricted by limited information on airleakage characteristics of many
components that make up structure
Pressure differences are also difficult to
predict because wind conditions vary
Calculation of Infiltration -cont.
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Air Change Method Will be Used in
Presentation
Auxiliary Heat Sources
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y
Heat energy supplied by people, lights and
equipment should be estimated, but anyactual allowance for these heat sources
requires careful consideration
Intermittently Heated Structures
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y
Enlarge Heating Equipment Capacity
when structure is not heated oncontinuous basis
In order to raise temperature to
comfortable level within reasonable period
of time
Supply Air Requirement
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Supplied Air quantity is computed from
qzone = 1.08 Qsa (tsa tr)
Supplied Air quantity for each room should be apportionedaccording to heating load
Qrn = Qsa (qrn/qzone)
This is necessary for design of duct system and selection of air outlets
Cooling Load
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Solar Radiation has important effect on coolingload of building
Solar Radiation
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Solar radiation effect depends on
location of sun in sky
Clearness of atmosphere Nature and orientation of building
Solar Radiation -cont.
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Earth rotates about its own axis
Fixed location on earth's surface goesthrough 24 hour cycle in relation to sun
Earth is divided by longitudinal lines
passing through poles into 360 degrees of
circular arc
Solar Radiation -cont.
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Point on earth's surface exactly 15
degrees west of another point will see sun
in exactly same position as first point after
one hour of time has passed
Universal time (Greenwich civil time) istime along zero longitude line passing
through Greenwich, England
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Solar Angles -cont.
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Solar Angles -cont.
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These quantities are latitude, sun'sdeclination and hour angle
Latitude angle represents location of
observer on the earth
Solar Angles -cont.
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Other angles:
Solar altitude
Solar azimuth
Wall azimuth
wall solar azimuth
Tilt angle Angle of incidence
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Solar Irradiation -cont.
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Mean solar constant
Gsc = 433 Btu/(hr-ft2)
Irradiations from sun vary (3.5% ) Due to
variation in distance
Solar Irradiation -cont.
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Precise value of solar constant is not used
in most HVAC calculations
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Solar Irradiation -cont.
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Normal Direct Radiation:
Part of radiation that is not scattered orabsorbed and reaches earth's surface
Solar Irradiation -cont.
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Diffuse Radiation:
Radiation that has been scattered orreemitted
Radiation may also be reflected onto
surface from nearby surfaces
Solar Irradiation -cont.
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Total irradiation on surface normal to sun's
rays is thus made up of normal direct
irradiation, diffuse irradiation and reflected
irradiation
Heat Gains Through Glass
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Total Heat Admission Through Glass is
equal
Radiation transmitted through glass
Inward flow of absorbed solar radiation
Conduction heat gain
Heat Gains Through Glass -cont.
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Heat Gains Through Glass -cont.
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Heat gain through simplest window is
complicated because of:
window is finite in size
It is framed
Sunlight striking it does so at varying
angles through day
Heat Gains Through Glass -cont.
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Simplified solar heat gain calculations aredone for double-strength sheet glass
(Reference Glass) It is called solar heat gain factor
Heat Gains Through Glass -cont.
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It takes into account combined effects of
transmitted solar heat gain and absorbed
solar heat gain conducted into space
Shading coefficient
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Solar Heat Gain through any given window
Is equal to
Solar Heat Gain Factor
Times
shading coefficient of given window
Shading coefficient -cont.
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Value of shading coefficient is between
zero and one
Value SC are given by glass manufacturer
Following Tables give SC
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Cooling Load
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Larger number of variables are considered in
making cooling load calculations than in heating
load calculations
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Cooling Load -cont.
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Design for cooling, transient analysis must
be used if satisfactory results are to be
obtained
Instantaneous heat gain is quite variable
with time because of hourly variation insolar radiation
Cooling Load -cont.
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Appreciable difference between heat gain ofstructure and heat removed by cooling
equipment
This difference is caused by storage and
subsequent transfer of energy from structureand contents to circulated air
Cooling Load -cont.
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If this is not taken into account
Cooling and Dehumidifying Equipment will
be grossly oversized
Cooling Load -cont.
It i i t t t diff ti t b t h t i
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It is important to differentiate between heat gain,cooling load, and heat extracted rate
Heat gain is rate at which heat is transferred toor generated within space
Cooling load is rate at which heat energy mustbe removed to maintain temperature andhumidity at design values
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Differences Between heat gains
and cooling loads
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Some of heat gains absorbed by structureand contents and appear as cooling load
later Interior surfaces and other objects are
cooled by convection when they attain
temperature higher than room air
Differences Between heat gains cooling
loads -cont.
H t t h t i ti f t t
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Heat storage characteristics of structure
govern relationship between heat gain and
cooling load
Radiant heat transfer inputs to structure
are delayed while convective and latentinputs are not
Heat Extracted
Heat extracted rate is equal to cooling load when space
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Heat extracted rate is equal to cooling load when spaceconditions are constant and equipment are operating
This is rarely true because of:
Some fluctuation in room temperature is necessary forcontrol system to operate
Cooling load is below design value most of time
Therefore, variable operation of cooling equipment isrequired
Sources of Heat Gains
Solar radiation
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Solar radiation
Heat conduction through exterior surfaces
Heat conduction through interior surfaces
Heat generated by occupants, light, and
equipment Ventilation and infiltration air
Other latent heat gains generated within space
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Steps for Computing Cooling Load -cont.
Obtain characteristics of structure from
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Obtain characteristics of structure fromplans and specifications
Determine building location, orientation,and external shading
Determine schedule of lighting,occupancy, and other sources of internalheat gains
Estimate time of day, day, and month forcalculations
Steps for Computing Cooling Load -cont.
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Computing Heat Gain due to
Solar heat gain for glass, walls, and roof
Conduction through interior partitions,
ceilings, and floors because of
temperature differential
Steps for Computing Cooling Load -cont.
Computing Heat Gain due to
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Computing Heat Gain due to
Heat sources within the conditioned space
Infiltration
Other latent loads
Convert heat gains to cooling loads
Heat balance Method
Estimation of cooling load for space involves
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Estimation of cooling load for space involvescalculating surface-by-surface conductive,
convective, and radiative heat balance for eachroom surface and convective heat balance forthe room air
This will ensure all energy flows in each zoneare balanced and involved solution of set of
energy balance equations for zone air and theinterior and exterior surfaces of each wall, roof,and floor
Representation of heat balance
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Heat Balance Method -cont.
These energy balance equations are
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These energy balance equations are
combined with equations for transient
conduction heat transfer through walls androofs and data for weather conditions
including outdoor air dry bulb temperature,wet bulb temperature, solar radiation, and
so on
Internal heat gains
Internal heat gains such as people, lights, and
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Internal heat gains such as people, lights, and
equipment are often significant component of
cooling load in commercial and institutionalbuildings
In fact, for many large office buildings, internal
heat gains are dominant source of cooling load
Heat Gain from People
Heat gain from people has two components:
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Heat gain from people has two components:
sensible and latent
Total and proportions of sensible and latent heat
vary depending on level of activity
Heat gain data from occupants in conditioned
spaces are given in following table
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Heat Gain from People -cont.
sensible heat gain for people generally is
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sensible heat gain for people generally is
assumed to be 30% convective (instant
cooling load) and 70% radiative (delayedportion)
Heat Gain from Lights
Source of heat from lighting comes from
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g g
lamps
Instantaneous rate of heat gain from
lighting may be calculated using
q = 3.41 W FWhere
W = Total installed light wattage
F = Special allowance factor
Heat Gain from Lights -cont.
Heat gain to space from fluorescent
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g p
fixtures is often assumed to be 59%
radiative and 41% convective
Heat gain to space from Tungsten fixturesis often assumed to be 80% radiative and
20% convective
Miscellaneous Equipment
Estimates of heat gain for miscellaneousequipment more subjective than for people and
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equipment more subjective than for people andlights
However, considerable data are available
sensible heat gain generally is assumed to be30% convective (instant cooling load) and 70%radiative (the delayed portion) similar to people
Several Examples to be Presented
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Several example will be presented to show
load calculations using computer software
Computer software computes cooling load
using heat balanced method
Description of Computer Software
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Computer software will be described and
through examples one can see how it
works
Users Guide is supplied with a copy of CD
Examples to be Presented
Example # 1: Three different west walls
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without windows
Example # 2: West wall with three different
windows
Example # 3: Three different Roof withoutskylights
Example # 3: Roof with three differentskylights
Examples to be Presented -cont.
Example # 5: People with different
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schedule
Example # 6: light with different schedule
Example # 7: Infiltration with different rates
Example # 8: Ventilation with different
rates
Examples to be Presented -cont.
Example # 9: Large mosque in Riyadh
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If Time Is Available Example # 10 Will Be
Presented: Small Commercial Office
Building
Steps Executed by Software
First: all zone parameters such as surface areas,thermal properties etc are determined
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thermal properties etc are determined
Second: all temperature-independent quantitiessuch as transmitted and incident solar radiation,internal loads, infiltration rates etc are
determined for each hour
Third: surface temperatures are determined
within nested loop that repeats the day untilsteady periodic solution is achieved
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Description of buildings usedin Examples
Example # 1
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Single Room Has:
Floor: 4 Concrete Slab, 200 ft2
West Wall, 200 ft2:
Case A: 8 Concrete Block (without insulation)Case B: 8 Concrete Block (with insulation)
Case C: Wood and 6 Insulation
Example # 2
Single Room Has:
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Floor: 4 Concrete Slab, 200 ft2
West Wall, 200 ft2, 8 Concrete Block: (un-
insulated)
Case A: Window Single sheet, 20 ft2
Case B: Window Single sheet, 100 ft2 Case
C: Window Single sheet, 180 ft2
Example # 3
Single Room Has:
Fl 4 C t Sl b 200 ft2
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Floor: 4 Concrete Slab, 200 ft2
Partition Wall, 200 ft2:
Roof 200 ft2:
Case A: Steel Decking (Un-insulation)Case B: Concrete Slab (with insulation)
Case C: Steel Decking (with insulation)
Example # 4
Single Room Has:
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Floor: 4 Concrete Slab, 200 ft2
Partition Wall, 200 ft2:
Roof 200 ft2: Conc. Slab insulation
Case A: Skylight Single sheet, 25 ft2
Case B: Skylight Single sheet, 50 ft2 Case
C: Skylight Single sheet,100 ft2
Example # 5
Single Room Has:
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Floor: 4 Concrete Slab, 200 ft2
Partition Wall, 200 ft2:
People, 20 persons:
Case A: 24 Hours Occupancy
Case B: 8 Hours Occupancy
Case C: 2 Hours Occupancy
Example # 6
Single Room Has:2
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Floor: 4 Concrete Slab, 200 ft2
Partition Wall, 200 ft2:
Lighting, 2.5 Watt/ft2:
Case A: 24 Hours, Turned on
Case B: 8 Hours, Turned on
Case C: 2 Hours, Turned on
Example # 7
Single Room Has:
C S f 2
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Floor: 4 Concrete Slab, 200 ft2
Partition Wall, 200 ft2:
Equipment, 35 Watt/ft2:
Case A: 24 Hours, Operating
Case B: 8 Hours, Operating
Case C: 2Hours, Operating
Example # 8
Single Room Has:
Floor: 4 Concrete Slab 200 ft2
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Floor: 4 Concrete Slab, 200 ft2
Partition Wall, 200 ft2:
Infiltration: Continuous
Case A: 6 Air change/hour Case B: 3 Air change/hour
Case C: 0.6 Air change/hourSpace Load
Example # 8
Ventilation : Continuous
C A 6 Ai h /h
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Case A: 6 Air change/hour
Case B: 3 Air change/hour
Case C: 0.6 Air change/hour
Coil Load
Example # 9
Large Mosque located in Riyadh
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One Story Building (One Room) Has: Floor: 4 Concrete Slab, 12000 ft2
North and South Walls, Each: 6000 ft2: 8
Concrete Block Un-insulated With 600 ft2Window Single Sheet
East and West Walls, Each: 8000 ft2: 8Concrete Block Un-insulated With 800 ft2
Window Single Sheet
Example # 9 -cont. : Plan of The Mosque
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Example # 9 -cont.
Roof: 12000 ft2: 6 Concrete Un-insulated With
300 ft2 Skylight Single Sheet
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300 ft Skylight Single Sheet
Lighting: 2.5 Watt/ft2 Fluorescent on for 24 Hoursand 1 Watt/ft2 Tungsten On for 24 Hours
People: 1200 persons 2 Hours Occupancy
Ventilation : 3 Air change/hour for 2 hours
Example # 10
One story small commercial building
located in Riyadh
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located in Riyadh
Adjoining buildings on north and west arenot conditioned (air temperature within
them is approximately equal to outdoor airtemperature at any time)
Example # 10 -cont. : Plan of Small OfficeBuilding
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E
S N
W
Example # 10 -cont.
South wall construction: 4 light-color face brick,
8 common brick, 0.625 plaster, 0.25 plywood
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p p y
panel glued on plaster East wall and outside north wall construction: 8
light-colored heavy concrete block, .625 plaster
on wall
West wall and adjoining north part wall
construction: 13 solid brick, no plaster
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Example # 10 -cont.
Door construction: 5x7 ft2 light-colored 1.75steel door with solid urethane core thermalbreak
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break
Occupancy: 85 office workers from 8:00 to 17:00
Lights: 17500 W, fluorescent, operatingfrom 8:00 to 17:00 hours daily, along with4000 W tungsten, operated continuously,lighting fixture are non-ventilated type
Equipment: none
Ventilation: 15 cfm/person
Example # 10 -cont. : Calculation ofInfiltration
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Infiltration through windows should beconsidered zero (windows are sealed)
Infiltration through wall surfaces should be
neglected
Infiltration through doors: calculation of door
infiltration require some judgment
Example # 10 -cont. : Calculation ofInfiltration
Assume that outside and inside doors are
frequently opened simultaneously
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Use of outside doors: 10 persons/hour Use of inside doors: 30 persons/hour
Assume 100 ft3 of air per person per door
passage
Qo =(30+10)(100 ft3)/(60 minutes) = 67 cfm
Example # 10 -cont. : Calculation ofVentilation
Ventilation = 85 persons x 15 cfm / person
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Ventilation 85 persons x 15 cfm / person
Qoa = 1275 cfm
Ventilation > Infiltration
Use only Ventilation
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