Calculation of Thermal Loads

8/13/2019 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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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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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



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Failure of system to maintain design

conditions during brief periods of severeweather is usually not critical



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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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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



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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



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Precise value of solar constant is not used

in most HVAC calculations


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Normal Direct Radiation:

Part of radiation that is not scattered orabsorbed and reaches earth's surface



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Diffuse Radiation:

Radiation that has been scattered orreemitted

Radiation may also be reflected onto

surface from nearby surfaces



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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 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



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Simplified solar heat gain calculations aredone for double-strength sheet glass

(Reference Glass) It is called solar heat gain factor



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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



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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




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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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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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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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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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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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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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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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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

Calculation of Thermal Loads

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