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INTRODUCTION
Cooling & heating load calculations are normally made to size HVAC (heating, ventilating, and
air conditioning) systems and their components. In principle, the loads are calculated to
maintain the indoor design conditions. The first step in any load calculation is to establish the
design criteria for the project that involves consideration of the building concept, construction
materials, occupancy patterns, density, office equipment, lighting levels, comfort ranges,
ventilations and space specific needs. Architects and other design engineers converse at early
stages of the project to produce design basis & preliminary architectural drawings.
The design basis typically includes information on:
1) Geographical site conditions (latitude, longitude, wind velocity, precipitation etc.).
2) Outdoor design conditions (temperature, humidity etc.).
3) Indoor design conditions.
4) Building characteristics (materials, size, and shape).
5) Configuration (location, orientation and shading).
6) Operating schedules (lighting, occupancy, and equipment).
7) Additional considerations (type of air-conditioning system, fan energy, fan location, duct heat
loss and gain, duct leakage, type and position of air return system…).
Every air conditioning application has its own special „needs‟ and provided its own challenges.
Shopping malls, office complexes, hotels, ATM’s, Airports and banks need uniform comfort
cooling in every corner of their sprawling spaces and activities involving computers, electronics,
aircraft products, precision manufacturing, communication networks and operation in
hospitals, infect many areas of programming will come to a halt, so air conditioning is no longer
a luxury but an essential part of modern part of modern living.
Heating and cooling load calculations are the primary design basis for most heating and air-
conditioning systems and components. These calculations affect the size of piping, ductwork,
diffusers, air handlers, boilers, chillers, coils, compressors, fans, and every other component of
the systems that condition indoor environments. Cooling and heating load calculations can
significantly affect the first cost of building construction, the comfort and productivity of
building occupants, and operating cost and energy consumption. Simply put, heating and
cooling loads are the rates of energy input (heating) or removal (cooling) required to maintain
an indoor environment at a desired temperature and humidity condition. Heating and air
conditioning systems are designed, sized, and controlled to accomplish that energy transfer.
The amount of heating or cooling required at any particular time varies widely, depending on
external (e.g., outside temperature) and internal (e.g., number of people occupying a space)
factors.
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Cooling load calculations may be used to accomplish one or more of the following objectives:
a) Provide information for equipment selection, system sizing and system design.
b) Provide data for evaluating the optimum possibilities for load reduction.
c) Permit analysis of partial loads as required for system design, operation and control.
Among six common ASHRAE methods: Equivalent Temperature Difference (ETD), Total
Equivalent Temperature Differential with Time Averaging (TETD/TA), Transfer Function Method
(TFM), Cooling Load Temperature Difference/Solar Cooling Load/Cooling Load Factor
(CLTD/SCL/CLF), Heat Balance Method (HBM), and Radiant Time Series Method (RTSM), in our
project we will use two of the most used methods in calculating cooling load 1)CLTD and 2)RTS.
CLTD (cooling load temperature difference), SCL (solar cooling load factor), and CLF (cooling
load factor): all include the effect of (1) time-lag in conductive heat gain through opaque
exterior surfaces and (2) time delay by thermal storage in converting radiant heat gain to
cooling load. CLTD is a theoretical temperature difference that accounts for the combined
effects of inside and outside air temp difference, daily temp range, solar radiation and heat
storage in the construction assembly/building mass. It is affected by orientation, tilt, month,
day, hour, latitude, etc. CLTD factors are used for adjustment to conductive heat gains from
walls, roof, floor and glass.
the Radiant Time Series Method (RTSM) is the latest ASHRAE method for calculating the cooling
load. RTSM is a simplified method that is “heat-balance based” but does not solve the heat
balance equations. The storage and release of energy in the zone is approximated by a set of
predetermined zone response factors, called radiant time factors (RTFs), Spitler, et al., 1997.
The radiant time series method (RTSM) has effectively replaced the manual load calculation
procedures and has attracted interest, Nigusse, 2007 due to:
1) Its amenability to spreadsheet implementations as opposed to the Transfer Function
Method, which requires iteration.
2) Captures and depicts the physics involved in the Conduction Time Series Factor (CTSF) and
Radiant Time Factor (RTF) coefficients, unlike the Transfer Function Method.
3) Has essentially the same accuracy as the TFM.
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2. CALCULATIONS USING CLTD METHOD
the total cooling load on a building consists of external as well as internal loads. The external
loads consist of heat transfer by conduction through the building walls, roof, floor, doors etc,
heat transfer by radiation through fenestration such as windows and skylights. All these are
sensible heat transfers.
2.1 ROOF
The basic conduction equation for heat gain is q = U A ∆T.
Where :
• q = Heat gain in Btu/hr
• U = Thermal Transmittance for roof in Btu/hr.ft².°F
• A = area of roof in ft2
• ∆T = Temperature difference in °F
The heat gain is converted to cooling load using the room transfer functions (sol-air
temperature) for the rooms with light, medium and heavy thermal characteristics. The equation
is modified as :
Q = U * A * (CLTD)
Where
• Q = cooling load, Btu/hr
• U = Coefficient of heat transfer roof or wall or glass, Btu/hr.ft².°F
• A = area of roof, ft2
• CLTD = cooling load temperature difference °F. The values are determined from tables
available in chapter 28 of AHSRAE fundamentals handbook.
the equation is
further adjusted to apply correction factors for conditions other than the mentioned base case.
Thus,
Q Roof = U * A * CLTD Roof Corrected
The typical steps to calculate the Roof load are as follows:
Step #1 Determine roof construction and overall heat transfer coefficient (U) (Chapter 28
ASHRAE 1997, Table A24-4, A29-5)
Step #2 Select roof no. from ASHRAE Table 31 or Text table 7-34 which is closest to matching
actual roof construction.
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Step #3 Select CLTD Roof for time of interest, typically on an hourly basis (Chapter 28 ASHRAE
Table A28-32, A28-34).
Step #4 Corrections: Values on tables are for :
a. Four latitudes on July or August
b. Indoor temperature of 78°F
c. Outdoor maximum temperature of 95°F with mean daily temperature of 85°F and daily
range of 21°F
CLTD ROOF Corrected= [CLTD Roof+ (78 – TR) +(TM – 85)]
Where
o (78 – TR) = indoor design temperature correction
o (TM – 85) = outdoor design temperature correction
o TR = Indoor room temperature
o Tm = Mean outdoor temperature
o Tmax = Maximum outdoor temperature
o Tm = Tmax – (Daily Range) / 2
Step # 5 Calculate roof area (A) from architectural plans.
Step #6 Q Roof = U * A * CLTD Roof Corrected.
2.2 WALLS
The cooling load from walls is treated in a similar way as roof:
Q Wall = U * A * CLTD Wall Corrected
Where
• Q Wall = Load through the walls in Btu/hr
• U = Thermal Transmittance for walls in Btu/ (h ft^2 F)
• A = area of walls in ft^2
• CLTD = Cooling Load Temperature Difference for walls in °F
The typical steps to calculate the Wall load are as follows:
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Step #1 Determine wall construction and overall heat transfer coefficient (U) (Chapter 24
ASHRAE 1997, Table A24-4, A29-5)
Step # 2 Select wall type from ASHRAE Table 33 which is closest to matching actual wall
construction. Pay attention to effect of mass distribution (inside insulation, outside insulation or
evenly distributed).
Step #3 Select CLTD Wall for time of interest, typically on an hourly basis (Chapter 28 ASHRAE
Table A28-32, A28-34).
Step #4 Corrections: Values on tables are for :
a. Four latitudes on July or August
b. Indoor temperature of 78°F
c. Outdoor maximum temperature of 95°F with mean daily temperature of 85°F and daily range
of 21°F
CLTD Wall Corrected= [CLTD Wall+ (78 – TR) +(TM – 85)]
Where
o (78 – TR) = indoor design temperature correction
o (TM – 85) = outdoor design temperature correction
o TR = Indoor room temperature
o Tm = Mean outdoor temperature
o Tmax = Maximum outdoor temperature
o Tm = Tmax – (Daily Range) / 2
Step #5 Calculate walls area (A). Use the architectural drawings to determine how much wall
area you have. Since you will be finding window and door losses separately, exclude windows
and door and count only wall area.
Step #6 Q Wall = U * A * CLTD Wall Corrected
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2.3 SOLAR LOAD THROUGH GLASS
Solar load through glass has two components: 1) Conductive and 2) Solar Transmission , The
absorbed and then conductive portion of the radiation through the windows is treated like the
roof & walls where CLTD values for standard glazing are tabulated in ASHARE fundamentals
handbook. For solar transmission, the cooling load is calculated by the cooling load SCL factor
and shading coefficient (SC).
The cooling load equations for glass are:
Conductive : Q Glass Conductive = U * A * CLTD Glass Corrected
Solar Transmission : Q Glass Solar = A * SC * SCL
Where
• Q Conductive = Conductive load through the glass in Btu/hr
• Q Solar = Solar transmission load through the glass in Btu/hr
• U = Thermal Transmittance for glass in Btu/ (h ft^2 F)
• A = area of glass in ft^2
• CLTD = Cooling Load Temperature Difference for glass in °F
• SC = Shading coefficient
• SCL = Solar Cooling Load Factor
STEPS FOR CONDUCTIVE CALCULATIONS
Step # 1 For the glass types used, select from ASHRAE tables the overall heat transfer
coefficient (U). The glass is the major contributor of heat gain in the commercial buildings. Pay
attention to effect of shading, reflective films, curtains, drapes etc (refer to example below)
Step # 2 Select CLTD Glass for time of interest, typically on an hourly basis (Chapter 28 ASHRAE
Table 34).
Step # 3 Corrections:
CLTD Glass Corrected= [CLTD Glass+ (78 – TR) +(TM – 85)]
Where
o (78 – TR) = indoor design temperature correction
o (TM – 85) = outdoor design temperature correction
o TR = Indoor room temperature
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o Tm = Mean outdoor temperature
o Tmax = Maximum outdoor temperature
o Tm = Tmax – (Daily Range) / 2
Step # 4 Calculate glass area (A) from architectural plans
Step # 5 Q Glass = U * A * CLTD Glass Corrected
STEPS FOR SOLAR TRANSMISSION CALCULATIONS
Step # 1 Determine shading coefficient (SC) from ASHRAE 1997 Chapter 27, Table 11
Step # 2 Determine zone type from ASHRAE 1997 Chapter 29, Table 35 B
Step # 3 Determine solar cooling load factor (SCL) from ASHRAE 1997 Chapter 28, Table A28-36
Step # 4 Calculate glass area (A) from architectural plans
Step # 5 Q Glass Solar = A * SC * SCL
2.4 PARTITIONS, CEILINGS AND FLOORS
Whenever a conditioned space is adjacent to a space with a different temperature, transfer of
heat through the separating physical section must be considered.
Q = U A (Ta - Trc)
Where
• U = coefficient of overall heat transfer between adjacent and conditioned space in Btu/ (h ft^2
F) See 1997 ASHRAE Fundamentals, Chapter 24 or 2001 ASHRAE Fundamentals, and Chapter 25.
• A = area of partition in ft^2, ceiling or floor calculated from building plans
• Ta = Temperature of adjacent space in °F (Note: If adjacent space is not conditioned and
temperature is not available, use outdoor air temperature less 5°F)
• Trc = Inside design temperature of conditioned space in °F (assumed constant)
Note that the temperature Ta may range widely from that in the conditioned space. The
temperature in a kitchen or boiler room, for example, may be as much as 15 to 50°F above the
outdoor air temperature. Actual temperatures in adjoining spaces should be measured when
possible. Where nothing is known, assume outdoor air temperature less 5°F.
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2.5 lights
The primary source of heat from lighting comes from light-emitting elements. Calculation of this
load component is not straightforward; the rate of heat gain at any given moment can be quite
different from the heat equivalent of power supplied instantaneously to those lights. Only part
of the energy from lights is in the form of convective heat, which is picked up instantaneously
by the air-conditioning apparatus. The remaining portion is in the form of radiation, which
affects the conditioned space only after having been absorbed and re-released by walls, floors,
furniture, etc. This absorbed energy contributes to space cooling load only after a time lag, with
some part of such energy still present and reradiating after the lights have been switched off.
Generally, the instantaneous rate of heat gain from electric lighting may be calculated from:
Q = 3.41 x W x FUT x FSA
Cooling load factors are used to convert instantaneous heat gain from lighting to the sensible
cooling load; thus the equation is modified to
Q = 3.41 x W x FUT x FSA x (CLF)
Where
• W = Watts input from electrical lighting plan or lighting load data
• FUT = Lighting use factor, as appropriate
• FSA = special ballast allowance factor, as appropriate
• CLF = Cooling Load Factor, by hour of occupancy. See 1997 ASHRAE Fundamentals, Chapter
28, Table 38
Note: CLF = 1.0, if operation is 24 hours or if cooling is off at night or during weekends. The total
light wattage is obtained from the ratings of all lamps installed, both for general illumination
and for display use.
The special allowance is the ratio of the wattage in use for the conditions under which the load
estimate is being made, to the total installed wattage. For commercial applications such as
stores, the use factor would generally be unity. The special allowance factor is for fluorescent
fixtures and/or fixtures that are either ventilated or installed so that only part of their heat goes
to the conditioned space. For fluorescent fixtures, the special allowance factor accounts for the
ballast losses, and can be as high as 2.19 for 32 W single lamp high-output fixtures on 277 V
circuits. Rapid start, 40 W lamp fixtures have special allowance factors that vary from a low of
1.18 for two lamps at 277 V to a high of 1.30 for one lamp at 118V, with a recommended value
of 1.20 for general applications. Industrial fixtures other than fluorescent, such as sodium
lamps, may have special allowance factors varying from 1.04 to 1.37, depending on the
manufacturer, and should be dealt with individually. For ventilated or recessed light fixtures,
manufacturers' or other data must be sought to establish the fraction of the total wattage that
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may be expected to enter the conditioned space directly (and subject to time lag effect), versus
that which must be picked up by return air or in some other appropriate manner.
2.6 power loads
The industrial and commercial applications use various equipment’s such as fans, pumps,
machine tools, elevators, escalators and other machinery, which add significantly to the heat
gain. There are 5 equations in use for different scenarios.
Case #1
If the motor and the machine are in the room the heat transferred can be calculated as
Q = 2545 * (P / Eff) * FUM * FLM
P = Horsepower rating from electrical power plans or manufacturer’s data
Eff = Equipment motor efficiency, as decimal fraction
FUM = Motor use factor (normally = 1.0)
FLM = Motor load factor (normally = 1.0)
Note: FUM = 1.0, if operation is 24 hours, In this situation the total power are transferred as heat
to the room.
Note! If the machine is a pump or a fan most of the power are transferred as energy to the
medium and may be transported out of the room. For such cases, case 4 shall be used.
Case #2
If the motor is outside and the machine is in the room the heat transferred can be calculated as
Q = 2545 * P * FUM * FLM
P = Horsepower rating from electrical power plans or manufacturer’s data
FUM = Motor use factor
FLM = Motor load factor
Note: FUM = 1.0, if operation is 24 hours
Case #3
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If the motor is belt driven and the motor and belt is outside and the machine is in the room the
heat transferred can be calculated as
Q = 2545 * P * Belt Eff * FUM * FLM
P = Horsepower rating from electrical power plans or manufacturer’s data
Belt Eff = Belt transmission efficiency, as decimal fraction
FUM = Motor use factor
FLM = Motor load factor
Note: FUM = 1.0, if operation is 24 hours
Case #4
If the motor is in the room and the machine is outside the heat transferred can be calculated as
Q = 2545 * [P/Eff - P] * FUM * FLM
P = Horsepower rating from electrical power plans or manufacturer’s data
Eff = Motor efficiency, as decimal fraction
FUM = Motor use factor
FLM = Motor load factor
Note: FUM = 1.0, if operation is 24 hours
Case #5
If the motor is belt driven and the motor and belt is in the room and the machine is outside the
heat transferred can be calculated as
Q = 2545 * [P/ (motor Eff) – P/ (belt Eff)] * FUM * FLM
P = Horsepower rating from electrical power plans or manufacturer’s data
FUM = Motor use factor
FLM = Motor load factor
The highest quantum of heat gain shall be from the case #1, when both the motor and driven
equipment are located inside the space. The physical location of equipment is not the only
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governing criteria on using a particular case. An exception is taken for pumps and fans to use
case# 4 equation in lieu case#1, even if both the driver & driven equipment is located inside the
space, provided the fluid is pumped outside to the conditioned space. This is due to the fact
that the heat energy is carried away by the fluid. It is important to use the discretion
judiciously, for example for the direct driven exhaust fans, the case #3 is applicable while for
the supply or intake ventilation fans, the case #1 is applicable.
Note that the heat output of a motor is generally proportional to the motor load, within the
overload limits. Because of typically high no-load motor current, fixed losses, and other
reasons, FLM is generally assumed to be unity, and no adjustment should be made for under-
loading or overloading unless the situation is fixed, can be accurately established, and the
reduced load efficiency data can be obtained from the motor manufacturer.
3. Cooling Load Example (CLTD method)
A school classroom is 6 m long, 6 m wide and 3 m high. There is a 2.5 m x 4 m window in the
east wall. Only the east wall/window is exterior. Assume the thermal conditions in adjacent
spaces (west, south, north, above and below) are the same as those of the classroom.
Determine the cooling load at 9:00 am, 12:00 noon on July 21. Other known conditions include:
Latitude = 40oN
Ground reflectance = 0.2
Clear sky with a clearness number = 1.0
Overall window heat transmission coefficient = 7.0 W/m^2K
Room dry-bulb temperature = 25.5°C
Permissible temperature exceeded = 2.5%
Schedule of occupancy: 20 people enter at 8:00 am and stay for 8 hours
Lighting schedule: 300 W on at 8:00 AM for 8 hours
Exterior wall structure:
• Outside surface, A0
• Face brick (100 mm), A2
• Insulation (50 mm), B3
• Concrete block (100 mm), C3
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• Inside surface, E0
Exterior window:
• Single glazing, 3 mm
• No exterior shading, SC = 1.0
Solution:
Cooling Load due to Exterior wall:
Layer Unit Resistance
(R value)m^2K/W
A0 0.059
A2 0.076
B3 1.173
C3 0.125
E0 0.121
total 1.554
U =1/R = 0.643 W/m^2 K
From Table A28-33A, find wall type 13.
From Table A28-32, CLTD9:00 = 9 and CLTD12:00 = 14 for east wall.
CLTD Corrected = CLTD + (25.5 – Ti ) + (Tm - 29.4)
Ti = inside temperature
Tm = mean outdoor temperature
Tm = (maximum outdoor temperature) - (daily range)/2
At 9:00 am CLTD Corrected = 9 + (25.5 - 25.5) + (31 -9/2 - 29.4) = 6.1 K
At 12:00 am CLTD Corrected = 14 + (25.5 - 25.5) + (31 -9/2 - 29.4) = 11.1 K
Q = U A (CLTD)
Q = 0.643 (W/m^2 K) x (6 x 3 - 4 x 2.5) m^2 x 6.1 = 34 W (at 9 am)
Q = 0.643 (W/m^2K) x (6 x 3 - 4 x 2.5) m^2 x 11.1 = 62 W (at 12 noon)
Window conduction:
From Table A28-34, CLTD9:00 = 1 and CLTD12:00 = 5
CLTD Corrected = CLTD + (25.5 - Ti) + (Tm - 29.4)
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CLTD corrected = 1 + 0 + (31 - 9/2 - 29.4) = -1.9 K (at 9 am)
CLTD Corrected = 5 + 0 + (31 - 9/2 - 29.4) = 2.1 K (at 12 noon)
Q = U A (CLTD)
Q = 7.0 W/m^2 K x 4 x 2.5 m^2 x (-1.9 K) = -133 W (at 9 am)
Q = 7.0 W/m^2 K x 4 x 2.5 m^2 x 2.1 K = 147 W (at 12 noon)
Solar load through glass:
From Table A28-35B, find zone type is A
From Table A28-36, find CLF = 576 at 9 am and CLF = 211 at 12 noon
Q = A (SC) (SCL)
Q = (2.5 x 4 m^2) x 1.0 x 576 = 5760 W (at 9 am)
Q = (2.5 x 4 m^2) x 1.0 x 211 =2110 W (at 12 noon)
Cooling load from partitions, ceiling, floors:
Q=0
People:
From Table A28-35B, find zone type is B
From Table A28-3, find sensible/latent heat gain = 70 W
From Table A28-37, find CLF9:00 (1) = 0.65 and CLF12:00 (3) =0.85
Q Sensible = N (sensible heat gain) CLF
Q Sensible = 20 x 70 W x 0.65 = 910 W (at 9 am)
Q Sensible = 20 x 70 x 0.85 = 1190 W (at 12 noon)
Q Latent = N (latent heat gain)
Q Latent = 20 x 45 = 900 W (at 9 am and 12 noon)
Lighting:
From Table A28-35B, find zone type is B
From Table A28-38, find CLF9:00 (1) = 0.75 and CLF12:00 (3) = 0.93
Q Lighting = W Ful Fsa (CLF)
Q Lighting = 300 W x 1 x 1 x 0.75 = 225 W (at 9 am)
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Q Lighting = 300 x 1 x 1 x 0.93 = 279 W (at 12 noon)
Appliances:
Q Sensible = 0
Q Latent = 0
Infiltration:
Since room air pressure is positive, we have:
Q Sensible = 0
Q Latent = 0
Total cooling load:
Component 9:00 AM 12:00 Noon
wall 34 62
Window conduction -133 147
Window solar transmission 5760 2110
partitions 0 0
people 910 1190
lights 225 275
appliances 0 0
infiltration 0 0
total 6796 W 3788 W
4. Cooling load using RTSM method
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The general procedure for calculating a cooling load for each load component is shown in Fig. 1
and includes:
1. Calculate 24 h profile of component heat gains for the design day as follows:
a. For conduction through walls, and roofs, first account for conduction time delay by
applying conduction time series. The Conduction Time Series (CTS) are series of 24 factors
tabulated in ASHRAE's issues for different construction types of roofs and walls and grouped
according to the thermal properties of structure (U value, mass per unit area, and thermal
capacity (m*c)). These factors are denoted by cf in the present study, and represent the hourly
percentage of converting the heat conduction across the external construction (walls and roofs)
to hourly heat gain. The conduction through exterior walls and roofs is calculated using
conduction time series (CTS) as follows:
Wall and roof conductive heat input at the exterior at n hours ago is defined by the familiar
conduction equation:
Qi,t-n = UA(Te,t-n – Ti) (1)
where Ti is the indoor temperature and Te,t-n is the sol-air temperature at n hours ago and is
expressed as:
Te,t-n = To,t-n + u/ho It,t-n - ε ΔR(t)/ ho (2)
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Conductive heat gain through walls or roofs can be calculated using conductive heat inputs for
the current hour and past 23 hours and conduction time series , ASHRAE, 2009:
Qt = cf0Qi,t + cf1Qi,t-1 + cf2Qi,t-2 + cf3Qi,t-3 + … + cf23Qi,t-23 (3)
cf0, cf1, etc. represent the conduction time factors. Multiplying of the conduction time factors
by the U value gives the periodic response factors, Pr and equation (3) may be rewritten as:
Qt = ProA(Te,t – Ti) + Pr1A(Te,t-1 - Ti) + ….. + Pr23A(Te,t-23 – Ti) (4)
b. For other components of heat gain (fenestration, ventilation and infiltration, internal,
and etc.), the same procedure is applied as in any other method as follows: i. Solar and
thermal heat gain through fenestration is calculated as (ASHRAE 2009):
Qf s = Ib * SHGC(0) * Af * IAC(θ) + (Ib + Ir ) * SHGCD * Af * IACD (5a)
Qf th = U * Af * (To – Ti ) (5b)
where I is the solar radiation. The subscripts b, d, and r refer to beam, diffuse and reflected
portions respectively. SHGC() and SHGCD are the solar heat gain coefficients as a function of
the incident angle and the diffuse radiation respectively. IAC() and IACD are the indoor solar
attenuation coefficient functions of the incident angle and diffuse radiation respectively. ii.
Total heat gain from infiltration or ventilation is :
Qinf or Qv = ⍴a * Va * Δ h (6)
where ∆h is the enthalpy difference.
iii. Heat gain due to lighting, occupancy and equipment are; For lighting,
QLi = WFulFsa (7)
For occupancy, occupants emit sensible and latent heat at a metabolic rate depending on the
state of activity. Tables of ASHRAE summarize design data for common conditions. For
equipment,
Q em = (P/Em) Fum Flm (8)
where P is the motor power rating, EM is the motor efficiency, FUM is the motor use factor, and
FLM is the motor load factor
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2. Heat gain through all components, are calculated for each hour, and then divided into two
portions , The hourly convective portion heat gain, which is converted directly to hourly
convective cooling load, and, the hourly radiant portion heat gain.
3. Apply appropriate radiant time series (table 19 and 20 in chapter 18 of, ASHRAE Handbook
of Fundamentals 2009 to the radiant heat gains to account for time delay in conversion to
cooling load. The radiant time series are the series of 24 factor denoted by r in the present
study and represent the hourly ratio of converting the radiant part of hourly heat gain to hourly
cooling load.
The radiant time series or Radiant Time Factors (RTF) are thus generated from heat balance
procedures between interior surfaces radiant heat gain and room air for different types of
structures, fenestrations, and furnishing. These factors are tabulated for specific cases, (as
indicated in table 19 and 20 in chapter 18 of ASHRAE Handbook of Fundamentals 2009) to use
them directly for the certain application instead of performing inside surface and room air heat
balances. Converting the radiant portion of hourly heat gains into hourly cooling loads is
accomplished by the following equation:
Qclr,t = r0Qr,t + r1Qr,t –1 + r2Qr,t –2 + r3Qr,t –3 + … + r23Qr,t –23 (9)
4. The hourly radiant portion cooling load calculated in 3 above is then added to the hourly
convective cooling load to obtain the total hourly cooling load for a certain component.
5. After calculating cooling loads for each component for each hour, sum them to determine
the total cooling load for each hour and select the hour with the peak load for design of the air
conditioning system.
Tables 19 and 20 (in chapter 18 of ASHRAE Handbook of Fundamentals 2009) of radiant time
factors introduce representative solar and non-solar radiant time series data for light, medium,
and heavyweight constructions. The two different radiant time series solar and non-solar are
used as follows:
a. Solar, for direct transmitted solar heat gain (radiant energy is assumed to be distributed to
the floor and furnishings only) and,
b. Non-solar, for all other types of heat gains (radiant energy assumed to be uniformly
distributed on all internal surfaces). Non-solar RTS apply to radiant heat gains from people,
lights, appliances, walls, roofs, and floors. Also, for diffuse solar heat gain and direct solar heat
gain from fenestration with inside shading (blinds, drapes, etc.), the non solar RTS should be
used.
The radiant time series representative zone construction for tables 19 and 20 is indicated in
table 21 in chapter 18 of, ASHRAE Handbook of Fundamentals 2009.
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