IPE Laboratory-Air Conditioning System Design

COOLING LOAD CALCULATIONS

LEARNING OUTCOME:
1. Explain the general procedure required to calculate the heat gain and
cooling load for the space to be air-conditioned in designing an air-
conditioning system.
2. Define and differentiate heat gain, cooling load and heat extraction
rate.

I. Heat Gain, Cooling Load, and Heat Extraction Rate

1. Heat Gain is the rate at which heat is transferred to or generated

within a space.

2. Cooling Load
The rate at which heat energy must be removed to maintain the
temperature and humidity at the design values. It is taken as the
sum of the heat gain and the system heat loads. The cooling load
will generally differ from the heat gain since some of the heat gains
such as solar radiation are absorbed by the structure and the
contents and do not appear as cooling load until sometime later.

3. Heat Extraction Rate

The rate at which heat is removed from the space by the cooling
and dehumidifying equipment. The heat extraction rate is equal to
the cooling load when the space conditions are constant and the
equipment is operating. However, this is very rare to happen
because of fluctuation of heat gain caused by the space
orientation and location.

General Procedure Required to Calculate the Heat Gain and Cooling

Load

1. Estimate time of day, day, and month for calculations.

- Select a particular day and month for peak load calculation.
- Assign several different times in a given day.
2. Select outdoor design conditions
3. Select indoor design conditions

Recommended for summer: 20-24OC, 50-60% RH and 4.6-7.6 m/min air movement within the living
zone. Source: PME Code 2012 edition, item 8.a page 53.

4. Obtain characteristics of structure from plans and specifications

5. Determine building location and orientation
 IPE Laboratory-Air Conditioning System Design

6. Determine the schedule of lighting, occupancy, and other source

of internal heat gain
7. Make instantaneous heat gain calculation
a. Solar heat gain for glass, walls and roofs
b. Conduction and convection through interior partitions,
ceilings, and floor because of temperature differential
c. Heat source within the conditioned space
d. Infiltration
e. Other sensible and latent loads
8. Compute the total Sensible Heat gain, QS, and total Latent Heat
gain, QL, separately.
9. Include the following system loads in addition to the computed QS
and QL above: See pages 679 – 681, Refrigeration & Air Conditioning by Arora 2nd edition or
pp 625-627 3rd edition.
a) Safety Factor – a factor for probable error in the estimation of the
loads. An additional 5% is added to both QS and QL.
b) Supply air duct heat gain – The supply duct may pass through an
unconditioned space having a temperature higher than the
supply air temperature passing through the supply duct itself.
As a rough estimate, a value in the order of 5% of QS if the whole
supply duct is installed outside the conditioned space, and
proportionately less if some are installed in the conditioned
space. A fraction of 5%, that is, equal to the ratio of the length
of the duct outside the space to the total length of supply duct
is most acceptable.
c) Supply air duct leakage loss – a 10% leakage loss is to be assumed
on both QS and QL if all ducts are outside the conditioned
space and proportionately less, as in item b) if some of it is within
the conditioned space.
d) Heat gain from air conditioning fan – The heat equivalent of the air
conditioning fan horsepower is added as the a sensible heat to
the system. There are two types of air supply systems:
i) Draw-through system – the fan is drawing air through the
cooling coil and supplying it to the space. In short, the fan is
located after the cooling coil. In this case, the fan heat is
added to the room sensible heat.
ii) Blow-through system – The fan blows air through the fan and is
located before the cooling coil. The fan heat is a load to the
cooling coil and therefore added to the grand total heat.
Assume a fan heat to be 2.5 to 7.5 of the QS. Designers usually
take 5%.
e) Heat gain from Dehumidifier Pump and Piping – The horsepower
required to pump water through the dehumidifier adds heat to
the system and it is to be considered like that of the other
motors. The heat gain of dehumidifier piping may be calculated
as a percentage of the Grand Total Heat (GTH) as follows:
i) Very little external piping: 1% of GTH.
 IPE Laboratory-Air Conditioning System Design

ii) Average external piping: 2% of GTH.

iii) Extensive external piping: 4% of GTH.

Add all of the sensible and latent loads discussed in item number nine,
the room sensible heat, RSH and the room latent heat, RLH are now
determined. Equate the two such as:

RSH = 0.0204Vs(tr – ts), kW

RLH = 50Vs(Wr – Ws), kW

RSH RSH
RSHF = =
RTH RSH + RLH

where: 0.0204 and 50 are constants.

(see item 15.2.1, page 514 to 517 2nd edition or 478-479 3rd edition)

RSH = ms(1.0216)(tr – ts), kW

ρVs, m3/min (1.2 kg/m3)(Vs, m3/min)
m , kg/sec = =
s
60 sec/min 60 sec/min
(1.2 kg/m3)(Vs, m3/min)
RSH = (1.0216)(tr – ts)
60 sec/min
RSH = 0.0204Vs(tr – ts), kW

Vs = Volume of supply air, m 3/min or cmm

ρ = Density of standard air, kg/m3
= 1.2 kg/m3 (std. air is at 20OC - 24OC & 50% RH)
tr = maintained room temperature, OC
ts = temperature of supply air, OC
W r = maintained room specific humidity, kgv/kga
W s = specific humidity of supply air, kgv/kga
 IPE Laboratory-Air Conditioning System Design

I. Heat Load Calculations:

1. Sensible Heat Gains, QS

a. Thermal Transmission – Refers to heat gain or heat transfer
through the structure opaque walls, fenestrations, roofs,
ceilings, floors and other partitions due to temperature
difference from high temperature (outdoor) to low
temperature (indoor).

i) Simple wall,
x
Outside Inside
air film air film

QST

to ti

Homogeneous Material
Non-homogeneous Material

AΔt
QST = UAΔt = UA(to - ti) =
RT
1
U =
RT

QST = Sensible heat due to transmission, W

RT = Total thermal resistance of wall material, m 2-K/W
U = Overall thermal coefficient, W/m 2-K
A = Surface area of wall, m2
to = Outdoor air temperature, OC
ti = Indoor air temperature, OC

ii) Composite Wall,

x1 x2

to
Outside ti
air film Inside
a air film
C K1 K2

QST

Non-homogeneous Air space

material Material 1 Material 1
 IPE Laboratory-Air Conditioning System Design

QST = UAΔt
1 1
U = =
RT 1 + 1 + x1 + 1 + x2 + 1
fo C k1 a k2 fi

Where: fo and fi = Outside and inside film coefficient accorded

for convection and radiation, W/m 2-k

k1 and k2 =Thermal conductivity of materials 1 and 2,

W/m-k

C = Conductance of non-homogeneous material

from surface to surface, W/m 2-k

a = Heat transmission from surface across an air

space. Convection, conduction and radiation,
W/m 2-k

For values of U, R, f, C and a use:

a) Table 18-1, 18-2 and 18-3 624 to 627, RAC by Arora 2nd edition
b) Table 18-1, 18-2 and 18-3 576 to 579, RAC by Arora 3rd edition
c) Table 4-4, page 68 to 69, RAC by Stoecker
d) MRII tables and Charts page 86 to 96.

Example:
Determine the heat transmission gain through a 12.0 m x 6.0 m external
wall with a cross section shown. Inside temperature is 24OC and
outside temperature is 34OC. The wall has a 1.5 m x 2.0 m flat glass
window.
Face
Air space Brick
12.0 m

to = 34OC

2.0 m Concrete ti = 24OC

Block

Window 1.5 m
6.0 m

Plaster
Q ST

Tile
Air space
 IPE Laboratory-Air Conditioning System Design

THICKNESS OF WALL MATERIALS:

Cement plaster - 1.5 cm
Hollow clay tile - 10 cm
Air space - 10 cm
Concrete block - 20 cm
Face brick - 10 cm
Glass - 6 cm

Assume an outside air velocity of 12.5 km/hr

QST = QSW + QSG

QSW = Sensible heat transmission through the wall.

Aw Δt
QSW =
R TW

RTW = Total resistances of wall, m2-k/W

Table 18-1, 18-2 or ASHRAE pp 24.2 and 18-3 pp 624 to 627,

RAC by Arora 2nd edition or 576-579 3rd edition,

1. fo = 23.3 w/m2-k - for outdoor air w/ wind velocity of 12.5 km/hr

R = 1/23.3 = 0.043 m2-k/w

2. kCP = 8.65 w/m-k at 0.015 m thickness.

x 0.015 m
R = = = 0.0017 m2-k/w
kCP 8.65 w/m-k

3. Ctile = 5.23 w/m2-k - for clay tiles at 10 cm thickness.

1 1
R = = = 0.191 m2-k/w
Ctile 5.23 w/m2-k

4. a = 6.9 w/m2-k - for air space at 10 cm. thickness & assume 32OC mean temperature.
1 = 1
R = = 0.145 m2-k/w
a 6.9 w/m2-k

5. CCB = 8.14 w/m2-k - for concrete block at 20 cm. thickness.

1 1
R = = = 0.191 m2-k/w
CCB 8.14 w/m2-k

6. kFB = 1.32 w/m-k - for face brick at 10 cm thickness.

x 0.1 m
R = = = 0.076 m2-k/w
kFB 1.32 w/m-k
 IPE Laboratory-Air Conditioning System Design

7. fi = 8.5 w/m2-k - for still air w/ vertical orientation and horizontal heat flow direction.

R = 1/8.5 = 0.118 m2-k/w

Total RTW = 0.7517 m2-k/w

Aw Δt
QSW =
RTW

Aw = (12)(6) – (2)(1.5) = 69m2

(69 m2)(34 – 24)k

QSW = = 917.92 W
0.7517 m2-k/W

QSG = Sensible heat transmission through window glass.

AG Δt
QSG =
RTG

1 x 1
RTG = + kG +
fo fi

Where fo and fi have the same values with that of the wall.

KG = 0.78 w/m-k - for clear glass at 0.06 m thickness.

x 0.06 m
R = = = 0.077 m2-k/w
KG 0.78 w/m-k

RTG = 0.043 + 0.077 + 0.118 = 0.238 m2-k/w

AG = 2 x 1.5 = 3 m2

(3 m2)(34-24) k
QSG =
0.238 m2-k/w

QSG = 126.05 w
 IPE Laboratory-Air Conditioning System Design

QST = QSW + QSG = 917.92 + 126.05

QST = 1,044 Watts

b. Sensible heat gains through solar radiation.

RELATIONSHIP OF SOLAR ANGLES:

The earth revolves around the sun in a period of approximately 365

¼ days in an elliptical orbit. It is closest to the sun on January 1 and
remotest on July 1 (about 3.3% farther away). The mean distance is
149,500,000 km. The intensity of the solar radiation is received by the
earth’s surface 7% more in January than in July. The solar radiation
intensity – normal to the sun’s rays incident upon a plane on the earth’s
surface - varies with the month of the year as the earth’s distance from
the sun varies. Its value when the earth is at mean distance is called
the solar constant and is equal to 1,353 W/m 2 at clear sky.
The earth’s axis is tilted 23.5O with respect to its orbit around the sun.
The earth rotates around its axis in a counter clockwise direction in 24
hours or 360O in one day. The solar constant varies with respect to the
month of the year, the time of the day, the location on the earth’s
surface and the building orientation.
N

W
12:00 MN

30O E
6:00 PM 12:00 NN
2:00 PM

Referring to the figure above, the earth takes 24 hours to rotate by

360O. The angle of rotation for every hour will be:
360O
= 15O/hour
24 hrs

Therefore the angle hour, h = (time, hrs)(15O/hr)

Example: At 2:00 PM, h = (2 hrs)( 15O/hr) = 30O

 IPE Laboratory-Air Conditioning System Design

At 2:00 AM, h = (14 hrs)( 15O/hr) = 210O

z

N ln2

P• Parallel
β

Ф
S
H

ln1

l d
o x
h

•A
y

In the figure above, x-y corresponds to the equatorial plane and

the z-axis corresponds to the earth’s axis. Vector ln1 representing the
sun’s radiation is in the x-z plane. It coincides with the line joining the
earth’s center and to the center of the sun. Line o-x is then a projection
of ln1 on the equatorial plane. The angle d therefore represents the sun’s
declination.
The projection of P on the equatorial plane x-y is A. The angle
PoA is, therefore, the latitude angle l and the angle xoA is the hour
angle h. The line oP represents the vertical through P, and the line PH
the horizontal. The line SN represents the south to north direction on the
earth’s surface, therefore line PN is pointing north. The latter, therefore,
is perpendicular to oP as well as the z-axis.
Altitude angle, ϐ is the angle in the vertical plane between the
sun’s rays, ln2 and the projection of the sun’s rays on the horizontal
plane represented by line PH.
Azimuth angle, ϕ is an arc on the horizon measured from the
south point S in a clockwise direction.
 IPE Laboratory-Air Conditioning System Design

Latitude angle, l is the angular distance north (+) or south (-) from
the earth’s equator to a point on the earth’s surface.

Sin ϐ = cosl cosh cosd + sinl sind

sinh
Tan ϕ =
sinl cosh + cosl tand

For values of d, use Table 17.5 page 579, Refrigeration and Air Conditioning by C. P.
Arora

i) Solar heat gain through transparent surfaces (glass)

Sunray
Glass
Absorbed solar heat

Inward flow
Outward flow of absorbed
of absorbed radiation (4%)
radiation (8%)

Reflected Transmitted (80%)

(8%)

 Total solar heat  Total solar heat

excluded (16%) admitted (84%)

Solar energy passing through window glass

QSG = (SHGFmax)(R)(A)

SHGF = Solar Heat Gain Factor

- For an ordinary glass, use Table 17-9 page 602-613 refrigeration and Air
Conditioning by C. P. Arora

R = Solar Factor
- Used to adjust the value of SHGF for other types of glass or internal
shading device.
- Table 17-8, page 601 by Arora

A = Total surface area of window glass if no external

shading, m2
 IPE Laboratory-Air Conditioning System Design

= Sunlit area of window glass if there is external

shading.
Note: External shadings are overhangs, fins or recessed windows

☼ Estimate of sunlit area of window glass with external shading:

Sun Ray

External
Shading

Window
Glass

y
δ
Sunlit
w
Area
x

Isometric View of the Window

Not Drawn to Scale
 IPE Laboratory-Air Conditioning System Design

E
w

x ψ D

Plan View of the Window

Not Drawn to Scale

Section View of the Window

Not Drawn to Scale
 IPE Laboratory-Air Conditioning System Design

Shadow cast by external shading

y

Sunlit Area

x
Elevation View of the Window
Not Drawn to Scale

δ = Wall azimuth angle – The angle between the plane

normal to the glass and the
projection of the sun ray on the
horizontal.
ϐ = Solar altitude angle.
ϕ = Solar azimuth angle – the angle between the sun ray
and the south.
ψ = Wall zenith angle - The angle between a plane normal
to the window glass and the south.
D = The depth in which the window glass recessed from
the wall surface.
y = The height of the shadow cast by horizontal projection
above the window.

x = The width of the shadow cast by vertical projection of

d.

y
Tan ϐ = Eq. 1, see isometric view.
w

D
Cos δ = Plan view.
w
 IPE Laboratory-Air Conditioning System Design

D
w = Eq. 2
Cos δ

From Eq. 1,

y = w Tan ϐ Eq. 3

Substitute Eq. 2 in Eq. 3

Tan ϐ
y = D Solve for y in meters.
Cos δ
-

Where: δ = ϕ + ψ
(+) - If the outer side of the window glass is opposite sides of south.
( ) - If the outer side of the window glass is same side of south.

x
Tan δ =
D

x = D Tan δ Solve for x in meters.

A = (L - x)(h - y)------------- Sunlit Area in m2

Example:
A window 2.4 m x 1.5 m is recessed 300 mm from the outer surface of
the wall facing 10O west of south. Other data are to be considered:
Design month-------------------------------------- March 21
Condition/Location-------------------------------16O35’35” North latitude
Solar time` 3:00 PM
Glass specifications:
Thickness ------------ 6 mm
Type --------------------- Heat absorbing
Internal Shading --- Medium Venetian blinds.
Estimate the sunlit area and the solar heat gain.
 IPE Laboratory-Air Conditioning System Design

Solution:

h = (3 hrs) (15O/hr) = 45O

From Table 17.5 page 579 by Arora, for the month of March
21, d = 0O21” = 0.35O

Sin ϐ = cos16.6 cos45 cos0.35 + sin16.6 sin0.35

ϐ = 42.79O = 43O

sinh
Tan ϕ =
sinl cosh + cosl tand

Sin45
Tan ϕ =
Sin16.6cos45 + cos16.6 tan0.35

ϕ = 73.62O = 74O

From the given, angle ψ = 10O.

D E

δ
ϕ

δ = ϕ - ψ = 74O - 63O

δ = 64O
 IPE Laboratory-Air Conditioning System Design

Tan ϐ Tan 43O

y = D = 0.3
Cos δ Cos 64O

y = 0.638

x = D Tan δ = 0.3 Tan 64O

x = 0.615 m

A = (L - x)(h - y)

A = (2.4 - 0.615)(1.5 - 0.638)

A = 1.54 m2--------------------------- Sunlit Area

QSG = (SHGF)(R)(A)

Table 17-9, page 610, by Arora, based on 40O North latitude,

for the month of March
21, SW, 2:00 PM
SHGF = 495 W/m2

Table 17-8, page 601, for heat absorbing, 6.35 mm thickness

glass with medium venetian blinds,
Assume R = 0.75 for heat absorbing glass – standard value

R = (0.94)(0.75)(0.65)

R = 0.46

QSG = (495 W/m2)(0.46)(1.54 m2)

QSG = 350.7 Watts

ii) Solar heat gain on opaque surfaces

 Portion of the solar energy is reflected and the remainder is

absorbed by the wall.
 Of the absorbed solar energy, some are convected and some
are radiated to the outside.
 The remainder of the absorbed solar energy is transmitted to
the inside by conduction and temporarily absorbed by the wall
itself and it will be convected to the inside at a later time.
 IPE Laboratory-Air Conditioning System Design

Convected
Absorbed solar heat
Sunray

Conducted
Radiated

Reflected

Solar energy passing through an opaque surface

QOW = Solar heat gain through opaque surfaces including heat gain due
to temperature difference of outside and inside air with outside
and inside film coefficients.

QOW = UwA(CLTD)

Uw = Heat transfer coefficient of the wall, W/m2-k

A = Wall sunlit area, m2
CLTD = Equivalent Temperature Difference (ETD) or Cooling
Load Temperature Difference (CLTD), OC
 Use Table 18.9 for walls and Table 18.10 for roofs
page 602-605, by Arora 3rd Ed.
 The values on these tables are based on the
following conditions:
i) Latitude 40ON, but normally suitable for 0 to 50ON, for hottest
summer period.
ii) An outdoor range (tomax – tomin) of 11.1OC (20OF)
iii) An outdoor and indoor temperature difference (tomax – ti) of 8.3OC
(15OF)
iv) Dark color walls and roofs with absorpsivity of 0.9
v) A sp. heat of the construction material of 1.005 kJ/kg-K.

 When there is departure from the above

conditions, add the correction factor, Δt’e to CLTD

Δt’e = Correction factor, OC

= (tomax – ti) – 8.3 ± [(tomax – tomin) – 11.1]C
(+) if (tomax – tomin) is less than 11.1OC
(–) if (tomax – tomin) is greater than 11.1OC
C = 0.25 for medium construction
= 0.50 for heavy construction
ti = Inside room temperature, OC
tomax = Maximum outside room temperature, OC
tomin = Minimum outside room temperature, OC
 No correction factor for light construction.
 IPE Laboratory-Air Conditioning System Design

Example:
Calculate the solar heat gain through an opaque wall 22m x 15 m
made up of 100mm concrete block with 50mm insulation (blanket &
batt mineral fiber) facing SW with solar time 2:00 PM. Outdoor
maximum and minimum daily range recorded are 38OC and 20OC. The
room is to be maintained at 24OC temperature.

Outside 100mm 50mm

Inside
air film air film
coefficient coefficient

38OC 24OC

Concrete
Insulation
block

From MRII Tables and Charts, page 93, average density of concrete block at
200 mm thickness is 2100 kg/m3. At 100 mm, density is (100/200 x 2100) is
1050 kg/m3.

Concrete block wall mass, m = density x thickness

m = 1050 kg/m3 x 0.100 m

m = 105 kg/m2

Table 18.9, Page 602 by Arora 3rd Ed., at SW, 2:00 PM, m = 105 kg/m2,

CLTD = 14. 4OC --- by interpolation.

MRII Tables and Charts, page 93, for concrete block, 200mm thickness,

R = 0.37 m2-k/W

0.37(100mm)
R1 = = 0.185 m2-k/W
200mm
 IPE Laboratory-Air Conditioning System Design

Table 4-4, page 68, RAC by Stoecker, for mineral fiber, 75-90 mm or an
average of 82.5 mm thickness,

R = 1.94 m2-k/W

1.94(50mm)
R2 = = 1.176 m2-k/W
82.5mm

For outside air film, cooling season, 3.4 m/s wind velocity,

Ro = 0.044 m2-k/W

For inside air film, vertical, heat flow horizontal,

Ri = 0.12 m2-k/W

RT = Ro + R1 + R2 + Ri = 1.525 m2-k/W
1 1
Uw = =
RT 1.525

Uw = 0.6557 W/m2-k

A = 22 x 15 = 330 m2

(tomax – ti) = (37 – 24) = 13 > 8.3

(tomax – tomin) = (37 – 27) = 10 < 11.1

Therefore correction factor is needed, assume medium construction.

Δt’e = (tomax – ti) – 8.3 ± [(tomax – tomin) – 11.1]C

Δt’e = (37 – 24) – 8.3 + [(37 – 27) – 11.1](0.25) = 3.6OC

Qsw = (0.6557 W/m2-k)(330 m2)(14.24 + 3.6)k

Qsw = 3,894.85 W

c. Heat gain through floors and roofs

 IPE Laboratory-Air Conditioning System Design

i) Heat gain through Floors, QFlrs

QFlrs = FPtEF

Where: F = Heat loss coefficient for ground floors

= 1.4 W/m K, This is constant for
concrete slab-on-grade uninsulated
floor.
0.9 W/m K for insulated floor.
Source: R&AC pp 67 by Stoecker
P = Floor perimeter, m
tEF = Equivalent temperature difference,
OC

ii) Heat gain through roofs, QRoofs

QRoofs = URAR(RETD)

Where: UR = Roof overall thermal coefficient,

W/m2-K
AR = Roof area, m2
RETD = Roof equivalent temperature
difference, OC.
Use Table 18.10 pp 604, Arora

d. Sensible heat gains through infiltration or air leakage in to

the conditioned space.

Infiltration is the uncontrolled entry of unconditioned

outside air directly into the building resulting from natural
forces such as wind and buoyancy due to temperature
difference of outside and inside air.

QIS = 1.23V(to – ti)

to = Outside air temperature, OC

ti = Inside air temperature, OC
V = Volumetric rate of infiltrating air, m3/sec

(L x W x H)(NC)
V =
3600

L = Room length, m
W = Room width, m
 IPE Laboratory-Air Conditioning System Design

H = Room height, m
NC = a + b(vel) + c(to – ti) = number of air changes
vel = outside wind velocity
= 3.40 m/sec for summer average value
a, b and c --- Infiltration constants, use Table 4-
5, page 70 by Stoecker.

e. Sensible heat emission through occupants

QOS = (qsp)(N), watts

qsp = Sensible heat given off per person

- Use MRII tables page 81
N = Number of occupants
- Based on the given maximum capacity
of the space.
- If maximum capacity of space is not
given,
Floor Area, m2
N =
Space occupied per person
- Use Table 4-8 page 73 by Stoecker for
space occupied per person.

Example:
Determine the heat gain from the occupants doing light works for an
office building 150m x 70m. Working time is 8 hrs.

Floor Area, m2
N =
Space occupied per person

A = 150 x 70 = 10,500 m

From Table 4-8, page 73 by Stoecker, Space occupied per

person for office is 10-15 m2 per occupant. Take the average
value, 12.5
10,500 m2
N = = 840 persons
12.5 m2/person

qsp = 70 W/person --- from MRII Tables and Charts page 81

QOS = (70 W/person)(840 person)

QOS = 58,800 watts

 IPE Laboratory-Air Conditioning System Design

f. Sensible heat gain from electric lights

QLS = (Lamp rating, watts)(Fu)(Fb)(CLF), watts

Fu = Utilization factor or fraction of installed

Actual hrs of usage
Fu =
24 hrs
Fb = Ballast factor for fluorescent lamps
= 1.2 for common fluorescent lamps

CLF = Cooling Load Factor

Use Table 4-6, page 72 Stoecker

g. Sensible heat gain from products

QPS = mpCp(tp - tr), kW

mp = weight of product, kg/sec

Cp = specific heat of product, kJ/kg-k
From tables and charts
tp = temperature of product, OC
tr = room temperature, OC

h. Sensible heat gain from motors and appliances

QMS = (Motor rating, kW)(Power factor or

efficiency)(N)

For motor ratings and efficiencies, use MRII Tables and

Charts page 82-83

N = Number of motors

2. Latent Heat Gains, QL

a. Latent heat gain from infiltrating air.

QIL = 3000V(Wo – Wi)

V = Volumetric rate of infiltrating air, m3/sec.

- The same computation as in QIS.

Wo = Humidity ratio of outdoor air, kgv/kga

Wi = Humidity ratio of room air, kgv/kga
 IPE Laboratory-Air Conditioning System Design

b. Latent heat from occupants

QOL = (qLP)(N)

qLP = Latent heat gain per person

- MRII Tables and Charts page 81

N = Number of occupants
- The same as in QOS

c. Latent heat from product, QPL

- Moisture of products or materials may be obtained
from tables.

d. Latent heat from other sources such as cooking, hot

baths, foods, and other vaporization processes in the
space.
- May be obtained from tables.
- Example is MRII Tables and Charts page 81-82.

3. Ventilation standard VO

VO = 60(VP)(N), m3/min or cmm

VP = Ventilation requirement per person, m3/sec-
person
- PME Code, 2012 edition, Table 4-1 pp 62-63
- MRII Tables and Charts page 84

N = Number of occupants
- The same as in QOS