UNIVERSITY OF WAH
HEATING, VENTILATION
&
AIR-CONDITIONING TECHNOLOGY
(MT-362)
Cooling Load Calculations- Numericals
LECTURE NO. 27, 28
Prep By: Engr. Shahid Iqbal
Cooling Load
The purpose of estimation of cooling load is to determine the size of the
A.C. equipment.
Components
Cooling Load Calculations
Internal Loads:
a.
b.
c.
d.
e.
Occupancy Load
Lighting Load
Appliances Load
Product Load
Process Load
Cooling Load Calculations
External Loads:
a) Heat transfer through Wall/Roof/Floor/Door surfaces
b) Heat transfer through windows
c) Heat transfer due to infiltration
Heat Transfer through Wall
Q = U  A  T
Where;
U = Overall heat transfer coefficient of wall
A = Area of wall
 T = Effective temperature difference for wall
Heat Transfer through Roof
Q = U  A  T
Where;
U = Overall heat transfer coefficient of roof
A = Area of roof
 T = Effective temperature difference for roof
Heat Transfer through Floor
Q = U  A  T
Where;
U = Overall heat transfer coefficient of floor
A = Area of floor
 T = Dry bulb temperature difference of outside and inside design
conditions
Heat Transfer through Glass
Q = [ U  A  T ] + U [ sensible heat gain through glass x Shading
coefficient of glass ]
Where;
U = Overall heat transfer coefficient of glass
A = Area of glass either shaded or un-shaded
 T = Dry bulb temperature difference of outside and inside design
conditions
Shading Coefficient (SC)
It is an expression used to define how much of the radiant solar energy,
that strikes the outer surface of the window, is actually transmitted
through the window and into the space.
Heat transfer due to Infiltration
Qs = m  Cp  T
Where;
Qs = sensible heat due to infiltration
m = mass flow rate of infiltrated air
Cp = specific heat of moist infiltrated air
 T = Dry bulb temperature difference of outside and inside design
conditions
Mass flow rate of Infiltrated Air
Mass flow rate of Infiltrated Air = m =
Where;
ρair = density of air (kg/m³)
ρair x
Air charges/ sec x volume of room
3600
Heat Gain from People
QS = Number of people x Sensible heat gain per person
QL = Number of people  Latent heat gain per person
RSHF-Room sensible heat factor
Room Sensible Heat factor =RSHF=
Where;
QS (Total) = Total room Sensible Heat
Qtotal = Total sensible heat + Total Latent Heat
QS (Total)
QTotal
Ton of Refrigeration TR
TR =
Q(Total) in KW
x FOS
3.516
Where;
Qtotal = Total sensible heat + Total Latent Heat
FOS = Factor of Safety
Problem
An air conditioned room that stands on a well-ventilated basement having 3m
Height and 6m length. One of 3m wall its faces west and contains a double glazed
wall window and having dimensions 1.5m x 1.5m.
3m
2D view of Wall
1.5 m
WINDOW
1.5 m
3m
Problem
It mounted flash with the wall and contains no extra shading. There is no heat gain
through the walls other than one facing the west.
Calculate;
a) Sensible heat
b) Latent heat
c) Total heat gained the room
d) Room sensible heat factor
e) Ton of Refrigeration
Problem
With following data;
Inside design conditions = 25 Co DBT with φ = 50%
Outdoor design conditions = 43 Co DBT 24 Co WBT
Overall Heat Transfer Coefficients:
U for wall = 1.78 W/ m2Co
U for roof = 1.316 W/ m2Co
U for floor = 1.2 W/ m2Co
U for glass = 3.12 W/ m2Co
Problem
Effective temperature difference for wall = 25 Co
Effective temperature difference for roof = 30 Co
Solar heat gain factor for glass = SHGF = 300 W/ m2
Internal shading coefficient for glass = SC= 0.86
Occupancy of Room= 12 persons; [90W Qs and 40W QL] per person
Lighting load = 37 W/ m2 of floor area
Appliances load = Qs + QL = 400 W + 100 W
Infiltration = 0.45 air charges/hr
Barometric pressure=101KPa
Solution
Volume of room = 6x3x3 = 54 m³
Area of west wall = 3x3 = 9 m2
Heat Transfer through Wall = Qw = Uw x Aw x ΔTw
Q = 1.78 [3x3 – (1.5x1.5)] x 25
Q = 300.38 watt ( sensible)
Solution
Heat Transfer through Roof = Qroof = Uroof x Aroof x ΔTroof
Q = 1.36 x (6 x 3) x 30
Q = 710.6 watt ( sensible)
Solution
Heat Transfer through Floor = Qfloor = Ufloor x Afloor x ΔTfloor
Q = 1.2 x (6 x 3) x (43 – 25)
Q = 388.8 watt ( sensible)
Solution
Heat Transfer through Glass = Q = [Uglass x Aglass x ΔT ] + Uglass
[sensible heat gain through glass x Shading coefficient of glass ]
Q = [2.25 x 3.12 x (43-25) ] + 2.25 ( 300 x 0.8)
Q = 126.36 + 580.5
Q = 706.86 watt ( sensible)
Solution
Heat transfer due to Infiltration = Qs = m  Cp  T
Now , m=?
So, m =
m=
ρair x
Air charges/ sec x volume of room
3600
1.095 x 0.5 x 54
3600
= 8.21 x 10-3 Kg / sec
Solution
Now;
Qs = m x Cp x Δ T
Qs = 8.21 x 10-3 x 1021.6 x (43-25)
Qs = 151 watt ( Sensible)
Solution
Internal Load:a) Occupancy Load:QS = Sensible heat = 4 x 90 = 360 watt
QL = Latent heat = 4 x 40 = 160 watt
QLightining = 33 x Floor area = 33 x 18 = 594 watt (Sensible)
Solution
Load through Appliances:-
QS = Sensible Heat = 600 watt
QL = Latent Heat = 300 watt
Solution
QS(Total) = Total Sensible heat = 300.88+710.6+388.6+706.9+151+360+594+600
QS(Total) = 3811.98 watt
QL(Total) = 160 + 300 = 460 watt
QTotal = QS(Total) + QL(Total)
QTotal = 3811.98 + 460 = 4271.98 watt
Solution
Now;
Room Sensible Heat factor = RSHF =
QS (Total)
QTotal
3811.98
Room Sensible Heat factor = RSHF = 4271.98 = 0.892
So, TR = ?
Solution
TR =
Q(Total) in KW
x FOS
3.516
Now ,
QTotal = 4271.98 watt / 1000 = 4.271 KW
Then ;
3811.98
TR =
x 1.25 = 1.5
3.52
So
TR = 1.5 Answer