Heat Loss Calculation in Air Conditioning System

This report discusses the process of air conditioning and the calculation of heat loss in the system. The results of two tests, A and B, are compared and analyzed. The report includes detailed calculations and assumptions made during the experiment. The subject is air conditioning system and the course code is not mentioned. The content is relevant for students studying mechanical engineering, HVAC systems, and energy management.

Abstract
The objective of this report has been achieved since from test A the heat loss between
temperature T2 and T4 together with the calculation of the heat loss between temperature T4 and
T5 where all determined and found to be 7.453 kJ and 7.453 kJ and 7.451 kJ and 7.064 kJ at 40%
and 60% of power of the heater, furthermore, the comparison was made between the two test of
A and B, in addition, the calculation of heat loss between temperature T5 and T1 and also
calculation of heat loss between T4 and T5 in test be were determined to be 14.167 kJ and 15.6
kJ and 14.168 kJ and 14.67 kJ at 40% and 60% of power of the heater

Introduction
The system of air condition is used to cool the air below its temperature of dew point in
evaporator coil. The air that is moist releases the collected moisture that collects at the drain pan
and consequently from the system which will help to reduce the moisture from the room, the
process of air conditioning involves mixing the hot and moisture from out of door with stale, the
moisture and warm air that is recirculating will return back to a space of air conditioning, the air
that is in a mixed form will be cooled and its moisture is removed by the cooling coil. The air is
then electrically reheated before it is supplied to the room. The heat and humid is collected from
the environs of the room and leaves as air in return, in which its nature will be warmer and more
moist as compared when it initially entered. A fraction of this air in return is mixed with fresh
out of door air and further is supplied back to the room, while the other fraction is lost to the
ambient. It is approximated that the air that will leave the cooling coil will be about 12 degree
Celsius to 138 degree Celsius and at 100% relative humidity, while the air in supply is about 16
degree Celsius to 188 degree Celsius at about 50% relative humidity, always supply of energy is
supposed to be applied on reheat coil to help moderate air at its comfort ability level before the
air is supplied to the room and also the energy is applied to the cooling coil to cool and
dehumidify the process.
The loss of cold air is used to cool again and again warm out of door air by applying a recovery
heat coil, that comprises of rotary heat fan that may have or lack desiccants, heat that is being
supplied to the coil for cooling will be reduced by the recovered energy in the heat exchanger,
noting that a coil for precooling will be fixed at the upstream while the coil that is used for
reheating will be fixed at downstream on the system, therefore the air that is precooled will be
hotter as compared to the air that is supplied after is reheated by reheat coil, and hence energy
that is recovered from the precooling will be used to counterbalanced the reheat load, this will
help in lowering the requirement of operating energy.

Methodology
Equipment
Recirculating duct, preheater and reheater, velocity meter, a room model, five temperature
sensors (T1 to T5), sensors RH1 to RH5 and three louvres.
Procedure
The air conditioning is mixed with the hot and moisture air from the room with stale, the moist
air and warm air that is recirculating will return back to a space of air conditioning, the air that is
in a mixed form will be cooled and its moisture is removed by the cooling coil. The air is then
electrically reheated by reheater before it is supplied to the room. The heat and humid is
collected from the environs of the room and leaves as air in return, in which its nature will be
warmer and more moist as compared when it initially entered. A fraction of this air in return is
mixed with fresh out of door air and further is supplied back to the room, while the other fraction
is lost to the ambient, the temperature will be recorded by temperature sensors while the
humidity will be recorded by RH1 and RH5.

Results of calculations
Assumptions
1. The air is assumed to be dry
2. The flow and temperature are taken to be uniform in the duct
3. The ideal gas law will be used to determine air density
4. The specific heat of air will be calculated based on the formula
Cp = 28.11+0.1967∗10−2∗T +0.4802∗10−5∗T 2−0.1966∗10−9∗T 3
28.97
Test A
a) Determination of heat loss between T2 and T4 at 40% power of the preheater
From the lab results
T2 = 27.50C = 300.5 K
T4 = 26.90C = 299.9 K
Change in temperature = T2 – T4 = 27.5 – 26.9 = 0.6 K
At T4 = 299.9 K
Cp4 = 28.11+0.1967∗10−2∗299.9+ 0.4802∗10−5∗299.92−0.1966∗10−9∗299.93
28.97
= 1.00540 kJ/kg.K
At T2 = 300.5
Cp2 = 28.11+0.1967∗10−2∗300.5+0.4802∗10−5∗300.52−0.1966∗10−9∗300.53
28.97
= 1.0055
∆ Q2-4 = ˙m(Cp2*T2 - Cp4*T4)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT

T = ∆T = 0.6 K
ρ = 101325/287*0.6
= 588.41 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 588.41 * 0.5 * 0.04
= 11.7682 kg
∆ Q2-4 = 11.7682*(1.0055 *300.5 – 1.00540*299.9) = 7.453 kJ
b) Determination of heat loss between T2 and T4 at 60% power of the preheater
From the lab results
T2 = 30.20C = 303.2 K
T4 = 29.40C = 302.4 K
Change in temperature = T2 – T4 = 30.2 – 29.4 = 0.8 K
At T4 = 302.4 K
Cp4 = 28.11+0.1967∗10−2∗302.4+ 0.4802∗10−5∗302.42−0.1966∗10−9∗302.43
28.97
= 1.00582 kJ/kg.K
At T2 = 303.2 K
Cp2 = 28.11+ 0.1967∗10−2∗303.2+0.4802∗10−5∗303.22 −0.1966∗10−9∗303.23
28.97
= 1.00595
∆ Q2-4 = ˙m(Cp2*T2 - Cp4*T4)
˙m = ρ*V*A
But,
P/ ρ = RT
ρ = P/RT

T = ∆T = 0.8 K
ρ = 101325/287*0.8
= 441.31 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 441.31 * 0.5 * 0.04
= 8.8262 kg
∆ Q2-4 = 8.8262*(1.00595 *303.2 – 1.00582*302.4) = 7.45 kJ
c) From the calculation it is found that when the temperature T2 is increased in the
cooling system unit the heat loss is reduced
d) Determination of heat loss between T4 and T5 at 40% power of the preheater
From the lab results
T5 = 25.70C = 298.7 K
T4 = 26.90C = 299.9 K
Change in temperature = T2 – T4 = 26.9 – 25.7 = 1.2 K
At T4 = 299.9 K
Cp4 = 28.11+0.1967∗10−2∗299.9+ 0.4802∗10−5∗299.92−0.1966∗10−9∗299.93
28.97
= 1.00540 kJ/kg.K
At T5 = 298.7
Cp5 = 28.11+0.1967∗10−2∗298.7+ 0.4802∗10−5∗298.72−0.1966∗10−9∗298.73
28.97
= 1.0052
∆ Q5-4 = ˙m(Cp2*T2 - Cp4*T4)
˙m = ρ*V*A
But,
P/ ρ = RT

ρ = P/RT
T = ∆T = 1.2 K
ρ = 101325/287*1.2
= 294.21 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 294.21 * 0.5 * 0.04
= 5.8842 kg
∆ Q2-4 = 5.8842*(1.00540*299.9 - 1.0052*298.7) = 7.451 kJ
e) Determination of heat loss between T5 and T4 at 60% power of the preheater
From the lab results
T5 = 27.50C = 300.6 K
T4 = 29.40C = 302.4 K
Change in temperature = T4 – T5 = 29.4 – 27.5 = 1.9 K
At T4 = 302.4 K
Cp4 = 28.11+0.1967∗10−2∗302.4+ 0.4802∗10−5∗302.42−0.1966∗10−9∗302.43
28.97
= 1.00582 kJ/kg.K
At T5 = 300.6 K
Cp5 = 28.11+0.1967∗10−2∗300.6+ 0.4802∗10−5∗300.62−0.1966∗10−9∗300.63
28.97
= 1.00552
∆ Q5-4 = ˙m(Cp4*T4 - Cp5*T5)
˙m = ρ*V*A
But,
P/ ρ = RT

ρ = P/RT
T = ∆T = 1.9 K
ρ = 101325/287*1.9
= 185.82 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 185.82 * 0.5 * 0.04
= 3.7164 kg
∆ Q2-4 = 3.7164*(1.00582*302.4 – 1.00552*300.6) = 7.064 kJ

Test B
a) Determination of heat loss between T1 and T5 at 40% power of the preheater
From the lab results
T1 = 28.20C = 301.2 K
T5 = 29.10C = 302.1 K
Change in temperature = T5 – T1 = 29.1 – 28.2 = 0.9 K
At T1 = 301.2 K
Cp1 = 28.11+ 0.1967∗10−2∗301.2+0.4802∗10−5∗301.22 −0.1966∗10−9∗301.23
28.97
= 1.00562 kJ/kg.K
At T5 = 302.1
Cp5 = 28.11+ 0.1967∗10−2∗302.1+0.4802∗10−5∗302.12 −0.1966∗10−9∗302.13
28.97
= 1.00577 kJ/kg.k
∆ Q1-5 = ˙m(Cp5*T5 - Cp1*T1)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT
T = ∆T = 0.9 K
ρ = 101325/287*0.9

= 392.28 kg/m3
Average velocity = (0.9 + 1.0)/2 = 0.95
˙m = 392.28 * 0.95 * 0.04
= 14.907 kg
∆ Q2-4 = 14.907*(1.00577 *302.1 – 1.00562*301.2) = 14.167 kJ
b) Determination of heat loss between T1 and T5 at 60% power of the preheater
From the lab results
T1 = 29.60C = 302.6 K
T5 = 30.80C = 303.8 K
Change in temperature = T5 – T1 = 30.8 – 29.6 = 1.2 K
At T1 = 302.6 K
Cp1 = 28.11+0.1967∗10−2∗302.6+ 0.4802∗10−5∗302.62−0.1966∗10−9∗302.63
28.97
= 1.00585 kJ/kg.K
At T5 = 303.8
Cp5 = 28.11+0.1967∗10−2∗303.8+ 0.4802∗10−5∗303.82−0.1966∗10−9∗303.83
28.97
= 1.00624 kJ/kg.k
∆ Q1-5 = ˙m(Cp5*T5 - Cp1*T1)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT
T = ∆T = 1.2 K
ρ = 101325/287*1.2

= 294.21 kg/m3
Average velocity = (0.9 + 1.1)/2 = 1
˙m = 294.21 * 1 * 0.04
= 11.7684 kg
∆ Q2-4 = 11.7684*(1.00624 *303.8 – 1.00585*302.6) = 15.6 kJ
c) Determination of heat loss between T5 and T4 at 40% power of the preheater
From the lab results
T4 = 31.40C = 304.4 K
T5 = 29.10C = 302.1 K
Change in temperature = T4 – T5 = 31.4 – 29.1 = 2.3 K
At T4 = 304.4 K
Cp4 = 28.11+0.1967∗10−2∗304.4+ 0.4802∗10−5∗304.42−0.1966∗10−9∗304.43
28.97
= 1.00615 kJ/kg.K
At T5 = 302.1
Cp5 = 28.11+ 0.1967∗10−2∗302.1+0.4802∗10−5∗302.12 −0.1966∗10−9∗302.13
28.97
= 1.00577 kJ/kg.k
∆ Q1-5 = ˙m(Cp5*T5 - Cp1*T1)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT
T = ∆T = 2.3 K
ρ = 101325/287*2.3

= 153.5 kg/m3
Average velocity = (0.9 + 1.0)/2 = 0.95
˙m = 153.5 * 0.95 * 0.04
= 5.833 kg
∆ Q5-4 = 5.833*(1.00615 *304.4 – 1.00577*302.1) = 14.168 kJ
d) Determination of heat loss between T4 and T5 at 60% power of the preheater
From the lab results
T4 = 33.90C = 306.9 K
T5 = 30.80C = 303.8 K
Change in temperature = T5 – T1 = 33.9 – 30.8 = 3.1 K
At T4 = 306.9 K
Cp4 = 28.11+0.1967∗10−2∗306.9+0.4802∗10−5∗306.92−0.1966∗10−9∗306.93
28.97
= 1.00657 kJ/kg.K
At T5 = 303.8
Cp5 = 28.11+0.1967∗10−2∗303.8+ 0.4802∗10−5∗303.82−0.1966∗10−9∗303.83
28.97
= 1.00624 kJ/kg.k
∆ Q4-5 = ˙m(Cp5*T5 - Cp1*T1)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT
T = ∆T = 3.1 K
ρ = 101325/287*3.1

= 113.887 kg/m3
Average velocity = (0.9 + 1.1)/2 = 1
˙m = 113.887 * 1 * 0.04
= 4.555 kg
∆ Q5-4 = 4.555*(1.00657 *306.9 – 1.00624*303.8) = 14.67 kJ
e) From the calculation it is found that when the temperature T2 is increased in the
cooling system unit the heat loss is reduced
f) Determination of the power supplied by preheater on test A and test B
Test A
1. Determination of heat loss between T1 and T2 at the first 40% power of the preheater
From the lab results
T1 = 22.60C = 295.6 K
T2 = 27.50C = 300.5 K
Change in temperature = T2 – T1 = 27.5 – 22.6 = 4.9 K
At T1 = 295.6 K
Cp1 = 28.11+0.1967∗10−2∗295.6+ 0.4802∗10−5∗295.62−0.1966∗10−9∗295.63
28.97
= 1.004693 kJ/kg.K
At T2 = 300.5 K
Cp2 = 28.11+0.1967∗10−2∗300.5+0.4802∗10−5∗300.52−0.1966∗10−9∗300.53
28.97
= 1.0055
∆ Q1-2 = ˙m(Cp2*T2 - Cp1*T1)
˙m = ρ*V*A
But,

P/ρ = RT
ρ = P/RT
T = ∆T = 4.9 K
ρ = 101325/287*4.9
= 72.051 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 762.051 * 0.5 * 0.04
= 1.4410 kg
∆ Q2-4 = 1.4410*(1.0055 *300.5 – 1.004693*295.6) = 7.443 kJ
Power = heat loss/time in seconds
Let assume the time taken by the heater to dissipated heat is 1.0 seconds
Power = 7.443/1.0 seconds
= 7.443 W
2. Determination of heat loss between T1 and T2 at second 60% power of the preheater
From the lab results
T1 = 22.80C = 295.8 K
T2 = 30.20C = 303.2 K
Change in temperature = T2 – T1 = 30.2 – 22.8 = 7.4 K
At T1 = 295.8 K
Cp1 = 28.11+ 0.1967∗10−2∗295.8+0.4802∗10−5∗295.82−0.1966∗10−9∗295.83
28.97
= 1.004726 kJ/kg.K
At T2 = 303.2
Cp2 = 28.11+ 0.1967∗10−2∗303.2+0.4802∗10−5∗303.22 −0.1966∗10−9∗303.23
28.97
= 1.00595
∆ Q1-2 = ˙m(Cp2*T2 - Cp1*T1)

˙m = ρ*V*A
But,
P/ ρ = RT
ρ = P/RT
T = ∆T = 7.4 K
ρ = 101325/287*7.4
= 47.709 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 47.709 * 0.5 * 0.04
= 0.95419 kg
∆ Q2-1 = 0.95419*(1.00595 *303.2 – 1.004726*295.8) = 7.448 kJ
Power = heat loss/time in seconds
Let assume the time taken by the heater to dissipated heat is 1.0 seconds
Power = 7.448/1.0 seconds
= 7.448 W
Test B
1. Determination of heat loss between T1 and T2 at the first 40% power of the preheater
From the lab results
T1 = 28.20C = 301.2 K
T2 = 32.40C = 305.4 K
Change in temperature = T2 – T1 = 32.4 – 28.2 = 4.2 K
At T1 = 301.2 K
Cp1 = 28.11+ 0.1967∗10−2∗301.2+0.4802∗10−5∗301.22 −0.1966∗10−9∗301.23
28.97
= 1.00562 kJ/kg.K

At T2 = 305.4 K
Cp5 = 28.11+0.1967∗10−2∗305.4+ 0.4802∗10−5∗305.42−0.1966∗10−9∗305.43
28.97
= 1.006317 kJ/kg.k
∆ Q1-5 = ˙m(Cp2*T2 - Cp1*T1)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT
T = ∆T = 4.2 K
ρ = 101325/287*4.2
= 84.0592 kg/m3
Average velocity = (0.9 + 1.0)/2 = 0.95
˙m = 84.0592 * 0.95 * 0.04
= 3.19425 kg
∆ Q2-4 = 3.19425*(1.006317 *305.4 – 1.00562*301.2) = 14.171 kJ
Power = heat loss/time in seconds
Let assume the time taken by the heater to dissipated heat is 1.0 seconds
Power = 14.171/1.0 seconds
= 14.171 W
2. Determination of heat loss between T1 and T2 at second 60% power of the preheater
From the lab results
T1 = 29.60C = 302.6 K
T2 = 25.30C = 298.3 K
Change in temperature = T2 – T1 = 29.6 – 25.3 = 4.3 K
At T1 = 302.6 K

Cp1 = 28.11+0.1967∗10−2∗302.6+ 0.4802∗10−5∗302.62−0.1966∗10−9∗302.63
28.97
= 1.00585 kJ/kg.K
At T2 = 298.3 K
Cp2 = 28.11+0.1967∗10−2∗298.3+ 0.4802∗10−5∗298.32−0.1966∗10−9∗298.33
28.97
= 1.005138 kJ/kg.k
∆ Q1-2 = ˙m(Cp2*T2 - Cp1*T1)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT
T = ∆T = 4.3 K
ρ = 101325/287*4.3
= 82.104 kg/m3
Average velocity = (0.9 + 1.1)/2 = 1.0
˙m = 82.104 * 1.0 * 0.04
= 3.2842 kg
∆ Q2-1 = 3.2842*(1.00585 *302.6 – 1.005138*298.3) = 14.902 kJ
Power = heat loss/time in seconds
Let assume the time taken by the heater to dissipated heat is 1.0 seconds
Power = 14.902/1.0 seconds
= 14.902 W
The variation in temperatures for tests A and test B resulted to different power dissipated by the
preheater.

Conclusion
The results from test A and test B at different percentage of 40% and 60% were determined
between temperature T2 and T4 in test A at 40% and 60% was 7.453 kJ and 7.453 kJ
respectively, while between T4 and T5 of test A was 7.451 kJ and 7.064 kJ, similarly on test B at
40% and 60% was 14.167 kJ and 15.6 kJ respectively, and finally for test A at T5 and T4 was
14.168 kJ and 14.67 kJ respectively.
The reason why the was errors in the measurement and calculation of heat loss was due to heat
loss that was been lost on walls of the air conditioning unit and surrounding room , which could
have been taken in to considerations.
The heat loss on test B was higher than the heat loss on test A because the temperature recorded
in test B were higher than those in test A, it should be known that temperature is directly
proportion to the heat loss.

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