Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Lecture 35
Analysis of Air Conditioning Processes
The Five HVAC Processes
• Heating
– With and without humidification
• Adiabatic humidification
• Cooling
– With and without dehumidification
• Adiabatic mixing of moist air streams
• Evaporative cooling
2
Heating Without Humidification
QH
1
Q H  m a  h #2  h1# 
2
h #2
1
2
#
1
h
1   2
1
2
T1
3
T2
• Constant humidity ratio
• Relative humidity
decreases
• Outlet relative humidity
may be too dry to be
comfortable
Heating with Humidification
QH
QH  ma h1#  m water hwater  ma h#2
1
2
1'
m w1  m water  m w 2
m water
h #2
h1'#
water
ma
2
2
h1#
 1'
1
1
1'
4
ma

m w1
ma
  2  1
2
1
T1

m w2
T1' T2
QH
ma
  h #2  h1#    2   1  h water
Adiabatic Humidification
ma h1#  m water hwater  m a h#2
1
2
m w1  m water  m w 2
m water
water
ma
2b
h
2c
2a
#
2

m w2
ma

m w1
ma
  2  1
 h1#    2   1  h water  0
1
1
1
T1
5
The location of state 2 (2a,
2b, or 2c) depends on the
state of the water being
injected for humidification
Cooling Without Dehumidification
QC
1
h1#
2
1
h #2
1   2
2
1
T2
6
Q C  m a  h1#  h #2 
2
T1'
T1
This process is not very
common in HVAC systems.
Often, the surface of the
heat exchanger is below
the dew point which causes
water to condense
Cooling With Dehumidification
QC
Qc  ma h#2  m water hwater  ma h1#
1
2
m w 2  m water  m w1
m water
water
ma
h1#
2
h
1
2
7
ma
#
2
T2
QC
1
1
Tdp
T1
2

m w1
ma

m w2
ma
 1   2
  h1#  h #2    1   2  h water
If Twater is not given, it is
common to assume
that it is equal to T2
ma1h1#  m a 2 h#2  m a3h3#
Adiabatic Mixing
 1m a1   2 m a 2   3m a 3
m a1  m a 2  m a 3
1 (cold)
m a1h1#  m a 2 h #2   m a1  m a 2  h3#
m a1 h #2  h3#
 #
m a 2 h3  h1#
3 (warm)
2 (hot)
h #2
h 3#
2
#
1
h
3
1
2
3
1
1m a1   2 m a 2   m a1  m a 2   3
m a1  2   3

m a 2  3  1
2  3

#
#
h3  h1  3  1
h #2  h3#
1-2-3 are on a straight line!
8
Evaporative Cooling
This is the adiabatic
humidification
1
process when the
water used for
humidification is
colder than T1
ma h1#  m water hwater  m a h#2
2
m w1  m water  m w 2
m water
water
ma
2
2
2
1
1
T2
9
T1
1
h
#
2

m w2
ma

m w1
ma
  2  1
 h1#    2   1  h water  0
This system works very well in hot,
dry climates. Notice that there can
be a significant increase in the
humidity levels (both relative
humidity and humidity ratio).
Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Example
Combined Cooling and Heating
Processes
Example
Given: In a combined cooling/heating system, moist air enters the
cooling section at 90°F,  = 50% at a volumetric flow rate of 5000 cfm.
Saturated, moist air and liquid condensate leave the cooling section at a
temperature that is 15 degrees below the dew point of the entering moist
air. After leaving the cooling section, the saturated, moist air enters the
heating section. After passing through the heater, the moist air leaves
the heating section at 68°F. The pressure throughout the system can be
assumed to be constant at normal sea-level pressure (29.921 inHg –
consistent with the psychrometric chart).
Find:
(a) The volumetric flow rate of the condensate (gpm)
(b) The required refrigeration capacity of the cooling section (tons)
(c) The relative humidity of the air leaving the heating section
(d) The heat transfer rate required in the heating section (Btu/hr)
11
Example
A sketch of the system and a psychrometric chart showing
the processes is shown below.
QC
1
T1  90F
1  50%
V1  5000 cfm
QH
3
2
h1#
T3  68F
water T  T2  Tdp1  15 R
h
h #2
12
3
1
2
#
3
1
2
3
T2
T3
1
Tdp
T1
 2  3
Properties from the Chart and Tables
h1#  38.4 Btu/lbma
 3  61%
 1  50%
2
h3#  26.0 Btu/lbma
h#2  22.5 Btu/lbma
2
 1  0.0152
3
1
 2  3  0.0089
v1  14.19 ft 3 /lbma
Tdp  69F
T1  90F
T3  68F
T2  Tdp 15°F   69 15 F  54°F
T2  Twater  54F  h water  22.1 Btu/lbmw
Table C.1a
13
vw  0.01605ft 3 / lbm
Table C.1a
QC
Example
1
T3  68F
water T  T2  Tdp1  15 R
h1#  38.4 Btu/lbma
QC
ma
  h1#  h #2    1   2  h water
3
2
T1  90F
1  50%
V1  5000 cfm
Cooling section analysis …
QH
 3  61%
 1  50%
2
h3#  26.0 Btu/lbma
h#2  22.5 Btu/lbma
2
 1  0.0152
3
1
3
ft
5000
V
min 60 min  21141.6 lbm a
ma  1 
ft 3
v1
hr
hr
14.19
lbm a
 2  3  0.0089
v1  14.19 ft 3 /lbma
Tdp  69F
T1  90F
T3  68F
T2  Tdp 15°F   69 15 F  54°F
T2  Twater  54F  h water  22.1 Btu/lbm w
Table C.1a
Q C  m a  h1#  h #2    1   2  h water 
lbm a

Q C   21141.6
hr

Q C  333, 208
14
lbm w 
Btu
Btu  

38.4

22.5

0.0152

0.0089
22.1





 
lbm
lbm
lbm

a
a 
w 
Btu
 27.8 tons 
hr
QC
Example
1
The condensate flow is determined
by conservation of mass around
the cooling section,
QH
3
2
T3  68F
T1  90F
1  50%
V1  5000 cfm
water T  T2  Tdp1  15 R
h1#  38.4 Btu/lbma
 3  61%
 1  50%
2
h3#  26.0 Btu/lbma
h#2  22.5 Btu/lbma
m w 2  m water  m w1
m water
ma

m w1
ma

m w2
ma
2
 1  0.0152
3
1
 2  3  0.0089
v1  14.19 ft 3 /lbma
 1   2
Tdp  69F
T1  90F
T3  68F
T2  Tdp 15°F   69 15 F  54°F
T2  Twater  54F  h water  22.1 Btu/lbm w
m water  m a  1   2 
Table C.1a
lbm a 
lbm w
lbm w

m water   21141.6
0.0152

0.0089

133.2



hr 
lbm a
hr

V water
lbm w  
ft 3   hr 
gal


 m wvw  133.2
0.01605
 0.267 gpm 



3 
hr  
lbm   60 min  0.13368 ft 

Table C.1a
15
QC
Example
The relative humidity leaving the
heating section can be read from
the psychrometric chart,
1
QH
3
2
T3  68F
T1  90F
1  50%
V1  5000 cfm
water T  T2  Tdp1  15 R
h1#  38.4 Btu/lbma
 3  61%
 1  50%
2
h3#  26.0 Btu/lbma
h#2  22.5 Btu/lbma
3  61% 
Heating section analysis …
Q H  m a  h3#  h #2 
2
 1  0.0152
3
1
v1  14.19 ft 3 /lbma
Tdp  69F
T1  90F
T3  68F
T2  Tdp 15°F   69 15 F  54°F
T2  Twater  54F  h water  22.1 Btu/lbm w
lbma 
Btu
Btu

Q H   21141.6
26.0

22.5

73,996


hr 
lbm a
hr

16
 2  3  0.0089
Table C.1a

Example
h1#  38.4 Btu/lbma
QC
QH
T1  90F
1  50%
V1  5000 cfm
3
2
T3  68F
 1  50%
2
h3#  26.0 Btu/lbma
1
 3  61%
h#2  22.5 Btu/lbma
2
 1  0.0152
3
1
 2  3  0.0089
v1  14.19 ft 3 /lbma
water T  T2  Tdp1  15 R
Tdp  69F
T1  90F
T3  68F
T2  Tdp 15°F   69 15 F  54°F
T2  Twater  54F  h water  22.1 Btu/lbm w
Table C.1a
EES Solution (Key Variables)
These are a bit different due to reading the psychrometric chart
17