Energy Efficient Buildings
Psychrometrics
Introduction
The air around us is mixture of dry air and water vapor, and can be modeled as a
mixture of these two ideal gasses. The study of moist air is called psychrometry. The
properties of moist air at a given pressure are displayed on a psychrometric chart. The
psychrometric chart contains a rich density of information, and learning to visualize air
heating and cooling properties as they occur on the psychrometric chart is very useful.
The two cardinal dimensions on the psychrometric chart are dry bulb temperature, T (F),
on the horizontal axis and specific humidity, w (lbw/lba), on the vertical axis. Extensive
properties, such as enthalpy and volume are expressed in terms of pounds of dry air.
The enthalpy of air increases with both temperature and humidity. The relative
humidity of air is the faction of water vapor the air can hold at a given temperature. The
relative humidity of air is bounded at 100% along the left side of the psychrometric
chart. The specific volume of air is the reciprocal of the density, and increases with
temperature. Any two independent properties fix the state.
1
For many years, the most common method for measuring humidity was by using a “sling
psychrometer”. A sling psychrometer has two thermometers attached to a handle so
that the thermometers can rotate around the handle. The bulb of one thermometer is
wrapped in wet cloth and is called the wet bulb. The other bulb is called the dry bulb.
The temperature of the web bulb thermometer will decrease due to evaporation as the
thermometers are swung about. The quantity of evaporation depends on the humidity
of the air; thus the difference between wet and dry bulb temperatures is a measure of
ambient humidity.
Wet
Dry
Conceptually, the wet-bulb temperature is the same as the adiabatic saturation
temperature, which is the exit temperature of air traveling over water through an
infinitely long chamber. Because the evaporation process occurs at constant enthalpy,
lines of constant enthalpy on the psychrometric chart are essentially parallel to lines of
constant wet-bulb temperature.
Twb  Tadiabatic saturation
Adiab Sat Chamber
T1
Tad,sat<T1
Constant h
Tw=C
Cooling
The total heat removed from air during a cooling process can be calculated from mass
and energy balances on a system in which air enters and leaves a cooling chamber and
heat is removed from the chamber.
Qtot

ma1

ma2
Conservation of mass on a control volume states that the sum of all the mass flows in
minus the sum of all mass flows out equals the change in mass stored in the system. At
2
steady state conditions, no mass is accumulated in the cooling chamber. Thus, for this
system, conservation of mass gives:
Min - Mout = Mstored
ma1 – ma2 = 0
ma1 = ma2
where ma1 and ma2 are the mass flow rates of moist air entering and leaving the
chamber.
Conservation of energy on a control volume states that the sum of all energy flows in
minus the sum of all energy flows out equals the change in energy stored in the system.
At steady state conditions, no energy is accumulated in the cooling chamber. The
energy of non-reacting mass flows is the sum of the kinetic, potential and internal flow
energies. In most cooling applications, the change in velocity (V) and elevation (z)
between the entrance and outlet of the chamber are negligible. Thus, for this system,
conservation of energy gives:
Ein - Eout = Estored
ma1 (V12/2 + gz1 + h1) – ma2 (V22/2 + gz2 + h2) – Qtot = 0
Qtot = ma1 (h1 – h2)
where Qtot is the total heat removed from the moist air and h1 and h2 are the
enthalpies of moist air entering and leaving the chamber.
Sensible Cooling
When warm air is cooled as it passes over a cooling coil, the temperature begins to
decrease. As the temperature decreases, the air can hold less moisture before
becoming 100% saturated. If the air leaves the coil before becoming saturated, then no
condensation occurs. On a psychrometric chart, the exiting air is at a lower temperature
than the incoming air, while the humidity ratio remains constant since no moisture is
condensed from the air. Reducing the temperature of air without changing the quantity
of water in the air is called sensible cooling, Qsen, and is shown schematically below.
Qtot
2

ma1
1
Δω  0

ma2
3
Example
20,000 cfm of air is sensibly cooled from a temperature of 90 F and a relative humidity
of 60% to a temperature of 75 F. Determine the total heat removed from air.
V 1  20,000 cfm from T = 90°F, φ = 60% to T = 75°F
 ft 3 
 Btu 
 lbw 
v

14.26
  w 1  .01827 
1


 lba 
 lba 
 lba
1: T  90F φ  60% h1  41.74 
2: T  75F w 1  w 2 h2  38.02 
Btu 

 lba 
M-balance:
 a1  m
 a2
m
 a1 
and m
V 1
v1
Note
V 1  V 2 because v 1  v 2 
E-balance:
 a1 h1  Q tot  m
 a2 h2  0 (SS)
m
 ft 3 
V
1  lba 
 min
 Btu 
 a1 h1  h2   1 h1  h2   20,000 
Q tot  m

 41.74  38.02 
  60 



3
v1
 hr  14.26  ft 
 lba 
 min
 Btu 
 313,705 

 hr 
Sensible cooling can also be calculated from the change in air temperature, rather than
the change in air enthalpy. Specific heat at constant pressure is defined as:
 dh 
cp   
 dT  p
Thus the change in enthalpy is then given by the following differential equation.
 dh   c dT
p
Most cooling operations occur at constant pressure. At atmospheric conditions, the
specific heat of water is about 1.0 Btu/lb-F and the specific heat of dry air is about 0.24
Btu/lb-F. In sensible cooling applications, the quantity of water in the air does not
change. Further, the change in temperature during the cooling process is small enough
that the specific heats are nearly constant. Thus, the specific heat of moist air can be
4
assumed to be constant, and can be taken outside the integral. Thus, in the case of
sensible cooling, the change in enthalpy is:
h2  h1   cp T2  T1 
The specific heat, cp, of moist air is approximately 0.26 Btu/lb-F. The density, p, of
moist air is approximately 0.075 lb/ft3. Thus, the product of the density and specific
heat of moist air, pcp, is approximately 0.018 Btu/ft3-F. Using this approximation,
sensible cooling, Qsen, in which the quantity of water in the air is unchanged, can be
calculated from the volume flow rate of air and the change in air temperature.
Qsen = V1 pcp (T1 – T2)
Example
20,000 cfm of air is sensibly cooled from a temperature of 90 F and a relative humidity
of 60% to a temperature of 75 F. Determine the sensible cooling using the
approximation that pcp = 0.018 Btu/ft3-F. Also determine the error between the
enthalpy-based calculation and this approximate method.
V
 a h1  h2   m
 c p T1  T2   c p T1  T2   V ρcp T1  T2 
Q sen  m
v
3
 ft 
 min
 Btu 
 Btu 
 20,000 
 .018  3
 90  75 F  324,000 
  60 



 hr 
 ft  F 
 hr 
 min
The error between the exact and approximate methods 
324,000  313,705
 3.3%
313,705
Sensible and Latent Cooling
The dew point temperature Tdewpoint is the temperature when air is 100% saturated and
water begins to condense out of the air. When air is cooled below the dew point
temperature, Tdewpoint, condensation occurs and moisture leaves the air stream. The
exiting air stream is at a lower temperature and humidity ratio than the incoming air
stream. This process is shown in the figure below. The cooling to reduce the
temperature of the air is called sensible cooling. The cooling to condense water from
the air is called latent cooling. Thus, this process includes both sensible and latent
cooling. The total cooling, Qtot, is the sum of the latent cooling, Qlat, and the sensible
cooling, Qsen.
5

Q tot
h1
h2'
1
2
h2
1
2

Q lat
2p
T2 Tdp

T2
T1
Q sen
T1
Qtot = ma (h1 – h2) = ma (h1 – h2p) + ma (h2p – h2) = Qlat + Qsen
The water removed from the air is:
mw = ma (w1 – w2)
Example
Determine the total, sensible and latent cooling required to cool 20,000 cfm of air from
a temperature of 90 F and a relative humidity of 60% to a temperature of 55 F and 100%
relative humidity. Also determine the mass flow rate of water removed from the air.

Q tot
h1
h2'
1
h2
2

Q lat
2p
T2

Q sen
T1
Cooling 20,000cfm from 90F, φ  60% to 55F, φ  100%
 ft 3 
 Btu 
 lbw 
v

14.26
  w 1  .01827 
1


 lba 
 lba 
 lba 
1: T1  90F φ1  60% h1  41.74 
 Btu 
 lbw 
w 2  .009186 


 lba 
 lba 
2: T2  55F φ2  100% h2  23.2
6
 a1 h1  Q tot  m
 ah2  0 (SS)
m
V
 a h1  h2   1 h1  h2 
Q tot  m
v1
 20,000  60  ft3  lba  min
 Btu  
 kBtu
Q tot  
 41.74  23.20     1,560 



3
 14.26
 lba  
 hr 
 min  ft  hr 

V
 a h1  h2p   1 h1  h2p' 
Q lat  m
v1
 20,000  60  ft3  lba  min
 Btu  
 kBtu
Q lat  
 41.74  31.74     841.3 



3
 14.26

 lba  
 hr 
 min  ft  hr 

V
 a h2p  h2   1 h2p  h2' 
Q sen  m
v1
 20,000  60  ft3  lba  min
 Btu  
 kBtu


Q sen  

31.74

23
.
20
 719.2 





3
 14.26

 lba  
 hr 
 min  ft  hr 

 w m
 a w 1  w 2 
m

V 1
w 1  w 2   20,000
v1
14.26
 ft 3  lba 
 lbw 
 min
 lbw 
 .01827  .009186 
 60 
 764.1 




3
 lba 
 hr 
 hr 
 min  ft 
Latent cooling can also be calculated from the volume flow rate of air, V, and the
enthalpy of evaporation of water. The density, p, of moist air is approximately 0.075
lb/ft3. The enthalpy of evaporation (condensation) of water, hfg, at atmospheric
pressure is approximately 1,075 Btu/lbw. Using these approximations, latent cooling,
Qlat, can be calculated as:
Qlat = V p (w1-w2) hfg
Example
Determine the latent cooling using the enthalpy of evaporation of water and an
approximate value of air density required to cool 20,000 cfm of air from a temperature
of 90 F and a relative humidity of 60% to a temperature of 55 F and 100% relative
humidity. Also determine the error between the enthalpy-based calculation and this
approximate method.
7
Q lat  V ρ w 1  w 2  h fg

 ft 3 
 min
 lba 
 lbw 
 Btu  
  20,000 
 60 
 .075 3   .01827  .009186 
 1,075 





 hr 
 ft 
 lba 
 lbw  
 min

 kBtu 
 878.8 

 hr 
Error 
878.8  841.3
 4.4%
841.3
Cooling Done by Cooling Coil
In HVAC systems, air is typically cooled by passing it over a cooling coil. Most cooling
coils are finned-tube heat exchangers in which cool water or refrigerant flows through
tubes, and the tubes have external fins to increase heat transfer area. If the air is cooled
below the dewpoint temperature, water will condense and must be drained from the
bottom of the cooling coil. A schematic of a cooling coil is shown below.
(1) Warm wet air
(2) Cool dry air
In this situation, Qc is that part of the total heat cooling (Qtot) that leaves with coolant
in the cooling coil. Qw is that part of the total cooling (Qtot) that leaves with
condensate draining from the bottom of the cooling coil. Thus,
Qtot = Qc + Qw
The energy carried away by the condensing water is equal to product of the the mass
flow rate of the condensing water, mw, and the enthalpy of the saturated water, hw.
The enthalpy of the saturated water is the enthalpy or air at 100% RH at the
temperature of the cooling coil. The process is shown schematically below.
8

Qc


ma2

mω 2
ma1

mω1
1
2

mω
T2 Tdp
T1
The process can be modeled using mass balances on the air and water, and an energy
balance on the entire process. The steady state mass and energy balances are shown
below.
M-balance (air):
 a1  m
 a2
m
M-balance (water):
 w1  m
 w2  m
w
m
 w m
 w1  m
 w2  m
 a1 w 1  m
 a2 w 2  m
 a w 1  w 2 
m
E-balance:
 a1h1  Q c  m
 ah2  m
 whw  0 (SS)
m
 a h1  h2   m
 whw
Q c  m
Example
Determine the energy removed by the cooling coil and condensate when cooling 20,000
cfm of air from a temperature of 90 F and a relative humidity of 60% to a temperature
of 55 F and 100% relative humidity. The fluid inside the cooling coil is at 55 F.

Q tot
1
2
T2

Q sen

Q lat
T1
Cooling 20,000cfm from 90F, φ  60% to 55F, φ  100%
9
 ft 3 
 Btu 
 lbw 
v

14.26
  w 1  .01827 
1


 lba 
 lba 
 lba 
1: T1  90F φ1  60% h1  41.74 
2: T2  55F φ2  100% h2  23.2
Btu 
 lbw 
w 2  .009186 


 lba 
 lba 
Btu
w: hw Tc  55 F, φ2  100%   23.2 
 lba 
Mass balance:
 w m
 a w 1  w 2 
m

V 1
w 1  w 2   20,000
v1
14.26
 ft 3  lba 
 lbw 
 min
 lbw 
 .0183  .0092 
 60 
 764.1 




3
 lba 
 hr 
 hr 
 min  ft 
Energy balance:
 a h1  h2   m
 whw
Q c  m
 20,000  60  ft 3  lba  min
 Btu   
 lbw 
 Btu  


Q c  

41.74

23.2
  764.1 
 23.2 





 
3
 14.26

 lba   
 hr 
 lbw  
 min  ft  hr 

 kBtu 
 kBtu 
 kBtu 
Q c  1,560 
- 17.73 
 1,543 



 hr 
 hr 
 hr 
Note that the energy carried away by the condensate (17.73 kBtu/hr) is much less than
the total latent cooling (841.3 kBtu/hr). Most of the latent cooling is done by the
cooling coil. Because the energy removed by the condensate is typically very small
compared to the energy removed by the cooling coil, the energy transferred to the
cooling coil, Qc, is frequently approximated as the total cooling, Qtot.
Bypass Factor
In most cooling applications, the air leaving the cooling coil is not entirely saturated
since some air does not come in contact with the cooling coil. The fraction of air that
misses the coil is called the bypass factor, BF. The bypass factor can be determined
from the temperature of water supplied to the cooling coil and from the states of
incoming and exiting air. For example, consider the cooling process shown below.
10
Tcooling coil
1
1
2
3
2
Tcc
T2
T1
The bypass factor, BF, can be determined from the temperatures of water supplied to
the cooling coil and the incoming and exiting air temperatures as:
BF 
T2  Tcc
T1  Tcc
Example
Air enters a cooling coil at 90F ,   60% , while the coil is at 55°F and the BF = 0.2.
What is the leaving air temperature?
1
3
2
Tcc=55 T2=62
BF 
T1=90
T2  Tcc
T1  Tcc
T2  BFT1  Tcc   Tcc  .290  55  55  62F
Adiabatic Mixing
Air streams at different conditions are frequently mixed together. The condition of the
exiting air stream can be determined from the conditions of incoming streams by
applying mass and energy balances to the system. In most cases, heat loss from the
system is negligible and the system can be modeled at adiabatic. On a psychrometric
chart, the condition of the exiting air stream must be on the line connecting the
incoming air streams. The distance along that line is determined by the ratio of the
mass flow rates of the incoming streams.
11
1
1
3
3
2
2
Air Mass Balance
 1 m
2 m
3
m
V 1 /v 1  V 2 /v 2  V 3 /v 3
Total Energy Balance:
 1h 1  m
 2h 2  m
 3h 3
m
 h m
 2h 2
m
h3  1 1
 1 m
2
m
V h  V 2h2
if ρ 1  ρ 2 
h3  1
V 1  V 2
Sensible Energy Balance:
 1 c p T1  m
 2 c p T2  m
 3 c p T3
m
 1 T1  m
 2 T2
m
 1 m
2
m
V T  V 2 T2
if ρ1  ρ 2 
T3  1 1
V 1  V 2
T3 
Sensible Heat Ratio, Latent Cooling and Air Flow Rate
The sensible heat ratio, SHR, is the ratio of sensible cooling to total cooling.
SHR 
Q sen
Q sen

.
Q tot Q sen  Q lat
On a psychrometric chart, the total cooling line can be decomposed into sensible and
latent cooling lines, as shown below.
12
1
2
A
Example
Determine the sensible heat ratio for cooling 20,000 cfm of air from a temperature of 90
F and a relative humidity of 60% to a temperature of 55 F and 100% relative humidity.

 a h1  hA   m
 a h A  h2 
Q tot  ma h1  h2   m

Q lat

Q sen
SHR 
 h  h2  h A  h2 31.74  23.2
Q sen m
 a A


 46%
 a h1  h2  h1  h2 41.74  23.2
Q tot m
For a given cooling coil, SHR varies with the volume of air flowing across it, V cc . As V cc
increases, the total cooling capacity, Q tot , increases. However, the increased volume
flow rate of air also increases the quantity of moist air that does not come in contact
with the cooling coil; which increases the bypass factor, BF. The increased bypass
factor, BF, causes latent cooling, Q lat , to increase less than total cooling, Q tot . The net
result is that latent cooling decreases with higher air flow rates.
Thus, for dehumidification, it is advantageous to run air conditioners with low air flow
rates for longer periods of time. It also follows that oversized air conditioning systems
that meet total cooling loads by cycling on for short periods with high air flows and then
remaining off for long periods are less able to provide adequate dehumidification than
properly sized units which run more continuously.
13
Example
Performance specifications from an air conditioning system are shown below. Consider
the case when the condenser temperature is 120 F and the evaporator air wet bulb
temperature is 72 F. Determine the SHR of the evaporator coil when operating with
9,000 cfm and 15,000 cfm of air across the evaporator. Note that the SHR increases,
and latent cooling decreases.
Cond Temp
(F)
9,000 / 0.15
Evap Air cfm / BF
12,000 / 0.18
15,000 / 0.21
Evap Air Entering Wet BulbTemp (F)
62
72
67
62
72
67
320.6
410.7
374.9
340.1
424.4
388.0
280.0
213.9
270.6
325.9
233.0
301.2
28.5
30.7
30.0
29.0
31.0
30.2
403.6
499.7
461.7
424.5
514.1
475.7
305.9
391.5
356.8
323.6
404.1
368.6
273.0
206.7
263.3
316.2
226.2
293.8
30.7
33.3
32.4
31.4
33.6
32.8
395.2
488.0
450.8
414.7
501.3
463.6
284.7
365.1
332.1
300.7
376.2
342.8
263.1
197.8
253.7
299.5
216.6
283.9
32.9
36.1
34.9
33.7
36.5
35.3
380.2
468.5
431.6
398.4
481.5
443.0
270.8
345.8
315.3
285.4
355.6
324.8
257.0
190.9
246.9
285.4
209.8
277.1
35.0
38.5
37.2
35.9
38.9
37.6
372.3
457.5
423.0
389.5
468.3
433.7
261.2
335.3
304.9
274.8
344.7
314.2
252.7
187.3
243.0
274.8
206.1
273.2
35.9
39.7
38.1
36.7
40.1
38.6
365.7
450.2
381.2
381.2
460.8
426.0
72
67
62
TC
389.8 354.2
353.1
SHC
192.5 236.8
353.1
110
KW
30.3
29.5
29.4
THR
477.6 439.7
438.5
TC
372.1 338.1
335.5
SHC
185.6 229.9
335.5
120
KW
32.9
31.9
31.8
THR
467.3 430.6
427.8
TC
348.0 315.6
311.2
SHC
176.8 220.6
311.2
130
KW
35.5
34.3
34.1
THR
448.8 414.6
409.9
TC
330.9 300.3
294.9
SHC
170.5 214.3
294.9
140
KW
37.9
36.5
36.3
THR
440.7 406.3
400.3
TC
320.9 290.2
283.8
SHC
167.0 210.4
283.8
145
KW
39.0
37.5
37.1
THR
433.7 398.7
391.4
TC - Total Cooling (KBtu / hr)
SHC - Sensible Cooling (KBtu / hr)
KW - Electric Power Input (KW); Electric Power Input = Compressor Power + Evaporator Fan Power
THR - Total Heat Rejected (KBtu / hr)
Qtot
Qsen
Qlat
SHR = Qsen/Qtot
V=9,000 cfm
372.1
185.6
186.5
0.50
V=15,000 cfm
404.1
226.2
177.9
0.56
14