Conditioning of Moist Air




 a T2  T1 
q  1.02m
Properties of moist air
Equations
– Conservation of mass
– Conservation of energy Change using bulk properties at inlet and outlet
 a h 2  h1 
qm
Classic moist air processes
– Sensible heating and cooling
– Cooling and dehumidifying
h
h

w
– Adiabatic humidifying
W
– Heating and humidifying
– Adiabatic mixing
Space conditioning



m
32 m
32 m
13



– Sensible heat factor (SHF) m 13 m 12 m 12
– Cooling coil bypass (b)
– Evaporative cooling
– Economizer cycle
2
3
– Effect of fans
1
– Designing indoor RH
a1
a1
a2
a2
a3
a3
h
q

 hw
w
W m
2
1
Adiabatic mixing - Example
– Two thousand cubic feet per minute (cfm) of air at 100 F
db and 75 F wb (state 1) are mixed with 1000 cfm of air
at 60 F db and 50 F wb (state 2). The process is
adiabatic at a steady flow rate and at standard sea level
pressure. Find the condition of the mixed streams.
– Convert to SI:
o
C  (o F  32) / 1.8
100o F  37.8o C
1 cfm  0.4719 Ls
75o F  23.9 o C
2000 cfm  944 L / s
60o F  15.6o C
1000 cfm  472 L / s
50o F  10o C
Adiabatic mixing - Solution
v1  0.9 m3 / kg
 a1  0.944 / 0.9  1.05 kg / s
m
v 2  0.825 m3 / kg
 a 2  0.472 / 0.825  0.57 kg / s
m
partial graphical method
 a2
m
W2  W1   13.0  0.57 5.3  12.9  10.3 g / kg
W3  W1 
 a3
m
1.62
T3  30o C
fully graphical method
 a1 23
m

 a 3 12
m
3  38%
Twb3  19.9o C
1
3
1.05
48.5 units 
23 
1.62
 31.4 units
2
Typical air handling system
23°C
Exhaust air
Return air
recirculation air
energy
recovery
system
-20°C
Outdoor air
economizer
damper
22°C
filter
heating coil
cooling coil
Supply air
15°C - 35°C
Sensible heat factor
SHF 
q
sensible heat trans fer q s
 s
total heat trans fer q s  q l q
 a (h z  h 2 )
qs  m
 a1 , T1 , W1 , h1
m
1
 a ( h1  h z )
ql  m
q
coil
 a 2 , T2 , W2 , h 2
m
2
 w , Tw , h w
m
 a1  m
 a2  m
a
m
1
2
z
Sensible heat factor - Example
– Conditioned air is supplied to a space at T=15C and
Twb=14C at a flow rate of 0.5 kg/s. The sensible heat
factor for the space is 0.70 and the space is to be
maintained at 24C . Find the sensible and latent
cooling loads for the space.
2
T  24o C
SHF  0.7
If the cooling load is 6.6kW,
what is m1 if state 1 is here?
1
 1  0.5 kgs
m
T1  15o C
Twb1  14 o C
 a ( h 2  h1 )
qm
parallel
2
1
 0.5(52.5  39.3)  6.6 kW (i.e., the supply air is heated and the space air is cooled)
q s  q SHF  6.60.7  4.6 kW
q l  q  q s  2.0 kW
Cooling coil bypass
 a 3 , T3 , W3
m
 a1 , T1 , W1
m
1
3
 w , Tw , h w
m
T5 , 5
6
 a 3 , T3 , W3
m
 a1 , T1 , W1
m
1
4
 w , Tw , h w
m
3
1
2
5
3
bypass factor
4
T4 = apparatus dew point
b
 a 6 43
m

 a1 14
m
T5 , 5
Evaporative cooling - 1
3
 a 2 , T2 , W2 , h 2
m
 a1 , T1 , W1 , h1
m
– Energy balance
1
 a h1  m
 whw  m
 ah2
m
h1  h 2
 w , hw
m
qs
ql
2
negligible for water
SHF  0.6
– Given air at outdoor
conditions (1) and the
SHF of the space we can
use evaporative cooling to
cool from state 1 to state 2
– The flow rate of air that
is required to cool the
space depends on
cooling load
2
3
1
Is this a more favourable outdoor
condition for evaporative cooling?
Evaporative cooling - 2
– Low cost alternative to refrigerant systems


Requires less energy for cooling
Requires less capital investment
– Cooling potential  (Twb,design – Twb,outdoors)


If the cooling potential is small, the air flow rates become very large and
system is not economical because of fan and duct costs
If cooling load is large, air flow rates become large
– Evaporative systems are subject
to mold and bacteria growth
– Freezing may be a problem
during swing seasons
(spring/fall) in colder climates
no potential
OA
design target
good potential
OA
Economizer cycle
– Using outdoor air to condition the indoor space
– Can be used during “off-design” conditions to save
energy (usually cooling) by increasing the amount of
outside (ventilation) air (usually at night-time)
– May be limited in amount of outdoor air if RH must be
controlled to a specific value (often RH must remain
below some maximum value)
– must measure the enthalpy (T & ) to properly control
4
5
OA
1
2
3
Economizer cycle
4
4
OA
– Known:
1
2
3
1.
2.
3.
4.
5.
–
Outdoor design conditions
Indoor design conditions
Loads -> SHF -> condition line
Often T3,min -> state 3 -> ma3
Knowing the ventilation rate
ma1, mix 1 and 4 to get 2
6. Cool (+bypass) from 2 to 3
Increase ventilation rate
for “off-design”
conditions of outdoor air
want Td  8 C
o
1
1
2
2
Td
3
Td
1
4
Effect of fans
– Large HVAC systems have both a supply and exhaust
air fan to keep the building at a desired P compared to
outdoors.
– All of the power is ultimately degraded to sensible
energy in the airstream.
– We will assume all of the energy rise occurs at the fan
(most does)
4’
Power 
Q P
1000f m
4
5
Q  volumetric air flow rate (L/s)
P  fan pressure rise (Pa)
3
f  fan efficiency ( 60 to 80%)
OA
m  motor efficiency ( 80 to 90%)
 100% if motor is outside the airstream
1
2
3’
4’
Effect of fans - summer
4
5
3
1
2
3’
Where are state points 3’ and 4’ ?
1
2
3’
Td
4’
3
Exhaust fan increases the
load on the cooling coil
4
Supply fan increases Td or reduces reheat
Effect of fans - winter
6
qs
ql
7
qh
5
1
Both fans reduce the heating coil load
4
5
6
7
2
1
3
2
3
4
Designing indoor RH - cooling
– Given:







qs - load calculation
ql - load calculation
T4 - comfort
T3 - comfort, energy
m1 - ventilation
T1 - weather data/design conditions
1 - design conditions
4
qs
ql
qc
1
2
3
1
1
SHF 
qs
qs  ql
Only state point 1 and the slope
of the load line are specified
The condition line is not fixed
vertically
T3
T4
T1
Humid climate
– Assume condensation at the coil
– We can now design 4 and control
4
it with the apparatus dew point
1
2
4
4
3
d
T3
T4
2
1
1
d
qs
ql
qc
T1
3
Dry climate
– Assume no condensation at the coil
– We must use outdoor air to control
the indoor humidity (4)
– More difficult because we may not
be able to control the outdoor air
1
flow rate over a large range
4
qs
ql
qc
2
3
 w h fg m
 w  q l / h fg
ql  m
Since there is no condensation at
the coil, this moisture must be
removed by outdoor air
 w m
 1 W4  W1 
m
4
4
3
1
2
1
T3
T4
T1
Horizontal line for no condensation
W4 
w
m
 W1

m1
Dry climate - 2
– If we can’t increase the outdoor air flow
rate to reach the indoor design humidity,
the indoor humidity will decrease
– Usually ok to allow indoor RH to
decrease in summer
1
– In this case the indoor humidity can
be
calculated as follows
 w h fg
ql  m
4
qs
ql
qc
2
3
 w  q l / h fg
m
Since there is no condensation at the coil, this
moisture must be removed by outdoor air
4
 w m
 1 W4  W1 
m
W4
4
Solving for W4
W4 
w
m
 W1

m1
3
1
2
1
This locates state point 4 and determines 4
T3
T4
T1