conditioning of moist air l properties of moist air l equations – conservation of mass –...
TRANSCRIPT
Conditioning of Moist Air Properties of moist air Equations
– Conservation of mass– Conservation of energy
Classic moist air processes– Sensible heating and cooling– Cooling and dehumidifying– Adiabatic humidifying– Heating and humidifying– Adiabatic mixing
Space conditioning– Sensible heat factor (SHF)– Cooling coil bypass (b)– Evaporative cooling– Economizer cycle– Effect of fans– Designing indoor RH
Change using bulk properties at inlet and outlet
23
1 1
2
12a TTm02.1q
12a hhmq
W
hhw
ww
hm
q
W
h
31
23
m
m
2a
1a
21
23
m
m
3a
1a
21
31
m
m
3a
2a
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:8.1/)32F(C oo
s/L472cfm1000
C8.37F100 oo
sL4719.0cfm1
C9.23F75 oo C6.15F60 oo C10F50 oo
s/L944cfm2000
3
1
2
kg/m9.0v 31
Adiabatic mixing - Solution
kg/m825.0v 32
s/kg05.19.0/944.0m 1a
s/kg57.0825.0/472.0m 2a
kg/g3.109.123.562.1
57.00.13
C30T o3
123a
2a13 WW
m
mWW
C9.19T o3wb %383
21
32
m
m
3a
1a
510
0
20
30
40
50
units4.31
units5.4862.1
05.132
method graphical partial
method graphicalfully
Typical air handling system
Outdoor air
Exhaust air
filter
heating coil
cooling coil
Return air
energyrecoverysystem
recirculation air
damper
-20°C
22°C
23°C
15°C - 35°C
Supply air
economizer
Sensible heat factor
a2a1a mmm
q
q
q
ferheat trans total
ferheat trans sensibleSHF s
ls
s
1
2
)hh(mq 2zas
)hh(mq z1al
z
sensib
le
laten
t
1 2
coil q 2222a h,W,T,m1111a h,W,T,m
www h,T,m
Sensible heat factor - Example
1
2
kW6.6)3.395.52(5.0
)hh(mq 12a
kW0.2qqq sl
– Conditioned air is supplied to a space at T=15C and Twb=14C 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 24C . Find the sensible and latent cooling loads for the space.
condition lineskg
1 5.0m C15T o
1 C14T o
1wb
1
2
C24T o7.0SHF
kW6.47.06.6SHFqqs
parallel
If the cooling load is 6.6kW,what is m1 if state 1 is here?
(i.e., the supply air is heated and the space air is cooled)
1
3
Cooling coil bypass
2
4
41
34
m
mb
1a
6a
bypass factor
condition line
1 3
333a W,T,m111a W,T,m
www h,T,m55 ,T
5
T4 = apparatus dew point
1 3
333a W,T,m111a W,T,m
www h,T,m
6
4
55 ,T
Evaporative cooling - 1
1 2
1111a h,W,T,m
ww h,m
1
2
– Energy balance
– Given air at outdoorconditions (1) and theSHF of the space we canuse evaporative cooling tocool from state 1 to state 2
– The flow rate of air thatis required to cool thespace depends on cooling load
2aww1a hmhmhm negligible for water
2222a h,W,T,m
21 hh
3
condition line
3
6.0SHF
Is this a more favourable outdoorcondition for evaporative cooling?
sq
lq
Evaporative cooling - 2
OA
OAdesign target
good potential
no potential
– 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 subjectto mold and bacteria growth
– Freezing may be a problemduring swing seasons(spring/fall) in colder climates
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
OA1 2 3
4
5
– Known:1. Outdoor design conditions2. Indoor design conditions3. Loads -> SHF -> condition line
4. Often T3,min -> state 3 -> ma3 5. Knowing the ventilation rate
ma1, mix 1 and 4 to get 26. Cool (+bypass) from 2 to 3
– Increase ventilation rate for “off-design”conditions of outdoor air
Economizer cycle
OA1 2 3
4
4
1
4
2
condition line
1
3
Td
1
2
Td
C8Twant od
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)
OA1 2 3’
4
5mf1000
PQPower
(L/s) rate flowair volumetricQ
(Pa) rise pressurefan P
80%) to60( efficiencyfan f
airstream theoutside ismotor if 100%
90%) to80( efficiencymotor m
3
4’
1
4
2
condition line3
Td
1 2 3’
4
5
3
4’
3’
4’
Supply fan increases Td or reduces reheat
Where are state points 3’ and 4’ ?
Exhaust fan increases theload on the cooling coil
Effect of fans - summer
1
condition line
1 2 4
6
7
5
Both fans reduce the heating coil load
Effect of fans - winter
qh
qs
ql
3
64
7
2 3
5
condition line
Designing indoor RH - cooling
1 2
4
qcqs
ql
3
1
– Given: qs - load calculation
ql - load calculation
T4 - comfort
T3 - comfort, energy
m1 - ventilation
T1 - weather data/design conditions
1 - design conditions
T1
1
T4T3
ls
s
qSHF
Only state point 1 and the slope of the load line are specifiedThe condition line is not fixed vertically
condition line
Humid climate
1 2
4
qcqs
ql
3
1
– Assume condensation at the coil
– We can now design 4 and control it with the apparatus dew point
T1
1
T4T3
4
4
2
3
d
d
4
condition line
Dry climate
1 2
4
qcqs
ql
3
1
– 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 flow rate over a large range
T1
1
T4T3
4
23Horizontal line for no condensation
fgwl hmq fglw h/qm
Since there is no condensation at the coil, this moisture must be removed by outdoor air
141w WWmm
11
w4 W
m
mW
4
Dry climate - 2
1 2
4
qcqs
ql
3
1
– 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
– In this case the indoor humidity can be calculated as follows
T1
1
T4T3
4
23
fgwl hmq fglw h/qm
Since there is no condensation at the coil, this moisture must be removed by outdoor air
141w WWmm
11
w4 W
m
mW
Solving for W4
This locates state point 4 and determines 4
W4