MECH 430: psychrometrics 5 - Queen's Universitymy.me.queensu.ca/Courses/MECH4301/430 Psychro 5 plus...
Transcript of MECH 430: psychrometrics 5 - Queen's Universitymy.me.queensu.ca/Courses/MECH4301/430 Psychro 5 plus...
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MECH 430: psychrometrics 5
Humidification
Steam Injection Liquid Injection
See Ex. 12.14 in text. Alternative solution recommended with Psychrometric Chart.
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MECH 430: psychrometrics 5
http://www.condair.com/isothermal-steam-humidifiers
http://steamix.com/humidification-conditioned-steam-stainless-steel-direct-steam
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MECH 430: psychrometrics 5
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MECH 430: psychrometrics 5
Evaporative Cooling
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MECH 430: psychrometrics 5
Suitable for cooling in hot, relatively dry climates
Low moist air enters and readily “soaks” up liquid via evaporation.
This evaporation accomplished by Q extracted from moist air intake.
Results in a decrease in the temperature of the moist air
Liquid either in soaked pad or sprayed in air stream.
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MECH 430: psychrometrics 5
Mass and Energy Balances
Assuming 0cvW and Q with surroundings = 0.
2 2 2 2 1 1 1 1( ) ( ) ( )a g f a gh h h h h Eq. 30.1
where,
All of the water is assumed to evaporate into air stream
2 1( ) fh = Energy carried in with the injected liquid
water. Typically, this is much less than the enthalpy of moist air in or out.
Thus mixH constant and we can represent process as a
constant mixH process. Since mix
a
H
m and WBT lines coincide, it is
also a constant WBT process.
In reality, the process deviates from a line of constant mix
a
H
m to
a slightly higher mix
a
H
m.
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MECH 430: psychrometrics 5
Example Text Problem 12.86 Air at 35oC, 1 bar and 10% enters an evaporative cooler
(operating at steady state) at 50 m3/min. Liquid water at 20oC enters the cooler and fully evaporates. Moist air leaves the
cooler at 25oC, 1 bar. Assume Q with surroundings = 0.
Find ,w exitm
wm at T3 = 20oC
T1 = 35oC T2 = 25oC P1 = 1 bar P2 = 1 bar
2 ? 10%
3
1( ) 50min
mAV
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MECH 430: psychrometrics 5
Solution
1 2a a am m m
2 1w v vm m m
Since v
a
m
m , then 2 1
w
a
m
m
Therefore, to find wm , we need: 1 2, , am
To find 1 :
11
1
0.622 v
v
P
P P
where
1 1 1
1
1
(0.1)(0.05628)
0.005628
v g
v
v
P P
P
P bar
1
1
(0.622)(0.005628)
(1 0.005628)
0.00352 ( ) / ( )kg v kg a
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MECH 430: psychrometrics 5
To find 2 :
Recall equation 30.1:
2 2 2 2 1 3 1 1 1( ) ( ) ( )a g f a gh h h h h
Isolating 2w :
1 2 1 1 3
2
2 3
2
2
( ) ( )
(308.23 298.18) 0.00352(2565.3 83.96)
2547.2 83.96
( )0.00763
( )
a a g f
g f
h h h h
h h
kg v
kg a
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MECH 430: psychrometrics 5
To find am :
1 1
1
3 5 2
( )
(50 / min)(0.994373 10 / )
8314 28.97 (308 )
56.25 ( ) / min
aa
a
a
a
AV Pm
RT
M
m x N mm
N m kgK
kmol K kmol
m kg a
Thus
2 1( )
56.25 ( ) / min 0.00763 0.00352
0.231 / min
w a
w
w
m m
m kg a
m kg
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MECH 430: psychrometrics 5
2( ) (25 )o
g gP T P C
To find 2 :
22
2
2
2
0.622
(0.00763)(1 )
0.00763 0.622
0.01212
v
v
v
PP
barP
P bar
22
2
2
2
0.01212
0.03169
0.382 or 38.2%
v
g
P
P
bar
bar
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MECH 430: psychrometrics 5
Cooling Towers
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MECH 430: psychrometrics 5
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MECH 430: psychrometrics 5
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MECH 430: psychrometrics 5
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MECH 430: psychrometrics 5
Used, for example, when:
Sufficient cooling unattainable from natural sources e.g., lakes, rivers.
Environmental concerns limit Tmax of cooling water returned to environment.
Can be operated by Natural or Forced Convection Configurations:
Counterflow
Cross-flow
Usually neglect FANW and Q with surroundings as these are
usually much smaller than other quantities considered in energy balances on cooling towers.
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MECH 430: psychrometrics 5
Analysis Mass Rate Balances
3 4a am m (dry air)
1 5 3 2 4v vm m m m m (water)
Since 1 2m m :
then 5 4 3v vm m m
Since 4 4v am m and 3 3v am m :
then 5 4 3( )am m Eq. 30.2
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MECH 430: psychrometrics 5
Energy Rate Balance
1 1 3 3 3 5 5 2 2 4 4 40 ( ) ( )w a a v v w w a a v vm h m h m h m h m h m h m h
Let each v gh h at given T
Let each w fh h at given T
with
1 2m m , 5 4 3( )am m , 3 3v am m , 4 4v am m
We can re-write the above equations as:
1 1 2
4 4 4 3 3 3 4 3 5
( )
( ) ( ) ( )
f f
a
a g a g f
m h hm
h h h h h
Eq. 30.3
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MECH 430: psychrometrics 5
Example Cooling water leaves the condenser of the power plant and enters a cooling tower at a rate of 100 kg/s. The water is cooled to 22oC in the cooling tower by air which enters the
tower at 1 atm, 20oC, 60% and leaves saturated at 30oC.
Neglect the power input to the fan. Find: ) )a makeup watera m b m
4
4
30
100%
oT C
1
1
35
100 /
oT C
m kg s
2
2
22
100 /
oT C
m kg s
5
5
22
?
oT C
m
3
3
3
1
60%
20
?
o
a
P atm
T C
m
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MECH 430: psychrometrics 5
Solution To solve, evaluate Eq. 30.3 and then 30.2 2 approaches:
1) Tables 2) Psychrometric Chart
* Review text Ex. 12-17 to familiarize yourself with tabular approach. In this example, we will use a psychrometric chart.
1
1
2
5 2 5 2
100 /
(35 ) 146.68 /
(22 ) 92.33 /
(in this example since )
o
f f
o
f f
f f
m kg s
h h C kJ kg
h h C kJ kg
h h T T
Mixture enthalpies, 4 3, from psychrometric chart:
4
3
4 4 4
4
3 3 3
3
0.0273
0.0087
( ) 100.0 / ( )
( ) 42.2 / ( )
mixa v
a
mixa v
a
Hh h kJ kg a
m
Hh h kJ kg a
m
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MECH 430: psychrometrics 5
2T 4T
3
4w
3
mix
a
H
m
4
mix
a
H
m
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MECH 430: psychrometrics 5
Evaluate Eq. 30.3
1 1 2
4 4 4 3 3 3 4 3 5
( )
( ) ( ) ( )
(100 / )(146.68 92.33) /
(100 42.2) / (0.0273 0.0087)(92.33) /
96.9 /
f f
a
a g a g f
a
a
m h hm
h h h h h
kg s kJ kgm
kJ kg kJ kg
m kg s
Evaluate Eq. 30.2
5 4 3
5
5
( )
(96.9 / )(0.0273 0.0087)
1.80 /
am m
m kg s
m kg s
Note: , 100 /in waterm kg s
, 100 1.80 98.2 /returned waterm kg s
That is, only 1.8% of condensate lost from cooling tower.
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MECH 430: psychrometrics 5
What if the cooling effect was provided by circulating cooling water coming in at 5oC and leaving at 30oC? ?cwm
( (35 ) (22 )) ( (30 ) (5 ))
(146.68 92.33)100 /
(125.79 20.98)
51.86 /
o o o o
cond f f cw f f
cw
cw
m h C h C m h C h C
m kg s
m kg s
Recall for a cooling tower, 1.80 /makeup waterm kg s
35oC 35oC
22oC 5oC
100 /condm kg s ?cwm
1
2
3
4