Answer:
−
39.87633
k
J
/k
g
Explanation:
Air at 35 C, 1 atm, and 50% relative humidity enters a dehumidifier operating at steady state. Saturated moist air and condensate exit in separate streams, each at 15 C.
Neglecting kinetic and potential energy effects, determine:(a) the heat transfer from the moist air, in kJ per kg of dry air.
b) the amount of water condensed, in kg per kg of dry air.
(c) Check your answers using data from the psychrometric chart.
Using conservation of mass
˙mv1=mv2+mw
mw=mv1-mv2
mv1=w1mair
mv2=w2mair
w=humidity ratio
m=mass
v=velocity
rate of condensation
mv1-mv2
mw=w1mair-w2mair
mw/mair=w1-w2
humidity ratios , we know is
w=0.662pv1/(p-pv1)
p=atmosphjeric pressure 1.013bar
Using the the psychometric table
pv1=∅*pg
Pv1=50%(0.05629)
Pv1=0.028145
b
a
r
from w=0.662pv1/(p-pv1)
w=0.662*0.028145/(1.013-0.028145)
w=0.0189185
at T=15C
pv2=p sat=0.1705
w2=0.662*0.01705/(1.013-0.01705)
0.011332
Water condenses at
mw/mair=w1-w2
0.0189185
−
0.011332
=
0.0007598
b.Heat transfer
Q
c
v
=
m
(
h
a
1
−
h
a
2
)
−
w
1
h
g1
+
w
2
h
g
2
+
(
w
1
−
w
2
)
h
f
2
Using table
h
f
1
=
146.68
h
f
2
=
62.9
h
g
1
=
2565.3
h
g
2
=
2528.9
For air
T
=
35
0
C
h
a
1=
308.2
T
=
15
0
C
h
a
2
=
288.15
Thus
Q
/m
a
i
r
=
(
288.15−308.2−
0.0189185
(
2565.3)+
0.01132
(
2528.9
)
+
0.00075985
(
62.9
)
=
−
39.87633
k
J
/k
g
