Question

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.

Answer 1

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