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V=320

P=latin

Air flowing at 320 m/min is to be conditioned in a constant pressure process at 1 atm from 43°C&

and 40 percent relative humidity to 25°C dry bulb and 18.5°C wet bulb. The air is first passed ove

T= 43°C

coils to remove all the moisture necessary to achieve the final moisture content and then is

heating coils to achieve the final state. Fill in the properties table [8%) and determine the following:

=40%

passed

T₁=25°C

Twbs=18.5°C

1

2

3

Name:

Problem 3: [30%]

Tab

(°C)

43

-Y = Pv

8.71

25

Pu

117057

P- sat

_=W3

Pu

= Pesat

Heating coils

teleeeeeeeeeeeee

W₂ =W₂

from

then ermine

$

(%)

40

too

2

chart

(0

2

state 2

$/2=100%.

0.0714

1.635

1.635

a) [3%] Sketch the process on the provided psychrometric chart.

b) [2%] The mass flow rate of air in "kg/s"

→Pv₁ = 3.484 kpa

Cooling coils

P=1₁7057

15°C

h

(kJ/kga)

llllllle

1 atm

Condensate

Condensate

removal

(m³/kga)

T=43

Pat= 8.71 kpap

W 0.622 Pu

P-PU

= 0.622 (3.484)

8.71 (3,484)

= 0.0714

T₁=43°C

₁-40%

State

Twb

(°C)

اسپنسه

18.5

|Tz=15°C1

Sot = 1.7057

0,622(1.7057)

1.50

1176037-(1.1257)

التي

1/2 = 1.639

8=.

12

2

1.55

T=25°C

=316=

0.60

065

03/nName:

c) [5%] The rate of condensation in cooling section in "kg/s"

d) [7%] The rate of heat removal in the cooling section in "kW"

e) [5%] The rate of heat input in the heating section in "kW"

Fig: 1

Fig: 2