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Closed Tank - Chemical Engineering - Previous Solved Exam, Exams of Engineering Chemistry

Main points of this exam paper are: Closed Tank, Liquid, Large Initial Pressure, Final Pressure, Atmospheric, Balance Reads, Ideal

Typology: Exams

2012/2013

Uploaded on 04/01/2013

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Chemical Engineering 140 October 1, 2010
Midterm #1 Solutions
1 a. As liquid drains, air expands, and the pressure in the tank declines. One of two cases
arises. Eventually, the liquid may completely drain, and the air pressure in the tank
matches atmospheric. However, it is also possible for the gas to expand to atmospheric
pressure while some liquid remains in the tank. For large initial gas volume (large α) and
for large initial pressure (large β) we expect complete drainage. For small α and small β,
air remains in the tank.
b. The final pressure is atmospheric, Pa
c. Let n represent the moles of gas. Mass balance reads
dn/ dt 0 so n(t) =n(0)= constant
(1)
d. Since the initial pressure and gas volume are known and since the gas is ideal and
isothermal, we can re-express Eqn. 1 as
Ga
PV P V

(2)
where VG is the gas volume. Since the final pressure is atmospheric, the final gas volume
is
G
V ( ) V


(3)
Clearly, the maximum gas volume is when the tank is completely empty of liquid.
Thus, if
1

, gas remains in the tank. If
1

, liquid completely drains
e. Let VL represent the volume of liquid in the tank. Since liquid is incompressible, mass
balance reads
La
dV / dt k( P P )
(4)
f. We must re-express Eqn. 4 in terms of pressure. Since
GL
V V V
, Eqn. 4 is
combined with Eqn 2 to read
Gaa
2
dV 1
PV dP / dt k( P P )
dt P

(5)
or
pf3
pf4

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Chemical Engineering 140 October 1, 2010

Midterm #1 Solutions

1 a. As liquid drains, air expands, and the pressure in the tank declines. One of two cases arises. Eventually, the liquid may completely drain, and the air pressure in the tank matches atmospheric. However, it is also possible for the gas to expand to atmospheric pressure while some liquid remains in the tank. For large initial gas volume (large α) and for large initial pressure (large β) we expect complete drainage. For small α and small β, air remains in the tank.

b. The final pressure is atmospheric, Pa

c. Let n represent the moles of gas. Mass balance reads

dn / dt0 so n(t) =n(0)= constant (1)

d. Since the initial pressure and gas volume are known and since the gas is ideal and isothermal, we can re-express Eqn. 1 as

PVG   PaV (2)

where VG is the gas volume. Since the final pressure is atmospheric, the final gas volume is

V (G  )   V (3)

Clearly, the maximum gas volume is when the tank is completely empty of liquid.

Thus, if   1 , gas remains in the tank. If   1 , liquid completely drains

e. Let VL represent the volume of liquid in the tank. Since liquid is incompressible, mass balance reads dV / dtL   k( PP )a (4)

f. We must re-express Eqn. 4 in terms of pressure. Since VVGVL , Eqn. 4 is

combined with Eqn 2 to read

G a 2   a

dV 1 PV dP / dt k( P P ) dt P

or

Chemical Engineering 140 Midterm solution continued October 1, 2010

2

dP / dt   kP ( P  P ) /(a  PV )a (6)

with P(0)=βPa. Eqn 6 is integrable (see # 41 of Standard Math Table of integrals). You are not asked for the solution. However, the answer is

1 a a a a a

( 1)( P / P ) [1 ( P / P )]

ln ktP /( V ) [( P / P ) 1] P / P

   

   ^ 

 ^ 

This result teaches that it takes an infinite time to reach the final state. Of course, to get almost there, say 95 %, requires a finite time. Note that Eqn. 7 is implicit in pressure, quite an acceptable result.

g. To arrive at the characteristic time without solving the problem, we rewrite Eqn 6 in nondimensional form

a 2 a a a

d( P / P ) ( P / P ) [( P / P ) 1]kP /( V ) dt

Inspection gives the characteristic time

chac a

V

t kP

temperature. The CO 2 and water with a small amount of MEA is fed into a gas-liquid separator in order to further purify CO 2. The water and MEA solution from the separator is recycled and sent back to the stripper.