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HOMEWORK SET 4
4.41
Ammonia (0.5 kg) is in a piston cylinder at 200 kPa, 10
o
C is heated in a process
where the pressure varies linear with the volume to a state of 120
o
C, 300 kPa.
Find the work the ammonia gives out in the process.
Solution:
Take CV as the Ammonia, constant mass.
Continuity Eq.:
m
2
= m
1
= m
;
Process:
P = A + BV
(linear in V)
State 1: Superheated vapor
v
1
= 0.6193 m
3
/kg
State 2: Superheated vapor
v
2
= 0.63276 m
3
/kg
Work is done while piston moves at increasing pressure, so we get
1
W
2
=
∫
P dV = area = P
avg
(V
2
−
V
1
) =
1
2
(P
1
+ P
2
)m(v
2
−
v
1
)
= ½(200 + 300)
×
0.5 (0.63276
−
0.6193) =
1.683 kJ
4.49
Helium gas expands from 125 kPa, 350 K and 0.25 m
3
to 100 kPa in a polytropic
process with n = 1.667. How much work does it give out?
Solution:
Process equation:
PV
n
= constant = P
1
V
n
1
= P
2
V
n
2
Solve for the volume at state 2
V
2
= V
1
(P
1
/P
2
)
1/n
= 0.25
×
⎝
⎜
⎛
⎠
⎟
⎞
125
100
0.6
= 0.2852 m
3
Work from Eq.4.4
1
W
2
=
P
2
V
2
 P
1
V
1
1n
=
100
×
0.2852  125
×
0.25
1  1.667
kPa m
3
=
4.09 kJ
P
P
2
P
v
1
1
2
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View Full DocumentThe actual process is
on a steeper curve than
n = 1.
4.67
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 Fall '09
 DennisBlumenfield

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