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THERMODYNAMICS II: TUTORIAL PROBLEMS
THE WORKING FLUID:
1.1 A certain fluid at 10 bar is contained in a cylinder behind a piston, the initial volume
being 0.05 m3. Calculate the work done by the fluid when it expands reversibly:
(i) at constant pressure to a final volume of 0.2 m3;
(ii) according to a linear law to a final volume of 0.2 m3 and final pressure of 2 bar;
(iii) according to a law pV = constant to a final volume of 0.1 m3;
(iv) according to a law p3 = constant to a final volume of 0.06 m3;
(v) according to a law, p = (A/V2) – (B/V), to a final volume of 0.1 m3 and a final
pressure of 1 bar, where A and B are constants.
Sketch all processes on a p- diagram.
Solution:
(i) Consider the constant pressure process:
The work done during this process is the area under the curve:
Work = area of the rectangle
= p x (v2 – v1)
= 10 x 105 x (0.2 – 0.05)
= 150000 N m
(ii) Consider linear law:
1
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, lOMoARcPSD|5769186
The work done is the sum of the two areas, the area of the triangle and the area of
the rectangle:
Consider the triangle, Work = 0.5 x base x height
= 0.5 x (v2 – v1) x (p1 – p2)
= 0.5 x (0.2 – 0.05) x (10 bar – 2 bar)
= 60000 N m
Consider the rectangle, Work = p2. (v2 – v1)
= 2 bar x (0.2 – 0.05)
= 30000 N m
The total work done = 60000 + 30000
= 90000 N m
(iii) pV = constant, where v2 = 0.1 m3 and v1 = 0.05 m3
W mp dv
pV = c
p = c/V c = 10 x 105 x 0.05
= 50000 bar (m3)
c
V
W dV
v2 dV
W c
v1 V
W cln v2 ln v1
W 50000xln 0.1 ln 0.05
W 34657Nm
2
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, lOMoARcPSD|5769186
(iv) pv3 = constant
p = c/v3 c = 10 x 105 x 0.05(3)
= 125 bar (m3)
W mp dv
c
v3
W dv
dv
W c
v3
1 1
W 0.5c 2 2
v 2 v1
1 1
W 0.5 x125
0.06
2
0.052
W 7639Nm
1.2 1 kg of fluid is compressed reversibly according to a law pv = 0.25, where p is in bar
and v is in m3/kg. The final volume is ¼ of the initial volume. Calculate the work done by
the gas on the fluid and sketch the processes on a p-v diagram.
c = 0.25 x105 bar (m3/kg)
c
p
V
c
V
W dV
v2 dV
c
v1 V
cln v2 ln v1
0.25x105 ln1 ln 0.25
W 34657Nm
3
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