Thermodynamics answers

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Thermodynamics 2 (Vaal University of Technology)




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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 p3 = 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:




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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  cln v2  ln v1 


W  50000xln 0.1  ln 0.05


W  34657Nm




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(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

 cln v2  ln v1 


 0.25x105 ln1  ln 0.25 
W  34657Nm



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