Answer:
a) 90 kJ
b) 230.26 kJ
Explanation:
The pressure at the first point = 10 bar —> 10 x 102 = 1020 kPa
The volume at the first point = 0.1 m^3
The pressure at the second point = 1 bar —> 1 x 102 = 102 kPa
The volume at the second point = 1 m^3
Process A.
constant volume V = C from point (1) to P = 10 bar.
Constant pressure P = C to the point (2).
Process B.
The relation of the process is PV = C
Required
For process A & B
(a) Sketch the process on P-V coordinates
(b) Evaluate the work W in kJ.
Assumption
Quasi-equilibrium process
Kinetic and potential effect can be ignored.
Solution
For process A.
V=C
There is no change in volume then
The work is defined by
║
V║limit 1--0.1
90 kJ
Process B
PV=C
By substituting with point (1) C = 10^2 x 1= 10^2
The work is defined by
║
ln(V)║limit 1--0.1
=230.26 kJ
