Answer:
The shaft work generated per kilogram is [tex]-1.3 \frac{MJ}{kg}[/tex]
Explanation:
Given:
Temperature [tex]T = 1273.15[/tex] K
Initial Pressure [tex]P_{1} = 1.8[/tex] MPa
Final pressure [tex]P_{2} = 0.1[/tex] MPa
From the table superheated,
[tex]h_{i} = 4635[/tex] [tex]\frac{K J}{Kg}[/tex] and [tex]h_{f} = 2706.54[/tex] [tex]\frac{K J}{Kg}[/tex]
Work done by shaft is,
[tex]W = h_{f} - h_{i}[/tex]
[tex]W = 2706.54 - 4635[/tex]
[tex]W = -1928.46 \frac{kJ}{kg}[/tex]
But here efficiency is 0.56,
So work generated per kg is,
Work = [tex]0.56 \times(- 1928.46)[/tex]
Work = [tex]-1.3[/tex] [tex]\frac{MJ}{kg}[/tex]
Therefore, the shaft work generated per kilogram is [tex]-1.3 \frac{MJ}{kg}[/tex]
