Answer:500,551.02
Explanation:
Given
Initial enthaly of pump \left ( h_1\right )=500KJ/kg
Final enthaly of pump \left ( h_2\right )=550KJ/kg
Final enthaly of pump when efficiency is 100%=[tex]h_2^{'}[/tex]
Now pump efficiency is 98%
[tex]\eta [/tex]=[tex]\frac{h_2-h_1}{h_2^{'}-h_1}[/tex]
0.98=[tex]\frac{550-500}{h_2-500}[/tex]
[tex]h_2=551.02KJ/kg[/tex]
therefore initial and final enthalpy of pump for 100 % efficiency
initial=500KJ/kg
Final=551.02KJ/kg
