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3 years ago

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The enthalpy of the water entering an actual pump is 500 kJ/kg and the enthalpy of the water leaving it is 550 kJ/kg. The pump h

as 98% efficiency, what would have been the enthalpies at the inlet and outlet if the pump was 100% efficient?

1 answer:

n200080 [17]3 years ago

3 0

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%= $h_2^{'}$

Now pump efficiency is 98%

$\eta$ = $\frac{h_2-h_1}{h_2^{'}-h_1}$

0.98= $\frac{550-500}{h_2-500}$

$h_2=551.02KJ/kg$

therefore initial and final enthalpy of pump for 100 % efficiency

initial=500KJ/kg

Final=551.02KJ/kg

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