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A pump operating at steady state receives liquid water at 508C, 1.5 MPa. The pressure of the water at the pump exit is 15 MPa. The magnitude of the work required by the pump is 18 kJ per kg of water flowing. Stray heat transfer and changes in kinetic and potential energy are negligible. Determine the isentropic pump efficiency

Respuesta :

okpalawalter8

Answer:

[tex]n=76\%[/tex]

Explanation:

From the question we are told that:

Temperature [tex]T=50 \textdegree C[/tex]

Intake Pressure [tex]P=1.5MPa[/tex]

Exit Pressure [tex]P=15MPa[/tex]

Work [tex]W=18kJ[/tex]

Therefore

From Table

V_f=1.01*10^{-4}

Generally the equation for isentropic Work is mathematically given by

[tex]W_{iso}=V(P_2-P_1)[/tex]

[tex]W_{iso}=1.01*10^{-4}(15-1.5)[/tex]

[tex]W_{iso}=13.622[/tex]

Since

Efficiency n

[tex]n=\frac{W_{iso}}{W}[/tex]

[tex]n=\frac{W_{iso}}{18}[/tex]

[tex]n=0.76[/tex]

Therefore

[tex]n=76\%[/tex]

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