Question

In a power plant, water (density $=1000kg/m^3$) is pumped from $80kPa$ to $3MPa$. The pump has an isentropic efficiency of $0.85$. ssuming that the temperature of the water remains the same, the specific work (in $kJ/kg$) supplied to the pump is

• A. $0.34$
• B. $2.48$
• C. $2.92$
• D. $3.43$

Answer 1

The correct option is D $3.43$

Ideal work input: $w=v_f\Delta p$
$\frac{1}{{\rho }_{water}}(p_2-P_1)=\frac{(3000-80)}{1000}=\frac{2920}{1000}$
$=2.92kJ/kg$
Actual input work $=\frac{w}{\eta }=\frac{2.92}{0.85}=3.43kJ/kg$