Question

A pump handling a liquid raises its pressure from 1 bar to 30 bar. Take the density of liquid as 990 kg/m³. The isentropic specific work done by the pump in kJ/kg is
1. 3.93
2. 1.93
3. 4.93
4. 2.93

Answer 1

Correct Answer - Option 4 : 2.93

Concept:

Work done by the pump is given as,

\(\int_{1}^{2}dW=v\int_{1}^{2}dP\)

\(\rm{W}=\rm v(P_2-P_1)\)

But we know that \(ρ=\frac{m}{V}\) and for unit mass, \(ρ=\frac{1}{V}\) 

means \(V=\frac{1}{ρ}\)

\(\therefore W=\frac{P_2\;-\;P_1}{ρ}\) kJ/kg

Calculation:

Given:

P₁ = 1 bar = 10⁵ Pa, P₂ = 30 bar = 30 × 10⁵ Pa, ρ = 990 kg/m³

\({\rm{W}} = \frac{{{{\rm{P}}_2} - {{\rm{P}}_1}}}{{\rm{\rho }}} = \frac{{\left( {30 - 1} \right){{\times10}^5}}}{{990}} = 2.93\;kJ/kg\)

The isentropic specific work done by the pump is 2.93 kJ/kg.