Question

0.8 kg of air flows through a compressor under steady state conditions. The properties of air at entry are : pressure 1 bar, velocity 10 m/s, specific volume 0.95 m³/kg and internal energy 30 kJ/kg. The corresponding values at exit are 8 bar, 6 m/s, 0.2 m³/kg and 124 kJ/kg. Neglecting change in potential energy, the power input will be
1. 117 kW
2. 127 kW
3. 137 kW
4. 147 kW

Answer 1

Correct Answer - Option 2 : 127 kW

Concept:

The input and output conditions of compressor are given. By equating input energy equal to output energy,

\(Q + m{h_1} + {\left( {K.E.} \right)_1} + {\left( {P.E.} \right)_1} = W + m{h_2} + {\left( {K.E.} \right)_2} + {\left( {P.E.} \right)_2}\)

\(m\left( {{U_1} + {P_1}{\nu _1}} \right) + \left( {\frac{1}{2}m{V_1}^2 \times {{10}^{ - 3}}} \right) = W + m\left( {{U_2} + {P_2}{\nu _2}} \right) + \left( {\frac{1}{2}m{V_2}^2 \times {{10}^{ - 3}}} \right)\)

where,

Q = heat transfer to the system

W = work transfer from the system

Calculation:

Given:

m = 0.8 kg, P₁ = 100 kPa, V₁ = 10 m/s, v₁ = 0.95 m³/kg, U₁ = 30 kJ/kg, P₂ = 800 kPa, V₂ = 6 m/s, v₂ = 0.2 m3/kg, U₂ = 124 kJ/kg

Q = 0, compression process is adiabatic and \(\Delta P.E. = 0\), neglecting change in potential energy

Therefore, \(0.8 \times \left( {30 + 100 \times 0.95} \right) + \left( {\frac{1}{2} \times 0.8 \times {{10}^2} \times {{10}^{ - 3}}} \right) = W + 0.8\left( {124 + 800 \times 0.2} \right) + \left( {\frac{1}{2} \times 0.8 \times {6^2} \times {{10}^{ - 3}}} \right)\)

\(\begin{array}{l} 100 + 0.04 = W + 227.2 + 0.0144\\ W = - 127.2104\;kW \end{array}\)

-ve sign shows that 127.2104 kW power should be supplied to compressor. Because compressor is work absorbing device.