A single-acting, single-cylinder reciprocating air compressor is compressing 20 kg/min of air from 110 kPa and 300K to 660 kPa according to PV1.25 = Constant. Mechanical efficiency is 80%. What is the power input to the compressor? [R = 0.287 kJ/kg-K, , Neglect clearance, leakage and cooling]?

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  1. 39.23 kW
  2. 58.54 kW
  3. 27.38 kW
  4. 71.75 kW

Answer (Detailed Solution Below)

Option 4 : 71.75 kW
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Concept:

We use the polytropic process equations and mechanical efficiency to determine the power input required for compressing air in a reciprocating compressor.

Given:

  • Mass flow rate of air, \( \dot{m} = 20 \, \text{kg/min} = 0.333 \, \text{kg/s} \)
  • Inlet pressure, \( P_1 = 110 \, \text{kPa} \)
  • Inlet temperature, \( T_1 = 300 \, \text{K} \)
  • Outlet pressure, \( P_2 = 660 \, \text{kPa} \)
  • Polytropic index, \( n = 1.25 \)
  • Gas constant, \( R = 0.287 \, \text{kJ/kg·K} \)
  • Mechanical efficiency, \( \eta_{\text{mech}} = 80\% = 0.8 \)
  • Given: \( (6)^{0.2} = 1.43 \)

Step 1: Calculate the Polytropic Work Done

The work done per kg of air for a polytropic process is:

\( W_{\text{polytropic}} = \frac{n}{n-1} \times R \times T_1 \times \left[ \left( \frac{P_2}{P_1} \right)^{\frac{n-1}{n}} - 1 \right] \)

Substitute the values:

\( W_{\text{polytropic}} = \frac{1.25}{0.25} \times 0.287 \times 300 \times \left[ (6)^{0.2} - 1 \right] \)

\( W_{\text{polytropic}} = 5 \times 0.287 \times 300 \times (1.43 - 1) \)

\( W_{\text{polytropic}} = 184.515 \, \text{kJ/kg} \)

Step 2: Calculate the Indicated Power

The indicated power is the work done multiplied by the mass flow rate:

\( P_{\text{indicated}} = \dot{m} \times W_{\text{polytropic}} \)

\( P_{\text{indicated}} = 0.333 \times 184.515 = 61.5 \, \text{kW} \)

Step 3: Calculate the Power Input

The power input accounts for mechanical efficiency:

\( P_{\text{input}} = \frac{P_{\text{indicated}}}{\eta_{\text{mech}}} \)

\( P_{\text{input}} = \frac{61.5}{0.8} = 76.875 \, \text{kW} \)

 

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