# Step-by-Step Solution ## **Given Data** - **Inlet Pressure, \( P_1 \) = 100 kPa** - **Inlet Temperature, \( T_1 \) = 285 K** - **Outlet Pressure, \( P_2 \) = 500 kPa** - **Outlet Temperature, \( T_2 \) = 550 K** - **Mass flow rate, \( \dot{m} = 4 \) kg/min** - **Heat loss, \( q_{out} = 25 \) kJ/kg** - **Find:** Required power input \( \dot{W}_{in} \) to the compressor. --- ### **Step 1: Apply the Steady-Flow Energy Equation** The steady-flow energy equation (First Law of Thermodynamics for control volume) is: \[ \dot{W}_{in} = \dot{m} \left[ (h_2 - h_1) + q_{out} \right] \] **Explanation:** - \( h_2 - h_1 \) is the enthalpy change of air. - \( q_{out} \) is heat lost per kg (added since it is lost to surroundings). - Kinetic and potential energy changes are negligible. --- ### **Step 2: Substitute Values and Calculate** For an ideal gas (air), the change in enthalpy is: \[ h_2 - h_1 = c_p (T_2 - T_1) \] Where \( c_p \) for air ≈ 1.005 kJ/kg·K. \[ h_2 - h_1 = 1.005 \times (550 - 285) = 1.005 \times 265 = 266.325 \text{ kJ/kg} \] Total energy per kg needed: \[ \text{Total} = (h_2 - h_1) + q_{out} = 266.325 + 25 = 291.325 \text{ kJ/kg} \] Mass flow rate in kg/s: \[ \dot{m} = \frac{4}{60} = .0667 \text{ kg/s} \] Required power input: \[ \dot{W}_{in} = .0667 \times 291.325 = 19.42 \text{ kW} \] **Explanation:** We calculated the required energy per kg, then multiplied by the mass flow (in kg/s) to get power in kW. --- ## **Final Answer** \[ \boxed{19.4~\text{kW}} \] is the necessary power input to the compressor.