Question

Air enters a compressor with a stagnation pressure of \(100 \mathrm{kPa}\) and a stagnation temperature of \(35^{\circ} \mathrm{C},\) and it is compressed to a stagnation pressure of 900 kPa. Assuming the compression process to be isentropic, determine the power input to the compressor for a mass flow rate of \(0.04 \mathrm{kg} / \mathrm{s}\)

Short answer

Answer: The power input to the compressor is 40.4 kW.

Step-by-step solution

Define isentropic compression

An isentropic process is a reversible adiabatic process. This means that during the compression process, there is no exchange of heat, and all the work done goes into compressing the air. For an ideal gas, we can relate the initial and final stagnation pressures and temperatures with the following isentropic relation: $$(\frac{P_2}{P_1})^{\frac{k-1}{k}} = \frac{T_2}{T_1}$$ where \(P_1\) = initial stagnation pressure (\(100 \mathrm{kPa}\)) \(P_2\) = final stagnation pressure (\(900 \mathrm{kPa}\)) \(T_1\) = initial stagnation temperature (\(35^\circ \mathrm{C}\)) \(T_2\) = final stagnation temperature \(k\) = specific heat ratio (Assuming air is an ideal gas, \(k \approx 1.4\))

Calculate the final stagnation temperature

Convert the initial stagnation temperature to Kelvin $$T_1 = 35^\circ \mathrm{C} + 273.15 = 308.15 \mathrm{K}$$ Now use the isentropic relation to find the final stagnation temperature: $$T_2 = T_1 (\frac{P_2}{P_1})^{\frac{k-1}{k}} = 308.15 \times (\frac{900}{100})^{\frac{1.4-1}{1.4}} = 308.15 \times 4.273198 \approx 1315 \mathrm{K}$$

Determine the specific heat and enthalpy change

We can now find the change in specific enthalpy by using the specific heat at constant pressure, \(C_p\), which can be approximated for air as \(C_p = 1.005 \frac{\mathrm{kJ}}{\mathrm{kg \cdot K}}\). The change in specific enthalpy, \(\Delta h\), can be calculated as follows: $$\Delta h = C_p (T_2 - T_1) = 1.005 \times (1315 - 308.15) \approx 1010 \ \frac{\mathrm{kJ}}{\mathrm{kg}}$$

Calculate the power input to the compressor

The power input to the compressor, \(W\), can be found by multiplying the mass flow rate, \(\dot{m}\), by the change in specific enthalpy, \(\Delta h\): $$W = \dot{m} \Delta h = 0.04 \,\mathrm{kg/s} \times 1010 \, \frac{\mathrm{kJ}}{\mathrm{kg}} = 40.4 \, \mathrm{kW}$$ Solution: The power input to the compressor is \(40.4 \, \mathrm{kW}\).