Question

Steam is compressed by an adiabatic compressor from \(0.2 \mathrm{MPa}\) and \(150^{\circ} \mathrm{C}\) to \(2.5 \mathrm{MPa}\) and \(250^{\circ} \mathrm{C}\) at a rate of \(1.30 \mathrm{kg} / \mathrm{s} .\) The power input to the compressor is \((a) 144 \mathrm{kW}\) (b) \(234 \mathrm{kW}\) \((c) 438 \mathrm{kW}\) \((d) 717 \mathrm{kW}\) \((e) 901 \mathrm{kW}\)

Short answer

Answer: The power input to the adiabatic compressor is approximately (b) 234 kW.

Step-by-step solution

Convert given temperatures to kelvin

To work with the temperatures in this exercise, it is necessary first to convert them from Celsius to Kelvin. To convert degrees Celsius to Kelvin, add 273.15 to the given value: Inlet temperature: \(150^{\circ} \mathrm{C} + 273.15 = 423.15 \mathrm{K}\) Outlet temperature: \(250^{\circ} \mathrm{C} + 273.15 = 523.15 \mathrm{K}\)

Calculate specific enthalpy changes using steam tables

For an adiabatic compressor, the power input is equal to the specific enthalpy change for the compressed steam. Using the steam tables, find the specific enthalpy values for the given points: Inlet: \(0.2 \mathrm{MPa}\) and \(423.15 \mathrm{K} \Rightarrow h_1 = 2775.2 \mathrm{kJ/kg}\) Outlet: \(2.5 \mathrm{MPa}\) and \(523.15 \mathrm{K} \Rightarrow h_2 = 2949.3 \mathrm{kJ/kg}\)

Calculate the specific enthalpy change

Subtract the specific enthalpy at the inlet from the specific enthalpy at the outlet: \(\Delta h = h_2 - h_1 = 2949.3 \mathrm{kJ/kg} - 2775.2 \mathrm{kJ/kg} = 174.1 \mathrm{kJ/kg}\)

Calculate the power input

To find the power input, multiply the specific enthalpy change by the mass flow rate: \(P = \Delta h \times m = 174.1 \mathrm{kJ/kg} \times 1.30 \mathrm{kg/s} = 226.33 \mathrm{kW}\)

Choose the correct answer

Compare the calculated power input with the given options: (a) 144 kW (b) 234 kW (c) 438 kW (d) 717 kW (e) 901 kW Our calculated value of \(226.33 \mathrm{kW}\) is closest to (b) \(234 \mathrm{kW}\). The correct answer is (b) 234 kW.