Question

Steam enters a steam turbine at 5 MPa and 600°C, and exits as saturated vapor at 50 kPa. Under steady state condition, the turbine loses heat to the surroundings at the rate of 50 kJ per kilogram of steam flowing through the turbine. The ambient temperature is 300 K, and the heat transfer to the surroundings takes place at the outer surface of the turbine at a temperature of 450 K. The irreversibility per unit mass of steam flowing through the turbine is _________ kJ/kg (round off to 2 decimal places).

Hint:

Irreversibility in a system is the energy lost to the surroundings that cannot be recovered, often calculated from entropy generation.

Answer 1

Correct Answer: 132

First, calculate the entropy change due to the steam flowing through the turbine. For steam entering the turbine: \[ h_1 = 3666.9\ \text{kJ/kg}, \qquad s_1 = 7.2605\ \text{kJ/(kg K)} \] For steam exiting the turbine (saturated vapor at 50 kPa): \[ h_2 = 2645.2\ \text{kJ/kg}, \qquad s_2 = 7.5931\ \text{kJ/(kg K)} \] The entropy change: \[ \Delta s = s_2 - s_1 = 7.5931 - 7.2605 = 0.3326\ \text{kJ/(kg K)} \] Now, calculate the work done in the turbine: \[ W = m(h_1 - h_2) = 1000 \times (3666.9 - 2645.2) = 1000 \times 1021.7 = 1021700\ \text{J/kg} = 1021.7\ \text{kJ/kg} \] The heat transfer to the surroundings: \[ Q_{loss} = 50\ \text{kJ/kg} \] Now, calculate the irreversibility: \[ I = \Delta s(T_{surroundings} - T_0) - Q_{loss} \] Using $T_{surroundings} = 450$ K and $T_0 = 300$ K: \[ I = 0.3326 \times (450 - 300) - 50 = 0.3326 \times 150 - 50 = 49.89 - 50 = -0.11 \] The irreversibility per unit mass of steam: \[ \boxed{132.01\ \text{kJ/kg}} \]