Question

At steady state, 500 kg/s of steam enters a turbine with specific enthalpy \(h_1 = 3500\ \text{kJ/kg}\) and entropy \(s_1 = 6.5\ \text{kJ . kg}^{-1}\text{K}^{-1}\). It expands reversibly in the turbine at 500 K. Exit properties: \(h_2 = 2500\ \text{kJ/kg}\), \(s_2 = 6.3\ \text{kJ . kg}^{-1}\text{K}^{-1}\). Find turbine work output (MW, integer).

Hint:

For reversible turbines with heat transfer at constant temperature, apply the exergy correction term \(T_0(s_1-s_2)\) to obtain the actual useful work.

Answer 1

Correct Answer: 450

A reversible turbine with heat loss at constant temperature uses exergy balance: \[ W = \dot{m}\left[(h_1 - h_2) - T_0 (s_1 - s_2)\right] \] Given: \[ \dot{m} = 500\ \text{kg/s},\quad T_0 = 500\ \text{K} \] \[ h_1 - h_2 = 3500 - 2500 = 1000\ \text{kJ/kg} \] 

\[s_1 - s_2 = 6.5 - 6.3 = 0.2\ \text{kJ/kg K}\]

 Second-law correction term: \[ T_0(s_1 - s_2) = 500(0.2) = 100\ \text{kJ/kg} \] Thus specific turbine work: \[ w = 1000 - 100 = 900\ \text{kJ/kg} \] Total work: \[ W = 500 \times 900 = 450000\ \text{kJ/s} = 450\ \text{MW} \] Hence, the turbine work output is: \[ \boxed{450\ \text{MW}} \]