Air at 400 kPa and 970 K enters a turbine operating at the steady-flow condition

Question

Air at 400 kPa and 970 K enters a turbine operating at the steady-flow condition and exits at 100 kPa and 670 K. Heat loss occurs at an average outer surface temperature of 315 K at the rate of 30 kJ per unit mass (kg) of air flow. Air can be treated as ideal gas with a constant Cp = 1.1 kJ/kg K. The kinetic and potential effects are negligible, determine The rate of power developed per unit mass of air flow, in kJ/kg The rate of entropy generation within the turbine. kJ/kg K The whether the process is reversible or not

Explanation / Answer

q = Cp (delta T) q = 1.1 * (970-670) q = 330 kj/kg q (total) = q + q (loss) q (total) = 330 + 30 = 360 kj/kg total power developed is 360 kj/kg rate of entropy generation is s = q (total) / T = 360 / 315 = 1.1428 kj/kgk as there is heat loss, there is dq so there is also ds so the process is not reversible

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