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Energy Analysis of Control Volumes at Steady
Chapter 4, Problem 41P(choose chapter or problem)
Steam enters a well-insulated turbine operating at steady state at \(4 \mathrm{MPa}\) with a specific enthalpy of \(3015.4 \mathrm{~kJ} / \mathrm{kg}\) and a velocity of \(10 \mathrm{~m} / \mathrm{s}\). The steam expands to the turbine exit where the pressure is \(0.07 \mathrm{MPa}\), specific enthalpy is \(2431.7 \mathrm{~kJ} / \mathrm{kg}\), and the velocity is \(90 \mathrm{~m} / \mathrm{s}\). The mass flow rate is \(11.95 \mathrm{~kg} / \mathrm{s}\). Neglecting potential energy effects, determine the power developed by the turbine, in kW.
Questions & Answers
QUESTION:
Steam enters a well-insulated turbine operating at steady state at \(4 \mathrm{MPa}\) with a specific enthalpy of \(3015.4 \mathrm{~kJ} / \mathrm{kg}\) and a velocity of \(10 \mathrm{~m} / \mathrm{s}\). The steam expands to the turbine exit where the pressure is \(0.07 \mathrm{MPa}\), specific enthalpy is \(2431.7 \mathrm{~kJ} / \mathrm{kg}\), and the velocity is \(90 \mathrm{~m} / \mathrm{s}\). The mass flow rate is \(11.95 \mathrm{~kg} / \mathrm{s}\). Neglecting potential energy effects, determine the power developed by the turbine, in kW.
ANSWER:Step 1 of 3:
Steam enters into an insulated turbine and expands. The pressure, velocity, and specific enthalpy got changed during the process. We are going to find the power developed by the turbine.
The initial pressure p1 = 4 MPa = 4 x 106 Pa
The final pressure p2 = 0.07 MPa = 0.07 x 106 Pa
The initial velocity v1 = 10 m/s
The final velocity v2 = 90 m/s
The initial specific enthalpy h1 = 3015.4 kJ/kg
The final specific enthalpy h2 = 2431.7 kJ/kg
The mass flow rate = 11.95 kg/s