Please anyone help me to solve the following QUESTION? Thank you.
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Please anyone help me to solve the following QUESTION? Thank you.

[From: ] [author: ] [Date: 11-04-30] [Hit: ]
If heat is lost from the turbine to the surroundings at a rate of 150 kJ/s, the power output of the gas turbine is:-Pi = 0.m_dot = 2.Pe = 0.Energy balance for control volume at steady state,W = 2.......
Hot combustion gases (assumed to have the properties of air at room temperature) enter a gas turbine at 0.8 MPa and 1500 K at a rate of 2.1 kg/s, and exit at 0.1 MPa and 800 K. If heat is lost from the turbine to the surroundings at a rate of 150 kJ/s, the power output of the gas turbine is:

-
Pi = 0.8 Mpa
Ti = 1500 K
m_dot = 2.1 kg/s
Pe = 0.1 Mpa
Te = 800 K
Q = - 150 kW (heat loss)

Energy balance for control volume at steady state,
dEcv/dt = 0 = Q - W + m_dot(hi - he)
W = Q + m_dot(hi - he)
W = 2.1 * (1635.97 - 821.95) - 150
W = 1559.442 kW (air-standard analysis)

If air assumed to have properties the same as air at room temp.,
W = Q + m_dot (hi - he)
W = m_dot * Cp (Ti - Te) - 150
W = 2.1 * 1.005*700 - 150
W =1327.35 kW (cold air-standard analysis)
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