Hot combustion gases (assumed to have the properties of air at room temperature, Cp=1.005 kJ/kg-K) 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 isQuestion 10Select one:a.(a) 1477 kWb.(b) 1677 kWc.(c) 1327 kWd.(d) 1124 kWe.(e) 872 kWf.Insufficient information to answer question
Question
Hot combustion gases (assumed to have the properties of air at room temperature, Cp=1.005 kJ/kg-K) 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 isQuestion 10Select one:a.(a) 1477 kWb.(b) 1677 kWc.(c) 1327 kWd.(d) 1124 kWe.(e) 872 kWf.Insufficient information to answer question
Solution
To solve this problem, we need to use the first law of thermodynamics for a steady-flow system, which can be written as:
Q_in - W_out = ΔH + ΔKE + ΔPE
In this case, we can neglect the changes in kinetic and potential energy (ΔKE and ΔPE), so the equation simplifies to:
Q_in - W_out = ΔH
We know that the heat transfer Q_in is negative because heat is lost from the system, so Q_in = -150 kJ/s. We're trying to find the work output W_out.
The change in enthalpy ΔH can be calculated using the specific heat capacity at constant pressure Cp and the change in temperature, as follows:
ΔH = m_dot * Cp * (T_out - T_in)
where m_dot is the mass flow rate, T_out is the outlet temperature, and T_in is the inlet temperature. Substituting the given values:
ΔH = 2.1 kg/s * 1.005 kJ/kg-K * (800 K - 1500 K) = -1471.05 kJ/s
Substituting Q_in and ΔH into the first law equation gives:
-150 kJ/s - W_out = -1471.05 kJ/s
Solving for W_out gives:
W_out = 1471.05 kJ/s - 150 kJ/s = 1321.05 kJ/s = 1321.05 kW
So, the power output of the gas turbine is approximately 1321 kW. The closest answer is (c) 1327 kW.
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