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Air enters an adiabatic turbine at 2.8 MPa and 400°C and expands to a lower pressure of 150 kPa. Assume an isentropic efficiency of 90% for the turbine. Determine the actual outlet temperature of the turbine.

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Answer:

The outlet temperature of turbine is 327.51 K.

Step-by-step explanation:

We know that

in adiabatic process


PV^\gamma =C


(T_2)/(T_1)=\left((P_2)/(P_1)\right)^{(\gamma -1)/(\gamma )}

Now by putting the values


(T_2)/(673)=\left((150)/(2800)\right)^{(1.4 -1)/(1.4 )}


T_2=289.39 K

The efficiency of turbine is given as


\eta =\frac{T_1-\acute{T_2}}{T_1-T_2}

By putting the values


0.9=\frac{673-\acute{T_2}}{673-289.39}


\acute{T_2}=327.51 K

So the outlet temperature of turbine is 327.51 K.

Air enters an adiabatic turbine at 2.8 MPa and 400°C and expands to a lower pressure-example-1
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