## Related questions with answers

A gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The air enters the turbine at 120 psia and 2000 R and leaves at 15 psia and 1200 R. Heat is rejected to the surroundings at a rate of 6400 Btu/s, and air flows through the cycle at a rate of 40 lbm/s. Assuming the turbine to be isentropic and the compresssor to have an isentropic efficiency of 80 percent, determine the net power output of the plant. Account for the variation of specific heats with temperature.

Solution

VerifiedThe enthalpies at temperatures 3 and 4 are determined from the given temperatures and the data from A-17E:

$\begin{align*} &h_{3}=504.71\:\dfrac{\text{Btu}}{\text{lbm}}\\ &h_{4}=291.3\:\dfrac{\text{Btu}}{\text{lbm}} \end{align*}$

The enthalpy at state 1 is determined from the energy balance and the rate of heat output in 4-1:

$\begin{align*} h_{1}&=h_{4}-\dfrac{\dot Q_{\text{out}}}{\dot m}\\ &=\bigg(291.3-\dfrac{6400}{40}\bigg)\:\dfrac{\text{Btu}}{\text{lbm}}\\ &=131.3\:\dfrac{\text{Btu}}{\text{lbm}} \end{align*}$

The relative pressure at this state is determined from this value with data from A-17E using interpolation:

$\begin{align*} P_{r1}=1.474 \end{align*}$

The relative pressure at state 2 is determined from the pressure ratio across the cycle:

$\begin{align*} P_{r2}&=r_{p}P_{r1}\\ &=\dfrac{120}{15}\cdot1.474\\ &=11.792 \end{align*}$

The isentropic enthalpy at this state is then determined from this value with data from A-17E:

$\begin{align*} h_{2s}=238.08\:\dfrac{\text{Btu}}{\text{lbm}} \end{align*}$

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