Question 9.5: The Simple Ideal Brayton Cycle A gas-turbine power plant ope...

A power plant operating on the ideal Brayton cycle is considered.
The compressor and turbine exit temperatures, back work ratio, and the thermal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible.
4 The variation of specific heats with temperature is to be considered.
Analysis The T-s diagram of the ideal Brayton cycle described is shown in Fig. 9–35. We note that the components involved in the Brayton cycle are steady-flow devices.
(a) The air temperatures at the compressor and turbine exits are determined from isentropic relations:
Process 1-2 (isentropic compression of an ideal gas):

\begin{gathered}T_{1}=300 K \rightarrow h_{1}=300.19 kJ / kg \\\quad P_{r 1}=1.386 \\P_{r 2}=\frac{P_{2}}{P_{1}} P_{r 1}=(8)(1.386)=11.09 \rightarrow T_{2}=540 K \quad \text { (at compressor exit) } \\h_{2}=544.35 kJ / kg\end{gathered}

Process 3-4 (isentropic expansion of an ideal gas):

\begin{aligned}  &T_{3}=1300 K \rightarrow h_{3}=1395.97 kJ / kg \\ &\qquad P_{r 3}=330.9 \\ &P_{r 4}=\frac{P_{4}}{P_{3}} P_{r 3}=\left(\frac{1}{8}\right)(330.9)=41.36 \rightarrow T_{4}=770 K \quad \text { (at turbine exit) } \\ &h_{4}=789.37 kJ / kg \end{aligned}

(b) To find the back work ratio, we need to find the work input to the compressor and the work output of the turbine:

\begin{aligned} &w_{\text {comp,in }}=h_{2}-h_{1}=544.35-300.19=244.16 kJ / kg \\ &w_{\text {turb,out }}=h_{3}-h_{4}=1395.97-789.37=606.60 kJ / kg \end{aligned}

Thus,

r_{ bw }=\frac{w_{\text {comp,in }}}{w_{\text {turb,out }}}=\frac{244.16 kJ / kg }{606.60 kJ / kg }=0.403

That is, 40.3 percent of the turbine work output is used just to drive the compressor.
(c) The thermal efficiency of the cycle is the ratio of the net power output to the total heat input:

\begin{aligned} q_{\text {in }} &=h_{3}-h_{2}=1395.97-544.35=851.62 kJ / kg \\ w_{\text {net }} &=w_{\text {out }}-w_{\text {in }}=606.60-244.16=362.4 kJ / kg \end{aligned}

Thus,

\eta_{ th }=\frac{w_{ net }}{q_{ in }}=\frac{362.4 kJ / kg }{851.62 kJ / kg }=0.426 \text { or } 42.6 \%

The thermal efficiency could also be determined from

\eta_{ th }=1-\frac{q_{ out }}{q_{ in }}

where

q_{\text {out }}=h_{4}-h_{1}=789.37-300.19=489.2 kJ / kg

Discussion Under the cold-air-standard assumptions (constant specific heat values at room temperature), the thermal efficiency would be, from Eq. 9–17,

\eta_{ th , \text { Brayton }}=1-\frac{1}{r_{p}^{(k-1) / k}}

\eta_{ th , \text { Brayton }}=1-\frac{1}{r_{p}^{(k-1) / k}}=1-\frac{1}{8^{(1.4-1) / 1.4}}=0.448

which is sufficiently close to the value obtained by accounting for the variation of specific heats with temperatur