The Simple Ideal Brayton Cycle A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas temperature is 300 K at the compressor inlet and 1300 K at the turbine inlet. Utilizing the air-standard assumptions, determine (a) the gas temperature at the exits of

Question

The Simple Ideal Brayton Cycle A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas temperature is 300 K at the compressor inlet and 1300 K at the turbine inlet. Utilizing the air-standard assumptions, determine (a) the gas temperature at the exits of the compressor and the turbine, (b) the back work ratio, and (c) the thermal efficiency.

Accepted Answer

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):

[latex]\begin{gathered}T_{1}=300 K ightarrow 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 ightarrow T_{2}=540 K \quad ext { (at compressor exit) } \h_{2}=544.35 kJ / kg\end{gathered}[/latex]

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

[latex] \begin{aligned} &T_{3}=1300 K ightarrow 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} ight)(330.9)=41.36 ightarrow T_{4}=770 K \quad ext { (at turbine exit) } \ &h_{4}=789.37 kJ / kg \end{aligned} [/latex]

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

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

Thus,

[latex] r_{ bw }=\frac{w_{ ext {comp,in }}}{w_{ ext {turb,out }}}=\frac{244.16 kJ / kg }{606.60 kJ / kg }=0.403 [/latex]

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:

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

Thus,

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

The thermal efficiency could also be determined from

[latex] \eta_{ th }=1-\frac{q_{ out }}{q_{ in }} [/latex]

where

[latex] q_{ ext {out }}=h_{4}-h_{1}=789.37-300.19=489.2 kJ / kg [/latex]

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

[latex] \eta_{ th , ext { Brayton }}=1-\frac{1}{r_{p}^{(k-1) / k}} [/latex]

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

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

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

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