512 | Thermodynamics
gas temperature at the exits of the compressor and the turbine, (b) the back
work ratio, and (c) the thermal efficiency.Solution A power plant operating on the ideal Brayton cycle is considered.
The compressor and turbine exit temperatures, back work ratio, and the ther-
mal efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assump-
tions 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-sdiagram 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):Process 3-4 (isentropic expansion of an ideal gas):(b) To find the back work ratio, we need to find the work input to the com-
pressor and the work output of the turbine:Thus,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:Thus,hthwnet
qin362.4 kJ>kg
851.62 kJ>kg0.426 or 42.6%wnetwoutwin606.60244.16362.4 kJ>kgqinh 3 h 2 1395.97544.35851.62 kJ>kgrbwwcomp,in
wturb,out244.16 kJ>kg
606.60 kJ>kg0.403wturb,outh 3 h 4 1395.97789.37606.60 kJ>kgwcomp,inh 2 h 1 544.35300.19244.16 kJ>kgh 4 789.37 kJ>kgPr 4 P 4
P 3Pr 3 a1
8b1330.9 2 41.36ST 4 770 K¬¬ 1 at turbine exit 2
Pr 3 330.9T 3 1300 KSh 3 1395.97 kJ>kgh 2 544.35 kJ>kgPr 2 P 2
P 1Pr 1 1821 1.386 2 11.09ST 2 540 K¬¬ 1 at compressor exit 2
Pr 1 1.386T 1 300 KSh 1 300.19 kJ>kgsT, K2341P = const.P = const.wturbwcomprp = 81300300qoutqinFIGURE 9–35
T-sdiagram for the Brayton cycle
discussed in Example 9–5.