Modern Engineering Thermodynamics

13.16 Brayton Cycle 497
Equation (7.38) also allows us to write
so that
T 3 /T 4s = ðp 4s /p 3 Þ ð1 − kÞ/k = ðv 3 /v 4s Þ 1 − k
ðη T Þ Brayton
= 1 − T 3 /T 4s = 1 − PR ð1 − kÞ/k = 1 − CR 1 − k (13.23)
cold ASC
where PR is the isentropic pressure ratio p 4s /p 3 , and CR is the isentropic compression ratio v 3 /v 4s . Thus, the thermal
efficiency of the Brayton cold ASC can be written as a function of the isentropic pressure or compression
ratio and the specific heat ratio of the working fluid.
EXAMPLE 13.12
The average power cylinder and compression cylinder pressures for the early 1878 Brayton cycle engine shown in Figure 13.45
were p 1 = p 4s = 0.210 MPa and p 2s = p 3 = 0.190 MPa, respectively. For this engine, determine
a. The isentropic pressure ratio PR.
b. The isentropic compression ratio CR.
c. The Brayton cold ASC thermal efficiency.
Solution
Using the Brayton cycle diagram shown in Figure 13.44, we can carry out
the following analysis.
a. The isentropic pressure ratio of a Brayton cycle engine is given by
PR = p 4s
p 3
=
0:210 MPa
0:190 MPa = 1:11
b. The isentropic compression ratio of a Brayton cycle engine is given by
CR ¼ðPRÞ 1/k = ð1:11Þ 1/1:40 = 1:07
c. Equation (13.23) gives the Brayton cold ASC thermal efficiency as
ðη T Þ Brayton
= 1− T 3
= 1−PR ð1−kÞ/k = 1−ð1:11Þ ð1−1:40Þ/1:40 = 0:0294 = 2:94%
T 4s
cold ASC
Exercises
34. Determine the isentropic pressure ratio for the Brayton cycle engine in
Example 13.12 if the power piston pressure is increased from 0.210 MPa
to 0.300 MPa. Assume all the other variables remain unchanged.
Answer: PR= 1.58.
FIGURE 13.45
Example 13.12.
35. If the compression pressure of the Brayton cycle engine in Example 13.12 is lowered from 0.190 MPa to 0.100 MPa,
determine the corresponding isentropic compression ratio of the engine. Assume all the other variables remain
unchanged. Answer: CR= 1.70.
36. If operating conditions of the Brayton cycle engine in Example 13.12 are altered such that the isentropic temperature at
the exit of the compressor is T 4s = 35.0°C when the inlet air temperature is 22.0°C, determine the Brayton cycle cold
ASC thermal efficiency of this engine under these operating conditions. Answer: (η T ) Brayton cold ASC = 4.22%.
Since T 3 = T L but T 4s < T 1 = T H , the Brayton cold ASC thermal efficiency is less than that of the Carnot cold ASC
working between the same temperature limits (T 1 and T 3 ). However, for fixed values of the temperature limits
T 1 and T 3 , there is an optimum value for the compressor outlet temperature T 4s that maximizes the net output
work. This can be determined as follows for an isentropic turbine and compressor:
ð _W out Þ net
isentropic
= _W pm − j _W c j = _mc p ðT 1 − T 2s − T 4s + T 3 Þ
Now, holding T 1 and T 3 fixed and replacing T 2 with its isentropic equivalent, T 2s = T 1 T 3 /T 4s , we get
ð _W out Þ net
isentropic
= _mc p ðT 1 − T 1 T 3 /T 4s − T 4s + T 3 Þ