Transcript Reflection of Buddhism in Contemporary Cinema

ENGR 2213 Thermodynamics
F. C. Lai
School of Aerospace and Mechanical
Engineering
University of Oklahoma
Brayton Cycles
George Brayton (1870)
Applications of Gas-Turbine Engines
• Propulsion
• Power Generation
- The first gas turbine for electric power generation
was installed in 1949 in Oklahoma.
- Before 1980s, gas power plants were mainly used
for peak-load power production.
- It is forecast that more than half of all power plants
to be installed in the future are gas power plants.
Brayton Cycles
George Brayton (1870)
Open cycle
Closed Cycle
Brayton Cycles
Brayton Cycles
Staedy-flow q – w = he – hi
qin = h3 – h2 = cp (T3 – T2)
qout = h4 – h1 = cp (T4 – T1)
qout
  1
qin
 T4

T1   1
T1
T4  T1


 1
 1
 T3

T3  T2
T2   1
1
 T2

 1
rp(k 1) / k
3
T
Qin
2
4
1
Qout
k 1
k
T2  p2
 
T1  p1 
p2
rp 
p1
k 1
k
 p3
 
 p4 
S
T3

T4
Brayton Cycles
• The thermal efficiency
increases with the
pressure ratio.
• The highest temperature
in the cycle is limited by
the maximum temperature
that the turbine blades
can withstand.
Example 1
A gas power plant operating on an ideal Brayton cycle
has a pressure ratio of 8. The gas temperature is 300 K
At the compression 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,
(c) the thermal efficiency of this cycle.
Example 1 (continued)
(a) State 1: air at the inlet of compressor, T1 = 300 K
Table A-17 h1 = 300.19 kJ/kg
pr1 = 1.386
Process 1-2: Isentropic compression
pr2 p2
pr2  rp pr1  8(1.386)  11.09

 rp
pr1 p1
Table A-17 T2 = 540 K
h2 = 544.35 kJ/kg
Example 1 (continued)
State 3: air at the inlet of turbine, T3 = 1300 K
Table A-17 h3 = 1395.97 kJ/kg
pr3 = 330.9
Process 3-4: Isentropic expanion
pr 4 p4 1


pr3 p3 rp
pr 4
pr3 330.9


 41.36
rp
8
Table A-17 T4 = 770 K
h4 = 789.11 kJ/kg
wcomp = h2 – h1 = 544.35 – 300.19 = 244.16 kJ/kg
Example 1 (continued)
wturb = h3 – h4 = 1395.97 – 789.11 = 606.86 kJ/kg
(b) rbw 
w comp
w turb
244.16

 0.402
606.86
qin = h3 – h2 = 1395.97 – 544.35 = 851.62 kJ/kg
wnet = wturb – wcomp = 606.86 – 244.16 = 362.7 kJ/kg
w net
362.7
(c)  

 0.426
qin
851.62
  1
1
rp(k 1) / k
= 1 – rp(1-k)/k = 1 – (8)-(0.4/1.4) = 0.448
Brayton Cycles with Regeneration
Brayton Cycles with Intercooling,
Reheating, and Regeneration