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HEAT ENGINE CYCLES
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Lecture notes of 26 pages for the course energy at University Of Manchester HISTORY 10101 (HEAT ENGINE CYCLES)
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1CHAPTER 1
HEAT ENGINE CYCLES
1.1 Carnot cycle
This cycle consists of two isothermal processes joined by two
adiabatic processes. It is most conveniently represented on a T-s and
p-v diagrams as follows:
Process 1-2 = isentropic expansion from T 1 to T2
Process 2-3 = isothermal heat rejection
Process 3-4 = isentropic compression from T 2 to T1
Process 4-1 = isothermal heat supply.
The cycle is completely independent of the working substance used.
T p
p4 4
P1 1
T1 4 1
P3
p2 3
T2 2
3 2
C D v
A B s
The cycle efficiency is given by
net work output
heat supplied
heat supplied heat rejected
heat supplied
, 2
T1 s B s A T2 s B s A
T1 s B s A
T1 T2 sB s A T
1 2
(1.1)
T1 sB s A T1
There is no attempt to use the Carnot cycle with gas as working
substance in practice because of two reasons:
1. The pressure of the gas changes continuously from p 4 to p1
during the isothermal heat supply and from p2 to p3 during the
isothermal heat rejection. But in practice it is much more
convenient to heat a gas at approximately constant pressure or at
constant volume.
2. The Carnot cycle, despite its high thermal efficiency, has a
small work ratio. [Work ratio is the ratio of the net work output
(area 12341) to the gross work output of the system (area
412DC4); the work done on the gas is given by 234CD2.]
Example 1.1
What is the highest possible theoretical efficiency of a heat engine
operating with a hot reservoir of furnace gases at 2000 oC when the
cooling water is available at 10oC?
Solution
T2
From Eq. (1.1) Carnot 1
T1
10 273 283
So C 1 1 0.8754
2000 273 2273
, 3
Example 1.2
A hot reservoir at 800oC and a cold reservoir at 15oC are available.
Calculate the thermal efficiency and the work ratio of a Carnot cycle
using air as the working fluid, if the maximum and minimum
pressures in the cycle are 210 bar and 1 bar.
Solution
The cycle is shown below on T-s and p-v diagrams.
Using Eq. (1.1)
T2 288
Carnot 1 1 0.732
T1 1073
In order to find the work output and the work ratio it is necessary to
find the entropy change (s1 – s4).
For an isothermal process from 4 to A,
p 210
s A s 4 R ln 4 0.287 ln = 1.535 kJ/kg K
2
p 1
At constant pressure from A to 2,
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