Lec 21: Isentropic efficiencies, air
standard cycle, Carnot cycle, Otto cycle
1
• For next time:
– Read: § 8-6 to 8-7
– HW11 due Wednesday, November 12, 2003
• Outline:
– Isentropic efficiency
– Air standard cycle
– Otto cycle
• Important points:
– Realize that we already know how to analyze
all these new cycles, we just need to define
what the cycle steps are
– Know the difference between the air standard
cycle and the cold air approximations
– Know how to solve cycles using variable
specific heats and constant specific heats
2
Isentropic Efficiencies
We can use the isentropic process as an
ideal by which to compare real processes
in different engineering devices.
COMPRESSORS
Ws
ws
c 

Wact w act
Actual compressors take more work than
isentropic compressors. The efficiency
will vary between zero and one.
3
Compressor Isentropic Efficiency
For a steady-state, adiabatic compressor
q  w  h 2  h1  KE  PE
If q, KE, and PE are all zero, then:
w  h1  h 2
General expression
w s  h 1  h 2s
Isentropic
compressor
Actual compressor
w a  h1  h 2a
4
Compressor Isentropic Efficiency
The compressor efficiency is then:
w s h1  h 2s
c 

w a h1  h 2a
For an ideal gas with constant specific heats,
w s  h 1  h 2 s  c p ( T1  T2 s )
w a  h 1  h 2a  c p (T1  T2a )
Thus:
T1  T2s
c 
T1  T2a
5
Compressor Isentropic Efficiency
Note that the work is directly
proportional to T with constant specific
heats. Real gases will also have
dependence on P.
The work can then be represented by
a change on the T-axis of a Ts
diagram.
6
Compressor Isentropic
Efficiencies
T
2a
2s
T1–T2a
T1–T2s
1
S
7
Isentropic Efficiencies
TURBINES
With turbines, we’re interested in the
work/power output not input. An isentropic
turbine will produce the maximum output.
Efficiency is given by:
Wact w act
T 

Ws
ws
h1  h 2a
T 
h1  h 2s
Again,
isentropic
efficiency will
vary between
zero and one.
8
Isentropic Efficiencies
T
1
2a
2s 2a 2a
S
Actual path
will vary
depending on
amount of
irreversibilities
...
9
Isentropic Efficiencies
For nozzles, the isentropic efficiency is given by
 Vexit  


2

a
N 
2
 Vexit  


2

s
2
10
TEAMPLAY
• Work problem 7-89
11
Chapter 8, Gas Cycles
• Carnot cycle is the most efficient cycle
that can be executed between a heat
source and a heat sink.
TL
  1TH
• However, isothermal heat transfer is
difficult to obtain in reality--requires large
heat exchangers and a lot of time.
12
Gas Cycles
• Therefore, the very important (reversible)
Carnot cycle, composed of two reversible
isothermal processes and two reversible
adiabatic processes, is never realized as a
practical matter.
• Its real value is as a standard of
comparison for all other cycles.
13
Gas Cycles
• Assumptions of air standard cycle
• Analyze two cycles in detail
– Otto
– Brayton
14
Assumptions of air standard cycle
• Working fluid is air
• Air is ideal gas
• Combustion process is replaced by heat
addition process
• Heat rejection is used to restore the fluid
to its initial state and complete the cycle
• All processes are internally reversible
• Constant or variable specific heats can be
used
15
Gas cycles have many engineering
applications
• Internal combustion engine
– Otto cycle
– Diesel cycle
• Gas turbines
– Brayton cycle
• Refrigeration
– Reversed Brayton cycle
16
Some nomenclature before
starting internal combustion
engine cycles
17
More terminology
18
Terminology
•
•
•
•
•
Bore = d
Stroke = s
 d 2 

Displacement volume =DV = s
 4 
Clearance volume = CV
Compression ratio = r
VBDC
DV  CV

r
CV
VTDC
19
Mean Effective Pressure
Mean Effective Pressure (MEP) is a fictitious
pressure, such that if it acted on the piston
during the entire power stroke, it would
produce the same amount of net work.
Wnet
MEP 
Vmax  Vmin
20
The net work output of
a cycle is equivalent to
the product of the
mean effect pressure
and the displacement
volume
21
Real Otto
cycle
22
Real and Idealized Cycle
23
Idealized Otto cycle
24
Idealized Otto cycle
• 1-2
- ADIABATIC COMPRESSION (ISENTROPIC)
• 2-3
- CONSTANT VOLUME HEAT ADDITION
• 3-4
- ADIABATIC EXPANSION (ISENTROPIC)
• 4-1
- CONSTANT VOLUME HEAT REJECTION
25
Performance of cycle
Efficiency:
w net

q in
Let’s start by getting heat input:
q in  u 3  u 2
26
Cycle Performance
Get net work from energy balance of
cycle:
net
in
out
q q
w
Substituting for qin and qout:
w net  (u 3  u 2 )  (u 4  u1 )
Efficiency is then:
w net

q in
27
Cycle Performance
Substituting for net work and heat input:
(u 3 - u 2 ) - (u 4 - u1 )

(u 3 - u 2 )
We can simplify the above expression:
(u 4 - u1 )
  1
(u 3 - u 2 )
28
Teamplay
Problem 8-36
29
Cold air standard cycle
cp, cv, and k are constant at ambient
temperature ( 70 °F) values.
Assumption will allow us to get a quick
“first cut”approximation of performance
of cycle.
30
Cycle performance with cold air
cycle assumptions
If we assume constant specific heats:
c v (T4 - T1 )
(u 4 - u1 )
  1
1
(u 3 - u 2 )
c v (T3 - T2 )
(T4 - T1 )
  1
(T3 - T2 )
31
Cycle performance with cold air
cycle assumptions
Because we’ve got two isentropic processes
in the cycle, T1 can be related to T2, and T3
can be related to T4 with our ideal gas
isentropic relationships….
Details are in the book!
T2  V1 
  
T1  V2 
k 1
r
k 1
T4  V4 
  
T3  V3 
T4 T3

Thus
T1 T2
k 1

1
r k 1
32
Cycle performance with cold air
cycle assumptions
T1
1
 1
 1  k 1
T2
r
This looks like the Carnot efficiency, but it
is not! T1 and T2 are not constant.
What are the limitations for this
expression?
33
Differences between Otto and
Carnot cycles
T
2
33
2
44
1
3
s
34
Effect of compression ratio on Otto
cycle efficiency
35
Sample Problem
The air at the beginning of the compression
stroke of an air-standard Otto cycle is at 95
kPa and 22C and the cylinder volume is
5600 cm3. The compression ratio is 9 and
8.6 kJ are added during the heat addition
process. Calculate:
(a) the temperature and pressure after the
compression and heat addition process
(b) the thermal efficiency of the cycle
Use cold air cycle assumptions.
36
Draw cycle and label points
3
r = V1 /V2 = V4 /V3 = 9
P
Q23 = 8.6 kJ
2
4
T1 = 299 K
1 P = 95 kPa
1
v
37
Major assumptions
•
•
•
•
Kinetic and potential energies are zero
Closed system
1 is start of compression
Ideal cycle: 1-2 isentropic compression,
2-3 const. volume heat addition, etc.
• Cold cycle const. properties
38
Carry through with solution
Calculate mass of air:
P1V1
m
 6.29 x 10-3 kg
RT1
Compression occurs from 1 to 2:
 V1 
T2  T1  
 V2 
k 1
 isentropic compressio n
T2  22  273K 9
1.4 1
T2  705.6 K
But we need T3!
39
Get T3 with first law:
Q23  W  m(u  ke  pe)  mc v T3  T2 
Solve for T3:
q
8.6 kJ 6.29x10 3 kg
 705.6 K
T3   T2 
cv
0.855 kJ
kg
T3  2304.7 K
40
Thermal Efficiency
  1
1
r
k 1
1 
1
1.4 1
9
  0.585
41
Let’s take a look at the Diesel cycle.
42
Idealized Diesel cycle
1-2
- ADIABATIC COMPRESSION (ISENTROPIC)
2-3
- CONSTANT PRESSURE HEAT ADDITION
3-4
- ADIABATIC EXPANSION (ISENTROPIC)
4-1
- CONSTANT VOLUME HEAT REJECTION
43
Performance of cycle
Efficiency:
w net

q in
Heat input occurs from 2 to 3 in
constant pressure process:
q in  h 3  h 2
Why enthalpies?
44
TEAMPLAY
• Work problem 8-16
45