Thermodynamic Cycles
Look at different cycles that approximate real
processes
You can categorize these processes several
different ways
Power Cycles vs. Refrigeration
Gas vs. Vapor
Closed vs. open
Internal Combustion vs. External Combustion
Power Cycles
Otto Cycle
Spark Ignition
Diesel Cycle
Brayton Cycle
Gas Turbine
Rankine Cycle
These are all heat engines. They convert heat to work, so
the efficiency is:
Wnet
th =
Qin
Ideal Cycles
Well be using ideal cycles to analyze
real systems, so lets start with the only
ideal cycle weve studied so far
Carnot Cycle
Q-W=0 Q=W
In addition, we know that the efficiency for a
Carnot Cycle is:
th , Carnot
TL
= 1
TH
Carnot Cycle is not a good model for most real
processes
For example
Internal combustion engine
Gas turbine
We need to develop a new model, that
is still ideal
Air-Standard Assumptions
Air continuously circulates in a closed
loop and behaves as an ideal gas
All the processes are internally
reversible
Combustion is replaced by a heataddition process from the outside
Heat rejection replaces the exhaust
process
Also assume a constant value for Cp,
evaluated at room temperature
Terminology for Reciprocating Devices
Compression Ratio
V max VBDC
r=
=
V min VTDC
Mean Effective Pressure
W = PdV
1
W = PV
1-2
2-3
3-4
4-1
Isentropic Compression
Constant Volume Heat Addition
Isentropic Expansion
Constant Volume Heat Rejection
Thermal Efficiency of the Otto Cycle
Wnet Qnet Qin Qout
Qout
th =
= 1
=
=
Qin
Qin
Qin
Qin
Apply First Law Closed System to Process 2-3, V = Constant
Qnet , 23 Wnet , 23 = U 23
3
Wnet , 23 = Wother , 23 + Wb , 23 = 0 + PdV = 0
2
Qnet , 23 = U 23
Qnet , 23 = Qin = mCv (T3 T2 )
Apply First Law Closed System to Process 4-1,V = Constant
Qnet , 41 Wnet , 41 = U 41
1
Wnet , 41 = Wother , 41 +W b , 41= 0 + PdV = 0
4
Qnet , 41 = U 41
Qnet , 41 = Qout = mCv (T1 T4 )
Qout = mCv (T1 T4 ) = mCv (T4 T1 )
th , Otto
Qout
= 1
Qin
th , Otto
mCv ( T4 T1 )
= 1
mCv ( T3 T2 )
(T4 T1 )
= 1
(T3 T2 )
T1 ( T4 / T1 1)
= 1
T2 ( T3 / T2 1)
Recall processes 1-2 and 3-4 are isentropic, so
T2 v1
=
T1 v2
k 1
and
T3 v4
=
T4 v3
k 1
T2 T3
=
T1 T4
v3 = v2 and v4 = v1
or
T4 T3
=
T1 T2
th , Otto
(T4 T1 )
= 1
(T3 T2 )
T1 (T4 / T1 1)
= 1
T2 (T3 / T2 1)
1
th , Otto
T1
= 1
T2
Is this the same as the Carnot efficiency?
Efficiency of the Otto Cycle vs. Carnot Cycle
There are only two temperatures in the
Carnot cycle
Heat is added at TH
Heat is rejected at TL
There are four temperatures in the Otto
cycle!!
Heat is added over a range of temperatures
Heat is rejected over a range of temperatures
Since process 1-2 is isentropic,
T1 V2
=
T2 V1
th , Otto
k 1
1
r
T1
= 1
T2
Increasing Compression Ratio
Increases the Efficiency
k 1
th , Otto = 1
1
r
k 1
Typical Compression
Ratios for Gasoline
Engines
Why not use higher compression Ratios?
Premature Ignition
Causes Knock
Reduces the Efficiency
Mechanically need a better design
Diesel Engines
No spark plug
Fuel is sprayed into hot compressed air
State Diagrams for the Diesel Cycle
Diesel Cycle
Otto Cycle
The only
difference
is in
process
2-3
Consider Process 2-3
This is the step where heat is transferred into the
system
We model it as constant pressure instead of constant
volume
qin , 23 wb ,out = u = u3 u2
qin , 23 = u + Pv = h = C p (T3 T2 )
Consider Process 4-1
This is where heat is rejected
We model this as a constant v process
That means there is no boundary work
q41 w41 = u
q41 = qout = u = Cv (T1 T4 )
qout = Cv (T4 T1 )
As for any heat engine
wnet
qout
th =
= 1
qin
qin
qout = Cv (T4 T1 ) and qin = C p (T3 T2 )
nth ,diesel
Cv (T4 T1 )
(
T4 T1 )
= 1
= 1
k (T3 T2 )
C p (T3 T2 )
Rearrange
nth ,diesel
T1 (T4 T1 1)
= 1
kT2 (T3 T2 1)
PV
PV
3 3
= 2 2 where P3 = P2
T3
T2
T3 V3
=
= rc
T2 V2
rc is called the cutoff ratio its the ratio of
the cylinder volume before and after the
combustion process
nth ,diesel
T1 (T4 T1 1)
= 1
kT2 (rc 1)
PV
PV
4 4
= 1 1 where V4 = V1
T4
T1
T4 P4
=
T1 P1
nth ,diesel
T1 (TP44 TP11 1)
= 1
kT2 (rc 1)
Since Process 1-2 and Process 3-4
are both isentropic
PV = P V and
k
1 1
k
2 2
P V = PV
k
4 4
k
3 3
V3
P3 V
P4 V
k
= = rc
=
P1 V
P2 V
V2
k
3
k
2
k
4
k
1
nth ,diesel
T1 r 1
= 1
kT2 (rc 1)
k
c
Finally, Since process 1-2 is
isentropic
T1 v2
=
T2 v1
T1 1
=
T2 r
k 1 The volume ratio
k 1
from 1 to 2 is the
compression ratio, r
th ,diesel
rc 1
1
= 1 k 1
r k (rc 1)
k=1.4
The efficiency of the Otto cycle is always higher than the
Diesel cycle
Why use the Diesel cycle?
Because you can use higher compression ratios
