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
η th =
Qin
Ideal Cycles
• We’ll be using ideal cycles to analyze
real systems, so lets start with the only
ideal cycle we’ve studied so far
Carnot Cycle
Q
W
Q-W=0 Æ Q=W
In addition, we know that the efficiency for a
Carnot Cycle is:
η 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
2
W = ∫ PdV
1
W = P∆V
v
v
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
η 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 )
η th , Otto
Qout
= 1−
Qin
η th , Otto
mCv ( T4 − T1 )
= 1−
mCv ( T3 − T2 )
Æ
(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
Æ
v3 = v2 and v4 = v1
or
T4 T3
=
T1 T2
η th , Otto
(T4 − T1 )
= 1−
(T3 − T2 )
T1 (T4 / T1 − 1)
= 1−
T2 (T3 / T2 − 1)
1
η 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 ⎠
η th , Otto
k −1
=
1
r
T1
= 1−
T2
Increasing Compression Ratio
Increases the Efficiency
k −1
Æ
η 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 + P∆v = ∆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
ηth =
= 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 – it’s 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
k
⎛ V3 ⎞
P3 V
P4 V
k
= ⎜⎜ ⎟⎟ = rc
=
P1 V
P2 V
⎝ V2 ⎠
k
3
k
2
k
4
k
1
1
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
ηth ,diesel
k
⎡
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