EGR 334 Thermodynamics
Chapter 9: Sections 3-4
Lecture 32:
Gas Power Systems:
The Diesel Cycle
Quiz Today?
Today’s main concepts:
• Understand common terminology of a Diesel engine
• Be able to explain the processes of the Diesel Cycle
• Be able to perform a 1st Law analysis of the Diesel Cycle and
determine its thermal efficiency.
• Be able to discuss limitations of the Diesel cycle compared to
real Diesel engines.
Reading Assignment:
Read Chapter 9, Sections 5-6
Homework Assignment:
Problems from Chap 9: 17, 20, 24, 38
Sec 9.1 : Introducing Engine Terminology
Two types of internal combustion engine
• Spark Ignition (lower power & lighter)
• Compression Ignition (spontaneous combustion)
Terminology
Stroke : The distance the piston moves in
one direction
Top Dead Center : The piston has
minimum volume at the top of the
stroke.
Bottom Dead Center : The piston has
maximum volume at the bottom
of the stroke.
Clearance Volume : Min vol
Displacement Volume : Max vol. – Min vol.
Compression Ratio: Max vol. / min vol.
3
Sec 9.3 : Air-Standard Diesel Cycle
4
The Diesel cycle can operate using either the 4 or 2 stroke engine.
Has a lower efficiency than the Otto cycle for the same compression ratio.
Compression Ignition
• More difficult to start when cold
• Can have higher compression ratio
• 16 <rDiesel <25 vs r≈11 for Otto cycle to avoid pre-ignition
• Produces less CO, because excess air, but produces soot.
Diesel Engine Animation at
http://www.animatedengines
.com/diesel.html
Sec 9.3 : Air-Standard Diesel Cycle
Process 1 – 2 : Isentropic compression of air (compression stroke).
Process 2 – 3 : Constant pressure heat transfer to the air from an
external source while moves down (ignition and part of power stroke)
Process 3 – 4 : Isentropic expansion (remainder of power stroke)
Process 4 – 1 : Completes cycle by a constant volume process in which
heat is rejected from the air while piston is at bottom dead center
For the Otto cycle, the
Qin occurs at constant
volume (compressed
cylinder)
For the diesel cycle,
the fuel is injected
while the cylinder is
expanding, so the Qin
occurs at constant
pressure
5
Sec 9.3 : Air-Standard Diesel Cycle
6
Diesel Cycle analysis:
Process 2 – 3 : Constant pressure heat transfer to the air from an
external source while moves down (ignition and part of power stroke)
dW23  pdV
3
W23
  pdv  p  v3  v2 
2
m
Use W23 with energy balance to find Q23.
Q23  W23  m  u3  u2 
U  Q  W
Q23
 p  v3  v2    u3  u2   h3  h2
m
Remaining processes analysis is W12  U12  mu2  u1 
the same as for Otto cycle.
W34  U34  mu3  u4 
Q41  U 41  mu4  u1 


Q41 m
u4  u1 
 1
 1
h3  h2 
m
Q23 m
Wcycle m
Qin
Sec 9.3 : Air-Standard Diesel Cycle
7
Find State 1 Properties
Given T1 and r  use table to find u1 & h1.
Find state 2: for isentropic process.
vr 2
V2 vr1
 vr1

V1
r
or
pr 2
p2  p1
pr1
Find state 3: Use ideal gas law with p3 = p2.
V3
T3  T2
 rcT2
V2
were rc is the cutoff ratio:
Find state 4.
vr 4
V3
rc 
V2
V4 V4 V2 V1 V2
r



V3 V2 V3 V2 V3 rc
V4
r
 vr 3
 vr 3
V3 rc
For Cold-Air Standard analysis: (isentropic processes 1-2 and 3-4)
For state 2.
For state 4.
 T2

 T1
  V1
  
  V2



k 1
r
k 1
 T4

 T3
  V3
  
  V4



k 1
 rc 
 
r
k 1
8
A couple words about how and when to use
relative pressure and relative spec. volume.
only for use
between isentropic
processes.
9
Example (9.26): An air-standard Diesel cycle has a compression ratio of 16
and a cut-off ratio of 2. At the beginning of the compression, p1= 14.2 psi,
V1= 0.5 ft3, T1= 520°R. Calculate
(a)
(b)
(c)
(d)
The
The
The
The
State
heat added, in Btu.
T (R)
max T.
p (psi)
thermal efficiency.
mean effective pressure, u (Btu/lbm)
Given info: Diesel Cycle
State 1: p1=14.2 psi, T1=520 R
V1 = 0.5 ft3
State 2: s2 = s1
State 3: p3 = p2
State 4: s4 =s 3 and v4=v1
h (Btu/lbm)
vr
pr
Compression ratio: r = V1/V2 =16
Cutoff ratio: rc = V3/V2 = 2
1
520
14.2
2
3
4
10
Example (9.26): An air-standard Diesel cycle has a compression ratio of 16
and a cut-off ratio of 2. At the beginning of the compression, p1= 14.2 psi,
V1= 0.5 ft3, T1= 520°R. Calculate
State
State
(a)
(b)
(c)
(d)
The
The
The
The
heat added, in Btu.
TT (R)
(R)
max T.
pp (psi)
(psi)
thermal efficiency.
uu (Btu/lb
(Btu/lbmm))
mean effective pressure,
Determine State Properties:
For State 1:
Using Table A-22E: for T1
Read u 1 = 88.62 Btu/lbm
h1 = 124.27 Btu/lbm
pr1 = 1.2147
and vr1 = 158.58
11
22
33
44
520
520
14.2
14.2
88.62
hh (Btu/lb
(Btu/lbmm))
vvrr
158.58
pprr
1.2147
Also using Ideal Gas Equation:
m
pV  mRT
pV
RT
(14.2lb f / in 2 )(0.5 ft 3 )
Btu
144in 2

(0.06855 Btu / lbm  R)(520 R) 778lb f  ft ft 2
 0.03687lbm
11
Example (9.26):
For State 2:
(process 1-2 is isentropic)
Using compression ratio, r
State
1
2
T
T (R)
(R)
520
520
1502.5
p
p (psi)
(psi)
14.2
14.2
657.8
u
u (Btu/lb
(Btu/lbm) )
88.62
88.62
266.84
h
h (Btu/lb
(Btu/lbm) )
124.27
124.27
369.85
m
m
3
4
vvr
158.58
9.911
158.58
V1 vr1
r
r

p
1.2147
56.27
prr
1.2147
V2 vr 2
vr1 158 .58
V1 0.5 ft 3
vr 2 

 9.911
V2 

 0.03125 ft 3
r
16
r
16
Next use vr2 and Table A-22E to find other state properties:
T2 = 1502.5°R
u2= 266.84 Btu/lbm
pr2= 56.27
h2 = 369.85 Btu/lbm
Then:
p2
pr 2

p1
pr1
pr 2
 56.27 
p2  p1
 14.2 psi 
  657.8 psi
pr1
 1.2147 
12
Example (9.26):
For State 3:
Using ideal Gas pV  mRT
process 2-3 constant pressure
T2 T3

V2 V3
T3  T2
V3
 T2 rc
V2
State
State
11
22
33
TT (R)
(R)
520
520
1502.5
1502.5
3005
pp (psi)
(psi)
14.2
14.2
657.8
657.8
657.8
uu (Btu/lb
(Btu/lbmm))
88.62
88.62
266.84
266.84
586.16
hh (Btu/lb
(Btu/lbmm))
124.27
124.27
369.85
369.85
792.03
vvr
r
158.58
158.58
9.911
9.911
1.174
ppr
r
1.2147
1.2147
56.27
56.27
948.36
 rcT2   2 1502.5 R   3005 R
V3  rcV2  2(0.03125 ft 3 )  0.0625 ft 3
Then using Table A-22E, use T3 to find properties
u3 = 586.16 Btu/lbm
h3 = 792.03 Btu/lbm
p3r = 948.36
v3r = 1.174
44
Example (9.26):
For State 4:
(process 3-4 is isentropic
and V4 = V1)
V4 vr 4

V3 vr 3
therefore:
vr 4
State
State
T (R)
T (R)
p (psi)
p (psi)
u (Btu/lbm)
u (Btu/lbm
)
h (Btu/lbm)
h (Btu/lbm
)
v
vrr
p
prr
p4  p3
2
2
3
3
520
520
14.2
14.2
88.62
88.62
124.27
124.27
158.58
158.58
1.2147
1.2147
1502.5
1502.5
657.8
657.8
266.84
266.84
369.85
369.85
9.911
9.911
56.27
56.27
3005
3005
65.65
657.8
586.16
586.16
792.03
792.03
1.174
1.174
948.36
948.36
 0.5 ft 3 
V4
V1
 vr 3
 vr 3
 1.174 
 9.392
3 
V3
V3
 0.0625 ft 
Then using Table A-22E, use vr4 to find properties
u4 = 272.38Btu/lbm
h4 = 377.47 Btu/lbm
pr4 = 60.46
T4 = 1530.8 R
then
1
1
pr 4
 60.46 
 657.8 psi 
  41.9 psi
pr 3
 948.36 
13
4
4
1530.8
41.9
272.38
377.47
9.392
60.46
14
Example (9.26):
(a)
(b)
(c)
(d)
The heat added, in BTU.
The max T.
The thermal efficiency.
The mean effective pressure,
Heat is added during
process 2-3
State
1
2
3
4
T (R)
520
1502.5
3005
1530.8
p (psi)
14.2
657.8
657.8
41.9
u (Btu/lbm)
88.62
266.84
586.16
272.38
h (Btu/lbm)
124.27
369.85
792.03
377.47
vr
158.58
9.911
1.174
9.392
pr
1.2147
56.27
948.36
60.46
Applying the 1st Law
Q23  W23  m  u3  u2 
Q23  p V3  V2   m  u3  u2   m(h3  h2 )
Q23   0.0369lbm  792.03  369.85  Btu / lbm  15.58 Btu
Maximum Temperature is at Sate 3: Tmax = 3005 oR
15
Example (9.26):
(a)
(b)
(c)
(d)
1
2
3
4
T (R)
520
1502.5
3005
1530.8
14.2
657.8
657.8
41.9
88.62
266.84
586.16
272.38
124.27
369.85
792.03
377.47
vr
158.58
9.911
1.174
9.392
pr
1.2147
56.27
948.36
60.46
The heat added, in BTU.
p (psi)
The max T.
u (Btu/lbm)
The thermal efficiency.
The mean effective pressure, h (Btu/lbm)
Thermal Efficiency:

State
Wcycle
Qin
272.38  88.62 

u4  u1 

 1
 0.565
 1
 792.03  369.85 
 h3  h2 
Mean Effective Pressure:
where: Qin  Q23  15.58 Btu
mep 
Wcycle
V1 1  1r 

Qin  Qout
V1 1  1r 
Qout  Q41  m(u4  u1 )   0.0369lbm  272.38  88.62  Btu / lbm  6.78Btu
15.58  6.78 Btu
mep 
 0.5 ft 3  1  161 
778 ft  lb ft 2
 101.4 psi
2
Btu 144in
16
End of Slides for Lecture 34