Transcript Reflection of Buddhism in Contemporary Cinema

ENGR 2213 Thermodynamics
F. C. Lai
School of Aerospace and Mechanical
Engineering
University of Oklahoma
Gas Power Cycles
Otto Cycle
- Ideal cycle for spark-ignition engines
Diesel Cycle
- Ideal cycle for compression-ignition
engines
Brayton Cycle
- Ideal cycle for gas-turbine engines
Gas Power Cycles
Cold Air-Standard Assumptions
1. The working fluid is air which continuously circulates
in a closed loop and always behaves as an ideal gas.
2. All the processes which make up the cycle are internally
reversible.
3. The combustion process is replaced by a heat addition
process from an external source.
4. The exhaust process is replaced by a heat rejection
process which restores the air to its initial state.
► Air has constant specific heats which are evaluated at
room temperature (25 ºC or 77 ºF).
Gas Power Cycles
Vmin
TDC
Stroke
Vmax
Displacement
volume
BDC
Bore
Gas Power Cycles
Compression Ratio
Vmax VBDC
r

Vmin VTDC
Compression ratio is a volume ratio and should not be
confused with the pressure ratio.
Mean Effective Pressure (MEP)
W = MEP • AP • Stroke = MEP • Displacement volume
W
w
MEP 

Vmax  Vmin vmax  vmin
Otto Cycles
Nikolaus A. Otto (1876) – four-stroke engine
Beau de Rochas (1862)
Otto Cycles
Two-stroke vs. Four-stroke
Advantages
Disadvantages
1. Less efficiency
1. Simple
- Incomplete expulsion of
2. Inexpensive
the exhausted gases
3. High power-to-weight
and power-to-volume - Partial expulsion of the
ratios
air-fuel mixture with the
exhausted gases
Otto Cycles
First Law
q – w = Δu
qin = u3 – u2 = cv (T3 – T2)
qout = u4 – u1 = cv (T4 – T1)
qout
  1
qin
 T4

T1   1
T1
T4  T1


 1
 1
 T3

T3  T2
T2   1
1
 T2

 1
r k 1
3
T
Qin
2
4
1
Qout
k 1
T1  v 2 
 
T2  v1 
v1
r
v2
S
k 1
 v3 
 
 v4 

T4
T3
Otto Cycles
• The increase in thermal
efficiency is not as
pronounced at high
compression ratios.
• At high compression
ratios, autoignition may
occur.
• Autoignition in gasoline
engines hurts performance
and cause engine damage.
Example 1
An ideal Otto cycle has a compression ratio of 8. At
the beginning of the compression process, air is at
100 kPa and 17 ºC, and 800 kJ/kg of heat is transferred
to air during the constant-volume heat addition process.
Accounting for the variation of specific heats of air with
temperature, determine
(a) the maximum temperature and pressure which
occur during the cycle,
(b) the net work output,
(c) the thermal efficiency of this cycle, and
(d) the mean effective pressure for the cycle.
Example 1 (continued)
(a) State 1: air at p1 = 100 kPa and T1 = 290 K
Table A-17 u1 = 206.91 kJ/kg
vr1 = 676.1
Process 1-2: Isentropic compression
v r2 v 2 1
v r1 676.1


v r2 

 84.51
v r1 v1 r
r
8
Table A-17 T2 = 652.4 K
u2 = 475.11 kJ/kg
Example 1 (continued)
p2 v 2 p1v1

T2
T1
 T2  v1 
 652.4 
p2  p1     100 
(8)

 290 
 T1  v 2 
= 1799.7 kPa
Process 2-3: constant-volume heat addition
qin = u3 – u2
u3 = qin + u2 = 800 + 475.11 = 1275.11 kJ/kg
Table A-17 T3 = 1575.1 K
vr3 = 6.108
Example 1 (continued)
p3 v 3 p 2 v 2

T3
T2
 T3   v 2 
 1575.1 
p3  p2      1799.7 
(1)

 652.4 
 T2  v 3 
= 4347 kPa
(b) Process 3-4: Isentropic expansion
vr 4 v 4

r
vr4  rvr3  8(6.108)  48.864
v r3 v 3
Table A-17 T4 = 795.6 K
u4 = 588.74 kJ/kg
qout = u4 – u1 = 588.74 – 206.91 = 381.83 kJ/kg
Example 1 (continued)
wnet = qnet = qin – qout = 800 – 381.83 = 418.17 kJ/kg
w net
418.17
(c)  

 0.523
qin
800
1
  1  k 1 = 1 – r1-k = 1 – (8)1-1.4 = 0.565
r
RT1 (0.287)(290)
3
v



0.832
m
/ kg
(d) 1
p1
100
w net
w net
418.17
MEP 


= 574.4 kPa
v1  v 2 v1(1  1/ r) 0.832(1  1/ 8)
Diesel Cycles
Rudolph Diesel (1890)
Diesel Cycles
Gasoline Engines
1. Spark ignition (SI)
2. Constant-volume
heat addition
3. Low compression-ratio
Diesel Engines
Compression ignition (CI)
Constant-pressure heat
addition
High compression-ratio
4. High fuel cost
Low fuel cost
5. Low efficiency
High Efficiency
Diesel Cycles
First Law
q – w = Δu
Qin
T
qin = w + Δu = pΔv + Δu
= h3 – h2 = cp(T3 – T2)
2
qout = u4 – u1 = cv (T4 – T1)
1
qout
  1
qin
 T4

T1   1
T1
T4  T1


 1
 1
 T3

k  T3  T2 
kT2   1
 T2

3
p = const
4
v = const
Qout
S
Diesel Cycles
Cutoff Ratio
v 3 T3
rc 

v 2 T2
k 1
T1  v 2 
1
    k 1
T2  v1 
r
k 1
k 1
k 1
k 1
k 1
 v3 v2 
 v3   v 2 
T4  v 3 
 rc 
  
      
T3  v 4 
r
 v2 v4 
 v 2   v1 
k 1
T4 T4 T3 T2  rc 

   rc r k 1  rck
T1 T3 T2 T1  r 
Diesel Cycles
 T4

T1   1
T1


 1
 T3

kT2   1
 T2

qout
  1
qin
k

rc  1 
1
 1  k 1 

r  k  rc  1 
► ηD > ηO
► ηD increases as rc decreases
► ηD → ηO as rc → 1
Example 2
An ideal Diesel cycle has a compression ratio of 18
and a cutoff ratio of 2. At the beginning of the
compression process, air is at 100 kPa, 27 ºC and
0.0018 m3. Utilizing the cold-air-standard assumptions,
determine
(a) the temperature and pressure of air at the end of
each process,
(b) the net work output,
(c) the thermal efficiency of this cycle, and
(d) the mean effective pressure for the cycle.
Example 2 (continued)
(a) State 1: air at p1 = 100 kPa, T1 = 300 K and
V1 = 0.0018 m3
V1 0.0018
V2 

 0.0001m3
r
18
V3 = rc V2 = 2 (0.0001) = 0.0002 m3
V4 = V1 = 0.0018 m3
Process 1-2: Isentropic compression
k 1
 V1 
1.41
T2  T1    300(18)
 953.3 K
 V2 
Example 2 (continued)
k
 V1 
p2  p1    100(18)1.4  5719.8 kPa
 V2 
Process 2-3: Constant-pressure heat addition
p3 = p2 = 5719.8 kPa
 V3 
p3 v 3 p 2 v 2
T3  T2    953.3(2)

T3
T2
 V2 
= 1906.6 K
Process 3-4: Isentropic expansion
k 1
 V3 
T4  T3  
 V4 
0.4
 0.0002 
 1906.6 

0.0018


= 791.7 K
Example 2 (continued)
k
 V3 
0.0002 1.4
p4  p3    5719.8(
)  263.9 kPa
0.0018
 V4 
p1V1 100(0.0018)

 0.0021kg
(b) m 
RT1
0.287(300)
Qin = m(h3 – h2) = mcp(T3 – T2)
Table A-2
cp = 1.005 kJ/kg·K
cv = 0.718 kJ/kg·K
= 0.0021(1.005)(1906.6 – 593.3) = 2.77 kJ
Qout = m(u4 – u1) = mcv(T4 – T1)
= 0.0021(0.718)(791.7 – 300) = 0.74 kJ
Example 2 (continued)
Wnet = Qnet = Qin – Qout = 2.77 – 0.74 = 2.03 kJ
w net 2.03
(c)  

 0.733
qin
2.77
Wnet
2.03
= 1194.1 kPa

(d) MEP 
V1  V2  0.0018  0.0001