Transcript Lecture 33: Otto Cycle

EGR 334 Thermodynamics
Chapter 9: Sections 1-2
Lecture 33:
Gas Power Systems:
The Otto Cycle
Quiz Today?
Today’s main concepts:
• Understand common terminology of gas power cycles.
• Be able to explain the processes of the Otto Cycle
• Be able to perform a 1st Law analysis of the Otto Cycle and
determine its thermal efficiency.
• Be able to discuss limitations of the Otto cycle compared to
real spark ignition power systems.
• Be able to state the assumptions of standard air analysis.
Reading Assignment:
Read Chapter 9, Sections 3-4
Homework Assignment:
Problems from Chap 9: 1, 4, 11, 14
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 9.1 : Introducing Engine Terminology
Four Stroke Cycle : Two revolutions : Combusts mix of hydrocarbons + O2
1.Intake stroke : fill cylinder
•Spark cycle : fill with fuel and air mixture
•Compression cycle : fill with air
2.Compression stroke : p , T , V , Win
•Spark cycle : spark near end of stroke
•Compression cycle : inject fuel
3. Power stroke  : gas expands
4. Exhaust stroke  : spent gas
is exhausted
U--Tube video of 4 stroke:
http://www.youtube.com/
watch?v=2Yx32F1cncg
4
Sec 9.1 : Introducing Engine Terminology
Two Stroke Cycle : Two revolutions
1. Power/Exhaust:
•The piston is forced down
•@ exhaust port, spent gas leaves
•Piston continues down and
compresses air/fuel in crank case
•Compressed charge enters cylinder
2.Intake/Compression
•Piston moves up compressing charge
•Draws vacuum in crank case.
Two Stroke Animation:
http://library.thinkquest.org/C00
6011/english/sites/2_taktmotor.p
hp3?v=2
5
Sec 9.1 : Introducing Engine Terminology
6
Mean Effective Pressure (mep)
net work from one cycle
mep 
displacement volume
Air –standard Analysis
(A simplification used to allow
for thermodynamic analysis)
Assumptions:
-- Fixed amount of air modeled as
closed system
-- Air is treated as Ideal Gas
-- Constant cp (cold air-standard)
-- Combustion is modeled as a heat
transfer to system, Exhaust as
heat flow out of system
-- All processes internally reversible
Ideal
Relations:
Chap Gas
3: Model Quality
7
Polytropic Process
Ideal Gas Model Review
State Equation:
pv  RT
pV  mRT
pV  nRT
Energy Relationships:
H  m  c p dT  mc p T
U  m  cv dT  mcv T
where
kR
cp 
k 1
cp  cv  R
cv 
R
k 1
or look up values for k, cv, and cp on Table A-20
Entropy Relationships:
s(T2 , p2 )  s(T1 , p1 )  s o (T2 )  s o (T1 )  R ln
where so values are found on Table A-22
p2
p1
s(T2 , p2 )  s(T1 , p1 )  c p ln
s(T2 , v2 )  s(T1 , v1 )  cv ln
Special case: isentropic process where s1 = s2 then
k 1
T2  p2 
T2  v1 
p2  v1 
 
 
 
T1  v2 
T
p1  v2 
1
 p1 
( assuming constant specific heats)
k
T2
p
 R ln 2
T1
p1
( k 1) / k
v2 vr 2

v1 vr1
T2
v
 R ln 2
T1
v1
p2 pr 2

p1 pr1
( vr and pr for use with Table A-22)
Sec 9.2 : Air-Standard Otto Cycle
8
Assumption: At top dead center, heat addition occurs instantaneously
Otto Cycle: comprised of 4 internally reversible processes
Process 1 – 2 : Isentropic compression of air (compression stroke).
Process 2 – 3 : Constant volume heat transfer to the air from an
external source while piston is at top dead center (ignition)
Process 3 – 4 : Isentropic expansion (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
ignition
power
ignition
compression
compression
exhaust
power
exhaust
Sec 9.2 : Air-Standard Otto Cycle
9
Otto Cycle analysis
Closed system energy balance :
U  Q  W
Processes 1–2: ∆s = 0 and Q = 0
W12  m  u2  u1 
Process 3–4: ∆s = 0 and Q = 0
W34  m  u4  u3 
Processes 2–3 :
∆V = 0 and W = 0
Processes 4-1 :
∆V = 0 and W = 0
Wcycle  W12  W34
 m  u2  u1    u4  u3  
Q23  m  u3  u2 
Q41  m  u1  u4 
Qin  Q23  m  u3  u2 
Qout  Q41  m  u4  u1 
Sec 9.2 : Air-Standard Otto Cycle
10
Otto Cycle Thermal Efficiency:
Qin  Qout


Qin
Qin
u3  u2    u4  u1 


 u3  u2 
Wcycle
u4  u1 

 1
 u3  u2 
Thermal Efficiency can also be related
to the compression ratio:
clearance
displacement volume  clearance volume
r
clearance volume
displacement
V1 V4
r

V2 V3
  1
1
r k 1
As the compression ration, r, , the efficiency, η, 
11
Example (9.11): An air-standard Otto cycle has a compression ratio of 7.5.
At the beginning of compression, p1 = 85 kPa and T1 = 32°C. The mass of air
is 2 g, and the maximum temperature in the cycle is 960 K. Determine
(a)
(b)
(c)
(d)
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure,
in kPa.
State
1
2
T (K)
305
p (kPa)
85
3
4
960
u (kJ/kg)
v
T3= 960 K
m  2 g  0.002kg
p1=85 kPa
T1=305 K
12
Example (9.11):
(a)
(b)
(c)
(d)
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure,
in kPa.
State 1: Using Ideal Gas Law:
v1 
State
1
T (K)
305
p (kPa)
85
2
3
960
u (kJ/kg)
v
pv  RT
and Table A-22:
RT1 (0.287 kJ / kg  K )(305 K ) kPa kN  m
3


1.0298
m
/ kg
2
p1
(85kPa )
kN / m
kJ
u1  217.67kJ / kg
State 2: Using compression ratio, ideal gas law, and Table A-22:
v1
r
v2

v1 1.0298
v2  
 0.1373m3 / kg
r
7.5
4
13
Example (9.11):
(a)
(b)
(c)
(d)
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure,
in kPa.
State
1
T (K)
305
p (kPa)
85
u (kJ/kg)
217.67
v
1.0298
2
3
960
0.1373
State 3: using v3 = v2, ideal gas law, and Table A22:
v3  0.1373m3 / kg
p3 
RT3 (0.287 kJ / kg  K )(960 K ) kPa kN  m

 2006.7 kPa
3
2
v3
(0.1373m / kg )
kN / m
kJ
u3  725.02kJ / kg
State 4: Using v4=v1:
v4  v1  1.0298m3 / kg
4
14
Example (9.11):
(a)
(b)
(c)
(d)
(e)
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure
, in kPa.
State
1
T (K)
305
960
p (kPa)
85
2006.7
u (kJ/kg)
217.67
725.02
v
1.0298
T v 
also knowing for isentropic processes 2   1 
T1  v2 
2
k 1
0.1373
3
0.1373
p2  v1 
 
p1  v2 
4
1.0298
k
then for isentropic process 1-2 using k = 1.361 and Table A-22:
T2  T1  r 
k 1
 305K  7.5 
0.361
 631K
p2  p1  r   85(7.5)1.361  1319kPa
k
u2  458.55kJ / kg
and for isentropic process 3-4:
 r
T4  T3 1
k 1

 960 K 1
u4  328.5kJ / kg
7.5

0.361
 464 K p4  p3
 r
1
k
 2006.7( 1
)1.4  129.3kPa
7.5
15
Example (9.11):
(a)
(b)
(c)
(d)
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure,
in kPa.
State
1
2
3
4
T (K)
305
631
960
464
p (kPa)
85
1319
2006.7
129.3
u (kJ/kg)
217.67
458.55
725.02
328.5
v
1.0298
0.1373
0.1373
1.0298
Heat added during process 2-3:
Q23  U 23  m  u3  u2 
kJ
  0.002kg  725.02  458.55 
 0.5329kJ
kg
Heat rejected during process 4-1:
Q41  U 41  m  u1  u4 
kJ
  0.002kg  217.67  328.50 
 0.2217 kJ
kg
U  Q  W
16
Example (9.11):
(a)
(b)
(c)
(d)
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure,
in kPa.
State
1
2
3
4
T (K)
305
631
960
464
p (kPa)
85
1319
2006.7
129.3
u (kJ/kg)
217.67
458.55
725.02
328.5
v
1.0298
0.1373
0.1373
1.0298
Net Work over the cycle:
Wcycle  Qin  Qout  Q23  Q41
 (0.5329)  (0.2217)  0.3112kJ
Cycle Efficiency:

Wcycle
Qin
Compare to
  1
0.3112kJ

 0.5839
0.5329kJ
1
r k 1
1
 1
 0.553
0.4
7.5
17
Example (9.11):
(a)
(b)
(c)
(d)
The heat rejection, in kJ.
The net work, in kJ.
The thermal efficiency.
The mean effective pressure,
in kPa.
State
1
2
3
4
T (K)
305
631
960
464
p (kPa)
85
1319
2006.7
129.3
u (kJ/kg)
217.67
458.55
725.02
328.5
v
1.0298
0.1373
0.1373
1.0298
Mean Effective Pressure:
net work from one cycle Wcycle
mep 

displacement volume
V1  V2
where
V1  mv1  0.002kg (1.0298m3 / kg )  0.0020596m3
V2  mv2  0.002kg (0.1373m3 / kg )  0.0002746m3
0.3112kJ
kN  m kPa
mep 
 174.34kPa
3
2
(0.0020596 - 0.0002746)m
kJ kN / m
18
End of Slides for Lecture 33