gas cycles otto,diesel,dual cycles
TRANSCRIPT
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
1/43
1
Gas Cycles
Assumptions of air standard cycle Analyze three cycles
Otto
Diesel
Brayton
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
2/43
2
Assumptions of Air Standard Cycle
Working fluid is air
Air is ideal gas
Combustion process is replaced by heat addition
process Heat rejection is used to restore the fluid to its initial
state and complete the cycle
All processes are internally reversible
Constant or variable specific heats can be used
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
3/43
3
Gas cycles have many
engineering applications
Internal combustion engine Otto cycle
Diesel cycle
Gas turbines
Brayton cycle
Refrigeration
Reversed Brayton cycle
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
4/43
4
Some nomenclature before starting internal
combustion engine cycles
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
5/43
5
More Terminology
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
6/43
6
Terminology
Bore = d
Stroke = s
Displacement volume =DV =
Clearance volume = CV
Compression ratio = r
4
ds
2
TDC
BDC
V
V=
CV
CVDVr
+=
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
7/43
7
Mean Effective Pressure
Mean Effective Pressure (MEP) is a fictitiouspressure, that if acted on the piston during the
entire power stroke, would produce the same
amount of net work.
minmax VV
W
MEP
net
=
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
8/43
8
The net work output ofa cycle is equivalent tothe product of the meaneffect pressure and thedisplacement volume
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
9/43
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
10/43
10
Real and Idealized Cycle
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
11/43
11
Idealized Otto Cycle
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
12/43
12
Idealized Otto Cycle
1-2 - Adiabatic Compression (Isentropic)
2-3 - Constant Volume Heat Addition
3-4 - Adiabatic Expansion (Isentropic)
4-1 - Constant Volume Heat Rejection
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
13/43
13
Performance of Cycle
in
net
q
w
=Efficiency:
Lets start by getting heat input:
23in uuq =
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
14/43
14
Cycle Performance
Get net work from energy balance of cycle:
outinnet qqw =Substituting for qin and qout:
=netw )uu()uu( 1423
Efficiency is then:
in
net
q
w=
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
15/43
15
Cycle Performance
Substituting for net work and heat input:
)u-(u
)u-(u-)u-(u
23
1423=
We can simplify the above expression:
)u-(u)u-(u1
23
14=
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
16/43
16
Cold Air Standard Cycle
cp, cv, and k are constant at ambienttemperature (
70 F) values.
Assumption will allow us to get a quick first
cutapproximation of performance of cycle.
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
17/43
17
Cycle performance with cold air
cycle assumptions
If we assume constant specific heats:
)T-(Tc
)T-(Tc
)u-(u
)u-(u
1 23v
14v
23
14
==
)T-(T
)T-(T1
23
14=
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
18/43
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
19/43
19
Cycle performance with cold air
cycle assumptions
1k
2
1
r11
TT1 ==
This looks like the Carnot efficiency, but it is
not! T1 and T2 are not constant.
What are the limitations for this expression?
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
20/43
20
Differences between Otto and
Carnot Cycles
T
s
1
2
3
4
T
s
1
2
3
4
2
3
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
21/43
21
Effect of compression ratio on
Otto cycle efficiency
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
22/43
22
Sample ProblemThe air at the beginning of the compression stroke of
an air-standard Otto cycle is at 95 kPa and 22C and
the cylinder volume is 5600 cm3. The compression
ratio is 9 and 8.6 kJ are added during the heat
addition process. Calculate:
(a) the temperature and pressure after the
compression and heat addition process(b) the thermal efficiency of the cycle
Use cold air cycle assumptions.
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
23/43
23
Draw cycle and label points
P
v
1
2
3
4
T1 = 299 K
P1 = 0.95 bar
r = V1 /V2 = V4 /V3 = 9
Q23 = 8.6 kJ
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
24/43
24
Major assumptions
Kinetic and potential energies are zero Closed system
1 is start of compression
Ideal cycle: 1-2 isentropic compression, 2-3
constant volume heat addition, etc.
Cold cycle constant properties
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
25/43
25
Carry through with solutionCalculate mass of air:
kg10x29.6RT
VPm 3-
1
11 ==
Compression occurs from 1 to 2:
ncompressioisentropicV
VTT
1
2
112
=
k
( ) ( ) 11.42 9K27322T +=
K705.6T2= But we need T3!
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
26/43
26
Get T3 with first law:
( )23v23 TTmcpe)keu(mWQ =++=Solve for T3:
2
v
3 Tc
qT += K705.6
kgkJ0.855
kg6.29x10kJ8.6
3
=
K2304.7T3=
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
27/43
27
Thermal Efficiency
11.41k 911
r11 ==
585.0=
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
28/43
28
Lets take a look at the Diesel cycle.
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
29/43
29
Idealized Diesel cycle1-2 - Adiabatic Compression (Isentropic)
2-3 - Constant Pressure Heat Addition
3-4 - Adiabatic Expansion (Isentropic)
4-1 - Constant Volume Heat Rejection
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
30/43
30
Performance of cycle
in
net
q
w
=Efficiency:
Heat input occurs from 2 to 3 inconstant pressure process:
23in hhq = Why enthalpies?
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
31/43
31
Diesel Cycle Performance
Get net work from energy balance of cycle:
outinnet qqw =Substituting for qin and qout:
=netw )uu()hh( 1423
As we did with the Otto cycle:
in
net
q
w=
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
32/43
32
Diesel Cycle Performance
Substituting for net work and heat input:
)h-(h)u-(u-)h-(h
23
1423=
We can simplify the above expression:
)h-(h)u-(u1
23
14=
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
33/43
33
For cold cycle analysis
)T-(Tc
)T-(Tc1
23p
14v= )T-k(T)T-(T
123
14=
Its possible to rewrite this in a simpler form if we
define a new term:
ratiocutoffV
Vr
2
3c ==
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
34/43
34
Cold cycle efficiency
( )
= 1
1111
c
k
c
krk
r
r
Efficiency is dependent on compressionratio and cutoff ratio.
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
35/43
35
k = 1.4
Th th l ffi i f th id l Di l l
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
36/43
36
The thermal efficiency of the ideal Diesel cycle as afunction of compression and cutoff rates (k=1.4)
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
37/43
37
Comparison between Otto and
Diesel cyclesFor same compression ratio
OTTO > DIESELFor same combustion temperature
DIESEL > OTTO
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
38/43
38
Sample ProblemA Diesel air-standard cycle has a compression ratio
of 15:1. The pressure and temperature at thebeginning of the compression are 100 kPa and 17 C,
respectively. If the maximum temperature of the
cycle is 2250 K, determine:
1. the cutoff ratio2. the thermal efficiency
3. the mean effective pressure
Draw cycle
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
39/43
39
Draw cycle
Apply assumption as before with
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
40/43
40
pp y assu pt o as be o e w t
Otto cycle
Cold cycle assumptions
Kinetic and potential energies neglected
Ideal cycle
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
41/43
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
42/43
42
More analysis
Cutoff ratio is then:
62.27.8562250 ==cr
Thermal efficiency:
( )
= 111
1 1c
k
c
krk
r
r
-
8/10/2019 Gas Cycles Otto,Diesel,Dual cycles
43/43
43
More analysis
Plug numbers into thermal efficiency:
( )
=
162.24.1
162.2
15
11
4.1
4.0
574.0=