Transcript Otto Cycle Derivation

TEKNIK
PERMESINAN
KAPAL II
(Minggu – 3)
LS 1329 ( 3 SKS)
Jurusan Teknik Sistem Perkapalan
ITS Surabaya
Gas Cycles
Carnot Cycle
T2
Heat
Q
2
3
Work
W
T1
1
s1
4
s2
1-2 - ADIABATIC COMPRESSION
(ISENTROPIC)
2-3 - HEAT ADDITION
(ISOTHERMAL)
3-4 - ADIABATIC EXPANSION
(ISENTROPIC)
4-1 - WORK
(ISOTHERMAL)
Carnot Cycle

Carnot cycle is the most efficient cycle
that can be executed between a heat
source and a heat sink.
T1
  1T2

However, isothermal heat transfer is
difficult to obtain in reality--requires large
heat exchangers and a lot of time.
Carnot Cycle

Therefore, the very important (reversible)
Carnot cycle, composed of two reversible
isothermal processes and two reversible
adiabatic processes, is never realized as a
practical matter.

Its real value is as a standard of
comparison for all other cycles.
Gas cycles have many engineering
applications



Internal combustion engine
 Otto cycle
 Diesel cycle
Gas turbines
 Brayton cycle
Refrigeration
 Reversed Brayton cycle
Some nomenclature before starting
internal combustion engine cycles
More terminology
Terminology





Bore = d
Stroke = s
 d 2 

Displacement volume =DV = s
 4 
Clearance volume = CV
Compression ratio = r
VBDC
DV  CV

r
CV
VTDC
Mean Effective Pressure
Mean Effective Pressure (MEP) is a fictitious
pressure, such that if it acted on the piston
during the entire power stroke, it would
produce the same amount of net work.
Wnet
MEP 
Vmax  Vmin
The net work output of
a cycle is equivalent to
the product of the
mean effect pressure
and the displacement
volume
Real Otto
cycle
Real and Idealized Cycle
Otto Cycle P-V & T-s Diagrams
Pressure-Volume
Temperature-Entropy
Otto Cycle Derivation

Thermal Efficiency:
QH - QL
QL
=1th =
QH
QH
 For
a constant volume heat addition (and
rejection) process;
Qin = m C v T
 Assuming
QRej = m C v T
constant specific heat:
T4 
T 1  - 1
m Cv ( T 4 - T 1 )
T
= 1-  1 
 th = 1 m Cv ( T 3 - T 2 )
T3 
T 2  - 1
T2 
Otto Cycle Derivation

For an isentropic compression (and expansion)
process:
 -1
 -1
T 2 =  V 1  =  V 4  = T 3
 
 
T1 V 2 
T4
V 3 


where:
γ = Cp/Cv
Then, by transposing,
T3 = T4
T2 T1
Leading to
T
1
 th = 1 T2
Differences between Otto and
Carnot cycles
T
2
33
2
44
1
3
s
Otto Cycle Derivation
The compression ratio (rv) is a volume ratio
and is equal to the expansion ratio in an otto
cycle engine.

Compression Ratio
rv =
V1 V4
=
V2 V3
where Compression ratio is defined as
Total volume
v s + vcc
=
rv =
Clearance volume
vcc
rv =
vs
+1
vcc
Otto Cycle Derivation

Then by substitution,
1-
T 1 =  V 2  = ( 1-
 
rv )
T2 V1 
The air standard thermal efficiency of the Otto cycle
then becomes:
1
 th = 1 - ( r v ) = 1  -1
( rv )
1-
Otto Cycle Derivation

Summarizing
th =
QH - QL
Q
=1- L
QH
QH
Q = m C v T
where
T4 
T 1  - 1
T
 th = 1 -  1  and
T3 
T 2  - 1
T2 
T3 = T4
T2 T1
then
T1
 th  1 
T2
1-
Isentropic
behavior
T 1 =  V 2  = ( 1-
 
rv )
T2 V1 
1
 th = 1 - ( r v ) = 1 ( r v ) -1
1-
Otto Cycle Derivation

Heat addition (Q) is accomplished through fuel
combustion

Q = Lower Heat Value (LHV) BTU/lb, kJ/kg
Qin
F
Q fuel
cycle = ma
A
also
Qin = m C v T
Effect of compression ratio on Otto
cycle efficiency
Sample Problem – 1
The air at the beginning of the compression
stroke of an air-standard Otto cycle is at 95
kPa and 22C 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.
Draw cycle and label points
3
r = V1 /V2 = V4 /V3 = 9
P
Q23 = 8.6 kJ
2
4
T1 = 295 K
1 P = 95 kPa
1
v
Carry through with solution
Calculate mass of air:
P1V1
m
 6.29 x 10-3 kg
RT1
Compression occurs from 1 to 2:
 V1 
T2  T1  
 V2 
k 1
 isentropiccompression
T2  22  273K 9
1.41
T2  705.6 K
But we need T3!
Get T3 with first law:
Q23  mcv T3  T2 
Solve for T3:
q
8.6 kJ 6.29x103 kg
 705.6K
T3   T2 
cv
0.855 kJ
kg
T3  2304.7K
Thermal Efficiency
  1
1
r
k 1
1 
  0.585
1
1.41
9
Sample Problem – 2
3
Solution
P
2
4
1
v
Diesel Cycle P-V & T-s Diagrams
Sample Problem – 3
Gasoline vs. Diesel Engine