Otto Cycle, ideal for spark ignition engines
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Transcript Otto Cycle, ideal for spark ignition engines
MT 313 IC ENGINES
LECTURE NO: 04
(24 Feb, 2014)
Khurram
[email protected]
Yahoo Group Address: ICE14
Air Standard Cycle
β’ The air as the working fluid follows the perfect
gas law
ππ = ππ
π
β’ The working fluid is homogeneous throughout
and no chemical reaction takes place
β’ Specific heats of air do not vary with temperature
β’ The mass of air in the cycle remains fixed
β’ The exhaust process is replaced by an equivalent
heat rejection process
β’ The combustion process is replaced by an
equivalent heat addition process
β’ All processes are internally reversible
Air Standard Cycle
β’ Thermal efficiency
ππππ π·πππ
Ξ·π‘β =
π»πππ‘ ππ’ππππππ
π»πππ‘ ππ’ππππππ β π»πππ‘ π
πππππ‘ππ
=
π»πππ‘ ππ’ππππππ
ππ β ππ
=
ππ
ππ β ππ
=
ππ
β’ Thermal efficiency is also called air standard
efficiency Ξ·a
Important Formulas
β’ Swept Volume
π 2
ππ = π πΏ
4
β’ Clearance Volume
π = ππ + ππ
β’ Compression Ratio
ππ + ππ
π=
ππ
β’ Clearance ratio
ππ
π=
ππ
Ideal cycles are simplified
Otto Cycle, ideal for spark ignition engines
OTTO CYCLE
β’ Process No 1-2 β Isentropic Expansion
1
pVΞ³ = c
P1
2
P2
Ξ³=
ππ
ππ£
V1
V2
pβ V diagram
β’
OTTO
CYCLE
Process No 1-2 β Reversible Adiabatic or
Isentropic Expansion
1
T1
T2
2
T β S diagram
No Heat is added or rejected Q 1-2 = 0
S1, S2
OTTO CYCLE
β’ Process No 2-3 β Constant volume cooling process
2
P2
P3
3
V2
pβ V diagram
Heat is rejected by air getting cooled from T2 to T3
β’
OTTO
CYCLE
Process No 2-3 β Constant volume cooling process
T2
2
T3
S3 , S4
T β S diagram
S1, S2
Heat is rejected by air getting cooled from T2 to T3
OTTO CYCLE
β’ Process No 3-4 β Isentropic Compression
4
pVΞ³ = c
P4
P3
Ξ³=
ππ
ππ£
3
V4
pβ V diagram
No heat is added or rejected
V3
OTTO CYCLE
β’ Process No 3-4 β Reversible Adiabatic or Isentropic Expansion
T4
4
T3
3
S3 , S4
T β S diagram
No Heat is added or rejected Q 3-4 = 0
OTTO CYCLE
β’ Process No 4-1 β Constant volume heating process
1
P1
P4
4
V2
pβ V diagram
Heat is absorbed by air getting heated from T4 to T1
β’
OTTO
CYCLE
Process No 4-1 β Constant volume heating process
1
T1
T4
4
S3 , S4
S1, S2
T β S diagram
Heat is absorbed by air getting heated from T4 to T1
OTTO Cycle
β’ Process 1-2
No heat is added or rejected
π1β2 = 0
β’ Process 2-3 Heat is rejected by air getting
cooled from temperature T2 to T3
π1β2 = πππ£ π2 β π3
β’ Process 3-4 No heat is added or rejected
π3β4 = 0
β’ Process 4-1 Heat is absorbed by air getting
heated from temperature T4 to T1
π4β1 = πππ£ π1 β π4
OTTO Cycle
β’ Work Done = Heat absorbed β Heat rejected
β’ Work Done = π4β1 β π1β2
β’ Work Done = πππ£ π1 β π4 - πππ£ π2 β π3
β’Ξ·=
β’Ξ·=
ππππ π·πππ
π»πππ‘ π΄ππ πππππ
πππ£ π1 βπ4 β πππ£ π2 βπ3
πππ£ π1 βπ4
OTTO Cycle
β’Ξ·=
πππ£ π1 βπ4 β πππ£ π2 βπ3
πππ£ π1 βπ4
β’ Ξ·=1β
β’ Ξ·=1β
β’ Ξ·=1β
πππ£
πππ£
π2 βπ3
π1 βπ4
π2 βπ3
π1 βπ4
π2
π3
π3
π1
π4
π4
β1
β1
OTTO Cycle
β’ For reversible adiabatic expansion process
1-2
β’
π2
π1
=
π£1 Ξ³β1
[ ]
π£2
β’ where expansion ratio =
π£2
π£1
OTTO Cycle
β’ For reversible adiabatic expansion process
3-4
β’
π3
π4
=
π£4 Ξ³β1
[ ]
π£3
β’ where expansion ratio =
π£3
π£4
β’
π£3
π£4
=
OTTO Cycle
π£2
π£1
β’ Ξ·=1β
β’ Ξ·=1 β
β’ Ξ·=1β
π3
π4
π2
π1
1
π Ξ³β1
OTTO CYCLE
β’ Process No 1-2 β Isentropic Expansion
1
pVΞ³ = c
P1
2
P2
Ξ³=
ππ
ππ£
V1
V2
pβ V diagram
OTTO Cycle
2
πππ
1
β’ π1β2 =
β’ ππ Ξ³ = πΆπππ π‘πππ‘
β’ π1β2 = πΆ
β’ π1β2 =
β’ π1β2 =
2 ππ
1 πΞ³
π1 π1 βπ2 π2
Ξ³β1
π
(π1 βπ2 )
Ξ³β1
Problem 1
β’ Calculate the air standard efficiency of a four
stock Otto cycle engine with the following
data
Piston diameter (bore)= 13.7 cm
Length of stock
= 13.0 cm
Clearance volume
= 14.6 %
β’ Diagram
Solution
β’ Swept Volume
β’ ππ =
β’
=
π
4
π
4
π·2πΏ
13.72 β 13
β’
= 1916 cm
β’ Clearance Volume
β’ ππ =
β’
π
4
β ππ
= 297.7 cm3
Solution
β’ Compression ratio
β’ ππ =
ππ +ππ
ππ
β’
= 7.85
β’ Air Standard efficiency
β’ Ξ·=1β
β’
1
π
Ξ³β1
= 56.2%
Problem 2
β’ In an Otto cycle the compression ratio is 6 .
The initial pressure and temperature of the air
are 1 bar and 100ΛC. the maximum pressure in
the cycle is 35 bar. Calculate the parameter at
the salient points of the cycle. What is the ratio
of heat supplied to heat rejected
β’ How does air standard efficiency of the cycle
compares with that of a Carnot cycle working
within the same extreme temperature limits?
Explain the difference between the two values
Problem 2
β’ If the engine has a relative efficiency of 50 %
determine the fuel consumption per kWh.
Assume the fuel used has a calorific value of
42,000 kJ/kg
Problem 3
β’ An Otto cycle working on air has a
compression ratio of 6 and starting condition
are 40ΛC and 1 bar. The peak pressure is 50
bar. Draw the cycle on p-v and T-S coordinates
if compression and expansion follow the law
pV1.25 = C. Calculate mean effective pressure
and heat added per kg of air.
Problem 4
β’ An Otto cycle has compression ratio of 8 and
initial conditions are 1 bar and 15ΛC. Heat
added during constant volume process is 1045
kJ/kg. Find :
β’ Maximum cycle temperature
β’ Air standard efficiency
β’ Work done per kg of air
β’ Heat rejected
β’ Take cv = 0.7175 kJ/kg-K and Ξ³ = 1.4
Problem 5
β’ Find out the compression ratio in an Otto for
maximum work output
β’ An Otto cycle engine has the following data.
Calculate compression ratio, air standard efficiency
and specific fuel consumption.
Piston diameter
=
13.7
Length of stock
=
13 cm
Clearance volume
=
280 cm3
Relative efficiency
=
60 %
Lower calorific volume of petrol =
41900kJ/kg