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the accurate analysis of internal combustionengine is very complicated. In order to understand them it
is advantageous to analyze the performance of anidealized closed cycle that closely approximates the realcycle. One such approach is the air standard cycle
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WHY DO WE NEED AIR STANDARD CYCLES1.Accurate analysis is a very expensive affair.
2. Accurate analysis consumes a lot of time
3. Accurate analysis is complex phenomena and cannotbe modeled easily.
4. Theoretical analysis gives us the power to analyseengine performance without actually building it.
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ASSUMPTIONS Working medium is a perfect gas
There is no change in mass of the medium
All the process are reversible
Some heat is rejected to a constant low temp. Sink
There are no heat losses from system to thesurroundings
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DIESEL CYCLE
It was developed by RUDOLF DIESEL in the year 1892 as acycle in which the heat is added at a constant volume and itforms the basis for todays compression ignition engines.
PROCESSES
0-1 SUCTION
1-2 ISENTROPIC COMPRESSION
2-3 HEAT ADDITION3-4 ISENTROPIC EXPANSION
4-1 HEAT REJECTION
1-0 EXHAUST
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DIESEL CYCLE
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OTTO CYCLE
It was proposed by Nicholas Otto in the year 1876 as a
constant volume heat addition cycle which forms the
basis for the working of todays spark igntion engines
PROCESSES
0-1 SUCTION
1-2 ISENTROPIC COMPRESSION
2-3 HEAT ADDITION
3-4 ISENTROPIC EXPANSION
4-1 HEAT REJECTION
1-0 EXHUAST
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DIAGRAMATIC REPRESENTATION OF THE
OTTO CYCLE
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DUAL CYCLE
It is the hybrid form of diesel and otto cycle. Dual cycleconsists of cycles of both constant volume andconstant pressure heat addition
PROCESSES0-1 .. SUCTION
1-2 .. ISENTROPIC COMPRESSION
2-3 AND 3-4 .. HEAT ADDITION
4-5 .. ISENTROPIC EXPANSION
5-1 .. HEAT REJECTION
1-0 .. EXHAUST
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DUAL CYCLE
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COMPARISON OF OTTO,DIESEL
AND DUAL CYCLES Important variable factors used for the basis of
comparison COMPRESSION RATIO , PEAKPRESSURE , HEAT ADDITION, HEATREJECTION , NET WORK
To compare the performance of these cyclessome of the variables must be fixed
The analysis will show which cycle is more
efficient for a given set of operating conditions
5 different cases are taken for comparison
COMPRESSIONRATIO
PEAKPRESSURE
HEATADDITION
HEATREJECTION
NET WORK
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CASE 1:SAME COMPRESSION RATIO AND HEAT
ADDITION Figure below shows the Otto cycle 1-2-3-4-1, the Diesel
cycle 1-2-3'-4'-1 and the Dual cycle 1-2-2-3-4-1 in p-v andT-s diagrams respectively
From T-s diagram, Area 5-2-3-6= Area 5-2-3'-6 = Area 5-2-2"-3"-6 same heat addition
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CASE A
SAME COMPRESSION RATIO AND HEAT
ADDITION All the cycles start from the same initial state point 1 andthe air is compressed from state 1 to 2 thecompression ratio is same
Observations from T-s diagram:Heat rejection in Otto cycle (area 5-1-4-6) is minimumHeat rejection in Diesel cycle (5-1-4'-6') is maximum
Inferences :Otto cycle has the highest work output and efficiencyDiesel cycle has the least efficiencyDual cycle has the efficiency between the two
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CASE B
SAME COMPRESSION RATIO AND HEAT REJECTION
Heat rejection QR is same for 3 cycles Heat supplied in otto cycle(QS )=area area under the curve2-3
Heat supplied in diesel cycle(QS )=area area under the curve2-3
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CASE B
same compression ratio and heat
rejection(contd.) Efficiency of Otto cycle : Otto=1-QR/QS
Efficiency of diesel cycle: diesel
=1-QR/Q
s
Observations from T-s diagram: QS>Qs
Inference: The efficiency of the otto cycle is greater thanthat of the diesel cycleOtto>diesel
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CASE C
SAME PEAK PRESSURE,PEAK TEMPERATURE AND
HEAT REJECTION
Peak pressure and temperature and the amount of heat rejected(QR)are same for 3 cycles
Heat supplied in Otto cycle(QS )=area area under the curve2-3
Heat supplied in diesel cycle(QS )=area area under the curve2-3
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CASE C
SAME PEAK PRESSURE,PEAK TEMPERATURE AND
HEAT REJECTION(contd.) Efficiency of Otto cycle : Otto=1-QR/QS
Efficiency of diesel cycle: diesel
=1-QR/Q
s
Observations from T-s diagram: The heat heat absorbed incase of otto cyle will be greater than that of the dieselcycleQS>Qs
Inference: diesel>Otto
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CASE D
SAME MAXIMUM PRESSURE AND HEAT INPUT
Figures show Otto cycle (1-2-3-4-1) and Diesel cycle(1-2'-3'-4'-1) shown in figure for same max. pressure and heat input
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CASE D
SAME MAXIMUM PRESSURE AND HEAT INPUT Observation from T-s diagram:
Heat rejection for Otto cycle (area 1-5-6-4) is morethan the heat rejected in Diesel cycle (1-5-6'-4')
Inference: The efficiency for diesel cycle will be greaterthan that for otto cycle diesel>Otto
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CASE E
SAME MAXIMUM PRESSURE AND WORK OUTPUT Area 1-2-3-4 (work output of Otto cycle) = area 1-2'-3'-4'
(work output of Diesel cycle)
Efficiency()= Work done/ Heat supplied= Workdone/(Work done+ Heat rejected)
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CASE E
SAME MAXIMUM PRESSURE AND WORK
OUTPUT(contd.) For the same work output, we find thatentropy S3>entropy
S3
Observations from T-s diagram:Heat rejection for Otto cycle is more than that of
diesel cycle
Inference: The efficiency of diesel cycle is greater than thatof the otto cycle that is diesel>Otto
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SOLUTION:-
Consider the process 1-2,
p2/p1 = r= 9.5^1.4 = 23.378
p2 = 23.378*1*10^5 = 23.378*10^5.Ans
T2/T1 = r(-1) = 9.5^0.4 = 2.461
T2 = 2.461*298 = 733.34 K..Ans
TO FIND:- Pressure & Temperature at the salient points
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Consider the process 2-3,
p3 = 50*10^5 N/m2
T3/T2 = p3/p2 = 50/23.378 = 2.139
T3 = 2.139*733.34 = 1568.7 K
Consider the process 3-4,
p3/p4 = (v4/v3) = (v1/v2)
r = 9.5^1.4 = 23.378
p4 = p3/23.378 = 2.139 N/m2.Ans
T3/T4 = r(-1) = 9.5^0.4 = 2.461
T4 = T3/2.461 = 1568.7/2.461 = 637.42 KAns
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SOLUTION:-Consider the process 1-2,
P2V2n = P1V1
n
P2= P1(V1/V2)n = 1*6^1.3 = 10.27 bar
T2 = T1(P2V2/P1V1)
330*(10.27*(1/6)) = 565 K
To Find:- Maximum Pressure, Compare this value with
obtained when Cv = 0.717 KJ/kg K
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