94225879 otto cycle with animation

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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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p3 = 50*10^5 N/m2

T3/T2 = p3/p2 = 50/23.378 = 2.139

T3 = 2.139*733.34 = 1568.7 K


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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94225879 otto cycle with animation

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