diesel cycle presentation lec(3)

7/27/2019 Diesel Cycle Presentation Lec(3)

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Air-Standard

Diesel Cycle


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Air-Standard Diesel Cycle

The air-standard Diesel cycle is an ideal cycle that assumes the heat

addition occurs during a constant-pressure process that starts with the

piston at top dead center. The Diesel cycle is shown onpv and Ts

diagrams. The cycle consists of four internally reversible processes in

series.

The first process from state 1 to state 2 is the same as in the Otto

cycle: an isentropic compression.

Heat is not transferred to the working fluid at constant volume as inthe Otto cycle, however. In the Diesel cycle, heat is transferred to the

working fluid at constant pressure. Process 23 also makes up the first

part of the power stroke.

The isentropic expansion from state 3 to state 4 is the remainder of

the power stroke.

As in the Otto cycle, the cycle is completed by constant-volume

Process 41 in which heat is rejected from the air while the piston is at

bottom dead center. This process replaces the exhaust and intake

processes of the actual engine.


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Since the air-standard Diesel cycle is composed of

internally reversible processes, areas on the Ts and

p

v diagrams of Fig. 9.5 can be interpreted as heatand work, respectively.

On the Ts diagram, area 23ab2 represents the

heat added per unit of mass andarea 14ab1 is

the heat rejected per unit of mass. On thepv

diagram, area 12ab1 is the work input per unit

of mass during the compression process. Area 234

b

a

2 is the work done per unit of mass as the pistonmoves from top dead center to bottom dead center.

The enclosed area of each figure is the net work

output, which equals the net heat added.


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CYCLE ANALYSIS.

In the Diesel cycle the heat addition takes place at

constant pressure. Accordingly, Process 23 involves

both work and heat. The work is given by

The heat added in Process 23 can be found by applying

the closed system energy balance


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solving for the heat transfer


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where the specific enthalpy is introduced to simplify the

expression. As in the Otto cycle, the heat rejected in Process

4

1 is given by

The thermal efficiency is the ratio of the net work of

the cycle to the heat added


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EFFECT OF COMPRESSION RATIO ON

PERFORMANCE.

As for the Otto cycle, the thermal efficiency ofthe Diesel cycle increases with increasing

compression ratio. This can be brought out

simply using a cold air-standard analysis. On acold air-standard basis, the thermalefficiency of

the Diesel cycle can be expressed as


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where r is the compression ratio and rc the

cutoff ratio. The derivation is left as an exercise.

This relationship is shown in Fig. 9.6 for k =1.4.Equation 9.13 for the Diesel cycle differs from

Eq. 9.8 for the Otto cycle only by the term in

brackets, which for rc >1 is greater than unity.Thus, when the compression ratio is the same,

the thermal efficiency of the cold air-standard

Diesel cycle would be less than that of the coldair-standard Otto cycle.


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E X A M P L E 9 . 2

Analyzing the Diesel Cycle

At the beginning of the compression process

an air-standard Diesel cycle operating with

compression ratio of 18, the temperature is 300

and the pressure is 0.1 MPa. The cutoff ratio f

the cycle is 2. Determine

(a) the temperature and pressureat the end of

each process of the cycle,

(b) the thermal efficiency,

(c) the mean effective pressure, in MPa.


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S O L U T I O N

Known:An air-standard Diesel cycle is executed with

specified conditions at the beginning of the

compression stroke. The compression andcutoff ratios are given.

Find:Determine the temperature and pressure at

the end of each process, the thermal efficiency,

and mean effective pressure.


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Assumptions:

1. The air in the pistoncylinder

assembly is the closed system.2. The compression and expansion

processes are adiabatic.3. All processes are internally

reversible.

4. The air is modeled as an ideal gas.

5. Kinetic and potential energy effects

are negligible


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Analysis:

(a)The analysis begins by determining properties at each principal

state of the cycle.

With T1 =300 K, u1 =214.07 kJ/kg , use equation of state

111RTvP

RTPv

mRTPV


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18

2

1

v

v

tconspv k tan

For the isentropic compression process 1

2 get p2,T2 and hence u2

And


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tconsPvk

tan


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we get u4 =664.3 kJ/kg and T4 =887.7 K. The pressure at state 4

can be found using the isentropic relationship \or the ideal gas

equation of state applied at states 1 and 4. With V4 =V1, the ideal

gas equation of state gives

34v

r

r

v

c


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diesel cycle presentation lec(3)

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