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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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