The operating conditions at the beginning of compression in an air-standard Diesel cycle are: P, 200 kPa, TI = 380 K. The compression ratio is VI/V2 = 20 and the maximum temperature is T3 = 1879.5 K. Neglect kinetic and potential energy variations and conside a steady state operation for all processes. %3D %3D (a) Plot the PV diagram. Determine (b) the temperatures T2 & T4, in K (Note: V/Vy T:/T). (c) the thermal efficiency of the Diesel cycle.

Elements Of Electromagnetics
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The operating conditions at the beginning of compression in an air-standard Diesel cycle
are: P, = 200 kPa, T, = 380 K. The compression ratio is Vi/V2 = 20 and the maximum
%3D
temperature is T3 = 1879.5 K. Neglect kinetic and potential energy variations and consider
a steady state operation for all processes.
(a) Plot the PV diagram.
Determine
(b) the temperatures T2 & T4, in K (Note: V/Vy=T;/Ty).
(c) the thermal efficiency of the Diesel cycle.
Transcribed Image Text:The operating conditions at the beginning of compression in an air-standard Diesel cycle are: P, = 200 kPa, T, = 380 K. The compression ratio is Vi/V2 = 20 and the maximum %3D temperature is T3 = 1879.5 K. Neglect kinetic and potential energy variations and consider a steady state operation for all processes. (a) Plot the PV diagram. Determine (b) the temperatures T2 & T4, in K (Note: V/Vy=T;/Ty). (c) the thermal efficiency of the Diesel cycle.
Analysis:
(a) The analysis begins by determining properties at each principal state of the cycle. With T = 300 K, Table A-22 gives
u, = 214.07 kJ/kg and v, = 621.2. For the isentropic compression process 1-2
%3D
V2
U2 =
621.2
%3D
34.51
%3D
%3D
18
Interpolating in Table A-22, we get T = 898.3 K and h, = 930.98 kJ/kg. With the ideal gas equation of state
%3D
%3D
T V
P: = P = (0.1)|
= (0.1)
898.3
(18) 5.39 MPa
%3D
%3D
300
The pressure at state 2 can be evaluated altematively using the isentropic relationship, p2 P1(P2/P1).
Since Process 2-3 occurs at constant pressure, the ideal gas equation of state gives
%3D
T3 =T
V2
Introducing the cutoff ratio, r, = V3/V2
T3 = rT, = 2(898.3) = 1796.6 K
%3D
%3D
%3D
From Table A-22, h = 1999.1 kJ/kg and va = 3.97.
For the isentropic expansion process 3–4
%3D
VA
U4 =
V V2
V2 V3
%3D
%3D
Transcribed Image Text:Analysis: (a) The analysis begins by determining properties at each principal state of the cycle. With T = 300 K, Table A-22 gives u, = 214.07 kJ/kg and v, = 621.2. For the isentropic compression process 1-2 %3D V2 U2 = 621.2 %3D 34.51 %3D %3D 18 Interpolating in Table A-22, we get T = 898.3 K and h, = 930.98 kJ/kg. With the ideal gas equation of state %3D %3D T V P: = P = (0.1)| = (0.1) 898.3 (18) 5.39 MPa %3D %3D 300 The pressure at state 2 can be evaluated altematively using the isentropic relationship, p2 P1(P2/P1). Since Process 2-3 occurs at constant pressure, the ideal gas equation of state gives %3D T3 =T V2 Introducing the cutoff ratio, r, = V3/V2 T3 = rT, = 2(898.3) = 1796.6 K %3D %3D %3D From Table A-22, h = 1999.1 kJ/kg and va = 3.97. For the isentropic expansion process 3–4 %3D VA U4 = V V2 V2 V3 %3D %3D
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