The figure below is the basic plot of pressure vs. volume for the Dual Standard Cycle. Constant pressure heat addition 3 4 PV = constant Heat IN Adiabatic expansion Heat OUT (Constant volume heat rejection) Constant volume heat addition 1 Adiabatic compression Volume (V) Using the Dual Air Standard Cycle and the following data: • Assume constant properties of air. R= 287 J/kgK; Cp = 1005 J/kgK; Cv = 718 J/kgK; y = 1.4 • Cycle compression ratio = 14 (v1/v2) • Air conditions prior to compression: Pressure = 1.3 bar; Temperature = 293K • Heat added at constant volume = 280000 J/kg • Heat added at constant pressure = 662000 J/kg Pressure (P) 2.

The figure below is the basic plot of pressure vs. volume for the Dual Standard Cycle. Constant pressure heat addition 3 4 PV = constant Heat IN Adiabatic expansion Heat OUT (Constant volume heat rejection) Constant volume heat addition 1 Adiabatic compression Volume (V) Using the Dual Air Standard Cycle and the following data: • Assume constant properties of air. R= 287 J/kgK; Cp = 1005 J/kgK; Cv = 718 J/kgK; y = 1.4 • Cycle compression ratio = 14 (v1/v2) • Air conditions prior to compression: Pressure = 1.3 bar; Temperature = 293K • Heat added at constant volume = 280000 J/kg • Heat added at constant pressure = 662000 J/kg Pressure (P) 2.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

The figure below is the basic plot of pressure vs. volume for the Dual Standard Cycle.
Constant pressure
heat addition
3
4
PV = constant
%3D
Heat
IN
Adiabatic
expansion
2
Heat OUT
(Constant volume
heat rejection)
Constant
volume heat
addition
1
Adiabatic
compression
Volume (V)
Using the Dual Air Standard Cycle and the following data:
• Assume constant properties of air. R= 287 J/kgK; Cp = 1005 J/kgK; Cv = 718 J/kgK; y = 1.4
• Cycle compression ratio = 14 (V1/v2)
• Air conditions prior to compression: Pressure = 1.3 bar; Temperature = 293K
• Heat added at constant volume = 280000 J/kg
• Heat added at constant pressure = 662000 J/kg
Pressure (P)

d)
Calculate the maximum cycle pressure.
Pmax
bar
||
e)
Calculate the maximum cycle temperature.
T
K
тах
f)
Calulate the temperature at point 5.
T5 =
K

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