The figure below is the basic plot of pressure vs. volume for the Dual Standard Cycle. Constant pressure heat addition 3 4 PV = constant 2 5 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) Heat IN Adiabatic expansion

The figure below is the basic plot of pressure vs. volume for the Dual Standard Cycle. Constant pressure heat addition 3 4 PV = constant 2 5 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) Heat IN Adiabatic expansion

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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Fluid Dynamics

The study of the flow of gases and liquids, usually in and around solid surfaces, is known as fluid dynamics in physics and engineering. Fluid dynamics, for example, can be used to evaluate the movement of air through an airplane wing or the surface of a vehicle. It can also be used in ship design to increase the speed at which ships pass through water.

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)

g)
Calculate the pressure at point 5.
P5
bar
h)
Calculate the cycle efficiency as a percentage.

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