The figure below is the basic plot of pressure vs. volume for the Dual Standard Cycle.Constant pressureheat addition34PV = constant%3DHeatINAdiabaticexpansion2Heat OUT(Constant volumeheat rejection)Constantvolume heataddition1AdiabaticcompressionVolume (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/kgPressure (P) a)Calculate the pressure at the end of the compression process.P2 =barb)Calculate the temperature at the end of the compression process.T2 =Kc)Calculate the temperature at the end of the constant volume heat addition process.T3 =K

(a) To find: The pressure at at the end of compression process. Given: The gas constant is R=287…

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

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

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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IC engines

The term "engine" indicates a mechanical device/machine used to convert input energy to a specific mechanical output. The input energy may be in the form of chemical energy, thermal energy, or so on and mechanical output is in the form of mechanical energy (work done). Generally, the engines classify into two groups which are internal combustion type and external combustion type, and further, these two types can be classified into two types that are reciprocating and rotary engines.

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)

a)
Calculate the pressure at the end of the compression process.
P2 =
bar
b)
Calculate the temperature at the end of the compression process.
T2 =
K
c)
Calculate the temperature at the end of the constant volume heat addition process.
T3 =
K

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