We give: AB: adiabatic trigger such as VB = 2 VA AC: isothermal compression PA =1bar=105 Pa , VA = 10^-2 m3 Gas temperature at point A: TA= 300 Kelvin Perfect gas constant: R = 8,31 J.mol-1.K-1 Molar thermal capacity at constant pressure:Cp = 29 J.mol-1.K-1 Calorific capacity ratio of this gas: Cp/Cv = 1,40 4 – Calculate the amount of QBC heat and WBC work received by the system for the transformation that changes the system from state B to state C. 5 – a – Why is the UCA internal energy variation of the gas zero during the CA transformation? b – Apply the first principle to the ABCA cycle and deduce the WAB work received by the gas during the adiabatic relaxation AB. c – Interpret the WAB sign.
We give: AB: adiabatic trigger such as VB = 2 VA AC: isothermal compression PA =1bar=105 Pa , VA = 10^-2 m3 Gas temperature at point A: TA= 300 Kelvin Perfect gas constant: R = 8,31 J.mol-1.K-1 Molar thermal capacity at constant pressure:Cp = 29 J.mol-1.K-1 Calorific capacity ratio of this gas: Cp/Cv = 1,40 4 – Calculate the amount of QBC heat and WBC work received by the system for the transformation that changes the system from state B to state C. 5 – a – Why is the UCA internal energy variation of the gas zero during the CA transformation? b – Apply the first principle to the ABCA cycle and deduce the WAB work received by the gas during the adiabatic relaxation AB. c – Interpret the WAB sign.
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
We give:
AB: adiabatic trigger such as VB = 2 VA AC: isothermal compression
PA =1bar=105 Pa , VA = 10^-2 m3
Gas temperature at point A: TA= 300 Kelvin
Perfect gas constant: R = 8,31 J.mol-1.K-1
Molar thermal capacity at constant pressure:Cp = 29 J.mol-1.K-1
Calorific capacity ratio of this gas: Cp/Cv = 1,40
4 – Calculate the amount of QBC heat and WBC work received by the system for the transformation that changes the system from state B to state C.
5 – a – Why is the UCA internal energy variation of the gas zero during the CA transformation? b – Apply the first principle to the ABCA cycle and deduce the WAB work received by the gas during the
adiabatic relaxation AB.
c – Interpret the WAB sign.
Transcribed Image Text:PA
A
Pc
PB
B
VA
VB
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