INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
8th Edition
ISBN: 9781260940961
Author: SMITH
Publisher: INTER MCG
Question
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Chapter 3, Problem 3.28P

(a)

Interpretation Introduction

Interpretation:

To find the values of Q, W, ΔU and ΔH for each step in a cyclic process occur in closed system. The cyclic process consists a ideal gas initially at 30oC and 100KPa compressed adiabatically to pressure 500KPa then cooled at constant pressure of 500KPa to temperature 30oC and finally expanded to its original state.

Concept Introduction:

In order to find all the properties of cyclic process we must find out properties of individual steps.

For adiabatic compression from point 1 to point 2, the following formulas are applicable:

  Q12=0 since there is no heat transfer in adiabatic process.

  T2=T1(P2P1)γ1γ

  ΔU12=CV(T2T1)

  ΔH12=CP(T2T1)

  W12=ΔU12

For cooling at constant pressure from point 2 to point 3, the following formulas are applicable:

  ΔU23=CV(T3T2)

  W23=ΔU23Q23

  ΔH23=Q23

  ΔH23=CP(T3T2)

For isothermal expansion from point 3 to point 1, the following formulas are applicable:

  ΔU31=ΔH31=0

  W31=RT3ln(P1P3)

  Q31=3W1

(b)

Interpretation Introduction

Interpretation:

To find the values of Q, W, ΔU and ΔH for each step in a cyclic irreversible process with an efficiency 80% occur in closed system. The cyclic process consists a ideal gas initially at 30oC and 100KPa compressed to pressure 500KPa then cooled at constant pressure of 500KPa to temperature 30oC and finally expanded to its original state.

Concept Introduction:

In order to find all the properties of cyclic process we know that each step has 80% efficiency. So, all the properties will be same except Q and W .

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Chapter 3 Solutions

INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<

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