EBK INTRODUCTION TO CHEMICAL ENGINEERIN
EBK INTRODUCTION TO CHEMICAL ENGINEERIN
8th Edition
ISBN: 9781259878091
Author: SMITH
Publisher: MCGRAW HILL BOOK COMPANY
Question
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Chapter 3, Problem 3.21P

(a)

Interpretation Introduction

Interpretation:

For a process in which an ideal gas with CP=52R, CV=32R is changed from P1=1bar, V1=12m3 to P2=12bar, V2=1m3 by mechanical reversible isothermal compression process, calculate values of Q, W, ΔU and ΔH .

Concept Introduction:

For isothermalcompressionprocess,

PV=constant

And

W=nRT1ln(V2V1)

ΔU=0

Q=W=nRT1ln(V2V1)

ΔH=nCP(T2T1)=0

(a)

Expert Solution
Check Mark

Answer to Problem 3.21P

The values of values of Q, W, ΔU and ΔH for a process are:

ΔU=0

ΔH=0

Q=2982kJ

W=-2982kJ

Explanation of Solution

Given information:

It is given that CP=52R, CV=32R is changed from P1=PA=1bar, V1=VA=12m3 to P2=12bar, V2=1m3 and γ=CPCV=52R32R=53

For isothermal compression process,

W=nRT1ln(V2V1)=P1V1ln(P1P2)W=1bar×12m3×ln(112)W=29.82bar.m3×101.325kPa1.01325bar=2982kJ

Q=W=-2982kJ

“-” sign shows that heat is releasing, and volume is decreased on applying of work done on it.

(b)

Interpretation Introduction

Interpretation:

For a process in which an ideal gas with CP=52R, CV=32R is changed from P1=1bar, V1=12m3 to P2=12bar, V2=1m3 by mechanical reversible adiabatic compression followed by cooling at constant pressure process, calculate values of Q, W, ΔU and ΔH .

Concept Introduction:

The total amount of Q, W, ΔU and ΔH would be calculated using individual value of both process adiabatic compression followed by cooling at constant pressure. So, for these types of problem we usually calculate value for each process and then add them to find total value of path.

For step 1B adiabatic compressionprocess,

PVγ=constant

And

W=P1V1PBVBγ1

ΔU=nCV(T2T1)=PBVBP1V1

Q=0

ΔH=ΔU+(PBVBP1V1)

For step B2 constant pressureprocess,

VBTB=V2T2

And

W=P(V2VB)

ΔU=nCV(T2TB)=P2V2PBVB

Q=nCP(T2TB)=P2V2PBVB

ΔH=ΔU+(P2V2PBVB)

(b)

Expert Solution
Check Mark

Answer to Problem 3.21P

The values of values of Q, W, ΔU and ΔH for a process are:

ΔU=0

ΔH=0

W12=5105kJ

W=-5105kJ

Explanation of Solution

Given information:

It is given that CP=52R, CV=32R and P1=1bar, V1=12m3 to P2=PB=12bar, V2=1m3, and γ=CPCV=52R32R=53 .

Since initial and final temperature is same in the process because P1V1=P2V21×12=12×1 this implies T1=T2

So ΔH12=0 and ΔU12=0 and Q=W

T1=T2

So ΔH12=0 and ΔU12=0 and Q=W

So, we only need to do calculations for Q and W .

For step 1B adiabatic compressionprocess,

P1V1γ=PBVBγ

VB=V1(P1PB)1γ

VB=12(112)35VB=2.702m3

W1B=P1V1PBVBγ1=1×1212×2.702531W1B=30.636bar.m3×101.325kPa1.01325bar=3063kJ

Q1B=3063

For step B2 constant pressure process,

WB2=P(V2VB)=12×(12.702)WB2=20.424bar.m3×101.325kPa1.01325bar=2042kJ

QB2=2042kJ

For whole process,

Q12=Q1B+QB2Q12=30632042Q12=5105kJ

W12=W1B+WB2W12=3063+2042W12=5105kJ

(c)

Interpretation Introduction

Interpretation:

For a process in which an ideal gas with CP=52R, CV=32R is changed from P1=1bar, V1=12m3 to P2=12bar, V2=1m3 by mechanical reversible adiabatic compression followed by cooling at constant volume process, calculate values of Q, W, ΔU and ΔH .

Concept Introduction:

The total amount of Q, W, ΔU and ΔH would be calculated using individual value of both process adiabatic compression followed by cooling at constant volume. So, for these types of problem we usually calculate value for each process and then add them to find total value of path.

For step 1C adiabatic compressionprocess,

PVγ=constant

And

W=P1V1PCVCγ1

ΔU=nCV(T2T1)=PCVCP1V1

Q=0

ΔH=ΔU+(PCVCP1V1)

For step C2 constant volume process,

W=0

ΔU=nCV(T2TB)=P2V2PCVC

Q=nCV(T2TC)=P2V2PCVC

ΔH=ΔU+(P2V2PCVC)

(c)

Expert Solution
Check Mark

Answer to Problem 3.21P

The values of values of Q, W, ΔU and ΔH for a process are:

ΔU=0

ΔH=0

W12=7635kJ

Q12=-7635kJ

Explanation of Solution

Given information:

It is given that CP=52R, CV=32R and P1=1bar, V1=12m3 to P2=12bar, V2=VC=1m3, and γ=CPCV=52R32R=53 .

Since initial and final temperature is same in the process because P1V1=P2V21×12=12×1 this implies T1=T2

So ΔH12=0 and ΔU12=0 and Q=W

T1=T2

So ΔH12=0 and ΔU12=0 and Q=W

So, we only need to do calculations for Q and W .

For step 1Cadiabatic compressionprocess,

P1V1γ=PCVCγ

PC=P1(V1VC)γ

PC=1×(121)53PC=62.898bar

W=P1V1PCVCγ1=1×121×62.898531W1C=76.347bar.m3×101.325kPa1.01325bar=7635kJ

Q1C=W=7635kJ

For step C2 constant volume process,

WC2=0QC2=0

For whole process,

Q12=Q1C+QC2Q12=7635kJ

W12=W1C+WC2W12=7635+0W12=7635kJ

(d)

Interpretation Introduction

Interpretation:

For a process in which an ideal gas with CP=52R, CV=32R is changed from P1=1bar, V1=12m3 to P2=12bar, V2=1m3 by mechanical reversible heating at constant volume followed by cooling at constant pressure, calculate values of Q, W, ΔU and ΔH .

Concept Introduction:

The total amount of Q, W, ΔU and ΔH would be calculated using individual value of both process. So, for these types of problem we usually calculate value for each process and then add them to find total value of path.

For step 1Dheating at constant volumeprocess,

W=0

Q=PDVDP1V1

For step D2 cooling at constant pressureprocess,

W=P(V2VD)

(d)

Expert Solution
Check Mark

Answer to Problem 3.21P

The values of values of Q, W, ΔU and ΔH for a process are:

ΔU=0

ΔH=0

W12=13200kJ

Q12=-13200kJ

Explanation of Solution

Given information:

It is given that CP=52R, CV=32R and P1=1bar, V1=VD=12m3 to P2=PD=12bar, V2=1m3, and γ=CPCV=52R32R=53 .

Since initial and final temperature is same in the process because P1V1=P2V21×12=12×1 this implies T1=T2

So ΔH12=0 and ΔU12=0 and Q=W

T1=T2

So ΔH12=0 and ΔU12=0 and Q=W

So, we only need to do calculations for Q and W .

For step 1D heating at constant volumeprocess

W1D=0Q1D=0

For step D2 cooling at constant pressure process,

W=P(V2VD)

WD2=12(VDV2)=12×(121)WD2=132bar.m3×101.325kPa1.01325bar=13200kJ

QD2=WD2=13200kJ

For whole process,

Q12=Q1D+QD2Q12=013200Q12=13200kJ

W12=W1D+WD2W12=0+13200W12=13200kJ

(e)

Interpretation Introduction

Interpretation:

For a process in which an ideal gas with CP=52R, CV=32R is changed from P1=1bar, V1=12m3 to P2=12bar, V2=1m3 by mechanical reversible cooling at constant pressure followed by heating at constant volume, calculate values of Q, W, ΔU and ΔH .

Concept Introduction:

The total amount of Q, W, ΔU and ΔH would be calculated using individual value of both process. So, for these types of problem we usually calculate value for each process and then add them to find total value of path.

For step 1Ecooling at constant pressureprocess, volume decreases, heat will be released and negative and work done is positive.

W=P(VEV1)

For step E2 heating at constant volumeprocess, pressure increases

W=0

(e)

Expert Solution
Check Mark

Answer to Problem 3.21P

The values of values of Q, W, ΔU and ΔH for a process are:

ΔU=0

ΔH=0

W12=1100kJ

Q12=-1100kJ

Explanation of Solution

Given information It is given that CP=52R, CV=32R and P1=PE=1bar, V1=12m3 to P2=12bar, V2=VE=1m3, and γ=CPCV=52R32R=53 .

Since initial and final temperature is same in the process because P1V1=P2V21×12=12×1 this implies T1=T2

So ΔH12=0 and ΔU12=0 and Q=W

T1=T2

So ΔH12=0 and ΔU12=0 and Q=W

So, we only need to do calculations for Q and W .

For step 1E cooling at constant pressureprocess,

W=P(VEV1)

W1E=P(VEV1)=1×(112)W1E=11bar.m3×101.325kPa1.01325bar=1100kJ

Q1E=W=1100kJ

For step E2 heating at constant volume process,

P1=PE=1bar and V2=VE=1m3

WE2=0QE2=0

heating at constant volume causes increase in pressure inside vessel

For whole process,

Q12=Q1E+QE2Q12=11000Q12=1100kJ

W12=W1E+WE2W12=1100+0W12=1100kJ

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Chapter 3 Solutions

EBK INTRODUCTION TO CHEMICAL ENGINEERIN

Ch. 3 - Prob. 3.11PCh. 3 - Prob. 3.12PCh. 3 - Prob. 3.13PCh. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Prob. 3.20PCh. 3 - Prob. 3.21PCh. 3 - Prob. 3.22PCh. 3 - Prob. 3.23PCh. 3 - Prob. 3.24PCh. 3 - Prob. 3.25PCh. 3 - Prob. 3.26PCh. 3 - Prob. 3.27PCh. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - Prob. 3.30PCh. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - Prob. 3.33PCh. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - Prob. 3.38PCh. 3 - Prob. 3.39PCh. 3 - Prob. 3.40PCh. 3 - Prob. 3.41PCh. 3 - Prob. 3.42PCh. 3 - Prob. 3.43PCh. 3 - Prob. 3.44PCh. 3 - Prob. 3.45PCh. 3 - Prob. 3.46PCh. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Prob. 3.66PCh. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Prob. 3.70PCh. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Prob. 3.74PCh. 3 - Prob. 3.75PCh. 3 - Prob. 3.76PCh. 3 - Prob. 3.77PCh. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Prob. 3.86PCh. 3 - Prob. 3.87PCh. 3 - Prob. 3.88PCh. 3 - Prob. 3.89PCh. 3 - Prob. 3.90PCh. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - Prob. 3.95P
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