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 3.85P

(a)

Interpretation Introduction

Interpretation:

The value of ψ,ΩandZC for Redlich/Kwong equation of state should be calculated.

Concept Introduction :

To find the values of ψ,ΩandZC, first write the general cubic equation of state at critical conditions PC,VC,TC and write it in form of (VVC)3=V3+3VVC23V2VCVC3 since at critical conditions (VVC)=0 and so (VVC)3=0 and compare both equations to find three general derived equation in the terms of critical temperature, pressure and volume. Three equations and we will have five unknowns PC,VC,TC,a(TC)andb . Replace them with ψ,ΩandZC and find value of ψ,ΩandZC using three equations. The general cubic equation of state is:

  P=RTVba(T)(V+εb)(V+σb)......(1)

For critical conditions,

  PVT,cr=2PV2T,cr=0 and P=PC,V=VCand T=TC

For Redlich/Kwong equation of state,

  ε=0,σ=1

(a)

Expert Solution
Check Mark

Answer to Problem 3.85P

  ψ=0.42748,Ω=0.08664andZC=1/3

Explanation of Solution

Given information:

The general expressions for ψ,ΩandZC are:

  ψ=aPCR2TC2

  Ω=bPCRTC

And ZC=PCVCRTC

The general cubic equation of state may itself be written for the critical conditions, from equation (1) at critical conditions,

  PC=RTCVCba(TC)(VC+εb)(VC+σb)

Now replace five unknowns PC,VC,TC,a(TC)andb . Replace them with ψ,ΩandZC using ψ=aPCR2TC2, Ω=bPCRTC and ZC=PCVCRTC,

  ZC=PCVCRTCRTCPC=VCZC

  Ω=bPCRTCb=RTCΩPCb=VCΩZCb=VCΩZC

  ψ=aPCR2TC2a=ψR2TC2PC

  PC=RTCVCba(TC)(VC+εb)(VC+σb)

  PC=RTCVCba(TC)[VC2+VCσb+VCεb+εσb2]

Taking L.C.M

   P C = R T C [ V C 2 + V C σb+ V C εb+εσ b 2 ]a( T C )( V C b ) ( V C b )[ V C 2 + V C σb+ V C εb+εσ b 2 ]

   P C ( V C b )[ V C 2 + V C σb+ V C εb+εσ b 2 ]= R T C [ V C 2 + V C σb+ V C εb+εσ b 2 ]a( T C )( V C b )

   P C [ V C 3 + V C 2 σb+ V C 2 εb+εσ b 2 V C b V C 2 V C σ b 2 V C ε b 2 εσ b 3 ]= [R T C V C 2 +R T C V C σb+R T C V C εb+R T C εσ b 2 ]( a( T C ) V C a( T C )b )

   P C V C 3 + P C V C 2 σb+ P C V C 2 εb+ P C εσ b 2 V C P C b V C 2 P C V C σ b 2 P C V C ε b 2 P C εσ b 3 = [R T C V C 2 +R T C V C σb+R T C V C εb+R T C εσ b 2 ]a( T C ) V C +a( T C )b

Divide the whole equation by PC

  VC3+VC2σb+VC2εb+εσb2VCbVC2VCσb2VCεb2εσb3=[RTCVC2PC+RTCVCσbPC+RTCVCεbPC+RTCεσb2PC]a(TC)VCPC+a(TC)bPC

Rearrange,

  VC3+(σb+εbbR T C P C)VC2+(εσb2σb2εb2R T Cσb P CR T Cεb P C+a( T C) P C)VC(εσb3+R T Cεσ b 2 P C+a( T C)b P C)=0

In terms of non-critical condition

  V3+(σb+εbbRTP)V2+(εσb2σb2εb2RTσbPRTεbP+a(T)P)V(εσb3+RTεσ b 2P+a(T)bP)=0

The above equation is similar as (VVC)3=V3+3VVC23V2VCVC3, Now compare both equations and at critical conditions

  σb+εbbRTCPC=3VC

  εσb2σb2εb2RTCσbPCRTCεbPC+a(TC)PC=3VC2

And

  (εσb3+RTCεσb2PC+a(TC)bPC)=VC3

Now, the three general derived equations are the term of critical temperature, pressure and volume. Three equations and we will have five unknowns PC,VC,TC,a(TC)andb . Replace them with ψ,ΩandZC .

From equation σb+εbbRTCPC=3VC

Divide by b

  σ+ε1RTCbPC=3VCb

Since, Ω=bPCRTC and b=VCΩZC

Gives

  σ+ε11Ω=3ZCΩOrσε+1+1Ω=3ZCΩ1+ΩσΩεΩΩ=3ZCΩOr,

  1+(1σε)Ω=3ZC ......(2)

From equation εσb2σb2εb2RTCσbPCRTCεbPC+a(TC)PC=3VC2

Divide by b2

  εσσεRTCσbPCb2RTCεbPCb2+a(TC)PCb2=3VC2b2OrεσσεRTCσPCbRTCεPCb+a(TC)PCb2=3VC2b2

Since, Ω=bPCRTC and b=VCΩZC

  a=ψR2TC2PC

Gives

  εσσεσΩεΩ+ψR2TC2PCPCR2TC2Ω2PC2=3VC2b2

Or,

  εσσεσΩεΩ+ψΩ2=3ZC2Ω2

Or,

  εσΩ2σΩ2εΩ2σΩεΩ+ψ=3ZC2

Or

  εσΩ2Ω(ε+σ)(Ω+1)+ψ=3ZC2......(3)

From equation (εσb3+RTCεσb2PC+a(TC)bPC)=VC3

Divide by b3

  (εσ+RTCεσbPC+a(TC)b2PC)=VC3b3

Since, Ω=bPCRTCb=ΩRTCPC and b=VCΩZC

  a=ψR2TC2PC

Gives

  (εσ+R T Cεσ ΩR T C P C P C+ ψ R 2 T C 2 P C Ω2 R2 TC PC 2 2 P C)=ZC3Ω3Or,(εσ+εσΩ+ψ Ω 2)=ZC3Ω3

Or,

  εσΩ3+εσΩ2+Ωψ=ZC3Or,

Or

  εσΩ2Ω(ε+σ)(Ω+1)+ψ=3ZC2

Or,

  εσΩ2(Ω+1)+Ωψ=ZC3......(4)

For Redlich/Kwong equation of state,

  ε=0,σ=1

Put values in equation (2) (3) and (4),

  1+(110)Ω=3ZCZC=13

Put values in equation (3),

  εσΩ2Ω(ε+σ)(Ω+1)+ψ=3ZC2

  0×1×Ω2Ω×(0+1)(Ω+1)+ψ=3×132Ω(Ω+1)+ψ=13

Put values in equation (4),

  εσΩ2(Ω+1)+Ωψ=ZC30×1×Ω2×(Ω+1)+Ωψ=133Or,ψ=127Ω

From equation Ω(Ω+1)+ψ=13, Put ψ=127Ω

  Ω(Ω+1)+127Ω=13Ω2Ω+127Ω=1327Ω327Ω2+1=9Ω27Ω327Ω29Ω+1=0

After solving cubic equation, roots of Ω are

  Ω=0.08664,0.5433±0.364i

Since value of Ω cannot be complex number so, value of Ω is:

  Ω=0.08664

And ψ=127Ω

  ψ=127Ω=127×0.08664=0.42748So,ψ=0.42748

Hence proved.

(b)

Interpretation Introduction

Interpretation:

The value of ψ,ΩandZC for Soave/Redlich/Kwong equation of state should be calculated.

Concept Introduction :

To find the values of ψ,ΩandZC, first write the general cubic equation of state at critical conditions PC,VC,TC and write it in form of (VVC)3=V3+3VVC23V2VCVC3 since at critical conditions (VVC)=0 and so (VVC)3=0 and compare both equations to find three general derived equation in the terms of critical temperature, pressure and volume. Three equations and we will have five unknowns PC,VC,TC,a(TC)andb . Replace them with ψ,ΩandZC and find value of ψ,ΩandZC using three equations. The general cubic equation of state is:

  P=RTVba(T)(V+εb)(V+σb)......(1)

For critical conditions,

  PVT,cr=2PV2T,cr=0 and P=PC,V=VCand T=TC

For Soave/Redlich/Kwong equation of state,

  ε=0,σ=1

(b)

Expert Solution
Check Mark

Answer to Problem 3.85P

  ψ=0.42748,Ω=0.08664andZC=1/3

Explanation of Solution

Given information:

The general expressions for ψ,ΩandZC are:

  ψ=aPCR2TC2

  Ω=bPCRTC

And ZC=PCVCRTC

For Soave/Redlich/Kwong equation of state,

  ε=0,σ=1

Put values in general equation (2) (3) and (4) found in subpart (1),

  1+(110)Ω=3ZCZC=13

Put values in equation (3),

  εσΩ2Ω(ε+σ)(Ω+1)+ψ=3ZC2

  0×1×Ω2Ω×(0+1)(Ω+1)+ψ=3×132Ω(Ω+1)+ψ=13

Put values in equation (4),

  εσΩ2(Ω+1)+Ωψ=ZC30×1×Ω2×(Ω+1)+Ωψ=133Or,ψ=127Ω

From equation Ω(Ω+1)+ψ=13, Put ψ=127Ω

  Ω(Ω+1)+127Ω=13Ω2Ω+127Ω=1327Ω327Ω2+1=9Ω27Ω327Ω29Ω+1=0

After solving cubic equation, roots of Ω are

  Ω=0.08664,0.5433±0.364i

Since value of Ω cannot be complex number so, value of Ω is:

  Ω=0.08664

And ψ=127Ω

  ψ=127Ω=127×0.08664=0.42748So,ψ=0.42748

Hence proved.

(c)

Interpretation Introduction

Interpretation:

The value of ψ,ΩandZC for Peng/Robinson equation of state should be calculated.

Concept Introduction :

To find the values of ψ,ΩandZC, first write the general cubic equation of state at critical conditions PC,VC,TC and write it in form of (VVC)3=V3+3VVC23V2VCVC3 since at critical conditions (VVC)=0 and so (VVC)3=0 and compare both equations to find three general derived equation in the terms of critical temperature, pressure and volume. Three equations and we will have five unknowns PC,VC,TC,a(TC)andb . Replace them with ψ,ΩandZC and find value of ψ,ΩandZC using three equations. The general cubic equation of state is:

  P=RTVba(T)(V+εb)(V+σb)......(1)

For critical conditions,

  PVT,cr=2PV2T,cr=0 and P=PC,V=VCand T=TC

For Peng/Robinson equation of state,

  ε=12,σ=1+2

(c)

Expert Solution
Check Mark

Answer to Problem 3.85P

  ψ=0.45724,Ω=0.07779andZC=0.30740

Explanation of Solution

Given information:

The general expressions for ψ,ΩandZC are:

  ψ=aPCR2TC2

  Ω=bPCRTC

And ZC=PCVCRTC

For Peng/Robinson equation of state,

  ε=12,σ=1+2

Put values in general equation (2) (3) and (4) found in subpart (1),

  1+(1σε)Ω=3ZC

  1+(1121+2)Ω=3ZC1Ω=3ZC

Put values in equation (3),

  εσΩ2Ω(ε+σ)(Ω+1)+ψ=3ZC2

  (12)×(1+2)×Ω2Ω×(12+1+2)(Ω+1)+ψ=3×ZC2

Put 1Ω=3ZC in above equation,

  (12)×(1+2)×Ω2Ω×(12+1+2)(Ω+1)+ψ=3×ZC2Ω22Ω(Ω+1)+ψ=3×( 1Ω3)2Ω22Ω22Ω+ψ=( 1Ω)233Ω22Ω+ψ=( 1Ω)239Ω26Ω+3ψ=1+Ω22Ω

  10Ω24Ω+(3ψ1)=0......(4)

Put values in equation (4),

  εσΩ2(Ω+1)+Ωψ=ZC3(12)×(12)×Ω2×(Ω+1)+Ωψ=( 1Ω3)3Or,Ω2×(Ω+1)+Ωψ=( 1Ω3)327Ω327Ω2+27Ωψ=1Ω3+3Ω23Ω

  26Ω330Ω2+(27Ωψ+3Ω1)=0......(5)

From equation 4

  10Ω24Ω+(3ψ1)=0multiply by 330Ω212Ω+(9ψ3)=0

  30Ω212Ω+(9ψ3)=030Ω2+12Ω+3=9ψψ=30Ω2+12Ω+39

Put value in equation 5

  26Ω330Ω2+(27Ωψ+3Ω1)=026Ω330Ω2+{27Ω×( 30Ω2 +12Ω+39)+3Ω1}=026Ω330Ω2+{3Ω×(30Ω2+12Ω+3)+3Ω1}=026Ω330Ω2+{90Ω3+36Ω2+9Ω+3Ω1}=064Ω3+6Ω2+12Ω1=0

After solving cubic equation, roots of Ω are

  Ω=0.07779,0.08577±0.43988i

Since value of Ω cannot be complex number so, value of Ω is:

  Ω=0.07779

And ψ=30Ω2+12Ω+39

  ψ=30Ω2+12Ω+39=30×0.077792+12×0.07779+39So,ψ=0.45724

And

  1Ω=3ZC10.07779=3ZCZC=0.30740

Hence proved.

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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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