sm_3a

8/19/2019 SM_3A

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Work

V1

V2

VP⌠ ⌡

d−= a bit of algebra leads to

Work c

P1

P2

PP

P b+

⌠

⌡

d⋅:=Work 0.516

J

gm= Ans.

Alternatively, formal integration leads to

Work c P2 P1− b lnP2 b+

P1 b+

⋅−

⋅:= Work 0.516J

gm= Ans.

3.5 κ a b P⋅+= a 3.9 10 6−⋅ atm

1−⋅:= b 0.1− 10 9−⋅ atm

2−⋅:=

P1 1 atm⋅:= P2 3000 atm⋅:= V 1 ft3

⋅:= (assume const.)

Combine Eqs. (1.3) and (3.3) for const. T:

Work V

P1

P2

Pa b P⋅+( ) P⋅⌠ ⌡

d⋅:= Work 16.65 atm ft3⋅= Ans.

Chapter 3 - Section A - Mathcad Solutions

3.1 β 1−ρ T

ρd

d

⋅= κ 1

ρ Pρd

d

⋅=

P T

At constant T, the 2nd equation can be written:

dρρ

κ dP⋅= lnρ2

ρ1

κ ∆P⋅= κ 44.18 10

6−⋅ bar 1−⋅:= ρ2 1.01 ρ1⋅=

∆Pln 1.01( )

κ := ∆P 225.2bar = P2 226.2 bar ⋅= Ans.

3.4 b 2700 bar ⋅:= c 0.125cm

3

gm⋅:= P1 1 bar ⋅:= P2 500 bar ⋅:=

Since

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P2 1 bar ⋅:= T1 600 K ⋅:= CP7

2R ⋅:= CV

5

2R ⋅:=

(a) Constant V: W 0= and ∆U Q= CV ∆T⋅=

T2 T1

P2

P1⋅:= ∆T T2 T1−:= ∆T 525− K =

∆U CV ∆T⋅:= Q and ∆U 10.91− kJ

mol= Ans.

∆H CP ∆T⋅:= ∆H 15.28− kJ

mol= Ans.

(b) Constant T: ∆U ∆H= 0= and Q W=

Work R T1⋅ lnP2

P1

⋅:= Q and Work 10.37− kJ

mol= Ans.

(c) Adiabatic: Q 0= and ∆U W= CV ∆T⋅=

3.6 β 1.2 10 3−⋅ degC

1−⋅:= CP 0.84 kJ

kg degC⋅⋅:= M 5 kg⋅:=

V1

1

1590

m3

kg⋅:= P 1 bar ⋅:= t1 0 degC⋅:= t2 20 degC⋅:=With beta independent of T and with P=constant,

dV

Vβ dT⋅= V2 V1 exp β t2 t1−( )⋅⋅:= ∆V V2 V1−:=

∆Vtotal M ∆V⋅:= ∆Vtotal 7.638 10 5−× m

3= Ans.

Work P− ∆Vtotal⋅:= (Const. P) Work 7.638− joule= Ans.

Q M CP⋅ t2 t1−( )⋅:= Q 84 kJ= Ans.

∆Htotal Q:= ∆Htotal 84kJ= Ans.

∆Utotal Q Work +:= ∆Utotal 83.99kJ= Ans.

3.8P1 8 bar ⋅:=

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Step 41: Adiabatic T4 T1

P4

P1

R

CP

⋅:= T4 378.831K =

∆U41 CV T1 T4−( )⋅:= ∆U41 4.597 10

3

×

J

mol=∆H41 CP T1 T4−( )⋅:= ∆H41 6.436 10

3× J

mol=

Q41 0 J

mol:= Q41 0

J

mol=

W41 ∆U41:= W41 4.597 103× J

mol=

P2 3bar := T2 600K := V2

R T2⋅P2

:= V2 0.017 m3

mol=

Step 12: Isothermal ∆U12 0 J

mol:= ∆U12 0

J

mol=

γ CP

CV:= T2 T1

P2

P1

γ 1−γ

⋅:=T2 331.227K = ∆T T2 T1−:=

∆U CV ∆T⋅:= ∆H CP ∆T⋅:=

W and ∆U 5.586− kJ

mol= Ans. ∆H 7.821−

kJ

mol= Ans.

3.9 P4 2bar := CP7

2

R := CV5

2

R :=

P1 10bar := T1 600K := V1

R T1⋅

P1:= V1 4.988 10

3−× m

3

mol=

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T4 378.831K = V4

R T4⋅

P4

:= V4 0.016m

3

mol

=

Step 34: Isobaric ∆U34 CV T4 T3−( )⋅:= ∆U34 439.997− J

mol=

∆H34 CP T4 T3−( )⋅:= ∆H34 615.996− J

mol=

Q34 CP T4 T3−( )⋅:= Q34 615.996− J

mol=

W34 R − T4 T3−( )⋅:= W34 175.999J

mol=

3.10 For all parts of this problem: T2 T1= and

∆U ∆H= 0= Also Q Work −= and all that remains is

to calculate Work. Symbol V is used for total volume in this problem.

∆H12 0J

mol⋅:= ∆H12 0

J

mol=

Q12 R − T1⋅ ln

P2

P1

⋅:= Q12 6.006 10

3

× J

mol=

W12 Q12−:= W12 6.006− 103× J

mol=

P3 2bar := V3 V2:= T3

P3 V3⋅

R := T3 400K =

Step 23: Isochoric ∆U23 CV T3 T2−( )⋅:= ∆U23 4.157− 103× J

mol=

∆H23 CP T3 T2−( )⋅:= ∆H23 5.82− 103× J

mol=

Q23 CV T3 T2−( )⋅:= Q23 4.157− 103× J

mol=

W23 0J

mol:= W23 0

J

mol=

P4 2bar =

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Work 5106kJ= Ans.

(c) Step 1: adiabatic compression to V2

Pi P1

V1

V2

γ

⋅:= (intermediate P) Pi 62.898bar =

W1

Pi V2⋅ P1 V 1⋅−

γ 1−:= W1 7635kJ=

Step 2: No work. Work W1:= Work 7635kJ= Ans.

(d) Step 1: heat at const V1 to P2 W1 0=

Step 2: cool at const P2 to V2

W2 P2− V2 V1−( )⋅:= Work W2:= Work 13200kJ= Ans.

P1 1 bar ⋅:= P2 12 bar ⋅:= V1 12 m3⋅:= V2 1 m

3⋅:=

(a) Work n R ⋅ T⋅ ln

P2

P1

⋅= Work P1 V1⋅ ln

P2

P1

⋅:=

Work 2982kJ= Ans.

(b) Step 1: adiabatic compression to P2

γ 5

3:= Vi V1

P1

P2

1

γ

⋅:= (intermediate V) Vi 2.702m3=

W1

P2 Vi⋅ P1 V 1⋅−

γ 1−:= W1 3063kJ=

Step 2: cool at const P2 to V2

W2 P2− V2 Vi−( )⋅:= W2 2042kJ=

Work W1 W2+:=

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Work P2 V2⋅ lnV1

V2

⋅:= Work 878.9kJ= Ans.

3.18 (a) P1 100 kPa⋅:= P2 500 kPa⋅:= T1 303.15 K ⋅:=

CP7

2R ⋅:= CV

5

2R ⋅:= γ

CP

CV:=

Adiabatic compression from point 1 to point 2:

Q12 0 kJ

mol⋅:= ∆U12 W12= CV ∆T12⋅= T2 T1

P2

P1

γ 1−γ

⋅:=

∆U12 CV T2 T1−( )⋅:= ∆H12 CP T2 T1−( )⋅:= W12 ∆U12:=

∆U12 3.679 kJ

mol= ∆H12 5.15

kJ

mol= W12 3.679

kJ

mol= Ans.

(e) Step 1: cool at const P1 to V2

W1 P1− V2 V1−( )⋅:= W1 1100kJ=

Step 2: heat at const V2 to P2 W2 0=

Ans.Work W1:= Work 1100kJ=

3.17 (a) No work is done; no heat is transferred.

∆Ut ∆T= 0= T2 T1= 100 degC⋅= Not reversible

(b) The gas is returned to its initial state by isothermal compression.

Work n R ⋅ T⋅ lnV1

V2

⋅= but n R ⋅ T⋅ P2 V2⋅=

V1 4 m3⋅:= V2

4

3m

3⋅:= P2 6 bar ⋅:=

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Q31 4.056 kJ

mol= Ans.

FOR THE CYCLE: ∆U ∆H= 0=

Q Q12 Q23+ Q31+:= Work W12 W23+ W31+:=

Q 1.094− kJ

mol= Work 1.094

kJ

mol=

(b) If each step that is 80% efficient accomplishes the same change of state,

all property values are unchanged, and the delta H and delta U values are

the same as in part (a). However, the Q and W values change.

Step 12: W12

W12

0.8

:= W12 4.598 kJ

mol

=

Q12 ∆U12 W12−:= Q12 0.92− kJ

mol=

Step 23: W23

W23

0.8:= W23 1.839

kJ

mol=

Cool at P2 from point 2 to point 3:

T3 T1:= ∆H23 CP T3 T2−( )⋅:= Q23 ∆H23:=

∆U23 CV T3 T2−( )⋅:= W23 ∆U23 Q23−:=

∆H23 5.15− kJ

mol= ∆U23 3.679−

kJ

mol= Ans.

Q23 5.15− kJ

mol= W23 1.471

kJ

mol= Ans.

Isothermal expansion from point 3 to point 1:

∆U31 ∆H31= 0= P3 P2:= W31 R T3⋅ lnP

1P3 ⋅:=

Q31 W31−:=

W31 4.056− kJ

mol=

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(a) Isothermal: Work n R ⋅ T1⋅ ln V1V2

⋅= P2 P1 V1

V2⋅:=

T2 T1:= T2 600K = P2 200kPa= Ans.

Work P1 V1⋅ lnV1

V2

⋅:= Work 1609− kJ= Ans.

(b) Adiabatic: P2 P1

V1

V2

γ

⋅:= T2 T1

P2

P1

⋅ V2

V1

⋅:=

T2 208.96K = P2 69.65kPa= Ans.

Work P2 V2⋅ P1 V1⋅−

γ 1−:= Work 994.4− kJ= Ans,

Q23 ∆U23 W23−:= Q23 5.518− kJ

mol=

Step 31: W31 W31 0.8⋅:= W31 3.245− kJmol

=

Q31 W31−:=Q31 3.245

kJ

mol=

FOR THE CYCLE:

Q Q12 Q23+ Q31+:= Work W12 W23+ W31+:=

Q 3.192− kJ

mol

= Work 3.192kJ

mol

=

3.19 Here, V represents total volume.

P1 1000 kPa⋅:= V1 1 m3⋅:= V2 5 V1⋅:= T1 600 K ⋅:=

CP 21 joule

mol K ⋅⋅:= CV CP R −:= γ

CP

CV:=

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∆U12 0kJ

mol⋅:=

If r V1

V2

=V1

V3

= Then r T1

T3

P3

P1⋅:= W12 R T1⋅ ln r ( )⋅:=

W12 2.502− kJ

mol= Q12 W12−:= Q12 2.502

kJ

mol=

Step 23: W23 0kJ

mol⋅:= ∆U23 CV T3 T2−( )⋅:=

Q23 ∆U23:= ∆H23 CP T3 T2−( )⋅:=

Q23 2.079− kJ

mol= ∆U23 2.079−

kJ

mol= ∆H23 2.91−

kJ

mol=

Process: Work W12 W23+:= Work 2.502− kJ

mol= Ans.

Q Q12 Q23+:= Q 0.424kJ

mol= Ans.

(c) Restrained adiabatic: Work ∆U= Pext− ∆V⋅=

Pext 100 kPa⋅:= Work Pext− V2 V1−( )⋅:= Work 400− kJ= Ans.

nP1 V1⋅R T1⋅

:= ∆U n CV⋅ ∆T⋅=

T2Work

n CV⋅ T1+:= T2 442.71K = Ans.

P2 P1

V1

V2⋅

T2

T1⋅:= P2 147.57kPa= Ans.

3.20 T1 423.15 K ⋅:= P1 8 bar ⋅:= P3 3 bar ⋅:=

CP7

2R ⋅:= CV

5

2R ⋅:= T2 T1:= T3 323.15 K ⋅:=

Step 12: ∆H12 0kJ

mol⋅:=

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3.22 CP7

2R ⋅:= CV

5

2R ⋅:= T1 303.15 K ⋅:= T3 403.15 K ⋅:=

P1 1 bar ⋅:= P3 10 bar ⋅:=

∆U CV T3 T1−( )⋅:= ∆H CP T3 T1−( )⋅:=

∆U 2.079 kJ

mol= Ans. ∆H 2.91

kJ

mol= Ans.

Each part consists of two steps, 12 & 23.

(a) T2 T3:= P2 P1 T2T1

⋅:=

W23 R T2⋅ lnP3

P2

⋅:= Work W23:=

Work 6.762 kJ

mol= Ans.

∆H ∆H12 ∆H23+:= ∆H 2.91− kJ

mol= Ans.

∆U ∆U12 ∆U23+:= ∆U 2.079− kJ

mol= Ans.

3.21 By Eq. (2.32a), unit-mass basis: molwt 28 gm

mol:= ∆H

1

2∆u

2⋅+ 0=

But ∆H CP ∆T⋅= Whence ∆Tu2

2u1

2−( )−

2 CP⋅=

CP 72

R molwt

⋅:= u1 2.5 ms

⋅:= u2 50 ms

⋅:= t1 150 degC⋅:=

t2 t1

u22

u12−

2 CP⋅−:= t2 148.8degC= Ans.

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T2 T1:= P2 P3:= W12 R T1⋅ lnP2

P1

⋅:=

∆H23 CP T3 T2−( )⋅:= Q23 ∆H23:=

∆U23 CV T3 T2−( )⋅:= W23 ∆U23 Q23−:=

Work W12 W23+:= Work 4.972 kJ

mol= Ans.

Q ∆U Work −:= Q 2.894− kJ

mol= Ans.

For the second set of heat-capacity values, answers are (kJ/mol):

∆U 1.247= ∆U 2.079=

(a) Work 6.762= Q 5.515−=

Q ∆U Work −:=

Q 4.684− kJ

mol

= Ans.

(b) P2 P1:= T2 T3:= ∆U12 CV T2 T1−( )⋅:=

∆H12 CP T2 T1−( )⋅:= Q12 ∆H12:=

W12 ∆U12 Q12−:= W12 0.831− kJ

mol=

W23 R T2⋅ lnP3

P2

⋅:= W23 7.718

kJ

mol

=

Work W12 W23+:= Work 6.886 kJ

mol= Ans.

Q ∆U Work −:= Q 4.808− kJ

mol= Ans.

(c)

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Q12 W12−:= Q12 5.608− kJ

mol=

Step 23: W23 0kJ

mol⋅:= Q23 ∆U:=

For the process: Work W12 W23+:=

Q Q12 Q23+:= Work 5.608kJ

mol= Q 3.737−

kJ

mol= Ans.

3.24 W12 0= Work W23= P2 V3 V2−( )−= R − T3 T2−( )⋅=

But T3 T1= So... Work R T2 T1−( )⋅=

Also W R T1⋅ ln PP1

⋅= Therefore

lnP

P1

T2 T1−

T1

= T2 350 K ⋅:= T1 800 K ⋅:= P1 4 bar ⋅:=

(b) Work 6.886= Q 5.639−=

(c) Work 4.972= Q 3.725−=

3.23 T1 303.15 K ⋅:= T2 T1:= T3 393.15 K ⋅:=

P1 1 bar ⋅:= P3 12 bar ⋅:= CP7

2R ⋅:= CV

5

2R ⋅:=

For the process: ∆U CV T3 T1−( )⋅:= ∆H CP T3 T1−( )⋅:=

∆U 1.871kJ

mol= ∆H 2.619

kJ

mol= Ans.

Step 12: P2 P3

T1

T3⋅:= W12 R T1⋅ ln

P2

P1

⋅:=

W12 5.608kJ

mol=

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TA final( ) TA= TB final( ) TB=

nA nB= Since the total volume is constant,

2 nA⋅ R ⋅ T1⋅

P1

nA R ⋅ TA TB+( )⋅

P2

= or2 T1⋅

P1

TA TB+

P2

= (1)

(a) P2 1.25 atm⋅:= TB T1

P2

P1

γ 1−γ

⋅:= (2)

TA 2 T1⋅ P2

P1⋅ TB−:= Q nA ∆UA ∆UB+( )⋅=

Define qQ

nA

= q CV TA TB+ 2 T1⋅−( )⋅:= (3)

TB 319.75K = TA 430.25K = q 3.118kJ

mol= Ans.

P P1 expT2 T1−

T1

⋅:= P 2.279bar = Ans.

3.25 VA 256 cm3⋅:= Define:

∆P

P1r = r 0.0639−:=

Assume ideal gas; let V represent total volume:

P1 V B⋅ P2 VA VB+( )⋅= From this one finds:

∆P

P1

VA−

VA VB+

= VB

VA− r 1+( )⋅

r

:= VB 3750.3cm3= Ans.

3.26 T1 300 K ⋅:= P1 1 atm⋅:= CP7

2R ⋅:= CV CP R −:= γ

CP

CV:=

The process occurring in section B is a reversible, adiabatic compression. Let

P final( ) P2=

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TA 2 T1⋅ P2

P1

⋅ TB−:= (1) TA 469K = Ans.

q CV TA TB+ 2 T1⋅−( )⋅:= q 4.032 kJ

mol= Ans.

(d) Eliminate TA TB+ from Eqs. (1) & (3):

q 3 kJ

mol⋅:= P2

q P1⋅

2 T1⋅ CV⋅ P1+:= P2 1.241atm= Ans.

TB T1

P2

P1

γ 1−

γ ⋅:= (2) TB 319.06K = Ans.

TA 2 T1⋅ P2

P1⋅ TB−:= (1) TA 425.28K = Ans.

(b) Combine Eqs. (1) & (2) to eliminate the ratio of pressures:

TA 425 K ⋅:= (guess) TB 300 K ⋅:=

Given TB T1

TA TB+

2 T1⋅

γ 1−γ

⋅= TB Find TB( ):=

TB 319.02K = Ans.

P2 P1

TA TB+

2 T1⋅

⋅:= (1) P2 1.24atm= Ans.

q CV TA TB+ 2 T1⋅−( )⋅:= q 2.993

kJ

mol= Ans.

(c) TB 325 K ⋅:= By Eq. (2),

P2 P1

TB

T1

γ γ 1−

⋅:= P2 1.323atm= Ans.

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W R − T⋅

P1

P2

P1−

PC' P⋅+

⌠ ⌡

d⋅:=dV R T⋅ 1−

P2

C'+

⋅ dP⋅=

Eliminate dV from Eq. (1.3) by the virial equation in P:(b)

Ans.Work 12.62 kJ

mol=Work R − T⋅

V1

V2

V1 B

V+

C

V2

+

1

V⋅

⌠ ⌡

d⋅:=

Eliminate P from Eq. (1.3) by the virial equation:

V2 241.33 cm

3

mol=V2 Find V2( ):=

P2 V2⋅

R T⋅ 1

B

V2+

C

V22

+=Given

V2R T⋅P2

:=Guess:

Solve virial eqn. for final V.

V1 30780 cm

3

mol=

3.30 B 242.5− cm

3

mol⋅:= C 25200

cm6

mol2

⋅:= T 373.15 K ⋅:=

P1

1 bar

⋅:= P

2 55 bar

⋅:=

B' B

R T⋅:=

B' 7.817− 10 3−× 1

bar =

C' C B

2−

R 2

T2⋅

:=C' 3.492− 10

5−× 1

bar 2

=

(a) Solve virial eqn. for initial V.

Guess: V1 R T⋅P1:=

GivenP1 V1⋅

R T⋅ 1

B

V1+

C

V12

+= V1 Find V1( ):=

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Z 0.929= Ans.

(b) B0 0.0830.422

Tr 1.6

−:= B0 0.304−=

B1 0.1390.172

Tr 4.2

−:= B1 2.262 10 3−×=

Z 1 B0 ω B1⋅+( ) Pr

Tr ⋅+:= Z 0.932= V

Z R ⋅ T⋅P

:= V 1924cm

3

mol= Ans.

(c) For Redlich/Kwong EOS:

σ 1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1

α Tr ( ) Tr 0.5−:= Table 3.1 q Tr ( )

Ψ α Tr ( )⋅

Ω Tr ⋅:= Eq. (3.51)

W 12.596kJ

mol= Ans.

Note: The answers to (a) & (b) differ because the relations between the two

sets of parameters are exact only for infinite series.

3.32 Tc 282.3 K ⋅:= T 298.15 K ⋅:= Tr T

Tc:= Tr 1.056=

Pc 50.4 bar ⋅:= P 12 bar ⋅:= Pr P

Pc:= Pr 0.238=

ω 0.087:= (guess)

(a) B 140− cm3

mol⋅:= C 7200

cm6

mol2

⋅:= VR T⋅

P:= V 2066

cm3

mol=

Given P V⋅R T⋅

1B

V+

C

V2

+=

V Find V( ):= V 1919cm

3

mol= Z

P V⋅R T⋅

:=

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Eq. (3.51) β Tr Pr ,( ) Ω Pr ⋅

Tr := Eq. (3.50)

Calculate Z Guess: Z 0.9:=Given Eq. (3.49)

Z 1 β Tr Pr ,( )+ q Tr ( ) β Tr Pr ,( )⋅ Z β Tr Pr ,( )−

Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅⋅−=

Z Find Z( ):= Z 0.928= VZ R ⋅ T⋅

P:= V 1918

cm3

mol= Ans.

(e) For Peng/Robinson EOS:

σ 1 2+:= ε 1 2−:= Ω 0.07779:= Ψ 0.45724:= Table 3.1

Table 3.1α Tr ω,( ) 1 0.37464 1.54226ω+ 0.26992ω2

−( ) 1 Tr

1

2−

⋅+

2

:=

β Tr Pr ,( ) Ω Pr ⋅

Tr := Eq. (3.50)

Calculate Z Guess: Z 0.9:=Given Eq. (3.49)


Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅⋅−=

Z Find Z( ):= Z 0.928= VZ R ⋅ T⋅

P:= V 1916.5

cm3

mol= Ans.

(d) For SRK EOS:

σ 1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1

Table 3.1α Tr ω,( ) 1 0.480 1.574ω+ 0.176ω

2−( ) 1 Tr

1

2−

⋅+

2

:=

q Tr ( ) Ψ α Tr ω,( )⋅

Ω Tr ⋅:=

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Pr P

Pc:= Pr 0.308=

ω 0.100:=(guess)

(a) B 156.7− cm

3

mol⋅:= C 9650 cm

6

mol2⋅:= V

R T

⋅P:= V 1791 cm

3

mol=

Given P V⋅R T⋅

1 B

V+

C

V2

+=

V Find V( ):= V 1625 cm

3

mol= Z

P V⋅R T⋅

:= Z 0.907= Ans.

(b) B0 0.083 0.422

Tr 1.6

−:= B0 0.302−=

B1 0.139 0.172

Tr 4.2

−:= B1 3.517 10 3−×=

q Tr ( ) Ψ α Tr ω,( )⋅

Ω Tr ⋅:= Eq. (3.51) β Tr Pr ,( )

Ω Pr ⋅

Tr := Eq. (3.50)

Calculate Z Guess: Z 0.9:=

Given Eq. (3.49)


Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅⋅−=

Z Find Z( ):= Z 0.92= V Z R ⋅ T⋅

P:= V 1900.6

cm3

mol= Ans.

3.33 Tc 305.3 K ⋅:= T 323.15 K ⋅:= Tr T

Tc:= Tr 1.058=

Pc 48.72 bar ⋅:= P 15 bar ⋅:=

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Z Find Z( ):= Z 0.906= V Z R ⋅ T⋅

P:= V 1622.7

cm3

mol= Ans.

(d) For SRK EOS:

σ 1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1

Table 3.1α Tr ω,( ) 1 0.480 1.574ω+ 0.176ω2−( ) 1 Tr

1

2−

⋅+

2

:=

q Tr ( ) Ψ α Tr ω,( )⋅

Ω Tr ⋅:= Eq. (3.51) β Tr Pr ,( )

Ω Pr ⋅

Tr := Eq. (3.50)


Given Eq. (3.49)


Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅⋅−=

Z 1 B0 ω B1⋅+( ) Pr

Tr ⋅+:= Z 0.912= V

Z R ⋅ T⋅P

:= V 1634 cm

3

mol= Ans.


σ 1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1

α Tr ( ) Tr 0.5−:= Table 3.1 q Tr ( )

Ψ α Tr ( )⋅

Ω Tr ⋅:= Eq. (3.51)


Tr := Eq. (3.50)


Given Eq. (3.49)


Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅⋅−=

34

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Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅⋅−=

Z Find Z( ):= Z 0.896= VZ R ⋅ T⋅

P

:= V 1605.5cm

3

mol

= Ans.

3.34 Tc 318.7 K ⋅:= T 348.15 K ⋅:= Tr T

Tc:= Tr 1.092=

Pc 37.6 bar ⋅:= P 15 bar ⋅:= Pr P

Pc:= Pr 0.399=

ω 0.286:=(guess)

(a) B 194− cm

3

mol⋅:= C 15300

cm6

mol2

⋅:= VR T⋅

P:= V 1930

cm3

mol=

Given P V⋅R T⋅

1B

V+

C

V2

+=

Z Find Z( ):= Z 0.907= VZ R ⋅ T⋅

P:= V 1624.8

cm3

mol= Ans.


σ 1 2+:= ε 1 2−:= Ω 0.07779:= Ψ 0.45724:= Table 3.1

Table 3.1α Tr ω,( ) 1 0.37464 1.54226ω+ 0.26992ω2−( ) 1 Tr

1

2−

⋅+

2

:=

q Tr ( ) Ψ α Tr ω,( )⋅

Ω Tr ⋅

:= Eq. (3.51) β Tr Pr ,( ) Ω Pr ⋅

Tr

:= Eq. (3.50)


Given Eq. (3.49)

35

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Table 3.1 q Tr ( ) Ψ α Tr ( )⋅

Ω Tr ⋅:= Eq. (3.51)


Tr

:= Eq. (3.50)


Given Eq. (3.49)


Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅⋅−=

Z Find Z( ):= Z 0.888= V Z R ⋅ T⋅P

:= V 1714.1 cm3

mol= Ans.

(d) For SRK EOS:

σ 1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1

V Find V( ):= V 1722cm

3

mol= Z

P V⋅R T⋅

:= Z 0.893= Ans.

(b) B0 0.0830.422

Tr 1.6

−:= B0 0.283−=

B1 0.1390.172

Tr 4.2

−:= B1 0.02=

Z 1 B0 ω B1⋅+( ) Pr

Tr ⋅+:= Z 0.899= V

Z R ⋅ T⋅P

:= V 1734cm

3

mol= Ans.


σ 1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1

α Tr ( ) Tr 0.5−:=

36

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Ω 0.07779:= Ψ 0.45724:= Table 3.1

Table 3.1α Tr ω,( ) 1 0.37464 1.54226ω+ 0.26992ω2−( ) 1 Tr

1

2−

⋅+

2

:=

q Tr ( ) Ψ α Tr ω,( )⋅

Ω Tr ⋅:= Eq. (3.51) β Tr Pr ,( )

Ω Pr ⋅

Tr := Eq. (3.50)


Given Eq. (3.49)


Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅

⋅−=

Z Find Z( ):= Z 0.882= V Z R ⋅ T⋅

P:= V 1701.5

cm3

mol= Ans.

Table 3.1α Tr ω,( ) 1 0.480 1.574ω+ 0.176ω2−( ) 1 Tr

1

2−

⋅+

2

:=

q Tr ( ) Ψ α Tr ω,( )⋅Ω Tr ⋅

:= Eq. (3.51) β Tr Pr ,( ) Ω Pr ⋅Tr

:= Eq. (3.50)


Given Eq. (3.49)


Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅⋅−=

Z Find Z( ):= Z 0.895= V Z R ⋅ T⋅

P:= V 1726.9

cm3

mol= Ans.


σ 1 2+:= ε 1 2−:=

37

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Tr 0.808= Pr 0.082= B0 0.51−=

B1 0.139 0.172

Tr 4.2

−:= B1 0.281−= Z 1 B0 ω B1⋅+( ) Pr

Tr ⋅+:=

V Z R ⋅ T⋅

P:= Z 0.939= V 2268

cm3

mol= Ans.

(c) Table F.2: molwt 18.015 gm

mol⋅:= V 124.99

cm3

gm⋅ molwt⋅:=

or V 2252 cm

3

mol= Ans.

3.37 B 53.4− cm

3

mol⋅:= C 2620

cm6

mol2

⋅:= D 5000 cm

9

mol3

⋅:= n mol:=

3.35 T 523.15 K ⋅:= P 1800 kPa⋅:=

(a) B 152.5

−

cm3

mol⋅:= C 5800

−

cm6

mol2⋅:= V

R T⋅

P:=(guess)

Given P V⋅

R T⋅ 1

B

V+

C

V2

+= V Find V( ):=

Z P V⋅

R T⋅:= V 2250

cm3

mol= Z 0.931= Ans.

(b) Tc 647.1 K ⋅:= Pc 220.55 bar ⋅:= ω 0.345:=

Tr T

Tc:= Pr

P

Pc:= B0 0.083

0.422

Tr 1.6

−:=

38

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Note that values of Z from Eq. (3.38) are not physically meaningful for

pressures above 100 bar.

Pi

1·10 -10

20

40

60

80

100

120

140

160

180

200

bar =

Z2i

1

0.951

0.895

0.83

0.749

0.622

0.5+0.179i

0.5+0.281i

0.5+0.355i

0.5+0.416i

0.5+0.469i

=Z1i

1

0.953

0.906

0.859

0.812

0.765

0.718

0.671

0.624

0.577

0.53

=Zi

1

0.953

0.906

0.861

0.819

0.784

0.757

0.74

0.733

0.735

0.743

=

Z2i1

2

1

4

B Pi⋅

R T

⋅

++:=Eq. (3.37)Z1i 1B Pi⋅

R T

⋅

+:= Eq. (3.38)

Eq. (3.12)Zif Pi Vi,( ) Pi⋅

R T⋅:=

(guess)ViR T⋅

Pi:=Pi 10

10−20 i⋅+( ) bar ⋅:=i 0 10..:=

f P V,( ) Find V( ):=P V⋅R T⋅ 1

B

V+ C

V2+ D

V3+=Given

T 273.15 K ⋅:=

39

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Z 0.01:=Guess:Calculate Z for liquid by Eq. (3.53)

Eq. (3.50)β Tr Pr ,( ) Ω Pr ⋅

Tr :=

Eq. (3.51)q Tr ( )

Ψ α Tr ( )⋅

Ω Tr ⋅:=Table 3.1

α Tr ( ) Tr

0.5−

:=

Table 3.1Ψ 0.42748:=Ω 0.08664:=ε 0:=σ 1:=

For Redlich/Kwong EOS:

Pr 0.323=Pr P

Pc:=Tr 0.847=Tr

T

Tc:=

P 13.71 bar ⋅:=T 313.15 K ⋅:=

ω 0.152:=Pc 42.48 bar ⋅:=Tc 369.8 K ⋅:=(a) Propane:3.38

0 50 100 150 2000.5

0.6

0.7

0.8

0.9

1

Zi

Z1i

Z2i

Pi bar 1−⋅

40

8/19/2019 SM_3A


Ans.V 1.538 103×

cm3

mol=V

R T⋅P

R B0 ω B1⋅+( )⋅ Tc

Pc⋅+:=

B1 0.207−=B1 0.139 0.172

Tr 4.2

−:=

B0 0.468−=B0 0.083 0.422

Tr 1.6

−:=

For saturated vapor, use Pitzer correlation:

Ans.V 94.17 cm

3

mol=V Vc Zc

1 Tr −( )0.2857

⋅:=

Zc 0.276:=Vc 200.0 cm3

mol⋅:=

Tr 0.847=Tr T

Tc:=Rackett equation for saturated liquid:

Given

Z β Tr Pr ,( ) Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅ 1 β Tr Pr ,( )+ Z−

q Tr ( ) β Tr Pr ,( )⋅

⋅+=

Z Find Z( ):= Z 0.057= V Z R ⋅ T⋅

P:= V 108.1

cm3

mol= Ans.

Calculate Z for vapor by Eq. (3.49) Guess: Z 0.9:=

Given

Z 1 β Tr

Pr

,( )+ q Tr

( ) β Tr

Pr

,( )⋅ Z β Tr Pr ,( )−

Z Z β Tr Pr ,( )+( )⋅⋅−=

Z Find Z( ):= Z 0.789= V Z R ⋅ T⋅

P:= V 1499.2

cm3

mol= Ans.

41

8/19/2019 SM_3A


Parts (b) through (t) are worked exactly the same way. All results are

summarized as follows. Volume units are cu.cm./mole.

R/K, Liq. R/K, Vap. Rackett Pitzer

(a) 108.1 1499.2 94.2 1537.8

(b) 114.5 1174.7 98.1 1228.7

(c) 122.7 920.3 102.8 990.4

(d) 133.6 717.0 109.0 805.0

(e) 148.9 1516.2 125.4 1577.0

(f) 158.3 1216.1 130.7 1296.8

(g) 170.4 971.1 137.4 1074.0

(h) 187.1 768.8 146.4 896.0

(i) 153.2 1330.3 133.9 1405.7

(j) 164.2 1057.9 140.3 1154.3

(k) 179.1 835.3 148.6 955.4

(l) 201.4 645.8 160.6 795.8

(m) 61.7 1252.5 53.5 1276.9

(n) 64.1 1006.9 55.1 1038.5

(o) 66.9 814.5 57.0 853.4

(p) 70.3 661.2 59.1 707.8

(q) 64.4 1318.7 54.6 1319.0

(r) 67.4 1046.6 56.3 1057.2

(s) 70.8 835.6 58.3 856.4

(t) 74.8 669.5 60.6 700.5

42

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Eq. (3.50)

Calculate Z for liquid by Eq. (3.53) Guess: Z 0.01:=

Given


q Tr ( ) β Tr Pr ,( )⋅

⋅+=

Z Find Z( ):= Z 0.055= VZ R ⋅ T⋅

P:= V 104.7

cm3

mol= Ans.


Given


Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅⋅−=

Z Find Z( ):= Z 0.78= VZ R ⋅ T⋅

P:= V 1480.7

cm3

mol= Ans.

3.39 (a) Propane Tc 369.8 K ⋅:= Pc 42.48 bar ⋅:= ω 0.152:=

T 40 273.15+( ) K ⋅:= T 313.15 K = P 13.71 bar ⋅:=

Tr T

Tc:= Tr 0.847= Pr

P

Pc:= Pr 0.323=

From Table 3.1 for SRK:

σ 1:= ε 0:= Ω 0.08664:= Ψ 0.42748:=

α Tr ω,( ) 1 0.480 1.574ω+ 0.176ω2−( ) 1 Tr

1

2−

⋅+

2

:=

q Tr ( ) Ψ α Tr ω,( )⋅

Ω Tr ⋅:= Eq. (3.51) β Tr Pr ,( )

Ω Pr ⋅

Tr :=

43

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SRK, Liq. SRK, Vap. Rackett Pitzer

(a) 104.7 1480.7 94.2 1537.8

(b) 110.6 1157.8 98.1 1228.7

(c) 118.2 904.9 102.8 990.4

(d) 128.5 703.3 109.0 805.0

(e) 142.1 1487.1 125.4 1577.0

(f) 150.7 1189.9 130.7 1296.8

(g) 161.8 947.8 137.4 1074.0

(h) 177.1 747.8 146.4 896.0

(i) 146.7 1305.3 133.9 1405.7

(j) 156.9 1035.2 140.3 1154.3

(k) 170.7 815.1 148.6 955.4

(l) 191.3 628.5 160.6 795.8

(m) 61.2 1248.9 53.5 1276.9

(n) 63.5 1003.2 55.1 1038.5

(o) 66.3 810.7 57.0 853.4

(p) 69.5 657.4 59.1 707.8

(q) 61.4 1296.8 54.6 1319.0

(r) 63.9 1026.3 56.3 1057.2

(s) 66.9 817.0 58.3 856.4

(t) 70.5 652.5 60.6 700.5

44

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Tr := Eq. (3.50)

Calculate Z for liquid by Eq. (3.53) Guess: Z 0.01:=

Given


q Tr ( ) β Tr Pr ,( )⋅

⋅+=

Z Find Z( ):= Z 0.049= V Z R ⋅ T⋅

P:= V 92.2

cm3

mol= Ans.


Given

Z 1 β Tr Pr ,( )+ q Tr ( ) β Tr Pr ,( )⋅ Z β Tr Pr ,( )−Z ε β Tr Pr ,( )⋅+( ) Z σ β Tr Pr ,( )⋅+( )⋅⋅−=

Z Find Z( ):= Z 0.766= V Z R ⋅ T⋅

P:= V 1454.5

cm3

mol= Ans.

3.40 (a) Propane Tc 369.8 K ⋅:= Pc 42.48 bar ⋅:= ω 0.152:=

T 40 273.15+( ) K ⋅:= T 313.15 K = P 13.71 bar ⋅:=

Tr T

Tc:= Tr 0.847= Pr

P

Pc:= Pr 0.323=

From Table 3.1 for PR:

α Tr ω,( ) 1 0.37464 1.54226ω+ 0.26992ω2−( ) 1 Tr

1

2−

⋅+

2

:=

σ 1 2+:= ε 1 2−:= Ω 0.07779:= Ψ 0.45724:=

q Tr ( ) Ψ α Tr ω,( )⋅

Ω Tr ⋅:= Eq. (3.51)

45

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PR, Liq. PR, Vap. Rackett Pitzer

(a) 92.2 1454.5 94.2 1537.8

(b) 97.6 1131.8 98.1 1228.7

(c) 104.4 879.2 102.8 990.4

(d) 113.7 678.1 109.0 805.0

(e) 125.2 1453.5 125.4 1577.0

(f) 132.9 1156.3 130.7 1296.8

(g) 143.0 915.0 137.4 1074.0

(h) 157.1 715.8 146.4 896.0

(i) 129.4 1271.9 133.9 1405.7

(j) 138.6 1002.3 140.3 1154.3

(k) 151.2 782.8 148.6 955.4

(l) 170.2 597.3 160.6 795.8

(m) 54.0 1233.0 53.5 1276.9

(n) 56.0 987.3 55.1 1038.5

(o) 58.4 794.8 57.0 853.4

(p) 61.4 641.6 59.1 707.8

(q) 54.1 1280.2 54.6 1319.0

(r) 56.3 1009.7 56.3 1057.2

(s) 58.9 800.5 58.3 856.4

(t) 62.2 636.1 60.6 700.5

46

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Tr T

Tc:= Tr 1.145= Pr

P

Pc:= Pr 2.282=

From Tables E.3 & E.4: Z0 0.482:= Z1 0.126:=

Z Z0 ω Z1⋅+:= Z 0.493= n P Vtotal⋅Z R ⋅ T⋅:= n 2171 mol=

mass n molwt⋅:= mass 60.898kg= Ans.

3.42 Assume validity of Eq. (3.37).

P1 1bar := T1 300K := V1 23000cm

3

mol:=

Z1P1 V1⋅R T1⋅

:= Z1 0.922= BR T1⋅

P1Z1 1−( )⋅:= B 1.942− 10

3× cm3

mol=

With this B, recalculate at P2 P2 5bar :=

3.41 (a) For ethylene, molwt 28.054gm

mol:= Tc 282.3 K ⋅:= Pc 50.40 bar ⋅:=

ω 0.087:= T 328.15 K ⋅:= P 35 bar ⋅:=

Tr T

Tc:= Pr

P

Pc:= Tr 1.162= Pr 0.694=

From Tables E.1 & E.2: Z0 0.838:= Z1 0.033:=

Z Z0 ω Z1⋅+:= Z 0.841=

n18 kg⋅

molwt

:= VtotalZ n⋅ R ⋅ T⋅

P

:= Vtotal 0.421m3= Ans.

(b) T 323.15 K ⋅:= P 115 bar ⋅:= Vtotal 0.25 m3⋅:=

47

8/19/2019 SM_3A


VR T⋅

P:= V 1044

cm3

mol=

3.44 T 320 K ⋅:= P 16 bar ⋅:= Tc 369.8 K ⋅:= Pc 42.48 bar ⋅:=

ω 0.152:= Vc 200cm

3

mol⋅:= Zc 0.276:=

molwt 44.097gm

mol:=

Tr T

Tc:= Tr 0.865= Pr

P

Pc:= Pr 0.377=

Vliq Vc Zc1 Tr −( )

0.2857

⋅:= Vliq 96.769cm

3

mol=

Vtank 0.35 m3⋅:= mliq

0.8 Vtank ⋅Vliq

molwt

:= mliq 127.594kg= Ans.

B0 0.0830.422

Tr 1.6

−:= B0 0.449−=

Z2 1B P2⋅

R T1⋅+:= Z2 0.611= V2

R T1⋅ Z2⋅

P2:= V2 3.046 10

3× cm

3

mol= Ans.

3.43 T 753.15 K ⋅:= Tc 513.9 K ⋅:= Tr T

Tc:= Tr 1.466=

P 6000 kPa⋅:= Pc 61.48 bar ⋅:=Pr

P

Pc:= Pr 0.976=

ω 0.645:=B0 0.083

0.422

Tr 1.6

−:= B0 0.146−=

B1 0.1390.172

Tr

4.2−:= B1 0.104=

VR T⋅

PB0 ω B1⋅+( ) R ⋅

Tc

Pc⋅+:= V 989

cm3

mol= Ans.

For an ideal gas:

48

8/19/2019 SM_3A


B1 0.139 0.172

Tr

4.2−:= B1 0.624−=

V R T⋅

PB0 ω B1⋅+( ) R ⋅

Tc

Pc⋅+:= V 9.469 10

3× cm

3

mol=

mvap

Vvap

V

molwt

:= mvap 98.213kg= Ans.

3.46 (a) T 333.15 K ⋅:= Tc

305.3 K ⋅:= Tr

T

Tc:= T

r 1.091=

P 14000 kPa⋅:= Pc 48.72 bar ⋅:= Pr P

Pc:= Pr 2.874=

ω 0.100:= Vtotal 0.15 m3⋅:= molwt 30.07

gm

mol:=

B1 0.139 0.172

Tr 4.2

−:= B1 0.177−=

Vvap R T⋅P

B0 ω B1⋅+( ) R ⋅ Tc

Pc⋅+:= Vvap 1.318 103× cm

3

mol=

mvap

0.2 Vtank ⋅Vvap

molwt

:=mvap 2.341kg= Ans.

3.45 T 298.15 K ⋅:= Tc 425.1 K ⋅:= Tr T

Tc:= Tr 0.701=

P 2.43 bar ⋅:= Pc 37.96 bar ⋅:= Pr

P

Pc:= Pr 0.064=

ω 0.200:= Vvap 16 m3⋅:= molwt 58.123

gm

mol⋅:=

B0 0.083 0.422

Tr 1.6

−:= B0 0.661−=

49

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This equation giving Tr as a function of Z and Eq. (3.54) in conjunction with

Tables E.3 & E.4 are two relations in the same variables which must be

satisfied at the given reduced pressure. The intersection of these two

relations can be found by one means or another to occur at about:

Tr 1.283:= and Z 0.693:=

Whence T Tr Tc⋅:=

T 391.7 K = or 118.5 degC⋅ Ans.

3.47 Vtotal 0.15 m3⋅:= T 298.15 K ⋅:=

Tc 282.3 K ⋅:= Pc 50.40 bar ⋅:= ω 0.087:= molwt 28.054 gm

mol:=

VVtotal

40 kg⋅molwt

:= P V⋅ Pr Pc⋅ V⋅= Z R ⋅ T⋅=

or Pr α Z⋅= where α R T⋅

Pc V⋅:= α 4.675=

From tables E.3 & E.4: Z0 0.463:= Z1 0.037−:=

Z Z0 ω Z1⋅+:= Z 0.459= V Z R ⋅ T⋅

P:= V 90.87

cm3

mol=

methane

Vtotal

V

molwt

:= methane 49.64kg= Ans.

(b)V

Vtotal

40 kg⋅:= P 20000 kPa⋅:= P V⋅ Z R ⋅ T⋅= Z R ⋅ Tr ⋅ Tc⋅=

or Tr α

Z

= where α P V⋅

R Tc⋅:= α 29.548

mol

kg

=

Whence Tr 0.889

Z= at Pr

P

Pc:= Pr 4.105=

50

8/19/2019 SM_3A


Tr1

T1

Tc:= Tr1 0.977=

P1 2200 kPa⋅:= Pc 48.72 bar ⋅:= Pr1

P1

Pc:= Pr1 0.452=

Vtotal

0.35 m3⋅:= ω 0.100:=

From Tables E.1 & E.2: Z0 .8105:= Z1 0.0479−:=

Z Z0 ω Z1⋅+:= Z 0.806= V1

Z R ⋅ T1⋅

P1:= V1 908

cm3

mol=

T2 493.15 K ⋅:= Tr2

T2

Tc:= Tr2 1.615=

Assume Eq. (3.37) applies at the final state.

B0 0.0830.422

Tr21.6

−:= B0 0.113−=

B1 0.1390.172

Tr24.2

−:= B1 0.116=

Whence Pr 4.675 Z⋅= at Tr T

Tc:= Tr 1.056=

This equation giving Pr as a function of Z and Eq. (3.54) in conjunction with

Tables E.3 & E.4 are two relations in the same variables which must be

satisfied at the given reduced temperature. The intersection of these two

relations can be found by one means or another to occur at about:

Pr 1.582:= and Z 0.338:= P Pc Pr ⋅:= P 79.73bar = Ans.

3.48 mwater 15 kg⋅:= Vtotal 0.4 m3⋅:= V

Vtotal

mwater := V 26.667

cm3

gm=

Interpolate in Table F.2 at 400 degC to find: P 9920 kPa⋅= Ans.

3.49 T1 298.15 K ⋅:= Tc 305.3 K ⋅:=

51

8/19/2019 SM_3A


PR T⋅

V B0 ω B1⋅+( ) R ⋅

Tc

Pc⋅−

:=P 10.863bar = Ans.

3.51 Basis: 1 mole of LIQUID nitrogen

Tn 77.3 K ⋅:= Tc 126.2 K ⋅:= Tr

Tn

Tc:= Tr 0.613=

P 1 atm⋅:= Pc 34.0 bar ⋅:= Pr P

Pc

:= Pr 0.03=

ω 0.038:= molwt 28.014gm

mol⋅:=

Vliq 34.7 cm3⋅:=

B0 0.0830.422

Tr 1.6

−:=B0 0.842−=

P2

R T2⋅

V1 B0 ω B1⋅+( ) R ⋅ Tc

Pc⋅−

:=P2 42.68bar = Ans.

3.50 T 303.15 K ⋅:= Tc 304.2 K ⋅:= Tr T

Tc:= Tr 0.997=

Vtotal 0.5 m3⋅:= Pc 73.83 bar ⋅:= ω 0.224:= molwt 44.01

gm

mol⋅:=

B0 0.0830.422

Tr 1.6

−:= B0 0.341−=

B1 0.1390.172

Tr 4.2

−:= B1 0.036−=

VVtotal

10 kg⋅molwt

:=V 2.2 10

3× cm

3

mol=

52

8/19/2019 SM_3A


aΨ α Tr ( )⋅ R

2⋅ Tc2⋅

Pc:= (3.42) b

Ω R ⋅ Tc⋅

Pc:= (3.43)

a 0.901 m3 bar cm

3⋅

mol2

= b 26.737

cm3

mol=

P R T⋅

V b−a

V V b+( )⋅−:= (3.44) P 450.1bar = Ans.

3.52 For isobutane: Tc 408.1 K ⋅:= Pc 36.48 bar ⋅:= V1 1.824 cm

3

gm⋅:=

T1 300 K ⋅:= P1 4 bar ⋅:= T2 415 K ⋅:= P2 75 bar ⋅:=

B1 0.139 0.172

Tr 4.2

−:= B1 1.209−=

Z 1 B0 ω B1⋅+( ) Pr

Tr ⋅+:= Z 0.957=

nvapor

P Vliq⋅

Z R ⋅ Tn⋅:= nvapor 5.718 10

3−× mol=

Final conditions:

ntotal 1 mol⋅ nvapor +:= V2 Vliq⋅

ntotal:= V 69.005

cm3

mol=

T 298.15 K ⋅:= Tr T

Tc:= Tr 2.363=

PigR T⋅

V:= Pig 359.2bar =

Use Redlich/Kwong at so high a P.

Ω 0.08664:= Ψ 0.42748:= α Tr ( ) Tr .5−:= α Tr ( ) 0.651=

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3.53 For n-pentane: Tc 469.7 K ⋅:= Pc 33.7 bar ⋅:= ρ1 0.63 gm

cm3

⋅:=

T1 291.15 K ⋅:= P1 1 bar ⋅:= T2 413.15 K ⋅:= P2 120 bar ⋅:=

Tr1

T1

Tc:= Pr1

P1

Pc:= Tr2

T2

Tc:= Pr2

P2

Pc:=

Tr1 0.62= Pr1 0.03= Tr2 0.88= Pr2 3.561=

From Fig. (3.17): ρr1 2.69:= ρr2 2.27:=

By Eq. (3.65), ρ2 ρ1

ρr2

ρr1

⋅:= ρ2 0.532 gm

cm3

= Ans.

Tr1

T1

Tc:= Pr1

P1

Pc:= Tr2

T2

Tc:= Pr2

P2

Pc:=

Tr1 0.735= Pr1 0.11= Tr2 1.017= Pr2 2.056=

From Fig. (3.17): ρr1 2.45:=

The final T > Tc, and Fig. 3.17 probably should not be used. One can easily

show that

with Z from Eq. (3.54) and

Tables E.3 and E.4. Thusρr

P Vc⋅

Z R ⋅ T⋅=

Vc 262.7 cm3

mol⋅:= ω 0.181:= Z0 0.3356:= Z1 0.0756−:=

Z Z0 ω Z1⋅+:= Z 0.322= ρr2

P2 Vc⋅

Z R ⋅ T2⋅:= ρr2 1.774=

Eq. (3.65): V2 V1

ρ r1

ρ r2

⋅:= V2 2.519 cm

3

gm= Ans.

54

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For ammonia:

Tc 405.7 K ⋅:= T 293.15 K ⋅:= Tr T

Tc:= Tr 0.723=

Pc 112.8 bar ⋅:= P 857 kPa⋅:= Pr P

Pc:= Pr 0.076=

Vc 72.5cm

3

mol⋅:= Zc 0.242:= ω 0.253:=

Eq. (3.63): Vliquid Vc Zc1 Tr −( )

0.2857

⋅:= Vliquid 27.11cm

3

mol=

B0 0.0830.422

Tr 1.6

−:=B0 0.627−=

B1 0.1390.172

Tr 4.2

−:= B1 0.534−=

3.54 For ethanol: Tc 513.9 K ⋅:= T 453.15 K ⋅:= Tr T

Tc:= Tr 0.882=

Pc 61.48 bar ⋅:= P 200 bar ⋅:= Pr PPc

:= Pr 3.253=

Vc 167cm

3

mol⋅:= molwt 46.069

gm

mol⋅:=

From Fig. 3.17: ρr 2.28:= ρ ρr ρc⋅=ρr

Vc

=

ρ ρr

Vc

molwt

:= ρ 0.629gm

cm3

= Ans.

3.55

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P 1 atm⋅:=

V 1400 ft3⋅:= n

P V⋅R T⋅

:= n 3.689lbmol=

For methane at 3000 psi and 60 degF:

Tc 190.6 1.8⋅ rankine⋅:= T 519.67 rankine⋅:= Tr T

Tc:= Tr 1.515=

Pc 45.99 bar ⋅:= P 3000 psi⋅:= Pr P

Pc

:= Pr 4.498=

ω 0.012:=

From Tables E.3 & E.4:

Z0 0.819:= Z1 0.234:= Z Z0 ω Z1⋅+:= Z 0.822=

Vvapor R T⋅

PB0 ω B1⋅+( ) R ⋅

Tc

Pc⋅+:= Vvapor 2616

cm3

mol=

∆V Vvapor Vliquid−:= ∆V 2589cm3

mol= Ans.

Alternatively, use Tables E.1 & E.2 to get thevapor volume:

Z0 0.929:= Z1 0.071−:= Z Z0 ω Z1⋅+:= Z 0.911=

Vvapor Z R ⋅ T⋅

P:= Vvapor 2591

cm3

mol=

∆V Vvapor Vliquid−:= ∆V 2564cm3

mol= Ans.

3.58 10 gal. of gasoline is equivalent to 1400 cu ft. of methane at 60 degF and 1

atm. Assume at these conditions that methane is an ideal gas:

R 0.7302ft

3atm⋅

lbmol rankine⋅= T 519.67 rankine⋅:=

56

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B0 0.083 0.422

Tr 1.6

−:= B0 0.495−= B1 0.139 0.172

Tr 4.2

−:= B1 0.254−=

Z 1 B0 ω B1⋅+( ) Pr

Tr ⋅+:= Z 0.823= Ans. Experimental: Z = 0.7757

For Redlich/Kwong EOS:

σ 1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1

α Tr ( ) Tr 0.5−:= Table 3.1 q Tr ( )

Ψ α Tr ( )⋅

Ω Tr ⋅:= Eq. (3.51)

Vtank Z n⋅ R ⋅ T⋅

P:= Vtank 5.636ft

3= Ans.

3.59 T 25K := P 3.213bar :=

Calculate the effective critical parameters for hydrogen by equations (3.55)

and (3.56)

Tc43.6

1 21.8K

2.016T+

K ⋅:= Tc 30.435K =

Pc

20.5

1 44.2K

2.016T+

bar ⋅:= Pc 10.922bar =

ω 0:=

Pr P

Pc:= Pr 0.294= Tr

T

Tc:= Tr 0.821=

Initial guess of volume: V R T⋅

P

:= V 646.903 cm

3

mol

=

Use the generalized Pitzer correlation

57

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Pr 0.022=

B0 0.083 0.422

Tr 1.6

−:= B0 0.134−= B1 0.139 0.172

Tr 4.2

−:= B1 0.109=

Z0 1 B0 Pr

Tr ⋅+:= Z0 0.998= Z1 B1 Pr

Tr ⋅:= Z1 0.00158=

Z Z0 ω Z1⋅+:= Z 0.998= V1Z R ⋅ T⋅

P:= V1 0.024

m3

mol=

(a) At actual condition: T 50 32−( ) 5

9⋅ 273.15+

K := P 300psi:=

Pitzer correlations:T 283.15 K =

Tr T

Tc:= Tr 1.486= Pr

P

Pc:= Pr 0.45=

B0 0.083 0.422

Tr 1.6

−:= B0 0.141−=


Tr := Eq. (3.50)

Calculate Z Guess: Z 0.9

:=Given Eq. (3.49)


Z Z β Tr Pr ,( )+( )⋅⋅−=

Z Find Z( ):= Z 0.791= Ans. Experimental: Z = 0.7757

3.61 For methane: ω 0.012:= Tc 190.6K := Pc 45.99bar :=

At standard condition: T 60 32−( ) 5

9⋅ 273.15+

K := T 288.706K =

Pitzer correlations: P 1atm:=

Tr T

Tc:= Tr 1.515= Pr

P

Pc:=

58

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Ans.u 8.738m

s=u

q2

A:=

A 0.259 m2=A

π4

D2:=D 22.624in:=(c)

Ans.n1 7.485 103×

kmol

hr =n1

q1

V1:=(b)

Ans.q2 6.915 106×

ft3

day=q2 q1

V2

V1⋅:=q1 150 10

6⋅ ft

3

day:=

V2 0.00109m

3

mol=V2

Z R ⋅ T⋅P

:=Z 0.958=Z Z0 ω Z1⋅+:=

Z1 0.0322=Z1 B1

Pr

Tr ⋅:=Z0 0.957=Z0 1 B0

Pr

Tr ⋅+:=

B1 0.106=B1 0.1390.172

Tr 4.2

−:=

59

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3.62 Use the first 29 components in Table B.1

sorted so that ω values are in ascending

order. This is required for the Mathcad

slope and intercept functions.

ω

0.012

0.087

0.1

0.140

0.152

0.181

0.187

0.19

0.191

0.194

0.196

0.2

0.205

0.21

0.21

0.212

0.218

0.23

0.235

0.252

0.262

0.28

0.297

0.301

0.3020.303

0.31

0.322

:= ZC

0.286

0.281

0.279

0.289

0.276

0.282

0.271

0.267

0.277

0.275

0.273

0.274

0.273

0.273

0.271

0.272

0.275

0.272

0.269

0.27

0.264

0.265

0.256

0.266

0.2660.263

0.263

0.26

:=m slope ω ZC,( ):= 0.091−( )=

b intercept ω ZC,( ):= 0.291( )=

R corr ω ZC,( ):= 0.878−( )= R 2 0.771=

0 0.1 0.2 0.3 0.40.25

0.26

0.27

0.28

0.29

ZC

m ω⋅ b+

ω

The equation of the line is:

Zc 0.291 0.091ω−=Ans.

sm_3a

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