07 residual propeties

Chemical Engineering 317 – 2015: Additional notes: Residual Properties and Examples

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ChemicalEngineering317

AdditionalNotes:ResidualPropertiesandExamples

1 TheNeedforDepartureFunctions

When designing chemical processes, chemical engineers often need to calculate the change of a

specific state property between one state and another state (e.g. ∆U, ∆S, ∆H, ∆A, ∆G, etc.). The

state (“toestand”) of a pure component in a specific phase can be fully described by any two of

the following parameters: T, P and V. In other words, the state property can be expressed as

f(T,P), f(T,V) or f(P,V). For example if U = U(T,V) then ∆U = U(T2,V2) – U(T1,V1).

In principle, departure functions fulfill the need to have a general approach, applicable to any

fluid, whereby a specific change (∆) in a state property can be calculated by following a specific

type of pathway while describing the property either in terms of {T,V} or {T,P}. Departure

functions make use of residual properties and it is therefore important that the definition and

nature of residual properties are well-understood.

In practice, the parameters T and P are easy to measure and they are thus convenient to use.

However, many equations of state (EOS) are pressure-explicit, in other words P = P(T,V), and it is

not easy to re-arrange these equations to volume-explicit functions such as V = V(T,P).

Therefore, in order to utilize the most widely used equations of state, the best approach is quite

often to develop the state properties in terms of {T,V}.

2 TheDepartureFunctionPathway

For simplicity sake, let’s consider internal energy, u, as the state property under investigation.

To avoid very complex calculations of a parameter such as cv we rather follow a pathway

consisting of three steps:

Step1: Calculate ΔU when isothermally changing from the real fluid of state 1 to an “ideal gas”

at the same {T,V} or { , }T P . In other words, the measured parameters remain constant, but the

component is changed from a real fluid to an ideal gas.

∆��→�� = �� − �� = −�� −�� ≡ −��

Step2: Calculate ΔU when changing the “ideal gas” from state 1 to state 2 (T and v can now

change), while remaining an ideal gas.

∆��→�� = �� − �� = ∆��

Note: Under ideal gas conditions, we know that cv is only a function of temperature and we can

thus change the temperature of the system while ignoring the volume-dependence of cv.

Step 3: Calculate ΔU when isothermally changing between the state 2 “ideal gas” and the

equivalent real fluid of state 2 (i.e. the final state).

∆��→�� = �� − �� = �� − �� ≡ ��

Following Simpson’s rule, we now combine the above three steps to provide:

∆��→�� = �� − �� = ∆��→��+ ∆��→��

+ ∆��→��

Therefore: ∆��→�� = �� − �� + ∆��


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This equation can be generalized to any fundamental or derived state property M, where M

could be u, s, h, a, or g. In general form:

∆� = �� −�� + ∆��

In addition, we know that molecular attraction and repulsion forces are, per definition, “ignored”

for an ideal gas. Furthermore, when a real fluid is at a state where � → ∞ or � → 0, the

molecules are so far apart that attraction and repulsion forces are completely insignificant.

Therefore, it follows that:

��!", �$ = �!", � → ∞$ or ��!", �$ = �!", � → 0$ In other words, to assist with solving the problem, the internal energy of an equivalent “ideal

gas” can be calculated at the same temperature as the real fluid, but at zero pressure (or at

infinite molar volume).

To summarise:

The advantage of the pathway proposed above (including two departure functions) is that all

calculations that relate to changes in T are performed in the ideal gas state, where ig ig

vPc c R= + .

Ideal gas heat capacities are independent of P and V, as is verified below:

%&'(&) *+ = " &&+ %,&�&)-+*) = " %&./&+.*) since 0) ≡ %&1&+*) and 0) = " %&�&+*)

For an ideal gas: � = �+) therefore %&./&+.*) = &

&+ %�)*) = 0

3 Residual Internal Energy

(Not considered in class)

Consider ( )U U T V= ,

R igT V T V T V

T v T

V V V

T T

U U U

U U

U SdU dV T P dV

V V∞ ∞

∞

∞

≡ −

= −

∂ ∂ = = = − ∂ ∂ ∫ ∫ ∫

, , ,

, , when isothermal

See Smith & VN, eq.6.31, p.205. By applying Maxwell, it follows that:

VRT V

V

PU T P dV

T∞

∂ = − ∂ ∫, Residual internal energy, UR, developed in terms of T V{ , }

Alternatively, one can simply apply equation 6.32 in Smith & VN (p.205), as follows:

RT V T V T

V T

VT

U U U

dU C dT

∞

∞

= −

= =∫ ∫

, , ,

V

V

PT P dV

T∞

∂ + − ∂ ∫

when isothermal

Now, the departure function pathway to calculate ∆U between states 1 and 2 results in:


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( ) ( ) ( )R R

ig ig ig ig

T

VV V T

U U

V Vig

U U U U U U U U U

P PT P dV T P dV C dT

T T∞ ∞

∆ = − = − − + −

∂ ∂ = − − − + ∂ ∂ ∫ ∫ ∫

64748 647482 1

1 22

1

2 1 2 12 1 2 1−−−−

4 Residual Enthalpy

4.1 Residual Entropy in terms of P, V and T

Consider an enthalpy change:

∆2 = 2� −2� = 2�� +2�� −2�� +2�� = 2�� −2�� + 3 0/4"+.+5

Therefore need an expression for HR:

2+,/� = 2+,/ −2+,/� = �2+,/ −2+,/67�� − �2+,/� −2+,/67��

But enthalpy of an ideal gas is pressure INDEPENDENT

Therefore 2+,/� = �2+,/ −2+,/67��

Previously: 42 = 0�4" + ,� − " %&)&+*/- 4�

Therefore at constant temperature: 42 = ,� − " %&)&+*/- 4�

Therefore: 2+� = 3 ,� − " %&)&+*/- 4�//67

4.2 Residual Enthalpy for Generalised correlation charts: Residual Enthalpy in terms of

Z, Tr and Pr

Above we derived: 2+� = 3 ,� − " %&)&+*/- 4�//67

� = 8�+/ therefore %&)&+*/ = 8�

/ + �+/ %&8&+*/

Therefore 2+� = 3 ,8�+/ − 8�+/ − �+

/ %&8&+*/- 4�//67

Therefore 2+� = 3 ,− �+./ %&8&+*/- 4�

//67

Divide each side by Tc and multiply Pc above and below the line on the RHS:

9:;<;=+> = 3 ?− �+;.

/; %&8&+;*/;@ 4��/;/;67

Remember A = A7!"��$ + BA�!"��$ Therefore−%&8&+;*/; = −%&8C&+;*/; −B %&85&+;*/;

Substitute above into 9:=+> expression:

9:;<;=+> = 3 ?�+;./; D%&8C&+;*/; +B %&85&+;*/;E@ 4��/;/;67

Therefore: 9:;<;=�+> = 3 ?+;./; %&8

C&+;*/;@ 4��

/;/;67 +B3 ?+;./; %&85

&+;*/;@ 4��/;/;67


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9:;<;=�+> = F9:;<;=�+> G7 +B F9:;<;=�+> G�

See tables E5 – E8 p700-703

Generalised correlation charts for HR(0) and HR(1) from Koretsky (reproduced with permission)


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4.3 Residual Enthalpy from EOSs

For a volume explicit EOS

Previously Residual Enthalpy: 2+,/� = 3 ,� − " %&)&+*/- 4�//67

Volume explicit EOS e.g. Virial EOS: � = %�+/ * !1 + IJ� + 0J�� +⋯$ • Determine %&)&+*/ and substitute into residual functions

• Remember B’ and C’ are temperature dependent, therefore also need to determine their

derivatives!

For a pressure explicit EOS

Pressure explicit EOS: � = LM", �N e.g. cubic EOSs

Residual enthalpy: 2+,/� = 3 ,� − " %&)

&+*/

- 4�//67

• Not that easy to work with for pressure explicit EOSs

If � = LM", �N then 4� = %&/&+*

)4" + %&/

&)*+

4�

At constant temperature: 4� = %&/&)*

+4�

Substitute above into residual enthalpy equation: 2+,/� = 3 ,� − " %&)

&+*/

- %&/&)*

+4� /

/67

Therefore: 2+,/� = 3 ,� %&/

&)*+

− " %&)&+*

/%&/

&)*+

- 4� //67

From cyclic relations: %&)&+*

/%&/

&)*+

= − %&/&+*

)

Therefore 2+,/� = 3 ,� %&/

&)*+

+ " %&/&+*

)- 4� /

/67

5 Residual Entropy

5.1 Residual Entropy in terms of P, V and T

Ideal gas entropy change: ∆O� = 3 'PQR

+ 4"+.+5

− STU %/./5

*

Consider an entropy change:

∆O = O� − O� = O�� + O�

� − O�� + O�

� = O�� − O�

� + 3 'PQR

+ 4"+.+5

− STU %/./5

*

Therefore need an expression for SR:

O+,/� = O+,/ − O+,/

� = �O+,/ − O+,/67�� − �O+,/

� − O+,/67��

From Maxwell: 4O+ = − %&)&+*

/4�

For an ideal gas: �� = S" or � = �+/ therefore%&)

&+*/

= �/

Therefore 4O+� = −S V/

/

Therefore: O+,/� − O+,/67

� = − 3 S V//

//67


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Back to above using Maxwell applied to a real gas: O+,/ − O+,/67� = −3 %&)&+*/ 4�//67

Substitute into above: O+,/� = −3 %&)&+*/ 4�//67 + 3 S V/

///67

Simplified: O+,/� = 3 ,�/ − %&)&+*/- 4�//67

5.2 Residual Entropy for Generalised correlation charts: Residual Enthalpy in terms of

Z, Tr and Pr

Remember from above: O+,/� = 3 ,�/ − %&)&+*/- 4�//67 as well as � = 8�+

/ and %&)&+*/ = 8�/ +

�+/ %&8&+*/

Therefore O+,/� = 3 ,�/ − 8�/ + �+

/ %&8&+*/- 4�//67

Therefore O+,/� = S 3 ,�W8/ + +/ %&8&+*/- 4�

//67

Divide RHS above and below the line with Tc and Pc: �:;<;=� = 3 F�W8/; + +;

/; %&8&+;*/G 4��//67

Similar to above it can be proven that: �:;<;=� = F�:;<;=� G7 +B F�:;<;=� G�

See tables E5 – E8 p704-707

Generalised correlation charts for SR(0) and SR(1) from Koretsky (reproduced with permission)


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5.3 Residual Entropy from an EOS

For a volume explicit EOS:

Previously Residual Entropy: O+,/� = 3 ,�/ − %&)&+*/- 4�//67

Volume explicit EOS e.g. Virial EOS: � = %�+/ * !1 + IJ� + 0J�� +⋯$ • Determine %&)&+*/ and substitute into residual functions

• Remember B’ and C’ are temperature dependent, therefore also need to determine their

derivatives!

For a pressure explicit EOS

Residual entropy: O+,/� = 3 ,�/ − %&)&+*/- 4�//67

Substitute 4� = %&/&)*+ 4� into above: O+,/� = 3 ,�/ − %&)&+*/- %&/&)*+ 4�

//67

Therefore: O+,/� = 3 ,�/ %&/&)*+ − %&)&+*/ %&/&)*+- 4�

//67

From cyclic relations: %&)&+*/ %&/&)*+ = −%&/&+*)

Therefore: O+,/� = 3 ,�/ %&/&)*+ + %&/&+*)- 4�//67

6 Class Examples

Parts from Koretsky reproduced with permission


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6.1 Class Example 5.3: Koretsky Example 5.5


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6.2 Class Example 5.4: Koretsky Example 5.4


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6.3 Class Example 5.5: Tutorial Test 2012 Question

Question

A mixture of carbon dioxide (75.2 mass %) and methane must be compressed on a plant near

the sea from a pressure of 2.02 bar(gauge) and temperature of 20.9 oC to a pressure of

120.2 bar(gauge) at a temperature of 52.2 oC. You want to calculate the energy difference

(in J/mol) between these two states. However, you are not sure which equation of state will

describe the gas behaviour adequately and therefore perform the calculations by

a. Firstly applying the Lee-Kesler correlations and

b. By using the simple van der Waals equation of state.

c. To what extent do the answers differ and what is your opinion regarding the accuracy of

the two methods?


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Solution

General

Let CO2 = component a ; Let CH4 = component b

Firstly remember that we have to work with mole fractions, instead of mass fractions.

0.752 / 44.010.525

0.752 / 44.01 0.248 / 16.0430.475

a

b

y

y

= =+

=

Furthermore, temperatures must be in Kelvin and pressures must be absolute. Therefore:

At State 1: T1 = 294.05 K; P1 = 2.02+1.01325 = 3.03325 bar (303325 Pa)

At State 2: T2 = 325.35 K; P2 = 120.2+1.01325 = 121.21325 bar (12121325 Pa)

The difference in enthalpy between State 1 and State 2 can be calculated by following a useful

hypothetical path that includes two residual enthalpy values (i.e. a departure function pathway).

2

11 2 1 2

TR R ig R R igPT

H H H H H H C dT∆ = − + + ∆ = − + + ∫

The two residual enthalpy value can be calculated according to Lee Kesler correlations (section a

of the question), or the van der Waals equation of state (section b of the question).

(a) Using Lee-Kesler Correlations

Obtain physical property data for CO2 and CH4, from Appendix in the textbook. For the gas

mixture, we need to calculate pseudo properties according to Kay’s rule, as follows:

( )

( )

( ) , , etc.

c mix a ca b cb

c mix a ca b cb

mix a a b b

ig ig iga b mix a a b b mix a a b bPa PbP mix

P y P y P

T y T y T

y y

C y C y C A y A y A B y A y A

ω ω ω

= +

= +

= +

= + ⇒ = + = +

Kay’s rule is not necessarily correct, but it is the best mixing rule that you have at this stage when

applying the Lee-Kesler correlations. The physical property data from the textbook, and the

calculated pseudo-properties, are summarised in the following table.

In order to use the Lee-Kesler correlations in the Appendix, we require reduced temperatures

and pressures for the gas mixture at each state, i.e.

( ) ( )( ) ( )

;r mix r mixc mix c mix

T PT P

T P= =

The calculated pseudo-reduced temperatures and pressures, as well as the correlation values

obtained from the Appendix, are indicated in the table below. Note that one has to interpolate

between Tr = 1.15 and Tr = 1.2 to obtain correct values for State 1. Furthermore, for each state:

( ) ( ) ( ) ( )0 1 0 1

( )( ) ( ) ( ) ( ) ( )

, or: R R R RR

Rc mix

c mix c mix c mix c mix c mix

H H H HHH RT

RT RT RT RT RTω ω

= + ⋅ = + ⋅

MMass

fraction

Mole

fractionTc Pc ωωωω

g/mol K bar A B x 103

C x 106

D x 10-5

CO2 44.010 0.752 0.5250 304.2 73.83 0.224 5.457 1.045 -1.157

CH4 16.043 0.248 0.4750 190.6 45.99 0.012 1.702 9.081 -2.164

Pseudo properties of mixture 1 1 250.242 60.607 0.1233 3.6734 4.8619 -1.0279 -0.60745

Heat capacity parameters for ideal gas C P /R


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We now have the two residual values and only need to calculate the enthalpy change in the ideal

gas condition (igH∆ ) when changing the temperature from T1 to T2.

( )

[ ]

2 2

1 1

2

1

2

1

2

11

( ) 2 2

2 3

52 3

3 6

( )

where , , etc. are for the mixture.

2 3

4.8619 1.0279 0.60745 103.6734

2 10 3 10

139.15

....

igT TP mix

T T

T

T

T

T

T TigP mix TT

CdT A BT CT DT dT A B

R

BT CT DAT

T

T T TT

C dT R

−= + + +

= + + −

×= ⋅ + − + × ×

=

∴ = ⋅

∫ ∫

∫2 8.314 139.15 1156.9 J/mol= ⋅ =

Therefore:

( ) ( )

2

11 2

83.8 3291.3 1156.9

2051 J/mol

TR R ig

PTH H H C dT∆ = − + +

= − − + − +

= −

∫

Note that this value is negative, i.e. energy has to be removed from the gas mixture.

(b) Using van der Waals

2 2

2

27; ;

64 8c c

c c

R T RTRT aP a b

V b V P P= − = =

−

Since this is a mixture, we need to calculate the a and b parameters for the mixture as follows:

2 2

2 2

5

2 2

5

55

5

2

27 8.314 304.20.36550

64 73.83 1027 8.314 190.6

0.2303564 45.99 10

0.2974

8.314 304.24.282 10

8 73.83 108.314 190.6

4.307 108 45.99 10

mix a a a b a b b b

mix a a b b

a

b

mix

a

b

a a y y y a a a y

b y b y b

a

a

a

b

b

−

= + +

= +

⋅ ⋅= =⋅ ×

⋅ ⋅= =⋅ ×

∴ =

⋅= = ×⋅ ×

⋅= = ×⋅ ×

5

54.294 10mixb

−

−∴ = ×

Derive the equation for residual enthalpy (for van der Waals) from first principles:

Tr (pseudo) Pr (pseudo) (HR)

0/RTc (H

R)

1/RTc H

R/RTc H

R

J/mol

State 1 1.1751 0.0500 -0.0375 -0.0225 -0.04027 -83.79

State 2 1.3001 2.0000 -1.56 -0.178 -1.58195 -3291.27


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( , )

under isothermal conditions ( 0)

( )

T V

VR ig

TV

H H V T

H HdH dV dT

V T

HH H H H dV dT

V

H U PV

dH dU d PV

dW dQ PdV VdP

PdV TdS PdV VdP

VdP TdS

=∞

=∂ ∂ = + ∂ ∂

∂ = − = ∆ = = ∂

= += += + + += − + + += +

∫

Maxwell

since represents an ideal gas conditionV

R

T TV

V

T VV

P SH V T dV V

V V

P PV T dV

V T

=∞

=∞

∂ ∂ ∴ = + = ∞ ∂ ∂

∂ ∂ = + ∂ ∂

∫

∫123

Now determine the partial derivatives from van der Waals, and then integrate:

( )

( )

( ) ( )

2

23

22

van der Waals

2

2

2ln ln

2

2

T

V

VR

V

V

V

R

RT aP

V b V

P a RT

V V V b

P R

T V b

a VRT TRH dV

V V bV b

a bRT V b RT V b

V V b

a RTb

V V b

RTb aH

V b V

=∞

∞

∞

= −−

∂ = − ∂ −

∂ = ∂ −

∴ = − +

−−

= − − − − + − −

= − + −

∴ = −−

∫

For each state, calculate the specific volume (V) from the van der Waals equation by iteration,

while using bmix and amix. Then apply the equation above (in block) to calculate the residual

enthalpy for each state.

( ) ( )

2

1

31

32

1

2

1 2

0.007981 m /mol

0.000149 m /mol

61.3 J/mol

2901.3 J/mol

61.3 2901.3 1156.9

1683 J/mol

R

R

TR R igPT

V

V

H

H

H H H C dT

=

=

= −

= −

∆ = − + +

= − − + − +

= −

∫


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(c) Comparison

There is a 22% difference between the answers from the Lee-Kesler and van der Waals

calculations. Van der Waals is the most simple, non-ideal equation of state and generally not

very accurate. One would expect the Lee-Kesler equation, which is a 4-parameter virial equation

of state, to be notably more accurate. Furthermore, one has to remember that the mixing rules,

as applied here, may not be very accurate. In short, approximately 2000 J/mol energy needs to

be removed from the gas mixture.

07 residual propeties

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