ChE 122 Christina May Tolentino

Michael Vicente

Eduardo Dungo Jr.

USER MANUAL

SHEET 1:

SHEET 2:

STEP 1: Select the name of the pure species from the drop-down list.

STEP 2: Click the equation of state to be used in the calculation.

The instructions are also available in this button.

A table of properties at saturation, i.e. P, V, U, H, S, from the triple to critical point will be generated.

SHEET 3:

STEP 1: Select the name of the pure species from the drop-down list.

The instructions are also available in this button.

Button that clears all values in Sheet 3.

Properties at

saturation, i.e. P, V, U,

H, S, at the specified

temperature will then

be shown in the table.

STEP 2: Type the temperature in K.

STEP 3: Click the equation of state to be used in the calculation.

The allowed temperature range (Ttriple to Tc) for the selected pure species is shown here.

EQUATIONS

Wagner equation

Generic cubic equation of state

=

()

+ ( + )

= ()

22

=

Hlv at the normal boiling point

Riedel Method

=

1.092(ln 1.013)

0.930

Variation of Hlv with temperature

Watson relation

=

1 1

0.38

= /

From cubic equations of state

= 1 +

ln ()

ln

= ln +

ln ()

ln

=

=()

=

If =

1

+

+

If = =

+

van der Waals equation of state ln ()

ln = 0

Redlich/Kwong equation of state ln ()

ln =

1

2

Soave/Redlich/Kwong equation of state

ln ()

ln =

1/2

()1/2 0.480 + 1.547 0.1762

Peng/Robinson equation of state ln ()

ln =

1/2

()1/2 0.37464 + 1.54226 0.269922

=

Ideal-Gas Heat Capacity

Evaluation of the Sensible Heat Integral

= ( ) +1

2( ) + 1 +

1

3( )

2 2 + + 1

+1

4( )

3 3 + 2 + + 1

=

=

Entropy Changes due to Temperature

= ( )

+ ( ) +1

2( )

2 + 1 +1

3( )

3 2 + + 1 1

ln

=

Using the Riedel method and the Watson relation, a value of Hlv can be calculated. The vapor

pressure correlations can be also be used to estimate enthalpies of vaporization. From the equation

ln = Hlv

1

we can define a dimensionless group as

=Hlv

=

ln

(1

)

By differentiating the vapor pressure equation used, in this case the Wagner equation, we can

obtain an expressions for .

= + 0.5 0.5 1.5 + 2 2 3 + 5(5 6)

= 1

Solving for Eq. C and substituting this value to Eq. B together with a calculated value for Hlv

from the Riedel and Watson equations, we can solve for the unknown . Furthermore,

=

In the volume calculation for the cubic equation of state, at least one real root is always

calculated, corresponding to the vapor or vapor-like volume. can be directly calculated from

=

Therefore, we could easily solve for from Eq. D knowing and . By

manipulation of Eq. E we could arrive at a value for .

(Eq. D)

(Eq. C)

(Eq. E)

(Eq. B)

(Eq. A)

SAMPLE SOLUTION

Estimate P, V, U, H and S for saturated n-butane liquid and vapor at 200 K if H and S are set equal to

zero for saturated liquid at the triple point temperature. Use Soave/Redlich/Kwong equation of

state.

The data available are:

Molar Mass = 58.13

Tc = 425.1 K Pc = 37.96 bars = 0.200

Ttr = 134.6 K

Tn = 272.7 (normal boiling point)

Wagner Constants:

VP A= -6.88709 VP B = 1.15157 VP C = -1.99873 VP D = -3.13003

Cp, vap constants:

Cp, vap A = 9.487 Cp, vap B = 0.3313 Cp, vap C = -1.0001108 Cp, vap D = -2.822*10-9

The vapor pressure for the saturated n-butane at 200 K is calculated directly from the Wagner

equation.

ln(Pvp/Pc) = (1-x)-1[(VP A)x + (VP B)x1.5 + (VP C)x3 + (VP D)x6]

where x = 1-T/Tc

x=1-(300 K/425.1 K) = 0.29428

ln(Pvp/Pc) = (1-0.29428)-1[(-6.88709)(0.29428) + (1.15157)(0.29428)1.5

+ (-1.99873)(0.29428)3 + (-3.13003)(0.29428)6]

ln(Pvp/Pc) = -7.58570

Pvp = Pce-7.58570

Pvp = 2.58585 bar

Pvp = 258.585 kPa

The volume of the saturate n-butane liquid and vapor is calculated by solving explicitly for the three

roots of the generic cubic equation of state.

P = RT

V b

a(T)

V + b (V + b)

For the Soave/Redlich/Kwong equation of state, the parameter assignments are:

Tr, )= [1 + (0.480 + 1.574 0.1762)(1- Tr1/2)]2

= 1 = 0 = 0.08664 = 0.42748

For the reduced conditions,

Tr = T/Tc = 300 K/425.1 K = 0.70572

Tr, )={1+[0.480 + (1.574)(0.200) (0.176)(0.200)2](1- (0.70572)1/2)}2 = 1.26785

a T = Tr , R

2Tc2

Pc

a T = 0.42748 1.26785 83.14

cm 3bar

mol K

2

(425 .1 K)2

37.96 bar = 1.78344*107 cm6bar/mol2

b = RTc

Pc

b = 0.08664 83.14

cm 3bar

mol K (425.1 K)

37.96 bar = 80.66653 cm3/mol

Substituting into the generic cubic equation of state,

2.58585 bar = 83.14

cm3 barmol K (300 K)

V 80.66653 cm3/mol

1.78344x107 cm6bar/mol2

V (V + 80.66653 cm3/mol)

Solving for V, the equation exhibits three real roots

V1 = 8971.13673 cm3/mol

V2 = 564.57570 cm3/mol

V3 = 109.84452 cm3/mol

The smallest root is a liquid or liquid-like volume, and the largest us a vapor or vapor-like

volume. The third root, lying between the other values, is of no significance.

The specific volume of the saturated liquid and vapor can then be solved as:

V(sat. vap.) = 8971 .1376

c m 3

mol

58.123g

mol

V(sat. vap.) = 154.347 cm3/g

V(sat. liq.) = 109.84452

c m 3

mol

58.123g

mol

V(sat. liq.) = 1.88986 cm3/g

Reference state:

Satd liquid n-

butane at 134.6

K, 0.50598 bar

H(sat. liquid)

S(sat. liquid)

H(sat.vapor)

S(sat. vapor)

Final state of satd n-

butane liquid at 300 K,

2.58585 bar

Final state of satd n-

butane vapor at 300 K,

2.58585 bar

(a) Hlv

Slv

Hvl

Svl

(e)

(d) H2R

S2R

-H1R

-S1R

(b)

(c)

Hig Sig

Satd vapor n-butane

at 134.6 K, 0.05098

bar

n-butane in ideal gas-

state at 134.6 K,

0.50598 bar

n-butane in ideal gas-state

at 300 K, 2.58585 bar

Figure 1. Calculation Path

For H and S, use a calculational path leading from an initial state of saturated liquid n-butane at the

triple point temperature 134.6 K, where H and S are zero, to the final state of interest. In this case,

an initial vaporization step is required, leading to the four step path for saturated vapor and up to

the fifth step for the saturated liquid as shown in the figure. These steps are:

(a) Vaporization at T1 and P1 = Psat (b) Transition to ideal gas state at (T1,P1) (c) Change to (T2,P2) in the ideal gas state (d) Transition to the actual state at (T2,P2) (e) Condensation at T2 and P2 = Psat

It is also necessary to calculate P1, the saturation pressure at the triple point temperature. Using the

Wagner equation:

ln(Pvp/Pc) = (1-x)-1[(VP A)x + (VP B)x1.5 + (VP C)x3 + (VP D)x6]

x = 1-T/Tc = 1-(134.6 K/425.1 K) = 0.68337

ln(Pvp/Pc) = (1-0.68337)-1[(-6.88709)(0.68337) + (1.15157)(0.68337)1.5

+ (-1.99873)(0.68337)3 + (-3.13003)(0.68337)6]

ln(Pvp/Pc) = 1.33292*10-7

Pvp = Pc e1.33292 107 = (37.96 bar)e1.33292 10

7

Pvp = 0.50598 bar

Step (a): Vaporization of saturated liquid n-butane at 134.6 K. Here the latent heat of vaporization is

required. The equation proposed by Riedel provides an estimate at the normal boiling point.

Hnlv

RTn= 1.092(

ln Pc 1.013

0.930 Trn)

Where Tr,n = Tn/Tc = 272.7 K / 425.1 K = 0.64150

Hnlv

RTn= 1.092(

ln 37.96 1.013

0.930 0.64150)

Whence, Hnlv = (9.93019)(8.314 J/mol-K)(272.7 K) = 22514 J/mol

The latent heat at 134.6 K, or Tr = T/Tc = 134.6 K/ 425.1 K = 0.31663 is given by the method

proposed by Watson:

Hlv

Hnlv

= 1 Tr

1 Tr n

0.38

Or Hlv= (22514 J/mol) 10.31663

10.64150

0.38

=28768.28070 J/mol

By H= TS

Slv= Hlv/T = (28768.28070 J/mol)/134.6 K = 213.73165 J/mol-K

Step (b): Transformation of saturated vapor n-butane into an ideal gas at the initial conditions

(T1,P1). The values of H1R and S1R are estimated by the cubic equations of state

HR

RT= Z 1 +

dln Tr

dlnTr 1 qI

SR

RT= ln Z +

dln Tr

dlnTr qI

For the given conditions:

Tr = 0.31663

Pr = P/Pc = 0.50598 bar/ 37.96 bar = 0.013329

Tr , = { 1 + [0.480 + 1.574 0.200) (0.176)(0.200)2][1-0.316631/2]}2

= Pr

Tr =

0.08664 0.013329

0.31663= 0.0036473

q = Tr,

Tr=

0.42748 1.80765

0.08664 0.31663 = 19.75651

Preliminary to application of these equations, one must find Z by solution of Z =PV

RT using V for

vapor phase.

Using Soave/Redlich/Kwong equation of state, V of saturated vapor at the triple point was

determined to be

V (saturated vapor) = 2.21170*109 cm3/mol

Therefore Z = 0.50598 bar 2.21170

109c m 3

mol

83.14c m 3bar

mol K 134 .6 K

= 1.00000

Since 0

I =1

ln(

Z+

Z+)

I =1

10ln

1.00000 0.0036473

1.00000 = 3.64727 108

With

dln (Tr)

dlnTr=

Tr0.5

0.5(0.480 + 1.574 0.1762)

dln (Tr)

dlnTr=

0.316630.5

1.807650.5(0.480 + 1.574(0.200) 0.176(0.200)2)

dln (Tr)

dlnTr= 0.32970

Substitution of , q, =0, and =1 into the equations for HR and SR

HR

T= 1.00000 1 + 0.32970 1 19.75651 3.64727 108 = 2.35698 106

SR

T= ln 1.00000 0.0036473 + 0.32970 19.75651 3.64727 108 = 1.36608 106

Whence,

H1R = (-2.35698*10-6)(8.314 J/mol-K)(134.6K) = -2.63761*10-3 J/mol

S1R = (-1.36608*10-6)(8.314 J/mol-K) = 1.13576*10-5 J/mol

As indicated in figure 1, the property changes for this step are - H1R and - S1R, because the change is

from the real to ideal gas state.

Step (c): Changes in the ideal gas state from (134.6K, 0.50598 bar) to (300 K, 2.58585 bar). Here

Hig and Sig are given by:

Hig = H (T-Ttr)

H = A + BTtr +1 + 1/3 CTtr2 2++1 + DTtr3 3+2++1

Sig = HlnT

Tr R ln

P

Ptr

s = A + [BTtr + CTtr2 +1 +1/3 D Ttr3 2++1 ] 1

ln

Where = T

Ttr=

300 K

134 .6 K= 2.22883

H = 75.96417 J/mol-K

Hig = (75.96417 J/mol-K)(300 K 134.6 K) = 12564.47246 J/mol

s = 72.85996 J/mol-K

Sig = 72.85996J

mol K ln

300

134.6 8.314

J

mol K ln

2.58585

0.50598 = 50.88609

J

mol K

Step (d): Transformation of n-butane from the ideal gas state to the real gas state at T2 and P2. The

final reduced conditions are:

Tr = 0.705726

Pr = P/Pc = 2.58585/37.96 = 6.81205*10-2

Following the same procedure as in step (b),

H2R = -488.62093 J/mol

S2R = -1.06546 J/mol-K

Step (e): Sublimation of saturated n-butane vapor at 300 K. The latent heat at 300 K is calculated

similarly as in step (a)

Hlv = 20887.00416 J/mol

Whence, Slv = 69.62335 J/mol-K

Hvl =- Hlv =-20887.00416 J/mol

Svl =- Slv = -69.62335 J/mol-K

The sum of the enthalpy and entropy changes for the five steps give the total changes for the

process leading from the initial reference state (where H and S are set equal to zero) to the final

state of the saturated liquid is:

H = H = 2876028070 (-2.63761*10-3) + 12564.47247 488.62093 20887.00416

= 19957.13250 J/mol

S = S = 213.73165 (-1.36608*10-6) 50.88609 1.06546 69.62335

= 92.15677 J/mol*K

The specific enthalpy and entropy of the saturated liquid is:

H =19957 .13250 J/mol

58.123 g/mol

H = 343.630 kJ/kg

S =92.15677

J

mol K

58 .123g

mol K

S = 1.58555 kJ/kg*K

The internal energy is

U= H PV

U = 342.872 kJ/kg

H and S for saturated vapor is calculated by summing up the enthalpy and entropy change,

respectively, for the first four steps.

H = 702.719 kJ/kg

S = 2.78341 kJ/kg*K

Also, the internal energy is

U = 662.807 kJ/kg

REFERENCES

Molar mass, , Tc, Pc, Tn

J. M. Smith, H. G. Van Ness, and M. M. Abbott, Introduction to Chemical Engineering Thermodynamics, 7th ed., App. B, Table B.1, McGraw-Hill, New York, 2005

Wagner and Cp,vap constants

R. C. Reid, J. M. Prausnitz, and B. E. Poling, The Properties of Gases and Liquids, 4th ed., App. A, McGraw-Hill, New York, 1987

Ttr (triple point temperature)

http://webbook.nist.gov/chemistry/name-ser.html

Volume, Enthalpy and Entropy of Vaporization, Residual Properties, Internal Energy

J. M. Smith, H. G. Van Ness, and M. M. Abbott, Introduction to Chemical Engineering Thermodynamics, 7th ed., App. B, Table B.1, McGraw-Hill, New York, 2005

Vapor Pressure, Ideal-Gas Enthalpy and Entropy, Adjustment for imaginary values of the volume

R. C. Reid, J. M. Prausnitz, and B. E. Poling, The Properties of Gases and Liquids, 4th ed., App. A, McGraw-Hill, New York, 1987