about the program
DESCRIPTION
visual basic program documentationTRANSCRIPT
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ChE 122 Christina May Tolentino
Michael Vicente
Eduardo Dungo Jr.
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USER MANUAL
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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.
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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.
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EQUATIONS
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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
=
=
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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)
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SAMPLE SOLUTION
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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)
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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
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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
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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
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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.
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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:
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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
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REFERENCES
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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