1 of 5
Similar to the "Continuous Mixing Example" on Page 7-18, except in this problem
you are solving for n2 and n3 given T3, instead of solving for T3 and n3 given n2.
Complete the numerical solution using E-Z Solve.
gmol/h
v07.08.04
© 2007, Michael E. Hanyak, Jr., All Rights Reserved
Page 7-23
2 of 5
Exact same math model used in the "Continuous Mixing Example" on Page 7-18.
Equation 7.5d-a with no
heat-of-reaction, heat,
and shaft work terms
Equations 1, 2, and 3 are linear;
each term in each equation has one
unknown variable raised to the 1
power.
Equations 4, 5, and 6 are nonlinear;
at least one term in each equation
has an unknown variable times an
unknown variable.
Equation 9 is nonlinear because of
the functional form "hmix[ ... ]".
In a NSOLVE construct, at least one
equation must be nonlinear. If all
equations were linear, than a linear
SOLVE construct would be used.
v07.08.04
© 2007, Michael E. Hanyak, Jr., All Rights Reserved
Page 7-24
Select the "click here" web link at the bottom of this page to view an Excel
"EZ Setup" math model that replaces the "E-Z Solve" model given below.
3 of 5
/* Quest P4.W1W.B - Simultaneous Material and Energy Balance Example */
CHEG 200, Michael Hanyak, October 27, 2005
E-Z SOLVE solution using the math model above
and equations for the three "hmix" functional forms.
Mathematical Model
/* total */
n1 +
/* pt = pentane */
/* hx = hexane */
= 0
0.40*n1 + 0.60*n2 - n3pt = 0
0.60*n1 + 0.40*n2 - n3hx = 0
/* pt mol frac */
/* hx mol frac */
/* energy bal'n */
n2 - n3
n3pt = n3*x3pt
n3hx = n3*x3hx
n1*h1 + n2*h2 - n3*h3 = 0
/* molar enthalpies h1, h2, and h3 are in kJ/gmol */
/* S1 mixture */
/* S2 mixture */
h1 = 0.40*dHhat("n-Pentane_l",Tpt,T1) + 0.60*dHhat("n-Hexane_l",Thx,T1)
h2 = 0.60*dHhat("n-Pentane_l",Tpt,T2) + 0.40*dHhat("n-Hexane_l",Thx,T2)
/* S3 mixture */
h3 = x3pt*dHhat("n-Pentane_l",Tpt,T3) + x3hx*dHhat("n-Hexane_l",Thx,T3)
/* Givens: */
n1
T1
T2
T3
/* Ref. Temp: */
/* Ref. Temp: */
Tpt = 20
Thx = 20
=
=
=
=
45
20
40
31
//
//
//
//
gmol/h
C, inlet Stream 1
C, inlet Stream 2
C, outlet Stream 3
// reference state for pure pentane:
// reference state for pure hexane:
E-Z Solve Numerical Solution
T1
T2
T3
20
40
31
C
C
C
n2
n3
57.128
102.128
gmol/h
gmol/h
n3hx
n3pt
49.8512
52.2768
gmol/h
gmol/h
x3hx
x3pt
0.488125
0.511875
mol frac
mol frac
h1
h2
h3
0
3.7524
2.099
kJ/gmol
kJ/gmol
kJ/gmol
Tpt, C & Ppt = 1 atm
Thx, C & Phx = 1 atm
In this problem (P4.W1M.B), you are solving for n2
and n3 given n1 = 45 gmol/h and T3 = 31°C. The
calculated values for the two unknown total flow rates
are shown here for this problem.
In the problem on Page 7-20 (P4.W1M.A), you are
solving for T3 and n3 given n1 = 45 gmol/h and n2 = 55
gmol/h. The calculated values for T3 = 30.8092°C
and n3 = 100 gmol/h.
If you were to change T3 = 30.8092°C in the math
model on this page, you would get these values:
n2 = 54.9885 ≈ 55 gmol/h
n3 = 99.9885 ≈ 100 gmol/h
which are the answers for the problem on Page 7-20.
Click here to view the Excel "EZ Setup" mathematical model in Worksheet "P4.W1M.B".
v07.08.04
© 2007, Michael E. Hanyak, Jr., All Rights Reserved
Page 7-25
4 of 5
Modify original problem on Page 7-23 by
adding the sentence: "The third stream
contains 51 mole percent of n-pentane."
Complete the numerical solution using E-Z Solve.
0.51
0.49
for two possible math models.
An example of a problem being over specified; that is, too much
information is given. In this case, the composition of Stream 3.
With this over specification, you could solve for the unknown flows
in Streams 2 and 3, one of two ways:
1. using the total material and energy balances, or
2. using the two component material balances.
The E-Z Solve model on the next page presents these two solutions.
v07.08.04
© 2007, Michael E. Hanyak, Jr., All Rights Reserved
Page 7-26
Select the "click here" web link at the bottom of this page to view an Excel
"EZ Setup" math model that replaces the "E-Z Solve" model given below.
5 of 5
/* Quest P4.W1W.C - Simultaneous Material and Energy Balance Example */
CHEG 200, Michael Hanyak, October 27, 2005
First Way to Solve the Problem
Mathematical Model
Solve for n2 and n3 using total material and energy balances only
/*
/* total */
n1
+ n2
- n3
/* energy bal'n */
n1*h1 + n2*h2 - n3*h3 = 0
*/
= 0
/* molar enthalpies h1, h2, and h3 are in kJ/gmol */
/* S1 mixture */
/* S2 mixture */
h1 = 0.40*dHhat("n-Pentane_l",Tpt,T1) + 0.60*dHhat("n-Hexane_l",Thx,T1)
h2 = 0.60*dHhat("n-Pentane_l",Tpt,T2) + 0.40*dHhat("n-Hexane_l",Thx,T2)
/* S3 mixture */
h3 = x3pt*dHhat("n-Pentane_l",Tpt,T3) + x3hx*dHhat("n-Hexane_l",Thx,T3)
/* Givens:
n1
T1
T2
T3
*/
=
=
=
=
45
20
40
31
//
//
//
//
gmol/h
C, inlet Stream 1
C, inlet Stream 2
C, outlet Stream 3
x3pt = 0.51
x3hx = 0.49
/* Ref. Temp: */
/* Ref. Temp: */
Tpt = 20
Thx = 20
// reference state for pure pentane:
// reference state for pure hexane:
E-Z Solve Numerical Solution
n2
n3
= 57.1914
= 102.191
Tpt, C & Ppt = 1 atm
Thx, C & Phx = 1 atm
similar answers to values on Page 7-25 for Problem P4.W1W.B
gmol/h
gmol/h
from total material and energy balances
Second Way to Solve the Problem
Solve for nc2 and nc3 using two component balances only
/*
/* pt = pentane
/* hx = hexane
*/
*/
0.40*n1 + 0.60*nc2 - x3pt*nc3 = 0
0.60*n1 + 0.40*nc2 - x3hx*nc3 = 0
E-Z Solve Numerical Solution
nc2
nc3
=
55
= 100
*/
This solution is fortuitous
since the effect of T3 is
not part of the solution.
similar values to those on Page 7-20 for Problem P4.W1W.A
gmol/h
gmol/h
from two component balances ONLY
Click here to view the Excel "EZ Setup" mathematical model in Worksheet "P4.W1M.C".
v07.08.04
© 2007, Michael E. Hanyak, Jr., All Rights Reserved
Page 7-27