1.
If we let propane be component 1, and let butane be component 2 then
the equations for the system are as follows:
(1) P1* = exp((11(1 - (TB1/T)))
(2) P2* = exp((11(1 - (TB2/T)))
(3) Py1 = x1P1*
(4) Py2 = x2P2*
(5) y1 + y2 = 1
where:
Pi* = vapour pressure of component i
TBi = atmospheric boiling point of component i
T = temperature
yi = vapour mole fraction of component i
xi = liquid mole fraction of component i
P = pressure
The unknowns here are: P1*, P2* y1, y2 and P. There are five equations and five unknowns, therefore in principle the system may be solved.
TB1, TB2 and T have been given. We are told that there is an equilmolar mixture of the species. As we are calculating the bubble point, this means that there is only liquid present. Thus x1 = x2 = 0.5
The incidence table for the above set of equations is:
| Equation | P1* | P2* | y1 | P | y2 |
| (1) | 1 | ||||
| (2) | 1 | ||||
| (3) | 1 | 1 | 1 | ||
| (4) | 1 | 1 | 1 | ||
| (5) | 1 | 1 |
Looking at this incidence table, it should be obvious that it will not be possible to rearrange this into a lower triangular form. There is a head, equations (1) and (2), and a partition, equations (3), (4) and (5). The spreadsheet solver may only solve equations arranged into a lower triangular format, thus some more work must be done.
There are two methods for solving this. The first involves rewriting equations (3) and (4) using partial pressures instead of vapour mole fractions and pressure. This would leave the set of equations as used in the revision section 1.1.2.2. This would mean extra manipulation to answer the question. This approach will be left as an exercise for the reader.
The other approach is to define a tear variable, and to give a value to one of the unknowns. The "solver" function in the spreadsheet may then be used to obtain the result. This approach will be used here.
The model for the spreadsheet may be viewed here, and is used to generate this spreadsheet.
model question1 parameter Tb1 = 231 !1 atm boiling point of propane Tb2 = 272.6 !1 atm boiling point of butane x1 = 0.5 !liquid mole fraction of propane x2 = 0.5 !liquid mole fraction of butane T = 293 !temperature of system P = 1 !pressure of system end parameter variable P1star !vapour pressure of propane P2star !vapour pressure of butane y1 !vapour mole fraction of propane y2 !vapour mole fraction of butane tear !tear varible end variable equation P1star = exp(11*(1-(Tb1/T))) P2star = exp(11*(1-(Tb2/T))) y1 = x1*P1star/P y2 = x2*P2star/P tear = y1 + y2 - 1 end equation steadystate end model question1
In order to have solved the set of equations, the tear variable must be equal to zero. This is achieved by adjusting the pressure. It may either be altered manually, or the solver function of the spreadsheet (if present) may be used. On Excel this may be found under the Tools tab.
Using this gives the following results:
Pressure = 6.2 bar
y1 = 0.827
y2 = 0.173
2.
This situation is very similar to the previous question. This time however the vapour composition has been specified. The equation set is:
(1) P1* = exp((11(1 - (TB1/T)))
(2) P2* = exp((11(1 - (TB2/T)))
(3) Py1 = x1P1*
(4) Py2 = x2P2*
(5) x1 + x2 = 1
Puting this into the spreadsheet modelling format gives the following model:
model question2 parameter Tb1 = 231 !1 atm boiling point of propane Tb2 = 272.6 !1 atm boiling point of butane y1 = 0.5 !vapour mole fraction of propane y2 = 0.5 !vapour mole fraction of butane T = 293 !temperature of system P = 1 !pressure of system end parameter variable P1star !vapour pressure of propane P2star !vapour pressure of butane x1 !liquid mole fraction of propane x2 !liquid mole fraction of butane tear !tear varible end variable equation P1star = exp(11*(1-(Tb1/T))) P2star = exp(11*(1-(Tb2/T))) x1 = P*y1/P1star x2 = P*y2/P2star tear = x1 + x2 - 1 end equation steadystate end model question2
generating this spreadsheet.
Following the same procedure as in question 1, the following answers were obtained:
Pressure = 3.56 bar
x1 = 0.173
x2 = 0.827
3.
The equation set here is the same as for question 1:
(1) P1* = exp((11(1 - (TB1/T)))
(2) P2* = exp((11(1 - (TB2/T)))
(3) Py1 = x1P1*
(4) Py2 = x2P2*
(5) y1 + y2 = 1
However in this case the pressure is given, and the temperature is an unknown. The incidence table for the equation set is:
| Equation | P1* | T | P2* | y1 | y2 |
| (1) | 1 | 1 | |||
| (2) | 1 | 1 | |||
| (3) | 1 | 1 | |||
| (4) | 1 | 1 | |||
| (5) | 1 | 1 |
Tearing P enables the set to be solved. The spreadsheet model used to do this is here.
model question3 parameter Tb1 = 231 !1 atm boiling point of propane Tb2 = 272.6 !1 atm boiling point of butane x1 = 0.5 !liquid mole fraction of propane x2 = 0.5 !liquid mole fraction of butane T = 293 !temperature of system P = 2 !pressure of system end parameter variable P1star !vapour pressure of propane P2star !vapour pressure of butane y1 !vapour mole fraction of propane y2 !vapour mole fraction of butane tear !tear varible end variable equation P1star = exp(11*(1-(Tb1/T))) P2star = exp(11*(1-(Tb2/T))) y1 = x1*P1star/P y2 = x2*P2star/P tear = y1 + y2 - 1 end equation steadystate end model question3
This generates this spreadsheet, which after using the solver, gives the answers:
Temperature = 260 K
y1 = 0.853
Y2 = 0.147
4.
The equations here are:
(1) P1* = exp((11(1 - (TB1/T)))
(2) P2* = exp((11(1 - (TB2/T)))
(3) P = P1* + P2*
(4) Py1 = P1*
(5) Py2 = P2*
Unknowns: P1*, T, P2*, y1, y2
There are five equations and five unknowns, therefore in principle the set may be solved.
The incidence table for the set is:
| Equation | P1* | T | P2* | y1 | y2 |
| (1) | 1 | 1 | |||
| (2) | 1 | 1 | |||
| (3) | 1 | 1 | |||
| (4) | 1 | 1 | |||
| (5) | 1 | 1 |
model question4 parameter Tb1 = 373 !1 atm boiling point of water Tb2 = 309.2 !1 atm boiling point of pentane T = 293 !temperature of system P = 2 !pressure of system end parameter variable P1star !vapour pressure of water P2star !vapour pressure of pentane y1 !vapour mole fraction of water y2 !vapour mole fraction of pentane tear !tear varible end variable equation P1star = exp(11*(1-(Tb1/T))) P2star = exp(11*(1-(Tb2/T))) y1 = P1star/P y2 = P2star/P tear = P1star + P2star - P end equation steadystate end model question4The spreadsheet generated using this model gives as the azeotrope conditions the following:
Temperature = 326.5 K
y1 = 0.104
y2 = 0.896
5.
If we let butane be component 1 and pentane component 2, then the equation set here is:
(1) Py1 - x1P1* = 0
(2) Py2 - x2P2* = 0
(3) x1 + x2 - 1 = 0
(3) y1 + y2 - 1 = 0
Both P, P1* and P2* are known. Therefore the equation set is linear in the mole fractions. Solving this set gives the following values:
x1 = 0.638
x2 = 0.871
y1 = 0.362
y2 = 0.129
Next - Section 3.7.1.4: Flash Calculation Problems
Return to Section 3.7.1.2: Questions
Course Organiser Last Modified 2/9/00