Now to balances:steadystate, no generation, consumption or accumulation.
Example: An aqueous potassium nitrate (KNO3) solution containing 60wt% KNO3 at 80oC is fed to a crystalliser where the empera ure s re uce o . a o the KNO3 in the feed forms crystals ? o
Data: solubility of KNO3 in H2O at 40 C = 63 kg KNO3 /100kg H2O.
IN
= OUT
KNO3 : 0.60 × 100 = Cr + P 2
(kg)
=
.
3 equations and 3 unknowns
other relations : solubility
Draw a diagram:
63 100
=
P W
Solving equations: = Feed BASIS:
Crystalliser
Sat’d solution
100 kg feed
From 3: P = 63x40/100 = 25.2 kg From 1: Cr = 6025.2 = 34.8 kg % of entering KNO3 in the crystals =
crystals
KNO3 in crystals/KNO3 entering x 100% (34.8 ÷ 60) x 100
Example: Strawberries contain about 15 wt% solids and 85 wt% water. water. To make jam, strawberries and sugar are mixed in the ratio 45:55 by mass, and the residue (jam) contains 1/3 of water by mass. What amount of strawberries produces 1 kg of jam?
= 58%
Solution: i)
Defi Define ne our our comp compon onen ents ts  we hav havee 3 materials: Strawberry solids
(St)
Water
(W)
Sugar
(Su)
ii) ii) Draw Draw a flow flowch char art: t:
Draw a diagram:
Sugar 1 Strawberries
Water 4 Jam
2 Crusher
5 Evaporator
Define a BASIS:
1
∴ Input = output (steadystate)
Now, equations about system: JAM PROCESS =
: u
g
1 H 2O : 0.85 × St = W + × 1 3 Solids : 0.15 × St = xsolids × 1 iv) Write your equations with simplifying assumptions: Assumptions: No components are generated, consumed or accumulate in any unit operation:
Steadystate nonreactive process
(kg) (kg)
other relations : strawberri es sugar
=
45 55
=
St Su
∴ Generation = accumulation
= consumption = 0
4 equations and 4 unknowns: solvable!!!
Mass Balance: IN = OUT
Solve the equations: From 4: Su = 55St/45 Sub into 1: St + 55St/45 = W +1 Rearrange: W = 2.222St – 1 Sub into 2: 0.85St = 2.222St – 1 + 1/3 Solve for St = 0.4859 kg
There is no need to solve for the other
Another Example Wet air containing 4.0 mole % water vapor is passed through a column of calcium chloride pellets. The pellets absorb 97.0% of the water and none of the other constituents of air. The column ackin was initiall dr and had a mass of 3.40 kg. Following 5.0 hours of operation, the pellets are reweighed and found to have a mass of 3.54 kg.
Calculate the molar flow rate of the feed gas and the mole fraction of water vapor in the product gas.
I.e. Efficient solving for only what is required !!
2
Solution: i)
Our components in this case are water (w) and bone dry air (BDA).
ii) Draw a diagram:
Wet Air: stream 1
Dried Air: stream 2
Dryer
Note: in this problem we will take a BASIS: 5 hours of operation.
iii) Mole balances over the system: For the bone dry air: IN = OUT
Dry Air: nBDA, 1 = nBDA, 2 Water accumulates within this system
Labeling stream variables:
(mole) …. Eq.1
For the water, there is accumulation over time in the dr er call this nacc : ACCUMULATION = IN  OUT
nBDA,i = no of moles of bone dry air in stream ‘i’ nw,i = number of moles of water in stream ‘i’
H2O: nacc = nw, 1 nw,2
(mole)…. Eq.2
So far 5 unknowns and only 2 equations: need 3 more
From equation 2
Other relations:
nw,1 nw,1 + n BDA,1
8.02 = 7.78 + n w,2 = 0.04
....Eq. 3
nacc = 0.97 nw,1 nacc = (3.54  3.40 )
(mole) 1000 18
....Eq. 4
⇒ nw,2 = 0.24 mol
From equation 3 8.02
(mole) ....Eq.5
= 7.78 mol ∴ From a basis of 5 hours
8.02 + nBDA,1
= 0.04
⇒
0.04 n BDA,1 = (1  0.04) × 8.02
⇒
n BDA,1 = 192.5 mol n BDA,2 = 192.5 mol
Now to the solution
∴ Molar feed rate of gas per hour is
192.5 + 8.02
subbing into 4 :
nw,1 =
7.78
0.97 = 8.02 mol
= 40.1 mol/hr 5 Water composition of outlet stream =
0.24 0.24 + 192.5
× 100 = 0.12%
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DEGREES OF FREEDOM
In all these problems we had to have the same number of independent equations as unknowns
To date 

Degrees of Freedom (DOF)
Material balances performed on one or two unit operations with only a few components. Enough information is provided to solve problems.
n rea n ustr es an t e env ronment 
How is it possible to know whether all the information is there to solve our material balances?
Tens, hundreds, even thousands of unit operations and different components.
2x + y = 5
______ 1
x  y =1
______ 2
Two unknown variables (x,y) with two independent equations ( 1 , 2 ). Solving x = 2, y = 1. This is the only solution to the equations. We say DOF = 0 [same number of equations as unknowns]
Another example:
Another example:
z + 3x + y = 9
________ 1
z +3x+y= 9
________ 1
2z + 4x  y = 11
________ 2
2z + 4x – y = 11
________ 2
3z + 7x
________ 3
z
= 2
3 variables (x, y, z) 2 independent equations ( 1 , 2 )
= 20
ree var a es x, y, z on y two independent equations (equation 3 = 2 + 1 )
1 specified variable (z). Must have independent equations Solving gives: z = 2, x = 2, y = 1 DOF = 0; x, , z are uni ue solution for x, y, z. However, if 1 extra independent equation were available, or if one of x, y, z were specified, then we could solve: ∴ DOF = 1
4
In summary:
Another Example:
2x + y = 5
______ 1
x  y =1
______ 2
DOF = Nv – Ne Nv = Number of unknown variables
x=1
Ne = Number of independent equations
From 1 , x = 1 ⇒ y = 3 2 ,x=1⇒ y=0 Not possible – this problem is over specified. Need either:
If DOF > 0 We need to find either: extra equations or
x not specified
extra variables
or One less equation DOF = 1
Look at last example: Absorption of water in a dryer i)
Components: Water (w) and bone dry air (BDA).
ii) Flow chart
iii) Mole balance equations: nBDA, 1 = nBDA, 2
_____ 1
nw, 1 = nacc + nw,2
_____ 2
iv) Other equations:
nw,1
accumulates 1
2
Wet Air
Dryer Air Dryer
iii) Variables:
n BDA, i
= moles of bone dry in stream ' i'
n w, i
= moles of water in stream ' i'
n acc,w
= moles of water that accumulate in bed
Total number of variables = 5
nw,1 + n BDA,1
= 0.04
____ 3
n acc = 0.97 n w,1 n acc = 0.14
____ 4 (kg)
____ 5
Total number of independent equations = 5
Therefore: DOF = 5  5 = 0
(nBDA,1, nBDA,2, nw,1, nw,2, nacc)
5
Another Example:
A mixture of propane in air containing 4.03% C3H8 (fuel gas) is in the feed to a combustion furnace. If there is a problem in the furnace, a stream of pure air(dilution air) is added to the fuel mixture to make sure that ignition is not ossible.. If ro ane flows at a rate of 150 mol C3H8 /s in the fuel gas, what is the minimum molar flow rate of the dilution air?
If the percentage of fuel in a fuel/air mixture falls below the lower flammability limit (LFL) then the mixture cannot ignite. , is 2.05 mole % C3H8
3
8
If the % of propane in the propane/air mix is greater than 2.05% then the mixture can be ignited. If the % is less than 2.05% propane than the mixture cannot be i nited.
Flow Diagram
Will perform material balances using equations 1 and 3.
Fuel Gas
Diluted gas
1 4.03% C3H8
Mixer
3
2.05% C3H8
Other relations:
nC3H8 , 1 = 0.0403 nTot, 1
____ 4
.
____
C3 H8 , 3 2
Tot, 3
nC3H8 , 1 = 150 mol / s
____ 6
Dilution Air
Material Balance Equations (mole):
nC3H8 , 1 = nC3H8 , 3 nair, 1
nTot, 1 + nair, 2 = nTot, 3
Note :
2(stream 1) + 1(stream 2) + 2(stream 3) = 5 = N v
____ 1
+ nair, 2 = nair, 3
∴ Number of variables is
Number of independent equations
____ 2
Ne = Eq’ns 1 , 3 , 4 , 5, 6 = 5
____ 3
Only 2 are independent equations
DoF = 55 = 0
Problem can be solved !!!
No. components = no. independent equations
6
Degrees of Freedom – Another example
Solution from
A liquid mixture containing 30.0 mole % benzene (B), 25.0 mole % toluene (T) and the balance xylene (X) is fed to a distillation column.
1
150 + 0 = n C3H8 , 3
e ottoms pro uct conta ns ‘X’ and no ‘B’.
n C3H8 , 3 = 150 mol/s from
4
n Tot, 1 = from
Tot, 3 = from
150 0.0403 5 150 0.0205 3
96.0% of the ‘X’ in the feed is recovered in this the bottoms product.
= 3720 mol/s
The overhead product is fed to a second column. The overhead product from the second column contains 94% ‘B’ and the ‘ ’.
=
97% of ‘B’ entering the second column is recovered in the overhead of the second column.
n Tot, 2 = 7317  3722 = 3600 mol/s dilution air
Flow Diagram 4
B, T, X 3
0.25 T 0.45 X 1
Perform the DOF analysis to see whether this problem can be solved ??
What are the variables:
0.94 B 0.06 T
0.30 B
. mo e
Tower 1
Tower 2
B, T, X 2 0.98 X 0.02 T
Stream 1 Stream 2 Stream 3 Stream 4 Stream 5
Total flow1
Total flow2
Total flow3
Total flow4
Total flow5
xB,1
xB,2
xB,3
xB,4
xB,5
xT,1
xT,2
xT,3
xT,4
xT,5
xx,1
xx,2
xx,3
xx,4
xx,5
5
20 variables in the table But we know some of these already: 9 are known
Unknown variables = 20  9 = 11
7
What are the equations ?? : Unit
Mass balance Equations
A few more relations we haven’t used yet:
Tower 1
3 equations
[96.0% of the ‘X’ in the feed is recovered in this the bottoms product].
Tower 2
3 equations
0.96 x Total flow 1 x Xx,1 = Total flow2 x Xx,2
Sum of the mole fractions in the unknown streams= 1 3
3
∑ xi ,3 = 1
∑ xi ,5 = 1
i =1
i =1
∴ 2 more equations
[97% of ‘B’ entering the second column is recovered in the overhead of the second column] 0.97 x Total flow 3 x XB,3 = Total flow4 x XB,4
∴ 2 more equations
.
=
=
Total no. unknown Variables = 11 DoF = 1 ∴ we must specify one more variable
BASIS = 100 mole feed [now solvable!!]
8