Mass Balance Examples4

1

Example:

An aqueous potassium nitrate (KNO3) solution containing 60wt% KNO3 at 80oC is fed to a crystalliser where the t t i d d t 40oC Wh t % ftemperature is reduced to 40oC. What % of the KNO3 in the feed forms crystals ?

Data: solubility of KNO3 in H2O at 40oC = 63 kg KNO3 /100kg H2O.

Draw a diagram:

CrystalliserFeed

crystals

Sat’d solutionBASIS:

100 kg feed

(kg)100400: (kg) 10060.0:3

WOHPCrKNO

=×+=×

Now to balances:steady-state, no generation, consumption or accumulation.

IN = OUT

P10063 solubility

:relationsother

(kg) 10040.0:2

W

WOH

=

=×

Solving equations:

W = 40 kg

3 equations and 3 unknowns

W 40 kg

From 3: P = 63x40/100 = 25.2 kg

From 1: Cr = 60-25.2 = 34.8 kg

% of entering KNO3 in the crystals =

KNO3 in crystals/KNO3 entering x 100%

(34.8 ÷ 60) x 100 = 58%

Example:

Strawberries contain about 15 wt% solids and 85 wt% water. To make jam, strawberries and sugar are mixed in the ratio 45:55 by mass, and the mixture is heated to evaporate water until themixture is heated to evaporate water until the residue (jam) contains 1/3 of water by mass.

What amount of strawberries produces 1 kg of jam?

Draw a diagram:

Define a BASIS:

Solution:

i) Define our components - we have 3 materials:

Strawberry solids (St)

( )Water (W)

Sugar (Su)

ii) Draw a flowchart:

341

Sugar Water

3

52Crusher Evaporator

Strawberries Jam

2

JAM PROCESS

iv) Write your equations with simplifying assumptions:

Assumptions: No components are generated, consumed or accumulate in any unit operation: Steady-state non-reactive process

∴ Generation = accumulation

= consumption = 0

Mass Balance: IN = OUT

∴ Input = output (steady-state)

Now, equations about system:

(kg)1W: StSuTOTAL ++

St45esstrawberri:relationsother

(kg) 1 15.0:

(kg) 131 85.0:

(kg) 1W :

solids

2

xStSolids

WStOH

StSuTOTAL

×=×

×+=×

+=+

55sugar Su

==

4 equations and 4 unknowns: solvable!!!

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 amounts as we have answered the question!!amounts as we have answered the question!!

I.e. Efficient solving for only what is required !!

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 packing was initially dry and had acolumn packing was initially dry 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.

3

Solution:

i) Our components in this case are water (w) and bone dry air (BDA).

ii) Draw a diagram:

Dryer

Dried Air: stream 2

Wet Air: stream 1

Wateraccumulates

i hi hi

Labeling stream variables:

nBDA,i = no of moles of bone dry air in stream ‘i’

nw,i = number of moles of water in stream ‘i’

within this system

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 (mole) …. Eq.1

For the water, there is accumulation over time in the dryer (call this n ) :over time in the dryer (call this nacc) :

ACCUMULATION = IN - OUT

H2O: nacc = nw, 1- nw,2 (mole)…. Eq.2So far 5 unknowns and only 2 equations: need 3 more

Other relations:

4 ....Eq. (mole) n 0.97 n

3 ....Eq. nn

n

w,1acc

BDAw

w

=

=+

04.01,1,

1,

( )

unknowns)5andequations(5 solutionthe to Now

hours 5 of basis a From mol 7.78

....Eq.5 (mole) 18

1000 3.40 - 3.54 nacc

∴=

=

mol 8.02 0.977.78 n :4 into subbing

unknowns) 5 and equations (5

w,1

=

=

From equation 2

8.02 = 7.78 + nw,2

⇒ nw,2 = 0.24 mol

From equation 3

ishour per gas of ratefeed Molar mol 192.5 n mol 192.5 n

8.02 0.04)-(1 n 0.04

0.04 02.8

02.8

BDA,2

BDA,1

BDA,1

BDA,1

∴

=

=⇒

×=⇒

=+ n

0.12% 100 192.50.24

0.24

streamoutlet of ncompositioWater

mol/hr 40.1 5

8.02 192.5

pg

=×+

=

=+

4

DEGREES OF FREEDOM

In all these problems we had to have the same number of independent equations as unknowns

To date - Material balances performed on one or two unit operations with only a few components.

- Enough information is provided to solve problems.

I l i d t i d th i tIn real industries and the environment

- Tens, hundreds, even thousands of unit operations and different components.

How is it possible to know whether all the information is there to solve our material balances?

Degrees of Freedom (DOF)

Simple mathsSimple maths

Two unknown variables (x,y) with two independent equations ( 1 , 2 ). Solving gives:

2 ______ 1 y - x 1 ______ 5 y 2x

==+

gives:x = 2, y = 1. This is the only solution to the equations.

We say DOF = 0

[same number of equations as unknowns]

Another example:

2 z2 ________ 11 y -4x 2z1 ________ 9 y 3x z

==+=++

3 variables (x, y, z)

2 independent equations ( 1 , 2 )

1 specified variable (z).

Solving gives: z = 2, x = 2, y = 1

DOF = 0; x, y, z are unique; , y, q

Another example:

z + 3 x + y = 9 ________ 1

2z + 4x – y = 11 ________ 2

3z + 7x = 20 ________ 3

Th i bl ( ) lThree variables (x, y, z) only twoindependent equations (equation 3 = 2 + 1 )

Must have independent equations

∴We cannot obtain an explicit unique∴We cannot obtain an explicit unique 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

5

2 ______ 1 y - x 1 ______ 5 y 2x

==+

Another Example:

x = 1

From 1 , x = 1 ⇒ y = 3

2 , x = 1 ⇒ y = 0

Not possible – this problem is over specified. Need either:

x not specified

or

One less equation

DOF = -1

In summary:

DOF = Nv – Ne

Nv = Number of unknown variables

Ne = Number of independent equations

If DOF > 0

We need to find either:

extra equations or

extra variables

Look at last example: Absorption of water in a dryeri) Components: Water (w) and bone dry air

(BDA).

ii) Flow chart

WaterWater accumulates

DryerDryer AirWet Air

1 2

iii) Variables:

i'' streamin dry bone of moles n i BDA, =

bedin accumulatet water thaof moles ni'' streamin water of moles n

wacc,

i w,

=

=

Total number of variables = 5

(nBDA,1, nBDA,2, nw,1, nw,2, nacc)

iii) Mole balance equations:

nBDA, 1 = nBDA, 2 _____ 1

nw, 1 = nacc + nw,2 _____ 2

iv) Other equations:

5 ____ (kg) 14.0n4 ____ n 0.97 n

3 ____ 04.0

acc

w,1acc

1,1,

1,

=

=

=+ BDAw

w

nnn

Total number of independent equations = 5

Therefore: DOF = 5 - 5 = 0

6

Another Example:

If the percentage of fuel in a fuel/air mixture falls below the lower flammability limit (LFL) then the mixture cannot ignite.

For example the LFL for propane (C H )For example, the LFL for propane (C3H8) is 2.05 mole % C3H8

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 ignited.g ted.

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 possible. If propane flows at a rate of 150possible. If propane flows at a rate of 150 mol C3H8/s in the fuel gas, what is the minimum molar flow rate of the dilution air?

Flow Diagram

MixerFuel Gas

4.03% C3H8

Diluted gas

2.05% C3H8

31

2

Dilution Air

1 ____ 3 ,HC1 ,HC 8383nn =

Material Balance Equations (mole):

equationst independen are 2nly :Note3 ____

2 ____

3 Tot,2 air,1 Tot,

3 air,2 air,1 air,

Onnn

nnn=+

=+

No. components = no. independent equations

Will perform material balances using equations 1 and 3.

Other relations:

502050

4 ____ 0403.0 1 Tot,1 ,HC 83

nn

nn

=

=

6 ____ / 150

5____ 0205.0

1 ,HC

3 Tot,3,HC

83

83

smoln

nn

=

=

∴ Number of variables is2(stream 1) + 1(stream 2) + 2(stream 3) = 5 = Nv

Number of independent equationsp q

Ne = Eq’ns 1 , 3 , 4 , 5, 6 = 5

DoF = 5-5 = 0 Problem can be solved !!!

7

1from

n 0 150 3 ,HC 83=+

Solution

mol/s7320150n

5 from

mol/s 3720 0.0403

150 n

4 from

mol/s 150 n

3T

1 Tot,

3 ,HC 83

==

==

=

air dilution mol/s 3600 3722 - 7317 n

3 from

mol/s 7320 0.0205

n

2 Tot,

3 Tot,

=

=

Degrees of Freedom – Another example

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.

Th b tt d t t i 98 0 l %The bottoms product contains 98.0 mole% ‘X’ and no ‘B’.

96.0% of the ‘X’ in the feed is recovered in this the bottoms product.

The overhead product is fed to a second column. The overhead product from the second column contains 94% ‘B’ and the balance ‘T’balance T .

97% of ‘B’ entering the second column is recovered in the overhead of the second column.

Perform the DOF analysis to see whether this problem can be solved ??

Flow Diagram

Tower 2

4

B, T, X

0.94 B0.06 T

2

Tower 1

3

1 2

0.30 B0.25 T0.45 X

B, T, X5

0.98 X0.02 T

What are the variables:

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

20 variables in the table

But we know some of these already: 9 are known

Unknown variables = 20 - 9 = 11

8

Unit Mass balance Equations

Tower 1 3 equations

Tower 2 3 equations

What are the equations ?? :

∑∑==

==3

15,

3

13, 1 1

ii

ii xx

Sum of the mole fractions in the unknown streams= 1

∴ 2 more equations

A few more relations we haven’t used yet:

[96.0% of the ‘X’ in the feed is recovered in this the bottoms product].0.96 x Total flow1 x Xx,1 = Total flow2 x Xx,2

[97% of ‘B’ entering the second column is recovered in the overhead of the second column] 0.97 x Total flow3 x XB,3 = Total flow4 x XB,4

∴ 2 more equations

Total no Equations = 6+2+2 = 10Total no. Equations 6+2+2 10

Total no. unknown Variables = 11

DoF = 1 ∴ we must specify one more variable

BASIS = 100 mole feed [now solvable!!]