Chap_4-_mat_bal_03-1

Mass and Energy Balance Lecture Series

Dr. Raja Razuan Raja DerisFaculty of Applied Science, UiTM, Shah Alam

Office: 03-55444604 email: [email protected]; [email protected]

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You do not learn much just sitting in classes listening to teachers, memorizing prepackaged assignments, and spitting out answers. You must talk about what you are learning, write reflectively about it, relate it to past experiences, and apply it to your daily lives. You must make what you learn part of yourselves.”

-Source:"Implementing the Seven Principles: Technology as Lever" by Arthur W. Chickering and Stephen C. Ehrmann

“Learning is not a spectator sport.

“Education is the kindling of a flame, not the filling of a vessel” - Socrates.

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Objectives/Intended Learning Outcome:

Learning

1. Explain in your own word the meaning of the following terms: batch, semi batch, continuous, transient and steady-state processes.

2. Explain the following process terms: recycle, purge, by-pass, limiting reactant and combustion reaction.

3. Draw and fully label a flowchart based on given process description.

4. Solve a simple material balance calculations.

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CHAPTER

3

FUNDAMENTALS OF MATERIAL BALANCES:DIFFERENTIAL MATERIAL BALANCE

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Material Balance- Step-by-Step


EXAMPLE 3: Batch mixing processThe house special at Oswald’s Oasis is a mixture containing 75 % C2H5OH bymass, and the balance is water. The cost of alcohol has been increasing, however,Oswald has decided that perhaps a 60 % blend would be just as effective. He hason hand vat containing 350 gallons of the 75 % mixture (SG = 0.877) and canpurchase any desired amount of a 40 % mixture (SG = 0.952). How many gallonsof the latter mixture (40 % blend) must he buy?

SOLUTION:

Draw and label the flow chart diagram:Since material balance cal. Can only be performed as mole or mass balance, we need to change V = 350 gal to mass.

3

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mixingprocess

P Ibm 60% blend sol.

0.60 C2H5OH0.40 Water

2560.16 Ibm 75% blend sol.(V = 350 gal, SG = 0.877

B Ibm 40% blend sol.(V = ?, SG = 0.877)

0.40 C2H5OH 0.60 Water

0.75 C2H5OH 0.25 Water

Known basis: 2650.16 Ibm 75% blend solution.

DOF

Number of unknowns = 2 (B, P)Number of independent equations = 2 (2 material balances + 0 information)∴ Unknown can be solved.



Total mass balance: input = output2560.16 + B = P ---------------(1)

Component balance:Bal. on alcohol: 0.75(2560.16) + 0.40(B) = 0.60(P)Divided by 0.60: 3200.20 + 0.667(B) = P ----------(2)(1) – (2): -640.04 + 0.333(B) = 0∴B = 1922.04 Ibm

Substitute B in equation (1): 2560.16 + 1922.04 = 4482.20

∴ P = 4482.20 Ibm

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Converting B to volume: . . . /

. .

. % .

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DIY


1. Vinegar with a strength of 4.63% (by weight) acetic acid is mixed with 1000 kg of 36.0% acetic acid solution to produce a mixture of 8.5% acid. How much of this 8.5% acid solution is present?

2. Fish cake which contains 80% water and the remainder is dry cake, is being dried in a rotary drum dryer in which 100 kg of water is removed. It is found that the fish cake is then 40% water. Calculate the mass of the fish cake originally put into the dryer.

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DIY


3. A concentrated sulfuric acid solution containing 20 wt.% sulfuric acid is distilled water to produce a solution containing 5 wt.% of acid. Determine how much the distilled water and the concentrated solution must be mixed to produce 600 Ibm of the diluted solution.

4. A cellulose solution contains 5.2 wt.% cellulose in water. How many kilograms of 1.2% solution are required to dilute 100 kg of 5.2 % solution to 4.2 %.

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DIY


5. If 100 g of Na2SO4 is dissolved in 200 g of water and the solution is cooled until 100 g of Na2SO4.10H2O crystallised out, determine:

a) The composition of the remaining solution (mother liquor).

b) The grams of crystals recovered per 100 g of initial solution.

6. 85 Ibmol gas mixture inside a tank consist of 80 % O2, 15 % CO2 and N2. The mixture is then diluted with air in order to reduce the concentration of O2 to 60%. Determine the amount of air needed and the final composition of the mixture after the dilution takes place. Assume that air consists of 21% O2 and 79% N2.

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DIY


7. A drilling mud consists of 60.0% water and 40.0% special clay. The driller wishes to increase the density of the mud and a curve shows that 48% water will give the desired density. Calculate the mass of bone-dry clay that must be added per metric ton of the original mud to give the desired composition.

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MATERIAL BALANCES


DIFFERENTIAL MATERIAL BALANCES ON ASTEADY-STATE BATCH PROCESS WITH NOCHEMICAL REACTIONS:Recall:

Continuous process- is a process where all input andoutput process streams cross into and out of theprocess system boundary continuously.

A process is at steady state when the values of all ofits process variables do not change with time exceptfor very minor fluctuations.

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MATERIAL BALANCES


Therefore a steady-state continuous process is aprocess where all input and output process streamscross into and out of the process system boundarycontinuously and the values of all of its processvariables do not change with time. For this type ofprocess, differential material balances are used.

Differential material balances for steady-statecontinuous processes with no chemical reactions areanalogous to integral balance for steady-state batchprocesses with no chemical reactions as discussed inprevious chapter, except that differential balance useflow rates instead of amounts.

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MATERIAL BALANCES


Differential balances for steady-state continuousprocesses with no chemical reactions can be writtenas:

Please note that there are no generation andconsumption terms since there are no chemicalreactions and the accumulation term is droppedbecause the system is at steady-state.

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Differential Material Balances


EXAMPLE 1: Balance on a mixing processOne method of determining the volumetric flow rate of a turbulently flowingprocess stream is to inject small metered amounts of some easily dispersed fluidand then to measure the concentration of this fluid in a sample of the mixedstream withdrawn at a suitable distance downstream. Suppose a stream containing95.0 mole% butane and 5.0 mole% O2 is injected with 16.3 gmol/h O2. Adownstream sample analyses 10 mole%O2. Calculate the flow rate of the processstream.

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SOLUTION Draw and label the flow chart diagram

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mixingprocess

D gmol/h downstream

0.90 Butane0.10 O2

U gmol/h process stream

16.3 gmol/h O2

0.95 Butane0.05 O2

Known basis: 16.3 gmol/h O2.

DOF

Number of unknowns = 2 (U, D)Number of independent equations = 2 (2 material balances + 0 information)∴ Unknown can be solved.


SOLUTION

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Total mole balance: input = outputU + 16.3 = D ---------------(a)

Component balance:Balance on Butane 0.95(U) = 0.90(D) Divide by 0.90: 1.056(U) = D -------------(b)(a) – (b): -0.056(U) + 16.3 = 0

∴ U = 293.17 gmol/h



EXAMPLE 2: Balance on a distillation unitIn a distillation column, an equimolar mixture of ethanol, propanol and butanol isseparated into an overhead stream containing 66.33 mole% ethanol and nobutanol and a bottom stream containing no ethanol. Calculate the following:

a) The ratio of the overhead flow rate to the bottom flow rate.b) The compositions of the bottom product.

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SOLUTION Draw and label the flow chart diagram

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Since all streams are unknowns, we must assign an assumed basis of calculation. Assume that the basis of cal. = 100 kgmol/h gas mixture (G).

DOF

Number of unknowns = 3 (B, D, x)Number of independent equations = 3 (3 material balances + 0 information)∴ Unknown can be solved.

Distillation unit

D kgmol/h Overhead

0.6633 ethanol0.3367 propanolG kgmol/h gas mixture

1/3 etanol1/3 propanol1/3 butanol B kgmol/h Bottom

x butanol1-x propanol


SOLUTION

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Total mole balance: input = output100 = D + B

Component balance:Ethanol mole bal.: 100/3 = 0.6633(D)

∴ D = 50.25 kgmol/hB = 49.75 kgmol/h

Butanol mole bal.: 100/3 = 49.75(x) kgmol/h∴ x = 0.67

a) Ratio of D/B = ..

b) Composition of bottom: 67% mole butanol and 33% mole propanol.



EXAMPLE 3: Balance on a humidification unitLiquid water and air flow into a humidification chamber in which the waterevaporates completely. The entering air contains 1.0 mole% H2O (vapour), 20.8%O2 and the balance N2. the humidification air contains 10.0 mole% H2O. Calculatethe volumetric flow rate (ft3/min) of liquid water required to humidify 200Ibmol/min of the entering air as shown in the flow chart below.

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Dehumidificationprocess

H Ibmol/min humidified air

0.10 H2Ox O20.9 – x N2

W Ibmol/min liquid

200 Ibmol/min air

0.010 H2O0.208 O20.782 N2


SOLUTION

Basis of cal. = 200 Ibmol/min air

DOF

Number of unknowns = 3 (W, H, x)Number of independent equations = 3 (3 material balances + 0 information)∴ Unknown can be solved.

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SOLUTIONTotal Mole Balance: Input = Output

W + 200 = H ---------------(a) Component bal.:Water mole bal.: W + 0.010(200) = 0.1(H) ---------(b)(a) – (b): 198 = 0.9 (H)

∴ H = 220 Ibmol/minW = 20 Ibmol/min

Convert molar flow rate of liquid water to its volumetric flow rate.

. /= 5.77 ft3/min liquid

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MATERIAL BALANCES


MATERIAL BALANCES WITH CHEMICALREACTIONS:Idea:

This section involves analyse and solve materialbalance which involves chemical reactions .

The main idea is that the generation andconsumption terms can come into play in makingcomponent mole balances.

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MATERIAL BALANCES


MATERIAL BALANCES WITH CHEMICALREACTIONS:Main Concepts:

From general Mat. Bal.: [accumulation within thesystem]=[input through system boundaries]-[outputthrough system boundaries]+[generation within thesystem]-[consumption within the system]Ideally, moles balance should be used when analysingfor element balances.But, often in component and total balances, the moleswill not necessarily balance unless the generation andconsumption terms are taken into account.

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MATERIAL BALANCES


MATERIAL BALANCES WITH CHEMICALREACTIONS:Example:

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Combustion

1 C

1 O2

1 CO2

C + O2 = CO2

Analysis: Total moles in are = 2Total moles out are = 1The mole of O2 in are = 1The moles of O2 out (as O2) are = 0

If we take into account the Rxn. Equation and assume complete reaction of the O2, then the generation term for O2 = 0And the consumption term has a value of 1.

MATERIAL BALANCES



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Combustion

1 C

1 O2

1 CO2

C + O2 = CO2

Thus the mat bal for O2 in moles would be:Accu. = in – out + gen - consumption

O2: 0 1 0 0 1What would the mole balances on C and CO2 as compounds be?

Accu. = in – out + gen - consumptionC: 0 1 1 0 0O: 0 2 2 0 0

MATERIAL BALANCES



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Combustion

1 C

1 O2

1 CO2

C + O2 = CO2

Compare these two balances with the mass balances for the elements. How do the mass balances on the elements differ from the mole bal. on the elements?


MATERIAL BALANCES



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Combustion

1 C

1 O2

1 CO2

C + O2 = CO2

Often for convenience the element bal in moles might be made on one C and 2 oxygens (O2). Not meaning the compound O2.


MATERIAL BALANCES- Multiple Unit Processes


It is very rare that raw materials undergo only asingle process. Often these materials have to gothrough a series processes before final products areobtained. This series of processes is called a multiple-unit processes.

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Process 1Unit 1

Process 2Unit 2

Process 3Unit 3



Take cooking oil for example.Producing cooking oil from soybean takes quite along series of processes. Soybeans have to be cleaned,crushed, rolled into flakes before oil could beextracted using hexane solvent. Then, the oil has tobe separated from hexane by heating. The oil is thenrefined, deodorised, decolourised or bleached beforeit is ready for packaging.

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Multiple Unit Processes

• Industrial process is rarely a one unit process. The reactants have to go through several units before becoming finished products

Unit 1 Unit 2


• For multiple unit calculations, DOF analysis has to be done for each boundary to see which one that can be solved first

• This is critical before starting any calculation, so that you will not waste time solving balances that can’t be solved without solving others


Unit 1 Unit 2

100 g/s A

800 g/s

0.2 A0.8 B

M g/s

z A(1-z) B

200 g/s C

N g/s

m An B(1-m-n) C

475 g/s

x Ay B(1-x-y) C

P g/s

0.012 A0.588 B0.400 C

Figure 4.1: Example of multiple-unit process problem


Unit 1 Unit 2

100 g/s A

800 g/s

0.2 A0.8 B

M g/s

z A(1-z) B

200 g/s C

N g/s

m An B(1-m-n) C

475 g/s

x Ay B(1-x-y) C

P g/s

0.012 A0.588 B0.400 C

Overall boundary

Figure 4.2: Overall boundary

Single boundary

Multiple Unit Processes• The most important thing to do is to select a boundary at

which material balance calculation can be performed.

• A boundary is real or imaginary border that separate a system that we want to study the material balance calculation.

• Everything outside the boundary is called surrounding and thus is not involved in the material balance calculation.

Multiple Unit Processes• When a boundary is drawn, it becomes a black box

which means that everything that occurs inside the boundary is ignored.

• We do not care what happen inside the box. What matters is how many inputs and output that cross the boundary.

• The flow chart diagram boundary is now similar to that of a single stage.

Multiple Unit Processes• Proper analysis of unknowns and independent

equations (DOF) must be done before attempting to solve material balance calculations for a boundary.

• Start with the overall boundary. • Overall boundary is a boundary that covers the

entire process (Figure 4.2).• Follow the same procedure as in differential

balance.



EXAMPLE 4.2a: Condition 1 (Overall boundary)

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Overall boundary

475 g/sx Ay B1 – x – y C

200 g/s C

800 g/s

0.2 A0.8 B

P g/s

0.012 A0.588 B0.400 C100 g/s ADOF

No of unknowns = 3 (x, y, P)No of ind. equa. = 3 (A, B, C)∴ Has a unique solution- can be solved


Unit 1 Unit 2

100 g/s A

800 g/s

0.2 A0.8 B

M g/s

z A(1-z) B

200 g/s C

N g/s

m An B(1-m-n) C

475 g/s

x Ay B(1-x-y) C

P g/s

0.012 A0.588 B0.400 C

Unit 1 + mixing boundary

Figure 4.2b: Overall boundary



EXAMPLE 4.2b: Condition 2 (Unit 1 + Mixing point)

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Unit 1 + mixing boundary

N g/s

m An B1 – m – n C

200 g/s C

800 g/s

0.2 A0.8 B

100 g/s ADOFNo of unknowns = 3 (m, n, N)No of ind. equa. = 3 (A, B, C)∴ Has a unique solution- can be solved


Unit 1 Unit 2

100 g/s A

800 g/s

0.2 A0.8 B

M g/s

z A(1-z) B

200 g/s C

N g/s

m An B(1-m-n) C

475 g/s

x Ay B(1-x-y) C

P g/s

0.012 A0.588 B0.400 C

Mixing + Unit 2 boundary

Figure 4.2c: Overall boundary



EXAMPLE 4.2c: Condition 3 (Mixing point + Unit 2)

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Mixing + Unit 2 boundary

N g/s


200 g/s C

M g/s

z A(1-z) B

100 g/s ADOFNo of unknowns = 6 (m, n, z, M, N, P)No of ind. equa. = 3 (A, B, C)∴ No unique solution – can not be solved

P g/s



Unit 1 Unit 2

100 g/s A

800 g/s

0.2 A0.8 B

M g/s

z A(1-z) B

200 g/s C

N g/s

m An B(1-m-n) C

475 g/s

x Ay B(1-x-y) C

P g/s

0.012 A0.588 B0.400 C

Unit 1 boundary

Figure 4.2d: Overall boundary



EXAMPLE 4.2d: Condition 4 (Unit 1 only)

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Unit 1 boundary

M g/s

z A(1-z) B

800 g/s

0.2 A0.8 B

100 g/s ADOFNo of unknowns = 2 (z, M)No of ind. equa. = 2 (A, B)∴ Has a unique solution- can be solved


Unit 1 Unit 2

100 g/s A

800 g/s

0.2 A0.8 B

M g/s

z A(1-z) B

200 g/s C

N g/s

m An B(1-m-n) C

475 g/s

x Ay B(1-x-y) C

P g/s

0.012 A0.588 B0.400 C

Unit 2 boundary

Figure 4.2e: Overall boundary



EXAMPLE 4.2e: Condition 5 (Unit 2 only)

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Unit 2 boundary

475 g/s

x Ay B(1 – x – y) C

N g/s

DOFNo of unknowns = 6 (m, n, x, y, N, P)No of ind. equa. = 3 (A, B, C)∴ No unique solution – can not be solved

P g/s

0.012 A0.588 B0.400 C

m An B(1 – m – n) C


Unit 1 Unit 2

100 g/s A

800 g/s

0.2 A0.8 B

M g/s

z A(1-z) B

200 g/s C

N g/s

m An B(1-m-n) C

475 g/s

x Ay B(1-x-y) C

P g/s

0.012 A0.588 B0.400 C

Mixing boundary

Figure 4.2f: Overall boundary



EXAMPLE 4.2f: Condition 6 (Mixing point)

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Mixing point boundary

N g/s


200 g/s C

M g/s

z A(1-z) B

DOFNo of unknowns = 5 (m, n, z, M, N)No of ind. equa. = 3 (A, B, C)∴ No unique solution – can not be solved

Multiple Unit Processes• Picking the right boundary for material bal. is the

most important key to solving mat. bal. for multiple-unit processes.

• Solve for unknowns for boundary which has the number of unknowns equal to the number of independent equations or DOF = 0.

•


Unit 1 Unit 2

100 g/s A

800 g/s

0.2 A0.8 B

M g/s

z A(1-z) B

200 g/s C

N g/s

m An B(1-m-n) C

475 g/s

x Ay B(1-x-y) C

P g/s

0.012 A0.588 B0.400 C

Example 4.1.

Determine the values of ALL unknown in the steady-state two-unit processes below.

Multiple Unit Processes• Step-by-step solution

– Firstly, make sure the block and arrow diagram is completely labeled and every unknown is identified.

– Analyses ALL possible boundary by determining the number of unknowns and the number of independent equations or DOF analysis.

• Start with the overall boundary• If DOF = 0 then solve for the unknowns.• If DOF ≠ 0 then go to other boundary.• Solve for any boundary that contains the number of unknowns

equal to the number of independent equations or the value of DOF = 0.


Unit 1 Unit 2

100 g/s A

800 g/s

0.2 A0.8 B

M g/s

z A(1-z) B

200 g/s C

N g/s

m An B(1-m-n) C

475 g/s

x Ay B(1-x-y) C

P g/s

0.012 A0.588 B0.400 C

Overall boundary

Example 4.1.



EXAMPLE 4.2a: Condition 1 (Overall boundary)

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Overall boundary

475 g/sx Ay B1 – x – y C

200 g/s C

800 g/s

0.2 A0.8 B

P g/s

0.012 A0.588 B0.400 C100 g/s ADOF

No of unknowns = 3 (x, y, P)No of ind. equa. = 3 (A, B, C)∴ Has a unique solution- can be solved

Multiple Unit Processes• Since the number of unknowns (x, y, P) are equal to

the number of independent equations, we can proceed to solve the problem.

SOLUTIONMaterial balance for overall boundary

input = outputTotal mat. bal: 800 + 200 = 100 + 475 + P

∴ P = 425 g/s

Multiple Unit ProcessesComponent balance:

input = outputA balance: 0.2(800) = 100 + 475x + 0.012(425)

∴ x = 0.116

B balance: 0.8(800) = 475y + 0.588(425)∴ y = 0.821



Material balance at unit 1 boundary

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Unit 1 boundary

M g/s

z A(1-z) B

800 g/s

0.2 A0.8 B

100 g/s ADOFNo of unknowns = 2 (z, M)No of ind. equa. = 2 (A, B)∴ Has a unique solution- can be solved

Multiple Unit ProcessesTotal material balance:

input = output800 = M + 100∴ M = 700 g/s

Component balance: A balance: 0.2(800) = 100 + z(700)

∴ z = 0.086



Material balance at mixing point

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Mixing point boundary

N g/s


200 g/s C

700 g/s

0.086 A0.914 B

DOFNo of unknowns = 3 (m, n, N)No of ind. equa. = 3 (A, B, C)∴ Has unique solution – can be solved

Multiple Unit ProcessesTotal material balance:

input = output700 +200 = N∴ N = 900 g/s

Component balance: A balance: 0.086(700) = m(900)

∴ m = 0.067

B balance: 0.914(700) = n(900)∴ n = 0.711

Multiple Unit ProcessesExample 4.2.

Acetone is used in the manufacture of many chemicals and also as a solvent. In its latter role, many restrictions are placed on the release of acetone vapour to the environment. As a chemical engineer you are assigned to design an acetone recovery system having the flowchart as shown in the figure below. All the concentrations shown are specified in weight percent. Determine the values of A, F, W, B and D per hour.


Absorber column

Distillation column

100 kg/h gas in

W kg/h water

0.03 acetone0.02 water0.95 air

F kg/h

0.19 acetone 0.81 water

A kg/h gas out

D kg/h Distillate0.995 air0.005 water Condenser


B kg/h Bottom



Absorber column

Distillation column

100 kg/h gas in

W kg/h water


F kg/h


A kg/h gas out

D kg/h Distillate0.995 air0.005 water

Condenser 0.99 acetone 0.01 water

B kg/h Bottom




DOFAnalysing the overall boundary No of unknowns = 4 (W, A, D, B)No of ind. equa. = 3 (3 material balance + 0 information)∴ No unique solution – can not be solved


Absorber column

Distillation column

100 kg/h gas in

W kg/h water


F kg/h


A kg/h gas out

D kg/h Distillate0.995 air0.005 water

Condenser 0.99 acetone 0.01 water

B kg/h Bottom




DOFAnalysing the absorber boundary No of unknowns = 3 (W, A, F)No of ind. equa. = 3 (3 material balance + 0 information)∴ Has unique solution – can be solved

SOLUTION:

Multiple Unit ProcessesMaterial balance at the absorber boundary: Component balance: input = outputAir mass balance: 0.95(100) = A(0.995)

∴ A = 95.48 kg/h

Acetone balance: 0.03(100) = (0.19)F∴ F = 15.79 kg/h

Total mass balance: 100 + W = A + F100 + W = 95.48 + 15.79

∴ W = 11.27 kg/h

Multiple Unit ProcessesMaterial balance at the overall boundary: Total mass balance: input = output

100 + 11.27 = 95.48 + D + B -------(a)

Component balance:Water mass balance:

0.02(100) + 11.27 = 0.005(95.48) + 0.01D + 0.96B(÷ 0.96) yields: 13.326 = 0.01042D + B ----------------------(b)(a) – (b) yields: 2.464 = 0.9896D

∴ D = 2.490 kg/h∴ B = 13.30 kg/h



DIY1. Sea water containing 3.5 wt.% salt passes through a series of 10 evaporators.

Roughly equal quantities of water are vaporised in each of the 10 units andare then condensed and combined to obtained a product stream of freshwater. The brine solution (concentrated salt solution) at the outlet of the lastevaporator contains 5.00 wt.% salt. If 30,000 kg/h of sea water are fed to theprocess, calculate:a) The fraction of the fresh water obtained from the sea water.b) The weight percent of salt in the solution leaving the forth evaporator.

2. A liquid mixture containing 30 mole% benzene (B), 25 mole% toluene (T)and 45 mole% xylene (X) is fed at a rate of 1275 kmol/h to a distillationsystem which consists of two distillation columns. The bottom product fromthe first column is to contain 99 mole% X and no B, and 98 % of the X is tobe recovered in this stream. The overhead product from the second columncontains 99 mole% B and no X. The benzene recovered in this streamrepresents 96 % of the B in the feed of this column. Calculate:



a) The molar flow rates (kmol/h) in each product stream from bothcolumns.

b) The mole fractions of each components in each product stream.



3) A two-stage separation unit is shown below. Given the input stream F1 is1000 Ib/h. Calculate the value and composition of F2.

Unit 1 Unit 2

0.4 toluene0.4 benzene0.2 xylene F2 Ib/h

0.01 toluene 0.99 benzene

0.1 toluene 0.9 xylene

P1 Ib/h

F1 Ib/h

P2D Ib/h

P2B Ib/h

0.95 toluene 0.05 benzene



4) The following flow chat diagram shows a sytem for an extraction-distillationprocesses. Determine the values of ALL unknowns if 40.16 % of acetone fedto the distillation column is recovered in the overhead stream.

Extractor E1

Extractor E2

0.5 A0.5 W

V kg Overhead0.97 A0.02 M0.01 W

F kg MG kg AH kg W

100 kg M

100 kg feed43.1 kg

E2 kg

0.053 A0.016 M

0.0931 W

Distillation column

75 kg M

P kg MQ kg AR kg W

0.09 A0. 88 M0.03 W

E1 kg

0.275 Ax M0.725 - x W

C kg MD kg AE kg W

MATERIAL BALANCES


DIY5. 100 kg aqueous feed solution containing 25.8 wt.% acetic acid

is to extracted with isopropyl ether in a series of cross-flowextractor which consists of three single stage extractors asshown in the diagram below. In each stage, 100 kg isopropylether is used and assuming that equilibrium is attained ineach stage. Determine:a) The composition of the combined extract.b) The percent acid extracted by the cross-flow extractor.

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MATERIAL BALANCES


Stage 1 Stage 2 Stage 3100 kg feed

100 kg ether 100 kg ether 100 kg ether

RaffinateRaf

finat

e

Raf

finat

eExtract Extract Extract

Combined extract

MATERIAL BALANCES


DIY6. A labeled flowchart of a continuous steady-state two-unit

distillation process is shown below. Determine the unknownflow rates M, N, P and their compositions.

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Unit 1 Unit 2Feed

50 kg/h A50 kg/h B

M kg/h

36 kg/h A4 kg/h B

9 kg/h A21 kg/h B

N kg/h

30 kg/h

60 wt.% A40 wt.% B

P kg/h

MATERIAL BALANCES


DIY7. Sea water containing 3.5 wt.% salt passes through a series of 5 evaporators.

Roughly equal quantities of water are vaporised in each of the 5 units and arethen condensed and combined to obtained a product stream of fresh water. If20,000 kg/h of sea water are fed to the process and the fraction of the freshwater obtained from the sea water is 0.2, calculate:a) The composition of brine solution at the outlet of the final evaporator.b) The amount of water produced per evaporator.

12/14/2013 Copyright Dr.RR 201312/14/2013

MATERIAL BALANCES


DIY8. Several streams are mixed as shown in the flowchart below. All compositions

are in weight percent. Calculate the values of ALL unknowns and thecomposition of stream C.

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Uni

t 1

Uni

t 2

4 % NaCl5 % HCl4% H2SO487 % H2O

A kg/min

C kg/min

D kg/min

E kg/min

290 kg/min

HClH2SO4

96 % H2O

1.5 % HCl1.5 % H2SO497 % H2O

1.38 % NaCl2.55 % HCl2.21% H2SO492.32 % H2O

Inert solid

9 % inert solid91 % H2O

B kg/min

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Chap_4-_mat_bal_03-1