Che 341 notes

Material balances

introduction

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INDUSTRIAL PROCESS CALCULATIONS

• Intro to equip of chemical plant; Equipment for movt and storage of

material. Heat transfer equip. Mass transfer equipment and equipment for

physical processes. The chemical equation and stoichiometry.; limiting

reactant, selectivity and yield xs rxt, conversion. Material balances: calculations

for steady state systems involving inerts, recycle, by pass and purges.

Course outline contd

• Energy balance for a chemical system. Heat capacities. Calculation of enthalpy changes; heat of fussion, vapourisation, rxn, formation and combustion, soln and mixing. Combined material and. Enthalpy conc charts application and cnstruction.

What is the course all abt?

• Intro to principles and tech used in the field of chem, petroleum and environ

engr

• Intro to systematic prob solving skills

• What are mat and energy balances and, how are they applied in industrial

pocesses

requirement

• Like every other course taken so far, 75 % attendance is compulsory to be

eligible to write the final exam.

Recommended txt

• Basic principles and calculations in Chem Engr by Himmeblau

• Elementary principles of chemical processes

• Felder and Rousseau

• Chemical Eng series Vol I by Coulson et al

• Unit operations of Chemical Engr by McCabe and Smith

Introduction to process calculation

•

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Conversion between mass and moles

Given that the molecular weight of CO is 28 determine how many of the

of each of the following are contained in 88 g of CO

• mol CO

• lbmoles CO

• mol O

• g O

88 g CO 1 mol CO

28 g CO

= 3.14 mol CO

Check conversion table in steam tables for the second question 8 10/14/2014

From tables 1 lbm = 453.6 g therefore

3.14 mol CO 1 lb- mole

453.6mol

= 0.0069 lb mole CO

Continue from here

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Introduction to process calculation

The stream into a vessel contains 25 % A by mass and 40 mole % B, determine the

mass of A in 300 kg of solution.

•

300kg solution

Kg solution

0.25 kg A

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A Set of mass fractions may be converted to an equivalent mole fractions by

• Assuming a basis

• Using the known mass fractions to calculate the mass of each component

in the basis quantity, and converting these masses to moles

• Taking the ratio of the moles of each component to the total number of

moles

Choose a basis: 100 g mixture

Next determine the moles of each substance present

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Average Molecular Mass

•

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Calculation of average mwt

•

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cont •

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Class work

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Determine

• Mass fraction

• Mole fraction of Br in pure

HBr given that the mwt of H is 1

And mwt of Br is 80

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Concentration

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Corresponding molar conc

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example

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Determine in terms of V, n, and

M,

a. The molar conc of A

b. Mass conc of A

of a solution, of volume V litres

containing n mol of a solute A

whose molecular weight is M g A

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Conversion between mass, molar and

volumetric flow rates of a solution

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c. The mass fraction of sulphuric

acid

Of a 0.50 molar aqueous solution

of shulphuric acid which flows

into a process unit at a rate of

1.25 m3 / min. given that the

specific gravity of the solution is

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Material balances are important first step when designing a new process or analyzing an existing one. They are almost

always prerequisite to all other calculations in the solution of process engineering problems.

They represent the application of the law of conservation of mass which suggests Input = output

Suppose propane is a component of both the input and output streams of a continuous process unit shown below,

these flow rates of the input and output are measured and found to be different.

Qin (kg propane/h) Q out(kg propane/h)

If there are no leaks and the measurements are correct, then the other possibilities that can account for this difference

are that propane is either being generated, consumed, or accumulated within the unit. A balance (or inventory) on a

material in a system(a single process unit, a collection of units, or an entire process) may

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Input + generation − output − consumption = accumulation

This general balance equation may be applied to any material that enters a process system ( total mass, or to any

molecular or atomic specie involved in the process);

Modification to the general equation

• For steady state continuous process accumulation = 0

• For physical process ( no chemical reaction) therefore generation and consumption terms = 0

• So for a steady state physical process Input = output

• Steps to follow in material balance calculations

i. Draw and label the process flow chart ( block diagram) indicating all known streams and assigning

symbols to unknown streams

ii. Select a basis for calculation which is usually the given stream amount or flow rates if given otherwise

assume a value which is usually better in multiples of 10 for a stream with known composition.

iii. Write material balance equations ( Overall and component material balance equations) with the number

of independent equations for the system equal to the number of species in the input and output

streams of the system.

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It is desired to 10 % solution of KOH from a stream of 30% KOH using water . Determine the following

• The ratio g water/g feed solution

• g product solution / g feed solution

• The feed rate of 20 % solution and diluting water required to produce 1500 Ibm/min of the 10 % solution

From steps to follow: draw up the process flow diagram

100 g 70% water(0.7 g water /g)

30%KOH (0.3g KOH/g) y g ( 10% KOH , 90 % water)

x g water 0.10g KOH /g and 0.90 g water/g

Since the known stream amount is in g it is convenient to label all unknown streams in the same unit

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Tower

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Integral balances on batch processes

• Balances on batch mixing process

Two ethanol – water mixtures are contained in separate flasks. The first mixture

contains 80 wt % ethanol, and the second contains 60 wt % ethanol . If 120 g

of the first mixture is combined with 180 g of the second, what are the mass

and composition of the product?

Solution

Draw the flow diagram

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120g

M (g)

.8 g E/g ; .2 g W/g

180 g ; 0.6 g E/g,, 0.4 g W/g

X g M/g

(1-x)g W/g

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0.25 mol air/min

H (mol/min)

Since air neither dissolves in the liq accm=0

No rxn implies generation = consumption= 0

Input = output

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Notice variation with time. Input and output are given per unit of time. Doing an integral balance from t = 0 to t =t