Transcript
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
/ mol 23 10/14/2014
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
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