Resumen CH03 Felder
Resumen CH03 Felder
Resumen CH03 Felder
Objectives
At the end of this chapter you should be able to:
Recognize each term in the general balance equation.
Recognize process unit basic functions.
Explain the difference between an open and closed
system, batch, semibatch and continuous processes.
Draw and label a process flowchart.
Choose a suitable basis.
Define a system and draw its boundaries.
Solve steady state material balance problems with a
single process unit and no chemical reactions (i.e.
Accumulation = generation = consumption = 0.0).
Principles of Chemical Engineering
Processes
Accumulation
within the
system
(buildup)
Input through
system
boundary
Generation
within the
system
Output through
system
boundary
Consumption
within the
system
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Notations:
The use of consistent notation is generally
advantage.
For example:
m: mass, m : mass flow rate
n : moles, n : mole flow rate
v : volume, V : volumetric flow rate
x : component fractions (mass or mole) in liquid
streams.
y : component fractions in gas streams.
Principles of Chemical Engineering
Processes
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Solution:
1,A m
2,A m
3,A
A: m
1,B m
2,B m
3,B
B: m
1,C m
2,C m
3,C
C: m
3,A 0 (no A exists in stream 3)
m
1,Total m
2,Total m
3,Total
Total mass balance : m
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Solution:
The process labeled flow sheet is shown in
figure E3.2
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Solution:
D is one tenth of the feed
0.1 (1000) 100 kg/h
Total mass balance:
1000 kg/h B 100 kg/h, B 900 kg/h
Component balance (Ethanol) :
100 kg/h 60 kg/h (X E ) (900 kg/h) x E 0.044
Check water :
900 kg 40 kg (0.956)(900 kg ) 900 kg OK
Principles of Chemical Engineering
Processes
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Solution:
Basis :100 kg / min of the feed stream
The deg ree of freedom; DF
2 (unknown) 2 (independent equations ) 0 ( solvable)
Total mass balance :100 kg / min 40 kg / min m o
Component mass balance ( NaOH ) 20 kg / min 5 kg / min m 1
Answer :
m o 60 kg / min
m 1 15 kg / min NaOH in the bottom stream
m 2 60 15 45 kg / min H 2O in the bottom product stream
m 3 40 5 35 kg / min H 2O in the top stream
Principles of Chemical Engineering
Processes
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We have to learn:
the process variables that we need to
describe the chemicals in a process
stream.
How to specify a process stream, process
unit.
Do mass balance on a single process
unit, and on sequence of process units.
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Process classification
A. Based on how the process varies with time.
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Process classification
B. Based on how the process was designed
to operate:
Continuous processes: material is
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Process classification
Batch processes: no material is transferred
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Types of Balance
1. Differential Balances: written in terms of
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Types of Balance
2. Integral Balances : written in terms of the
amounts of a specified quantity over a period
of time.
Example: a water storage tank at the start
contains 5 liters of water, after 30 minutes of
water flowing to the tank, it is found to contain
50 liters of water.
In this case the accumulated water after 30
minutes is 45 liters of water.
Accumulation =(final output initial input )=
50 - 5 =45 liters
Principles of Chemical Engineering
Processes
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xO2 0.2
xN 2 ?
xCH 4 1 0.2 x N 2
Chemical Engineering
.8 x Nof2Processes
0Principles
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mO2 200 kg
mN 2 ?
mCH 4 1000 200 mN 2
800 mN 2
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Scaling
Change value of all stream amounts of a certain process by a proportion
amount while leaving the stream compositions unchanged.
Scaling
down
Scaling
up
The final stream quantities are
larger than the original quantities.
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Basis of calculation
It is an amount or flow rate of one the process stream.
It is recommended to consider that:
1.If a stream amount or flow rates is given in the problem
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7.
8.
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Solution
1. The general material balance equation:
Accumulation = (In Out) + (Generation
Consumption)
Accumulation: change in quantity of material
inside system.
In: material that enters system by crossing
system boundary.
Out: material that exist system by crossing
system boundary.
Principles of Chemical Engineering
Processes
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Solution
502 lb
Polystyrene mass percent
100% 11 wt %
(502 lb 4060 lb)
502 lb
There are
0.0166 lb mol polystyrene and
30200 lb / lb mol
502 lb
39.04 lb mol styrene.
104 lb / lb mol
Polystyrene mole percent
0.0166 lb mol
100% 0.0425 lb mol %
0.0166 lb mol 39.04 lb mol
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Example
You have 180 g of a 12 wt%
solution of glucose (C6H12O6) in
water. Calculate the g-mole of
glucose and of water.
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Solution
180 g solution
0.88 g water
180 g solution
1g mol water/18 g water 8.8 g mol water
g solution
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453.59 g 1 gmol
27.8 gmol CO 2
lb
44 g
470,000 cm 3
P
1.42 atm
Principles of Chemical Engineering
Processes
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Separation Process
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Solution:
Basis:1000 Kg/h of feed
Assumptions: steady state, no reaction.
Material balance : In = out
B balance: 0.5 (1000)=450 Kg B/h +m3B
m3B = 50 Kg B/h
T balance:
0.5(1000) = m2T+475 Kg T/h
m2T =25 Kg T/h
Principles of Chemical Engineering
Processes
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Solution
Basis: 200g of the 40%wt methanol (i.e. stream 1).
Assumptions: Steady state, no reaction.
Material balance:
m3 =350
Total balance: 200 +150 = m3
M balance:
x3M=0.53
0.4(200)+0.7(150)=x3M(350)
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Solution :
Basis: 1 ton feed
Assumption: Steady state, no reaction
Total mass balance
m1 = m2 + m3
1 ton/min = 0.7 ton/min + m3
m3 = 0.3 ton/min
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