lecture 1 Balance de materia

Prévia do material em texto

Basic principle II Second class Dr. Arkan Jasim Hadi 
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Basic Principles and Calculations in Chemical Engineering 
 
 
Second Class 
 
Dr. ARKAN JASIM HADI 
 
 
DEPARTMENT OF CHEMICAL ENGINEEING 
COLLEGE OF ENGINEERING 
UNIVERSITY OF TIKRIT 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Basic principle II Second class Dr. Arkan Jasim Hadi 
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Objectives 
• To impart the basic concepts of Chemical Engineering 
• To develop understanding about energy balance for analysis of unit processes and unit 
operations 
Learning outcomes 
Abilities, and Skills Students Should Gain From This Course 
 
1. An ability to define heat, work, energy, enthalpy, etc. 
2. An ability to make estimations of heat capacity and to calculate enthalpy changes, 
with and without phase changes. 
3. An ability to use the steam tables. 
4. An ability to solve energy balances for closed and open systems, with and without 
chemical reaction. 
5. An ability to solve simple combined material and energy balances (with and without 
chemical reaction). 
6. An ability to distinguish and make calculations for adiabatic and non-adiabatic 
processes 
 
 
Text book: 
1- David M Himmelblau, Basic principles and calculations in chemical engineering, Prentice 
Hall. 
References: 
1- Hougen A, Watson K M, Ragatz R A, Chemical Process principles, John Wiley 
2- Richard M Felder & Ronald W. Rousseau Elementary Principles of Chemical Processes, 
Wiley India. 
 
 
 
 
 
 
 
 
 
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1. DEFINITIONS 
A closed system consists of a fixed amount of mass and no mass may cross the system 
boundary. The closed system boundary may move. 
An open system, or control volume, has mass as well as energy crossing the boundary, called 
a control surface. Examples of open systems are pumps, compressors, turbines, valves, and 
heat exchangers. 
An isolated system is a general system of fixed mass where no heat or work may cross the 
boundaries. 
 
 
 
Steady-state: means all the conditions in the process (temperature, pressure, mass of 
the material, flow rate , etc.) remain constant with time. A continuous process is one 
in which material enters and/or leaves the system without interruption 
Unsteady-state: means not all the conditions in the process (temperature, pressure, 
mass of the material, etc.) remain constant with time, and/or the flows in and out of 
the system can vary with time. 
 
 
 
 
 
 
 
 
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Semi-batch process: A process in which material enters the system but product is not 
removed during operation. 
Transient system A system for which one or more of the conditions (e.g., temper entre, 
pressure, amount of material) of the system vary with time. Also known as an unsteady-state 
system. 
The Reversible Process: A process is reversible when its direction can be reversed at any 
point by an infinitesimal change in external conditions. 
Irreversible Process: A process in which it is impossible to return both the system and 
surroundings to their original states. 
Isothermal Process – process that carried out at constant temperature 
Isochoric Process: A process is one in which the volume does not change 
Isobaric Process: A process is one which the pressure is kept constant 
Adiabatic Process: An adiabatic process is one in which no heat is gained or lost by the 
system. The first law of thermodynamics with Q=0 shows that all the change in internal 
energy is in the form of work 
Polytropic Process: Since polytropic means "turning many ways: It is defined as a process 
represented by empirical equation 
 
 
Throttling process: produces no shaft work, and in the absence of heat transfer, and the 
process therefore occurs at constant enthalpy. 
 
1. Material Balance 
General Procedure for Material Balance Problems 
 
Most of the literatures on problem solving view a problem as a gap between some initial 
information (the initial state) and the desired information (the final state). Problem solving is 
the activity of closing the gap between these two states. 
 When solving problems, either academic or industrial, you should always use 
“engineering judgment” even though much of your training to hate treats problems. 
 
 
The strategy for solving problems 
1. Read and understand the problem statement. This mean read the problem carefully so 
you know what is given and what is to be accomplished. 
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2. Draw a sketch of the process and specify the system boundary. A simple box or circle 
drawn by hand to denote the system boundary with some arrows to designate flows of 
material will he fine You can also state what the system is M words or by a label. 
3. Place labels (symbols, numbers, and units) on the diagram for all of the known flows, 
materials, and compositions. For the unknown flows, martials, and compositions 
insert symbols and units. Add any other useful relations or information 
4. Obtain any data you need to solve the problem, but are missing. 
5. Choose a basis: 
You discussed the topic of basis in Chapter 3 where we suggested three ways of 
selecting a basis: 
(1) What do I have 
(2) What do I want to find, 
(3) What is convenient? 
 
6. Determine the number of variables whose values are unknown (the unknowns). 
If you put symbols on the process diagram as described in steps 2,3 and 4 above, or 
make a list of them, determining the number of unknowns is easy. 
 
7. Determine the number of independent equations and carry out a degrees of freedom 
analysis 
The number of variables whose values are unknown equals the number of 
independent equations you formulate to solve a problem. 
 
Materail balance 
Matrail balance without chemical reaction: 
The processes which is often need a materail balance without chemical reaction are 
(distillation, drying, evaporating, mixing and crystilazation. 
 
 
 
 
 
 
 
 
To covert mole fraction to weight fraction 
 
 
∑ 
 
To convert weight fraction to mole fraction 
 
 
∑ 
 
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Example 1. Separation of Gases Using a Membrane Membranes represent a relatively new 
technology for the separation of gases. One use that has attracted attention is the separation of 
nitrogen and oxygen from air. Figure 1.1 illustrates a nanoporous membrane that is made by 
coating a very thin layer of polymer on a porous graphite supporting layer. What is the 
composition of the waste stream if the waste stream amounts to 80% of the input stream? 
 
 
 
 
 
 
 
 
 
 
 
 
Figure 1.1 
Solution 
Step 1 
This is an open, steady-state process without chemical reaction. The system is the membrane 
as depicted in Figure 1. Let be the mole fraction of oxygen as depicted in Figure 1, be 
the mole fraction of nitrogen, and let and be the respective moles. 
 
 
 
 
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Step 2,3,4 
All of the data and symbols have been placed in Figure 1. 
Step 5 Pick a convenient basis. 
Basis. 100 gmole = F 
A degree of freedom analysis that includes all of the variables comes next. You could either 
use moles or mole fractions as the unknowns. 
Steps 6 and 7 
Number of variables: 9 F, P, W and 6 n, 
Number of equations: 9 
Basis: 100 gmole of feed 
M.B. 
F = P + W (1) 
100 = P + 80 P = 20 
no2
F = 0.21 (100) = 21 
nN2
F = 0.79 (100) = 79 
An Oxygen M.B 
0.21 (100) = 0.25 (20) + nO2
W
 
21 = 5 + nO2
W nO2
W = 16 
An Nitrogen M.B 
0.79 (100) = 0.75 (20) + nO2
W79 = 15 + nN2
W nN2
W = 64 
Mass balance-multiple units: 
Example 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. You are asked to design an acetone recovery system having the flow-sheet 
illustrated in Figure E2.1. All the concentrations shown in E2.1 of both the gases and liquids 
are specified in weight percent in this special case to make the calculations simpler. 
Calculate, A, F, W, B, and D per hour. G = 1400 kg/hr. 
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Solution 
This is an open, steady-state process without reaction. Three subsystems exist. 
Steps 1, 2, 3, and 4 
All the stream compositions are given. All of the unknown stream flows are designated by 
letter symbols in the figure. 
Step 5 
Pick 1 hr as a basis so that G = 1400 kg. 
Steps 6 and 7 
We could start the analysis of the degrees of freedom with overall balances, but since the 
subsystems are connected serially, we will start the analysis with the absorber column, Unit 
1, and then proceed to Unit 2, and then to Unit 3. 
Unit I (Absorber) Variables: 16 
W, G, F, A (4 flow streams); species mass fractions in each stream = 3 so that 3 x 4 = 12 
more variables 
 Equations: 16 
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Basis: G 
Species material balances: 3 (one for each species) 
 
 
 
 
 
 
 
 
Implicit equations (∑ ) all redundant given the specifications Degrees of freedom: 0 
Before proceeding to calculate the degrees of freedom for Unit 2 (the distillation column), 
you should note the complete lack of information about the properties of the stream going 
from the distillation column to Unit 3 (the condenser). In general it is best that you avoid 
making material balances on systems that include such streams, as they contain no useful 
information, Thus, the next system and degree-of-freedom analysis we select will be for the 
system composed of Units 2 and 3 combined. 
Units 2 and 3 (Distillation Column plus Condenser) Variables: 9 F, D, B (3 streams); species 
mass fractions in each stream = 2 so that 2 X 3 = 6 more variables 
Equations: 9 Species material balances: 2 (one for each species) Specifications: 6 
 
Over all MB. 
G + W = A+B+D 
1400 + W = A+B+D 
MB on unit 1 (Absorber column) 
G + W = F+A 
1400 + W= F+A (1) 
MB/ air 
(0.95) 1400 +(0) W = (0) F +(0.995) A (2) 
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1330 = 0.995A 
MB/Acetone 
(0.03) 1400 + (0) W = (0.19)F +(0) A (3) 
42 = 0.19F 
MB/ Water 
(0.02) 1400 + (1) W = (0.005) A + (0.81) F (4) 
28 + W = 0.005A +0.81 F 
 Then from Eq. 2: A= 1330/0.994 = 1336.7 kg/hr 
From Eq. 3: F = 42/0.19 = 221.05 kg/hr 
Substitute the values of A and F in Eq. 4 
W = (0.005)(1336.7) + (0.81)(221.05) -28 
W= 157.734 kg/hr 
Now check the overall for unit 1 
G + W = F+A 
1400 +157.734 = 221.05 + 1336.7 
1557.734 = 1557.75 
The mass balances for combined units 2 plus 3 are 
 
F = D + B (1) 
221.05 = D + B 
MB/Acetone 
(0.19) 221.05 = (0.99)D + (0.04) B (2) 
42 = 0.99D + 0.04 B 
MB/ Water 
(0.81) 221.05 = (0.01) D + (0.96) B (3) 
179.0505 = 0.01D + 0.96 B 
Solve Eqs. 2, 3 
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Divide Eq. 2 by 0.99 
42.42 = D + 0.0404 B 
Then 
D = 42.42 -0.0404 B (4) 
Substitute Eq.4 in Eq. 3 
 179.0505 = 0.01(42.42 - 0.0404 B) + 0.96 B 
179.0505 = 0.4242 - 0.000404 B + 0.96 B 
179.0505- 0.4242 = 0.959596B 
Then B = 186.1 kg/r 
From Eq. 1 
D = 34.90 kg/hr 
Now check the overall MB Eq. for all the system 
G + W = A+B+D 
1400 + 157.734 =1336.7 + 186.1 + 34.90 
1557.734 = 1557.7 
 
 
Mass balance with chemical reaction: 
In = out – generated by reaction + consumed by reaction 
Or 
Out = In –reacted + generated 
This equation which the used in this type of mass balance 
- When you do a balance on any unit, the balance must be in mass not in mole 
- When the chemical equation is used in the calculation must be in moles not in mass. 
 
Example (1) Reaction in Which the Fraction Conversion Is Specified 
The chlorination of methane occurs by the following reaction 
CH4 + Cl2 CH3Cl + HCl 
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Determined the product composition if the conversion of the limiting reactant is 67%, 
and the feed composition in mole % is given as: 40% CH4, 50% Cl2, and 10% N2. 
Solution 
Steps 1,2,3 and 4 
Assume the reactor is an open, steady-state process. 
 
 
 
 
 
 
Basis: 100 gmole of feed 
Basis: F=100 
∑ 
 
∑ 
 
The conversion is 67% then 
MB on CH4: nCH4 out = in – reacted 
 Reacted CH4= conversion percentage * entering CH4 
 = 0.67 * 40 = 26.8 gmole 
Same for all. 
nCH4 out = 40- 1(26.8) = 13.2 gmole 
nCl2 out = 50 - 1(26.8) = 23.2 gmole 
nCH3Cl out = 0 +1(26.8) = 26.8 gmole 
nHCl out = 0 +1(26.8) = 26.8 gmole 
nN2 out = 10 – 0(26.8) = 10 gmole 
 
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Example 2. Material Balances for a Process in Which Two Simultaneous Reactions 
Occur using element 
Formaldehyde (CH2O) is produced industrially by the catalytic oxidation of methanol 
(CH3OH) according to the following reaction: 
CH3OH + ½ O2 CH2O + H2O (1) 
Unfortunately, under the conditions used to produce formaldehyde at a profitable rate, a 
significant portion of the formaldehyde reacts with oxygen to produce CO and H2O, that is, 
CH2O + ½ O2 CO + H2O (2) 
Assume that methanol and twice the stoichiometric amount of air needed for complete 
conversion of the CH3OH to the desired products (CH2O and H2O) are fed to the reactor. 
Also assume that 90% conversion of the methanol results, and that a 75% yield of 
formaldehyde occurs based on the theoretical production of CH2O by Reaction 1. Determine 
the composition of the product gas leaving the reactor. 
 
Solution 
Steps 1, 2, 3, and 4 
Figure E2.1 is a sketch of the process with yi, indicating the mole fraction of the respective 
components in P (a gas). 
 
 
 
 
 
 
 
 
 
 
 
 
n1 
n2 
n3 
n4 
n5 
n6 
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Basis: 1 gmole F 
nO2 entering the stream A = 2(1/2)F = 2(1/2) *1= 1 gmole 
Moles of A= nO2/0.21 = 4.76 gmole 
nN2 = 4.76-1= 3.76 gmole 
Element MB. 
C: 1*1= n1+ n4 + n6 (1) 
 1 = n1+ n4 + n6 (1) 
 
H: 1*4 = 4n1+2n4+2n5 
 4 = 4n1+2n4+2n5 (2) 
O: 1*1 + (0.21*4.76)*2 = n1+ 2n2 + n4 + n5+n6 
 3 = n1+ 2n2 + n4 + n5+n6 (3) 
N: (0.79*4.76)*2 = 2n3 
7.52 = 2n3 (4) 
n3 = 3.76 gmole 
 
The conversion of CH3OH is 90% 
Mean the reacted of CH3OH = 0.9 gmole 
Then the out of CH3OH = 1-0.9= 0.1 mole 
n1= 0.1 gmole 
The yield of CH2O = 75% 
Mean the out of CH2O = 0.75 mole 
n4= 0.75 gmole 
from Eq. (1) 
 
1= 0.1+ 0.75 + n6 
n6 = 0.15 gmole 
from Eq.(2) 
4 = 4(0.1)+2(0.75)+2n5 
n5= 1.05 gmole 
 
 from Eq. (3) 
 3 = 0.1+ 2n2 + 0.75 + 1.05+0.15 
n2=0.475 gmole 
 
∑ 
 
P= n1+n2+n3+n4+n5+n6 
P= 6.28 gmole 
 
Then 
yCH3OH= 0.1/6.28 =0.016 = 1.6% 
yO2= 0.475/6.28 =0.075 = 7.5% 
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yN2= 3.76/6.28 =0.598 = 59.8% 
yCH2O= 0.75/6.28 =0.119 = 11.9% 
yH2O= 1.05/6.28 =0.167 = 16.7% 
yCO= 0.15/6.28 =0.023 = 2.3%