Download Introduction to Material Balances in Chemical Engineering and more Cheat Sheet Stoichiometry in PDF only on Docsity!
Dr. Ahmed Faiq Al-Alawy
Chapter 6
Introduction to Material Balances
6.1 The Concept of a Material Balance
A material balance is nothing more than the application of the law of the conservation of mass: “Matter is neither created nor destroyed”
6.2 Open and Closed Systems
a. System By system we mean any arbitrary portion of or a whole process that you want to consider for analysis. You can define a system such as a reactor , a section of a pipe. Or, you can define the limits of the system by drawing the system boundary , namely a line that encloses the portion of the process that you want to analyze. b. Closed System Figure 6.1 shows a two-dimensional view of a three-dimensional vessel holding 1000 kg of H 2 O. Note that material neither enters nor leaves the vessel , that is, no material crosses the system boundary. Changes can take place inside the system , but for a closed system , no mass exchange occurs with the surroundings.
Figure 6.1 A closed system. c. Open System Figure 6.2 is an example of an open system (also called a flow system ) because material crosses the system boundary.
Figure 6.2 An open steady – state system.
Dr. Ahmed Faiq Al-Alawy
6.3 Steady-State and Unsteady-State Systems
a. Steady – State System Because the rate of addition of water is equal to the rate of removal, the amount of water in the vessel shown in Figure 6.2 remains constant at its original value ( 1000 kg ). We call such a process or system a steady – state process or a steady – state system because
- The conditions inside the process (specifically the amount of water in the vessel in Figure 6.2) remain unchanged with time , and
- The conditions of the flowing streams remain constant with time. Thus, in a steady-state process , by definition all of the conditions in the process (e.g., temperature, pressure, mass of material, flow rate, etc.) remain constant with time. A continuous process is one in which material enters and/or leaves the system without interruption.
b. Unsteady – State System Because the amount of water in the system changes with time (Figure 6.3) , the process and system are deemed to be an unsteady – state ( transient ) process or system. For an unsteady-state process , not all of the conditions in the process (e.g., temperature, pressure, mass of material, etc.) remain constant with time, and/or the flows in and out of the system can vary with time.
Figure 6.4 shows the system after 50 minutes of accumulation (Fifty minutes of accumulation at 10 kg/min amounts to 500 kg of total accumulation).
Figure 6.3 Initial conditions for an open unsteady – state system with accumulation.
Dr. Ahmed Faiq Al-Alawy The material balance for a single component process is
Equation 6.1 can apply to moles or any quantity that is conserved. As an example, look at Figure 6.7 in which we have converted all of the mass quantities in Figure 6.2 to their equivalent values in moles.
Figure 6.7 The system in Figure 6.2 with the flow rates shown in kg mol. If the process is in the steady state , the accumulation term by definition is zero , and Equation 6. simplifies to a famous truism What goes in must come out (In = Out) …6. If you are analyzing an unsteady-state process, the accumulation term over a time interval can be calculated as
The times you select for the final and initial conditions can be anything, but you usually select an interval such as 1 minute or 1 hour rather than specific times.
When you combine Equations 6.1 and 6.3 you get the general material balance for a component in the system in the absence of reaction
Dr. Ahmed Faiq Al-Alawy
Example 6. Will you save money if instead of buying premium 89 octane gasoline at $1.269 per gallon that has the octane you want, you blend sufficient 93 octane supreme gasoline at $l.349 per gallon with 87 octane regular gasoline at $1.149 per gallon? Solution Choose a basis of 1 gallon of 89 octane gasoline , the desired product. The system is the gasoline tank. For simplicity, assume that no gasoline exists in the tank at the start of the blending, and one gallon exists in the tank at the end of the blending. This arrangement corresponds to an unsteady-state process. Clearly it is an open system. The initial number of gallons in the system is zero and the final number of gallons is one. Let x = the number of gallons of 87 octane gasoline added, and y = the number of gallons of 93 octane added to the blend. Since x + y = 1 is the total flow into the tank, y = 1 – x According to Equation (6.4) the balance on the octane number is
The solution is x = 2/3 gal and thus y = 1 – x = 1/3 gal. The cost of the blended gasoline is (2/3) ($l.l49) + (1/3) ($l.349) = $ 1. A value less than the cost of the 89 octane gasoline ($l.269).
6.4 Multiple Component Systems
Suppose the input to a vessel contains more than one component , such as 100 kg/min of a 50% water and 50% sugar (sucrose, C 12 H 22 O 11 , MW = 342.3) mixture (see Figure 6.8). The mass balances with respect to the sugar and water , balances that we call component balances.
Figure 6.8 An open system involving two components.
Dr. Ahmed Faiq Al-Alawy
Example 6. Centrifuges are used to separate particles in the range of 0.1 to 100 μm in diameter from a liquid using centrifugal force. Yeast cells are recovered from a broth (a liquid mixture containing cells) using a tubular centrifuge (a cylindrical system rotating about a cylindrical axis). Determine the amount of the cell-free discharge per hour if 1000 L/hr is fed to the centrifuge, the feed contains 500 mg cells/L, and the product stream contains 50 wt.% cells. Assume that the feed has a density of 1 g/cm^3. Solution This problem involves a steady state , open (flow) system without reaction. Basis = 1 hour
Figure E6. M.B. on cells In (mass) = Out (mass)
P = 1000 g M.B. on fluid In (mass) = Out (mass)
D = (10^6 – 500) g
6.5 Accounting for Chemical Reactions in Material Balances
Chemical reaction in a system requires the augmentation of Equation 6.4 to take into account the effects of the reaction. To illustrate this point, look at Figure 6.10 , which shows a steady – state system in which HC1 reacts with NaOH by the following reaction:
NaOH + HCl NaC1 + H 2 O
Dr. Ahmed Faiq Al-Alawy
Figure 6.10 Reactor for neutralizing HC1 with NaOH.
Equation 6.4 must be augmented to include terms for the generation and consumption of components by the chemical reaction in the system as follows
6.6 Material Balances for Batch and Semi-Batch Processes
A batch process is used to process a fixed amount of material each time it is operated. Initially , the material to be processed is charged into the system. After processing of the material is complete, the products are removed. Batch processes are used industrially for speciality processing applications (e.g., producing pharmaceutical products), which typically operate at relatively low production rates. Look at Figure 6.11a that illustrates what occurs at the start of a batch process, and after thorough mixing, the final solution remains in the system (Figure 6.11b).
Figure 6.11a The initial state of a batch mixing process.
Figure 6.11b The final state of a batch mixing process.
Dr. Ahmed Faiq Al-Alawy
Example 6. A measurement for water flushing of a steel tank originally containing motor oil showed that 0. percent by weight of the original contents remained on the interior tank surface. What is the fractional loss of oil before flushing with water, and the pounds of discharge of motor oil into the environment during of a 10,000 gal tank truck that carried motor oil? (The density of motor oil is about 0.80 g/cm^3 ). Solution Basis: 10,000 gal motor oil at an assumed 77°F The initial mass of the motor oil in the tank was (10000 gal)(3.785 lit/1 gal)(1000 cm^3 /1 lit)(0.8 g/cm^3 )(1 lb/454 g) = 66700 lb The mass fractional loss is 0.0015. The oil material balance is Initial unloaded residual discharged on cleaning 66,700 = 66,700 (0.9985) + 66,700 (0.0015) Thus, the discharge on flushing is 66,700 (0.00 15) = 100 lb.
Questions
- Is it true that if no material crosses the boundary of a system, the system is a closed system?
- Is mass conserved within an open process?
- Can an accumulation be negative? What does a negative accumulation mean?
- Under what circumstances can the accumulation term in the material balance be zero for a process?
- Distinguish between a steady-state and an unsteady-state process.
- What is a transient process? Is it different than an unsteady-state process?
Figure 6.13a Initial condition for the semi-batch mixing process. Vessel is empty.
Figure 6.13b Condition of a semi-batch mixing process after 1 hour of operation.
Dr. Ahmed Faiq Al-Alawy
- Does Equation 6.4 apply to a system involving more than one component?
- When a chemical plant or refinery uses various feeds and produces various products, does Equation 6.4 apply to each component in the plant?
- What terms of the general material balance, Equation (6.5), can be deleted if a. The process is known to be a steady-state process. b. The process is carried out inside a closed vessel. c. The process does not involve a chemical reaction.
- What is the difference between a batch process and a closed process?
- What is the difference between a semi-batch process and a closed process?
- What is the difference between a semi-batch process and an open process?
Answers:
- Yes
- Not necessarily – accumulation can occur
- Yes; depletion
- No reaction (a) closed system, or (b) flow of a component in and out are equal.
- In an unsteady-state system, the state of the system changes with time, whereas with a steady-state system, it does not.
- A transient process is an unsteady-state process.
- Yes
- Yes
- (a) Accumulation; (b) flow in and out; (c) generation and consumption
- None
- A flow in occurs
- None, except in a flow process, usually flows occur both in and out
Problems
- Here is a report from a catalytic polymerization unit: Charge: Pounds per hour Propanes and butanes 15, Production: Propane and lighter 5, Butane 2, Polymer missing What is the production in pounds per hour of the polymer?
Dr. Ahmed Faiq Al-Alawy
Problem 2
Problem 3
Dr. Ahmed Faiq Al-Alawy
