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Dr. Ahmed Faiq Al-Alawy
Chapter 7
A General Strategy for Solving Material Balance Problems
7.1 Problem Solving
An orderly method of analyzing problems and presenting their solutions represents training in logical thinking that is of considerably greater value than mere knowledge of how to solve a particular type of problem.
7.2 The Strategy for Solving Problems
- Read and understand the problem statement.
- Draw a sketch of the process and specify the system boundary.
- Place labels for unknown variables and values for known variables on the sketch.
- Obtain any missing needed data.
- Choose a basis.
- Determine the number of unknowns.
- Determine the number of independent equations, and carry out a degree of freedom analysis.
- Write down the equations to be solved.
- Solve the equations and calculate the quantities asked for.
- Check your answer. Example 7. A thickener in a waste disposal unit of a plant removes water from wet sewage sludge as shown in Figure E7.l. How many kilograms of water leave the thickener per 100 kg of wet sludge that enter the thickener? The process is in the steady state.
Solution
Basis: 100 kg wet sludge The system is the thickener (an open system). No accumulation, generation, or consumption occurs. The total mass balance is
Figure E7.
100 kg Thickener Wet Sludge
70 kg Dehydrated Sludge
Water =?
Dr. Ahmed Faiq Al-Alawy In = Out 100 kg = 70 kg + kg of water Consequently, the water amounts to 30 kg. Example 7. A continuous mixer mixes NaOH with H 2 O to produce an aqueous solution of NaOH. Determine the composition and flow rate of the product if the flow rate of NaOH is 1000 kg/hr, and the ratio of the flow rate of the H 2 O to the product solution is 0.9. For this process,
- Sketch of the process is required.
- Place the known information on the diagram of the process.
- What basis would you choose for the problem?
- How many unknowns exist?
- Determine the number of independent equations.
- Write the equations to be solved.
- Solve the equations.
- Check your answer. Solution
- The process is an open one, and we assume it to be steady state.
- Because no contrary information is provided about the composition of the H 2 O and NaOH streams, we will assume that they are 100% H 2 O and NaOH, respectively.
Dr. Ahmed Faiq Al-Alawy
Degree of Freedom Analysis The phrase degrees of freedom have evolved from the design of plants in which fewer independent equations than unknowns exist. The difference is called the degrees of freedom available to the designer to specify flow rates, equipment sizes, and so on. You calculate the number of degrees of freedom (ND) as follows: Degrees of freedom = number of unknowns — number of independent equations ND = NU – NE When you calculate the number of degrees of freedom you ascertain the solve ability of a problem. Three outcomes exist: Case ND Possibility of Solution NU = NE 0 Exactly specified (determined); a solution exists NU > NE >0 Under specified (determined); more independent equations required NU < NE <0 Over specified (determined)
For the problem in Example 7.2 , NU = 4 NE = 4 So that ND = NU – NE = 4 – 4 = 0 And a unique solution exists for the problem.
Example 7. A cylinder containing CH 4 , C 2 H 6 , and N 2 has to be prepared containing a CH 4 to C 2 H 6 mole ratio of 1.5 to 1. Available to prepare the mixture is (l) a cylinder containing a mixture of 80% N 2 and 20% CH 4 , (2) a cylinder containing a mixture of 90% N 2 and 10% C 2 H 6 , and (3) a cylinder containing pure N 2. What is the number of degrees of freedom, i.e., the number of independent specifications that must be made, so that you can determine the respective contributions from each cylinder to get the desired composition in the cylinder with the three components?
Solution A sketch of the process greatly helps in the analysis of the degrees of freedom. Look at Figure E7.3.
Dr. Ahmed Faiq Al-Alawy
Do you count seven unknowns — three values of xi and four values of Fi? How many independent equations can be written? Three material balances: CH 4 , C 2 H 6 , and N 2 One specified ratio: moles of CH 4 to C 2 H 6 equal 1.5 or (XCH4/X (^) C2H6) = 1. One summation of mole fractions: x Fi^4 1
Thus, there are seven minus five equals two degrees of freedom (ND = NU – NE = 7 – 5 = 2). If you pick a basis, such as F 4 = 1, one other value has to be specified to solve the problem to calculate composition of F 4.
Questions
- What does the concept ―solution of a material balance problem‖ mean?
- (a) How many values of unknown variables can you compute from one independent material balance? (b) From three independent material balance equations? (c) From four material balances, three of which are independent?
- If you want to solve a set of independent equations that contain fewer unknown variables than equations (the over specified problem), how should you proceed with the solution?
- What is the major category of implicit constraints (equations) you encounter in material balance problems?
- If you want to solve a set of independent equations that contain more unknown variable than equations (the underspecified problem), what must you do to proceed with the solution?
Figure E7.
Dr. Ahmed Faiq Al-Alawy
Answers:
- (a) Two; (b) two of these three: acetic acid, water, total; (c) two; (d) feed of the 10% solution (say F) and mass fraction ω of the acetic acid in P; (e) 14% acetic acid and 86% water
- Not for a unique solution because only two of the equations are independent.
- F, D, P, ωD2, ωP
- Three unknowns exist. Because only two independent material balances can be written for the problem, one value of F, D, or P must be specified to obtain a solution. Note that specifying values of ωD2 or ωP1 will not help.
Supplementary Problems (Chapter Seven):
Problem 1
Dr. Ahmed Faiq Al-Alawy
Problem 2
