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4.3d Degree-of-Freedom Analysis. Everyone who has done material balance calculations has the frustrating experience of spending a long time deriving and attempting to solve equations for unknown process variables, only to discover that not enough information is available .
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4.3d Degree-of-Freedom Analysis Everyone who has done material balance calculations has the frustrating experience of spending a long time deriving and attempting to solve equations for unknown process variables, only to discover that not enough information is available. Before you do any lengthy calculations, you can use a properly drawn and labeled flowchart to determine whether you have enough information to solve a given problem. The procedure for doing so is referred to as degree-of-freedom analysis. To perform a degree-of-freedom analysis, draw and completely label a flowchart, count the unknown variables on the chart, then count the independent equations relating them, and subtract the second number from the first. The result is the number of degree of freedom of the process,
There are three possibilities: 1.If ndf = 0, there are n independent equations in n unknowns and the problem can in principle be solved. 2.If ndf > 0, there are more unknowns than independent equations relating them, and at least ndf additional variable values must be specified before the remaining variable values can be determined. Either relations have been overlooked or the problem is underspecified and has infinitely many solution; in either case, plunging into calculations is likely to be a waster of time. 3.If ndf < 0, there more independent equations than unknowns. Either the flowchart is incompletely labeled or the problem is overspecified with redundant and possible inconsistent relations. Again there is little point wasting time trying to solve it until the equations and unknowns are brought into balance.
Sources of equations relating unknown process stream variables include the following: 1.Material balances. For a nonreactive process, no more than nms independent material balances may be written, where nms is the number of molecular species (e.g., CH4, O2) involved in the process. For a reactive process, the procedure becomes more complicated. 2.An energy balance. If the amount of energy exchanged between the system and its surroundings is specified or if it is one of the unknown process variables, an energy balance provides a relationship between inlet and outlet material flows and temperatures. 3.Process specifications. The problem statement may specify how several process variables are related.
4.Physical properties and laws. Two of the unknown variables may be the mass and volume of a stream material, in which case a tabulated specific gravity for liquids and solids or an equation of state for gases would provide an equation relating the variables. In other instances, saturation or equilibrium conditions for one or more of the process streams may provide needed relations. 5.Physical constraints. 1-xA-xB 6.Stoichiometric relations. If chemical reactions occur in a system, the stoichiometric equations of the reactions (e.g., 2H2+O22H2O) provide relationships between the quantities of the reactants consumed and of the products generated.
Example 4.3-4 A stream of humid air enters a condenser in which 95% of the water vapor in the air is condensed. The flow rate of the condensate (the liquid leaving the condenser) is measured and found to be 225 L/h. Dry air may be taken to contain 21 mole% oxygen, while the balance nitrogen. Calculate the flow rate of the gas stream leaving the condenser and the mole fractions of oxygen, nitrogen, and water in this stream. Solution Basis: 225 L/h Condensate six unknowns three material balances O2, N2, H2O Determining from the given volumetric flow rate and the known specific gravity and molecular weight of liquid water. 95% of the water is condensed. The problem is therefore underspecified, and there is no point in attempting to solve it.
Suppose now that we had been given an additional piece of information – for example, that the entering air contains 10.0 mole% water. The flowchart would then appear as right: Density relationship 95% Condensation O2 Balance N2 Balance H2O Balance Total outlet gas flow rate Outlet gas composition
4.3e General Procedure for Single-Unit Process Material Balance Calculations Given a description of a process, the values of several process variables, and a list of quantities to be determined: 1.Choose as a basis of calculation an amount or flow rate of one of the process streams. 2.Draw a flowchart and fill in all known variable values, including the basis of calculation. Then label unknown stream variables on the chart. 3.Express what the problem statement asks you to determine in terms of the labeled variables. 4.If you are given mixed mass and mole units for a stream, convert all quantities to one basis or the other using the method of Section 3.3. 5.Do the degree-of-freedom analysis.
6.If the number of unknowns equals the number of equations relating them, write the equations in an efficient order (minimizing simultaneous equations) and circle the variable for which you will solve. 7.Solve the equations. 8.Calculate the quantities requested in the problem statement if they have not already been calculated. 9.If a stream quantity or flow rate ng was given in the problem statement and another value nc was either chosen as a basis or calculated for this stream, scale the balanced process by the ratio ng/nc to obtain the final result.
Example 4.3-5 A liquid mixture containing 45.0% benzene (B) and 55.0% toluene (T) by mass is fed to a distillation column. A product stream leaving the top of the column (the overhead product) contains 95.0 mole% B, and a bottom product stream contains 8.0% of the benzene fed to the column (meaning that 92% of the benzene leaves with the overhead product). The volumetric flow rate of the feed stream is 2000 L/h and the specific gravity of the feed mixture is 0.872. Determine the mass flow rate of the overhead product stream and the mass flow and composition (mass fractions) of the bottom product stream. Solution 1.Choose a basis. Having no reason to do otherwise, we choose the given feed stream flow rate (2000 L/h) as the basis of calculation. 2.Draw and label the flowchart. Every component mass flow rate in every process stream can be expressed in terms of labeled quantities and variables.
3.Write expressions for the quantities requested in the problem statement. In terms of the quantities labeled on the flowchart, the quantities to be determine are to be determined 4.Convert mixed units in overhead product stream. Basis: 100 kmol overhead 95.0 kmol B, 5.00 kmol T (95.0 kmol B) (78.11 kg B/kmol B) = 7420 kg B (5.00 kmol T) (92.13 kg T/kmol T) = 461 kg T (7420 kg B) + (461 kg T) = 7881 kg mixture yB2 = (7420 kg B)/(7881 kg mixture) = 0.942 kg B/kg 5.Perform degree-of-freedom analysis. 4 unknowns ( ) -2 material balances (since there are two molecular species in this nonreactive process) -1 density relationship (relating the mass flow rate to the given volumetric flow rate of the feed) -1 specified benzene (8% in bottom – 92% in overhead) 0 degree of freedom
6.Write system equations and outline a solution procedure. ~Volumetric flow rate conversion. ~Benzene split fraction. ~Benzene balance ~Toluene balance 7.Do the algebra 8.Calculate additional quantities requested in the problem statement.
4.4 Balances on Multiple-Unit Processes In general terms, a “system” is any portion of a process that can be enclosed within a hypothetical box (boundary). It may be the entire process, an interconnected combination of some of the process units, a single unit, or a point at which two or more process streams come together or one stream splits into branches. The inputs and outputs to a system are the process streams that intersect the system boundary. Figure 4.4-1 A flowchart for a two-units process is shown in Figure 4.4-1. Five boundaries drawn about portions of the process define systems on which balances may be written. Boundary A encloses the entire process (three inputs and three outputs) Boundary B encloses a feed stream mixing point (two inputs and one output) Boundary C encloses Unit 1 (one input and two outputs) Boundary D encloses a stream splitting point (one input and two outputs) Boundary E encloses Unit 2 (two inputs and one output) overall balances The difference is that with multiple-unit processes you may have to isolate and write balances on several subsystems of the process to obtain enough equations to determine all unknown variables. When analyzing multiple-unit processes, carry out degree-of-freedom analyses on the overall process and on each subsystem, taking into account only the streams that intersect the boundary of the system under consideration.
Example 4.4-1 A labeled flowchart of a continuous steady-state two-unit process is shown below. Each stream contains two components, A and B, in different proportions. Three streams whose flow rates and/or compositions are not known are labeled 1, 2, and 3. Calculate the unknown flow rates and compositions of stream 1, 2, and 3. Solution Basis – Given Flow Rates
Degree-of-Freedom Analysis We first outline the solution procedure by performing degree-of-freedom analyses on different systems. Remember that only variables associated with streams intersecting a system boundary are counted in the analysis of that system. Overall system: 2 unknowns ( ) – 2 balances (2 species) = 0 degree of freedom Determine Mixing point: 4 unknowns ( ) – 2 balances (2 species) = 2 degree of freedom Unit 1: 2 unknowns ( ) – 2 balances (2 species) = 0 degree of freedom Determine Mixing point: 2 unknowns ( ) – 2 balances (2 species) = 0 degree of freedom Determine overall system balances balances on Unit 1 balances on mixing point
Calculations Overall Mass Balance: Overall Balance on A:
Mass Balance on Unit 1: A Balance on Unit 1:
Mass Balance on Stream Mixing Point: A Balance on Stream Mixing Point:
Example 4.4-2 A mixture containing 50.0 wt% acetone and 50.0% wt% water is to be separated into two streams – one enriched in acetone, the other in water. The separation process of extraction of the acetone from the water into methyl isobutyl ketone (MIBK), which dissolves acetone but is nearly immiscible with water. The description that follows introduces some of the terms commonly used in reference to liquid extraction processes. The process is shown schematically below. The acetone (solute) – water (diluent) mixture is first contacted with the MIBK (solvent) in a mixer that provides good contact between the two liquids. A portion of the acetone in the feed transfers from the aqueous (water) phase to the organic (MIBK) phase in this step. The mixture passes into a settling tank, where the phases separate and are separately withdrawn. The phase rich in the diluent (water, in this process) is referred to as the raffinate, and the phase rich in the solvent is the extract. The mixer-settler combination is the first stage of this separation process.
The raffinate passes to a second extraction stage where it is contacted with a second stream of pure MIBK, leading to the transfer of more acetone. The two phases are allowed to separate in a second settler, and the raffinate from this stage is discarded. The extracts from the two mixer-settler stages are combined and fed to a distillation column. The overhead effluent is rich in acetone and is the process product. The bottom effluent is rich in MIBK and in a real process would be treated further and recycled back into the first extraction stage, but we will not consider recycle in this example. In a pilot-plant study, for every 100 kg of acetone-water fed to the first extraction stage, 100 kg of MIBK is fed to the first stage and 75 kg is fed to the second stage. The extract from the first stage is found to contain 27.5 wt% acetone. The second-stage raffinate has a mass of 43.1 kg and contains 5.3 wt% acetone, 1.6 wt% MIBK, and 93.1 wt% water, and the second-stage extract contains 9.0 wt% acetone, 88.0 wt% MIBK, and 3.0 wt% water. The overhead product from the distillation column contains 2.0 wt% MIBK, and 1.0 wt% water, and the balance acetone. Taking a basis of calculation of 100 kg acetone-water feed, calculate the masses and compositions (component weight percentages) of the Stage 1 raffinate and extract, the Stage 2 extract, the combined extract, and the distillation overhead and bottoms products.
Solution combining mixer and settler Overall process: 4 unknowns ( ) – 3 balances (3 species) = 1 degree of freedom First extractor: 5 unknowns ( ) – 3 balances (3 species) = 2 degree of freedom Second extractor: 4 unknowns ( ) – 3 balances (3 species) = 1 degree of freedom Extract mixing point: 6 unknowns ( ) – 3 balances (3 species) = 3 degree of freedom Distillation column: 7 unknowns ( ) – 3 balances (3 species) = 4 degree of freedom
Two extraction units: 3 unknowns ( ) – 3 balances (3 species) = 0 degree of freedom Balances Around Two-Extractor Subsystem
At this point, we may quickly determine that we can go no farther. There are four remaining unknowns – m5, mA6, mM6, and mW6. Whether we choose the overall process or the distillation column as our system, we will only have three independent equations and hence one degree of freedom, and so we will be unable to solve the problem. Moreover, since acetone, MIBK, and water all appear in both outlet streams, we cannot solve for any one of the individual unknowns. (If there were no water in the overhead product from the distillation column, for example, we could deduce that mW6 = mW4). The problem is thus underspecified; unless another piece of information is furnished, the amounts and compositions of the distillation column products are indeterminate.
