A flowchart is a convenient way of organizing process information for subsequent calculations, as mentioned in the previous chapter. To obtain maximum benefit from the flowchart in material balance calculations, one must write the values and units of all known stream variables at the loca-tions of the streams on the chart. Assign algebraic symbols to unknown stream variables and write these variable names and their associated units on the chart. The use of consistent notation is generally advantageous. The following notation is used. For example, m (mass), m (mass flow rate), n (moles), n (mole flow rate), V (volume), V (volumetric flow rate), x (compo-nent fractions (mass or mole) in liquid streams), and y (compo(compo-nent fractions in gas streams).
A systematic procedure will be outlined for solving single-unit processes where there are no reactions (consumption = generation = 0), and when processes are continuous and under steady state (accumulation = 0). The procedure will form the foundation for more complex problems involving multiple units and processes with reactions (described in later chapters).
For a stream to be fully specified, the flow rate and the composition of each component should be known. If any of these items are not given, then it will be considered as unknown. If the stream composition is unknown or if some of the component masses are known, represent the component masses directly and use a lower case letter for each chemical. If the stream composition is known from fractional compositions, represent the compo-nent masses directly and label them. If the stream composition is partially known with fractional compositions and the total is known, represent the component masses indirectly and use lowercase x, y, and z for each frac-tional composition [4,5].
3.3.1 Stream Specification
A material stream is fully specified if we can express the mass flow rates of each component in the stream. For example, the following stream is fully specified since the mass flow rate of components in the stream are known.
Stream F contains 30 kg/min of O2 and 70 kg/min of CH4. Note that the com-ponent masses must be added and be equal to the total mass. The total mass in F is 100 kg/min:
Stream kg/min) pro-cess. The stream contains 20% O2 by mass. The mass flow rate of component i in the stream, mi = F × xi.
In this case, the stream compositions are partially known, accordingly the stream is not fully specified. Note that the fractional compositions in a spe-cific stream must add to 1; thus, we can write two alternatives:
Stream kg/s)
or using component masses,
Stream kg/s)
Example 3.1 Material Balance on Distillation Column Problem
A mixture of three components (A, B, and C) enters a separation process.
The three components appear in the distillate with variable composi-tion. In contrast, only B and C appear in the bottom. Write a proper set of material balance equations.
Solution
Known quantities: No data is given.
Find: Possible material balance equation.
Analysis: The process-labeled flowchart is shown in Example Figure 3.1.1.
System: Distillation column
We can write four equations in total, but only three are independent, because the fourth one can be derived from the other three. So, in doing the math, we have our choice on which of the three equations we want to use.
Component balance (m: mass flow rate):
The number of independent material balance equations equals the num-ber of components. When writing down the independent equations, use the overall material balance around the system plus all component bal-ances less one, as shown here:
Independent material
where n is the number of components involved in the system.
Example 3.2 Ethanol–Water Separation Process Problem
A mixture containing 10% ethanol (E) and 90% H2O (W) by weight is fed into a distillation column at the rate of 100 kg/h. The distillate con-tains 60% ethanol and the distillate is produced at a rate of one tenth that of the feed. Draw and label a flowchart of the process. Calculate all unknown stream flow rates and compositions.
Solution
Known quantities: Feed and distillate flow rates and composition.
Find: Calculate all unknown stream flow rates and composition.
A
Schematic of a distillation column.
Analysis: The process-labeled flowchart is shown in Example Figure 3.2.1.
Assumptions: Continuous process, steady state, no reactions.
Basis: 100 kg/h of feed.
NDF = number of unknowns − number of independent equations
− number of relations
NDF = 3 − 2 − 1 = 0 (solvable)
Three independent material balance equations: Overall material balance and a component mass balance (one of the two components).
Relation (one relation only: distillate flow rate is one-tenth that of the feed).
D = 0.1 × F = 0.1 × 100 = 10 kg/h Overall total material in the system:
F = D + B
100 kg/h = 10 + B; B = 90 kg/h Component balance (ethanol):
0 1 100. × kg/h=0 6 10. × kg/h+xE,3×90kg/h⇒xE,3=0 044. Check your answer:
Perform a mass balance on the components (water) not used in the ear-lier solution procedure.
Mass of water in feed = mass of water in the distillate + mass of water in the bottom 90 kg = 4 kg + 0.956 × 90 kg = 90 kg
Example 3.3 Separation Process Problem
A feed stream is flowing at a mass flow rate of 100 kg/min. The stream contains 20 kg/min NaOH and 80 kg/min of water. The distillate flows
100 kg/h
10% E 1
2
3 Distillation column 90% W
D (kg/h)
Bottoms, B E xE,3 W 1– xE,3 60% E 40% W
EXAMPLE FIGURE 3.2.1
Schematic of the ethanol–water separation process.
at 40 kg/min and contains 5 kg/min NaOH. Determine bottom stream mass flow rate and composition.
Solution
Known quantities: Feed and distillate streams are fully specified (i.e., mass flow rate and compositions are known).
Find: Bottom stream flow rate and compositions.
Analysis: The process-labeled flowchart is shown in Example Figure 3.3.1.
Basis: 100 kg/min of the feed stream.
NDF = 2 (unknowns) − 2 (independent equations) − 0 (relation)
= 0 (problem is solvable) Total mass balance:
100kg/min =m m2+ 3
100kg/min=40kg/min+ m3
Component mass balance (NaOH):
20kg/min=5kg/min+ mNaOH
There are two simple linear algebraic equations with two unknowns (m3,mNaOH), solving for m3 and mNaOH:
m3=60kg/min mNaOH=15kg/min
mH O2 =m m3− NaOH=60 15 45− = kg/min 100 kg/min
20 kg/min NaOH 80 kg/min H2O
40 kg/min 5 kg/min NaOH
? kg/min H2O 2
2
3 Distillation column
m3 m3 = ? m2
mNaOH,3 mH2O,3 EXAMPLE FIGURE 3.3.1
Schematic of the sodium hydroxide–water separation process.