ERT 416/3 CHAPTER 3: FUNDAMENTALS OF MATERIAL AND ENERGY BALANCES IN BIOPROCESS PLANT. MISS. RAHIMAH BINTI OTHMAN (Email: email@example.com). COURSE OUTCOMES. OUTLINES. Engineering principle for the design process.
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ERT 416/3CHAPTER 3: FUNDAMENTALS OF MATERIAL AND ENERGY BALANCES IN BIOPROCESS PLANT
MISS. RAHIMAH BINTI OTHMAN
The general equations for material and energy balances.
Material out = Material in + Generation – Consumption - Accumulation
Total energy entering the system – Total energy leaving the system
= Change in the total energy of the system
∆(Energy of the system) = E in – E out =∆U+∆EK+ ∆EP
Balances on a Batch Mixing Process
Two methanol-water mixtures are contained in separate flasks. The first mixture contains 40.0 wt % methanol, and the second contains 70.0 wt % methanol. If 200 g of the first mixture is combined with 150 g of the second, what are the mass and composition of the product?
Input = Output
Yield of a Crystallization Process
A salt solution weighing 10000 kg with 30 wt % Na2CO3 is cooled
to 293 K (20 °C). The salt crystallizes as the decahydrate. What
will be the yield of Na2CO3•10H2O crystals if the solubility is 21.5
kg anhydrous Na2CO3/100 kg of total water? Do this for the
(a) Assume that no water is evaporated.
(b) Assume that 3% of the total weight of the solution is lost by
evaporation of water in cooling.
FIGURE3. Process flow for crystallization
The molecular weights are 106.0 for Na2CO3, 180.2 for
10H20, and 286.2 for Na2CO3 •10H2O. The process
flow diagram is shown in Fig. 3, with W being kg H2O
evaporated, S kg solution (mother liquor), and C kg
crystals of Na2CO3 •10H2O.Making a material balance
around the dashed line box for water for part
(a), where W = 0.
where (180.2)/(286.2) is wt fraction of water in the crystals.
Making a balance for Na2CO3,
Solving the two equations simultaneously, C = 6370 kg of Na2CO3•10H2O crystals and S = 3630 kg solution.
For part (b), W = 0.03(10000) = 300 kg H2O. Equation (13)
Equation (14) does not change, since no salt is in the W stream.
Solving Eqs. (14) and (15)simultaneously, C = 6630 kg of Na2CO3•10H2O crystals and S = 3070 kg solution.
The diagram shows the main steps in a process for producing a polymer. From the following data, calculate the stream flows for a production rate of 10, 000 kg/h.
Filter wash water approx. 1 kg/ 1 kg polymer
Recovery column yield 98 % (percentage recovered)
Dryer feed ~ 5 % water, product specification 0.5 % H2O
Polymer losses in filter and dryer ~ 1 %
Only those flows necessary to illustrate the choice of system boundaries and method of calculation are given in the solution.
Basis: 1 h
Take the first system boundary round the filter and dryer.
Take the next boundary round the reactor system; the feeds to the reactor can then be calculated.
Now consider recovery system;
Composition of effluent is 22 kg monomer, 54,461 kg water.
Consider reactor monomer feed
Material balances (mass balances) are based on the fundamental “law of conservation of mass (not volume, not moles)”. In particular, engineers are concerned with doing mass balances around the specific processes.
Waste easily assimilated by the environment
Inert waste always available
Plant, Process or Unit Operación
Reusable residues in other operation
Based on how the process varies with time.
Steady-state processis one that does not change with time. Every time we take a snapshot, all the variables have the same values as in the first snapshot.
Unsteady-state (Transient) processis one that changes with time. Every time we take a snapshot, many of the variables have different values than in the first snapshot.
∆(Energy of the system) = Q+W
∆(Energy of the system)
Differential Balance is a balance taken at a specific instant in time. It is generally applied to a continuous process. If the process is at steady state, a differential balance applied at any time gives the same result.
We will apply differential balances to steady-state continuous processes.
Each term in a differential balance represents a process stream and the mass flow rate of the chemical(s) in that stream.
Integral balanceis a balance taken at two specific instants in time. It describes what has happened over the time period between the two points. An integral balance is generally applied to the beginning and the end of a batch process. It accounts for what happens to the batch of chemicals.
We will apply integral balances to batch processes.
Each term in an integral balance represents a process stream and the mass of the chemical(s) in that stream.
Ammonia is produced from nitrogen and hydrogen in a batch reactor.
At time t = 0 there are n0 mol of NH3 in the reactor, and at a later time tf the reaction terminates and the contents of the reactor, which include nf mol of ammonia, are withdrawn.
Between to and tf no ammonia enters or leaves through the reactor boundaries, so the general balance equation is simply
generation = accumulation
Air is bubbled through a drum of liquid hexane at a rate of 0.100 kmol/min. The gas stream leaving the drum contains 10.0 mole% hexane vapor. Air may be considered insoluble in liquid hexane. Use an integral balance to estimate the time required to vaporize 10.0 m3 of the liquid.
Example: Integral Balances on Semibatch and Continuous Processes – Cont’
Example: A process unit involves 3 chemical components. How many mass balances can be written?
We can write 4 balances. We can write a total balance and 3 component balances.
Stream F contains 500 kg of O2 and 700 kg of CH4. Label the stream.
Solution: Note that the component masses must add to the total. The total mass in F is 1200 kg. Thus,
Stream F F=1200 kg
m O2 = 500 kg
m CH4 = 700 kg
1200 kg of a mixture of O2 , N2 and CH4 are fed to a process. The stream has 20% O2 by mass. Label the stream.
( Note the Mass of i in the stream is F xi )
Solution: The composition is partially known. Note that the fractional compositions must add to 1.0. Thus, we can write two alternatives
Using fractional composition
Feed Stream F, F=1200 kg
xO2 = 0.2
xN2 = ?
xCH4 = 1. – 0.2 - xN2 = 0.8 - xN2
Feed Stream F, F=1200 kg
mO2 = 240 kg
mN2 = ?
mCH4 = 1200 - 240 - mN2 = 960 - mN2
The catalytic dehydrogenation of propane is carried out in a continuous packed-bed reactor. One thousand kilograms per hour of pure propane is preheated to a temperature of 670 oC before it passes into the reactor. The reactor effluent gas, which includes propane, methane, and hydrogen, is cooled from 800 oC to 110 oC and fed to an adsorption tower, where the propane and propylene are dissolved in oil. The oil then goes to a stripping tower in which it is heated, releasing the dissolved gases; these gases are recompressed and sent to a distillation column in which the propane and propylene are separated. The propane stream is recycled back to join the feed to the reactor preheater.The product stream from the distillation column contains 98 % propylene, and the recycle stream is 97 % propane. The stripped oil is recycled to the adsorption tower. Draw the flowchart for this system.
Write the values and units of all known stream variables at the locations of the streams on the chart.
A stream containing 21 mole % O2, and 79 % N2 at 320 oC and 1.4 atm flowing at a rate of 400 mol/h might be labeled.
Assign algebraic symbols to unknown stream variables [such as (kg solution/min), x (lbm N2/lbm), and n (kmol C3H8)] and write these variable names and their associated units on the chart.
If you did not know the flow rate of the stream described in the first illustration of step 1, you might label the stream.
ndf (= nunknowns – nindep eqns)
If ndf = 0, there are n independent equations in n unknowns and the problem can in principle be solved.
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 solutions; in either case, plugging into calculations is likely to be a waste of time.
If ndf < 0, there are more independent equations than unknowns. Either the flowchart is incompletely labeled or the problem is overspecified with rebundant and possibly inconsistent relations. Again there is little point wasting time trying to solve it until the equations and unknowns are brought into balance.
2. Energy Balances
1. Material Balances
3. Process Specifications
Source Of Equations Relating Unknown Process Stream Variables
4. Physical Properties and Laws
6. Stoichiometric Relations
5. Physical Constrains
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, with 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.
Suppose we had given an additional piece of information- for example, that the entering air contains 10.0 mole% water. The flowchart would then appear as follows;
Choose as a basis of calculation an amount or flow rate of one of the process streams.
Draw flowchart and fill in all known variable values, including the basis of calculation. Then label unknown stream variables on the chart.
Express what the problem statement asks you to determine in terms of the labeled variables.
If you are given mixed mass and mole units for a stream (such as a total mass flow rate and component mole fractions or vice versa), convert all quantities to one basis.
Do the degree-of-freedom analysis.
If the number of unknowns equals the number of equations relating them (i.e: ndf =0), write the equations in an efficient order (minimizing simultaneous equations) and circle the variables for which you will solve.
Solve the equations.
Calculate the quantities requested in the problem statement if they have not already been calculated.
If the stream quantity or flow rate ngwas 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.
Material Balances on a Distillation Column
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 2000 L/h and the specific gravity of the fed mixture is 0.872. Determine the mass flow rate of the overhead product stream and the mass flow rate and composition (mass fractions) of the bottom product stream.
We will explicitly illustrate the implementation of the steps of the procedure just outlined.
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.
Draw and label the flowchart.
Material and Energy Balances on an Air Conditioner
Fresh air containing 4.00 mole% water vapor is to be cooled and dehumidified to a water content of 1.70 mole% H2O. A stream of fresh air is combined with a recycle stream of previously dehumidified air and passed through the cooler. The blended stream entering the unit contains 2.30 mole% H2O. In the air conditioner, some of the water in the feed stream is condensed and removed as liquid. A fraction of the dehumidified air leaving the cooler is recycled and the remainder is delivered to a room. Taking 100 mol of dehumidified air delivered to the room as a basis of calculation, calculate the moles of fresh feed, moles of water condensed, and moles of dehumidified air recycled.
An Extraction- Distillation Process
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 consists of extraction of the acetone from the water onto methyl isobutyl ketone (MIBK), which dissolves acetone but is nearly immiscible in water. The description the 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 liquid phases. 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 the process) is referred to as the raffinate, and the phase rich in the solvent (MIBK) 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 feed 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 to 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 MIBK is fed to the first stage and 75 kg is fed to the second stage. The extraction from the first stage is found to contain 27.5 wt% acetone. (All percentages in the remainder of this paragraph are weight percents.) The second-stage raffinate has a mass of 43.1 kg and contains 5.3% acetone, 1.6 % MIBK, and 93.1 % water, and the second extract contains 9.0% acetone, 88.0 % MIBK, and 3.0% water. The overhead product from the distillation column contains 2.0% MIBK, 1.0% 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.
= (2016830.989 - 2037730.989) kJ / batch
= - 20900 kJ / batch
Rajah 1 Graf aliran tunai tahunan melawan tahun bagi muatan 15000 tan/tahun
Rajah 2 Graf aliran tunai tahunan melawan tahun bagi muatan 25000 tan/tahun
Rajah 3 Graf aliran tunai tahunan melawan tahun bagi muatan 35000 tan/tahun
Rajah 4 Graf aliran tunai tahunan melawan tahun bagi muatan 100000 tan/tahun
MISS RAHIMAH OTHMAN