UNIT 3: MATERIAL BALANCE

UNIT 3: MATERIAL BALANCE
HUSNA BINTI ZULKIFLY

LEARNING OUTCOME At the end of Unit 3, students should be able to :
Describe chemical process classification Perform material balance calculation Explain chemical reaction stoichiometry Solve balance problem on various type of unit process

SUBUNIT FOR UNIT 3 3.1 Process Classification 3.2 Balances
3.3 Material Balances Calculation 3.4 Chemical Reaction Stoichiometry 3.5 Balance on various type of unit process 3.6 Phase System

3.1: PROCESS CLASSIFICATION
Learning outcome: Students should be able to describe chemical process classification

Chemical process categories
Chemical process may be classified based on: 1. How the process varies with time Steady state Transient / Unsteady state 2. How the process was built to operate Batch process Continuous process Semibatch process A transient event is a short-lived burst of energy in a system caused by a sudden change of state.

1. How the process varies with time Steady state process Values of all variables in a process (i.e. temperatures, pressures, volumes, flowrates) do not change with time, except possibly for minor fluctuation about constant mean values. Unsteady state process Process variables in the process change with time. Every time; the variables value are different from the first operation

Based on how the process was built to operate: Batch process The feed streams are fed to the process to get it start. The feed material is then processed through various process steps and the finished products are creating during one or more of the steps. The process is fed and products result only at specific times. No mass crosses the system boundaries between the time the feed is charged and the time the product is removed Batch processing is commonly used when relatively small quantities of a product are to be produces on any single occasion. Example: Rapidly add reactants to a tank and removed the products and unconsumed reactants sometime later when the system has come to equilibrium

Based on how the process was built to operate: Continuous process Inputs and outputs flow continuously throughout the duration of the process At every instant, the process is fed and product is produced Continuous processing is better suited to large production rates Example: Pump a mixture of liquids into distillation column at a constant rate and steadily withdraw product streams from the top and bottom of the column.

Based on how the process was built to operate: Semi batch process (Also called semi-continuous) Neither batch nor continuous Process that has some characteristics continuous and batch processes. Example: Some chemical in the process are handling batch process. Some chemical are processed continuously. Example: Allow the contents of a pressurized gas container to escape to the atmosphere

1. How the process varies with time Steady state Transient / Unsteady state 2. How the process was built to operate Batch process Continuous process Semibatch process Batch and semi-batch processes are unsteady state operation. Continuous processes may be either steady state or transient. **Continuous processing are usually run as close to steady state as possible. Unsteady state condition exist during the start-up of a process and changes in process operation condition.

RECAP!! Batch Process Continuous Process Semibatch Process
Feed is fed at the beginning of the process Continuous Process The input and outputs flow continuously throughout the duration of process Semibatch Process Any process neither batch nor continuous

RECAP!! Steady state Transient or unsteady state
Process variables not change with time Transient or unsteady state Process variables change with time

3.2: Balances Learning Outcome:
Student should be able to perform material balance calculation

General Material Balance
Material balances are based on conservation of mass. It accounts for the material, which enters, leaves, or accumulates in a system and also for the material that are generated or consumed by chemical reaction in the system. PROCESS

THE GENERAL BALANCE EQUATION
Suppose propane is a component of both the input and output streams of a continuous process unit shown below Flow rates of the input and output are measured and found to be different.

Several possible explanations for the observed difference between the measured flow rates:
Propane is being consumed as a reactant or generated as a product within the unit. Propane is accumulating in the unit possibly absorbing on the walls. Propane is leaking from the unit. The measurements are wrong.

If the measurement are correct and there are no leaks, the other possibilities generation or consumption in a reaction and accumulation within the process unit are all that can be account for a difference between the input and output flow rates. A balance on a conserved quantity in a system may be written in the following general ways:

Example Each year 50,000 people move into a city, 55,000 people move out, 30,000 are born and 19,000 die. Write a balance on the population of the city. Let P denote people Use the balances equation:

Answer: 50,000 People/year move in = input
30,000 People/year are born = generation 55,000 People/year move out = output 19,000 People/year die = consumption Thus,

Answer: Input+ Generation – Output – Consumption = Accumulation
50,000 P/y + 30,000 P/y – 55,000 P/y – 19,000 P/y = 6,000 P/y * Accumulation = 6,000 P/y * Each year the city’s population increase by 6,000 people.

TYPES OF BALANCES Differential Balances Integral Balances
Based on rates kg/hr, kmol/min, and Ibm/hr (quantity divided by time) Indicates what is happening in a system at an instant of time Integral Balances Based on amounts Ibm, kg, and kmol (quantity only) Indicates what is happening between two instant of time

The general balance equation
This general balance equation may be written for any material that enters or leaves any process system; it can be applied to the total mass of this material or to any molecular or atomic species involved in the process. The general balance equation may be simplified according to the process at hand. For example, by definition, the accumulation term for steady-state continuous process is zero. Thus the above equation becomes: Input + generation = output + consumption

The general balance equation
For physical process, since there is no chemical reaction, the generation and consumption terms will become zero, and the balance equation for steady-state physical process will be simply reduced to: Input = Output

EXAMPLE 1000 kg/h of a mixture of Benzene and Toluene containing 50% Benzene by mass is separated into two fractions. The mass flow rate of Benzene in the top stream is 450kg/h and that of Toluene in the bottom stream is 475kg/h. The operation is at steady state. Write balances on benzene and toluene to calculate unknown component flow rates in output stream.

Since there is no chemical reaction, the generation and consumption terms will become zero, and the balance equation for steady-state physical process will be simply reduced to: Input = Output Benzene Balance Mass Benzene IN = Mass Benzene OUT (1000 kg total/h)(0.5 kg B/kg total) = 450 kg B/h + FB2 Therefore, FB2 = 50 kg B/h Toluene Balance Mass Toluene IN = Mass Toluene OUT (1000 kg total/h)(0.5 kg T/kg total) = 475 kg T/h + FT1 Therefore, FT1 = 25 kg T/h

3.3: MATERIAL BALANCES CALCULATION
Learning Outcome: Student should be able to perform material balance calculation

INTRODUCTION Material balance problems are variations on single theme : Given values of some input and output stream variables Derive equations from a description of a process and collection of process data Solve equations for unknown process variables

a. FLOWCHARTS Organize the information in a way that is convenient for subsequent calculations Draw a flowchart of the process : Boxes to represent process units (reactors, mixers, separation units, etc) Lines with arrows to represent inputs and outputs Flowchart must be fully labelled for each input and output stream values of known process variables symbols for unknown variables

@ FLOWCHARTS Labelling of a flowchart
Write the values and units of all known stream variables at the locations of the streams on the chart Example: A stream containing 21 mole% oxygen and 79% nitrogen at 320oC and 1.4 atm flowing at a rate of 400 mol/hr 400 mol/hr 400 mol/hr @ 0.21 mol o2/mol 0.79 mol N2/mol T = 320oC, P= 1.4 atm 84 mol o2/hr 316 mol N2/hr T = 320oC, P= 1.4 atm

Write the values and units of all known stream variables at the locations of the streams on the chart Example: A stream containing 40 mole% oxygen and 60% nitrogen at flowing at a rate of 100 kmol/min 100 kmol/min 100 kmol/min @ 0.4 kmol o2/kmol 0.6 kmol N2/kmol 40 kmol o2/min 60 kmol N2/min

Write the values and units of all known stream variables at the locations of the streams on the chart Example: 10 lbm mixture containing 30% CH4 by mass, 40% C2H4 and 30% C2H6 enter a batch process 10 lbm 10 lbm @ o.3 lbm CH4/lbm 0.4 lbm C2H4/lbm 0.3 lbm C2H6/lbm 3.0 lbm CH4 4.0 lbm C2H4 3.0 lbm C2H6

FLOWCHARTS Labelling of a flowchart
Assign and write algebraic symbols to unknown stream variables and their associated units on the chart Mass, m mole, n Mass flow rate, molar flow rate, Example: A stream containing 21 mole% oxygen and 79% nitrogen at 320oC and 1.4 atm. ( ? Flowrate) (mol/hr) 0.21 mol o2/mol 0.79 mol N2/mol T = 320oC, P= 1.4 atm

FLOWCHARTS Labelling of a flowchart
Assign and write algebraic symbols to unknown stream variables and their associated units on the chart Mass, m mole, n Mass flow rate, molar flow rate, Example: A 400 mol/hr stream containing unknown composition of oxygen and nitrogen enters a process at 320oC and 1.4 atm . 400 (mol/hr) y mol o2/mol (1-y) mol N2/mol T = 320oC, P= 1.4 atm

3.3: MATERIAL BALANCES CALCULATION
Learning Outcome: Student should be able to perform material balance calculation

INTRODUCTION Material balance problems are variations on single theme : Given values of some input and output stream variables Derive equations from a description of a process and collection of process data Solve equations for unknown process variables

EXERCISE Calculate the indicated quantities in terms of the labelled stream variables 100 lbmole Calculate n (lbmole CH4) m (lbm C2H4) Answer 30.0 lbmole CH4 1120 lbm C2H4 0.3 lbmole CH4/lbmole 0.4 lbmole C2H4/bmole 0.3 lbmole C2H6/lbmole 250 (kg/hr) Calculate Mass flow rate of C7H8 (kg C7H8/ min) Answer 250(1-x)/60 kg C7H8/ min x kg C6H6/kg (1-x) kg C7H8/kg T = 320oC, P= 1.4 atm

b. FLOWCHART SCALING AND BASIS OF CALCULATION
The first step in balancing a process is to choose a basis of calculation; all unknown variables are then determined to be consistent with this basis A basis of calculation is an amount or flow rate of one stream (mass or moles) or stream component in a process. If no stream amounts / flow rates are known, YOU NEED TO ASSUME. TIPS - if mass fractions are known, choose total mass or mass flow rate of that stream as a basis (eg 100 kg or 100 kg/hr) - if mole fractions are known, choose total number of moles or molar flow rate

b. FLOWCHART SCALING AND BASIS OF CALCULATION
Procedure of changing values of all stream amounts or flow rates by a proportional amount while leaving the stream compositions unchanged is referred to as scaling the flowchart Scaling up : final stream quantities are larger than original quantities Scaling down: final stream quantities are smaller than original quantities Cannot scale masses or mass flow rates to molar quantities or vice versa by simple multiplication.

Procedure for scaling up
a) Scale factor b) Convert the amount/ flow rate c) Flow chart of scaled-up process

EXERCISE 1 Suppose a kilogram of benzene is mixed with a kilogram of toluene. The output from this process is 2 kg of a mixture that is 50% benzene by mass. Draw flowchart of the process Scale the flowchart accordingly if the desired output is 600 kg. Scale the flowchart accordingly if the desired output is 600 lbm/hr.

EXERCISE 1 Suppose a kilogram of benzene is mixed with a kilogram of toluene. The output from this process is 2 kg of a mixture that is 50% benzene by mass. FLOWCHART 1 kg C6H6 2 kg 1 kg C7H8 0.5 kg C6H6/kg 0.5 kg C7H8/kg

EXERCISE 1 Suppose a kilogram of benzene is mixed with a kilogram of toluene. The output from this process is 2 kg of a mixture that is 50% benzene by mass. Scale the flowchart accordingly if the desired output is 600 kg. a) Scale factor b) Convert the amount/ flow rate Feed Benzene stream: Toluene stream: Product

EXERCISE 1 Suppose a kilogram of benzene is mixed with a kilogram of toluene. The output from this process is 2 kg of a mixture that is 50% benzene by mass. c) Flow chart of scaled-up process 300 kg C6H6 600 kg 300 kg C7H8 0.5 kg C6H6/kg 0.5 kg C7H8/kg

EXERCISE 1 Suppose a kilogram of benzene is mixed with a kilogram of toluene. The output from this process is 2 kg of a mixture that is 50% benzene by mass. Scale the flowchart accordingly if the desired output is 600 lbm/hr. 300 lbm/hr C6H6 600 lbm/hr 300 lbm/hr C7H8 0.5 lbm C6H6/lbm 0.5 lbm C7H8/lbm

EXERCISE 2 It is desired to achieve the same separation with continuous feed of lbmoles/hr. Scale the flowchart accordingly 50.0 mol 0.95 mol A/mol 0.05 molB/mol 100.0 mol Calculate scale factor = 12.5 lbmoles/hr Feed 1250 Top product 50 x 12.5 = 625 lbmoles/hr Bottom 12.5 x 12.5 = 156 lbmole A/hr 37.5 x lbmole B/hr 0.60 mol A/mol 0.40 molB/mol 12.5 mol A 37.5 mol B

c. BALANCING A PROCESS

For nonreactive and steady state process; Input = Output
Overall mass balance Total moles IN = Total moles OUT We know IN = n1 and OUT = n2 + n3 Therefore, n1 = n2 + n3 Species Balance on A Moles A IN = Moles A OUT We know IN = n1yA1 and OUT = n2yA2 + n3yA3 Therefore, n1yA1 = n2yA2 + n3yA3 Species Balance on B Moles B IN = Moles B OUT We know IN = n1yB1 and OUT = n2yB2 + n3yB3 Therefore, n1yB1 = n2yB2 + n3yB3 This leaves us with 3 balance equations: n1 = n2 + n3 n1yA1 = n2yA2 + n3yA3 n1yB1 = n2yB2 + n3yB3

Consider following example
c. BALANCING A PROCESS Consider following example Suppose two separate streams of 3.0 kg/min of benzene and 1.0 kg/min of toluene are mixed. Calculate mass flow rate of product and the mass fractions in output stream. M = 4kg/min X = 0.75 kg benzene/kg 3.0 kg C6H6/min 𝒎 kg/min x kg C6H6/kg (1-x) kg C7H8/kg 1.0 kg C7H8/min

3.0 kg C6H6/min 𝒎 kg/min x kg C6H6/kg (1-x) kg C7H8/kg 1.0 kg C7H8/min Balancing the process From general material balance equation: Input+ Generation – Output – Consumption = Accumulation For nonreactive and steady state process; Input = Output

Total mass balance: Total mass in = Total mass out
3.0 kg C6H6/min 𝒎 kg/min x kg C6H6/kg (1-x) kg C7H8/kg 1.0 kg C7H8/min Input = Output Total mass balance: Total mass in = Total mass out 3.0 kg/min kg/min = 𝑚 𝒎 =𝟒.𝟎 𝐤𝐠/𝐦𝐢𝐧 Benzene balance mass of benzene in = mass of benzene out 3.0 kg/min = x x 𝑚 kg/min 3.0 = 4.0 x x = 0.75 kg C6H6/kg

3.0 kg C6H6/min 𝒎 kg/min x kg C6H6/kg (1-x) kg C7H8/kg 1.0 kg C7H8/min Thus, mass fraction of benzene xB = 0.75 kg C6H6/kg mass fraction of toluene xT = = 0.25 kg C7H8/kg

3.4: Chemical Reaction Stoichiometry
Learning Outcome: Student should be able to explain chemical reaction stoichiometry

TERMINOLOGY Stoichiometry Stoichiometric equation
Stoichiometric coefficient Stoichiometric ratio Limiting and Excess Reactant Fractional conversion Yield Selectivity

Stoichiometry Theory of the proportions in which chemical species combine with one another. Stoichiometric Equation Statement of the relative molecules/moles of reactants and products that participate in the reaction. Example: 2 SO2 + O2 → 2 SO3 Stoichiometric Coefficient Number that precede the formulas for each species taking part in a reaction. Example: SO2 + 1 O2 → 2 SO3 2, 1, and 2 are called as Stoichiometric coefficients of SO2, O2 and SO3 respectively.

Stoichiometric Ratio Ratio of species stoichiometry coefficients in the balanced reaction equation. Can be used as a conversion factor to calculate the amount of particular reactant (or product) that was consumed (or produced). Example: 2 SO2 + O2 → 2 SO3 The stoichiometric ratio is

EXAMPLE 1 C4H8 + 6O2 → 4CO2 + 4H2O What is the stoichiometric coefficient of CO2? Is this stoichiometric equation balanced? Check the no. of each elements on LEFT and RIGHT to make sure its EQUAL LEFT: C = 4, O = (6x2) = 12 and H = 8; RIGHT: C = 4, O = (2x4) + 4 = 12 and H = (4x2) = 8 The equation is balanced! Stoichiometric coefficient of CO2 is 4

C4H8 + 6O2 → 4CO2 + 4H2O What is the stoichiometric ratio of H2O to O2? How many lb-mol O2 react to form 400 lb-mol CO2?

Reactant that limits the amount of product that can be formed.
Limiting Reactant Reactant that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed. Limiting Reactant problems always involve : Identify the Limiting Reactant (LR) Calculate the amount of product obtained from the Limiting Reactant 6 steps to identify a limiting reactant and to calculate the amount of product obtained from the limiting reactant Write out balanced equation Convert reactant mass to moles Divide the moles of each reagent by its coefficient Identify the limiting reagent (refer to smallest number) Multiply limiting reagent by coefficient of product Convert moles of product back into mass

Excess Reactant The unfinished reactant after a reaction proceeded to completion. nA,feed = No. of moles of excess reactant A in the FEED nA,stoichiometry = No. of moles of the stoichiometric requirement of A (Amount needed to react completely with the limiting reactant Fractional conversion In many cases, chemical reactions do not proceed to completion and only a fraction will be converted. Fractional and percentage conversion are used. nA,feed = No. of moles of reactant A in the FEED nA,outlet = No. of moles of the reactant A in the OUTLET

EXAMPLE 2 Finely ground aluminum reacts with oxygen gas to produce aluminum oxide (Al2O3) as the only product. If 20.0 g of aluminum are mixed with 20.0g of oxygen gas; Identify Limiting Reactant Determine how many grams of aluminum oxide are present when the reaction is finished Calculate the % of the excess reactant. Given MW of O2 = 32 g/mol, Al = 27 g/mol, Al2O3 = g/mol.

Step 1: Balanced equation: 4Al(s) + 3O2(g)  2Al2O3(s)
Step 2: Convert reactant mass to moles Step 3: Divide the moles of each reactant by its coefficient (The coefficients are found in the balanced equation) Step 4: Identify the Limiting Reactant (The limiting reagent is the one that give the smallest number) 0.185 is smaller than 0.208, so Aluminum (Al) is the limiting reactant

Step 5: Multiply limiting reactant original number of moles with product stoichiometric coefficient.
Step 6: Convert moles of product back into grams/concentration to calculate the amount

To determine the % of excess reactant (O2)

EXAMPLE 3 If 100 moles of reactant are fed to the reactor and 90 moles react, calculate the fractional conversion of this reaction.

EXAMPLE 4 If 20 mol/min of reactant are fed and the conversion is 80%. Determine amount of reactant in the outlet after the reaction.

Yield Calculated to represent the maximum amount obtainable Most chemical reactions do not produce 100 % yield of product because of: Side reactions (unwanted reactions) Reversible reactions (products → reactant) Losses in handling and transferring Selectivity Sometimes when reactant has reacted, they can form desired main product and undesired side product. The selectivity of the desired reaction over the undesired reaction is given as:

EXAMPLE 5 For the balanced equation shown, 2C3H6 + 9O2  6CO2 + 6H2O If the reaction of 20.7 lbm of C3H6 produces 6.81 lbm of CO2, what is the percent yield? Given MW of C3H6 = 42.08, CO2 = 44.01

EXAMPLE 6 10 lb-mole/h of product C is produced by reacting reactant A with B. An undesirable product D of 30 lb-mole/h is produced by a side reaction. Determine the selectivity of the reaction. 2A + B  3C A + 2B  D

3.5 Balance on various type of unit process (single/multiple unit operation, reactive and non-reactive) Learning Outcome: Student should be able to solve balance problem on various type of unit process

Basis Flow chart Variables Balances Equation Scale up / down

GENERAL PROCEDURE Balances on Single-Unit Processes (Non Reactive)
Choose a basis of calculation ( amount or flow rate of one of the process streams) Draw a flowchart (fill all known variables value with unit, including basis of calculation, label unknown variables with unit) Identify problem statement – from labelled unknown variables Convert all quantities to one basis if you are given mixed mass and mole units for a stream. (example: mass flow rate and mole fractions) Balance the process to write equations * if the number of unknowns equals to number of independent equations relating them, you will be able to solve the problems Solve the equations – to solve problems statement Scale up or scale down if necessary

SINGLE UNIT

EXAMPLE 1 Given INPUT of a distillation column is 100 mol/min of a 40 mole% hexane (H) and 60 mole% benzene (B). Molar flow rate of H in the top product is 32 mol/min and the molar flow rate of B in the bottom product is 48 mol/min. Calculate 𝑛 𝐵 at the top and 𝑛 𝐻 at the bottom. Basis √ Flowchart

Problem statement : 𝑛 𝐵 at the top and 𝑛 𝐻 at the bottom
Balance process, write and solve equation For nonreactive and steady state process; Input = Output Hexane: 100 mol total/min (0.4 mol H/mol total) = (32 mol H/min + 𝑛 𝐻 ) 𝑛 𝐻 = 8 mol/min Benzene: 100 mol total/min (0.6 mol B/mol total) = (48 mol B/min + 𝑛 𝐵 ) 𝑛 𝐵 = 12 mol/min

EXAMPLE 2 1000 kg/h of a mixture of Benzene and Toluene containing 50% Benzene by mass is separated into two fractions. The mass flow rate of Benzene in the top stream is 450kg/h and that of Toluene in the bottom stream is 475kg/h. The operation is at steady state. Write balances on benzene and toluene to calculate unknown component flow rates in output stream.

F1 kg total/h 450 kg B/h FT1 1000 kg total/h 0.5 kg B/ kg 0.5 kg T/ kg F2 kg total/h 475 kg T/h FB2

Since there is no chemical reaction, the generation and consumption terms will become zero, and the balance equation for steady-state physical process will be simply reduced to: Input = Output Benzene Balance Mass Benzene IN = Mass Benzene OUT (1000 kg total/h)(0.5 kg B/kg total) = 450 kg B/h + FB2 Therefore, FB2 = 50 kg B/h Toluene Balance Mass Toluene IN = Mass Toluene OUT (1000 kg total/h)(0.5 kg T/kg total) = 475 kg T/h + FT1 Therefore, FT1 = 25 kg T/h

EXERCISE Skim milk is prepared by the removal of some of the fat from whole milk. This skim milk is found to contain 90.5% by mass of water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original milk contained 4.5% fat, calculate its composition (in %) assuming that fat only was removed to make the skim milk and that there are no losses in processing.

Solution Assume Basis: 100 kg of skim milk.
Whole milk mT, W kg Skim milk mT,S kg 0.045 kg f/kg Xw Xp Xc Xa 0.001 kg f/kg 0.905 kg w/kg 0.035 kg p/kg 0.051 kg c/kg 0.008 kg a/kg Fat removed Assume Basis: 100 kg of skim milk. Let the fat which was removed from it to make skim milk be x kg. Skim milk contains therefore, 0.1kg of fat.

Solution Assume Basis: 100 kg of skim milk.
Whole milk mT, W kg Skim milk mT,S = 100kg 0.045 kg f/kg Xw Xp Xc Xa 0.001 kg f/kg 0.905 kg w/kg 0.035 kg p/kg 0.051 kg c/kg 0.008 kg a/kg Fat removed, x Assume Basis: 100 kg of skim milk. Let the fat which was removed from it to make skim milk be x kg. Skim milk contains therefore, 0.1kg of fat.

Continue… For nonreactive and steady state process; Input = Output
Total mass balance: mT, W = (100 + x) kg Fat balance: Fat input= (x + 0.1) kg and as it is known that the original fat content was 4.5%, so: Mass fat ÷ total mass = mass fraction (x + 0.1) / (100 + x) = 0.045 x = (100 + x) x = 4.6 kg

Continue… So; the composition of the whole milk is Given fat = 4.5%
Balance on species (mass fraction) Water = (0.905 x 100) / = 0.865 Protein = (0.035x100) / = 0.033 Carbohydrate = (0.051x 100) / = 0.049 Ash = (0.008x100) /104.6 = 0.008

Continue… So; the composition of the whole milk is Given fat = 4.5%
Convert to composition in percent (x 100%) Water = 86.5% Protein = 3.3% Carbohydrate = 4.9% Ash = 0.8% * Checking using the percent: Total = 100%

Multiple Unit Consist more than one unit operation

GENERAL PROCEDURE Balances on Multiple-Unit Processes (Non Reactive)
The procedure is basically the same as single-unit. When analysing multiple-unit processes, carry out degree-of-freedom analyses on the overall process and on each subsystem. Only consider the streams that intersect the boundary of the system under consideration. Do not begin to write and solve equations for a subsystem until you have verified that it has zero degrees of freedom. Make sure to find same number of equations with the unknown variables!

Degrees of Freedom Draw and completely label the flow sheet.
For a nonreactive process that involves N species, up to N independent material balance equations may be written. Draw and completely label the flow sheet. Count the number of unknowns, N. Write down all the independent equations linking these unknowns. Count the number of equations, M. Degree of Freedom, DOF = N-M IF N=M, then the problem is well specified and the problem, in general, can be solved. IF N>M, then the problem is underspecified, and will, in general, have infinite solutions. IF N<M, then the problem is over specified, and will, in general, have no solutions.

DOF (EXAMPLE) Flow chart √ Number of unknown variables m1, m2 and V1
100 kg m2 kg 0.20 kg NaOH/kg 0.80 kg H2O/kg 0.080 kg NaOH/kg 0.920 kg H2O/kg m1 (kg H2O) V1 (liters H2O) Flow chart √ Number of unknown variables m1, m2 and V1 N=3

EXAMPLE Independent equation NaOH and H2O, can write 2 balance
100 kg m2 kg 0.20 kg NaOH/kg 0.80 kg H2O/kg 0.080 kg NaOH/kg 0.920 kg H2O/kg m1 (kg H2O) V1 (liters H2O) Independent equation NaOH and H2O, can write 2 balance V1 = m1 x density M= 3 For a nonreactive process that involves N species, up to N independent material balance equations may be written. Degree of Freedom, DOF = N-M DOF = 3-3=0, solvable problem

EXAMPLE For nonreactive and steady state process; Input = Output
100 kg m2 kg 0.20 kg NaOH/kg 0.80 kg H2O/kg 0.080 kg NaOH/kg 0.920 kg H2O/kg m1 (kg H2O) V1 (liters H2O) For nonreactive and steady state process; Input = Output Total mass balance: m1 = m2 NaOH balance: (o.20) (100)= (0.080) m2 m2 = 250 kg NaOH m1 = m2 – 100 m1 = 150 kg H2O Volume of water, V1 = m1 /ρ = 150 kg H2O ÷ 1000 kg/m3 = 0.15 m3 = 150 L

EXAMPLE (MULTIPLE UNIT)
A labelled 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 labelled 1,2 and 3. Calculate the unknown flow rates and compositions of streams 1,2 and 3. 40.0 kg/hr 0.90 kg A/kg 0.10 kg B/kg 30.0 kg/hr 0.60 kg A/kg 0.40 kg B/kg 100 kg/hr 1 2 3 0.50 kg A/kg 0.50 kg B/kg 30.0 kg/hr 0.30 kg A/kg 0.70 kg B/kg

Degree of freedom analysis
30.0 kg/hr 0.60 kg A/kg 0.40 kg B/kg 40.0 kg/hr 0.90 kg A/kg 0.10 kg B/kg 100 kg/hr 𝒎𝟐 kg/hr 𝒎𝟑 kg/hr 𝒎𝟏 kg/hr 0.50 kg A/kg 0.50 kg B/kg X1 kg A/kg 1-X1 kg B/kg X2 kg A/kg 1-X2 kg B/kg X3 kg A/kg 1-X3 kg B/kg 30.0 kg/hr 0.30 kg A/kg 0.70 kg B/kg Degree of freedom analysis Remember that only variables associated with streams intersecting a system boundary are counted in the analysis of the system. Overall system 2 unknowns (𝒎𝟑, X3 ) – 2 balances (2 species) = 0 DOF

30.0 kg/hr 0.60 kg A/kg 0.40 kg B/kg 40.0 kg/hr 0.90 kg A/kg 0.10 kg B/kg 100 kg/hr 𝒎𝟐 kg/hr 𝒎𝟑 kg/hr 𝒎𝟏 kg/hr 0.50 kg A/kg 0.50 kg B/kg X1 kg A/kg 1-X1 kg B/kg X2 kg A/kg 1-X2 kg B/kg X3 kg A/kg 1-X3 kg B/kg 30.0 kg/hr 0.30 kg A/kg 0.70 kg B/kg Unit 1 2 unknowns (𝒎𝟏, X1 ) – 2 balances (2 species) = 0 DOF Mixing point Since 𝒎𝟏, X1 considered as known from above, 2 unknowns (𝒎𝟏, X1 ) – 2 balances (2 species) = o DOF

You may start your calculation!
30.0 kg/hr 0.60 kg A/kg 0.40 kg B/kg 40.0 kg/hr 0.90 kg A/kg 0.10 kg B/kg 100 kg/hr 𝒎𝟑 kg/hr 𝒎𝟏 kg/hr 𝒎𝟐 kg/hr 0.50 kg A/kg 0.50 kg B/kg X1 kg A/kg 1-X1 kg B/kg X2 kg A/kg 1-X2 kg B/kg X3 kg A/kg 1-X3 kg B/kg 30.0 kg/hr 0.30 kg A/kg 0.70 kg B/kg Overall mass balance = 𝑚3 𝑚3 = 60.0 kg/hr Overall balance on A (0.5)(100)+(0.3)(30)=(0.9)(40)+(0.6)(30)+(X3)(60) X3= kg A/kg

30.0 kg/hr 0.60 kg A/kg 0.40 kg B/kg 40.0 kg/hr 0.90 kg A/kg 0.10 kg B/kg 100 kg/hr 𝒎𝟑 kg/hr 𝒎𝟏 kg/hr 𝒎𝟐 kg/hr 0.50 kg A/kg 0.50 kg B/kg X1 kg A/kg 1-X1 kg B/kg X2 kg A/kg 1-X2 kg B/kg X3 kg A/kg 1-X3 kg B/kg 30.0 kg/hr 0.30 kg A/kg 0.70 kg B/kg Mass Balance on Unit 1 100 = 𝑚 1 𝑚 1 = 60.0 kg/hr A balance on Unit 1 (0.5)(100) = (0.9)(40)+(X1)(60) X1= kg A/kg

30.0 kg/hr 0.60 kg A/kg 0.40 kg B/kg 40.0 kg/hr 0.90 kg A/kg 0.10 kg B/kg 100 kg/hr 60 kg/hr 60 kg/hr 𝒎𝟐 kg/hr 0.50 kg A/kg 0.50 kg B/kg 0.233 kg A/kg 0.767 kg B/kg X2 kg A/kg 1-X2 kg B/kg kg A/kg kg B/kg 30.0 kg/hr 0.30 kg A/kg 0.70 kg B/kg Mass Balance on mixing point = 𝑚 2 𝑚 2 = 90.0 kg/hr A balance on mixing point (0.233)(60) + (0.300)(30) = X2 𝑚2 X2= kg A/kg

MASS BALANCE ON REACTIVE PROCESS
Method 1: Extent of Reaction (ξ) Method 2: Atomic Balance

Method 1: Extent of Reaction (ξ) Extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction. ni = OUTPUT moles (batch) or molar flow rate (continuous) of species i ni0 = INLET (feed) moles (batch) or molar flow rate (continuous) of species i. vi = Stoichiometric coefficient of species i (+ve for product and –ve for reactant)

EXAMPLE Consider ammonia formation reaction: N2 + 3H2  2NH3
Suppose the feed to a continuous reactor consists of 100 mol/s of N2, 300 mol/s of H2, and 1 mol/s of an inert gas (argon). It is given that the unreacted N2 after the reaction is 30 mol/s. Find extent of reaction Determine the outlet molar flow rate of H2, Ar, and NH3.

N2 + 3H2  2NH3 Find extent of reaction Strategy 1: Write expression for EACH species molar flow rate in terms of ξ. Feed N2: nN2 = 100 – 1ξ (Equation 1) Feed H2: nH2 = 300 – 3ξ (Equation 2) Feed Ar: nAr = 1 (Equation 3) Product NH3: nNH3 = 0 + 2ξ (Equation 4)

N2 + 3H2  2NH3 Strategy 2: Substitute known feed or product values (based on information given) For N2: nN2 = 100 – 1ξ (Equation 1) Given from question, nN2 = 30 mol/s N2 Therefore, 30 = 100 – 1ξ Strategy 3: Solve and find the extent of reaction, ξ ξ = 70 mol/s

N2 + 3H2  2NH3 Determine the outlet molar flow rate of H2, Ar, and NH3. Strategy 4: Find the remaining unknown values to solve the material balance Substitute ξ = 70 mol/s into equation 2, and 4 Feed H2: nH2 = 300 – 3ξ = 300 – 3(70) = 90 mol/s H2 Feed Ar: nAr = 1 mol Ar/s (No change because it is an inert) Product NH3: nNH3 = 0 + 2ξ = 2(70) = 140 mol/s NH3

STRATEGIES TO SOLVE USING THIS METHOD
Write expression for EACH species molar flow rate (or no. of moles) in terms of extent of reaction, ξ. Substitute known feed or product values (based on information given) Solve and find the extent of reaction, ξ. Find the remaining unknown values to solve the material balance.

Method 2: Atomic Species Balance Atomic species balance generally lead to the most straightforward solution procedure. All balance on atomic species (Example C, H, O, etc) take the form INPUT = OUTPUT. Atomic balance is used for us to obtain equations that we can use to solve the balance. Atomic balance can be done for ALL atoms that is present in the system Always remember to identify the number of unknowns in the systems. Then find the same number of equations from atomic balance in order to solve the problem.

EXAMPLE Consider dehydrogenation of C2H6 C2H6 → C2H4 + H2
100 kmol/min of ethane is fed to the reactor. The molar flow rate of H2 in the product stream is 40 kmol/min. Find n1 and n2 STRATEGIES Identify the number of unknowns in the systems. Find the same number of equations from atomic balance in order to solve the problem.

We need 2 equations! We can do Atomic balance for Carbon and Hydrogen
C2H6 → C2H4 + H2 Number of unknowns is 2 (n1 and n2). Atomic balance for Carbon We need 2 equations! We can do Atomic balance for Carbon and Hydrogen

C2H6 → C2H4 + H2 Atomic balance for Hydrogen

Solve the equations

3.6 Phase System Learning Outcome:
Student should be able to determine physical properties of process materials

INTRODUCTION Problems in process analysis are rarely so conveniently self-contained Various physical properties of process materials must be determined to derive additional relations among system variables before carrying out complete material balance on a process (not all information given in problem statements) Matter also exhibits physical properties. Physical properties are used to observe and describe matter. Physical properties can be observed or measured without changing the composition of matter. These are properties such as mass, weight, volume, and density.

Density of Liquids and Solids
Density of liquid and solid are assumed constant with temperature and pressure → incompressible For solutions and mixtures, the density of a liquid or solid solution can be approximated by the following: xi is the mass fraction of component i. The accuracy of equation 1 over equation 2 depends on the system considered.

EXAMPLE What is the average density of a hexane/octane mixture comprised of 10 kg of hexane and 30 kg of octane? The density of hexane is 0.66 kg/m3 and that of octane is kg/m3. Mass fraction hexane = 0.25; octane = 0.75 Density =

Density of Gases/ Ideal Gas Equation
Gases are highly compressible, and therefore the incompressibility assumption does not apply. Equations of State (EOS): correlation or physically based equations that relate Pressure, Volume, and Temperature that exist inside a material (typically gas). These three thermodynamic variables are easily measured and make up what is known as a PVT relationship. The simplest model: Ideal Gas Equation of State PV = nRT EOS’s then must include the pressure, temperature, and volume to account for the variation of the density of a gas.

Density of Gases/Ideal Gas Equation
The ideal gas EOS is based on the primary assumptions that 1. There are no intermolecular forces between the gas particles. 2. The volumes of molecules are negligible (i.e. the gas particles can be treated as “point” particles). The equation can be rewritten in other forms. These alternate forms may be useful depending on the situation; they are given below:

Understanding these assumptions helps to understand the utility and limitations of this EOS. For example; Ideal gas assumption cannot be used at higher pressures because molecules will interact with each other. At lower temperatures there can be substantial error associated with the ideal gas equation, because molecules do not have enough thermal energy to overcome the interactions between one another. Generally, the ideal gas EOS works well at pressures of 1 atm or less and temperatures greater than O oC (both conditions should be satisfied).

EXAMPLE 100 g of nitrogen is stored in a container at 296K and 17.7 psia. Assuming ideal gas behavior, calculate the container volume in L. Given MW of nitrogen = 28 g/mol, R = liter.psi/mol.K PV =nRT n = 3.57 mol V = 72 L

Standard Temperature and Pressure (STP)
STP conditions are defined as 0oC and 1 atm. Standard cubic meters (SCM) and standard cubic feet (SCF) refer to volumes of gas evaluated at STP conditions. The volume of one mole of ideal gas at STP is L = m3. The volume of 1 lb-mole of ideal gas at STP is ft3.

Ideal Gas Mixtures For a gas mixture inside a container;
Partial Pressure: Pressure, Pi, that the moles of gas i in the gas mixture would exert if the moles of gas i was all that was present in the container (at the same temperature T). Pure Component Volume: This is the volume, vi, that the moles of pure gas i would occupy at the total pressure P and temperature T of the mixture.

In an ideal gas mixture, the various gas molecules do not interact
In an ideal gas mixture, the various gas molecules do not interact. This means that the pressure exerted by each component i is independent of the others; thus its partial pressure is given by PiV = niRT Dividing the above by the ideal gas EOS as applied to the total mixture (which is assumed to behave ideally), we get Pi V/ (PV) = niRT / (nRT) Therefore, the partial pressure of gas i in an ideal gas mixture can be calculated from the total pressure P and its mole fraction yi. Pi = yiP Similar series of calculation for pure component volume; vi = yiV

EXAMPLE An ideal gas mixture at 10 bar absolute in 100 m3 tanks contains 50 mole % H2 and 50 mole % N2. What is the partial pressure and pure volume of H2? ans: 5 bar, 50 m3

Non-ideal Gases In reality; molecules of a gas do interact → all real gases are nonideal!!! Ideal gas EOS may be a very good approximation at low pressures and high temperatures; but at higher pressures and/or lower temperatures, the impact of intermolecular interactions on gas behavior increases. Then, methods for nonideal gases must be used to account for the effect of these interactions on the relationship between P, V, and T (and related properties such as density) of a gas. Some of the methods for nonideal gas are: Virial Equation of State Van der Waals Equation of State Soave-Redlich-Kwong (SRK) Equation of State Compressibility Factor Equation of State Kay’s Rule (for non ideal gas mixtures)

Non-ideal Gases (DEFINITIONS)
Critical Temperature: This is the temperature above which a pure component can no longer coexist as a mixture of liquid and vapor phases. Critical Pressure: This is the pressure at the critical point. Critical Point (Critical State) of a Fluid: The critical point for a fluid refers to the state when the fluid is at Tc and the corresponding critical pressure, Pc. Such a fluid can still coexist, at Tc and Pc, as a liquid and a vapor mixture. However, if the temperature increases past Tc then coexistence is not possible and the fluid is referred to as a supercritical fluid EXAMPLE: Phase diagram of water

Non-ideal Gases (DEFINITIONS)
Reduced Pressure (Pr): This is the actual pressure of a fluid divided by its critical pressure Reduced Temperature (Tr): This is the actual temperature of a fluid divided by its critical temperature Law of Corresponding States: This is an empirically based principle that states that all fluids deviate from ideality in a similar fashion, when compared at the same reduced temperature T/Tc and reduced pressure P/Pc. Pitzer Accentric Factor: This factor (symbol w), is a parameter used in non-ideal equations of state that takes into account the geometry and polarity of a molecule.

THANK YOU

UNIT 3: MATERIAL BALANCE

Similar presentations

Presentation on theme: "UNIT 3: MATERIAL BALANCE"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

UNIT 3: MATERIAL BALANCE

Similar presentations

Presentation on theme: "UNIT 3: MATERIAL BALANCE"— Presentation transcript:

Similar presentations

About project

Feedback