Ert 455 manufacturing & production of biological product Material Balances

Material Balances Law of Conservation of Mass: Mass can neither be created nor destroyed (except in nuclear reactions). Because of this, we can write equations called "mass balances" or "material balances". Any process being studied must satisfy balances on the total amount of material, on each chemical component, and on individual atomic species. Later in the course, we'll use the Law of Conservation of Energy (1st Law of Thermodynamics) to write similar balance equations for energy. ERT 455 Session 2012/2013

Process Classification To determine in which categories process fall. Batch Process No mass crosses the system boundaries between the time the feed is charged and the product is removed. Small quantities are produce on single occasion. Continuous Process Input and output flow continuously throughout the duration of the process. Eg.: Pump a mixture of liquids into distillation column. Large production rate. Semibatch Process Any process neither batch nor continuous. Eg: Content of pressurized gas escape to atmosphere; slowly blend several liquid in a tank from which nothing is being withdrawn ERT 455 Session 2012/2013

Process Operation ERT 455 Session 2012/2013 Steady State Value of process variable do not change with time. Continuous process. Unsteady state/ Transient Process variable change with time. Batch and continuous process. ERT 455 Session 2012/2013

General Balance Equation Accumulation within system Flow in through system boundaries Flow out through system boundaries Generation (Produced within system) Consumption (consumed within system) ERT 455 Session 2012/2013

Example The General Balance Equation Each year 50,000 people move into a city, 75,000 people move out, 22 000 are born, and 19 000 die. Write a balance on the population of the city. Input + Generation – Output – Consumption = Accumulation ANSWER: Each year the city’s population decrease by 22 000 people ERT 455 Session 2012/2013

Forms of Balance Equations Differential Form Balances that indicate what is happening in a system at an instant in time. Balances equation terms = rate (rate of input, rate of generation). Units of balance = balances quantity unit/time unit (people/yr, g SO2/s, barrels/day, etc). Usually the best choice for a continuous process. When formulated for an instant in time, the result is an ordinary differential equation. Integral Form (Also called cumulative form) Written using total amounts as terms, so it describes the overall effect. Unit of balances = amount, g SO2, barrels, etc. Often a good choice for batch processes ERT 455 Session 2012/2013

Rules can be used to simplify the material balance equation. The system is any process or portion of a process chosen by the engineer for analysis. A system is said to be "open" if material flows across the system boundary during the interval of time being studied; "closed" if there are no flows in or out. Accumulation is usually the rate of change of holdup within the system -- the change of material within the system. It may be positive (material is increasing), negative (material decreasing), or zero (steady state). ERT 455 Session 2012/2013

If the system does not change with time, it is said to be at steady state, and the net accumulation will be zero. For our purposes, the generation and consumption terms are the consequence of chemical reaction. Note that while the total mass of a system and elements (or "atoms") are conserved, individual species are not. If there is no chemical reaction, the production and consumption terms are typically zero. ERT 455 Session 2012/2013

Or in a simple understanding… Balanced quantity = total mass Generation & Consumption = 0 Balanced substance is nonreactive species System at steady state Accumulation = 0 ERT 455 Session 2012/2013

EXAMPLE The figure describes a mixing tank problem. Find all flows and compositions. ERT 455 Session 2012/2013

Start with the general balance equation: Since we are doing a total material balance, the production and consumption terms are zero. We'll make a "differential" balance and use flow rates (lb/hr) for all terms. where the flowrates are in units of pounds per hour. If we assume the system is at steady state (and there is no indication not to), accumulation is zero, and: Accumulation = Flow in – Flow out + Production - Consumption Accumulation = (F1 + F2 + F3) – (F0) ERT 455 Session 2012/2013

Accumulation = Flow in – Flow out + Production - Consumption If we assume the system is at steady state (and there is no indication not to), accumulation is zero, and: Now we need to figure out the composition of the product stream. Start with the general equation and write a component balance on compound A: F0 = F1 + F2 + F3 = 100 + 40 + 150 = 290 lb/hr Accumulation = Flow in – Flow out + Production - Consumption ERT 455 Session 2012/2013

There is no reaction, so no production or consumption There is no reaction, so no production or consumption. We're assuming steady state, so accumulation is zero. Consequently: 0 = F1XA1 + F2XA2 + F3XA3 - F0XA0 = 100 (0.5) + 40(0) + 150 (0.75) - 290XA0 = 162.5 - 290XA0 XA0 = 162.5/290 = 0.56 lbs A/lbs product ERT 455 Session 2012/2013

Similar balances are done on compounds B and C: 0 = 100 (0.5) + 0 + 0 - F0XB0 XB0 = 50/290 = 0.17 0 = 0 + F2XC2 + F3XC3 - F0XC3 XC0 = 40 + 150(0.25)/290 = 0.27 ERT 455 Session 2012/2013

It is always smart to check answers for consistency It is always smart to check answers for consistency. Here, we do this by summing the mole fractions: XA0 + XB0 + XC0 = 0.56 + 0.17 + 0.27 = 1.0 ERT 455 Session 2012/2013

Balances on Continuous Steady state Processes Steady-state, accumulation = 0 If balances is on nonreactive species or on total mass, the generation and consumption = 0, equation reduces to Input + Generation = Output + Consumption Input = Output ERT 455 Session 2012/2013

Example Material Balances on a Continuous Distillation Process 450 kg B/h m1 kg T/h 500 kg B/h 500 kg T/h m2 kg B/h 475 kg T/h ERT 455 Session 2012/2013