PEMODELAN MATEMATIS PENGOLAHAN PANGAN

yusronsugiarto.lecturer.ub.ac.id

TUJUAN

Mengetahui perbedaan Keseimbangan Makroskopis yang meliputi hukum kekekalan massa, energi dan momentum dalam pemodelan matematis

Mengetahui penggunaan aplikasi pemodelan

KESEIMBANGAN MAKROSKOPIS

Keseimbangan Massa Keseimbangan Energi Keseimbangan Momentum

KESEIMBANGAN MASSA

Learning objectives

On successful completion of this topic you should be able to: Understand the basic concepts of mass

balance in the context of bioprocess Sketch simple block flow diagrams; and Conduct mass balance analysis on chemical

processes

• “Materials Balance analysis (MB) is a systematic reconstruction of the way in which a chemical element, a compound or material passes through a natural cycle and/or its economical benefits. An analysis of the material flow, usually is based on the origin of a physical balance.”

• German investigation Committe, 1993

INTRODUCTION

Conservation mass principle is used because it

indicates that:

• Input Material = Output Material

• Units of measurements given in kg or m3

Material Balance

Input and Output diagram Auxiliary materials

Unit Operation

Raw material

Product (expected)

byproduct (usable)

Waste

Water Energy

Waste easily assimilated by

the environment

Inert waste always available

toxic/dangerous waste

Plant, Process or Unit

Operación

Gaseous emissions Raw materials

Catalyst

Air/Water

Energy

Recycle

Reusable residues in other operation

Products

By-products

Wastewater

Liquid waste

Solid waste

Input and Output

Mass balances provide a very powerful tool in engineering analysis. Many complex situations are simplified by looking at the movement of mass and equating what comes out to what goes in.

Questions such as: what is the concentration of carbon dioxide in the fermenter off-gas? what fraction of the substrate consumed is not converted into products? how much reactant is needed to produce x grams of product? how much oxygen must be provided for this fermentation to proceed? can be answered using mass balances.

MASS BALANCE ON BIOPROCESS FIELD

Thermodynamic Preliminaries

System and Process

SYSTEM

Surroundings

System boundary

Law of Conservation of Mass

A mass balance for the system can be written in a general way to account for these possibilities:

Example

1000 kg of a mixture of benzene (B) and toluene(T) that contains 50% benzene by mass are separated by distillation into two fractions. The mass flow rate of benzene in the top stream is 450 kg B and that of toluene in the bottom stream is 475 kg T. The operation is at steady state. Write balances on benzene and toluene to calculate unknown component flow rate in the output streams.

Problem Process diagram

500 kg B 500 kg T

450 kg B q1 kg T

q2 kg Br 475 kg T

* Benzene balance 500 = 450 + q1

* Toluene Balance 500 = q2 + 475

Material Balance

Solution : Q1 = 50 kg T Q2 = 25 kg B

Benefits of using flowcharts

The catalytic dehydrogenation of propane is carried out in a continuous packed bed reactor. One thousand pounds per hour of pure propane are fed to preheater where they are heated to a temperature of 670 C before they pass into the reactor.The reactor effluent gas, which includes propane, propylene, methane and hydrogen, is cooled from 800 C to 100 C and fed to an absorption 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 high pressure distillation column in which the propane and propylene are separated. The product stream from the distillation column contains 98 % propane. The stripped oil is recycled to the absorption tower.

Process Description

Complex, not easy to understand

Flowchart

Preheater Reactor

(dehygdro -genation)

Cooler

Absorption Tower

Stripping Tower

High P Distillation

Recycle Stream (Propane)

100 lbm/h propane

600 C 800 C Propane Propylene Methane Hydrogen

110 C

Oil

Methane Hydrogen

Oil Propylene

Compact, easy to understand

Outline of a Procedure for Material Balance Calculations

• Draw a flow chart, and fill in all given values. • Choose as a basis of calculation an amount or flow rate

of one of the process streams. • Label unknown stream variables on the chart. • Do the problem bookkeeping. • Convert volume flow rates to mass or molar flow rates. • Convert mixed mass and molar flow rates to mass or

molar flow rates. • Translate given information to equations. • Write material balance equations. • Solve equations. • Scale up/down.

A continuous process is set up for treatment of wastewater. Each day, 105 kg cellulose and 103 kg bacteria enter in the feed stream, while 104 kg cellulose and 1.5 x 104 kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 7x104 kg d-1. The rate of bacterial growth is 2x104 kg d-l; the rate of cell death by lysis is 5 x 10 2 kg d -1. Write balances for cellulose and bacteria in the system.

Exercise 1

Humid air enriched with oxygen is prepared for a gluconic acid fermentation. The air is prepared in a special humidifying chamber. 1.5 I h- 1 liquid water enters the chamber at the same time as dry air and 15 gmol min- 1 dry oxygen gas. All the water is evaporated. The outflowing gas is found to contain 1% (w/w) water. Draw and label the flow sheet for this process.

Exercise 2

Density water 103 g/L Molecular weight of O2 32 g/gmol

Exercise 3

A fermentation slurry containing Streptomyces kanamyceticus cells is filtered using a continuous rotary vacuum filter. 120 kg h- 1 slurry is fed to the filter; 1 kg slurry contains 60 g cell solids. To improve filtration rates, particles of diatomaceous-earth filter aid are added at a rate of 10 kg h- 1. The concentration of kanamycin in the slurry is 0.05% by weight. Liquid filtrate is collected at a rate of 112 kg h-1; the concentration of kanamycin in the filtrate is 0.045% (w/w). Filter cake containing cells and filter aid is continuously removed from the filter cloth. (a) What percentage liquid is the filter cake? (b) If the concentration of kanamycin in the filter-cake liquid is the same as in the filtrate, how much kanamycin is absorbed per kg filter aid?

Exercise 3

Flowsheet for continuous filtration.

Exercise 4

Corn-steep liquor contains 2.5 % invert sugars and 50% water; the rest can be considered solids. Beet molasses containing 50% sucrose, 1% invert sugars, 18% water and the remainder solids, is mixed with corn-steep liquor in a mixing tank. Water is added to produce a diluted sugar mixture containing 2% (w/w) invert sugars. 125 kg corn-steep liquor and 45 kg molasses are fed into the tank. (a) How much water is required? (b) What is the concentration of sucrose in the final mixture?

ENERGY BALANCE

• Energy is expensive…. – Effective use of energy is important task for bioprocess

engineers.

• Topics of this chapter – Energy balance – Energy and energy transfer

• Forms of energy : Kinetic / Potential / Internal Energy • Energy transfer : Heat and Work

– Using tables of thermodynamic data – Mechanical energy balances

The law of conservation of energy means that an energy accounting system can be set up to determine the amount of steam or cooling water required to maintain optimum process temperatures. In this chapter, after the necessary thermodynamic concepts are explained, an energy-conservation equation applicable to biological processes is derived.

INTRODUCTION

SYSTEM

SYSTEM

Forms of Energy – The First Law of Thermodynamics

A balance on conserved quantity (i.e. mass, energy, momentum) in a process system may be written as:

Input + generation - output - consumption =accumulation

Forms of Energy – The First Law of Thermodynamics

• Forms of energy – Kinetic energy : due to the motion of the system

– Potential energy : due to the position of the system

– Internal energy : due to the motion of internal molecules

• U : from thermodynamic calculation

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Forms of Energy – The First Law of Thermodynamics

Change in kinetic energy:

Change in potential energy

Change in potential energy

How energy can be transferred between a system and its surroundings? - Heat – energy that flows as a result of temperature difference between a system and its surrounding ; heat is defined positive when it is transferred to the system from the surroundings. -Work – energy that flows in response to any driving force other than a temperature difference. ; work is defined positive when it is done by the system on the surroundings.

Forms of Energy – The First Law of Thermodynamics

Forms of Energy – The First Law of Thermodynamics

Types of Work • Flow work (Wfl) - energy carried across the boundaries of a system with the mass flowing across the boundaries (i.e. internal, kinetic & potential energy) • Shaft work (Ws) - energy in transition across the boundaries of a system due to a driving force other than temperature, and not associated with mass flow (an example would be mechanical work due to a piston, pump or compressor)

ENERGY – CONVERSION UNITS

Energy balance on closed systems There is no mass transfer into a closed system The only way energy can get into or out of a closed system

is by heat transfer or work

Q

Ws

a. Heat transfer (Q) b. Work (Ws) Note: * Work is any boundary interaction that is not heat (mechanical, electrical, magnetic, etc.)

1

The law of conservation of energy can be written as:

energy in through system boundaries

energy out through system boundaries

energy accumulated within the system - =

• Balance equation

(Final System Energy) – (Initial System Energy)

= (Net Energy Transfer)

• Initial System Energy

• Final System Energy

• Net Energy Transfer

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The first law of thermodynamics for closed systems

Energy balance on closed systems 1

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For a closed system what is ∆E equal to?

Is it adiabatic? (if yes, Q = 0) Are there moving parts, e.g. do the walls move? (if no, Ws = 0) Is the system moving? (if no, ∆KE = 0) Is there a change in elevation of the system? (if no, ∆PE = 0 ) Does temperature, phase, chemical composition change, or

pressure change less than a few atmospheres ? (if no to all, ∆U = 0)

Energy balance on closed systems 1

Energy Balance Procedures • Solve material balance Get all the

flow rate of streams

• Determine the specific enthalpies of each stream components – Using tabulated data

– Calculation

• Construct energy balance equation and solve it.

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Exercise

A closed system of mass 5 kg undergoes a process in which there is work of magnitude 9 kJ to the system from the surroundings. The elevation of the system increases by 700m during the process. The specific internal energy of the system decreases by 6 kJ/kg and there is no change in kinetic energy of the system. The acceleration of gravity is constant at g=9.6m/s2. Determine the heat transfer, in kJ!

Energy balance on open systems 2

How are open systems control volumes different from closed systems

What effect does this have on the energy balance?

Some common open system steady flow devices

Some common open system steady flow devices

Energy balance on open systems

*Steady State • Flow work and shaft work

– Flow work : work done on system by the fluid itself at the inlet and the outlet

– Shaft work : work done on the system by a moving part within the system

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in

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out

out

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2

Application of First Law - Conservation of energy for a control volume

The Steady-State Open System Energy Balance

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skp WQEEH

Water at 25oC enters an open heating tank at a rate of 10 kg h-1. Liquid water leaves the tank at 88 oC at a rate of 9 kg h-1; 1 kg h-1 water vapor is lost from the system through evaporation. At steady state, what is the rate of heat input to the system?

Exercise

Exercise 1. Assemble (i) Select units for the problem. kg, h, kJ, oC (ii) Draw the flow sheet showing all data and units. (iii) Define the system boundary by drawing on the flowsheet.

Exercise

(i) Select and state a reference state. The reference state for water is the same as that used in the steam tables: 0.010C and 0.6112 kPa. (ii) Collect any extra data needed. h (liquid water at 88 0C - 368.5 kJ kg- 1 (Table C. 1) h (saturated steam at 88 0C - 2656.9 kJ kg- 1 (Table C. 1) h (liquid water at 25 0C - 104.8 kJ kg- 1 (Table C. 1). (iii) Write down the appropriate energy-balance equation.

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