Physical Principles of Food Processing

Introduction

Ayes Rock , Uluru, 384 m high, 9.8 km circumference

Contents

Staff Aims Topics Activities Assessment

Staff

Dr Associate Professor Minh Nguyen Fellow, Australian Institute of Food

Science and Technology President, Australian Food Engineering

Association Mr Luong Hong Quang Lecturer, Food Engineering, Nong Lam

University

Aim: Develop working knowledge in Physical Principles of Food Processing

Objectives: To be familiar with units and dimensions To understand fundamental principles of mass and

energy conservation To understand and to apply basic engineering charts To use physical properties of foods in process

design To comprehend basics of fluid flow, Newtonian and

non Newtonian fluids To predict pressure drops, power and to select

pump

Why Study Physical Principles ?

Numerous ways to process foods Grouping into Unit Operations Fundamentals for each operation are

called Principles depending on aspects: physical, chemical or biological ...

Example: heat transfer principles cover cooking, drying, cooling, freezing etc

How to engineer a food process

What are the quantities involves? What is the rate of heat/ mass transfer ? What is the kinetics of chemical

changes? Quantitative approach needed for

analysis and design Review basic physical & chemical

concepts

Dimension

To define a physical entity: time, area... To measure the entity, dimension is

expressed as Units: hour, square metre... Primary dimensions: time, length... Secondary dimension: length/time=

velocity Equations must be consistent in dimension

Engineering SI units

Based SI units: length=meter, mass= kilogram, time= second, electric current= ampere, temperature= kelvin, amount of substance= mole, luminous intensity= candela

Supplementary SI units: plane angle= radian, solid angle= steradian

Derived SI unitsQuantity name symbol In SI base In other

unit

Force newton N m kg s-2

Power, radiant flux

watt W m2 kg s-3 J/s

Illuminance

lux lx m-2 cd sr lm/m2

Conductance

siemens S m-2 kg-1 s3 A2 A/V

Dynamic viscosity

pascal second

Pa s m-1 kg s-1

Surface tension

Newton per meter

N/m kg s-1

Exercise: Unit conversion British units are still used in countries

like the USA or in old machineriesSee Textbook table A1.2 for example 1.1

solution

foot ft 0.3048 m

British thermal unit BTU 1055 J

Pound lbm 0.454 kg

Pound per square inch absolute

psia 6.895 kPa lbf /in2 abs

System & state Region or quantity enclosed by a

boundary, away from surroundings Open, closed or isolated system Adiabatic or isothermal In equilibrium or changing state

process via path of many states State described by properties, extensive

or intensive

Density = mass/unit volume

Solid density, table 1.6 Particle density Bulk density, table 1.7 Porosity Interparticle porosity

Concentration & Moisture content

Mass per unit mass as % or unit volume

Molarity, mole fraction & molality Moisture content, wet & dry basis, See examples 1.4 & 1.5

Temperature

Ice point = 0 Celsius= 32 Farenheit

Boiling point= 100 oC= 212 oF Kelvin & Rankine scale ΔT (K) = ΔT (oC)

Pressure Pressure, atm=1.013 bar = 101.3 kPa Pabsolute =Pgauge + Patmosphere

Pvacuum =Pgauge – Patmosphere

normal stress, static head P=ρgh

See fig 1.9, atm water column= 10.2 m Static pressure & impact pressure Bourdon tube

Enthalpy

Sum of internal energy and product of pressure and volume, H=Ei + PV

Specific enthalpy is per unit mass Steam table reference state =

enthalpy of saturated water at 0 oC as zero!

Equation of State = Functional relationship between the

properties of a system Perfect gas equation of state PV’ = RTA P = ρRTA

PV = nRoTA Ro = M R R = universal gas cont 8314 m3 Pa/kg

mol K

Phase diagram of water

To study pressure –temperature relationship between various phases of pure water

Check fig 1.11 for saturated vapour, saturation pressure, saturated liquid, subcooled liquid, quality of the steam (vapour)

Conservation of Mass Principle

Matter can be neither created nor destroyed. However its composition can be altered from one form to another

(rate of mass entering boundary of system) – (rate of mass exiting...) = (rate of mass accumulating within...)

See fig 1.13 for open system Assume uniform flow of incompressible fluid at steady

state, mass flow rate Σinlet ρun dA = Σoutlet ρun dA Volumetric flow rate = Σinlet un dA = Σoutlet un dA Closed system, msystem = constant

Material balances Collect all known mass & composition of in & out

streams Draw process block diagram with streams. Draw

boundary Write all data on block diagram Select a convenient basis (mass or time) for calculations Write mass balance for each unknown ṁinlet - ṁexit = dmsystem / dt Solve material balances for each unknowns Check fig 1.14

Practicing material balances

In class, examples 1.6, 1.7, 1.8 At home, examples 1.9, 1.10

Start Reading the textbook

Begin working on part of the ASSIGNMENT

The END of Lecture One !

Revision