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Training in Food Engineering
Dr. Sirichai Songsermpong
Dept. of Food Science and Technology
Kasetsart University
Why we need to study food eng
• Design and model the process
• Design food processing equipment
• Evaluate new equipment and process
• Troubleshoot problems in plant processing
• Part of plant start up team
• Communicate effectively with engineers, designers, maintenance crews and plant operators which requires engineering methods and terminology
Objectives
• To understand the concept of material and
energy balances
• To understand the concept of thermodynamics
• To understand the concept of transport
phenomena in food processing
• To understand the engineering principles in
each unit operation
• To be able to calculate some parameters in the
process
References
Singh, R.P. and Heldman, D.R. 2001. Introduction to
Food Engineering. 3rd edition. Academic Press,
California.
Toledo, R.T. 2007. Fundamentals of Food Process
Engineering. 3rd edition. Springer Publisher, New
York.
Earle, R.L. 1966. Unit Operations in Food Processing.
Now available on line.
Heldman, D.R. and Singh, R.P. 1981. Food Process
Engineering. 2nd edition. AVI, Connecticut.
Others
Unit one: Material and Energy
Balance
Unit conversion
Conservation of mass and energy
Overall analysis
Unit analysis
A
B C
D
E
SI unit
• Meter
• Kilogram
• Second
• Kelvin
• Newton
• Joule
• Pascal
• Watt
• Ampere
Prefix
• Tera
• Giga
• Mega
• Kilo
• Milli
• Micro
• Nano
• Pico
Unit conversion
• Temperature conversion
R C KF
212 672 100 373boiling
freezing
32 492 0 273
0 460
F=1.8C+32 C=(F-32)/1.8 R=F+460 K=C+273Temp conversion
Temp. difference
conversion F=1.8C C=F/1.8 R=F K=C
180 100
Pressure
• Guage pressure based on atm. pressure
• Absolute pressure based on perfect
vacuum
Absolute pressure = Guage pressure + Atmospheric pressure
Absolute pressure (Pa)
Vacuum = difference between atmospheric and absolute pressure
1 atm=760mmHg=14.7 psi=406.8 in of water=101.3kPa
Pressure
Absolute pressure
Atmospheric
pressure
Guage pressure (Difference
between absolute and atm pressure
Vacuum (Diff between atm
and absolute Pressure)
Absolute pressure (less
than atmospheric)
0 kPa
101.3 kPa
100C
200 kPa
120C
Practice
• 150F = ?C
• 14.7 psig = ? kPa
• 1 lb/ft3 = ?kg/m3
• 1 kJ/kgK = ? Btu/lbF
• 1 Btu/hrftF = ? W/mC
• 1 centiPoise = ? mPas
Conservation of mass
Mass in = mass out + accumulation
Why we need to do mass balance?
• To see the loss
• To calculate the efficiency
• To plan how much raw material is needed
• To see how much product and waste are
generated
• To determine the size and number of machine
Conservation of mass in
blending process
• Flour A has amylose 15 %
• Flour B has amylose 30 %
• Blending of flour A and B to achieve
amylose 25% for 1000 kg
A
B
C
Conservation of mass in dilution
How many kg of a solution containing 10% sodium chloride can be obtained by diluting 15 kg of a 20% solution with water?
15 kg
20%NaCl
X =? kg
10% NaCl
Water
Conservation of mass in
dehydration
How much weight reduction would result
when a material is dried from 80% mc to
50% mc?
W
80% H20
20% solid
D
50% H20
50% solid
H2O
Conservation of mass in
concentration
How much water is removed from 1000 kg of juice?
Juice 10 BrixConcentrate
60 Brix
water
Multi components and various
compositions
Lean beef X kg
14%fat
67% water
19% protein
Pork fat Y kg
89% fat
8% water
3% protein
Water Z kg
Frankfuter 100 kg
20% fat
65% water
15% protein
Calculate mass of each component
Recycle
10,000 kg/h
Evaporator Crystallizer40% solid
Filtrated 45% solid
50% solid 95% Cake
5% (a 45% solution)
Energy balance
• Energy in = energy out + energy
accumulation
• Sensible heat Q=mCpΔT
• Latent heat Q=mL
energy balance: introduction
• Sensible heat
Q = m C ( T1 – T2)
C: heat capacity (p, v)
• Latent heat
Q = m L
L: latent heat of…
Solid Liquid Gas
Temperature
Solid Liquid Gas
Temperature
How to calculate Cp
• Water has specific heat of 4.18 kJ/kgK
• Air has specific heat of 1 kJ/kgK
• Specific heat of Food =
Mass fraction of water*Cp water +
Mass fraction of solid *Cp solid
Where Cp of solid = 0.837 kJ/kgK
Other equations
• Cp = 0.4F+0.2 SNF+1W
• Cp=1.424xcarbo+1.549xprotein+1.675xfat+0.8
37xash+4.187xwater kJ/kgK(Dickerson,
1969)
• Cp=xice*Cice+xsolid*Csolid where Cp of ice =
2.093 kJ/kgK
Latent heat
• Latent heat of fusion = 335 kJ/kg
• Latent heat of vaporization = 2257 kJ/kg
• Latent heat of sublimation
• Can be seen from steam table
Practice
• Calculate the heat removal from apple
juice 200 kg/hr at 25C to -20C
• Freezing point is -1C
• Latent heat of freezing = 281.5 kJ/kg
• Cp above freezing = 3.59 kJ/kgK
• Cp below freezing = 1.88 kJ.kgK
Steam table
• We can read enthalpy at state 1 and
enthalpy at state 2 from steam table and
then calculate the heat gain or loss
• Q = m (enthalpy at state 1- enthalpy at
state 2)
Steam table (saturated steam)
T
(C)
P
kPa
Spec
ific
(l)
volu
me
(g)
Enth
alpy
(l)
Enth
alpy
(g)
entro
py
0.01 0.61
100 101. 419 2676
120 198 503 2706
140 316 589 2734
Practice
• To heat mango juice 1000 kg/hr from 25 C
to 90 C, steam at 120 C is used and
condensate exit at 100 C. Calculate the
weight of steam needed.