Transcript Ch 1 Introduction to food Engineering

Dr. Muanmai Apintanapong
Email: [email protected]
Tel: 081-844-0799
Engineer Units
SI Base Units
Parameter
Symbol
Name
Unit
Symbol
length
l.
metre
m
mass
time
electric current
thermodynamic
temperature
amount of substance
luminous intensity
m
t
I
kilogram
second
ampere
kg
s
A
T
kelvin
K
n
mole
mol
Iv
candela
cd
The name Système International d'Unités (International System of Units) with the international
abbreviation SI is a single international language of science and technology first introduced in 1960
Density
The density of food sample is defined as its mass per unit
volume and is expressed as kg /m 3
Mass ( kg)
Density =
Volume ( m 3 )
Mass = Density x Volume
The density is influenced by temperature
Volumetric Flow rate Mass Flow rate
Q = Volumetric flow rate
A1 V1
A2 V2
Q = A1 V1 = A2 V2
m 3 /sec
o
m = mass flow rate
o
m = Density x Volume flow rate
o
m = Q
Kg /sec
Example 1. Determine volumetric and mass flow rate of water
( density = 1000 kg /m^3) , the diameter of pipe is 10 cm.
v = 20 m/s
A=

4
D2 =

4
2
0.1 = 0.0078 m 2
3
Q = A V = 20 x 0.0078 = 0.156 m / sec
m = Q = 1000 x 0.156 = 156 kg / sec
Temperature
The Kelvin and Celsius scales are related by following function
T ( oK )
=
T ( oC ) + 273.15
The Fahrenheit and Celsius scales are related by following
function
T ( oF )
=
5
[ T ( oC ) – 32 ]
9
Pressure
Pressure is the force on an object that is spread over a surface area. The
equation for pressure is the force divided by the area where the force is applied.
Although this measurement is straightforward when a solid is pushing on a
solid, the case of a solid pushing on a liquid or gas requires that the fluid be
confined in a container. The force can also be created by the weight of an
object.
Pressure =
F
A
Example2. How much 350 Kelvin degrees
would be in Fahrenheit degrees
Solution = 170.3 F
Example 3. How much 60 Fahrenheit degrees
would be in Kelvin degrees
Solution = 288.7 K
System
surroundings
System
Close system
System
Open system
Volume
Volumetric flow rate
Mass
Mass flow rate
Moisture Content
Moisture Content expresses the amount of water present
in a moist sample.
Two bases are widely used to express moisture content
Moisture content dry basis
MC db
Moisture content wet basis
MC wb
MC db =
MC wb
1 - MC wb
MC wb
MC db
Moisture content wet basis
Moisture content dry basis
MC wb =
MC db
1 + MC db
Example 4. Covert a moisture content of 85 % wet
basis to moisture content dry basis
MC wb =
From equation
0.85
MC db =
MC wb
1 - MC wb
MC db =
0.85
1 - 0.85
MC db =
5.67
=
567 % db
Example 5. A food is initially at moisture content of 90 % dry
basis . Calculate the moisture content in wet basis
MC db =
0.90
MC wb =
MC db
1 + MC db
MC wb =
0.90
1 + 0.90
MC wb =
0.4736
=
47.36 % wb
Food Sample = Food Solids + Food Liquid
Mass of product = Mass of water in food + Mass of dry solids
Mass of dry solid
Mass of water in food
Food Sample
Moisture Content , dry basis
% Dry basis =
mass of water
mass of dry solids
kg water
kg dry solids
Moisture Content , wet basis
% Wet basis =
=
mass of water
mass of water +mass of dry solids
mass of water
mass of product
kg water
kg product
Example 6. The 10 kg of food sample at a moisture contents of
75 % wet basis
% Wet basis =
=
mass of water
mass of water +mass of dry solids
0.75
1.00
10 kg of product = 7.5 kg water + 2.5 kg dry solids
10 kg of product = 7.5 kg water + 2.5 kg dry solids
at 75 % wet basis
75 % of total water
25 % of total Solids
% Dry basis = (75/25)*100 = 300%
Material Balance
The principle of conservation of mass states
that
Mass can be neither created nor destroyed. However, its composition can
altered from one from to another
Rate of mass
entering through the
boundary of system
-
Rate of mass exiting
through the
boundary of system
=
Rate of mass
Accumulation
through the
boundary of system
Antoine Laurent Lavoisier
(1743-1794)
Wastes
Feed in raw
product
Unit Operation
Mass in – Mass Out = Accumulation
F – (W+P) = Accumulation
Assumption: the accumulation = 0
F
=
W+P
Product
Example 10.
Wastes
20 kg/hr
Feed 100 Kg /hr
Unit Operation
Assumption : the accumulation = 0
F
=
W+P
100
=
20 + P
P
=
100 - 20
P
=
80 Kg / hr
Product
Example 7. 10 kg of food at a moisture content of 80 % wet
basis is dried to 30 % wet basis. The final product weight is 5 kg.
Calculate the amount of water removed.
Water removed
F = 10 kg of raw product
(80 % w.b.)
Drying
process
80 % of total water
20 % of total Solids
0.8 x 10 = 8 kg water
0.2 x 10 = 2 kg solid
Product = 2.86 kg
(30 % w.b.)
30 % of total water
0.3 x 2.86 = 0.86 kg water
Water removed
Mass of water of
raw product
= 8 kg water
Mass of water of
Drying
process
final product
= 0.86 kg water
8 = P+W
8 = 0.86 +W
W = 7.14 kg water
Example 8. The 20 kg of food at a moisture content of 80 %
wet basis is dried to 50 % wet basis. Calculate the amount of
water removed
Water removed
F = 20 kg of raw product
(80 % w.b.)
Drying
process
Product
(50 % w.b.)
80 % w.b.
80 % of total water
20 % of total Solids
Water = 20 kg product x 0.8 = 16 kg water
Solid = 20 kg product x 0.2 = 4 kg dry solid
Water removed
F = kg
20 water
kg of
16
Drying
process
product
% w.b.)
4
kg dry(80
solid
50 % w.b. =
A kg water
Product
(50
% dry
w.b.)solid
4 kg
A
A + 4 kg dry solids
A
0.5 =
A + 4 kg dry solids
0.5 A +(4 x 0.5) =
A
0.5 A + 2 =
A
0.5 A
=
2
A
=
2 = 4 kg water
0.5
Total mass of product = 4 +4 = 8 kg
Water removed
F = 20 kg
Drying
process
F = P+W
P = 8 kg
20 = 8 +W
W = 12 kg water
Example 9. The 10 kg of food at a moisture content of 320 %
dry basis is dried to 50 % wet basis. Calculate the amount of
water removed
Water removed
F = 10 kg of raw product
(320 % d.b.)
Drying
process
Product
(50 % w.b.)
% d.b. change to % w.b.
MC wb =
MC db
1 + MC db
3.20
=
1 + 3.20
= 0.7619
= 76.19 % w.b.
Water removed
F = 10 kg of raw product
(76.2 % w.b.)
Product
Drying
process
76.2 % of total water = 7.62 kg
23.8 % of total Solids = 2.38 kg
(50 % w.b.)
Mass
of total = A kg water + 2.38 kg
product
0.5
F = P +W
7.62 = 2.38 + W
W
= 7.62 -2.38 = 5.24 kg water
P
= 4.76 kg
=
A
A + 2.38
A = 2.38 kg water
Wastes
0.5 % Total solid
Example 11.
Feed 100 kg /hr
10 % Total solid
Product
Unit Operation
30 % Total solid
Assumption: the accumulation = 0
Step 2 Total Solid Balances
Step 1 Total mass Balances
F
=
W+P
100
=
W+P
P
=
100 - W
F (0.1)
=
W(0.005) + P (0.3)
100 kg /hr (0.1)
10 kg/hr
=
=
W(0.005) + P (0.3)
0.005W + 0.3 P
Equation 1
P =
10 – 0.005 W
0.3
Equation 2
Step 3 Determine Product rate
Equation 1 = Equation 2
100 - W
10 – 0.005 W
0.3
=
(0.3)(100) – 0.3 W = 10 – 0.005 W
30 - 10 = 0.3 W – 0.005W
20 = 0.295 W
W = 20 / 0.295 = 67.8 kg /hr
Step 4 Determine W
P
=
100 – 67.8
P
=
100 - W
P
=
32.2 kg / hr
Example 12. A membrane separation system is used to concentrate the liquid food
from 10 % to 30 % total solid (TS). The product is accomplished in two stages, in
the first stage, a low total solid liquid stream is obtained. In the second stage, there
are two streams, the first one is final product stream with 30% TS and the second is
recycled to the first stage. Determine the magnitude of the recycle stream when the
recycle contains 2 % TS , the waste stream from first stage contains 0.5 % TS and
the stream between stages 1 and 2 contains 25 % TS . The final product is 100
kg/min with 30 % TS.
W , 0.5 % TS
Feed
10 % TS
first stage
B
25 % TS
Second stage
100 kg/ min of
product
30 % TS
R
2 % TS
Total product balance
W , 0.5 % TS
Feed
first stage
Second stage
10 % TS
100 kg/ min of
product
30 % TS
F
0.1 F
= P+W
= 100 (0.3) + 0.005 W
0.1 ( 100+ W ) = 30 + 0.005 W
10 + 0.1 W = 30 + 0.005 W
W
= 210 .5 kg / min and F = 310.5 kg/min
W , 0.5 % TS
Feed
B
10 % TS
25 % TS
R
2 % TS
F +R = W +B
310.5 + R = 210.5 + B
B = 100 +R
0.1 F + 0.02 R = 0.005 W + 0.25 B
0.1 (310.5) + 0.02 R = 0.005 (210.5) + 0.25 B
31.05 + 0.02 R =
1.0525 + 0.25 (100+R)
R = 21.73 kg / min
Energy Balance
The first law of thermodynamic states that energy can
be neither created nor destroyed.
Total energy
entering the
system
Total energy
leaving the
system
=
Change in the
total energy of
system
Sensible Heat
o
Q = mCP T = mCP T
Close System
C
P
Open System
= Specific Heat at Constant pressure kJ/ kg K
Latent Heat
Q
L = latent heat
= mL
Relationship between sensible Heat and latent Heat
Relationship between Sensible Heat and Latent Heat
Q5 = sensible heat
Q3 = sensible heat
water at
water at 0 C
100 C
Q2 = Latent heat
ICE at -50 C
ICE at 0 C
Q1 = sensible heat
Of Fusion
vapor at
vapor at
100 C
150 C
Q 4 = Latent heat
Of vaporization
Overall View of an Engineering Process
Using a material balance and an energy balance, a food engineering
process can be viewed overall or as a series of units. Each unit is a
unit operation.
Wastes Energy
Raw
materials
By-products
Previous
Unit
Operation
Unit
Operation
Further Unit
Operation
Product
Wastes
Energy
Capital
Energy
Labor
Control
Example 13 . Steam is used for peeling of potatoes in a semi-continuous
operation . Steam is supplied at the rate of 4 kg per 100 kg of unpeeled potatoes.
The unpeeled potatoes enter system with a temperature of 17 C and the peeled
potatoes leave at 35 C . A waste stream from the system leaves at 60 C . The
specific heats of unpeeled potato, waste stream and peeled potatoes are 3.7 , 4.2
and 3.5 kJ/ (kg K ) , respectively. If the heat content of steam is 2750 kJ /kg ,
determine the quantities of the waste stream and the peeled potatoes from the
process
H s = 2750 kJ/kg
S = 4 kg
F = 100 kg
P = ?
o
TF = 17 C
TP = 35 C
W = ?
o
TF = 60 C
Solution
Select 100 kg of unpeeled potatoes as basis
Mass balance
F+S = W +P
100 + 4 = W + P
W = 104 - P
Energy balance
Q s = S H s = 4 kg x 2750 kJ /kg
= 11000 kJ
Q F = F C P (TF – 0 )
Q P = F C P (TP– 0 )
= 100 (3.7 kJ/kg K)( 17 -0)
= P (3.5 kJ/kg K)( 35 -0)
= 6290 kJ
= 122.5 P kJ
Q w = F C ( Tw– 0 )
P
= W (4.2 kJ/kg K)( 60 -0)
= 252 W kJ
Energy balance
Energy in from System = Energy out from system
QF + Qs
=
Qp + Qw
6290 + 11000
= 122.5 P + 252 W
17290
= 122.5 P + 252 W
W = 104 - P
17290
= 122.5 P + 252 W
Equation of mass balance
Equation of energy balance
17290 = 122.5 P + 252 ( 104 –P )
17291 =
122.5 P + 26208 – 252 P
P
68. 87 kg
=
W = 104 – 68.87
= 35.14 kg
Reference:

Prof. Athapol Noomhorm: