dr. muanmai apintanapong email: [email protected] [email protected] tel: 081-844-0799
TRANSCRIPT
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Dr. Muanmai ApintanapongDr. Muanmai Apintanapong
Email: Email: [email protected]
Tel: 081-844-0799 Tel: 081-844-0799
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Engineer Units
Parameter Symbol NameUnit Symb
ol
length l. metre m
mass m kilogram kg
time t second s
electric current I ampere A
thermodynamictemperature
T kelvin K
amount of substance n mole mol
luminous intensity Iv candela cd
SI Base Units
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
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Density
Density =Mass ( kg)
Volume ( m ) 3
Mass = Density x Volume
The density of food sample is defined as its mass per unit volume and is expressed as kg /m3
The density is influenced by temperature
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Volumetric Flow rate Mass Flow rate Q = Volumetric flow rate
m = mass flow rateo
m = Density x Volume flow rateo
A1V1
A2V2
A1V1
A2V2Q = = m /sec3
m =o
Q Kg /sec
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Example 1.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 0.1
2= 0.0078 m
2
Q = A V = 20 x 0.0078 = 0.156 m / sec3
m = Q = 1000 x 0.156 = 156 kg / sec
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Temperature
The Kelvin and Celsius scales are related by following function
T ( K ) o T ( C ) o + 273.15=
The Fahrenheit and Celsius scales are related by following function
T ( F ) o [ T ( C ) – 32 ] o 5
9=
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PressurePressure 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.
FA
Pressure =
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Example2. How much 350 Kelvin degrees would be in Fahrenheit degrees
Example 3. How much 60 Fahrenheit degrees would be in Kelvin degrees
Solution = 288.7 K
Solution = 170.3 F
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System
System
Open system
Volumetric flow rate
Mass flow rate
System
Close system
Volume
Mass
surroundings
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Moisture ContentMoisture Content expresses the amount of water present
in a moist sample.Two bases are widely used to express moisture content
Moisture content dry basis
MCdb
Moisture content wet basis
MCwb
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Moisture content dry basis
MCdb
Moisture content wet basis
MCwb
MCdb
MCdb1 +
MCwb =
MCwb
MCwb1 -
MCdb =
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Example 4.Example 4. Covert a moisture content of 85 % wet basis to moisture content dry basis
MCwb
MCwb1 -
MCdb =
0.851 -MC
db =0.85
MCdb = 5.67
= 567 % db
MCwb = 0.85
From equation
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Example 5.Example 5. A food is initially at moisture content of 90 % dry basis . Calculate the moisture content in wet basis
MCdb
MCdb1 +
MCwb =
0.901 +MC
wb =0.90
MCwb = 0.4736
= 47.36 % wb
MCdb = 0.90
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Food Sample =
Mass of product = Mass of water in food + Mass of dry solids
+ Food Liquid Food Solids
Mass of dry solid
Food Sample
Mass of water in food
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Moisture Content , dry basis
kg water
kg dry solids
mass of water
mass of dry solids% Dry basis =
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Moisture Content , wet basis
mass of water
mass of water +mass of dry solids
kg water
kg product
% Wet basis =
mass of water
mass of product=
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Example 6.Example 6. The 10 kg of food sample at a moisture contents of 75 % wet basis
10 kg of product = 7.5 kg water + 2.5 kg dry solids
mass of water
mass of water +mass of dry solids% Wet basis =
0.75
1.00=
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= 7.5 kg water + 2.5 kg dry solids10 kg of productat 75 % wet basis
25 % of total Solids 75 % of total water
% Dry basis = (75/25)*100 = 300%
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Material BalanceThe principle of conservation of mass states
that
Mass can be neither created nor destroyed. However, its composition can altered from one from to another
Antoine Laurent Lavoisier (1743-1794)
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
=-
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Unit Operation
Wastes
Mass in – Mass Out = Accumulation
F – (W+P) = Accumulation
Assumption: the accumulation = 0
F = W + P
Feed in raw product Product
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Unit Operation
Wastes 20 kg/hr
Assumption : the accumulation = 0
Feed 100 Kg /hr Product
F = W + P
100 = 20 + P
P = 100 - 20
P = 80 Kg / hr
Example 10.Example 10.
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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.
F = 10 kg of raw product
(80 % w.b.)
Product = 2.86 kg
(30 % w.b.)
Water removed
Drying process
20 % of total Solids 80 % of total water
0.8 x 10 = 8 kg water 0.2 x 10 = 2 kg solid
30 % of total water
0.3 x 2.86 = 0.86 kg water
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Mass of water of
raw product
= 8 kg water
Water removed
Drying process
Mass of water of
final product
= 0.86 kg water
8 = P + W
8 = 0.86 +W
W = 7.14 kg water
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Example 8.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
F = 20 kg of raw product
(80 % w.b.)
Product
(50 % w.b.)
Water removed
Drying process
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Water = 20 kg product x 0.8 = 16 kg water
Solid = 20 kg product x 0.2 = 4 kg dry solid
20 % of total Solids 80 % of total water
80 % w.b.80 % w.b.
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F = 20 kg of
product (80 % w.b.)
Product
(50 % w.b.)
Water removed
Drying process
16 kg water
4 kg dry solid
A kg water
4 kg dry solid
50 % w.b. = A
A + 4 kg dry solids
0.5 = A
A + 4 kg dry solids
0.5 A +(4 x 0.5) = A
0.5 A + 2 = A
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0.5 A = 2
A = 20.5
= 4 kg water
F = 20 kg P = 8 kg
Water removed
Drying process
Total mass of product = 4 +4 = 8 kg
F = P + W
20 = 8 +W
W = 12 kg water
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Example 9.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
F = 10 kg of raw product
(320 % d.b.)
Product
(50 % w.b.)
Water removed
Drying process
% d.b. change to % w.b.MC
db
MCdb1 +
MCwb =
3.201 +=
3.20= 0.7619
= 76.19 % w.b.
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F = 10 kg of raw product
(76.2 % w.b.)Product
(50 % w.b.)
Water removed
Drying process
23.8 % of total Solids = 2.38 kg
76.2 % of total water = 7.62 kg Mass of total product
= A kg water + 2.38 kg
0.5 = A
A + 2.38
A = 2.38 kg water
F = P + W
7.62 = 2.38 + W
W = 7.62 -2.38 = 5.24 kg water
P = 4.76 kg
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Unit Operation
Wastes0.5 % Total solid
Assumption: the accumulation = 0
Feed 100 kg /hr
10 % Total solid
Product
30 % Total solid
F = W + P
100 = W + P
P = 100 - W
Equation 1
Step 1 Total mass Balances
Example 11.Example 11.
F (0.1) = W(0.005) + P (0.3)
100 kg /hr (0.1) = W(0.005) + P (0.3)
10 kg/hr = 0.005W + 0.3 P
P = 10 – 0.005 W
0.3
Step 2 Total Solid Balances
Equation 2
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Equation 1 = Equation 2
= 10 – 0.005 W
0.3
100 - W
(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 3 Determine Product rate
Step 4 Determine W
P = 100 - W P = 100 – 67.8
P = 32.2 kg / hr
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Example 12Example 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. Feed
10 % TS
B
25 % TS
R
2 % TS
100 kg/ min of
product
30 % TS
W , 0.5 % TS
first stage Second stage
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Feed
10 % TS
100 kg/ min of
product
30 % TS
W , 0.5 % TS
first stage Second stage
F = P + W
0.1 F = 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
Total product balance
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Feed
10 % TS
B
25 % TS
R
2 % TS
W , 0.5 % 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
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Energy Balance
Total energy entering the system
Total energy leaving the system
Change in the total energy of system
=
The first law of thermodynamic states that energy can be neither created nor destroyed.
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Sensible Heat
CP = Specific Heat at Constant pressure kJ/ kg K
Close System Open System
Q = m C P T m C P To
=
Latent HeatQ = m L
L = latent heat
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Relationship between sensible Heat and latent Heat
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Relationship between Sensible Heat and Latent Heat
ICE at -50 C ICE at 0 C
water at 0 Cwater at
100 C
vapor at
100 C
vapor at
150 C
Q1 = sensible heat
Q3 = sensible heat
Q5 = sensible heat
Q2 = Latent heat
Of Fusion
Q 4 = Latent heat
Of vaporization
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Overall View of an Engineering Process
Using a material balance and an energy balance, a food engineeringprocess can be viewed overall or as a series of units. Each unit is aunit operation.
Raw
materials
Unit Operation
Further Unit Operation
Previous Unit Operation
By-products By-products
Product Product
WastesWastes
EnergyEnergy
Wastes Energy
Capital
Energy
Labor
Control
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Example 13Example 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
P = ?
T = 35 CP
F = 100 kg
T = 17 CF
o
W = ?
T = 60 CF
o
H = 2750 kJ/kg
S = 4 kg
s
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Solution Select 100 kg of unpeeled potatoes as basis
Mass balance
F + S = W + P
100 + 4 = W + P
W = 104 - P
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Energy balance
= 4 kg x 2750 kJ /kgQ s = S Hs= 11000 kJ
Q P = F C (T – 0 )P P
= P (3.5 kJ/kg K)( 35 -0)
= 122.5 P kJ
Q w = F C ( T – 0 )P w
= W (4.2 kJ/kg K)( 60 -0)
= 252 W kJ
Q F = F C (T – 0 )P F
= 100 (3.7 kJ/kg K)( 17 -0)
= 6290 kJ
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Energy balance
Energy in from System = Energy out from system
Q wQ p +Q sQ F + =
6290 + 11000 = 122.5 P + 252 W
17290 = 122.5 P + 252 W
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W = 104 - P Equation of mass balance
17290 = 122.5 P + 252 W 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
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Reference: