dr. muanmai apintanapong email: [email protected] [email protected] tel: 081-844-0799
TRANSCRIPT
Dr. Muanmai ApintanapongDr. Muanmai Apintanapong
Email: Email: [email protected]
Tel: 081-844-0799 Tel: 081-844-0799
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
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
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
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
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=
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 =
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
System
System
Open system
Volumetric flow rate
Mass flow rate
System
Close system
Volume
Mass
surroundings
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
Moisture content dry basis
MCdb
Moisture content wet basis
MCwb
MCdb
MCdb1 +
MCwb =
MCwb
MCwb1 -
MCdb =
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
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
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
Moisture Content , dry basis
kg water
kg dry solids
mass of water
mass of dry solids% Dry basis =
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=
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=
= 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%
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
=-
Unit Operation
Wastes
Mass in – Mass Out = Accumulation
F – (W+P) = Accumulation
Assumption: the accumulation = 0
F = W + P
Feed in raw product Product
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.
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
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
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
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.
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
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
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.
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
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
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
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
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
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
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.
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
Relationship between sensible Heat and latent Heat
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
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
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
Solution Select 100 kg of unpeeled potatoes as basis
Mass balance
F + S = W + P
100 + 4 = W + P
W = 104 - P
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
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
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
Reference: