Mass Balance
Nur Istianah, ST.,MT.,M.Eng
28/04/2017
Engineer Units
Parameter Symbol Name Unit
Symbol length l. metre m
mass m kilogram kg time t second s electric current I ampere A thermodynamic
temperature T kelvin K amount of substance n mole mol
luminous intensity I
vcandela 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 )
3Mass = Density x Volume
The density of food sample is defined as its mass per unit volume and is expressed as kg /m
3The density is influenced by temperature
Volumetric Flow rate Mass Flow rate
Q = Volumetric flow rate
m = mass flow rate
om = Density x Volume flow rate
oA
1V
1A
2V
2A
1V
1A
2
V
2Q = = m /sec
3m =
o Q Kg /sec
Equation of Continuity
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 D =
2
4 0.1
2= 0.0078 m
2Q = A V = 20 x 0.0078 = 0.156 m / sec
3m = 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 ( C )
o[ T ( F) – 32 ]
o1.8
=
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.
F
Pressure =
A
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 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
dbMoisture content wet basis
MC
wbFood 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
=
Moisture content dry basis
MC
dbMoisture content wet basis
MC
wbMC
dbMC
db1 +
MC
wb=
MC
wbMC
wb1 -
MC
db=
Example 3. Covert a moisture content of 85 % wet basis to moisture content dry basis
MC
wbMC
wb1 -
MC
db=
1 - 0.85
MC
db= 0.85 MC
db= 5.67
= 567 % db MC
wb= 0.85
From equation
Example 4. A food is initially at moisture content of 90 % dry basis . Calculate the moisture content in wet basis
MC
dbMC
db1 +
MC
wb=
0.90 1 +
MC
wb= 0.90
MC
wb= 0.4736
= 47.36 % wb
MC
db= 0.90
Example 5.
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 solids 10 kg of product
at 75 % wet basis
25 % of total Solids 75 % of total water
% Dry basis = (75/25)*100 = 300%
THANKS FOR YOUR ATTENTION
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28/04/2017
