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8/19/2019 3 Mass Balance Agro1
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Mass and Energy Balances 1
Supratomo
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Mass and Energy Balances 2
• The study of process engineering is an attempt to combine all forms of physical
processing into a small number of basic operations, which are called unit
operations
• The essential concept is therefore to divide physical food processes into basicunit operations, each of which stands alone and depends on coherent physical
principles
• Because of the dependence of the unit operation on a physical principle, or a
small group of associated principles, quantitative relationships in the form of
mathematical equations can be built to describe them. The equations can beused to follow what is happening in the process, and to control and modify the
process if required.
• Two very important laws which all unit operations obey are the laws of
conservation of mass and energy.
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Mass and Energy Balances 3
BASIC PRINSIPLES
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Mass and Energy Balances 4
CONSERVATION of MASS
Mass can be neither created nor destryed. However , its
composition can be altered from one form to another.
Rate of mass
entering
through the
boundary of asystem
Rate of mass
exiting through
the boundary
of a system
Rate of mass
accumulation
within the
system
- =
∑=
=n
1i
.
inlet
.
imm ∑=
=n
1i
.
outlet
.
ommdt
mm
system
onaccumulati
d =
dt
mmm
system
outletinlet
d =−
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Mass and Energy Balances 5
CONSERVATION of MASS for an Open System
Consider a section of pipe used in transporting a fluid.
dAvdm o ρ = ∫∫ ==v
o dVdAvm ρ ρ A
∫∫∫ =V
oo dVdt
d dAv-dAv ρ ρ ρ
outlet inlet A A
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Mass and Energy Balances 6
∫∑∑ = Vo
outleto
inletdVdt
d
dAv-dAv ρ ρ ρ
For a uniform flow :
For a steady state flow flow does not change with time
dAvdAv ooutlet
o
inlet∑∑ = ρ ρ
If we are dealing with and incompressible fluid there is no change in density.
dAvdAvoutlet
o
inlet
o ∑∑ =
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Mass and Energy Balances 7
CONSERVATION of MASS for a Closed System
A closed system mass cannot cross system bounderies. Therefore,
there is no time rate of change of mass in the system.
0
dt
msystem =d
msystem = constant
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Discusion
Mass and Energy Balances 8
Which of the following statements are true and which are false?1. The mass balance is based on the law of conservation of mass.
2. Mass balance may refer to total mass balance or component
mass balance.
3. Control volume is a region in space surrounded by a control
surface through which the fluid flows.
4. Only streams that cross the control surface take part in the mass
balance.
5. At steady state, mass is accumulated in the control volume.
6. In a component mass balance, the component generation termhas the same sign as the output streams.
7. It is helpful to write a mass balance on a component that goes
through the process without any change.
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Mass and Energy Balances 9
Laws of Thermodynamics
1st Law
Energy can be neither created nor detroyed but can
be transformed from one form to another
2nd Law by Rudolf Clausius and Lord Kelvin
No process is possible where sole result is the
removal of heat from a reservoir (system) at one
temperature and the absorption of an equal quantity
of heat by reservoir at a higher temperature.
The 2nd law of thermodynamics help explain why heat
always flow from a hot object to a cold object.
CONSERVATION of ENERGY
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Mass and Energy Balances 10
Forms of Energy
Potensial Energy of a system is by vitue of its location with respect to the
gravitational field.
mgh E PE =Kinetic Energy of an object is due to its velocity.
2
2
1
mv E KE =
Internal Energy is due to the microscopic nature of the system . They move in
random direction, collide each other, vibrate and rotate.
T mc E pi ∆=
Total Energy of a system can be written in the form of an equation as :
Etotal = EKE + EPE + Eelectrical + Echemical + …….. + Ei
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Mass and Energy Balances 11
ENERGY BALANCE
The first law of thermodynamics states that energy can be
neither created nor destryed.
Total energy
entering the
system
Total energy
leaving the
system
Change in the
total energy at
the system- =
systemoutin Ê ÊÊ ∆=−For a system is in a steady state, there is no change in the energy of the
system with time.
0ˆˆˆ =∆=− systemout in E E E
out in E E ˆˆ =
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Mass and Energy Balances 12
Energy Balance for an Open System
Movement of a liquid volume
Force to push a liquid through the system boundary is : F = P.A
A : cross-sectional area
P : pressure of the fluid
The work done on the fluid element is : Wmass flow = F.L = P.A.L = PV
The total energy of the fluid element :
ET = EPE + EKE + Ei + PV PV E mv
mgz E iT +++=2
2
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Mass and Energy Balances 13
The overall energy balance equation for a system with one inlet (point 1)
and one outlet (point 2) is:
The overall energy balance equation for a system at steady state with more
than two streams can be written as:
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Mass and Energy Balances 14
Where :
In most of the cases, the overall energy balance ends up as an enthalpy balancebecause the terms of kinetic and potential energy are negligible compared to the
enthalpy term, the system is assumed adiabatic (Q = 0), and there is no shaft work
(Ws = 0). Then:
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Discusion
Mass and Energy Balances 15
Which of the following statements are true and which are false?1. The energy in a system can be categorize as internal energy,
potentialenergy, and kinetic energy.
2. A fluid stream carries internal energy, potential energy, and
kinetic energy.
3. A fluid stream entering or exiting a control volume is doing PV
work.
4. The internal energy and the PV work of a stream of fluid make
up the enthalpy of the stream.
5. Heat and shaft work may be transferred through the control
surface to or from the control volume.
6. Heat transferred from the control volume to the surroundings isconsidered positive by convention.
7. For an adiabatic process, the heat transferred to the system is
zero.
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Mass and Energy Balances 16
Steps in problem solving a Material Balance
1. Collect all known data on mass and composition of all inlet and exit streamsfrom the statement of the problem.
2. Draw a block diagram, indicating the process, with inlet and exit streams
properly identified. Draw the system boundary.
3. Write all available data on the block diagram
4. Select a suitable basis (such as mass or time) for calculations. The selection
of basis depends on the convenience of the computations.
5. Using equation of mass balance, write material balance in term of the
selected basis for calculating unknowns. For each unknown, an independent
material balance is required.
6. Solve material balances to determine the unkowns.
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17
Example 1.
A wet food product contains 70 % water. After drying, it is found that 80 % of
original water has been removed. Determine (a) mass of water removed per
kilogram of wet food, and (b) composition of dried food.
Given :
Initial water content : F = 70 %
Water removed : R = 80 % of original water content
Feed (F)intial water
content 70 %
R : water removed 80 % oforiginal water content
Product (P)composition ?
Solution :
1. Select basis 1 kg wet food
2. Mass of water in inlet stream : Fwater = 70 % x 1 kg = 0.70 kg
3. Water removed in drying : R = 80 % x Fwater = 0.8 x 0.7 = 0.56 kg H2O / kg ofwet food material
4. Write material balance on water : Fwater – Pwater = R (a) Pwater = 0.14 kg
5. Write material balance on solid : Fsolid = Psolid Psolid = 0.3 kg
6. (b) the dried food contains 0.14 kg water and 0.3 kg solids
dryer
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18
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Mass and Energy Balances 19
Example 2.
Potato flakes (moisture content (mc) 75 % wet basis) are being dried in a
concurrent flow drier. The mc of the air entering the drier is 0.08 kg H2O/kg dry air.
The mc leaving the drier is 0.18 kg H2O/kg dry air. The air flow rate in the drier is
100 kg dry air per hour. At steady state, calculate the following :
a. What is the mass flow rate of dried potatoes ?
b. What is the mc, dry basis, of dried potatoes exiting the drier ?
Inlet air 0.08 kg H2O/kg dry air 100 kg/h dry air
Feed50 kg wet potato flakes/h
MC 75 % wet basis
Exit air 0.18 kg H2O/kg dry air
Productmoisture content ?
dryer
Given :
Weight of potato flakes entetring the dryer F = 50 kg with 75 % of MC wb
Moisture content of air entering the dryer is 0.08 kg H2O/kg dry air and the flow
rate is 100 kg dry air/h
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Mass and Energy Balances 20
Solution :
1. Basis = 1 hour
2. Mass of air entering the dryer = mass of dry air + mass of water
I = 100 + 0.08 x 100 = 108 kg
3. Mass of air leaving the dryer = mass of dry air + mass of water
E = 100 + 0.18 x 100 = 118 kg
4. Total balance on the dryer
I + F = E + P 108 + 50 = 118 + P (a) P = 40 kg/h
5. Solid balance on the dryer Mass of dry solids entering the dryer = mass of dry solid leaving the dryer
(1 – 0.75) F = y P
6. Therefore, moisture content (wet basis) of the product : 1 – 0.3125 =0.6875 kg
7. The dry basis MC of the Product
kg3125.040
5025.0==
x y
solidsdryO/kgHkg2.23125.0
6875.0
solidsdryof mass
water of mass2 _ ===basisdry MC
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21
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Mass and Energy Balances 22
Example 3.
A tubular water blancher is being used to process lima beans. The product
mass flow is 860 kg/h. it is found that the theoretical energy consumed
for blanching process amounts to 1.19 GJ/h. the energy lost due to lack
of insulation around the blancher is estimate to be 0.24 GJ/h. if the totalenergy input to blancher is 2.71 GJ/h,
a. Calculate the energy required to reheat water
b. Determine the percent energy associated with each stream
Water
blancher
Energy losses from
surface = 0.24 GJ/h
Energy input =
2.71 GJ/h
Energy leaving the
product = 1.19 GJ/h
Energy leaving with water ?
Given
Product mass flow rate = 860 kg/h
Theoritical energy consumed for blanching process = 1.19 GJ/h
Energy loss due to lack of insulation = 0.24 GJ/h
Energy input to blancher = 2.71 GJ/h
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Mass and Energy Balances 23
Solution
Select 1 hour as a basis
Write energy balance
E input to blancher = E out with product + E out with water + E losses fromsurfaces
2.71 = 1.19 + Eo_water + 0.24 Eo_water = 1.28 GJ/h
a. E required to reheat water = E input – E o_water = 2.71 – 1.19 = 1.43 GJ/h
b. Percent energy associate with :
- Product (out) = 1.19/2.71 x 100 % = 43.91 %
- Water (out) = 1.28/2.71 x 100 % = 47.23 %
- Losses = 0.24/2.71 x 100 % = 8.86 %
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Mass and Energy Balances 24
Example 4.
Steam is used for peeling of potatoes in a semicontinuous operation. Steam is
supplied at the rate of 4 kg per 100 kg of unpeeled potatoes. The unpeeled
potatoes enter the system with a temperature of 17 oC, and peeled pototoes
leave at 35o
C. A specific heat of unpeeled potatoes, waste stream, and peeledpotatoes are 3.7, 4.2 and 3.5 kJ/(kg.K), respectively. If the heat content
(assuming 0 oC reference temperature) of the steam is 2750 kJ/kg, determine
the quantities of the waste stream and peeled potatoes from the process.
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Mass and Energy Balances 25
Given : up unpeeled potatoes pp peeled potatoes waste steam wt
Mass flow of steam = 4 kg per 100 kgup
Tup = 17o
C Tpp = 35o
C Twt = 60o
CCpup = 3.7 kJ/kg-K Cppp = 3.5 kJ/kg-K Cpwt = 4.2 kJ/kg-K
Heat content of steam (H) = 2750 kJ/kg
Assuming 0 oC as temperature’s reference
Potato peeler
F = 100 kg
Tf = 17oC
W = ?
Tw = 60oC
Hs = 2750 kJ/kgS= 4 kg
P = ?
Tp = 35oC
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Mass and Energy Balances 26
Solution
Select 100 kg of up as a basis
From mass balance
F + S = W + P
100 + 4 = W + PW = 104 – P (I)
From energy balance
Einput_up + Einput_s = Eout_ws + Eout_pp
F Cpup ΔT + SH = W Cpwt ΔT + P Cppp ΔT
100 x 3700 x (17-0) + 4 x 2750 = W x 4200 x (60-0) + P x 3500 x (35-0)
6290 + 11000 = 252 W + 122.5 P
(I) 17290 = 252 (104 – P) + 122.5 P P = 68.87 kg
W = 35.14 kg
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Mass and Energy Balances 27
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Mass and Energy Balances 28
1. 10 kg of food at a moisture content of 320% dry basis is dried to 50% wet
basis. Calculate the amount of water removed.
2. A batch of 5 kg of food product has a moisture content of 150% dry basis.Calculate how much water must be removed from this product to reduce its
moisture content to 20% wet basis.
3. A liquid product with 10% product solids is blended with sugar before being
concentrated (removal of water) to obtain a final product with 15% product
solids and 15% sugar solids. Determine the quantity of final product obtained
from 200 kg of liquid product. How much sugar is required? Compute mass of
water removed during concentration.
4. 1000 kg/h of milk is heated in a heat exchanger from 45 oC to 72 oC. Water is
used as the heating medium. It enters the heat exchanger at 90 oC and leaves
at 75 oC. Calculate the mass flow rate of the heating medium, if the heat losses
to the environment are equal to 1 kW. The heat capacity of water is given equalto 4.2 kJ/kgoC and that of milk 3.9 kJ/kgoC.
5. How much saturated steam with 120.8 kPa pressure is required to heat 1000
g/h of juice from 5 oC to 95 oC? Assume that the heat capacity of the juice is 4
kJ/ kgoC.
Exercises