Supratomo
Mass and Energy Balances
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• 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 basic unit 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 be used 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. Mass and Energy Balances
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BASIC PRINSIPLES
Mass and Energy Balances
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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 a system .
n
.
m inlet = ∑ m i i =1
Rate of mass exiting through the boundary of a system
-
.
n
=
Rate of mass accumulation within the system
.
m outlet = ∑ m o m accumulation = i =1
m inlet − m outlet =
dm system dt
dm system
Mass and Energy Balances
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CONSERVATION of MASS for an Open System Consider a section of pipe used in transporting a fluid.
dm = ρ v o dA
m = ∫ ρ v o dA = ∫ ρ dV A
v
d ρ v o dA - ∫ ρ v o dA = ∫ ρ dV ∫ dt V Ainlet Aoutlet Mass and Energy Balances
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For a uniform flow :
d ρ v o dA - ∑ ρ v o dA = ∫ ρ dV ∑ dt V inlet outlet For a steady state flow flow does not change with time
∑ ρ v dA = ∑ ρ v dA o
o
inlet
outlet
If we are dealing with and incompressible fluid there is no change in density.
∑v inlet
o
dA =
∑v
o
dA
outlet
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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.
dm system dt
=0
msystem = constant
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Discusion 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 term has 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. Mass and Energy Balances
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CONSERVATION of ENERGY 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. Mass and Energy Balances
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Forms of Energy Potensial Energy of a system is by vitue of its location with respect to the gravitational field.
EPE = mgh
Kinetic Energy of an object is due to its velocity.
EKE
1 2 = mv 2
Internal Energy is due to the microscopic nature of the system . They move in random direction, collide each other, vibrate and rotate.
Ei = mc p ∆T Total Energy of a system can be written in the form of an equation as :
Etotal = EKE + EPE + Eelectrical + Echemical + …….. + Ei Mass and Energy Balances
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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
Eˆ in − Eˆ out = ∆Eˆ system For a system is in a steady state, there is no change in the energy of the system with time.
Eˆ in − Eˆ out = ∆Eˆ system = 0
Eˆ in = Eˆ out Mass and Energy Balances
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Energy Balance for an Open System A : cross-sectional area P : pressure of the fluid
Movement of a liquid volume
Force to push a liquid through the system boundary is : F = P.A The work done on the fluid element is : W mass flow = F.L = P.A.L = PV The total energy of the fluid element :
ET = EPE + EKE + Ei + PV
mv 2 ET = mgz + + Ei + PV 2
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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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Where :
In most of the cases, the overall energy balance ends up as an enthalpy balance because 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 (W s = 0). Then: Mass and Energy Balances
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Discusion 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 is considered positive by convention. 7. For an adiabatic process, the heat transferred to the system is zero. Mass and Energy Balances
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Steps in problem solving a Material Balance 1. Collect all known data on mass and composition of all inlet and exit streams from 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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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. R : water removed 80 % of Feed (F) Given :
intial water content 70 %
original water content
dryer
Product (P) composition ?
Initial water content : F = 70 % Water removed : R = 80 % of original water content 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 of wet 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
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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 Exit air
0.08 kg H2O/kg dry air 100 kg/h dry air
Feed
0.18 kg H2O/kg dry air
dryer Product
50 kg wet potato flakes/h MC 75 % wet basis
moisture content ?
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 Mass and Energy Balances
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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
y=
0.25 x50 = 0.3125 kg 40
6. Therefore, moisture content (wet basis) of the product : 1 – 0.3125 =0.6875 kg 7. The dry basis MC of the Product
MCdry _ basis =
mass of water 0.6875 = = 2.2 kg H 2 O/kg dry solids mass of dry solids 0.3125 Mass and Energy Balances
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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 total energy 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 Energy losses from surface = 0.24 GJ/h Energy leaving the product = 1.19 GJ/h Energy input = 2.71 GJ/h
Water blancher
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 Mass and Energy Balances
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Solution Select 1 hour as a basis Write energy balance E input to blancher = E out with product + E out with water + E losses from surfaces
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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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 35 oC. A specific heat of unpeeled potatoes, waste stream, and peeled potatoes 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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Given : up unpeeled potatoes pp peeled potatoes waste steam wt Mass flow of steam = 4 kg per 100 kgup Tup = 17 oC Tpp = 35 oC Twt = 60 oC Cpup = 3.7 kJ/kg-K Cppp = 3.5 kJ/kg-K Heat content of steam (H) = 2750 kJ/kg Assuming 0 oC as temperature’s reference
Cpwt = 4.2 kJ/kg-K
Hs = 2750 kJ/kg S= 4 kg F = 100 kg Tf = 17 oC
Potato peeler
P=? Tp = 35 oC
W=? Tw = 60 oC Mass and Energy Balances
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Solution Select 100 kg of up as a basis From mass balance F+S=W+P 100 + 4 = W + P W = 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 Mass and Energy Balances
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Mass and Energy Balances
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Exercises 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 equal to 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. Mass and Energy Balances
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