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Mass and Energy Balance Review 27 th July 2010

Mass and Energy Balance Review 27 th July 2010. Mass balance is the fundamental to control the process- control the yield of product When any changes occur in the process, the mass balance need to be determined again Energy balance is used at various stages of a process

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Mass and Energy Balance Review 27 th July 2010

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  1. Mass and Energy Balance Review27th July 2010

  2. Mass balance is the fundamental to control the process- control the yield of product • When any changes occur in the process, the mass balance need to be determined again • Energy balance is used at various stages of a process • The increasing cost of energy has caused the industries to examine means of reducing energy consumption in processing

  3. Products out mP1,mP2,mP3 Raw Materials in mR1 mR2 mR3 Waste products mW1,mW2,mW3 Unit Operation Stored Materials mS1 mS2 mS3 Stored Energy ES1 ES2 ES3 Energy in products EP1,EP2,EP3 Energy in ER1 ER2 ER3 Energy in Waste EW1,EW2,EW3 Energy losses To surroundings EL1,EL2,EL3

  4. Material and energy balances can be simple, at times they can be very complicated, but the basic approach is general. • Experience in working with the simpler systems such as individual unit operations will develop the facility to extend the methods to the more complicated situations • The increasing availability of computers has meant that very complex mass and energy balances can be set up and manipulated quite readily and therefore used in everyday process management to maximise product yields and minimise costs

  5. Centrifuging of milk If 35,000kg of whole milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream with 45% fat, what are the flow rates of the two output streams from a continuous centrifuge which accomplishes this separation? Basis 1 hour's flow of whole milk

  6. Answer: Mass in Total mass flow rate = 35000/6 = 5833 kg/h. Fat = 5833kg/h x 0.04 = 233 kg/h. Mass out Let the mass flow rate of cream be x kg then its total fat content is 0.45x. The mass flow rate of skim milk is (5833 - x) and its total fat content is 0.0045 (5833 – x) Material balance on fat: Fat in = Fat out 5833 x 0.04 = 0.0045(5833 - x) + 0.45x. and so x = 465 kg. So that the mass flow of cream is 465 kg / h and skim milk (5833 – 465) = 5368 kg/h

  7. Example: Autoclave heat balance in canning An autoclave contains 1000 cans of pea soup. It is heated to an overall temperature of 100 oC. If the cans are to be cooled to 40 oC before leaving the autoclave, how much cooling water is required if it enters at 15 oC and leaves at 35 oC? Info: The specific heats of the pea soup and the can metal are respectively 4.1 kJ/ kg oC and 0.50 kJ/ kg oC. The weight of each can is 60g and it contains 0.45 kg of pea soup. Assume that the heat content of the autoclave walls above 40 oC is 1.6 x 104 kJ and that there is no heat loss through the walls. Let w = the weight of cooling water required; and the datum temperature be 40oC, the temperature of the cans leaving the autoclave

  8. Answer: Heat entering Heat in cans = weight of cans x specific heat x temperature above datum = 1000 x 0.06 x 0.50 x (100-40) kJ = 1.8 x 103 kJ Heat in can contents = weight pea soup x specific heat x temperature above datum = 1000 x 0.45 x 4.1 x (100 - 40) = 1.1 x 105 kJ Heat in water = weight of water x specific heat x temperature above datum = w x 4.186 x (15-40) = -104.6 w kJ. Heat leaving Heat in cans = 1000 x 0.06 x 0.50 x (40-40) (cans leave at datum temperature) = 0 Heat in can contents = 1000 x 0.45 x 4.1 x (40-40) = 0 Heat in water = w x 4.186 x (35-40) = -20.9 w HEAT-ENERGY BALANCE OF COOLING PROCESS; 40oC AS DATUM LINE Heat Entering (kJ) Heat Leaving (kJ) Heat in cans 1800 Heat in cans 0 Heat in can contents 110000 Heat in can contents 0 Heat in autoclave wall 16000 Heat in autoclave wall 0 Heat in water -104.6 w Heat in water -20.9 W Total heat entering 127.800 – 104.6 w Total heat leaving -20.9 W Total heat entering =Total heat leaving 127800 – 104.6 w = -20.9 w w = 1527 kg

  9. Degree of freedom revisited Number of system variables whose values must be specified before the remaining variables can be calculated. Degree of freedom (ndf) of a system: ndf = nv – ne where, nv = variables; ne = independent eq If ndf = 0 (e.g. 3 unknowns & 3 independent eq), the unknown variables can be calculated. If ndf > 0 (e.g. 5 unknowns & 3 independent eq ndf = 2), specify the design variables(specified) & calculate the state variables(calculated). If ndf < 0 (independent eq > unknowns) process is overspecified.

  10. Degree of freedom revisited Unknown variables for a single unit: Unknown component amounts / flowrate for all inlet & outlet streams Unknown stream T & P Unknown rate of energy transfer (as heat & power) Equation to determine these unknowns: Material balances for each independent species Energy balance Phase & chemical equilibrium relations Additional specified relationship among process variables

  11. A heated mixer example ndf analysis: 6 variables (n1, …, n5, Q) – 3 eq (2 material balances & 1 energy balances) = 3 degrees of freedom Specify 3 design variables & solve the rest. n1 (kg O2) 40ºC Heated mixer n4 (kg O2) n5 (kg N2) 50ºC n2 (kg O2) n3 (kg N2) 25ºC Q (kJ)

  12. Degree of freedom revisited Given the following equations: x1 + 2x2 – x32 = 0 5x1 – x23 + 4 = 0 What is the ndf for this system? Which design variable to be chosen for an easier solution? Given the following equations: 5x – 3y = 7 10x – 3y – 6z = 14 y = 2z Why can’t you solve this equation? Choose a design variable, specify it & determine the rest of the state variables.

  13. Advantages vs. disadvantages Advantages of sequential modular: Block diagram looks like conventional process flowchart, engineers are comfortable to work with this approach Relatively simple & straight forward Disadvantages: Operated in forward calculation mode only, i.e. calculate product stream based on feed stream variables & process conditions. Tedious for the following situations (iterative calculation): Calculate feed stream based on process conditions & product stream variables Calculate process conditions based on feed & product stream variables.

  14. Equation solving Equations for all units are collected & solved simultaneously. Disadvantages: Decomposition of system into its constituent unit operation is lost. Large number of (nonlinear) equation, cumbersome & time-consuming problem. Powerful equation-solving commercial programs that make equation-based approach competitive with sequential modular: Maple, Mathematica, Matlab, MathCad & E-Z Solve.

  15. Consideration when solving problems Whether you do it manually, spreadsheet,or simulation you can only determine all unknown process variables if it has zero degree of freedom If you have d.o.f, select design variables, then solve state variables When you first construct the simulation, don’t believe any of the initial results it gives you. Check it manually.

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