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Premium member Presentation Transcript PowerPoint Presentation: RESENTATION ON KNOWLEDGE OF MATERIAL BALANCE OF PLANT PREPARED BY M.NAUMAN TALAT E-II UREA PLANTPROGRAM OUTLINE: PROGRAM OUTLINE MATERIAL BALANCE FUNDAMENTALS THE MASS BALANCE EQUATION STRATEGY FOR ANALYZING MATERIAL BALANCE PROBLEMS MATERIAL BALANCE OF UREA PLANTMaterial Balance Fundamentals : Material Balance Fundamentals Material balances (mass balances) are based on the fundamental “law of conservation of mass ” . In particular, chemical engineers are concerned with doing mass balances around chemical processes.PowerPoint Presentation: Doing a ‘mass balance’ is similar in principle to accounting . In accounting, accountants do balances of what happens to a Company’s money. Chemical engineers do a mass balance to account for what happens to each of the chemicals that is used in a chemical process.Classification of Processes : Classification of Processes A- Based on how the process varies with time. a. Steady-state process is one that does not change with time. Say a barrel of water had a hole in the bottom. Your job is to use a hose and keep filling the barrel with water. If the hose is discharging the same amount of water into the barrel as is draining out, then the water level in the barrel would not rise or fall. When the water level is constant it would be considered to be at steady state . b. Unsteady-state (Transient) process is one that changes with time.PowerPoint Presentation: B- Based on how the process was built to operate. A Continuous process is a process that has the feed streams and product streams moving chemicals into and out of the process all the time. At every instant, the process is fed and product is produced. A Batch process is a process where the feed streams are fed to the process to get it started. The feed material is then processed through various process steps and the finished products are created during one or more of the steps. The process is fed and products result only at specific times.The Mass Balance Equation : The Mass Balance Equation The law of ’conservation of mass’ states that mass cannot be created or destroyed. We will use this law in the form of a general mass balance equation to account for the total mass all of the chemicals that are involved in the process. The total mass balance equation can be written as I O = A INPUT OUTPUT = ACCUMULATIONPowerPoint Presentation: If the process is at steady-state, there is no accumulation of mass within the process. Thus INPUT = OUTPUT I = O When we apply this equation to a process, it is best to write it as Masses entering via feed streams = Masses exiting via product streamsPowerPoint Presentation: We understand that we must include the mass of every chemical in every stream. The above equation can applied to batch and continuous processes as Mass in = Mass out for a batch process, and Mass in by flow = Mass out by flow for a continuous process.PowerPoint Presentation: If the process involves chemical reaction(s), we must account for the formation of product chemicals and the consumption of feed chemicals. We must remind ourselves that the law of conservation of mass means total mass. For this case, we must write a mass balance for each chemical and account its formation and consumption as follows Mass in + Mass formed by reaction = Mass out + Mass used by reaction Or, written more simply as in + formed = out + consumedWhat balances can one write?: What balances can one write? A mass balance can be written using the total mass in each process stream. This is called a total balance. A separate mass balance can be written for each chemical component involved. These are called component balances. Example: A process unit involves 3 chemical components. How many mass balances can be written? Solution: We can write 4 balances. We can write a total balance and 3 component balances Thus, the number of independent balances we can write = the number of components.STRATEGY FOR ANALYZING MATERIAL BALANCE PROBLEMS: STRATEGY FOR ANALYZING MATERIAL BALANCE PROBLEMS Read the problem and clarify what is to be accomplished. Draw the sketch of the process. Label with symbols the flow of each stream and the associated compositions and other information that is unknown. Put all the known values of compositions and stream flows on the figure by each stream ; calculate the additional compositions and flow from given data as necessaryPowerPoint Presentation: Select a basis. Write down the names of appropriate set of balances to solve. Ascertain that a unique solution is possible. If not, look for more assumptions or check your assumptions. Solve the equations. Check your answers by introducing them, or some of them in any redundant material balancesMATERIAL BALANCE OF UREA PLANT: MATERIAL BALANCE OF UREA PLANT MATERIAL BALANCE Production Rate MTPD=370 Composition of finished product(%) Urea 99.7% Water .3% Raw material composition(%) Ammonia 99.7 Carbondioxide 97.5 Reactor conversion 60% Ammonia/cabondioxide 3.4 (Molar ratio) Stripper bottom composition(%) Urea 45.9 Ammonia 24.3 Carbondioxide 5.8 Water 24PowerPoint Presentation: MP Decomposer bottom composition(%) Urea 63.28 Ammonia 7.02 Carbondioxide 1.16 Water 28.54 LP Decomposer bottom composition(%) Urea 70.12 Ammonia 1.8 Carbondioxide 0.8 Water 27.28 Composition of V-6 outlet Urea(%) Urea 95.6 Ammonia 0.01 Carbondioxide 0.01 Water 4.38PowerPoint Presentation: Urea losses in V-7vapor 0.56% Urea losses in V-6vapor 0.23% Reactor outlet composition(%) Urea 31.9 Ammonia 33.14 Carbondioxide 15.6 Water 19.36 Process condensate to drain(%) Urea .0128 Ammonia .0128 Carbondioxide .0256 Water 99.948PowerPoint Presentation: Process condensate to V-8 (%) Urea 0 Ammonia 33.98 Carbondioxide 19.32 Water 46.7MATERIAL BALANCE AROUND PRILLING TOWER: MATERIAL BALANCE AROUND PRILLING TOWER BASIS: 370MTD of urea=15417Kg/hr Now composition of finished product Urea=99.7% Water=.3% So in finished product Urea=15417*.997=15371kg/hr Water=15417*.003=46kg/hrPowerPoint Presentation: Now we have to find the dust losses in prilling tower: As per designed figure urea losses in exhaust gases at the top of prilling tower is .2% According to SPL manual Air flow of 160*1000 Nm3/hr contains 20Kg/hr of urea dust. Air flow with four fans is 4*80000=320000 Nm3/hr So dust loss=83Kg/hr So urea melt entering in prilling tower = 15417+83=15500Kg/hrMaterial balance around V-7: Material balance around V-7 As per basis urea losses in V-07 is .56% of outlet urea in melt .56/100*15454=86Kg/hr So urea inlet to V-7 is 15454+86=15540Kg/hr Now we can calculate amount of A As A stream contains 95.6% urea so A=15540/.956=16255Kg/hrPowerPoint Presentation: Now applying overall balance to V-7 A=B+C So C=755Kg/hr Now applying component balance AMMONIA BALANCE : .0001(A)=C(xNH3) XNH3=.00215PowerPoint Presentation: CO2 BALANCE : .0001(A)=C(xCO2) XCO2=.00215 WATER BALANCE .0438(A)=.003(B)+C(XH2O) XH2O =.881 NOW XUREA=.1147PowerPoint Presentation: RESULTS A=16255 Kg/hr B=15500 Kg/hr C=755 Kg/hr COMPOSITION OF STREAM A:(Kg/hr ) UREA=15540 AMMONIA=1.625 CO2=1.625 WATER=711 COMPOSITION OF STREAM C:(Kg/hr) UREA=86 AMMONIA=1.6325 CO2=1.6325 WATER=655Material balance around V-6: Material balance around V-6 As per basis urea losses in V-6 is .23% of outlet urea in melt .23/100*15540=36Kg/hr Urea in A=urea in B+urea in C Urea in A=15540+36=15576Kg/hr So A=15576/0.7012=22213Kg/hrPowerPoint Presentation: Now A=B+C C=5958Kg/hr Now applying component balance AMMONIA BALANCE: .018(A)= B(.0001)+C(xNH3) XNH3=.0668PowerPoint Presentation: CO2 BALANCE: .008(A)=B(.0001)+C(xCO2) XCO2=.0295 WATER BALANCE: .2728(A)=.0438(B)+C(XH2O) XH2O =.8975 NOW XUREA=.0062PowerPoint Presentation: RESULTS A=22213 Kg/hr B=16255 Kg/hr C=5958Kg/hr COMPOSITION OF STREAM A:(Kg/hr) UREA=15575.7 AMMONIA=399.83 CO2=177.70 WATER=6059.7 COMPOSITION OF STREAM C:(Kg/hr) UREA=36 AMMONIA=397.77 CO2=175.76 WATER=5347.Material balance around LPD: Material balance around LPD Now applying overall balance to LPD A=B+C A=22213+C Now applying component balance UREA BALANCE: .6328(A)=B(0.7012) A=24614Kg/hrPowerPoint Presentation: SO C=A-B C=2401Kg/hr AMMONIA BALANCE: .0702(A)=B(.018)+C(xNH3) XNH3=.553 WATER BALANCE: .2854(A)=.2728(B)+C(XH2O) XH2O =.401 NOW XCO2=.046PowerPoint Presentation: RESULTS A=24614 Kg/hr B=22213 Kg/hr C=2401Kg/hr COMPOSITION OF STREAM A:(Kg/hr) UREA=15575.7 AMMONIA=1727.9 CO2=285.5 WATER=7024.8PowerPoint Presentation: COMPOSITION OF STREAM C:(Kg/hr) AMMONIA=1327.7 CO2=110.44 WATER=962.8 COMPOSITION OF B:(Kg/hr) UREA=15575.7 AMMONIA=399.83 CO2=177.7 WATER=6059.7Material balance around MPD: Material balance around MPD Now applying overall balance to MPD A=B+C A=24614+C Now applying component balance UREA BALANCE: .459(A)=B(0.6328) A=33934.07Kg/hrPowerPoint Presentation: Now C=A-B C=9320.07Kg/hr AMMONIA BALANCE: A(0.243)=B(.0702)+C(xNH3) XNH3=.699PowerPoint Presentation: Now C=A-B C=9320.07Kg/hr AMMONIA BALANCE: A(0.243)=B(.0702)+C(xNH3) XNH3=.699PowerPoint Presentation: CARBONDIOXIDE BALANCE .058(A)=.0116(B)+C(XCO2) XCO2=.181 NOW XH2O=.120PowerPoint Presentation: RESULTS A=33934.07 Kg/hr B=24614 Kg/hr C=9320.07Kg/hr COMPOSITION OF STREAM A:(Kg/hr) UREA=15575.7 AMMONIA=8245.97 CO2=1968.17 WATER=8144.17PowerPoint Presentation: COMPOSITION OF STREAM C:(Kg/hr) AMMONIA=6514.72 CO2=1686.93 WATER=1118.40Material balance around Stripper: Material balance around Stripper Now applying overall balance to Stripper A=B+C A=33934.07+C Now applying component balance UREA BALANCE: .319(A)=B(0.0459) A=48826.76Kg/hr SO C=A-B C=14892.69Kg/hrPowerPoint Presentation: AMMONIA BALANCE: .3314(A)=B(.243)+C(xNH3) XNH3=.532 CARBONDIOXIDE BALANCE .156(A)=.058(B)+C(XCO2) XCO2 =.379 NOW XH2O=.089PowerPoint Presentation: RESULTS A=48826.76 Kg/hr B=33934.07 Kg/hr C=14892.69Kg/hr COMPOSITION OF STREAM A:(Kg/hr) UREA=15575.73 AMMONIA=16181.1 CO2=7616.9 WATER=9452.8PowerPoint Presentation: COMPOSITION OF STREAM C:(Kg/hr) AMMONIA=7922.91 CO2=5644.32 WATER=1325.44TIME FOR HOME WORK : TIME FOR HOME WORK CARRTY OUT MATERIAL BALANCE OF WASTE WATER TREATMENT SECTIONOUTLINE FOR NEXT SESSIONS : OUTLINE FOR NEXT SESSIONS MATERIAL BALNCE OF WASTE WATER TREATMENT SECTION. MATERIAL BALANCE OF OFF GASES RECOVERIES SECTION. MATERIAL BALANCE OF REACTOR.PowerPoint Presentation: THANKS You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.