WINTER-15 EXAMINATION Model Answer

Transcription

1 Subject code :(735) Page of 9 Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may vary but the examiner may try to assess the understanding level of the candidate. 3) The language errors such as grammatical, spelling errors should not be given more Importance (Not applicable for subject English and Communication Skills. ) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn. 5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate s answers and model answer. 6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on candidate s understanding. 7) For programming language papers, credit may be given to any other program based on equivalent concept.

2 Subject code :(735) Page of 9 Q No. Answer marks Total marks -A Any 8 A-a Ideal Gas law: PV=nRT where P - pressure, V - volume, n- moles, K-absolute temperature and R universal gas constant A-b Principle involved in solving material balance problems without chemical reaction: Law of conservation of mass: It states that For any process input= output+accumulation A-c Amagat s law: It states that the total volume exerted by a gas mixture is equal to the sum of pure component volumes Mathematical Statement:V =V +V +V 3 where V is the total volume of gas mixture, V,V, V 3 are pure component volumes A-d Raoult s law: Raoult s law states that equilibrium partial pressure of a constituent at a given temperature is equal to the product of its vapour pressure in pure state and its mol fraction in the liquid phase. P A = P 0 A X A A-e Adiabatic reaction : It is the reaction which proceeds without loss or gain of heat. Adiabatic reaction temperature: Temperature of product under adiabatic condition is called adiabatic reaction temperature. A-f % Conversion: It is the ratio of amount of limiting reactant reacted to the amount of limiting reactant totally charged. Express in percentage. %Yield of desired product = (moles of limiting component reacted to form

3 Subject code :(735) Page 3 of 9 desired product/ total moles of limiting component reacted)* 00 -B Any B-a B-b B-c Basis: 00 kmoles of air Average mol.wt of air= 0.79*8+0.*3= 8.8 Density of air = P*Mav/ RT At NTP, P= 0.35 KPa, R= 8.3, T=73 K Density =.9 kg/m 3 Partial pressure of CO in gas phase P CO = H * x CO =7*0 6 **0-6 = 8 KPa Basis: 0.8m 3 fixed mass of gas at constant temperature P = V = 0.8 m 3 T = T= T P =.5 V =? m 3 P V /T = P V /T *0.8/T =.5*V /T Or V = m 3 Any 6 -a Steps involved in solving material balance calculations:.assume suitable basis of calculation as given in problem.. Adopt weight units in case of problem of process without chemical reaction. 3. Draw block diagram of process. Show input and output streams 5. Write overall material balance 6. Write individual material balance 7. Solve above two algebric equations ½ mark each

4 Subject code :(735) Page of 9 8. Get values of two unknown quantities. -b Basis: kg air Kg air 0. kg NH 3 absorption NH3 gas,0.00 Kg NH3 NH3 absorbed X Kg NH3 NH3 balance is 0.= X X= 0.96 Kg % recovery of ammonia = (NH3 absorbed/ NH3 original)*00 = (0.96/0.)*00 = 98% -c Excess component: It is the reactant which is in excess of the theoretical or stoichiometric requirement. Limiting component: It is the reactant which would disappear first if a rection goes to completion.or it is the reactant which decides the extent of a reaction. -d Basis: 00 of ethanol charged. C H 5 OH CH 3 CHO + H I kmol C H 5 OH reacted = kmol CH 3 CHO formed C H 5 OH reacted to produce 5 kmol CH 3 CHO = 5 kmol Conversion of C H 5 OH =( kmole of C H 5 OH reacted/ kmoles Of C H 5 OH fed)*00

5 Subject code :(735) Page 5 of 9 =( 5/00)*00 % conversionof ethanol = 5 % -e Basis 00Kmol SO charged O fed=75 kmoles Reaction SO + O = SO 3 Theorctical requirement of O Kmol SO 0.5 Kmol O theoretical 00 kmol SO =. 00 Theorctical O = 50 Kmol % excess of O used = = =50 00 % excess air used = 50% -f Basis: 00 Kg/hr C H 5 OH m=00 Kg/hr λ= 0 kcal/kg Latent heat = m. λ = 00 *0 = 000 Kcal / hr 3 Any 6 3-a SOLUTION : 8

6 Subject code :(735) Page 6 of 9 BASIS : 00 mol of ethylene Reaction I C H + ½ O C H O Reaction II CH + 3O CO + HO From reaction I, Kmol of CHO formed = Kmol CH reacted CH reacted to form 80 kmol CHO = x 80 = 80Kmol From reaction II, kmol of CO formed = Kmol CH reacted CH reacted to form 0 kmol CO = x0 = 5Kmol CH totally reacted = = 85 % conversion of CH = x 00 = 85% % yield of CHO = x00 = 9.% 3-b SOLUTION : BASIS : 00 kg of mixed acid H SO =5 Kg HNO 3 = Kg 8

7 Subject code :(735) Page 7 of 9 H SO 98 % HNO 3 =, H SO =5%,H O =39% HNO 3 Let, X be the kg of H SO and Y be the kg of HNO 3 x + y = 00 Material Balance on H SO :- 0.98x = 5 x= 5/0.98 x= 5.9 weight of H SO = 5.9 kg weight of HNO 3 = = 5.09 kg Let, y is strength of HNO y = y = /5.09 y = strength of HNO 3 = x 00 = 77.6 % Ans. Weight ratio of H SO 5.9 = Weight ratio of HNO

8 Subject code :(735) Page 8 of 9 = Ans. 3-c SOLUTION : BASIS : 0,000 kg /h of 5 % methanol feed solution to coloumn kg/h Feed 5 % methanol Distillate (d) 98 % methanol Distillation Coloumn Waste solution (w) % methanol (bottom product) Block diagram for distillation of 5 % methanol feed Let x and y be the mass flow rates of distillate / product and waste solution (bottom product) respectively. Overall Material Balance : x + y = 0000.(i) Material Balance of Methanol : 0.98 x y = 0.5 x x y = 5000.(ii) From equation(i),

9 Subject code :(735) Page 9 of 9 y = x.(iii) Put the value of y from equation (iii) in eqn (ii) and solve for x x (0000 x) = x 0.0 x = x = 800 x = 98.5 kg/h y = x = = kg/hr Mass flow rate of distillate = 98.5 kg/h Mass flow rate of waste solution = kg/hr.ans. Methanol in waste solution = 0.0 x = 50.5 kg /h Methanol in Feed = 0.5 x 0000 = 5000 kg /h % loss of methanol = x 00 =. x 00 = 3.0 % Ans Any 6 -a SOLUTION : Basis : mol of CH6 (liq) 8. C(s) + O(g) > CO(g) ΔH = KJ/mol. H (g) +/ O(g) > HO(l) ΔH = KJ/mol

10 Subject code :(735) Page 0 of 9 3. CH6 (liq.) +5.5 O(g) > CO(g)+3 HO(l) ΔH0c = kj/mol. C(s) + 3 H (g) > CH6 (liq.) ΔH 0 f =? ΔH 0 f = Standard heat of formation of gaseous ethyl alcohol at 98.5 K Reaction() = * Reaction () + 3* Reaction () Reaction (3) ΔH 0 f = *ΔH + 3*ΔH - ΔH 0 c = (-393.5) + 3*(-85.83) (-50.) = (-57.0) + (-857.9) = KJ/mol ΔH0f = KJ/mol Ans. -b SOLUTION : BASIS : 000 kg of wet solid kg 5 % H o Dried Product 99.5 % solid Weight of solid in feed = 000*0.85 = 850 kg Weight of water in feed = 50 kg Let, weight of dried producy = x kg

11 Subject code :(735) Page of 9 It contain 99.5 % solid and solid is unchanging component. solid balance, x = 850 x = 850/0.995 x = 85.7 kg weight of water in dried product = =.7 kg Water removed in dryer = 50.7 = 5.73 kg -c SOLUTION : original water removed =. BASIS : 00 kmol of SO SO + ½ O SO 3 kmol of SO ½ kmol O kmol O required = 00 * 0.5 O supplied = 50*.8 N supplied = x 90 = 50 kmol = 90 kmol = kmol SO reacted = 00 x 0.70 = 70 kmol O required = 70 x 0.5 = 35 kmol SO 3 formed = 70 kmol x 00 Composition of gases leaving reactor :- = 97.5 %.Ans. 8

12 Subject code :(735) Page of 9 Product Quantity,Kmo Mole % l SO = O = 55. N SO Total Any 6 5-a Basis: 00 kg gas Avg mol wt of gas = 30 * *0.35=39.8 Total moles= 00/39.8 =.5 kmoles 8 Kmoles of ethane=.638 Kmoles of butane= Reactions are: C H 6 + 7/ O CO + 3H O C H 0 + 3/ O CO + 5 O Oxygen theoretically required = 7/* / * =. O fed =. + 0.*. = 3.70 kmoles O unreacted = 3.70.=.8kmoles N fed= 5.5 kmoles CO formed = = kmoles H O formed = = 9.93 kmoles Weight of gas =.8* * *+9.93*8 = Kg

13 Subject code :(735) Page 3 of 9 5-b Case I: Basis: 00 Kg/hr of solid handling capacity of the evaporator Water evaporated 8 Weak liquor % solids Evaporator Thick liquor,50% solids Let X be the be the Kg/hr of weak liquor then we have X = 00 X = 000 Kg/hr Y be the be the Kg/hr of thick liquor then we have. 0.5 Y = 00 Y = 00 kg/hr Overall Material Balance: Kg/hr of weak liquor = Kg/hr of thick liquor + Kg/hr of Water evaporated 000 = 00 + Kg/hr of Water evaporated Kg/hr of Water evaporated = 800 Kg/hr Case II: Basis: 800 Kg/hr of Water evaporated Let A be the be the Kg/hr of weak liquor B be the be the Kg/hr of thick liquor Overall Material Balance: A = B Material Balance of Solids : 0.0 A = 0.35 B A = 8.75 B Putting in above equation

14 Subject code :(735) Page of B = B B = 3.6 Kg/hr A = = 03.6 Kg/hr Solid in weak liquor = 0.0 X 03.6 = 8.3 Kg/hr Solid handling Capacity = 8.3 Kg/hr Ans. 5-c Basis: mol of ethylene gas C H (g) + ½ O C H O(g) ΔH o R = Standard heat of reaction = [ Σ ΔH o c] product - [ Σ ΔH o c] reactant = [ x ΔH o c) C H O(g) ] - [ x ΔH o c C H (g) + x ΔH o c O ] = [ x (-.58)] [ x (.50) + / x (0.0) ] = = KJ per mol C H O(g) produced. Change in the enthalpy for 5 mol C H O(g) produced = X 5 = -5.0 KJ Ans Any 6 6-a )Stoichiometric Equation : The stoichiometric equation of a chemical reaction is the statement indicating relative moles of reactant and products that take part in the reaction. For example, the stoichiometric equation CO + H CH 3 OH Indicates that one molecule of CO react with two molecules of hydrogen to

15 Subject code :(735) Page 5 of 9 produce one molecule of methanol ) Stoichiometric Coefficient : It is the number that precedes the formula of each component involved in a chemical reaction. For example, the stoichiometric equation CO + H CH 3 OH In above example Stoichiometric Coefficient of CO is one, Stoichiometric Coefficient of H is Two and Stoichiometric Coefficient of methanol is one (Students may write other suitable example) 6-b Basis : 00 Kg of ground nut seeds Let X be the Kg of cake obtained Material balance of Solids : Solids in seeds = Solids in cake 0.5 * 00 = 0.8* X X = 56.5 Kg Material balance of Oil : Oil in seeds = Oil in cake + Oil recovered 0.5* 00 = 0.05* Oil recovered Oil recovered = =.9 Kg Oil recovered % recovery of Oil = * 00 Oil in Seeds

16 Subject code :(735) Page 6 of 9.9 % recovery of Oil = * 00 5 % recovery of Oil = 93.75% ans. 6-c Basis :000 Kg of desired acid Let X and y be the kg of dilute acid and Commercial Grade required to prepare desired acid. Overall Material balance Kg dilute acid + kg Commercial Grade = Kg of desired acid X + Y = () Material Balance of H SO H SO in dilute acid + H SO in Commercial Grade = H SO in desired acid 0.5 X Y = 0.65 * 000

17 Subject code :(735) Page 7 of 9 0.5X Y = X Y = Y= X put in equation () X + ( X ) = 000 X= 5 kg We have, X +Y = Y =000 Y = 58 Kg Dilute acid required = 5 Kg Commercial Grade Acid required =58 kg 6-d Basis: 50 kmol of butane fed to combustion reactor Reaction, C H 0 + 3/ O ---- CO +5H O From reaction, Kmol of C H 0 =6.5 kmol O

18 Subject code :(735) Page 8 of 9 50 Kmol of C H 0 = (50*6.5)/ Kmol O = 35 Kmol O required Theoretical O required = 35 Kmol % Excess O = 35 % Actual O fed = Theoretical O required [ + % Excess /00] Actual O fed = Theoretical O required [ + 35 /00] Actual O fed = 35* [ + 35 /00] Actual O fed = kmol Ans. 6-e Basis : 00 Kmol of feed Feed contains 60 kmol A, 30 kmol B and 0 kmol inerts Let X be the kmol of A reacted by reaction : A + B C From reaction kmol A = kmol B = kmol C B reacted = (/)* X = 0.5 X kmol C formed = (/)* X = 0.5 X kmol Material Balance of A give A unreacted = ( 60 X ) kmol Material Balance of Inerts : Inerts in feed = Inert in product = 0 kmol C formed = (/)* X = 0.5 X kmol B unreacted = ( X) kmol Total moles of product stream = (60-X) + (30-0.5X) + 0=0.5X = 00 X Kmol Mole % of A in product stream = % Kmol A in product stream Mole % of A = * 00

19 Subject code :(735) Page 9 of 9 Total kmol of product stream 60 - X = * X X = 59.8 kmol = amount of A reacted Kmol A reacted Conversion of A = * 00 Total kmol of A feed 59.8 Conversion of A = * 00 = 98.6 % Ans 60 6-f Sensible Heat : Sensible heat is the heat that must be transferred to raise or lower the temperature of a substance or mixture of substance. Latent Heat : When matter undergoes a phase change, the enthalpy change associated with unit amount of matter at constant temperature and pressure is known as Latent Heat of phase change.