MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION. Summer-13 EXAMINATION. Model Answer

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1 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION Subject &code:mto(97) Important instructions to examiners : Summer-3 EXAMINATION Model Answer. The answers should be examined by keywords 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 given more importance.. While assessing figures, examiner may give credit for principal components indicated in a 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 answer and model answer. 6. In case of some questions credit may be given by judgment of relevant answer based on candidates understanding. Q no: A i Answer mar k Tot al mar ks Steady state equimolar counter diffusion: MTO(97) Page of

2 ii. MTO(97) Page of

3 Boiling point diagram for minimum boiling azeotrope Equilibrium diagram for minimum boiling azeotrope iii. Wet Bulb Temperature:The temperature recorded by a thermometer whose bulb is kept wet by wrapping with a wet cotton in the open air is called dry bulb temperature. Dry bulb temperature: The temperature recorded by a thermometer whose bulb is kept dry is called dry bulb temperature. Critical Moisture content: The moisture content of a material at which the constant rate period ends and the falling rate period starts is called critical moisture content. Equilibrium moisture content: The moisture content of a solid material that is in thermodynamic equilibrium with its vapour in the gas phase under specified humidity and temperature of gas is called equilibrium moisture content. MTO(97) Page 3 of

4 iv 3 Packed column for continuous distillation. Names of tower packings:. Pall ring. Lessing ring 3. Raschig ring. Cross partition ring 5. Intalox saddle MTO(97) Page of

5 B Fick s law of diffusion: 6 i It states that the flux of a diffusing component A in z direction in a binary mixture of A and B is proportional to the molar concentration gradient. J A = -D AB dc A /dz Reynolds Analogy between heat and mass transfer h/cp.u.ρ = kc/u Complete Reynolds analogy h/cp.u.ρ = kc/u = f/.b Selection criteria for solvent selection in liquid-liquid extraction: 6.ii. Selectivity: The ratio of concentration ratio of solute to feed solvent mar in extract phase to that in raffinate phase is called selectivity factor. It k is the measure of effectiveness of solvent for separating the for constituents. any. Recoverability: As solvent should be recovered for reuse frequently six by distillation, it should not form an azeotrope with extracted solute poi and for low cost recovery, relative volatility should be high. nts 3. Distribution coefficient: Higher values are desirable as less solvent will then be required for given extraction duty.. Density: The difference in densities of saturated liquid phases should be larger for physical separation. 5. Insolubility of solvent: The solvent insoluble in original liquid solvent should be preferred and it should have high solubility for solute to be extracted, then small amounts of solvent are required. 6. Chemical Stability: The solvent should be stable chemically and inert towards other components and should not be corrosive. 7. Cost: The solvent should be cheap. 8. The solvent should be non toxic, non flammable. 9. Solvent should have low viscosity, freezing point, vapor pressure for ease in handling and storage. 0. Interfacial tension: It should be high for coalescence of emulsions to occur more readily, as the same is of greater importance than MTO(97) Page 5 of

6 a dispersion. Basis: 00 kgmol feed. X F =0., X D =0.9, X W =0. F X F =D X D +W X W D= 37.5 kmol W=6.5 kmol R=3= L/D But L=Ln Therefore 3=Ln/D Or Ln=3D=.5 kmol Rectifying section operating line Y n+ = 0.75Xn+0.5 Lm=Ln+F=.5kmol Stripping section operating line Y m+ =.5Xm-0.0 Yn=Xd=0.9 Draw X-Y diagram Calculate X values from equilibrium diagram and calculate Y values from equations till the X value is equal to or less than Xw. Number of stages required=7 8.b Basis: 00 kg feed solution X F = 0.8 Solvent balance F(- X F ) = L L=5 kg NaNO 3 balance 0.8*00 = C+L*80.8/00 C = Yield of crystals=6.3 kg % yield of crystals = 6.3*00/8 =3.% 8 MTO(97) Page 6 of

7 .c Area(A) =0.0m Weight of dry material(w)=0.5 kg Initial moisture contentx =00/50=0.667 Moisture content after 0.5 hours X = 80/50=0.533 Rate of drying R= W(X-X)/A*t =.0kg/m hr Moisture content after 0.75 hours X 3 = 65/50=0.33 Rate of drying R= W(X-X)/A*t =.68kg/m hr 3.a 8 Resistance to transfer in each phase is regarded as line in a thin film close to the interface. ii) The transfer in these films is by steady state process of molecular diffusion. iii) The concentration gradient is assume to be linear in this films and it is zero outside the films. iv) The theory assume that the turbulence in the bulk fluid dye out at the interface of the films. v) The film capacity is negligible.] The equation for this theory is, /K y = /k y + m/k x 3.b Point Packed column Plate column Pressure drop Less More Handling corrosive Preferable Not preferable mar k MTO(97) Page 7 of

8 liquid eac Handling very small Preferable Not preferable h volume of liquid cost Less more 3.c i)it consists of vertical tank incorporating a turbine or propeller agitator. ii)feed solution to be extracted to be taken into an agitated vessel, then the required amount of solvent is added, and whole mass is agitated for predetermined time. iii)at the end of this, two phases are formed, one is extract phase and other is raffinite phase which are then separated. 3.d Various methods of distillations are: i) Simple or differential distillation ii)flash or equilibrium distillation iii) Fractionation or Rectification Flash distillation is carried out in a continuous manner. In this method, a liquid mixture is partially vaporized the vapor and liquid are allowed to attained equilibrium and finally withdrawn separately MTO(97) Page 8 of

9 The equation for flash distillation: y= -((-f)/f)*x + x f /f Where, x= the molefraction of more volatile component in liquid phase y= the molefraction of more volatile component in vapor phase f= fraction of liquid/feed vaporized 3.e. solution: The flux for equimolar counter diffusion is given by, NA = D AB / RTz (pa-pa) Where, D AB = 6.75*0-5 m /s, R= 8.35 m 3. kpa/ (kmol. K) T = 98 K, z= 30mm=0.03m pa = 55kPa, pa= 5 kpa NA = 6.75*0-5 (55-5)/8.35*98*0.03 = 3.63*0-5 kmol/(m.s).ai Tray Dryer: MTO(97) Page 9 of

10 Working: The material to be dried is spread over the trays and put into the cabinet then it is closed. Steam is continuously passed through the coil and the fan is started. Air is heated by heating coils, so its relative humidity decreases and the hot air then passes over the trays. The moisture is evaporated from the wet feed, gets added in air and finally air leaves the dryer through the outlet. The process is continued until the solids are dried. Advantages (any two): i) It is cheap and easy to construct. ii) Low space requirement. iii) Ease of cleaning. iv) No loss of product during drying. MTO(97) Page 0 of

11 .aii Swenson-walker Crystallizer: i) It is the cooling type continuous jacketed trough crystallizer. ii) It is an example of scrapped surface crystallizer. iii) It contain slow speed long pitch agitator with a speed of 7 r.p.m. It helps to transport crystals from one point to another point and doesn t allow crystals to settle at the bottom. iv) Feed is admitted at one end and mixture of crystals and mother liquor is removed from other end..aiii Reflux ratio: The ratio of quantity of liquid returned to the liquid that has taken out as distillate is known as reflux ratio (L/D). Total reflux: It is the total overhead product which is returned back to the column. At total reflux no, of plates required is minimum with maximum purity. Minimum reflux: At minimum reflux infinite no. of plates are required and purity also low and no product is obtained. Optimum reflux: The optimum ratio occurs at a point where sum of the fixed cost and operating cost is minimum. The optimum reflux ratio usually lies in the range of. to.5 times the minimum reflux ratio. MTO(97) Page of

12 .A -iv Basis : 00kg/hr of air flow to absorption tower. X = NH3 composition at bottom = kg of NH 3 /kg of water X = NH3 composition at top = kg of NH 3 /kg of water Y = NH3 composition in inlet gas to tower = kg of NH 3 /kg of inert gas Y = NH3 composition in outlet gas from tower = 0.00 kg of NH 3 /kg of air L = mass flowrate of solute free solvent in kg/hr V = mass flowrate of solute free gas or air in kg/hr MTO(97) Page of

13 V (Y -Y ) = L (X -X ) 00 ( ) = L ( ) L = 336. kg/hr Mass flowrate of water = 336. kg/hr.b -i Triangular diagram: 3 6 Where A=dilutent B=solvent C=solute Binodial curve: It represents two phase region that will split up into two layers in equilibrium with each other. Tie line: Tie line is a line which connects two phases in equilibrium with each other. Plait point: The plait point on binodial solubility curve represent a single phase where the length of tie line is zero. MTO(97) Page 3 of

14 .b Note:(since the problem given in this question is same as Qc except wt of -ii dry solid and dimensions of tray students can assume the data for area and Wt from same problem and if they write right expressions they should be the marks) Solution: A= Area of tray= 0.*0.=0.0m W = wt of dry solids = 50gm=50/000= 0.5kg Moisture in saw dust initially = 50-50=00gm X= initial moisture content = 00/50=0.666 kg / kg of dry solids Moisture after 0.5 hr in wet saw dust = =80gm X = final moisture cotent = 80/50=0.533 Time of drying = t= W /A ((X-X)/R) Where, R= rate of drying in kg /m. Hr Rearranging above equation R= W /A t (X-X) = 0.5/0.0*0.5 ( ) =3.99 kg/m. Hr Wt of wet sample after 0.75 hr = 5gm Wt of moisture in wet solids after 0.75 hr =5-50=65gm X3= moisture content after 75 hrs = 65/50=0.33 Rate of drying = W (X-X3)/ A* t =0.5 ( )/0.0*0.75 =.66 kg/m. Hr.. 5.a n-heptane=0% n-octane=60% Basis: 00 kmol of feed. Given :60% of feed distilled. D=60 kmol F=D+W W=0 kmol. Xf= 0.( mole fraction of heptanes in feed.) Xw= mole fraction of heptanes in residue. 6 8 MTO(97) Page of

15 Yd,avg= mole fraction of heptanes in distillate. According to Rayleigh equation Ln(f/w)=int dx/y-x within limit xw to xf () LHS ln(f/w) =ln(00/0)= () Generation of x-y data with the help of α=.6 Y=σx/+(σ-)x As xf=0., generate the data upto x=0.5 X Y /y-x From eqn () & () 0.96= Area x (0. unit/ cm )X( units/ cm ) Area =9.6 cm So measure the area under the curve from X F =0.0 till we get the area=9.6 cm This will give the value of Xw=0. From material balance 5.b FX F =DX D +WX W And find out X D = 0.5 Equation of feed line or q-line y=(-q/-q)x+(x f /-q) where q is the measure of various thermal condition of feed. 8 MTO(97) Page 5 of

16 6 5.c Mier s supersaturation theory: 8 According to Mier s theory there is a definite relationship between the conc and temp at which crystals will spontaneously formed in a pure solution. This relationship is represented by the super solubility curve which is approximately parallel tp the solubility curve. The curve AB is the solubility curve and curve PQ is the super solubility curve. The curve AB represents maximium conc of solution which can be achieved by bringing solid-solute into eqm with liquid solvent. If a solution having the composition and temp indicated by point C is cooled in the direction shown by the arrow it first crosses the solubility curve AB and it is expected to start of crystallization. Actually if the process started with initially unseeded solution crystal formation will not begin until the solution is super cooled considerably passed the curve AB. According to Mier s theory, crystallization will start in the neighbourhood of the point D and the concentration of the solution then follows roughly along the curve DE.For an initially unseeded solution, the curve PQ represents the limit at which spontaneous nuclei formation begin and consequently, crystallization can start. Curve MTO(97) Page 6 of

17 Two methods of supersaturation: a) By cooling a concentrated,hot solution through indirect heat exchange. b) By evaporating a part of the solvent. 6.a a) i) Daltons law : it states that the total pressure exerted by gas/vapour mixture is equal to the sum of the partial pressures of components present in it, thus it expresses the additive nature of the partial pressure. Mathematically, for binary system P = p A + p B ii) Gibbs phase rule: according to gibbs phase rule F= C P + where, F= Degree of freedom C= No of component P= no of phases 6.b a) Oslo evaporative crystalliser MTO(97) Page 7 of

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19 6.c c) Higbie s penetration theory: ) As the time of exposure of fluid for mass transfer generally being short, development of the concentration gradient of film theory is not possible. ) The transfer is largely because of fresh material brought to the interface by the eddies. 3) A process of unsteady state transfer occurs for a fixed period at the freshly exposed surface. Each fluid element (eddy) resides for the same length of time period at the surface. According to this theory, the mass transfer coefficient is proportional to the square MTO(97) Page 9 of

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