King Saud University Department of Chemical Engineering May 31, 2012 Final Exam (Part 1) Closed Book ChE 316 Time: 30min

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1 King Saud University Department of Chemical Engineering May 31, 2012 Final Exam (Part 1) Closed Book ChE 316 Time: 30min Name: No: Total Pages: 2 Question 1: In the adjoining figure, a feed containing two components A and B enters a membrane separator. The semipermeable membrane provides easier passage to solute A (P MA > P MB ). Here, X F is feed concentrations. The permeate stream is P, and the retentate stream is R, then (1) (a) X FA < X RA (b) X FA = X RA (c) X FA > X RA (d) none of these (2) (a) X FA < Y PA (b) X FA = Y PA (c) X FA > Y PA (d) none of these where, X FA is the mole fraction of component A in the feed, X RA is the mole fraction of A in the retentate, and Y PA is the mole fraction of A in the permeate. Given X FA = 0.72, the following values are acceptable (3) α A,B = 0.99 (a) True (b) False (4) Y PB = 0.08 (a) True (b) False (5) X RB = 0.08 (a) True (b) False Semi-permeable membrane (easier passage to A) (6) A high value of separation factor means (a) good separation (b) bad separation (7) In order to increase the flux of the permeate through the semi-permeable membrane, one should use the following membrane thickness (a) l M = 100 µm (b) l M = 10 µm (c) l M = 1.0 µm (d) l M = 0.10 µm (8) For the separation of hydrogen and ethane in a gas mixture using a micro-porous membrane in Knudsen flow regime, the permeate will have a higher concentration of (a) hydrogen (b) ethane (c) almost equal concentrations of both (d) exactly equal concentrations of both (9) Consider following two membranes with permeability values given in barrer, which one will you chose for the separation? (a) P MA = 10X10-13 ; P MB = 2X10-13 (b) P MA = 20X10-13 ; P MB = 4X10-13 (10) Which of the following membrane module, you will recommend for the case of the dialysis (a) Plate and frame (b) Hollow fiber (11) The purpose of cascading the membrane modules in the membrane separation is (a) to increase the degree of separation (b) to recycle the permeate to feed (c) to recycle the retentate to feed (d) recycle both permeate and retentate (12) To obtain water for drinking from the seawater, the most common membrane separation process is (a) gas permeation (b) reverse osmosis (c) osmosis (d) dialysis

2 (13) For the separation of two components of a gas mixture using a micro-porous membrane, their molecular weight should be. (a) equal (b) closely similar (c) greatly different (d) can not say (14) To remove urea and other smaller solutes from the blood, the following membrane-based separation process is used (a) dialysis (b) osmosis (c) reverse osmosis (d) gas permeation (15) For the separation based on the reverse osmosis, the driving force is (a) difference in salt concentration on the feed (b) difference in solvent/water concentration and the permeate side on feed and permeate side (c) difference in the pressure on the feed and (d) none of these permeate side (16) The humidity of an air-water vapor mixture depends upon (a) enthalpy of the mixture (b) vapor pressure of water vapors (c) partial pressure of water vapors (d) none of these (17) The saturation humidity of an air-water vapor mixture depends upon (a) enthalpy of the mixture (b) vapor pressure of water vapors (c) partial pressure of water vapors (d) none of these (18) For which of the following system, the wet bulb temperature and the adiabatic saturation temperature are almost equal. (a) acetone-air system (b) benzene-nitrogen system (c) toluene-nitrogen system (d) water vapor- air system (19) For a unsaturated air-water vapor mixture (a) Tw > Td (b) Tw = Td (c) Tw < Td (d) none of these where Td is dry bulb and Tw is wet bulb temperature. (20) The cooling tower is used to (a) heat water (c) decrease air humidity (b) cool water (d) increase air humidity (21) The air used in cooling tower must be saturated (a) true (b) false (22) To decrease the height of the cooling tower, one should flow rate. (a) decrease the gas flow (b) increase the gas flow (c) increase water flow (d) none of these (23) For drying, one should have (a) dry air (c) high air velocity (b) hot air (d) all of these Q Answer (24) The mode of heat transfer in the tray dryer is (a) Convection (b) Conduction (c) Radiation (d) none of these (25) The presence of fan or blower in the dryer enhances the heat transfer by (a) Convection (b) Conduction (c) Radiation (d) none of these

3 King Saud University Department of Chemical Engineering May 31, 2012 Final Exam (Part 2) Closed Book ChE 316 Time: 150 min Name: No: Total Pages: 3 Question 2 Part A An air-water mixture at 1 atm pressure has a humidity of kg water vapor per kg dry air. Its dry bulb temperature is 60 C. Determine wet bulb, dew point, and the adiabatic saturation temperature using the chart. Saturation humidity Enthalpy using reference temperature 0 degree C. Part B An air-water mixture at 202 kpa pressure has a dry bulb temperature is 60 C. Its wet bulb temperature is 40 C. Determine its percentage humidity. Data: Specific heats in (kj/kg.k) of dry air = 1.005; of water vapor =1.88, latent heat of vaporization of water at 0 C = 2501 kj/kg Question 3 (PartA) The bone dry sample weight (dry solid only) of sample of wet solid is 5 kg. Its equilibrium moisture content (X*) is 0.02 kg water per kg dry solid. Its total moisture content (Xt) is 0.50 kg water per kg dry solid (Use this for next three questions). free moisture content of the solid sample total weight of the wet solid total weight of the solid sample after long time drying when equilibrium is reached Question 3 (Part B) A rotary dryer is fed with wet sand. The feed flow on the dry basis is 1 kg dry solid /s. The feed contains X 1 = 0.50 kg water per kg dry solid and the sand is discharged with X 2 = 0.03 kg water per kg dry solid moisture. The entering air is at 380 K and an absolute humidity of kg/kg. The feed enters at 294 K and leaves at 309 K. The air leaves at 310 K. The gas flow is G = 10 kg dry air per second. Calculate humidity of the air leaving dryer Q in kw Total wet sand feed rate in kg per second moisture content of feed on the wet basis (Data: Cp of sand = 0.88 kj/kg K, Cp of dry air = 1.005, Cp of water vapor = 1.88, Cp of water = 4.19, Moisture content on the dry basis, X) Air, T 1 =310K H 1 =? Ls = 1 kg dry solid/s X1=0.5 Dryer Air, T 2 =380K H 2 = X2=0.03

4 Question 4 (Part A) A membrane separation unit is being used to separate component A and B from a feed gas. The semipermeable membrane is more permeable to component A. As a result, permeate is richer in A, and the retentate is richer in component B. It is desired to obtain a product with a high purity of the component B using a two stage cascade system. Sketch the proposed cascade system. Question 4 (Part B) From a single stage membrane separation, compositions (mole%) of streams is as follows: Feed: CO 2 (A) =10, CH 4 (B) = 90 Permeate: CO 2 = 30, CH 4 = 70 Retentate: CO 2 = 4, CH 4 = 96 I. Determine the separation factor (α CO2, CH4 ). II. If Feed= 100 mol/h, Permeate= 10 mole/h and the Retentate= 90 mol/h. Determine %recovery of CH 4 in the retentate. Question 4 (Part C) A reverse osmosis membrane module is being used with the bulk condition on the feed side being 3.5 wt% NaCl, 25 C, and 1200 psia and the permeate side being 0.10 wt% NaCl, 25 C, and 50 psia. The permeance values are 15X10-6 g/cm 2 -s-atm for water and 20X10-6 cm/s for the salt. Ignore the mass transfer resistances on both sides of the membrane. Calculate Osmotic pressure in atm Flux of water in kg/m 2 -hr Flux of salt through the membrane in kg/m 2 -hr. (Data: MW of NaCl = 58.5, 1atm = 14.7 psi,) Formula 1:water flux = permeance X ( p π) Formula 2:π = 1.12 ΤΣmi, where π is in psi, temp. T in K and Σmi = 2XC NaCl in mol/l for the salt ( )( ) H = + H T T + λ H H G is the air enthalpy, h L a is the liquid heat transfer coefficient and K G a is the gas G 0 0 phase mass transfer coefficient in kg mol/s.m 3. Pa, M B (=29) is the air molecular weight, C L (=4.19 kj/kg.k) is the water specific heat. Also, latent heat of water vaporization at 0 C is about 2501 kj/kg. The humidity, saturation humidity, percent humidity, percent relative humidity, wet bulb temperature eqn and enthalpy are given as 0 p M p M 0 H-HW C = S ; H = G ( C PA + H C PB )( T T 0 ) + λ 0 H H = ; ; 0 ( s ) ( ) A A H = A A H = H H H = p P p A pa A MB P pa MB T-Tw λw T K ρ = +Η ν s p 100; R H Humid volume, m 3 /kg dry air = ( + Η ) ( ) Density, ( ) The heat transfer coefficient in W/m 2 K is given as h = G 0.8 for parallel flow and h = 1.17G 0.37 for perpendicular flow of air. Here, G = ρv is air flow-rate in (kg/m 2 h). ht ( Tw ) LS dx The rate of drying and the time of drying can evaluated using Rc = 3600 ; t = λw A R Specific heats in (kj/kg.k) of dry air = 1.005; of water vapor =1.88; Latent heat of vaporization of water at 0 C = 2501 kj/kg MW of NaCl = 58.5; 1atm = 14.7 psi; water flux = permeance X ( p π) ; Osmotic pr. π = 1.12 ΤΣmi, where π is in psia and m is the salt concentration in mol/litre. Assume any missing data giving proper justification

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