Pset TakeHomeFinalExam Part1_1stsem1617

ChE 415 Heat & Mass Transfer Take Home Exam – Problem Set

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DIRECTIONS: PROVIDE THE SOLUTION TO SELECT THE CORRECT ANSWER FROM THE GIVEN CHOICES. Solve the problems on long bond papers. Submit not later than December 16, 2016, 8am. Late problem sets will not be accepted. 1. Calculate the rate of heat transfer through a bimetallic tube consisting of an outer tube of low- carbon steel and an inner tube (“liner”) of 347 stainless steel. The outer tube has an outside diameter of 31.8 mm (1.25 in.) and an inside diameter of 19.0 mm (0.75 in.). The liner has an outside diameter of 18.9 mm (0.75 in.) and a wall thickness of 20 BWG (= 0.889 mm = 0.035 in.), giving a nominal inside diameter of 17.3 mm (0.680 in.). The inside wall temperature of the liner is 260°C (500°F) and the outside wall temperature of the outer tube is 100°C (212°F). Assume the liner and the outer tube are in perfect thermal contact. 1.1The rate of heat transfer on the inside tube is a. 1.18 x 106 W/m2 b. 2.25 x 106 W/m2 c. 4.13 x 106 W/m2 d. 6.41 x 105 W/m2 1.2 The rate of heat transfer on the outside tube is a. 1.18 x 106 W/m2 b. 2.25 x 106 W/m2

c. 4.13 x 106 W/m2

d. 6.41 x 105 W/m2

2. A rectangular steel cold box with 4.76 mm (3/16 in.) thick walls has interior dimensions of 1.0 m by 1.25 m by 2.0 m (3.28 ft by 4.10 ft by 6.56 ft). It is insulated on all sides by a 10 cm (3.94 in.) thick layer of rock wool with a thermal conductivity of 0.040 W/m ⋅K (0.023 Btu/h⋅ft⋅°F). The interior wall temperature is −20°C (−4°F), and the exterior surface of the insulation is at 40°C (104°F). Calculate the heat-transfer rate from the surrounding atmosphere to the interior of the box. a.125 W b. 298 W c. 425 W d. 618 W 3. Calculate the rate of heat loss by convection from the outer surface of an uninsulated 4-in. Schedule. 40 pipe (Do = 4.500 in. = 0.114 m) exposed to wind blowing at 8.0 m/s (26.3 ft/s) perpendicular to the pipe. The air is at −23°C (−9.4°F), and the pipe su rface is at 27°C (80.6°F). a. 438 W/m b. 544 W/m c. 637 W/m d. 759 W/m 4. A furnace is constructed with 0.20 m of firebrick, 0.10 m of insulating brick, and 0.20 m of building brick. The inside temperature is 1200 K and the outside temperature is 330 K. If the thermal conductivities are as shown in the figure, estimate the heat loss per unit f irebrick and the insulating brick. area and the temperature at the junction of the firebrick

4.1. The heat loss per unit area is a. 961 W/m2 b. 1056 W/m2

c. 1248 W/m2

d. 1567 W/m2

4.2. The temperature at the junction of the firebrick and the insulating brick is a.983 K b. 1063 K c. 1115 K

d. 1472 k

5. A triple effect evaporator is supplied with 10,000 lbs/hr of saturated steam at 240degF to the 1 st effect. The temperature at the last effect is 150oF. Assume U1=150, U2=200 and U3=250 BTU/(hr-ft2-oF) and the boiling point rises are 10, 18 and 27oF respectively. 5.1 Estimate the area needed for the first effect in ft2. a. 3320 b. 3645 c. 3980 d. 4260 5.2 Estimate the boiling point of the solution in effect number two in oF. a. 225 b. 204 c. 185

d.177

6. Glycerine is to be concentrated from 12% to 72% in a single effect evaporator. The inlet steam used is at 25 psig and comes out at 170degF. The vapour space in the evaporator has a pressure of 25 inches Hg vacuum. Ten metric tons of glycerine per hour is fed at 85oF. The specific heat of the feed is 0.75 BTU/lbm-oF. Assume Assume no BPR and U = 600 BTU/hr-ft2-oF. 6.1. Calculate the pounds of steam required per metric ton of the thick liquor. a. 13,587 b. 10,380 c. 9,340 d. 6,120 6.2. Estimate the area of the evaporator. a. 190 ft2 b. 225 ft2

c. 246 ft2

d. 295 ft2

7. A triple effect evaporator is concentrating a liquid that exhibit no BPR. The temperature of the steam to the 1 st effect is 227 degF. The boiling point of the solution in the last effect is 125 degF. If U1=500, U2=400 and U3=200 BTU/hr-ft2-oF. At what temperature will the liquid boil in the 2nd effect? a. 179oF b. 195oF c. 204oF d. 211oF 8. A single effect evaporator is required to concentrate a solution from 10% solids to 30% solids at the rate of 250 kg of feed per hour. If the pressure in the evaporator is 77 kPa absolute, and if steam is available at 200 kPa gauge, if the overall heat transfer coefficient is 1700 J m-2 s-1 °C-1. Assume that the temperature of the feed is 18°C and that the boiling point of the solution under the pressure of 77 kPa absolute is 91°C. Assume, also, that the specific heat of the solution is the same as for water, that is 4.186 x 103 J kg-1°C-1, and the latent heat of vaporization of the solution is the same as that for water under the same conditions. 8.1.Calculate the Heat available per kg of steam

ChE 415 Heat & Mass Transfer Take Home Exam – Problem Set

a. 1.44 x 106 J

b. 2.34 x 106 J

c. 3.34 x 10 6 J

d. 5.56 x 106 J

8.2. Calculate theHeat required by the solution a. 1.68 x 108 J h-1 b. 4.57 x 108 J h-1

c. 6.95 x 108 J h-1

d. 8.25 x 108 J h-1

8.3. Calculate the quantity of steam required per hour a. 105 kg h-1 b. 195 kg h-1

c. 385 kg h-1

d. 555 kg h-1

8.4. Calculate the quantity of steam/kg of water evaporated a. 1.17 b. 1.24

c. 1.56

d. 1.94

8.5. Calculate the area of heat transfer surface a. 1.74 m2 b. 1.86 m2

c. 1.97 m2

d. 2.04 m2

9. Estimate the requirements of steam and heat transfer surface, and the evaporating temperatures in each effect, for a triple effect evaporator evaporating 500 kg h-1 of a 10% solution up to a 30% solution. Steam is available at 200 kPa gauge and the pressure in the evaporation space in the final effect is 60 kPa absolute. Assume that the overall heat transfer coefficients are 2270, 2000 and 1420 J m2 s-1 °C-1 in the first, second and third effects respectively. Neglect sensible heat effects and assume no boiling-point elevation, and assume equal heat transfer in each effect. 9.1. The boiling point in the first effect a. 134˚C b. 130˚C c. 121˚C d. 109˚C 9.2. The boiling point in the second effect a. 121˚C b. 110˚C

c. 106.5˚C

d. 101˚C

9.3. The boiling point in the third effect a. 102˚C b. 98˚C

c. 86˚C

d. 82˚C

9.4.The total water evaporated a.450 kg

c. 333 kg

d. 310 kg

9.5. The steam consumption per kg of water evaporated a.0.15 b. 0.25

c. 0.35

d. 0.45

9.6. The heat transfer area per effect a. 20 m2 b. 24m2

c. 28m2

d. 32 m2

b. 405 kg

10. Tomato juice is to be concentrated from 12% solids to 28% solids in a climbing film evaporator, 3 m high and 4 cm diameter. The maximum allowable temperature for tomato juice is 57°C. The juice is fed to the evaporator at 57°C and at this temperature the latent heat of vaporization is 2366 kJ kg-1. Steam is used in the jacket of the evaporator at a pressure of 170 kPa (abs). If the overall heattransfer coefficient is 6000 J m -2 s-1 °C-1, estimate the quantity of tomato juice feed per hour. Take heating surface as 3 m long x 0.04 m diameter. 10.1. The Area of evaporator tube a. 0.15 m2 b. 0.24 m2 c. 0.342 m2 d. 0.42 m2 10.2 Heat required per kg of feed for evaporation a. 1.32 x 105 J b. 1.34 x 106 J

c. 9.32 x 10 5 J

d. 4.32 x 106 J

10.3. Rate of evaporation a. 250 kg h-1

c. 780 kg h-1

d. 940 kg h-1

b. 360 kg h-1

11. As solution of 28% wt Na2SO4 is to be crystallized to yield Glauber salt in a four section (one section is 10 ft in length) Swenson Walker Crystallizer. It is fed to the unit at 120oF and cooled to 60oF using cooling water which enters at 50oF and leaves at 70oF. The crystallizer has a cooling surface of 3 ft2 per foot of length. The feed rate is 500 lb/hr and the crystallizer is seeded with 50 lb/hr of 65 mesh crystals. The amount of the crystalline crop (C) produced in lb per hour is a. 285.3 b. 249.3 c. 255.65 d. 214.65 12. A saturated solution containing 1500 kg of potassium chloride at 360 K is cooled in an open tank to 290K. If the specific gravity of the solution is 1.2, the solubility of potassium chloride per 100 parts of water is 53.55 at 360K and 34.5 at 290K, calculate the amount of crystals obtained assuming that loss of water by evaporation is negligible. a. 4301 kg b. 1500 kg c.966 kg d. 534 kg 13. What is the yield of sodium acetate crystals ( CH COONa x 3H2 O) obtainable from a vacuum crystallizer operating at 1.33 kN/m² when it is supplied with 0.56 kg/s of a 40% a queous solution of the salt at 353 K? the boiling point elevation of the solution is 11.5 K. Data: Heat of crystallizer, q = 144 KJ/Kg trihydrate

ChE 415 Heat & Mass Transfer Take Home Exam – Problem Set

Heat capacity of the solution, Cp = 3.5 kJ/kg K Latent heat of water at 1.33 kN/m², = 2.46 MJ/kg Boiling point of water at 1.33 kN/m² = 290.7K Solubility of sodium acetate at 290.7K= 0.539 kg/kg water. a. 1.38 kg/s b. 0.183 kg/s

c. 0.318 kg/s

d.0.813 kg/s

14. Ammonium sulphate is to be crystallized from a solution containing 48% (NH4 )2 SO4  by cooling it in a counterflow crystallizer from 85 to 35°C. During cooling the amount of water that evaporates is 5% of the mass of the feed solution. Data: Feed rate = 1000 kg/h Solubility of (NH4 )2 SO4  at 35°C = 75 kg/ 100 kg of water Specific heat of 48% (NH4 )2 SO4  solution = 2.97 kJ/kg K Heat of crystallization of (NH4 )2 SO4 = 75.2 kJ/kg Latent heat of vaporization of water ay 35°C = 2414 kJ/kg 14.1. Calculate the rate of formation of crystals. a. 127.5 kg/h b. 678.3 kg/h

c. 822.5 kg/h

d. 321.7 kg/h

14.2 What is the cooling water rate, if it is heated from 18 to 29°C? a. 764 kg/h b. 505 kg/h

c. 813 kg/h

d. 427 kg/h

14.3 Determine the required cooling surface, if the overall heat transfer is 125 W/m²K. a. 3.13 m² b. 2.54 m² c. 1.95 m²

d. 4.02 m²

15. Carbon disulphide is to be absorbed from a dilute gas mixture of CS 2  – N 2 into a pure nonvolatile oil at atmospheric pressure in a counter current absorber. The mole fraction of CS2 in inlet gas stream is 0.05 and the flow rate of gas stream, G is 1500 k.mol/hr. The equilibrium relation is given by; y = 0.5x, Where x is the mole fraction of CS 2 in liquid stream. It is desired to reduce the mole fraction of CS2 in the gas stream is 0.005. Calculate the minimum value of L/G, where L s the liquid flow rate in kmol/hr. a. 0.40

b. 0.45

c. 0.50

d. 0.55

16. Equilibrium relationship for the system heptane-oil-air is given by Y = 2X (Y and X are kg –heptane/kg-air and kg-heptane/kg-oil respectively). Oil containing 0.005 kg-heptane/kg-oil is being used as solvent for reducing the heptane content of air from 0.10 to 0.20 kg-heptane/kg-air in continuous counter-current pact bed absorber. What column height is required to treat 1400 kg/hr.m 2  of empty tower cross section of pure air containing heptane if the overall mass transfer coefficient is 320 kg/hr. m 3 per unit gradient of Y? The oil rate employed is 5100 kg/hr.m2. Solve analytically. a.5.68 m b. 6.25 m c. 7.62 m d. 8.65 m 17. A mixture 5% butane and 95% air is absorbed in a bubble plate tower containing 8 ideal plates. The absorbing liquid is a heavy nonvolatile having a molecular weight of 250 and a specific gravity of 0.90. The absorption takes place at 1 atm and 60°F. the butane is to be recovered to the extent of 95%. The vapor pressure of butane at 60°F is 28lb force per sq. in and liquid butane has a density of 4.84 lb/gal at 60°F. Calculate the gallons of fresh absorbing oil gallon of butane recovered. 18. A tray is to be used for drying a solid material . Each tray is 3 ft. wide by 4 ft. long and the wet material has depth of 2 in. The density of the wet material is 50 lb/ft3. The wet material contains 3 lb of water per pound of dry solid. How many trays are necessary to obtain 3,000 lb of a product containing 1 lb of water per pound of dry solids? 19. A porous solid is dried in a batch dryer under constant drying conditions. Five hours were required to reduce the moisture content from 30 to 12 lb water/lb dry solid. The critical moisture content is 18 lb water/lb dry solid and the equilibrium moisture is 5 lb water / lb dry solid. If the drying rate during the FRP is a straight line through the origin, calculate the time required to dry the solid from 30 to 8 lb water/lb dry solid. 20. A 100 kg batch of granular solids containing 30 per cent moisture is to be dried in a tray drier to 15.5 per cent of moisture by passing a current of air at 350 K tangentially across its surface at a velocity of 1.8 m/s. If the constant rate of drying under these conditions is 0.0007 kg/s m2 and the critical moisture content is 15 per cent, calculate the approximate drying time. Assume the drying surface to be 0.03 m2/kg dry mass. -end of examPrepared by: Engr. Rejie C. Magnaye ChEFE Dept. Instructor I