Choppi

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh 1. A mixture of helium and nitrogen gas is contained in a pipe at 298K at 1 atm total pressure which is constant throughout. At one end of the pipe at point 1 and the partial pressure Pa, 1 of He is 0.60 atm and at the other end 0.2m, Pa,2=0.20 atm. Calculate the flux of the He. At steady state, Dab of the He-N2 mixture is 0.687x10 -4 m2/s. a. 7.7892 x 10-5 b. 6.1122 x 10-5

c. 5.6194 x 10-6

d. 4.8910 x 10-6

Solution: 𝐷𝑣 (𝑃 − 𝑃𝐴1 ) ∆𝑧𝑅𝑇 𝐴2 0.687𝑥10−4 𝑚 2 (0.20 − 0.60)𝑎𝑡𝑚 𝑠 𝐽𝐴 = − 𝐿 − 𝑎𝑡𝑚 1𝑚 3 0.2𝑚(0.08205 𝑚𝑜𝑙 − 𝐾 (298𝐾)(1000𝐿 ) 5.6194𝑥10 −9 𝑚𝑜𝑙 1𝑘𝑚𝑜𝑙 𝐽𝐴 = ( ) 𝑚2 − 𝑠 1000𝑚𝑜𝑙 𝟓. 𝟔𝟏𝟗𝟒𝒙𝟏𝟎−𝟔 𝒌𝒎𝒐𝒍 𝑱𝑨 = 𝒎𝟐 ∙ 𝒔 𝐽𝐴 = −

2. An Arnold cell is used to measure the diffusivity of acetone in air at 20°C and 100 kPa pressure. At time= 0, the liquid acetone surface is 1.10 cm from the top of the tube and after 8 hours of operation, the liquid surface drops to 2.05 cm. if the concentration of acetone in air that flows over the top of the tube is zero, what is the diffusivity of acetone in air? At 20°C, the vapor pressure of acetone is 24 kPa and density is 790 kg/m 3. a. 6.2737 x 10-6 b. 5.4518 x 10-6

c. 6.2737 x 10-5 d. 5.4518 x 10-5

Solution: Assume PA2=0 58.08𝑘𝑔 𝑘𝑚𝑜𝑙 𝑍𝐹 − 𝑍𝑜 (𝜌𝐴 𝑅𝑇(𝑃𝑇 − 𝑃𝐴 )𝐿 𝑡𝐹 = ( ) 2 𝑀𝐴 𝐷𝑉 𝑃𝑇 (𝑃𝐴1 − 𝑃𝐴2 ) (100 − 0) − (100 − 24) (𝑃𝑇 − 𝑃𝐴 )𝐿 = = 87.4518 100 − 0 ) ln ( 100 − 24 2 2.05 1.10 2 ( ) − ( ) 100 (790(8.314)(293.15)(87.4518)) 8(3600𝑠 ) = 100 2 58.08(𝐷𝑉 )(100)(24) −𝟔 𝟐 𝟔. 𝟐𝟕𝟓𝟖𝒙𝟏𝟎 𝒎 𝑫𝑽 = 𝒔 𝑀𝐴 =

3. An ethanol-water solution in the form of a stagnant film 2.0 mm thick at 293K is in contact at one surface with an organic solvent in which ethanol is soluble and water is insoluble. In point 1, the concentration of ethanol is 16.8 wt% and the solution density is 972.8 kg/m3. At point 2, the concentration of ethanol is 6.8 wt% and the solution density is 988.1 kg/m3. The diffusivity of ethanol is 0.740x10-9 m2/s. Calculate the steady-state flux of ethanol. a. 8.9553 x 10-7 b. 8.5993 x 10-7

c. 8.3959 x 10-7 d. 8.9355 x 10-7

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh Solution: 𝐷𝐴𝐵 𝐶𝑎𝑣𝑒 1 − 𝑋𝐴2 ln ( ) ∆𝑧 1 − 𝑋𝐴1 𝜌 𝜌 𝐶𝑎𝑣𝑒 = ( ) + ( ) 𝑀 1 𝑀 2 100 𝑘𝑔 𝑀1 = = 20.0526 16.8 83.2 𝑘𝑚𝑜𝑙 46.07 + 18 100 𝑘𝑔 𝑀2 = = 10.7780 6.8 93.2 𝑘𝑚𝑜𝑙 + 46.07 18 972.8 988.1 + 18.7880 𝑘𝑚𝑜𝑙 20.0526 𝐶𝑎𝑣𝑒 = = 50.5905 3 2 𝑚 6.8 46.07 𝑋𝐴2 = = 0.0277 6.8 93.2 + 46.07 18 16.8 46.07 𝑋𝐴1 = = 0.0731 16.8 83.2 + 46.07 18 0.740𝑥10−9 (50.5905) 1 − 0.0277 𝑁𝐴 = ln ( ) 0.002 1 − 0.0731 𝟖. 𝟗𝟓𝟎𝟗𝒙𝟏𝟎−𝟕 𝒌𝒎𝒐𝒍 𝑵𝑨 = 𝒎𝟐 ∙ 𝒔 𝑁𝐴 =

4. An adiabatic saturator is going to treat air having a temperature of 40°C and 20% relative humidity and after which, the treated air will be treated to 70°C having a 60% relative humidity. Determine the kilograms of water absorbed in the saturation if given 100 kg/hr of entering dry air. a. 0.1491 b. 12.9011 c. 1.8910 d. 13.0900 Solution: From Psychrometric Chart 40℃ & 20%𝑅𝐻 ; 𝐻 = 0.0092 (8.07131−

1730.63

𝑘𝑔 𝐻2 𝑂 𝑘𝑔 𝑑𝑎 )

233.426+70 = 233.1733 𝑚𝑚𝐻𝑔 𝑃𝐴 °𝐻20 = 10 𝑃𝐴 0.60 = ; 𝑃 = 139.9040 𝑚𝑚𝐻𝑔 233.17333 𝐴 139.9040 18 𝑘𝑔 𝐻2 𝑂 𝐻= ( ) = 0.1400 760 − 139.9040 29 𝑘𝑔 𝑑𝑎 𝐻2 𝑂 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 = (0.1400 − 0.0092)(100) 𝒌𝒈 = 𝟏𝟑. 𝟎𝟖𝟑𝟖 𝒉𝒓

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh 5. 2000 ft3 per hour of air at 110 F, saturated with vapor, is to be dehumidified. Part of the air is sent through a unit where it is cooled and some water condensed. The air leaves the unit saturated at 60 F. It is then remixed with air which by-passed the unit. The final air contains 0.02 lb H2O/ lb da. Determine the lbs dry air bypassed per lb of dry air sent to the dehumidifier. a. 0.1189 b. 0.2308 c. 0.1109 d. 0.2981 Solution:

0.0588𝑀𝑑𝑎 3 𝑀𝑑𝑎 𝑓𝑡 + 𝑓𝑡 3 62.3 0.0704 𝑀𝑑𝑎 = 140.80 𝑙𝑏; 𝑀𝐻20 = 8.2790 𝑙𝑏 𝐷2 𝑏𝑎𝑙: 140.80 = 𝐷2𝐸 𝐻2 𝑂 𝑏𝑎𝑙: 8.2790 = 𝑀𝑐 + 0.02𝐷2𝐸 𝑀𝑐 = 5.463 𝑙𝑏 𝑎𝑟𝑜𝑢𝑛𝑑 𝑑𝑒ℎ𝑢𝑚𝑖𝑑𝑖𝑓𝑖𝑒𝑟: 𝐻2 𝑂 𝑏𝑎𝑙: 𝑀𝐵 = 5.463 + 𝑀𝐷 𝑀𝐵 𝑀𝐷 = 0.0580 0.0110 𝑀𝐵 = 6.7202 𝑙𝑏; 𝑀𝐷 = 1.2572 𝑙𝑏 𝐴𝐵 = 114.2891 𝑙𝑏; 𝐴𝐷 = 114.2891 𝑙𝑏 𝐴𝐴 = 140.80 − 114.2891 = 26.5109 𝑙𝑏𝑠 𝑨𝑨 = 𝟎. 𝟐𝟑𝟐𝟎 𝑨𝑩 2000𝑓𝑡 2 =

6. Water is to be cooled in a packed tower from 330 to 295 K by means of air flowing counter currently. The liquid flows at the rate of 275 cm3/m2 s and the air at 0.7 m3/m2 s. The entering air has a temperature of 295 K and a humidity of 20%. Calculate the required height of tower and the condition of the air leaving at the top. The whole of the resistance to heat and mass transfer can be considered as being within the gas phase and the product of the mass transfer coefficient and the transfer surface per unit volume of column had may be taken as 0.2 s -1. a. 1m b. 1.5m c. 2m d. 2.75m Solution:

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh 𝐴𝑠𝑠𝑢𝑚𝑒: 𝑘𝐽 1.003𝑘𝐽 2.006𝑘𝐽 ; 𝐶𝑝𝑑𝑎 = ; 𝐶𝑝𝐻2𝑂 = 𝑘𝑔 𝑘𝑔 ∙ 𝐾 𝑘𝑔 ∙ 𝐾 𝑘𝑔 𝐻2 𝑂 𝑘𝑚𝑜𝑙 𝐻2 𝑂 𝐻@295𝐾 & 20%𝑅𝐻 = 0.003 = 0.005 𝑘𝑔 𝑑𝑎 𝑘𝑚𝑜𝑙 𝑑𝑎 𝑘𝑔 𝜌𝑎𝑖𝑟 @295𝐾 = 1.198 3 𝑚 𝐿𝐻 𝑜𝑓 𝐻2 𝑂 = 24.95

𝐻𝑖𝑛𝑙𝑒𝑡 = 1.003(295 − 273) + .003(2495 + 2.006(295 − 273)) = 29.68

𝑘𝐽 𝑘𝑔

0.697𝑚 3 𝑚2 ∙ 𝑠 0.835𝑘𝑔 𝐹𝑙𝑜𝑤 𝑜𝑓 𝑑𝑎 𝑖𝑛 𝑚𝑎𝑠𝑠 = 0.697(1.198) = 𝑚2 ∙ 𝑠 0.275𝑘𝑔 𝐹𝑙𝑜𝑤 𝑜𝑓 𝑑𝑎 𝑖𝑛 𝑤𝑎𝑡𝑒𝑟 = 275𝑥10−6 (1600) = 𝑚2 ∙ 𝑠 4.18 𝑆𝑙𝑜𝑝𝑒 = 0.275 ( ) = 1.38 0.835 𝑘𝐽 𝐶𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠: 𝑇1 = 295𝐾; 𝐻𝐺1 = 29.7 𝑘𝑔 𝑒𝑛𝑡ℎ𝑎𝑙𝑝ℎ𝑦 − 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑆𝑙𝑜𝑝𝑒 = 1.38 2𝑝𝑡(29.7,295) 𝑡𝑜𝑝 𝑝𝑜𝑖𝑛𝑡 𝑘𝐽 𝑇2 = 330𝐾; 𝐻𝐺2 = 78.5 𝑘𝑔 𝐻𝐺2 𝑑𝐻𝐺 𝐺 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑝𝑎𝑐𝑘𝑖𝑛𝑔 = 𝑍 = ∫ ( ) 𝐻𝐺1 𝐶𝐻𝐹 − 𝐻𝐺 ℎ𝐷𝑎𝜌 0.573 (0.835) = = 1.997𝑚 = 𝟐. 𝟎𝒎 0.2(1.198) 𝐹𝑙𝑜𝑤 𝑜𝑓 𝑑𝑎 = (1 − .005)(0.7) =

7. During an experiment conducted on the drying of copra, it was found out that copra dries at a rate proportional to its free moisture content and losses 60% of its free moisture in 2 hours. How many hours will it take to lose 90% under the same drying condition? a. 1.7612 hrs b. 12.0000 hrs c. 2.6712 hrs d. 5.0259 hrs Solutinon: 𝑙𝑒𝑡 𝑀 = 𝐻2 𝑂 @ 𝑎𝑛𝑦 𝑡𝑖𝑚𝑒 𝑡 𝑀𝑜 = 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑑𝑀 𝛼𝑀 𝑑𝑡 𝑑𝑀 − = 𝑘𝑀 𝑑𝑡 log(𝑀) = −𝑘𝑡 + 𝑐 𝑀 = 𝑀𝑜 @ 𝑡 = 𝑡𝑜 log 𝑀 = −𝑘𝑡 + 𝑙𝑜𝑔𝑀𝑜 𝑀 log = −𝑘𝑡 𝑏𝑢𝑡 𝑀 = 0.40𝑀𝑜 @𝑡 = 2ℎ𝑟𝑠 𝑀𝑜

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh 0.40𝑀𝑜 0.199 = −𝑘(2ℎ𝑟𝑠); 𝑘 = 𝑀𝑜 ℎ𝑟𝑠 𝑀 0.199 log = 𝑀𝑜 ℎ𝑟𝑠(−𝑡) 𝑀 = 0.10𝑀𝑜 𝑡𝑙𝑜𝑔(0.90) log(0.10) = − 2 𝒕 = 𝟓. 𝟎𝟐𝟓𝟗 𝒉𝒓𝒔 log

8. A piece of canvass dries in the open air at a rate approximately proportional to its moisture content, if a sheet hung to dry, the wind losses half of its free moisture content in an hour, when will it have loss 99% of its moisture assuming that conditions remain constant. a. 1.4427 hrs b. 4.4455 hrs c. 6.6439 hrs d. 10.9012 hrs Solution: 𝑙𝑒𝑡 𝑀 = 𝐻2 𝑂 @ 𝑎𝑛𝑦 𝑡𝑖𝑚𝑒 𝑡 𝑀𝑜 = 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑑𝑀 𝛼𝑀 𝑑𝑡 𝑑𝑀 − = 𝑘𝑀 𝑑𝑡 log(𝑀) = −𝑘𝑡 + 𝑐 𝑀 = 𝑀𝑜 @ 𝑡 = 𝑡𝑜 log 𝑀 = −𝑘𝑡 + 𝑙𝑜𝑔𝑀𝑜 𝑀 log = −𝑘𝑡 𝑏𝑢𝑡 𝑀 = 0.5𝑀𝑜 @𝑡 = 1ℎ𝑟 𝑀𝑜 𝑀 log ( )=𝑘 2𝑀𝑜 𝑀 log ( ) = log(2) 𝑥 (−𝑡) 𝑀𝑜 1 𝑀 = (1 − 99%)𝑀𝑜 = 𝑀 100 𝑜 1 log ( ) = −𝑡𝑙𝑜𝑔(2) 100 𝒕 = 𝟔. 𝟔𝟒𝟑𝟗𝒉𝒓𝒔 9. It is desired to recover 95% of the SO2 contained in waste flue gas with 5% SO2 and 95% air in a packed tower having a cross-sectional of 0.093m2. The absorbing liquid is water and is allowed to flow counter with the gas flow. How much minimum water rate(kg/hr) is needed? Assume 20°C temperature and entering gas rate is 1.36kg/min. a. 1173.9006 b. 1093.3075 c. 1035.7812 d. 1170.3260

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh Solution:

𝑌𝑏 = 0.05; 𝑋𝑎 = 0 0.05(0.05) 𝑌𝑎 = = 0.0026 0.05(0.05) + 0.95 @20℃; 𝑃𝑏 = 38 𝑚𝑚𝐻𝑔 𝑘𝑔 𝑆𝑂2 𝐶 = 0.6846 100 𝑘𝑔 𝐻2 𝑂 0.6846 64 𝑋𝑏∗ = = 0.00192 0.6846 100 + 64 18 1.36(60) 𝑘𝑚𝑜𝑙 𝑉= = 2.6537 0.05(64) + 0.95(29) ℎ𝑟 𝑘𝑚𝑜𝑙 𝑉 ′ = 0.95(2.6537) = 2.521 ℎ𝑟 0.05 0.0026 0.00192 2.521 ( − ) = 𝐿′ min ( ) 0.95 1 − 0.0026 1 − 0.00192 𝑘𝑚𝑜𝑙 𝒌𝒈 𝐿′ 𝑚𝑖𝑛 = 65.5575 = 𝟏𝟏𝟖𝟎. 𝟎𝟑 ℎ𝑟 𝒉𝒓 10. Sulfur dioxide is to be absorbed into water in a plate column. The feed gas (20%mol SO2) is to be scrubbed to 2mole% SO2. Water flow rate is 6,000 kg/hr.m2. The inlet air flow rate is 150 kg air/hr.m2. Tower temperature is 293 K. Find the number of theoretical plates. a. 1 b. 3 c. 2 d. 4

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh Solution:

𝑌𝑎 = 0.02: 𝑌𝑏 = 0.2; 𝑌𝑎∗ = 0 150 𝑘𝑚𝑜𝑙 𝑉= = 6.4655 ( ) 0.8 29 ℎ𝑟−𝑚 2 6000 𝑘𝑚𝑜𝑙 𝐿= = 333.3333 18 ℎ𝑟 − 𝑚 2 0.2 0.02 𝑋𝐷 0.8(6.4655) ( − ) = 333.3333 ( ) 0.8 0.98 1 − 𝑋𝐷 𝑋𝐷 = 0.00285 𝐶 𝑘𝑔𝑆𝑂2 64 0.00285 = ; 𝐶 = 1.0162 𝐶 100 100 𝑘𝑔𝐻2 𝑂 64 + 18 𝑃𝐴 = 60.0692𝑚𝑚𝐻𝑔 𝑃𝐴 𝑋𝑏 60.0692 𝑌𝑏 = = = 0.0790 𝑃𝑇 760 0.02 log (0.2 − 0.079) 𝑁= = 2.1858~𝟑𝒕𝒓𝒂𝒚𝒔 0.079 log (0.2 − 0.02)

11. Given that the Henry's Law constant for carbon dioxide in water at 25°C is 1.6 x 10-5 kPa (mole fraction)-1, calculate the percentage solubility by weight of carbon dioxide in water under these conditions and at a partial pressure of carbon dioxide of 200 kPa above the water. a. 78% b. 69% c. 31% d. 22% Solution: 1.6𝑥10 −5 𝑚𝑜𝑙 𝐾𝐻 = 𝐿 ∙ 𝑘𝑃𝑎 𝑃°𝐶02 = 200𝑘𝑃𝑎

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh (200) (44)(100%) 1.6𝑥10−5 % = 𝟑𝟐% %=

12. Determine the % error if the Antoine equation is used to estimate the normal boiling point of benzene. From literature value, the normal boiling point of benzene is 353.26 K. a. 100% b. 10% c. 1% d. 0.01% Solution: Using Antoine’s eqn: log(760 𝑚𝑚𝐻𝑔) = 15.9008 −

2788.51 −52.36 + 𝑇

𝑇 = 353.3 𝐾 353.26 − 353.3 (100) %𝑒𝑟𝑟𝑜𝑟 = 353.26 𝟎. 𝟎𝟏% 13. A mixture of 40 mol% benzene and 60% toluene are distilled in a column to give a product of 98% benzene and a waste containing 5% benzene. For a relative volatility of 2.4, calculate the minimum number of plates if the mixture is fed at its boiling point. a. 61290 b. 61541 c. 18901 d. 1.5500 Solution: log ( 𝑁𝑚𝑖𝑛 =

0.98(1 − 0.05) ) 0.05(1 − 0.98) log(2.4)

𝑁 = 7.8087 𝑅𝐷𝑚𝑖𝑛 0.98 − 0.4 2.4(0.4) = 𝑤ℎ𝑒𝑟𝑒 𝑦 ′ = = 0.6154 ′ 𝑅𝐷𝑚𝑖𝑛 + 1 0.98 − 𝑦 1 + 1.4(0.4) 𝑅𝐷𝑚𝑖𝑛 0.98 − 0.4 = = 𝟏. 𝟓𝟓𝟎𝟒 𝑅𝐷𝑚𝑖𝑛 + 1 0.98 − 0.6154 14. Tung meal containing 55 weight % oil is to be extracted at a rate of 4000 lb/h using n-hexane containing 5 weight % oil as solvent. A countercurrent multi-stage extraction sytem is to be used. The meal retains 2 lbs of solvent per lb of oil-free meal. The residual charge contains 0.11 lb oil per lb of oil-free meal while the product is composed of 15 weight % oil. Determine the number of ideal stages. a. 2.2417 b. 3.1205 c. 2.0890 d. 3.7811 Solution:

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh

0.15 = 𝑋𝑎 = 0.1765 0.85 0.05 𝑌𝑏 = = 0.0526 0.95 0.11 𝑙𝑏 𝑜𝑖𝑙 𝑙𝑏 𝑜𝑖𝑙 𝑓𝑟𝑒𝑒 𝑋𝑏 = 𝑌𝑏° = = 0.055 2 𝑙𝑏 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑙𝑏 𝑜𝑖𝑙 − 𝑓𝑟𝑒𝑒 𝑎𝑟𝑜𝑢𝑛𝑑 𝑒𝑥𝑡𝑟𝑎𝑐𝑡𝑜𝑟 1: 𝑉𝑏′ = 𝑉𝑎′ = 0.95(𝑉𝑏) = 0.95(22.417) 𝑙𝑏 = 21 296.15 ℎ 𝑌𝑎° =

𝑇𝑀𝐵: 𝐿𝑎 + 𝑉𝑏 = 𝐿𝑏 + 𝑉𝑎 4000 + 𝑉𝑏 = 𝐿𝑏 + 𝑉𝑎 𝑂𝑖𝑙 𝐵𝑎𝑙𝑎𝑛𝑐𝑒: 0.55(4000) + 0.05𝑉𝑏 = 0.11(0.45)(4000) + 0.15𝑉𝑎 𝑆𝑜𝑙𝑣𝑒𝑛𝑡 𝐵𝑎𝑙𝑎𝑛𝑐𝑒: 0 + 0.95𝑉𝑏 = 0.85𝑉𝑎 + 2(0.45)(4000) 𝑙𝑏 𝑙𝑏 𝑙𝑏 𝑉𝑎 = 20 819 ; 𝑉𝑏 = 22 417 ; 𝐿𝑏 = 5 598 ℎ𝑟 ℎ𝑟 ℎ𝑟 15 (0.45)(4000)(2) + 0.15(21296.15) 𝑆𝑜𝑙𝑢𝑡𝑒 𝑏𝑎𝑙: 0.55(4000) + 𝑌𝑎(21296.15) = 85 𝑌𝑎 = 0.0732 0.0526 − 0.055 ln (0.0732 − 0.1765 ) 𝑁−1= ; 𝑵 = 𝟑. 𝟏𝟏𝟗𝟗 0.0526 − 0.0732 ln ( 0.055 − 0.1765 ) 15. Roasted copper containing the copper as CuSO4 is to be extracted in a countercurrent stage extractor. Each hour a charge consisting of 10 tons of inert solids, 1.2 tons of copper sulfate, and 0.5 ton of water is to be treated. The strong solution produced is to consist of 90 percent water and 10% CuSO4 by weight. The recovery of CuSO4 is to be 98% of that in the ore. Pure water is to be used as the fresh solvent. After each 1 ton of inert solids retains 2 tons of water plus the copper sulfate dissolved in that water. Equilibrium is attained in each stage. How many stages are required? a. 10 b. 9 c. 8 d. 7 Solution:

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh

(1.2 + 0.5) + 𝑉𝑏′ =

2 𝑡𝑜𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (10 𝑡𝑜𝑛 𝑠𝑜𝑙𝑖𝑑 ) + 𝑉1 𝑡𝑜𝑛 𝑠𝑜𝑙𝑖𝑑

0.98(1.2) = 11.76 𝑡𝑜𝑛𝑠; 𝑉𝑏 = 30.06 𝑡𝑜𝑛𝑠 = 𝑉𝑎 0.10 𝐶𝑢𝑆𝑂4 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 (𝑎𝑟𝑜𝑢𝑛𝑑 𝑠𝑡𝑎𝑔𝑒 1) 1.2 + 𝑌𝑎𝑉𝑎 = 0.10(11.76) + 0.1(2 ∗ 10) 𝑌𝑎 = 0.0657 0 − 0.0012 ) ln ( 0.0657 − 0.1 + 1 𝑁= 0 − 0.0657 ln (0.0012 − 0.1) 𝑵 = 𝟗. 𝟐𝟏𝟕𝟕 𝒔𝒕𝒂𝒈𝒆𝒔 𝑏𝑢𝑡 𝑉1 =

16. We wish to extract nicotine from water using kerosene. If we have 100 lb of a 2% nicotine solution extracted once with 200 lb of kerosene, what percentage will be extracted? Equilibrium data: Y=0.90X a. 65.8900% b. 64.7500% c. 65.9800% d. 64.5700% Solution:

𝑁𝑖𝑐𝑜𝑡𝑖𝑛𝑒 𝑏𝑎𝑙: 0.02(100) + 0 = 𝑛𝑖𝑐𝑜𝑡𝑖𝑛𝑒 𝑖𝑛 𝑟𝑎𝑓𝑓𝑖𝑛𝑎𝑡𝑒 + 𝑛𝑖𝑐𝑜𝑡𝑖𝑛𝑒 𝑖𝑛 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 2 + 0 = 𝑥 (0.98)(100) + 𝑌(200)

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh 𝑏𝑢𝑡 𝑌 = 0.90𝑥 𝑌 𝑛𝑖𝑐𝑜𝑡𝑖𝑛𝑒 (0.98)(100) + 𝑌(200); 𝑌 = 6.4748𝑥10 −3 𝑘𝑔 2= 0.9 𝑘𝑔 𝑘𝑒𝑟𝑜𝑠𝑒𝑛𝑒 𝑛𝑖𝑐𝑜𝑡𝑖𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡𝑒𝑑 = 0.0062(200) = 1.295𝑘𝑔 1.295𝑘𝑔 (100) = 𝟔𝟒. 𝟕𝟓 % %𝑛𝑖𝑐𝑜𝑡𝑖𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡𝑒𝑑 = 2 17. An aqueous waste stream containing 3.25% by weight phenol is to be extracted with one-third its volume of methylene chloride to produce a raffinate without more than 0.2% phenol. How many stages are required? a. 6 b. 5 c. 4 d. 3 Solution: 3.25(100) 3.36 𝑝ℎ𝑒𝑛𝑜𝑙 = 96.75 100 𝑤𝑎𝑡𝑒𝑟 0.2(100) 0.2 𝑝ℎ𝑒𝑛𝑜𝑙 𝑋𝑅 = = 99.8 100 𝑤𝑎𝑡𝑒𝑟 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑝ℎ𝑒𝑛𝑜𝑙 𝑡𝑜 𝑏𝑒 𝑟𝑒𝑚𝑜𝑣𝑒𝑑: 3.36 − 0.2 = 3.16 𝑔𝑟𝑎𝑚𝑠 𝑤𝑠 1 1.31 = ( ) = 0.451 𝑤𝐹 3 1 − 0.0325 3.16 𝑝ℎ𝑒𝑛𝑜𝑙 𝑦𝐸 = = 7.01𝑔 0.451 100𝑔 𝑚𝑒𝑡ℎ𝑦𝑙 𝑐ℎ𝑙𝑜𝑟𝑖𝑑𝑒 3.36(0.451 − 0.2) log ( ( ) 0.2 0.451 − 3.36) 𝑁= + 1 = 𝟑. 𝟗𝟖𝟗𝟓 𝒔𝒕𝒂𝒈𝒆𝒔 log(0.451) 𝑋𝐹 =

18. A plate and frame filter press is used to filter a compressible sludge (S = 0.45) at 50 psia for 2 hours. Washing is done at 30 psia with wash water equal to 10% of the filtrate volume collected. The washing time is a. 127 min b. 136 min c. 156 min d. 178 min Solution: 𝑉 2 = 𝑘𝑡 = 𝑘 (2) 𝑉 2 = 2𝑘 0.1 1 1 = ( ) 𝑡𝑤 4 4 60𝑚𝑖𝑛 1 − 0.45 𝑡𝑤 = 1.6ℎ𝑟𝑠 ( )( ) ℎ𝑟 0.45 𝒕𝒘 = 𝟏𝟐𝟕 𝒎𝒊𝒏𝒔 19. A dilute slurry contains small solid food particles having a diameter of 5 x 10-2 mm which are to be removed by centrifuging. The particle density is 1050 kg/m3 and the solution density is 1000kg/m3. The viscosity of the liquid is 1.2 cP.A 60mm diameter bowl that is 100 mm deep operated at 50Hz gives a 25mm thick liquid layer. Calculate the expected flowrate in L/s just to remove these particles. a. 0.29 b. 0.11 c. 0.90 d. 0.34 Solution:

Pagsanjan, Sylvester S. Martinez, Marah Faye F. ChE 520L 8:30-11:30 TTh 𝜋𝐷2𝑝 𝑏(𝜌𝑝 − 𝜌)𝑤 2 (𝑟 22 − 𝑟 21 ) 2𝑟 18𝜇 ln (𝑟 −2𝑟 ) 1 2 𝜋(0.00005)2 (0.1)(1050 − 1000)(100𝜋 2 )(0.03)2 𝑞= 18 (0.0012) ln(2) 2.9𝑥10−4 𝑚 3 = 𝟎. 𝟐𝟗𝑳/𝒔 𝑠 𝑞=

20. Find the width of an apron conveyor without skirts whose capacity is 56 tons per hour at a speed of 50 fpm handling solids with average density of 50 pound per cubic foot. a. 36 in b. 42 in c. 48 in d. 54 in Solution: 𝑡𝑜𝑛𝑠 ℎ𝑟 𝑙𝑏 𝜌𝑏𝑢𝑙𝑘 = 50 3 𝑓𝑡 𝑆 = 50 𝑓𝑝𝑚 𝑇 = 56

Belt Width 30 inches 36 inches 42 inches

50 lb/ft3 79 115 165 79 𝑡𝑜𝑛𝑠 ) = 39.5 100 ℎ𝑟 115 𝑡𝑜𝑛𝑠 max 𝑐𝑎𝑝36 = 50 ( ) = 57.5 100 ℎ𝑟 165 𝑡𝑜𝑛𝑠 max 𝑐𝑎𝑝42 = 50 ( ) = 82.5 100 ℎ𝑟 max 𝑐𝑎𝑝30 = 50 (

With 57.5 tons/hr being the maximum capacity closest to the given capacity, the width of the conveyor is 36 inches.