Please show every single step thoroughly. And read the instruction to the questi

Question

Please show every single step thoroughly. And read the instruction to the question to answer them according and all parts to it. ***ONLY THE CIRCLED ONES! 14.1 AND 14.6

L 10 800 s (10 800 s)/(3 600 s/h) 3.0 h Problems blem, follow the steps of Section 14.4 -that is, (1) identify the system, (2) deter- mine whether the process is batch or rate, and so on 11.0 percent solids, the resulting in a mixture of 2 150 amounts of the liquids A and B? ids, A and B, are mixed together. A is 5.00 percent solids, the rest water. B is rest water. To the mixture is added 18.5 kg of bone-dry solids kg containing 8.25 percent solids. What were the original 17 Al the Grace Foods facility, a cereal product containing 53 percent water is produced at 14.2 a rate of 750 pounds per hour. A dried product containing 75 percent cereal is desired How much water must be removed in an eight-hour period and how much dried product is produced? 14.3 An order is placed with a plant to deliver 2 500 kg of a 9.5 percent solution of cellulose 144 A slurry (mix of water and insoluble material) of crushed mine coal is fed to a cyclone Pure coal is recovered at the top of the separator. The waste from the bottom of nitrate. The only product on hand is a 5.5 percent solution. How much dry cellulose nitrate must be added to the solution to meet the requirement? separator at a rate of 8.0 th. The mine coal averages 78 percent pure coal and 22 percent pyrite. the separator contains 3.0 percent pure coal and 35 percent pyrite; the rest is water. Ho much pure coal can be obtained in a 12 h time period? Casein, a dairy product used in making cheese, contains 25 percent moisture when A dairy sells this product for $40/100 kg. If requested they will dry the casein to 12 pr cent moisture. The drying costs are S5/100 kg of water removed. What should the dairy sell the dried product for in order realize the same margin of profit? 14.5 14.6 A vat holds 2.25(10) kg of a 22.0 percent solution of ethylene glycol and water. How much must be drained from the tank and replaced with an 85.0 percent ethylene glycol solution to obtain a 33.0 percent concentrate? Assume percentage values are based on mass. Three brine solutions, B1, B2, B3, are mixed. B1 is one-half of the total mixture Brine Bl is 2.5 percent salt, B2 is 4.5 percent salt, and B3 is 5.5 percent salt. To this mixture s added 35 Ibm of dry salt while 280 lbm of water is evaporated leaving 3 200 lbm of 14.7 percent brine. Determine the initial amounts of B1, B2, and B3 aker holds 962 g of a brine solution that is 6.20 percent salt. If 123 g of water are S. percent brine eva solution? solution is 20.0% salt and has 70.0 kg of water evaporated per hour. To produce kg of pure salt (0% moisture) per day, how long should the process operate each day 14.9 A brine and how much brine must be fed to the evaporator per hour? 14.10 An ethano alant distills alcohol from corn. The distiller processes 2.0 vh of feed con- ar the rest is inert material. The bottoms

Explanation / Answer

Let x= mass of liquid A and y= mass of liquid B.

After addition of 18.5kg of bone dry solids, the mass of mixture is 2150 kg

Hence   mass of A+ mass of B+ 18.5= 2150

Mass of A + mass of B= 2150-18.5=2131.5

x+y= 2131.5 (1)

Writing balance of solids

Solids in A + solids in B= Solids in the mixture

x*5/100+y*11/100 = 2150*8.25/100               or

5x+11y= 2150*8.25= 17737.5 or x+11/5y= 17737.5/5 , x+2.2y= 3547.5 (2)

Eq.2-Eq.1 gives 1.2y= 1416, y= 1416/1.2=1180 kg

Hence from Eq.1,y= 2131.5-1180 = 951.5 kg

2.

Let x= mass of liquid A and y= mass of liquid B.

After addition of 18.5kg of bone dry solids, the mass of mixture is 2150 kg

Hence   mass of A+ mass of B+ 18.5= 2150

Mass of A + mass of B= 2150-18.5=2131.5

x+y= 2131.5 (1)

Writing balance of solids

Solids in A + solids in B= Solids in the mixture

x*5/100+y*11/100 = 2150*8.25/100               or

5x+11y= 2150*8.25= 17737.5 or x+11/5y= 17737.5/5 , x+2.2y= 3547.5 (2)

Eq.2-Eq.1 gives 1.2y= 1416, y= 1416/1.2=1180 kg

Hence from Eq.1,y= 2131.5-1180 = 951.5 kg

2. Mass of solution= 2.25*1000 kg

Mass % of ethylene glycol= 22%

Mass of ethylene glycol= 2250*22/100 =495 kg

Mass of ethylene glycol= 85%, let this mass to be added = x             

Mass of ethylene glycol= x*85/100=0.85x

Total mass after addition = 2250+x

Final concentration of solution = 33%, mass of ethylene glycol = (2250+x)*33/100= (2250+x)*0.33

Mass of ethylene glycol before = mass of ethylene glycol after mixing

495+0.85x= (2250+x)*0.33

496+0.85x= 742.5+0.33x

(0.85-0.33)x= 742.5-495

Hence x= 247.5/0.52 =475.9 kg

water in the solution before concentration = 2250*0.78=1755 kg

water added from 85% ethylene glycol= 0.85*475.9 kg=404.5 kg

water in the final solution = (2250+475.9)*0.67= 1826.3 kg

water drained = water in fresh solution + water in 85% ethylene glycol-water In final solution

=1755+404.5-1826.3 =333.2 kg

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