Question

Engineers Sophia and Giles are modifying a process to make concentrated apple cider starting from a slurry - a slurry is a mixture of solids in a liquid that can normally be pumped. A slurry made with apples having a mass fraction of 0.422 and the balance water is fed to a juicer at a rate of 27.3 kg/min. Seventy percent of the apples entering the juicer leave as part of the solid waste stream (containing apples and water). The primary product from the juicer is cider (32.9 wt% apples and 67.1 wt% water). The cider is sent to an evaporator where concentrate - apple concentration increases to 49.1 wt% - is produced along with a stream of pure water that is recycled. The pure water stream leaving the evaporator is mixed with fresh water entering the process. The combined water stream mixes with pristine apples entering the process to create the slurry. Determine the following three unknowns:

1. Mass fraction of apple in the solid waste
2. Total flow rate of the concentrate (kg/min)
3. Flow rate of pure water entering the overall system (kg/min)

Answer 1

1) Mass fraction of apple in the solid waste = 0.644

2) Total flow rate of the concentrate = 12.89 kg/min

3) Flow rate of pure water entering the overall system = 9.434 kg/min

Step-by-step explanation:

Mass fraction of the slurry = 0.422

Rate at which the balance water is fed to a juicer, m = 27.3 kg/min

Stream fed to juice contains (0.422 * 27.3) = 11.5206 kg/min apple

Stream fed to juice contains (27.3 - 11.5206) = 15.7794 kg/min water

70% of the apples entering the juicer leave as part of the solid waste stream

Amount of Apples leaving the stream as part of the solid waste stream,

W = 0.7 * 11.5206

W = 8.06442 kg/min

Overall balance on juicer: m = p + W

p = m - W

p = 27.3 - 8.06442

p = 19.236 kg/min

p stream contains 32.9 wt% apples and 67.1 wt% water

1) Mass fraction of apple in the solid waste:

Let x = Mass fraction of apple in the solid waste

Apple in m stream = Apple in p stream + Apple in W stream

11.5206 = (0.329 * 19.236) + (x * 8.06442)

11.5206 = 6.33 + 8.06442x

8.06442x = 11.5206 - 6.33

x = 0.644

2) Total flow rate of the concentrate (kg/min)

Cider sent to evaporator where concentrate increases from 32.9 wt% to 49.1 wt%

Solid balance; 0.329 p = 0.491 c

0.329 * 19.236 = 0.491 c

c = 12.89 kg/min

Total flow rate of the concentrate = 12.89 kg/min

3 Flow rate of pure water entering the overall system (kg/min)

Overall balance for evaporator: p = v + c

Where P is the pure water from evaporator

v = p - c

v = 19.236 - 12.89

v = 6.346 kg/min

Rate of Water entering into the juicer = (1 - 0.422) *27.3 = 15.78 kg/min

Pure water from evaporator, v = 6.346 kg/min

Flow rate of pure water entering the overall system = 15.78 - 6.346

Flow rate of pure water entering the overall system = 9.434 kg/min