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• Balances de Materia en el Proceso de Remoción de Grasas de la Leche. Considere 50,000...

• Balances de Materia en el Proceso de Remoción de Grasas de la Leche. Considere 50,000 kg de leche entera que contenga 12% de grasa, mientras el resto es agua y materia no grasa. Se desea separarla en un periodo de 4 horas para obtener leche descremada con 0.50% grasa y en crema con 50% grasa. SUGERENCIA: Tome como base de cálculo 1 hora de flujo de leche entera y dibuje un diagrama de flujo de un proceso de separación indicando un proceso de separación.

(a) Calculate the total input mass (mTOT) per hour of process.

(b) Calculate the flows of fat and non-fatty components (including water) that take place per hour of process.

(c) Calculate the total cream flow at the exit of the process.

(d) Calculate the flow of skim milk at the exit of the process.

(e) Prepare a summary table with the mass flow of each stream and the amount of fat and non-fat matter that contain.

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Answer #1

Mass of whole milk = 50000 kg

Process time = 4 hours

Part a

Total input mass per hour = mass of whole milk / process time

= (50000 kg) / (4hours)

= 12500 kg/h

Part b

Mass of fat in = 12% of Total input mass per hour

= 0.12 x 12500

= 1500 kg/h

Mass of non fatty (including water) in =(100-12)% of Total input mass per hour

= 0.88 x 12500

= 11000 kg/h

Part c and part D

Let the mass of cream out = X kg/h

Total fat out = 0.50X

Mass of skim milk out = 12500 - X

Total fat in skim milk out = 0.50% of (12500 - X)

= 0.0050*(12500-X)

Fat Mass balance

12500*0.12 = 0.0050*(12500-X) + 0.50X

1500 = 62.5 - 0.0050X + 0.50X

X = 2904.04 kg/h

Fat out = 2904.04 kg/h

Skim milk out = 12500 - 2904.04

= 9595.96 kg/h

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