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natali 33 [55]
3 years ago
6

Se ha añadido un evaporador para una alimentación de 11500 kg/dia de zumo de pomelo de forma que evapore 3000 kg/dia de agua por

un lado y se obtenga una disolución concentrado de 50% por el otro ¿ con qué concentración porcentual inicial se deberá alimentar el zumo? (m1 = m2 + m3) (%m1 = %m2 + %m3)
Chemistry
1 answer:
mamaluj [8]3 years ago
6 0

Answer:

37 %.

Explanation:

¡Hola!

En este caso, para el problema descrito, conocemos la corriente de entrada y la de salida del agua, por lo que podemos obtener el flujo de la corriente que contiene el zumo a la salida una vez el agua fue evaporada:

F_{sol}=11500kg/dia-3000kg/dia=8500 kg/dia

Luego, por medio de un balance de zumo de limón en el evaporador en el cual la cantidad que entra es igual a la que sale con sus respectivas concentraciones:

x_z^{entra}*11500kg/dia=x_z^{sale}*8500kg/dia

Como la concentración del zumo a la salida es del 50 % (0.50), la de entrada es:

x_z^{entra}=\frac{x_z^{sale}*8500kg/dia}{11500kg/dia} =\frac{0.50*8500kg/dia}{11500 kg/dia}\\ \\x_z^{entra}=0.37

Que es igual al 37%.

¡Saludos!

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