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EJERCICIO, ALTERNATIVA B
En una placa se mezclan 4 corrientes de procesos, para dar una corriente única
de 2000Kg/h, las composiciones de las 4 corrientes de entrada y la salida se
muestran en la siguiente tabla:
Calcula las proporciones en que deberían de mezclarse las corrientes, si se
utilizan 2 Kg de la corriente A por 1Kg de la corriente B y 3 Kg de la corriente B
por 1Kg de la corriente D, para dar la misma mezcla de salida.
SOLUCION:
Datos:
mA
̇
mC
̇
=
2kg
1kg
→ mA
̇ = 2mC
̇
mB
̇
mD
̇
=
3kg
1kg
→ mB
̇ = 3mD
̇ #R=2
1)PASO: DIAGRAMA DE FLUJO DEL PROCESO
2)PASO: HACIENDO ANÁLISIS DEL NÚMERO DE LOS GRADOS DE
LIBERTAD
#VARIABLES #EB #CE #FE #R GRAD.LIBERT.
16 -4 -11 -1 -2 =-2
OBS: Si N.º (G.L) <0; entonces el problema está sobre especificado, es decir existen
datos por demás, datos redundantes o inconsistentes por lo que hay que descartarlos en
una forma lógica y luego encontrar la solución correcta del problema.
3)PASO: HACIENDO CUANTIFICACIONES
3.1. Balance global entorno a la mezcladora
El enunciado dice que existe una salida de:
2000 kg → 1h
m(entrada a la mezcladora)
̇ = m(salida de la mezcladora)
̇
A + B + C + D = F
mA
̇ + mB
̇ + mC
̇ + mD
̇ = mF
̇
2mC
̇ + 3mD
̇ + mC
̇ + mD
̇ = 2000kg
3mC
̇ + 4mD
̇ = 2000kg… … …… … … …… … …… … …(1)
3.2. Balance específico para componentes
Pueden expresarse a nivel de masa o moles de componentes
Balance con H2SO4:
m𝐇𝟐 𝐒𝐎𝟒
A + m𝐇𝟐𝐒𝐎𝟒
C + m𝐇𝟐 𝐒𝐎𝟒
D = m𝐇𝟐 𝐒𝐎𝟒
F
X𝐇𝟐𝐒𝐎𝟒
A
(mA
̇ )+ X𝐇𝟐𝐒𝐎𝟒
C
(mC
̇ ) + X𝐇𝟐𝐒𝐎𝟒
D
(mD
̇ ) = X𝐇𝟐𝐒𝐎𝟒
S
(mF
̇ )
80%(mA
̇ ) + 30%(mC
̇ ) + 10%(mD
̇ ) = 40%(2000kg)
0,8mA
̇ + 0,3mC
̇ + 0,1mD
̇ = 800kg
1,9mC
̇ + 0,1mD
̇ = 800 kg… … …… … … …… … …. . . . (2)
4)PASO: RELACIONANDO ECUACIONES
3mC
̇ + 4mD
̇ = 2000kg …… … …… … …… … … …… … (1)
1,9mC
̇ + 0,1mD
̇ = 800 kg… …… … …… … …… … …. . (2)
mC
̇ = 410,96 kg
mD
̇ = 191,78 kg
Entonces de los datos iniciales de proporción:
mA
̇ = 2mC
̇ = 2(410,96) kg = 821,92 kg
mB
̇ = 3mD
̇ = 3(191,78) kg = 575,3 kg
Por lo tanto, las proporciones en que deberían de mezclarse las corrientes A,
B, C y D son:
mA
̇ = 821,92 kg
mB
̇ = 575,30 kg
mC
̇ = 410,96 kg
mD
̇ = 191,78 kg
Tarea de balance materia y energia UNAC

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Tarea de balance materia y energia UNAC

  • 1. EJERCICIO, ALTERNATIVA B En una placa se mezclan 4 corrientes de procesos, para dar una corriente única de 2000Kg/h, las composiciones de las 4 corrientes de entrada y la salida se muestran en la siguiente tabla: Calcula las proporciones en que deberían de mezclarse las corrientes, si se utilizan 2 Kg de la corriente A por 1Kg de la corriente B y 3 Kg de la corriente B por 1Kg de la corriente D, para dar la misma mezcla de salida. SOLUCION: Datos: mA ̇ mC ̇ = 2kg 1kg → mA ̇ = 2mC ̇ mB ̇ mD ̇ = 3kg 1kg → mB ̇ = 3mD ̇ #R=2 1)PASO: DIAGRAMA DE FLUJO DEL PROCESO
  • 2. 2)PASO: HACIENDO ANÁLISIS DEL NÚMERO DE LOS GRADOS DE LIBERTAD #VARIABLES #EB #CE #FE #R GRAD.LIBERT. 16 -4 -11 -1 -2 =-2 OBS: Si N.º (G.L) <0; entonces el problema está sobre especificado, es decir existen datos por demás, datos redundantes o inconsistentes por lo que hay que descartarlos en una forma lógica y luego encontrar la solución correcta del problema.
  • 3. 3)PASO: HACIENDO CUANTIFICACIONES 3.1. Balance global entorno a la mezcladora El enunciado dice que existe una salida de: 2000 kg → 1h m(entrada a la mezcladora) ̇ = m(salida de la mezcladora) ̇ A + B + C + D = F mA ̇ + mB ̇ + mC ̇ + mD ̇ = mF ̇ 2mC ̇ + 3mD ̇ + mC ̇ + mD ̇ = 2000kg 3mC ̇ + 4mD ̇ = 2000kg… … …… … … …… … …… … …(1) 3.2. Balance específico para componentes Pueden expresarse a nivel de masa o moles de componentes Balance con H2SO4: m𝐇𝟐 𝐒𝐎𝟒 A + m𝐇𝟐𝐒𝐎𝟒 C + m𝐇𝟐 𝐒𝐎𝟒 D = m𝐇𝟐 𝐒𝐎𝟒 F X𝐇𝟐𝐒𝐎𝟒 A (mA ̇ )+ X𝐇𝟐𝐒𝐎𝟒 C (mC ̇ ) + X𝐇𝟐𝐒𝐎𝟒 D (mD ̇ ) = X𝐇𝟐𝐒𝐎𝟒 S (mF ̇ ) 80%(mA ̇ ) + 30%(mC ̇ ) + 10%(mD ̇ ) = 40%(2000kg) 0,8mA ̇ + 0,3mC ̇ + 0,1mD ̇ = 800kg 1,9mC ̇ + 0,1mD ̇ = 800 kg… … …… … … …… … …. . . . (2) 4)PASO: RELACIONANDO ECUACIONES 3mC ̇ + 4mD ̇ = 2000kg …… … …… … …… … … …… … (1) 1,9mC ̇ + 0,1mD ̇ = 800 kg… …… … …… … …… … …. . (2) mC ̇ = 410,96 kg mD ̇ = 191,78 kg Entonces de los datos iniciales de proporción: mA ̇ = 2mC ̇ = 2(410,96) kg = 821,92 kg mB ̇ = 3mD ̇ = 3(191,78) kg = 575,3 kg Por lo tanto, las proporciones en que deberían de mezclarse las corrientes A, B, C y D son: mA ̇ = 821,92 kg mB ̇ = 575,30 kg mC ̇ = 410,96 kg mD ̇ = 191,78 kg