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Problema
Se tiene los siguientes datos de secado utilizando un equipo de secador de bandejas (1
bandeja de 50 cm de largo 40 cm de ancho y un espesor de 1 cm) con flujo de aire
constante solo sobre la superficie superior expuesta. El peso de la muestra
totalmente seca fue de 3.700 Kg de sólido seco. La muestra húmeda en equilibrio pesó
3.900 (Kg agua + sólido). En el experimento de secado se obtuvieron los siguientes pesos
de las muestras en función del tiempo.
Tiempo
(Hr)
0 0.4 0.8 1.4 2.2 3 4.2 5 7 9 12
Masa
húmeda
(Kg)
4.944 4.885 4.808 4.699 4.554 4.404 4.241 4.15 4.019 3.978 3.900
a) Calcule el contenido de humedad libre X (Kg agua/Kg sólido seco) para cada punto y
construya la curva de X en función del tiempo.
b) Determine las pendientes, calcule las velocidades de secado NA en Kg agua/h*m2, y
grafique NA en función de X.
c) Empleando la curva de velocidad de secado, pronostique el tiempo total necesario
para secar la muestra desde X=0.20 hasta X=0.04. ¿Cuál es la velocidad de secado
NA en el periodo de velocidad constante y el valor de Xc?
Solución:
a) 𝑋 = 𝑋𝑇 − 𝑋∗
𝑋𝑇 =
𝑚𝐻−𝑚𝑠𝑠
𝑚𝑠𝑠
𝑋∗
=
𝑚∗−𝑚𝑠𝑠
𝑚𝑠𝑠
DONDE:
X= humedad libre
𝑋𝑇 = 𝐻𝑢𝑚𝑒𝑑𝑎𝑑 𝑡𝑜𝑡𝑎𝑙
𝑋∗
= 𝐻𝑢𝑚𝑒𝑑𝑎𝑑 𝑑𝑒 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑜
𝑚𝐻 = 𝑚𝑎𝑠𝑎 𝑑𝑒𝑙 𝑠𝑜𝑙𝑖𝑑𝑜 ℎ𝑢𝑚𝑒𝑑𝑜 𝑣𝑒𝑟 𝑡𝑎𝑏𝑙𝑎 𝑑𝑒 𝑑𝑎𝑡𝑜𝑠
𝑚𝑠𝑠 = 𝑚𝑎𝑠𝑎 𝑑𝑒𝑙 𝑠𝑜𝑙𝑖𝑑𝑜 𝑠𝑒𝑐𝑜 = 3,700 𝐾𝑔
𝑚∗
= 𝑚𝑎𝑠𝑎 𝑑𝑒𝑙 𝑠𝑜𝑙𝑖𝑑𝑜 ℎ𝑢𝑚𝑒𝑑𝑜 𝑒𝑛 𝑒𝑙 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑜 = 3,900 𝐾𝑔
Finalmente realizar la gráfica correspondiente humedad libre vs tiempo.
b) 𝑁𝐴 = −
𝑚𝑠𝑠
𝐴
∗
𝑑𝑥
𝑑𝑡
c) 𝑁𝐴 = −
𝑚𝑠𝑠
𝐴
∗
𝑑𝑥
𝑑𝑡
Discretizando
𝑁𝐴 = −
𝑚𝑠𝑠
𝐴
∗
𝛥𝑥
𝛥𝑡
== −
𝑚𝑠𝑠
𝐴
∗
(𝑥𝑓−𝑥𝑜)
(𝑡𝑓−𝑡𝑜)
y se debe sacar humedades
promedios
Enunciados 1 y 2

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Enunciados 1 y 2

  • 1. Problema Se tiene los siguientes datos de secado utilizando un equipo de secador de bandejas (1 bandeja de 50 cm de largo 40 cm de ancho y un espesor de 1 cm) con flujo de aire constante solo sobre la superficie superior expuesta. El peso de la muestra totalmente seca fue de 3.700 Kg de sólido seco. La muestra húmeda en equilibrio pesó 3.900 (Kg agua + sólido). En el experimento de secado se obtuvieron los siguientes pesos de las muestras en función del tiempo. Tiempo (Hr) 0 0.4 0.8 1.4 2.2 3 4.2 5 7 9 12 Masa húmeda (Kg) 4.944 4.885 4.808 4.699 4.554 4.404 4.241 4.15 4.019 3.978 3.900 a) Calcule el contenido de humedad libre X (Kg agua/Kg sólido seco) para cada punto y construya la curva de X en función del tiempo. b) Determine las pendientes, calcule las velocidades de secado NA en Kg agua/h*m2, y grafique NA en función de X. c) Empleando la curva de velocidad de secado, pronostique el tiempo total necesario para secar la muestra desde X=0.20 hasta X=0.04. ¿Cuál es la velocidad de secado NA en el periodo de velocidad constante y el valor de Xc? Solución: a) 𝑋 = 𝑋𝑇 − 𝑋∗ 𝑋𝑇 = 𝑚𝐻−𝑚𝑠𝑠 𝑚𝑠𝑠 𝑋∗ = 𝑚∗−𝑚𝑠𝑠 𝑚𝑠𝑠 DONDE: X= humedad libre 𝑋𝑇 = 𝐻𝑢𝑚𝑒𝑑𝑎𝑑 𝑡𝑜𝑡𝑎𝑙 𝑋∗ = 𝐻𝑢𝑚𝑒𝑑𝑎𝑑 𝑑𝑒 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑜 𝑚𝐻 = 𝑚𝑎𝑠𝑎 𝑑𝑒𝑙 𝑠𝑜𝑙𝑖𝑑𝑜 ℎ𝑢𝑚𝑒𝑑𝑜 𝑣𝑒𝑟 𝑡𝑎𝑏𝑙𝑎 𝑑𝑒 𝑑𝑎𝑡𝑜𝑠 𝑚𝑠𝑠 = 𝑚𝑎𝑠𝑎 𝑑𝑒𝑙 𝑠𝑜𝑙𝑖𝑑𝑜 𝑠𝑒𝑐𝑜 = 3,700 𝐾𝑔 𝑚∗ = 𝑚𝑎𝑠𝑎 𝑑𝑒𝑙 𝑠𝑜𝑙𝑖𝑑𝑜 ℎ𝑢𝑚𝑒𝑑𝑜 𝑒𝑛 𝑒𝑙 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑜 = 3,900 𝐾𝑔 Finalmente realizar la gráfica correspondiente humedad libre vs tiempo. b) 𝑁𝐴 = − 𝑚𝑠𝑠 𝐴 ∗ 𝑑𝑥 𝑑𝑡 c) 𝑁𝐴 = − 𝑚𝑠𝑠 𝐴 ∗ 𝑑𝑥 𝑑𝑡 Discretizando
  • 2. 𝑁𝐴 = − 𝑚𝑠𝑠 𝐴 ∗ 𝛥𝑥 𝛥𝑡 == − 𝑚𝑠𝑠 𝐴 ∗ (𝑥𝑓−𝑥𝑜) (𝑡𝑓−𝑡𝑜) y se debe sacar humedades promedios