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Ejercicios resueltos: Balance de materiales (masa) - Mecánica de Fluidos
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Guía de ejercicios resueltos paso a paso, incluye esquemas/gráficos.
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SERIE N°0: BALANCES DE MATERIALES (resolución)PROBLEMA Nº1.- A un secador de aire entra una masa de aire de refrigeración con 71% de
humedad. A la salida del secador se determina que se eliminó el 60% de la humedad original.
Para un caudal de 1.000 kg/hora de aire húmedo, calcular la composición del aire seco y el
caudal de humedad eliminada. (Las concentraciones de humedad están medidas en % m/m).
SECADOR DE AIRE
𝑘𝑔 𝑚̇2
𝑚̇1 = 1.000
ℎ 𝑥2
𝑥1 = 0,29 𝑚̇3
𝑥3 = 0
𝑘𝑔
𝐴𝑙 𝑠𝑒𝑐𝑎𝑑𝑒𝑟𝑜 𝑖𝑛𝑔𝑟𝑒𝑠𝑎𝑛 𝑎𝑖𝑟𝑒 𝑠𝑒𝑐𝑜 𝑦 𝑣𝑎𝑝𝑜𝑟 𝑑𝑒 𝑎𝑔𝑢𝑎 𝑒𝑛 𝑢𝑛 𝑡𝑜𝑡𝑎𝑙 𝑑𝑒 𝑚̇1 = 1.000 𝑑𝑜𝑛𝑑𝑒
ℎ
𝑘𝑔 𝑘𝑔
ℎ𝑎𝑦 𝑚̇1 ∗ 𝑦1 = 1.000 ∗ 0,71 = 710 𝑑𝑒 𝑣𝑎𝑝𝑜𝑟 𝑑𝑒 𝑎𝑔𝑢𝑎 (ℎ𝑢𝑚𝑒𝑑𝑎𝑑)
ℎ ℎ
𝐸𝑛 𝑒𝑙 𝑠𝑒𝑐𝑎𝑑𝑒𝑟𝑜, 𝑠𝑒 𝑒𝑙𝑖𝑚𝑖𝑛𝑎 𝑒𝑙 60% 𝑑𝑒 𝑙𝑎 ℎ𝑢𝑚𝑒𝑑𝑎𝑑 𝑞𝑢𝑒 𝑖𝑛𝑔𝑟𝑒𝑠𝑎:
𝑘𝑔 𝑘𝑔
𝑚̇3 = 710 ∗ 0,60 = 426 𝑒𝑙 𝑐𝑎𝑢𝑑𝑎𝑙 𝑚̇3 𝑒𝑠𝑡𝑎 𝑐𝑜𝑛𝑓𝑜𝑟𝑚𝑎𝑑𝑜 𝑠𝑜𝑙𝑜 𝑝𝑜𝑟 𝑎𝑔𝑢𝑎 (𝑦3 = 1)
ℎ ℎ
𝐿𝑎 𝑚𝑎𝑠𝑎 𝑑𝑒 𝑎𝑖𝑟𝑒 𝑠𝑒𝑐𝑜 𝑞𝑢𝑒 𝑖𝑛𝑔𝑟𝑒𝑠𝑎 𝑒𝑛 (1) 𝑠𝑎𝑙𝑒 𝑒𝑛 (2):
𝑘𝑔(𝑤) 𝑘𝑔(𝑎𝑠) 𝑘𝑔 𝑘𝑔(𝑎𝑠)
𝑚̇1 = 710 + 290 = 1.000 ⇒ 290 = 𝑚̇2 ∗ 𝑥2
ℎ ℎ ℎ ℎ
𝑘𝑔(𝑤)
𝐷𝑎𝑑𝑜 𝑞𝑢𝑒 𝑠𝑒 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑛 426 , 𝑒𝑙 𝑐𝑎𝑢𝑑𝑎𝑙 𝑑𝑒 𝑎𝑔𝑢𝑎 𝑒𝑛 (2) 𝑞𝑢𝑒𝑑𝑎 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑜 𝑝𝑜𝑟:
ℎ
𝑘𝑔(𝑎𝑠) 𝑘𝑔(𝑎𝑠)
𝑘𝑔(𝑤) 𝑘𝑔(𝑤) 𝑘𝑔(𝑤) 290 290
710 − 426 = 284 ⇒ 𝑥2 = ℎ = ℎ
ℎ ℎ ℎ 𝑘𝑔(𝑎𝑠) 𝑘𝑔(𝑤) 𝑘𝑔
290 + 284 574
ℎ ℎ ℎ
𝑥2 = 0,5052264808 ⇒ 𝑦2 = 0,4947735192
𝑘𝑔(𝑎𝑠)
𝐸𝑙 𝑐𝑎𝑢𝑑𝑎𝑙 𝑑𝑒 𝑎𝑖𝑟𝑒 𝑠𝑒𝑐𝑜 𝑠𝑒 𝑚𝑎𝑛𝑡𝑖𝑒𝑛𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 290
ℎ
𝐷𝑎𝑑𝑜 𝑞𝑢𝑒 𝑒𝑙 𝑎𝑖𝑟𝑒 𝑒𝑠 𝑠𝑒𝑐𝑎𝑑𝑜, 𝑙𝑎 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑐𝑖𝑜𝑛 𝑎𝑢𝑚𝑒𝑛𝑡𝑎 𝑑𝑒: 𝑥1 = 29% 𝑎 𝑥2 = 50,52%
𝑘𝑔 𝑘𝑔
𝑆𝑒 𝑟𝑒𝑡𝑖𝑟𝑎𝑛 𝑚̇3 = 710 ∗ 0,60 = 426
ℎ ℎ
, PROBLEMA Nº2.- A un tanque de mezclado de ácido sulfúrico para baterías ingresan un
caudal de 200 kg/min de ácido concentrado al 77,7% (el resto es agua) y un ácido muy diluido
al 12,4% (el resto es agua). Del tanque sale ácido preparado para su uso al 18,6% (el resto es
agua). Calcular el caudal de ácido preparado. (Las concentraciones están medidas en % m/m).
(1)
(3)
MEZCLADORA DE ACIDO
(2)
𝐶𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 (1) − 𝐴𝑐𝑖𝑑𝑜 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑑𝑜
𝑘𝑔
𝑚̇1 = 200 𝑥1 = 0,777 ⇒ 𝑦1 = 0,223
𝑚𝑖𝑛
𝐶𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 (2) − 𝐴𝑐𝑖𝑑𝑜 𝐷𝑖𝑙𝑢𝑖𝑑𝑜
𝑘𝑔
𝑚̇2 = ¿ ? 𝑥2 = 0,124 ⇒ 𝑦2 = 0,876
𝑚𝑖𝑛
𝐶𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 (3) − 𝐴𝑐𝑖𝑑𝑜 𝐷𝑜𝑠𝑖𝑓𝑖𝑐𝑎𝑑𝑜
𝑘𝑔
𝑚̇3 = ¿ ? 𝑥3 = 0,186 ⇒ 𝑦3 = 0,814
𝑚𝑖𝑛
𝐵𝑎𝑙𝑎𝑛𝑐𝑒 𝑑𝑒 𝑚𝑎𝑠𝑎 𝐺𝐿𝑂𝐵𝐴𝐿
𝑘𝑔
𝑚̇1 + 𝑚̇2 = 𝑚̇3 ⇔ 200 = 𝑚̇3 − 𝑚̇2
𝑚𝑖𝑛
𝐵𝑎𝑙𝑎𝑛𝑐𝑒 𝑑𝑒 𝑚𝑎𝑠𝑎 𝑆𝑂𝐿𝑈𝑇𝑂
𝑘𝑔
𝑚̇1 ∗ 𝑥1 + 𝑚̇2 ∗ 𝑥2 = 𝑚̇3 ∗ 𝑥3 ⇔ 200 ∗ 0,777 + 𝑚̇2 ∗ 0,124 = 𝑚̇3 ∗ 0,186
𝑚𝑖𝑛
𝑅𝑒𝑠𝑜𝑙𝑣𝑖𝑒𝑛𝑑𝑜 𝑒𝑙 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 𝑙𝑖𝑛𝑒𝑎𝑙 𝑑𝑒 𝑔𝑟𝑎𝑑𝑜 2:
𝑘𝑔 𝑘𝑔
𝑚̇2 = 1.906,451613 𝑚̇3 = 2.106,451613
𝑚𝑖𝑛 𝑚𝑖𝑛
𝐶𝑎𝑢𝑑𝑎𝑙 𝑑𝑒 𝐴𝐶𝐼𝐷𝑂 𝑃𝑅𝐸𝑃𝐴𝑅𝐴𝐷𝑂:
𝑘𝑔
𝑚̇3 = 2.106
𝑚𝑖𝑛
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