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Ejercicios resueltos: Balances macroscopicos - Mecánica de Fluidos
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Guía de ejercicios resueltos paso a paso, incluye esquemas/gráficos.
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SERIE N°4: BALANCES MACROSCOPICOS (resolución)PROBLEMA Nº1.- Circula agua a 20ºC en régimen turbulento a razón de 85 litros/min por una
cañería horizontal en U de 4 pulgadas de diámetro interno. Calcular la fuerza actuante sobre la
curva si la presión manométrica en un punto inmediato anterior a la curva es de 1,5 kgf/cm2 y
en un punto inmediatamente posterior a la misma es de 1,35 kgf/cm2.
𝐷 = 4"
𝑦(+)
⃗⃗⃗⃗
𝐴1
(1)
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝐹𝑢𝑒𝑟𝑧𝑎
𝑥(+)
⃗⃗⃗⃗
𝐴2
(2)
𝑦(+)
(1)
𝑥(+)
(2)
𝑧(+)
𝑥(+)
⃗⃗⃗⃗⃗
𝑘𝑔 ⃗⃗⃗⃗⃗
𝑘𝑔
∆𝑃1 = 1,50 = 147.105,0 𝑃𝑎 ∆𝑃2 = 1,35 = 132.394,5 𝑃𝑎
𝑐𝑚2 𝑐𝑚2
, 𝐷𝑖 2 (0,1016 𝑚)2
𝐴𝑟𝑒𝑎 𝑑𝑒 𝑓𝑙𝑢𝑗𝑜 = ‖⃗⃗⃗⃗⃗
𝐴1 ‖ = ‖⃗⃗⃗⃗⃗
𝐴2 ‖ = 𝜋 ∗ =𝜋∗ = 8,107319666 ∗ 10−3 𝑚2
4 4
⃗⃗⃗⃗1 ‖ = ‖𝐴
‖𝐴 ⃗⃗⃗⃗2 ‖ 𝑝𝑒𝑟𝑜 𝑙𝑜𝑠 𝑣𝑒𝑐𝑡𝑜𝑟𝑒𝑠 𝑡𝑖𝑒𝑛𝑒𝑛 𝑠𝑒𝑛𝑡𝑖𝑑𝑜𝑠 𝑜𝑝𝑢𝑒𝑠𝑡𝑜𝑠: ⃗⃗⃗⃗ ⃗⃗⃗⃗2
𝐴1 = −𝐴
𝑙𝑖𝑡𝑟𝑜𝑠 𝑚3 1 𝑚𝑖𝑛
𝑐𝑎𝑢𝑑𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐𝑜 (𝑉̇ ) 85 ∗ 0,001 ∗
𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑 (𝑣) = = 𝑚𝑖𝑛 𝑙𝑖𝑡𝑟𝑜𝑠 60 𝑠 = 0,1747392141 𝑚
𝐴𝑟𝑒𝑎 𝑑𝑒 𝑓𝑙𝑢𝑗𝑜 8,107319666 ∗ 10−3 𝑚2 𝑠
𝑁
𝐶𝑎𝑚𝑝𝑜 𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑐𝑖𝑜𝑛𝑎𝑙: 𝑔 (𝑥; 𝑦; 𝑧) = (0 ; 0 ; −9,807)
𝑘𝑔
𝐵𝐴𝐿𝐴𝑁𝐶𝐸 𝑀𝐴𝐶𝑅𝑂𝑆𝐶𝑂𝑃𝐼𝐶𝑂 𝐷𝐸 𝐶𝐴𝑁𝑇𝐼𝐷𝐴𝐷 𝐷𝐸 𝑀𝑂𝑉𝐼𝑀𝐼𝐸𝑁𝑇𝑂
𝜕
∭ 𝜌 ∗ 𝑣 ∗ 𝑑𝑉 = 𝜌1 ∗ 𝑣1 2 ∗ ⃗⃗⃗⃗⃗
𝐴1 − 𝜌2 ∗ 𝑣2 2 ∗ ⃗⃗⃗⃗⃗
𝐴2 + 𝑃1 ∗ ⃗⃗⃗⃗⃗
𝐴1 − 𝑃2 ∗ ⃗⃗⃗⃗⃗
𝐴2 + 𝐹⃗⃗ + 𝑃𝑒𝑠𝑜
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝜕𝑡
𝜕
∭ 𝜌 ∗ 𝑣 ∗ 𝑑𝑉 = 0 𝐸𝑛 𝑒𝑠𝑡𝑎𝑑𝑜 𝑒𝑠𝑡𝑎𝑐𝑖𝑜𝑛𝑎𝑟𝑖𝑜, 𝑒𝑠 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑖𝑒𝑛𝑡𝑒 𝑑𝑒𝑙 𝑡𝑖𝑒𝑚𝑝𝑜
𝜕𝑡
−𝐹 = 2 ∗ 𝜌 ∗ 𝑣 2 ∗ ⃗⃗⃗⃗
𝐴1 + ⃗⃗⃗⃗ 𝑃𝑒𝑠𝑜 = ⃗⃗⃗⃗
𝐴1 ∗ (𝑃1 − 𝑃2 ) + ⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝐴1 ∗ [2 ∗ 𝜌 ∗ 𝑣 2 + (𝑃1 − 𝑃2 )]
𝑘𝑔 𝑚 2
−𝐹 = ⃗⃗⃗⃗
𝐴1 ∗ [2 ∗ 1.000 3
∗ (0,1747392141 ) + 147.105,0 𝑃𝑎 − 132.394,5 𝑃𝑎]
𝑚 𝑠
⃗⃗⃗⃗
𝐴1 = (8,107319666 ∗ 10−3 ; 0 ; 0) 𝑚2 ⃗⃗⃗⃗
𝐴2 = (−8,107319666 ∗ 10−3 ; 0 ; 0) 𝑚2
2 ∗ 𝜌 ∗ 𝑣 2 + (𝑃1 − 𝑃2 ) = 14.771,56759 𝑃𝑎
−𝐹 = (8,107319666 ∗ 10−3 ; 0 ; 0) 𝑚2 ∗ 14.771,56759 𝑃𝑎
−𝐹 = (8,107319666 ∗ 10−3 ; 0 ; 0) 𝑚2 ∗ 14.771,56759 𝑃𝑎 = (119,7578204 ; 0 ; 0) 𝑁
𝐴𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑑𝑎𝑚𝑒𝑛𝑡𝑒, 𝑒𝑙 𝑎𝑔𝑢𝑎 𝑒𝑗𝑒𝑟𝑐𝑒 𝑢𝑛𝑎 𝑓𝑢𝑒𝑟𝑧𝑎 𝑑𝑒 120 𝑁 𝑐𝑜𝑛𝑡𝑟𝑎 𝑙𝑎 𝑐𝑢𝑟𝑣𝑎
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