EJERCICIO RESUELTO DE ANALISIS DIMENSIONAL
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PARA LOS ALUMNOS DE CIENCIAS E INGENIERIA LES DEJO UN PEQUEÑO EJERCICIO RELACIONADO AL TEMA DE ANALISIS DIMENCIONAL, EN EL CUAL APLICAMOS CONCEPTOS DE MATEMATICAS Y FISICA ESPERO LES SIRVA. SALUDOS. ATTE: ELMER LUIS KAPA CHOQUE
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![ANÁLISIS DIMENSIONAL:
Problema # 1: El caudal de masa o gasto másico a través de una tubería, según la
ecuación de continuidad es:
𝐺 = 𝜋 ∙ 𝑟𝑥
∙ 𝑣𝑦
∙ 𝜌𝑧
Donde:
𝐺 = 𝑔𝑎𝑠𝑡𝑜 𝑚𝑎𝑠𝑖𝑐𝑜 [
𝑔𝑟
𝑠𝑒𝑔
]
𝑟 = 𝑟𝑎𝑑𝑖𝑜 [𝑚𝑒𝑡𝑟𝑜𝑠]
𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑 [
𝑚𝑒𝑡𝑟𝑜𝑠
𝑠𝑒𝑔
]
𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑑𝑎𝑑 [
𝑔𝑟
𝑐𝑚3]
Resuelva la ecuación para “x”, “y”, “z”.
Solución:
Primero analizamos las unidades de la ecuación.
𝐺 = 𝜋 ∙ 𝑟𝑥
∙ 𝑣𝑦
∙ 𝜌𝑧
[
𝑔𝑟
𝑠
] = [𝜋] ∙ [𝑚]𝑥
∙ [
𝑚
𝑠
]
𝑦
∙ [
𝑔𝑟
𝑐𝑚3]
𝑧
Pero:
𝑔𝑟𝑎𝑚𝑜 = 𝑔𝑟 = 𝑚𝑎𝑠𝑎 = 𝑀
𝑠𝑒𝑔𝑢𝑛𝑑𝑜 = 𝑠 = 𝑡𝑖𝑒𝑚𝑝𝑜 = 𝑇
𝑚𝑒𝑡𝑟𝑜𝑠 = 𝑚 = 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑 = 𝐿
𝑛𝑢𝑚𝑒𝑟𝑜 𝑝𝑖 = 𝜋 = 𝑎𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 = 1
Luego realizamos el análisis dimensional:
𝑀
𝑇
= 1 ∙ 𝐿𝑥
∙ (
𝐿
𝑇
)
𝑦
∙ (
𝑀
𝐿3
)
𝑧
Aplicando la siguiente propiedad matemática:
1
𝑎𝑛
= 𝑎−𝑛
Tenemos lo siguiente:](https://image.slidesharecdn.com/introduccion1-210121234059/85/EJERCICIO-RESUELTO-DE-ANALISIS-DIMENSIONAL-1-320.jpg)
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EJERCICIO RESUELTO DE ANALISIS DIMENSIONAL
- 1. ANÁLISIS DIMENSIONAL: Problema # 1: El caudal de masa o gasto másico a través de una tubería, según la ecuación de continuidad es: 𝐺 = 𝜋 ∙ 𝑟𝑥 ∙ 𝑣𝑦 ∙ 𝜌𝑧 Donde: 𝐺 = 𝑔𝑎𝑠𝑡𝑜 𝑚𝑎𝑠𝑖𝑐𝑜 [ 𝑔𝑟 𝑠𝑒𝑔 ] 𝑟 = 𝑟𝑎𝑑𝑖𝑜 [𝑚𝑒𝑡𝑟𝑜𝑠] 𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑 [ 𝑚𝑒𝑡𝑟𝑜𝑠 𝑠𝑒𝑔 ] 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑑𝑎𝑑 [ 𝑔𝑟 𝑐𝑚3] Resuelva la ecuación para “x”, “y”, “z”. Solución: Primero analizamos las unidades de la ecuación. 𝐺 = 𝜋 ∙ 𝑟𝑥 ∙ 𝑣𝑦 ∙ 𝜌𝑧 [ 𝑔𝑟 𝑠 ] = [𝜋] ∙ [𝑚]𝑥 ∙ [ 𝑚 𝑠 ] 𝑦 ∙ [ 𝑔𝑟 𝑐𝑚3] 𝑧 Pero: 𝑔𝑟𝑎𝑚𝑜 = 𝑔𝑟 = 𝑚𝑎𝑠𝑎 = 𝑀 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 = 𝑠 = 𝑡𝑖𝑒𝑚𝑝𝑜 = 𝑇 𝑚𝑒𝑡𝑟𝑜𝑠 = 𝑚 = 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑 = 𝐿 𝑛𝑢𝑚𝑒𝑟𝑜 𝑝𝑖 = 𝜋 = 𝑎𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 = 1 Luego realizamos el análisis dimensional: 𝑀 𝑇 = 1 ∙ 𝐿𝑥 ∙ ( 𝐿 𝑇 ) 𝑦 ∙ ( 𝑀 𝐿3 ) 𝑧 Aplicando la siguiente propiedad matemática: 1 𝑎𝑛 = 𝑎−𝑛 Tenemos lo siguiente:
- 2. 𝑀𝑇−1 = 𝐿𝑥 ∙ (𝐿𝑇−1)𝑦 ∙ (𝑀𝐿−3)𝑧 Luego aplicamos la siguiente propiedad matemática: (𝑎 ∙ 𝑏)𝑛 = 𝑎𝑛 ∙ 𝑏𝑛 Y tenemos lo siguiente: 𝑀𝑇−1 = 𝐿𝑥 ∙ 𝐿𝑦 ∙ 𝑇−𝑦 ∙ 𝑀𝑧 ∙ 𝐿−3𝑧 Por definición de potenciación: 𝑎0 = 1 𝑦 𝑎𝑚 ∙ 𝑎𝑛 = 𝑎𝑚+𝑛 Lo que nos da como resultado: 𝑀 ∙ 𝐿0 ∙ 𝑇−1 = 𝑀𝑧 ∙ 𝐿𝑥+𝑦−3𝑧 ∙ 𝑇−𝑦 Por definición de potenciación: 𝑎𝑚 = 𝑎𝑛 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠: 𝑚 = 𝑛 Con lo que tendríamos el siguiente sistema de ecuaciones: 𝑧 = 1 … … … … … … … … … . (1) 𝑥 + 𝑦 − 3𝑧 = 0 … … … … … (2) −𝑦 = −1 … … … … … … … . (3) De donde resolviendo el sistema tenemos: 𝑥 = 2 ; 𝑦 = 1 ; 𝑧 = 1 Y reemplazando estos valores en la ecuación original tenemos: 𝐺 = 𝜋 ∙ 𝑟2 ∙ 𝑣 ∙ 𝜌