ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
EJERCICIOS DESARROLLADOS GRUPO N° 08 CÁTEDRA
: MECANICA DE FLUIDOS II
CATEDRÁTICO
: Ing. MARCO ANTONIO PALIZA ARAUJO
ALUMNOS
:
CCORA HUALLPA, NELSON HUAMAN PEÑA JAIR MITMA MAYTA WILBER POMA CASTELLANOS FRANKLIN QUISPE FLORES WILMER SALVATIERRA BOZA ELIAS SINCHE NOE FELIX
HUANCAVELICA - 2015
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
ALUMNO: CCORA HUALLPA NELSON E-MAIL: NECO.HVCA@HOTMAIL.
[email protected] COM Juan Saldarriaga 5.2 Como parte del sistema de riego de una finca se utilizan dos tuberías en serie para conectar la bocatoma con un tanque de almacenamiento. La diferencia de nivel entre estas dos estructuras es de 31.7m, estando la bocatoma por debajo del tanque. El caudal que debe llegar al tanque es de 87 l/s. la primera tubería, en acero tiene un diámetro de 8”. Una longitud de 184m y un coeficiente global de perdidas
menores de 7.1. al final de esta debe extraerse un caudal de 94l/s con el fin fi n de regar la parte baja de la finca. La segunda tubería en pvc, tiene una longitud de 393m un diametro de 6” y un coeficiente global de perdidas menores de 11.2 el cual incluye la
valvula de control. Calcular la potencia de bomba requerida para realizar el trabajo. El fluido es agua a 15°C. A partir del diagrama de flujo 8 se obtienen los siguientes resultados Para la 1ra tubería Caudal
+ + .
VELOCIDAD
Diagrama de flujo 8 cálculo para tuberías en serie Perdidas por friccion:
.. . . . . .. − .
Con estos dos datos se calcula el factor de fricción de acuerdo con los diagramas de
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
. . . . .. Perdidas menores:
∑ . .. . . . . . . .. . . .. − . .
Para la segunda tubería.
Velocidad:
Perdidas por friccion:
Nuevamente a utilizar los diagramas de flujo 2a o 2b
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
Perdidas menores
. . . . ∑ . .. . .
Calculo de las pérdidas totales:
∑= + ∑=
.+. + .+. . Calculo de la potencia de la bomba Antes de calcula la potencia requerida en la bomba es necesario sumar la cabeza topográfica a las pérdidas totales antes calculadas
+ .+ + .. . . . . . . .
Por consiguiente, consiguiente, la potencia potencia es:
Si se supone una eficiencia global para la bomba del 75%
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
ROBERT MOTT – 6TA 6TA EDICION 10.35 Determine la caída de presión a través de un codo roscado a 90°, en una tubería de acero de 2 ½” cedula 40, si existe un flujo de agua a 15°C a razón de 750 L/min
Solución:
1ro tabla 10.4 para para codo roscado a 90° Le/D= 50 2do tabla 8.2 Rugosidad de tuberías acero 1.5x10 -4 pies o 4.6x10 -5 m 3ro llevando a la tabla8.6 el diagrama de mody Fr=0,0187
Tons hallamos k=(le/D)*fr K=0.935
∗ V=Q/A
PARA AREA USAR TABLA F.1 A= 3.090x10-3 m2
/. − . /
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
DE LA ECUAC. DE LA ENERGIA.
+ + + + . + ……. . ∗ V1 =V2
DONDE
K=0.935 V=4.045 m/s
. .. ∗ . . .. . + . .
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
Tablas usadas
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
Quispe flores Wilmer Luis (2012151071) Elias-salvatierra-boza POMA CASTELLANOS, Franklin ……
[email protected]) MECANICA-DE-FLUIDOS-ROBERT-MOTT-6TA-EDICION (PARA-UNH) Capítulo 8 (problemas), no se considera perdidas menores MECANICA-DE-FLUIDOS-ROBERT-MOTT-6TA-EDICION (PARA-UNH) Capítulo 10 (problemas)
10.8.- Calcule la perdida de energía para expansiones graduales con ángulos de
cono que van de 20 a 60°, para los incrementos mostrados en la figura 10.5. En cada caso fluyen 85 gal/min de agua a 60°F, por una tubería de acero de 2 pulg cedula 40, que aumenta a otra de 6 pulg cedula 40.
Solución:
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
∗ / . . . ∗/ ∶ . . ∗ ∗ ∗ . ∗. . Remplazando en la formula
Luego se pude comprobar remplazando en la temperatura que se requiera y se podrá comprobar
Para temperatura 20° k seria 0.31 entonces remplazamos
Para temperatura 60° k seria 0.7 1 entonces remplazamos
. . . . . .
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
10. 17.- determine la pérdida de energía cuando fluye aceite con gravedad específica de 0.87 de un tubo de 4pulg a otro de 2pulg atreves de una contracción súbita. súbita. Si la velocidad de flujo en el tubo grande es de 4.00pies/s.
:
: / , ∗∗ ∗ ⇒ ∗ ∗ : ∗∗ / ⇒ ⇒ .
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
⟶ . ⟶ ⟶ . . . ⇒ . ↔ . : . ∗ ∗ ∗. . 10.26 En las figuras 10.10 y 10.11, observe que la energía mínima para una contracción gradual (K = .04, aproximadamente) ocurre cuando el ángulo del cono está en el rango de 15° a 40°. Elabore dibujos a escala de las contracciones en ambos extremos, para una reducción de 6 a 3 pulg de un tubo de hierro dúctil.
≥
FIGURA 10.10 Coeficiente de resistenc resistencia-contracción ia-contracción gradual con 6 15°
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
LO GRAFICAMOS A ESCALA EN EXCEL
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
MECANICA-DE-FLUIDOS-ROBERT-MOTT-6TA-EDICION (PARA-UNH) Capítulo 11 (problemas)
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
:
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
MECANICA-DE-FLUIDOS-ROBERT-MOTT-6TA-EDICION (PARA-UNH) Capítulo 12 (problemas)
12.8.- Por el si stema de tube tu berr ía r ami f i cado que se se apr apre ecia en l a fi f i gur a 12.7, en en el punto A cir culan 850 L /min de agua a 10 10 °C. Por u n a tuberí tu bería de 4 pul pu l gadas, cedul cedul a 40. El E l f l u j o se bif bi f u r ca en dos tuber ías de 2 pul pu l gadas, cedul cedul a 40. segú según n se observa. observa. \ >a, y vuelve a un i r se en el pun to B. B . Calcu l e (a) el f l uj o volu mé tr i co en en cada una u na de las r amas y (b) l a dif er encia nci a de presión presión pA — pg. I ncl uya el efecto de l as per per didas menore menor es en l a rama inf eri or del del
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
− ∗ . ∗ . . ∗ ∗. ∗ .∗ Para
.. ∗ ∗ ∗ ∗∗ ∗ ∗∗
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
JUAN G. SALDARRIAGA V. Capítulo 5 (problemas) Bombas en sistemas de tuberías 5.8) Una de las estructuras de enfriamiento del agua de producción de un campo petrolero iene la configuración mostrada en la figura P5.8. Si el agua se encuentra a 80 oC, ¿cuáles deben ser los diámetros de las tuberías si su material es acero comercial? La potencia de la bomba es de 325 kW, con una eficiencia de 78%. La presión mínima en el aspersor 3 debe ser 50 kPa manométrica.
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
. . . . . . ++ + + + + + .. + .. + .. − . . .+ . + . + − − . . . . . . + . . . + + . . − − . . . . . . . . + +
EN LA ECUACION DE BERNOULLI
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
POR ITERACIONES HALLAMOS VALORES PARA ‘f2’: VALORES PARA ‘ f 2’
ITERACION 1 ITERACION 2 ITERACION 3 ITERACION 4 ITERACION 5
0.0001 4.7231856x10^-2 4. 7231825x10^-2 4. 7229070x10^-2 4.72229071x10^-2
POR ITERACIONES HALLAMOS VALORES PARA ‘f3’: VALORES PARA ‘ f 3’
ITERACION 1 ITERACION 2 ITERACION 3 ITERACION 4 ITERACION 5 REEMPLAZAMOS:
0.0001 5.2802585x10^-2 5.2802578x10^-2 5.2802546x10^-2 5.2801982x10^-2
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
ROBERT MOT 8.8 Calcule el número de Reynolds para el flujo de 325 L/min de agua a 10 ºC en una tubería de acero estándar de 2 pulg, con espesor de pared de 0.065 pulg. ¿El flujo es laminar o turbulento? SOLUCION:
Para la solución del ejercicio tenemos que tener en cuenta sobre el número de Reynolds: 1. Flujo laminar: Reynolds menor a 2000 2. Flujo en transición: Reynolds Reynolds mayor a 2000 y menor menor a 4000 3. Flujo turbulento: Reynolds mayor a 4000
... ∗− .∗ ∗ . / ∗ . / / − ∗ / .
Para hallar la velocidad solo aplicamos la ecuación del caudal, convirtiendo las unidades de medida tenemos:
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
Juan Saldarriaga 4.8 en la figura P4.8ª se muestra el esquema de una tubería simple con bombeo. En la figura P48b se muestran las curvas de la bomba instalada en la tubería. Calcule el caudal que pasa por la tubería. ¿Cuál es la potencia requerida? La tubería tiene un diámetro de 200mm en acero comercial. Los coeficientes de perdida menores son: entrada k m =0.5, cheque k m= 4.5, codo k m = 0.8, válvula k m=1.2, salida k m =1.0, uniones k m=7.0
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
Entonces la altura de la bomba es:
+ + ∗ ++ + ∗ ++ + +++ + + + + ++ +
Por la ecuación de Bernoulli tenemos: t enemos:
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
De Quispe flores Wilmer Luis (2012151071) MECANICA-DE-FLUIDOS-ROBERT-MOTT-6TA-EDICION (PARA-UNH) Capítulo 8 (problemas), no se considera perdidas menores MECANICA-DE-FLUIDOS-ROBERT-MOTT-6TA-EDICION (PARA-UNH)
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
/ . . . ∗/ ∶ . Remplazando en la formula
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
MECANICA-DE-FLUIDOS-ROBERT-MOTT-6TA-EDICION (PARA-UNH) Capítulo 12 (problemas)
12.8.- Por el si stema de tube tu berr ía r ami f i cado que se se apr apre ecia en l a fi f i gur a 12.7, en en el punto A cir culan 850 L /min de agua a 10 10 °C. Por u n a tuberí tu bería de 4 pul pu l gadas, cedul cedul a 40. El El fl j bif bi f ca en dos tuber ías de 2 pul pu l gadas, cedul cedul a 40. segú segú e observa. observa. \
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
−
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
ROBERT MOTT CAPÍTULO 10
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
ALUMNO: MITMA MAYTA, WILBER
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ING CIVIL - MECANICA DE FLUIDOS II – GRUPO N°8
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