2.12. La esfera hueca.
Consideremos el caso particular de la esfera hueca, en estado estable y sin generación de calor, la
cual mantiene sus paredes interior y exterior a temperaturas y
respectivamente, como se ilustra en la figura.
y
respectivamente, y de radios
Si la conductividad térmica del material de la pared es constante, entonces la distribución de
temperaturas y el flujo de calor a través de la pared, puede obtenerse a partir del caso general de la conducción. Note que para este caso el área de la esfera
varia de acuerdo acuerdo al los los radios. Además, la transferencia de calor calor se realizará en la dirección radial. Así, de la ecuación general de la conducción, y después de eliminar los términos pertinentes se obtiene:
Multiplicando la ecuación por
En el caso de que
Y puesto que el radio es variable, e ntonces:
Dividiendo nuevamente la ecuación entre
Integrando la primera vez
Integrando nuevamente,
se se obtiene;
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Nuevamente, esta expresión es el perfil de temperaturas en su solución general para la esfera hueca sin generación de calor. Se le recuerda al estudiante que este mismo resultado se obtiene al extraer un elemento diferencial del sistema y aplicar la primera ley de la termodinámica para flujo estable. Desde luego para encontrar los valores de las constantes, las condiciones de frontera son:
Sustituyendo estas condiciones en la solución general, se llega al sistema de ecuaciones lineales siguiente;
Resolviendo el sistema, se obtiene:
Sustituyendo estos valores en la solución general, se tiene:
Solución particular para el perfil de tem peraturas de la esfera sin fuente de calor interna. Note que este perfil es de la forma exponencial. El flujo de calor que pasa a través de la esfera en dirección radial, puede ser calculado utilizando la Ley de Fourier.
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Sustituyendo este valor;
Ejemplo 2.12.1 Una esfera hueca es fabricada de aluminio y tiene un diámetro interior de 4 cm y uno exterior de 8 cm. Las temperaturas interior interior y exterior son 100 °C y 50 °C respectivamente. respectivamente. Calcule la la transferencia de calor desde desde el interior de la esfera. Tome Solución:
Para una esfera hueca en estado estacionario sin generación de calor, se tiene:
2.13. Resistencia térmica conductiva en esferas huecas. Si comparamos la ecuación para el cálculo del flujo de calor en la esfera hueca, con la Ley de Ohm, se obtiene la siguiente expresión:
En esta, se observa que el denominador es la resistencia térmica conductiva.
2.14. La esfera hueca compuesta.
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Puesto que todas las resistencias están conectadas en serie, el flujo de calor total a través de la esfera hueca compuesta puede calcularse como el cociente que resulta de dividir la diferencia de temperaturas interior y exterior entre la resistencia total, así:
∑
Si además consideramos que la esfera está expuesta tanto en su interior como en su exterior a fluidos que tienen temperaturas transferencia de calor
y
y
, respectivamente, al igual que los coeficientes de
el flujo de calor a través de ella se calcula añadiendo las dos
resistencias convectivas a la ecuación anterior. Donde:
∑
2.15. Radio crítico de aislamiento en una esfera hueca. En la práctica, es necesario cubrir con aislante ciertos depósitos esféricos que almacenan líquidos o gases calientes con el propósito propósito de proteger a las personas de quemaduras. quemaduras. Sin embargo, el hecho de cubrir estas paredes esféricas tiene también como meta principal reducir el flujo de calor que pasa a través de la pared, de adentro adentro hacia fuera o viceversa. La figura siguiente ilustra una geometría esférica cubierta con un material aislante y las resistencias térmicas que intervienen en el cálculo del flujo de calor. Las resistencias son:
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los radios
; puesto que esta diferencia es muy pequeña y la conductividad del metal muy
grande, esta resistencia también se desprecia. Y por consiguiente el flujo de calor por unidad depende solo de las resistencias tres y c uatro.
( )
Puesto que, la resistencia térmica total del circuito es la suma de estas e stas dos resistencias; entonces:
Y como al variar el radio del aislante que existe un valor para
en el cuál el flujo de calor sea máximo. Este valor de
denomina radio crítico de aislante y se denota por
, el flujo de c alor debe aumentar o disminuir, quiere decir .
Por tanto para encontrar este valor es necesario derivar con respecto a del circuito e igualar a cero. Así:
, se
la resistencia total
, se obtiene que:
Derivando, y resolviendo algebraicamente y despejando
Nota: Se deja al estudiante comprobar que este valor corresponde a un mínimo, aplicando la segunda derivada a la
y sustituir en esta nueva derivada el valor de:
2.16. Esfera sólida con generación de calor desnuda. Considere una esfera sólida sin recubrimiento de radio
con temperatura superficial
, y de
conductividad constante; la cuál posee posee una fuente de calor en su interior. Nuevamente podemos deducir el perfil de temperaturas y calcular el flujo de calor desde el centro de la misma hasta su superficie. Así, según la figura y utilizando la ecuación general de la conducción.
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Puesto que el área superficial de la esfera varía con el radio radio,
entonces multiplicando esta expresión por el área y dividiéndola a su vez entre
Integrando por primera vez,
, se obtiene:
Integrando nuevamente:
Esta expresión es el perfil de temperaturas en su solución solución general. Y las condiciones de frontera para este caso son;
Note que la primera condición se debe a que en el centro de la esfera, esta la fuente de calor y la temperatura debe ser por consiguiente la máxima. Sustituyendo estas condiciones de frontera en la solución general, se tiene:
Sustituyendo nuevamente los valores de las constantes, en la misma ecuac ión:
Factorizando,
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Esta ecuación permite calcular la temperatura para cualquier radio
en la geometría esférica y
define en forma gráfica la curva curva mostrada en la figura. La temperatura máxima en el conductor se evalúa con esta expresión solo que en
|
Asimismo para calcular el flujo de calor a través de la esfera, se aplica la Ley de Fourier y se evalúa la diferencial de la temperatura respecto al límite
, así:
No debe sorprender que el calor transferido en dirección radial, sea el que genera la fuente de calor en el interior de la esfera.
2.17. Esfera sólida desnuda con generación de calor, expuesta a un fluido.
Consideremos de nuevo el caso anterior solo que en lugar de conocer la temperatura superficial de la esfera, se conoce la temperatura del fluido circundante
. En este caso, como se
muestra en la figura; se agrega la resistencia convectiva y se modifica la segunda segunda condición de
frontera. De la solución general se tiene:
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Despejando
de la segunda condición,
[ ]
Esta expresión relaciona la temperatura superficial con la del fluido y el coeficiente de película.
Sustituyendo el valor de
, en la solución general, en e l límite cuando
se se tiene:
Y despejando
Sustituyendo los valores de
Puesto que
y
, en la solución general, se obtiene:
, agrupando y factorizando términos y teniendo en c uenta que:
Se obtiene:
[ ]
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y la conductividad térmica es de 8 w/m-k. ¿Cuál es la temperatura en la superficie de la esfera si la temperatura del ambiente circundante es de 25 °C y la del centro de la esfera es de 65 °C? Solución: La temperatura en al superficie de la esfera puede calcularse a partir de La ecuación,
Sin embargo, se desconoce el valor del calor generado por unidad de volumen calcularse a través de la expresión;
, el cuál puede
[ [ ] Despejando el valor de
, se obtiene:
Sustituyendo este valor en la primera ec uación, se obtiene:
2.18. Superficies extendidas. En muchas aplicaciones es necesario aumentar la razón de transferencia de calor en una
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transferencia. También cabe la posibilidad de aumentar el coeficiente de transferencia de calor, sin embargo esto representa mayor dificultad que la propuesta anterior, porque el coeficiente de película depende de las propiedades del fluido y eso equivale a modificarlas. Por último es posible y factible en la práctica, aumentar el área de transferencia de calor de la superficie expuesta, esto se logra a través dela adherencia de superficies extendidas o aletas de enfriamiento de diversas geometrías a la superficie base, las cuales contribuyen en forma por demás eficaz, para llevar al cabo dicho efe cto. En el análisis y diseño de una superficie ale tada, la cantidad de calor disipado por una sola aleta, se determina apoyándose en el gradiente de temperatura y el área transversal disponible para conducir el calor desde la base de la aleta. Así, el número total de aletas necesarias para disipar una cantidad de calor dada, se estimará en base a la acumulación de energía a transferir desde la superficie base. El fluido puede ser un líquido o un gas, sin embargo es común el uso de superficies extendidas cuando el fluido es un gas, porque el coeficiente de película para un gas es menor que el de un fluido. Ejemplos típicos de de superficies aletas expuestas a un fluido son: son: Algunos motores de combustión interna (motocicletas y Volkswagen), motores eléctricos, transformadores y algunos calentadores caseros.
2.19. Tipos de superficies extendidas. Las aletas pueden ser de secciones transversales y perfiles diferentes, tales como tiras anexadas a lo largo de un tubo, llamadas longitudinales; o bien discos anulares concéntricos alrededor de un tubo, llamadas aletas circunferenciales; o adheridas a paredes verticales como las llamadas triangulares o rectangulares. En todos los casos el espesor de de la aleta puede ser uniforme o variable según el perfil. La figura 2.19.1, ilustra algunos tipos de aletas como los mencionados. En el análisis de una superficie extendida se presenta tanto la conducción como la convección de calor, por lo
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2.20. Análisis general de una superficie extendida. Considere una aleta de sección transversal variable como se muestra en la figura siguiente,
estrictamente hablando el problema de conducción de una aleta hacia el fluido que la rodea, no es unidimensional, puesto que la conducción de calor es a lo largo del eje
mientras que la
pérdida por convección en las superficies implican un gradiente de temperatura diferente de cero en la dirección
para cualquier valor de
Pero como se menciono anteriormente, anteriormente, si la aleta es muy delgada, entonces la temperatura será constante sobre el área transversal la
Se supone
que la temperatura en la base de la aleta es uniforme, por lo tanto la temperatura no dependerá de o
Además, no existen fuentes fuentes de calor y prevalecen las condiciones de estado estacionario. estacionario.
Así que, la temperatura tura solo dependerá de la longitud
las propiedades del material de la las
aleta y el coeficiente convectivo del fluido fluido que la rodea. Así mediante un balance de energía por por la primera ley, se tiene:
De la ley de Fourier:
Donde:
Así, sustituyendo estas expresiones en la ecuación del balance de energía se tiene:
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Si la conductividad del material de la aleta es constante, se divide entre
Puesto que el área de la sección transversal es variable, se tiene:
Dividiendo la ecuación entre
finalmente se obtiene: finalmente
La ecuación diferencial anterior ordinaria, de segundo orden, se denomina ecuación general de una superficie extendida o aleta de enfriamiento, de sección variable en estado estacionario.
2.21. Superficie extendida de sección transversal constante. Si consideramos que la sección transversal de la alet a es constante en toda su longitud, la ec uación general se convierte en la ecuación diferencial siguiente:
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Esta nueva ecuación, se denomina ecuación o solución general de una superficie extendida de sección transversal constante, sin fuente de calor y en estado estacionario.
Luego de obtener la solución general, se requiere de dos condiciones de frontera, para determinar las constantes y Existen tres posibles conjuntos conjuntos de condiciones de frontera, estos estos son:
Caso 1. La aleta es infinitamente larga comparada su espesor, de tal suerte que que la temperatura en el extremo libre es igual a del fluido que es que la rodea. En la práctica una aleta de 3 mm de espesor y de varios cm de largo, satisface en forma aproximada esta condición, al igual que una barra de 3 mm de diámetro y varios cm de largo. Para una mejor comprensión al problema, consideremos una aleta muy larga de sección transversal rectangular constante como se indica en la figura, expuesta un fluido y con temperatura en en su base.
La solución general para esta aleta e s:
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Sustituyendo el valor de las constantes en la solución general, se tiene:
La ecuación anterior define el perfil de tempertauras a través de toda la longitud de la aleta, tal como se indica en la figura siguiente: El flujo de calor disipado por la aleta puede calcularse a partir de la ley de Fourier por derivación o por la ley de enfriamiento de Newton por integración, del perfil de temperatura. Así;
∫ En este apartado se hará por integración, invitando al estudiante a comprobar este mismo resultado por derivación: Integrando el segundo miembro de la igualdad, se tiene;
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Se exhorta al alumno a realizar r ealizar el análisis dimensional de esta ecuación.
Ejemplo 2.21.1 Una aleta larga de acero tiene una sección transversal de 6.25 mm x 12.5 mm y mantiene uno de sus extremos a 200 °C, mientras que el resto de la barra esta expuesta a un medio ambiente convectivo. El acero tiene una una conductividad conductividad térmica igual a 35 w/m-K, y el 2 coeficiente convectivo de transferencia de calor es 30 w/m -K. Determine el calor perdido perdido por la barra si la temperatura ambiente es de 25 °C. Solución: La barra es larga y de muy pequeño espesor, por tanto el problema puede resolverse a través del caso 1. Así, aplicando la ecuación que que calcula el calor perdido por la barra, se tiene;
√
Tomando en cuenta que la barra es de sección rectangular, el perímetro y el área de la sección serán;
Caso 2. Nos referimos a una aleta finita, donde el calor conducido conducido desde su extremo adherido a la la base; se pierde por convección desde su periferia hacia el fluido que que la rodea. Es decir, la
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Por lo tanto:
Derivando la solución general y sustituyendo la segunda condición.
Despejando
, de la ecuación (1)
Sustiruyendo este valor en la ecuación (2)
Operando algebraicamente y despejando
Así,
, se obtiene:
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Se obtiene finalmente:
+ * Esta expresión define nuevamente el perfil de temperaturas de la aleta considerada en el caso2.
Se sugiere al estudiante estudiante graficar este perfil para observar la variación variación de la
temperatura respecto a la longitud de la aleta. Asimismo, el flujo de calor disipado por la aleta puede calcularse a partir de la ley de Fourier por derivación o por la ley de enfriamiento de Newton por integración, del perfil de temperatura. Así;
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2
convectivo de transferencia de calor es de 15 w/m -°C. Calcúlese el calor perdido por la varilla. Haga una lista de suposiciones básicas. Solución.
Prevalecen condiciones de estado estacionario. Todo el calor disipado por la varilla entra a t ravés de su base Se considera finita a la varilla, puesto que el área en el extremo libre es mucho muy pequeña comparada con el área de su superficie, y por consiguiente la pérdida de calor se efectúa en su periferia. Caso del extremo aislado. aislado. La varilla es de aluminio con una conductividad térmica de 20 5 w/m-°C.
Aplicando la expresión para el cálculo del calor disipado por una aleta del tipo 2, se tiene:
√
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Puesto que:
Donde:
Sustituyendo estos valores en (2);
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Sustituyendo el valor de;
En el primer miembro de la igualdad, se t iene;
Sustituyendo los valores de
y
, en esta expresión; se tiene:
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2.21. Transferencia de calor en aletas rectangulares, triangulares y circunferenciales. En la sección 2.19 se presentaron varios tipos de aletas, de formas y secciones transversales
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Donde el calor máximo transferido, es la energía que la aleta debería transferir si toda ella estuviera a la temperatura de la base. Es decir;
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rendimiento
, para los diversos tipos de aletas antes mencionadas. En las figuras 2.23.1 y
2.23.2 se muestran las graficas para obtener los rendimientos de aletas rectangulares, triangulares y circunferenciales. Note que en la figura 2.23.1, se comparan los rendimientos de una aleta triangular de perfil
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Sustituyendo este valor en la ecuación del flujo de calor real, se obtiene:
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1. Una pared está construida con 2 cm de cobre (k= 385 w/m-°C), w/ m-°C), 3 mm de lámina de asbesto (K= 0.166 w/m-°C) y 6 cm de fibra de vidrio (K=0.038 w/m-°C). Calcúlese el flujo de calor calor por unidad de área para una diferencia de temperaturas total de 500 °C.
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