tarea receso_fen 15 ii
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FENÓMENOS DE TRANSPORTE (23572) – Grupos J1 y L1Tarea – Transferencia de cantidad de movimiento y calor
1. Se tiene el flujo de un fluido newtoniano incompresible entre dos cilindros coaxiales como se muestra en la figura. La
superficies de los cilindros interior y exterior se mantienen a una temperatura y , respectivamente. Se pued
esperar que sea una función exclusiva de .
2. To prevent heat loss from a fluid-filled vessel, it is decided to attach an electrical heater to the outside of the wall, ashown in figure. The passage of a current through the heater material yields a constant rate of energy generation The inside bulk temperature and heat transfer coefficient are and ℎ, respectively. The outside values are and ℎwhere < . Assume that the steady-state temperature is a function of only.
Calculate the heating rate that is just sufficient to eliminate any heat loss from the inside of the container.
3. A flat plate at =0 is in contact with a Newtonian fluid, initially at rest, which occupies the space >0. At =0 the plais suddenly set in motion in the direction at a velocity , and that plate velocity is maintained indefinitely. This problem
may be viewed, for example, as representing the early time (or penetration) phase in the start-up of a Couetviscometer.
A. Show that the non-dimensional formulation of the problem is (analyze the Navier-Stokes equation):
η + 2η
η = 0Where =/ and =/√ 4νt. Moreover is a function of .
B. Solve the ordinary differential equation and determine the fluid velocity as a function of time and position. Fthe mathematical solution, associate the boundary conditions with and .
A medida que el cilindro exterior gira, cada capa de fluido roza con la capa
adyacente. Esta fricción entre capas produce calor; es decir, la energía
mecánica se degrada en energía térmica. Cuando esto sucede, la fuente de
calor por unidad de volumen se conoce como disipación viscosa ; donde
es la función de disipación para fluidos newtonianos, la cual depende de
los gradientes de velocidad presentes en el fluido. Así mismo, no es posible
despreciar los efectos de curvatura.
Considerando que las propiedades fisicoquímicas del fluido no se alteran con
la elevación de la temperatura y permanecen constantes, determine una
expresión para el perfil de temperatura. ¿Qué pasaría si las paredes del
sistema permanecieran a igual temperatura ? Además, determine la
temperatura máxima del sistema en función de los parámetros conocidos.
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C. Plot the velocity profile for different times. Assume that the kinematic viscosity ( ν) is equal to 1 and is vesmall
4. Figure shows an example of a system with a gas-liquid interface of unknown shape, consisting of a liquid in an ope
container of radius that is rotated at an angular velocity . If the container is rotated long enough, a steady state
reached in which the liquid is in rigid-body rotation. It is desired to determine the steady-state interfase height, ℎ(assuming that the ambient air is at a constant pressure, . The viscous stress vanishes for rigid-body rotation. Theffects of surface tension will be neglected.
Show that interface height is (¿Newtonian or non-Newtonian fluid?):
ℎ() = ℎ + (ωR)
2 [
− 12]
Where ℎ≡/(2) is the liquid height under static conditions. is the volume of fluid.
5. Consider the steady-state temperature in a cylindrical wire or radius that is heated by passage of an electric curre
and cooled by convective heat transfer to the surrounding air. The local heating rate is assumed to be depende
of position ( = ). For steady conduction in a solid with such a heat source:
= −
Cylindrical coordinates are the natural choice for this problem. The convection boundary condition at the surface of thwire is the Newton’s law of cooling. For simplicity, assume that ℎ is independent of position. With , , ℎ and (ambient temperature) all assumed to be constant. The temperature field is axisymmetric. It is apparent that all of th
physical conditions can be satisfied by a temperature field which depends only on . Determine the temperature as
function of position.
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Ecuación de energía por conducción
Coordenadas cilíndricas
= − [1
()
+ 1
+ ] +
Coordenadas esféricas
= − 1 () + 1sin ( sin ) + 1sin +
Donde, es la fuente de calor por unidad de volumen.