10 Problemas Resueltos
Short Description
FENOMENOS DE TRANSPORTE...
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Fenómenos de Transporte
Problema 1 La pared de un cilindro está compuesta por dos capas de materiales con conductividad kA y kB. Ambos materiales están separados por una resistencia eléctrica muy delgada de muy muy alta alta co cond nduc ucti tivi vida dad. d. Por el inte interi rior or de la tube tuberí ría a cir circula cula un líqu líquid ido o a temperatura temperatura Ti y con un coefciente de película i. !n el e"terior la temperatura y el coefciente de película son respectivamente Te y e. A. #btener la temperatura de la resistencia eléctrica cuando el calor disipado por ésta es nulo. B. #btener la temperatura de la resistencia eléctrica cuando el calor disipado por ésta es q$$c %&'m().
Solución: Datos: *apa A+ kA , *apa B+ k B -esistencia eléctrica muy delgada de alta conductividad que genera+ / ( m ' & qc00 *ondici1n de contorno e"terior+ e, Te *ondici1n de contorno interior+ i, Ti
Incógnitas: A. Tc cuando q$c 2 3 B. Tc cuando q$c 4 3
Desarrollo: A+ •
5tili6ando la analogía eléctrica de conducci1n, podemos e"presar el 7u8o de calor desde la superfcie intermedia acia el interior y acia el e"terior+
Página 1
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Fenómenos de Transporte
T c −T i
qi =
ln 1
hi A
+
r
( c) ri
2 π k b L
T c −T i
qi =
ln 1
hi 2 π r i L
+
r
( c) ri
2 π k b L
T c − T e
q e= 1
he A e
ln
+
r
( e )
c 2 π k A L
T c −T i
qi =
ln 1
hi 2 π r i L
+
r
( c) ri
2 π k b L
B+ •
Para el caso de que e"ista una generaci1n de energía superfcial el balance de energía en esa superfcie sería el siguiente+
qi + qe =q
•
'' c2
π r c L
9 por tanto al despe8ar la temperatura tendríamos+
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Fenómenos de Transporte
T i
''
qc rc +
r
( c)
ln 1
hi r i
T c =
+
1
hi ri
+
1
k B r
he re
+
rc
k A
1
+
( c)
ln
ri
k B
r
( e)
ln
ri
1 ln
T e
+
1
he re
+
r
( e) rc
k A
Problema 2 !l muro de una cámara :rigorífca de conservaci1n de productos p roductos congelados, se constituirá del modo siguiente+ • • • • • •
-evoco de cemento de ; cm de espesor %k 2 3.< kcal'=m>*) 5n pie %;? cm) de ladrillo maci6o %k 2 3.@ kcal'=m>*) Pantalla antivapor de .; cm de espesor %k 2 3. kcal'=m>*) *orco e"pandido %k 2 3.3? kcal'=m>*) C cm de ladrillo ueco %k 2 . kcal'=m>*) -evoco de cemento de ; cm de espesor %k 2 3.< kcal'=m>*)
Diendo la temperatura interior E;?>* y la del e"terior F3>*. Di las pérdidas orarias por unidad de área del muro, se evalGan por motivos econ1micos en 3 kcal'=m(, determinar+ a. !l coefciente global de transmisi1n de calor del muro b. !l espesor de corco que debe colocarse c. La distribuci1n de temperaturas en el muro De tomarán como coefcientes de transmisi1n de calor por convecci1n e"terior e interior ;3 y ; kcal'=m(>*, respectivamente.
Solución:
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Fenómenos de Transporte
Capa 1
2
3
4
5
6
!spesor %cm)
;
;?
.;
"
C
;
*onductividad %kcal'=m>*)
3.<
3.@
3.
3.3?
.
3.<
∫ ¿=25 C o
o
Temperaturas+ Temperaturas+ T ext =30 C H
*oefcientes de película+
hext = 20
T ¿
H
KCal h . m . ° C H 2
' '
Ilu8o de calor por unidad de área+
q
∫ ¿ =12 h .KCal m . ° C h¿
KCal
=10
h.m
Incógnitas: a. *oefciente global de transmisi1n de calor+ 5 b. !spesor de la capa de corco+ e c. Jistribuci1n de temperaturas en el e l muro.
Esuema:
2
2
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Fenómenos de Transporte
Desarrollo: a. *oefciente global de transmisi1n de calor+
∫¿
T ext −T ¿
¿ ∫¿ T ext −T ¿ ¿ ¿ '' q =U ¿ U = 0.182
KCal 2
h . m .° C
b. !spesor de aislante+ 5tili6ando la analogía eléctrica en conducci1n+
∫¿
T ext −T ¿
¿ ¿ ' ' q =¿
∫¿
T ext − T ¿
¿
1
+
e1
hext k 1
+
e2 e3
+
k 2 k 3
+
e4 k 4
¿ q =¿
+
e5
+
e6
k 5 k 6
+
1
h∫ ¿
''
!n la ecuaci1n anterior la Gnica inc1gnita es el espesor de corco+ e 4= 24.03 cm .
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Fenómenos de Transporte
*apa
-esistencia %m(>*='kcal)
E! t"
1
2
3.3 ?
3.3 ;?
3. C
3
4
3.3 .< F
5
6
Int"
3.3 @
3.3 ;?
3.3 * %*oefciente convectivo interior ? &'m(O) *alcular+ a. Ilu8o de calor por unidad de área que atraviesa el muro. b. Temperatura Temperatura en las dos superfcies e"tremas y en la inter:ase inter: ase entre las dos capas
Solución+ Datos:
•
*apa + e 2 3. mH k 2 3.M %N3.33@T) &'m.O *apa ;+ e; 2 3.3? mH k ; 2 3.3 &'m.k
•
' ' *ontorno e"terno q sol =300 ; T ext = 40 ºC;hext = 10 W / / m ² K
•
•
*ontorno interno
∫ ¿=5 W / m ² K ∫ ¿ =20 ºC;h¿ T ¿
nc1gnitas+ • •
a. Ilu8o de calor por unidad de área que atraviesa el muro+ q$$ b. Temperatura de las superfcies+ T , T; , TF
!squema+
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Fenómenos de Transporte
Jesarrollo+ a. Ilu8o de calor por unidad de área que atraviesa el muro+ La ecuaci1n di:erencial en la capa será la siguiente+
(
d dT k ( T ) dx dx
)=
0 k ( T )
dT dT = cteq' ' =−k ( T ) dx dx
!l 7u8o de calor por unidad de área debe ser constante. La conductividad es variables con la temperatura siguiendo una ley lineal del tipo+ k%T) 2 k 3 %NbT). Di integramos la ecuaci1n anterior para toda la capa + e1
T 2
−∫ q dx =∫ dT ' '
0
T 1
−q' ' e =k ( T −T )+ k 1
0
2
1
0
b
b T −T )= k ( T − T ) + k ( T −T ) ( T + T ) ( 2 2 2
2
2
1
0
2
1
0
2
1
2
1
−q' ' e =k ( T −T ) ( 1 + bT med ) 1
0
2
1
−q' ' e =k ( 1 +b T med )( T −T )= k med ( T −T ) 1
0
2
1
2
1
Aora impondremos las dos condiciones de contorno+ x =0 hext ( T ext −T 1 ) + q sol =q ' ' ' '
!sta condici1n de contorno podemos e"presarla como si :uera una condici1n de contorno puramente convectiva contra una temperatura equivalente %Temperatura solEaire) de C3>*.
[(
) ]
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Fenómenos de Transporte
k 2
+
1
e 2 h∫ ¿ T ∫ ¿ T 2−
¿
' '
x =e 1 q
=¿
Tenemos Tenemos pues F ecuaciones con F inc1gnitas inc1gnitas %q$$, T , T;)+
−q' ' e =k 1
0
( ( 1+ b
T 2−T 1 2
)) (
q ' ' ¿ h ext ( T ext −T 1 ) + q sol ' '
k 2 e2
+
T 2 −
T 2 −T 1 )
%)
%;)
1
h∫ ¿ T ∫ ¿
%F)
¿ q =¿ ' '
gualando la ecuaci1n %;) con la %F) e introduciendo la %;) en la %) tenemos ; ecuaciones con ; inc1gnitas+ k 2
+
1
e 2 h∫ ¿ T ∫ ¿ T 2−
¿ ' ' hext ( T ext − T ) + qsol =¿ 1
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Fenómenos de Transporte
1
h∫ ¿ T ∫ ¿
T 3 −
T 3 =25.34 ºC
¿ q =¿ ' '
Di pintamos la distribuci1n de temperaturas será la siguiente+
P&'()E*+ 4: !l parabrisas de un autom1vil autom1vil se desempaRa desempaRa mediante el paso de aire caliente a T i 2 3 S* sobre su superfcie interna. !l coefciente de convecci1n en esta superfcie es i 2 F3 &'m; E SO/. La temperatura del aire e"terior es T in: 2 2 E3 S* y el coefciente de convecci1n es c 2 @? &'m; E SO/.
. *alcular *alcular las temperatura temperaturass de las superfcies superfcies intern interna a y e"terna e"terna del parabrisas parabrisas de vidrio que tiene mm/ de espesor. espesor. %k vidrio%a F33 SO) 2 , &'m E SO/). ;. Jibu8e perfles perfles %en %en :orma cualita cualitativa) tiva) de temperat temperatura ura si el parabrisa parabrisass tuviese+ tuviese+ a) Jobl Joble e vidr vidrio io con con air aire. e.
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Fenómenos de Transporte
Para la trans:erencia de calor a nivel global se tiene que+
q
=
T i − T inf
A
RT
, donde la -esistencia Total Total se calcula como sigue+
RT
=
1
+
hi
1
+
∆ x
hc
k w
!ntonces, RT
q A
=
1
+
hi
=
1 hc
T i − T inf RT
+
∆ x
k w
=
=
[
30 W
1 m2 −
] º K
( 40 − ( −10))º K 2 0,052 m − º K W
[
]
+
4 × 10−3 [ m] + 65 W 2 1,4 W m−º K m −º K 1
[
[
= 961,54 W
m2
]
[
]
Luego, se tiene en las inter:ases de aire en convecci1n+
]
=
[
0,052 m
2
− º K
]
W
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Fenómenos de Transporte
T wi
=
7,95º C
- Externa q q = hc × (T wo − T inf ) ⇒ T wo = T inf − A hc × A
T wo
= T inf +
q hc × A
961,54 W = −10º C +
[
65 W
m2
]
m 2 − º C
T wo
=
4,79º C
%;) Caso (a): Vidrio con aire Ti 2 TQi TQi Jentro del
Iuera del autom1vil TQo TQo
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Fenómenos de Transporte
Ti 2 Jentro del autom1vil
Iuera del
TQi TQi TQo TQo
autom1vil Tin: 2 2 E3S*
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Fenómenos de Transporte
5n alambre eléctrico está frmemente envuelto con una cubierta de plástico. De va a determinar la temperatura de la inter:ase. Dupondremos que+ . La trans:e trans:eren rencia cia de calor es estacion estacionaria aria ya que no ay indica indicaci1 ci1n n del algGn cambio con el tiempo. ;. La trans:erenc trans:erencia ia de calor es unidimen unidimensiona sionall dado que se se tiene simetría simetría térmica térmica con respecto a la línea central y no ay variaci1n en la direcci1n a"ial. F. Las conductivida conductividades des térmica térmicass son constantes constantes.. . La resisten resistencia cia térmica térmica por contacto contacto en en la inter:ase inter:ase es despreci despreciable. able. ?. !n el co coef efci cient ente e de trans: trans:er eren enci cia a de calor calor se incorp incorpora oran n los los e:ect e:ectos os de la radiaci1n, si los ay. Además se conocen las siguientes propiedades+ . La conduct conductividad ividad térmica térmica del plástico plástico es k 2 2 3,? &'mES*/.
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Fenómenos de Transporte
Rconv
=
R plás tico =
1
=
hA2
[
12 W
ln( r 2 r 1 ) 2π kL
1
=
] 0,110[m ] m 2
2
⋅
=
ln( 3,5 1,5) 2π (0,15 W ) ⋅ 5[ m] m
[
]
[
]
0,76 º C W
=
[
]
0,18 º C W
9, por lo tanto, Rtotal = R plá stico + Rconv
-
=
0,76 + 0,18 = 0,94 º C W
!ntonces, se puede determinar la temperatura en la inter:ase a partir de+
Q=
T 1 − T inf Rtotal
→ T 1 = T inf + Q ⋅ Rtotal
T 1 = T inf + Q ⋅ Rtotal
[
]
T 1 = 30º C + (80W ) ⋅ 0,94 º C W
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Fenómenos de Transporte
1" &gimen estacionario" 2" +nálisis e elocia: v r
2
v
2
/
v
z
3" +nálisis e es7uer8os cortantes:
4" 9luio incompresible" 5" r
g 2 g \
cos
g 2 g \
g z
sin\
2
/
\r
6
6 6 ]r \ r \ r
] rr
]
] \\
\
zz
2/
] zr
] z
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Fenómenos de Transporte
Dimplifque las ecuaciones de variaci1n que se muestran a continuaci1n para estudiar el movimiento del 7uido al circular por la rendi8a, despreciando los e:ectos de borde. ndique en el recuadro una relaci1n numerada de las ra6ones por las que se anulan los términos, y anote ba8o cada término tacado el nGmero correspondiente. -ecuadre fnalmente los términos que no se anulan. Admítase régimen estacionario. Doluci1n+
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Fenómenos de Transporte
Problema $ Jibu8e sobre las siguientes gráfcas+ a) !l perfl de velocidad en la rendi8a para la situaci1n representada en el esquema inicial, y b) el perfl de velocidad en el caso de que el 7uido siga cayendo, pero pe ro el cilindro gire en el sentido de las agu8as del relo8. !scriba en ambos casos las variables y límites representados en los e8es.
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Fenómenos de Transporte
%;) !s el transporte convectivo de c.d.m. (uesto que la velocidad permanece constante en la dirección de avance del fluido este término deber"a ser nulo, pero debido al cambio de dirección en θ no no lo es. $)& *uerza de rozamiento rozamiento o transporte transporte viscoso de de c.d.m. +o es nulo ya que que hay #radientes #radientes de velocidad velocidad en dirección r . $& *uerza ejercida ejercida por la presión. presión. -a presión presión es diferente a la entrada entrada y salida de de la rendija, lue#o lue#o este término s" eistir/. $0& *uerza de #ravedad. #ravedad. 1érmino 1érmino no2nulo, no2nulo, puesto que que tiene componentes componentes en las direcciones direcciones r y y θ.
Problema 9
3n l"quido circula entre dos l/minas planas de #ran anchura, separadas una distancia 4 5 5 la velocidad del fluido es muy baja debido a que se mantiene una diferencia de presiones constante y muy peque6a entre sus etremos. -a l/mina de arriba mantiene una temperatura constante, 1 %, y mayor que el fluido a la entrada, 1 7, la l/mina de abajo aporta al fluido una densidad de flujo de calor, q a, constante. a& Suponiendo Suponiendo ré#imen estacionario estacionario y 8 constante, simplificar todo lo que se pueda la ecuación de
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Fenómenos de Transporte
aislarlo eléctricamente. !ntre los e"tremos a"iales del conductor se establece una di:erencia de potencial [, circulando una intensidad eléctrica . a) !n régimen régimen estacionario estacionario,, dibu8ar dibu8ar el perfl de temperatura temperatura segGn segGn la coordenada coordenada radial, desde el centro asta un punto cualquiera en el aceite. b) ndicar ndicar c1mo se calcula calcula el 7u8o 7u8o de calor, calor, X, a partir del del perfl de temperat temperatura, ura, en -3. c) Al aumentar aumentar el radio radio c1mo a:ecta a:ecta al 7u8o 7u8o de calor calor y a la densida densidad d de 7u8o 7u8o de calor, en el aislante. Doluci1n+
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