Problemas de transferencia de calor y masa
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CONDUCCIÓN DE CALOR EN ESTADO ESTACIONARIO EN PAREDES PLANAS 1. Considere una pared de ladrillo ladrillo de 4 m de de alto, 6 m de ancho y 0.3 m de de espesor cuya conductividad térmica es k=0.8 W/m.°C. En cierto día, se miden las temperaturas de la superficie interior y exterior de la pared y resulta ser de 14°C y 6°C respectivamente. Determine la velocidad de la perdida de calor a través de la pared en ese día.
Q=
−
Rcond =
A=4m x 6 m =24m =24m 2 Rcond =
Q=
. = ./° ²
– ° =
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2. Considere una casa de ladrillos calentada eléctricamente (k=0.40 Btu/h.pie.°F), cuyas paredes tienen 9 pies de alto y 1 pie de espesor. Dos de las paredes tienen 40 pies de largo y las otras tienen 30 pies. La casa se mantiene a 70°F en todo momento, en tanto que la temperatura del exterior varía. En cierto día se mide la temperatura de la superficie interior de las paredes y resulta ser de 55°F, en tanto que se observa que la temperatura promedio de la superficie exterior permanece en 45°F durante el día por 10 h, y en 35°F en la noche por 14 h. Determine la cantidad de calor perdido por la casa ese día. También determine el costo de esa pérdida de calor para el propietario, si el precio de la electricidad es de 0.09 dólar/kWh.
Hipótesis La transferencia de calor a través de las paredes es constante desde las temperaturas de la superficie de las paredes se mantienen constantes a los valores especificados durante el período de tiempo tie mpo considerado.
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Propiedades La conductividad térmica de la pared de ladrillo se da a ser k = 0,40 Btu / h.pie. h.pie. ° F .
Análisis Consideramos que la pérdida de calor a través de sólo las paredes.
El área total de transferencia de calor es
A=2(50x9+35x9)=1530ft A=2(50x9+35x9)=1530ft 2
La tasa de pérdida de calor calor durante el día es
Q=k . A
Q = (0.40Btu/h.pie.°F)(1530pie 2 )
−
− = 6120 Btu/h
La tasa de pérdida de calor calor durante la noche es
Qday=k . A
Qday= (0.40Btu/h.pie.°F)(1530pie (0.40Btu/h.pie.°F)(1530pie 2 )
−
− = 12.240 Btu/h
La cantidad de pérdida de calor de la casa esa noche será
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Q =Qnigh Δt=10Qday+ 14Qnigh = 10h.6120Btu/h + 14h.12.240Btu/h= 232.560Btu 232.560Btu
A continuación, el coste de esta pérdida de de calor para que se convierte en día
Costo = (232.560/3412kWh)($0.09/kWh) (232.560/3412kWh)($0.09/kWh) = $6.13
3. Considere una persona parada en un cuarto a 20°C con un área superficial expuesta de 1.7 m2. La temperatura en la profundidad del organismo del cuerpo humano es 37°C y la conductividad térmica de los tejidos cercanos a la piel es alrededor de 0.3 W/m.°C. El cuerpo está perdiendo a razón de 150 W, por conducción natural y radiación hacia los alrededores. Se toma como 37°C la temperatura del cuerpo a 0.5 cm por debajo de la piel, determine la temperatura te mperatura de la epidermis de la persona. Hipótesis
Existen 1 condiciones de funcionamiento estable.
El calor coeficiente de transferencia es constante y uniforme sobre toda la expuesta superficie de la persona.
Las superficies circundantes están a la misma temperatura que que la temperatura del aire interior.
Generación de calor dentro de la capa externa gruesa 0,5 - cm del tejido es insignificante
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Análisis La temperatura de la piel se puede determinar directamente directamente a partir de
Q=k . A
−
T 2 = T 1-
. =35.5°C . ° .
T 2 = 37°C –
4. Está hirviendo agua en una cacerola de aluminio (k=237 W/m · °C) de 25 cm de diámetro, a 95°C. El calor se transfiere de manera estacionaria hacia el agua hirviendo que está en la cacerola a través del fondo plano de ésta de 0.5 cm de espesor, a razón de 800 W. Si la temperatura de la superficie interior del fondo es de 108°C, determine. a) El coeficiente de transferencia de calor de ebullición sobre esa superficie interior. b) La temperatura de la superficie exterior del fondo.
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Propiedades La conductividad térmica de la bandeja bandeja de aluminio se da a ser k = 237 237 W / m.°C.
Análisis a) El coeficiente de transferencia de calor de ebullición es
A =
A =
²
.² = 0.0491m2 Qconv=h . A s (T s - T ͚ )
h=
H =
.−° 1254W/m2 °C
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Q=k.A
T s = T interior interior +
−
= 108°C + . = 108.3°C /.° .
5. Se construye una pared pared de dos capas de tablaroca (k 0.10 Btu/h · ft · °F) de 0.5 in de espesor, la cual es un tablero hecho con dos capas de papel grueso separadas por una capa de yeso, colocadas con 7 in de separación s eparación entre ellas. El espacio entre los tableros de tablaroca está lleno con aislamiento de fibra de vidrio (k 0.020 Btu/h · ft · °F). Determine a) La resistencia térmica de la pared. b) El valor R del aislamiento en unidades inglesas.
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Las conductividades térmicas se les da a k sheetrock = = 0.10 Btu / h ⋅ ft ⋅ ⋅ ° ° F y k aislamiento aislamiento = 0,020 Btu / h⋅ ft ⋅ ⋅ ° ° F.
Análisis No se le da la superficie de la pared y por lo tanto consideramos una superficie por unidad (A = 1 m2 ) . Entonces, el valor R de aislamiento de la pared se vuelve equivalente a su resistencia termal , que se determina a partir de .
./ =0.583pie2.°F.h /Btu ./..° ./ =29.17 pie2.°F.h /Btu ./..°
R sheetrock = R1 = R3 = R fibra de vidrio = R2 =
Rtotal = 2R1 + R2 =2 x 0.583 + 29.17 =30.34pie 2.h.°F/Btu.
6. El techo de una casa consta de una losa de concreto (k =2 W/m · °C) de 3 cm de espesor, que tiene 15 m de ancho y 20 m de largo. Los coeficientes de transferencia de calor por convección sobre las superficies interior y exterior del techo son 5 y 12 W/m2 · °C, respectivamente. En una noche clara de invierno, se informa que el aire ambiente está a 10°C, en tanto que la temperatura nocturna del cielo es de 100 K. La casa y las superficies interiores de la pared se mantienen a una temperatura constante de 20°C. La emisividad de las dos superficies del techo de concreto es 0.9. Si se consideran las transferencias de calor tanto por radiación como por convección, determine la
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Propiedades La conductividad térmica del hormigón se da a ser k = 2 W / m⋅ ° ° C. La emisividad de ambos superficies de la azotea se da para ser 0.9
Análisis Cuando la temperatura de la superficie circundante es diferente que la temperatura ambiente , la red de resistencias térmicas enfoque se vuelve muy complicada en problemas que implican la radiación. radiación. Por lo tanto, voy a utilizar un enfoque diferente pero intuitivo. En funcionamiento constante, transferencia de calor desde la habitación a la techo ( por convección y radiación ) debe ser igual al calor transferir desde el techo hasta el entorno ( por convección y radiación) , que debe ser igual a la transferencia de calor a través del
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horno es de 30°C y el coeficiente combinado de transferencia de calor por convección y radiación es de 10 W/m 2. °C. Se propone aislar esta sección de pared del horno con aislamiento de lana de vidrio (k = 0.038 W/m. °C) con el fin de reducir la perdida de calor es 90%. Si se supone que la temperatura de la superficie exterior de la sección metálica todavía permanece alrededor de 80°C, determine el espesor del aislamiento que necesita usarse. El horno opera en forma continua y tiene una eficiencia de 785. El precio del agua natural es de 0.55 dólar/ therm ( 1therm = 105 kJ de contenido de energía). Si la instalación del aislamiento costara 250 dólares por los materiales y la mano de obra, determine cuanto tiempo tardara el aislamiento en pagarse por la energía que ahorra.
Solución:
A = 2m x 1.5m = 3m 2
R.T.: Para el Sistema de analisis analisis
L=? Hallando “q” para el 1 er caso:
q = h x A (T 2 – T T 2*) 2
RT =R =Rcond + + Rconv 2
q = 10W/m °C x 3m (80°C – 30°C) 30°C) q= 1500 W *
Como la velocidad de transferencia de calor es constante constante se cumple que:
∗ 1 ℎ
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150 50° 1 0.114 30 ° ° 1 50° 150 0.114 30 ° ° 50° 1 0.114 ° 150 30 ° ° 0.114 0. 3 ° 0.0342 b)Precio de H 2O natural= 0.55 dólar/therm x 1therm/105500 1therm/ 105500 kJ =5.21x10 -6 *Debo pagar 250 dólares, entonces la cantidad de calor que debo alcanzar para los 250 dólares es:
5.21250 10 47984644.91 − q= 1500W (0.90) q=1350W ahorro ---------------> 100 % Pero tiene una eficiencia de 78,5%
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t=12577.5589 horas= 524 dias
REDES GENERALIZADAS GENERALIZADAS DE RESISTENCIA RESISTENCIA TÉRMICA 1. Una pared de 4m de alto y 6m de ancho consiste de ladrillos con una sección transversal de 18 cm por 30 cm (K = 0.72 W/m °C) separados por capas de mezcla (K = 0.22 W/m °C) de 3 cm de espesor. También se tienen capas de mezcla de 2 cm de espesor sobre cada lado de la pared y una espuma rígida (K = 0.026 W/m2 °C) de 2 cm de espesor sobre el lado inferior de la misma. Las temperaturas en el interior y el exterior son de 22 °C y -4°C y los coeficientes de transferencia de calor por convección sobre los lados interior y exterior son h1 = 10 W/m2 °C y h2 = 20 W/m 2 °C, respectivamente. Si se supone una transferencia unidimensional de calor se descarta la radiación, determine la velocidad de transferencia de calor a través de la pared. Solución:
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T 1
h2 T 2
h1
T 3 T 4
T 5
∗ = -4°C RConV 1
RCon1
.
+
.
RCon2 RCon3 RConV 2 RConV 2
+
.
+
.
.
+
.
+
+
.
+
.
Anotamos la formula. Rtotal = =
.
+
.
+
.
+
.
Reemplazando a Rtotal Rtotal = 0.303 °C/w + 2.331 °C/w + 0.2755 °C/w + 0.8333 °C/w + 0.8333 °C/w + 0.1515 °C/w Rtotal = 4.145 °C/w
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Qtotal = 456.2 w
2. Una pared de 12 m de largo y 5 m de alto está constituida de dos capas de tabla roca (K = 0.17 W/m. °C) de 1 cm de espesor, espaciados 12 cm por montantes de madera (K = 0.11 W/m °C) cuya sección transversal es de 12 cm por 5cm. Los montantes están colocados verticalmente y separados 60 cm, y el espaciado entre ellos está lleno con aislamiento de fibra de vidrio (K = 0.034 W/m °C). La casa se mantiene a 20 °C y la temperatura ambiental en el exterior es de -5°C. Si se toma los coeficientes de transferencia de calor en las superficies interior y exterior de la casa como 8.3 y 3.4 W/m2 °C, respectivamente, determine. a) La resistencia térmica de la pared, si se considera una sección representativa de
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Anotando la gráfica
∗ = 20°C T 1
h2 T 2
h1
T 3 T 4
∗ = -5°C RConV 1
RCon1
RCon2
RCon3 RConV 2
RConV 2
.
+
.
+
Anotamos la formula. 1
1
.
+
.
+
.
+
.
T 5
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3. Se va construir una pared de 10 in de espesor, 30 pies de largo y 10 pies de alto, usando ladrillos solidos (K= 0.40BTU/h.pies.°F) con una sección transversal de 7 pulg. Por 7 pulg. ; o bien , ladrillos de idéntico tamaño con nueve orificios cuadrados llenos d aire (K= 0.015BTU/h.pies.°F) que tienen 9 pulg. De largo y una sección transversal de 1.5 pulg. Se tiene una capa de mezcla (K= (K= 0.10BTU/h.pies.°F) 0.10BTU/h.pies.°F) de 0.5 pulg de espesor entre dos ladrillos adyacentes, sobre los cuatro lados y sobre los dos de la pared. La casa se s e mantiene a 80°F y la temperatura ambiental en el exterior es de 30 °F. Si los coeficientes transferencia de calor en las superficies interior y exterior de la pared son 1.5 y 4 BTU/h.pie2.°F respectivamente. Determine la velocidad transferencia de calor a través de la pared construida de. a) Ladrillos a) Ladrillos sólidos y b) Ladrillo b) Ladrillo con orificios llenos de aire.
Solución:
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Datos: Sistema: Britanico
0.3906 A =7. 5 ×0. 5 3. 7 5 × 0.0260 A =7×0.53.5 × 0.02431 A =7×749 × 0.3403 A =91. 5 × ×1.55× 0.1406 A =49 × 0.14060.1997 A1= 2 3 4
espacios
ladrillos
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∗ ∗ ∗ ∗ ° ∗ ℎ∗ 1× 1.5 ×0.1 3906 1.7068 ×° 4166666667 1.0667 ° × 0.100.0 ×0. 3 906 ×° ° × 0.10 0.75×0.260 288.4615 ×°
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Si:
5.1054 0.3906 6 300 ×5.1054 300 0.3906 3921.1981
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3.817 13.80°30° ° 0992 Si:
3.817 0.3906 6 300 ×3.817 300 0.3906 2931.6436 4. Considere una pared de 5 m de de alto, 8 m de largo y 0.22 m de espesor cuya sección
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Á Á 80,04 0,032 Á Á 80,06 0,48
Calculando la velocidad de transferencia de calor desde la pared interior hasta la
pader exterior:
−
q
Entonces:
2 0,0,0196 5,20810− °C
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4,1664910− ° Entonces el es: 5,20810− ° 0.0325521 ° 4,1664910− ° 0.073 °C
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100 0.031252739,726027 185,6164393°
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0.13 ° 0.1 1.1 0.026 ° 1.5
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33° 0.0371 ° 889,4879 b) espesor de una tela de lana (k= 0.035W/m.°C) si la persona debe lograr el mismo nivel de comodidad térmica usando un grueso abrigo de lana en lugar de una chaqueta para esquiar de cinco capas.
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0.201×10− 0.03320.0020 0.03320.0020
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(0.02
)
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2.
Seis transistores de potencia idénticos con caja de aluminio están sujetos a uno
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ℎ 49.000
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