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3 years ago

7

Considere uma estufa usada em um laboratório qualquer. A câmara tem formado de um cubo perfeito com lado igual a 75 cm, desse mo

do deseja-se saber qual a espessura mínima requerida do isolante térmico de espuma de poliestireno (K = 0,025 W / m . K) que deve ser usado na estufa, para garantir que a taxa que entra nele seja inferior a 900 W.
Considere que suas temperaturas são: superfície interna 105 °C e externa 28°C.

1 answer:

levacccp [35]3 years ago

5 0

Responder:

0.0328m

Explicación:

La tasa de transferencia de calor se expresa como se muestra

Q / t = kA∆T / d

Q / t es la tasa de transferencia de calor o la energía consumida

k es la conductividad térmica

A es el área de superficie del cubo

∆T es el cambio de temperatura

d es el grosor Dado que

Q = Potencia × tiempo Potencia = Q / t = 900W

K = 0,025 W / m. K

Área de superficie del cubo A = 6a²

A = 6 (0.75) ²

A = 3.375m²

d es el grosor =?

∆T = T2-T1 ∆T = 105 ° C-28 ° C

∆T = 77 ° C ∆T = 77 + 273 = 350 Kelvin

Al sustituir los valores en la fórmula para obtener el grosor 'd' que tenemos;

900 = 0.025 × 3.375 (350) / día Multiplicación cruzada que tenemos; 900d = 29.53125

d = 29.53125 / 900

d = 0.0328m

El espesor es de 0.0328 m

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