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Una resistencia eléctrica de 100 W consiste en un alambre de 2 metros de longitud y 2 mm de diámetro.
Determine la temperatura de la superficie del alambre cuando fluye perpendicularmente a él, aire a 1
atm y 20 °C a una velocidad de 3 m/s.
v = 3 m/s
Gráfico pg. 207 Holman
Si T film = 30 °C; para el aire ρ = 1.17 kg/m3
; μ = 1.85 x 10-5
; k = 0.026 W/mK
𝑅𝑒 =
𝐷𝑣𝜌
𝜇
=
. 002(3)1.17
1.85𝑥10−5
= 379
En gráfica, a este valor corresponde un Nu = 9.5 (aprox.)
𝑁𝑢 =
ℎ𝐷
𝑘
=
ℎ(.002)
. 026
; ∴ ℎ = 124
𝑊
𝑚2𝐾
𝑞 = ℎ𝐴∆𝑇 = 100𝑊 = 124𝜋(2)(. 002)(𝑇𝑠𝑢𝑝 − 20); ∴ 𝑇𝑠𝑢𝑝 = 84°𝐶
𝑇𝑓𝑖𝑙𝑚 =
20 + 84
2
= 52°𝐶
2ª iteración, si T film = 52 °C; para el aire ρ = 1.09 kg/m3
; μ = 1.965 x 10-5
; k = 0.02735 W/mK
Re = 333; se lee aproximadamente Nu = 9;
𝑁𝑢 =
ℎ𝐷
𝑘
=
ℎ(.002)
. 02735
; ∴ ℎ = 123
𝑊
𝑚2𝑘
𝑞 = ℎ𝐴∆𝑇 = 100𝑊 = 123𝜋(2)(. 002)(𝑇𝑠𝑢𝑝 − 20); ∴ 𝑇𝑠𝑢𝑝 = 84°𝐶
𝑇𝑓𝑖𝑙𝑚 =
20+84
2
= 52°𝐶 concluye iteración y la respuesta es Tsup = 84 °C
b) Determine el espesor de la subcapa viscosa de aire que controla la transferencia de calor en el
sistema
𝑞 = 100𝑊 =
2𝜋𝑘𝐿∆𝑇
𝑙𝑛
𝑟2
𝑟1
⁄
=
2𝜋(.028)(2)(84 − 20)
𝑙𝑛
𝑟2
. 001
⁄
r2 = .00125 m por lo tanto el espesor es de .00025 m

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Cálculo de h convectiva cuando tfilm es desconocida

  • 1. Una resistencia eléctrica de 100 W consiste en un alambre de 2 metros de longitud y 2 mm de diámetro. Determine la temperatura de la superficie del alambre cuando fluye perpendicularmente a él, aire a 1 atm y 20 °C a una velocidad de 3 m/s. v = 3 m/s Gráfico pg. 207 Holman Si T film = 30 °C; para el aire ρ = 1.17 kg/m3 ; μ = 1.85 x 10-5 ; k = 0.026 W/mK 𝑅𝑒 = 𝐷𝑣𝜌 𝜇 = . 002(3)1.17 1.85𝑥10−5 = 379 En gráfica, a este valor corresponde un Nu = 9.5 (aprox.) 𝑁𝑢 = ℎ𝐷 𝑘 = ℎ(.002) . 026 ; ∴ ℎ = 124 𝑊 𝑚2𝐾 𝑞 = ℎ𝐴∆𝑇 = 100𝑊 = 124𝜋(2)(. 002)(𝑇𝑠𝑢𝑝 − 20); ∴ 𝑇𝑠𝑢𝑝 = 84°𝐶 𝑇𝑓𝑖𝑙𝑚 = 20 + 84 2 = 52°𝐶
  • 2. 2ª iteración, si T film = 52 °C; para el aire ρ = 1.09 kg/m3 ; μ = 1.965 x 10-5 ; k = 0.02735 W/mK Re = 333; se lee aproximadamente Nu = 9; 𝑁𝑢 = ℎ𝐷 𝑘 = ℎ(.002) . 02735 ; ∴ ℎ = 123 𝑊 𝑚2𝑘 𝑞 = ℎ𝐴∆𝑇 = 100𝑊 = 123𝜋(2)(. 002)(𝑇𝑠𝑢𝑝 − 20); ∴ 𝑇𝑠𝑢𝑝 = 84°𝐶 𝑇𝑓𝑖𝑙𝑚 = 20+84 2 = 52°𝐶 concluye iteración y la respuesta es Tsup = 84 °C b) Determine el espesor de la subcapa viscosa de aire que controla la transferencia de calor en el sistema 𝑞 = 100𝑊 = 2𝜋𝑘𝐿∆𝑇 𝑙𝑛 𝑟2 𝑟1 ⁄ = 2𝜋(.028)(2)(84 − 20) 𝑙𝑛 𝑟2 . 001 ⁄ r2 = .00125 m por lo tanto el espesor es de .00025 m