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Se debe enfriar aceite caliente en un intercambiador de calor en contre flujo de doble
tubo. los tubos interiores de cobre tienen un diámetro de 2 cm y un grusor insignificante.
el diámetro interno del tubo externo (coraza) es de 3 cm. Fluye agua a través del tubo a
una taza de 0.5 kg/s y aceite fluye a traves de la coraza a una tasa de 0.8 kg/s tomando las
temperturas promedio del agua y del aceite como 45 y 80°C respectivamente, determine
el coeficiente global de transferencia de calor de este intercambiador.




Datos:
Agua Aceite
ρ= 990.1 kg/m3 ρ= 852 kg/m3
k= 0.637 w/mic k= 0.138 W/m*k
Pr= 3.91 Pr= 499
V= 6.02E-07 m2/s V= 3.79E-05 m2/s
D0= 0.02 m Di= 0.03 m

𝑚̇ = 0.5 kg/s 𝑚̇ = 0.8 kg/s

A0= 0.00031416 m2 Ai= 0.00070686 m2




1/𝑈=1/ℎ_𝑖 =1/ℎ_0 𝑉=𝑚 ̇/(𝜌∗𝐴)


𝑁_𝑢=ℎ𝐷/𝑘 𝑁_𝑢=0.023𝑅_𝑒^0.8 𝑃𝑟^0.4

Aceite
Agua
V= 1.6074595588 m/s D= Di-D0 = 0.01

Re= 53403.97 V= 2.3910545866 m/s
Interpolación
Nu= 240.25 Re= 1890.66
Di/D0
hi= 7651.86 W/m2K Nu= 0.6666666667
0.5
h0= 63.112 W/m2K 0.66666667
1
U= 62.60 W/M2k
y=
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