An electrical cable with 20 mm in diameter and emissivity equal to 0.85 is installed inside a conduit whose inner surface and air in the its interior is at 30oC. The electrical resistivity of the cable, ρe (µΩ.m), is a function of its temperature, given by ρe=a[1+b(T-T0)], where a=0.0171 µΩ.m, b=0.00396 K-1 and T0= 25oC. The natural convection heat transfer coefficient is expressed by the relation h=cD-0.25(T-Tꚙ)0.25where c=1.21 W/(m1.75.K1.25) and D is the cable diameter. Electrical resistance per unit of cable length is R’e=ρe/Ac (Ac is the cross-sectional area). (a) For steady state operating conditions, estimate the maximum current that can be dissipated in the wire so that its temperature does not exceed 65oC;

An electrical cable with 20 mm in diameter and emissivity equal to 0.85 is installed inside a conduit whose inner surface and air in the its interior is at 30oC. The electrical resistivity of the cable, ρe (µΩ.m), is a function of its temperature, given by ρe=a[1+b(T-T0)], where a=0.0171 µΩ.m, b=0.00396 K-1 and T0= 25oC. The natural convection heat transfer coefficient is expressed by the relation h=cD-0.25(T-Tꚙ)0.25where c=1.21 W/(m1.75.K1.25) and D is the cable diameter. Electrical resistance per unit of cable length is R’e=ρe/Ac (Ac is the cross-sectional area). (a) For steady state operating conditions, estimate the maximum current that can be dissipated in the wire so that its temperature does not exceed 65oC;

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter1: Basic Modes Of Heat Transfer

Section: Chapter Questions

Problem 1.65P

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