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nickbone59
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Hey, have been working on this question for about 2 weeks now and can't crack it, I found a formula on the net which gives me a value of just over 30C, i am not confident i have the correct value though. It just seems too low. I anyone could point me in the right direction with this it would be greatly appreciated.
Q1)
A high conductivity electric wire, diameter 3 mm and length 10m, is tightly wrapped with a plastic cover of thickness 1mm and thermal conductivity 0.15W/(mK). Electrical measurements indicate that as a current of 10A passes through the wire, there is a voltage drop 6V along the wire. The insulated wire is exposed to ambient conditions at a temperature of 30C with a surface convective heat transfer coefficent of 12W/(m^2K).
Neglecting any temperature variation across the wire cross-section,
i)calculate the temperature at the interface of the wire and the plastic cover under steady state operation.
ii)measurements indicate that the rate of heat loss from the wire increases rather than decreases as the insulation thickness increases. an the measurements be correct. Explain.
Heat generation in a resistive wire
Per unit volume : I^2R/Pi r0^2 L
= 100x0.442/pi x1.5^(2)x10^-3 x 10
=44.2/70.685x10^-6
=625446.7909
Surface Temperature:
To = T infinity + (g ro^2)/4k
= 303 + 62544.7909 x ((3x10^-3)^2/4x0.15)
= 303.938k
= 30.94c
Q1)
A high conductivity electric wire, diameter 3 mm and length 10m, is tightly wrapped with a plastic cover of thickness 1mm and thermal conductivity 0.15W/(mK). Electrical measurements indicate that as a current of 10A passes through the wire, there is a voltage drop 6V along the wire. The insulated wire is exposed to ambient conditions at a temperature of 30C with a surface convective heat transfer coefficent of 12W/(m^2K).
Neglecting any temperature variation across the wire cross-section,
i)calculate the temperature at the interface of the wire and the plastic cover under steady state operation.
ii)measurements indicate that the rate of heat loss from the wire increases rather than decreases as the insulation thickness increases. an the measurements be correct. Explain.
Homework Equations
The Attempt at a Solution
Heat generation in a resistive wire
Per unit volume : I^2R/Pi r0^2 L
= 100x0.442/pi x1.5^(2)x10^-3 x 10
=44.2/70.685x10^-6
=625446.7909
Surface Temperature:
To = T infinity + (g ro^2)/4k
= 303 + 62544.7909 x ((3x10^-3)^2/4x0.15)
= 303.938k
= 30.94c