Example 4/ A 17-mm-diameter and 5-m-long electric wire is tightly wrapped with a 2-mm- thick plastic cover whose thermal conductivity is k = 0.15 W/m . °C. Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a medium at T = 30°C with a heat transfer coefficient of h = 12 W/m² . °C, de- termine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine whether doubling the thickness of the plastic

Elements Of Electromagnetics
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I have a problem similar to this in my case the problem is from the outside in. the heat is entering the interior and I don't know what temperature it will have inside since part of the heat will be lost in the thickness of the cylinder, I want to know how to fix the formula so that I can obtain the temperature inside the cylinder.

 

 

Example 4/
A 17-mm-diameter and 5-m-long electric wire is tightly wrapped with a 2-mm-
thick plastic cover whose thermal conductivity is k = 0.15 W/m °C. Electrical
measurements indicate that a current of 10 A passes through the wire and there
is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a
medium at T = 30°C with a heat transfer coefficient of h = 12 W/m² . °C, de-
termine the temperature at the interface of the wire and the plastic cover in
steady operation. Also determine whether doubling the thickness of the plastic
cover will increase or decrease this interface temperature.
k
[₁
h
Figure 2-18 Schematic for Example 4
Solution:
T₁
T₂
Too
wwwww
Rplastic
conv
Q=W= VI = (8 V) (10 A) = 80 W
(27₂) L = 2m (0.0035 m) (5 m) = 0.110 m²
1
0.76°C/W
hA₂ (12 W/m² °C) (0.110 m²)
In(r₂/r₁)
In(3.5/1.5)
2TkL 27 (0.15 W/m °C) (5 m)
=
Rplastic + Rconv 0.76 +0.18 0.94°C/W
and therefore
Az
Rcony
Rplastic
Rotal
0.18°C/W
Transcribed Image Text:Example 4/ A 17-mm-diameter and 5-m-long electric wire is tightly wrapped with a 2-mm- thick plastic cover whose thermal conductivity is k = 0.15 W/m °C. Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a medium at T = 30°C with a heat transfer coefficient of h = 12 W/m² . °C, de- termine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature. k [₁ h Figure 2-18 Schematic for Example 4 Solution: T₁ T₂ Too wwwww Rplastic conv Q=W= VI = (8 V) (10 A) = 80 W (27₂) L = 2m (0.0035 m) (5 m) = 0.110 m² 1 0.76°C/W hA₂ (12 W/m² °C) (0.110 m²) In(r₂/r₁) In(3.5/1.5) 2TkL 27 (0.15 W/m °C) (5 m) = Rplastic + Rconv 0.76 +0.18 0.94°C/W and therefore Az Rcony Rplastic Rotal 0.18°C/W
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