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Question : A 3-mm-diameter and 5-m-long electric wire is tightly wrappe...

A 3-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·K. 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 a T_{\infty}=30^{\circ} C with a heat transfer coefficient of h=12 W / m ^{2}.K determine 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.

 

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SOLUTION An electric wire is tightly wrapped with a plastic cover. The interface temperature and the effect of doubling the thickness of the plastic cover on the interface temperature are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. 5 Heat transfer coefficient incorporates the radiation effects, if any.

Properties The thermal conductivity of plastic is given to be k = 0.15 W/m·K.

Analysis Heat is generated in the wire and its temperature rises as a result of resistance heating. We assume heat is generated uniformly throughout the wire and is transferred to the surrounding medium in the radial direction. In steady operation, the rate of heat transfer becomes equal to the heat generated within the wire, which is determined to be \dot{Q}=W_{e}= V I=(8 V )(10 A )=80 W

The thermal resistance network for this problem involves a conduction resistance for the plastic cover and a convection resistance for the outer surface in series, as shown in Fig. 3–32. The values of these two resistances are

A_{2}=\left(2 \pi r_{2}\right) L=2 \pi(0.0035 m )(5 m )=0.110 m ^{2}

 

R_{\text {conv }}=\frac{1}{h A_{2}}=\frac{1}{\left(12 W / m ^{2} \cdot K \right)\left(0.110 m ^{2}\right)}=0.76^{\circ} C / W

 

R_{\text {plastic }}=\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi k L}=\frac{\ln (3.5 / 1.5)}{2 \pi(0.15 W / m \cdot K )(5 m )}=0.18^{\circ} C / W

 

and therefore

R_{\text {total }}=R_{\text {plastic }}+R_{\text {canv }}=0.76+0.18=0.94^{\circ} C / W R_{\text {total }}=R_{\text {plastic }}+R_{\text {conv }}=0.76+0.18=0.94^{\circ} C / W

Then the interface temperature can be determined from

\dot{Q}=\frac{T_{1}-T_{\infty}}{R_{\text {total }}} \longrightarrow T_{1} =T_{\infty}+\dot{Q} R_{\text {total }}=30^{\circ} C +(80 W )\left(0.94^{\circ} C / W \right)=105^{\circ} C

Note that we did not involve the electrical wire directly in the thermal resistance network, since the wire involves heat generation. To answer the second part of the question, we need to know the critical radius of insulation of the plastic cover. It is determined from Eq. 3–50 r_{ cr , \text { cylinder}}=\frac{k}{h}  to be

r_{ cr }=\frac{k}{h}=\frac{0.15 W / m \cdot K }{12 W / m ^{2} \cdot K }=0.0125 m =12.5 mm

which is larger than the radius of the plastic cover. Therefore, increasing the thickness of the plastic cover will enhance heat transfer until the outer radius of the cover reaches 12.5 mm. As a result, the rate of heat transfer \dot{Q} will increase when the interface temperature T_{1} is held constant, or T_{1} will decrease when \dot{Q} is held constant, which is the case here.

Discussion It can be shown by repeating the calculations above for a 4-mmthick plastic cover that the interface temperature drops to 90.6°C when the thickness of the plastic cover is doubled. It can also be shown in a similar manner that the interface reaches a minimum temperature of 83°C when the outer radius of the plastic cover equals the critical radius.