Question

A 2.2-mm-diameter and 10-m-long electric wire is tightly wrapped with a 1-mm-thick plastic cover whose thermal conductivity is k = 0.15 W/m.K. Electrical measurements indicate that a current of 13 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 = 24 W/m2.K, determine (a) the temperature (in °C) at the interface of the wire and the plastic cover in steady operation. Electrical wire To = 30°C Insulation 10 m (c). Continue the previous questions. Determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature. will increase interface temperature because of heat transfer from interface decrease will decrease interface temperature because of heat transfer towards interface decrease will decrease interface temperature because of heat transfer from interface increase will increase interface temperature because of heat transfer towards interface increase will have no effect on interface temperature

Answer 1

Doubling the thickness of the plastic cover increases the rate of heat loss and decreases the interface temperature.

Given:

The diameter of the electric wire t=1mm

The diameter of plastic cover ro=1+1.1=2.1

Internal conductivity of plastic cover k=0.15w/MK

the current passed through circle I=13A

voltage drop v=8v

temperature T=30c

Heat transfer fluid h=24w/m^2k

Let To be the temperature of the wire and T1 is the temperature of the plastic cover.

the heat generated  Q=>v*I=13*8=104 watt

Now the heat generated is connected and converted through wire

Q= To-T1/ln(ro/rc)/2pi*KL+1/hAo

104=To-30/ln(2.2/1.1)/2pi*0.15*10+1/24*20*0.00021*16

104=To-30/0.38439

To=69.9768c

To=70c

The temperature at the Interface of the wire is 70c

Now, the heat generated is connected to the wire

Q=To-T1/ln(ro/rc)/2piKL

104=70-T1/ln(2.1/1.1)/2pi*0.15*10

7.13536=10-T1

T1=6.2864c

The critical radius of Isolation:

rc=k/h=0.5/24

rc=6.25*10^-3m

rc=6.25mm

Now the thickness of isolation is increased to double

ro=1.1+2=3.1mm

Now,rc>ro ie critical radius of isolation is greater than the outer radius of the circle.

∴doubling the thickness of the plastic cover increases the rate of heat loss and decreases the interface temperature.

To know more about the critical radius of Isolation:

https://brainly.com/question/13646773

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