A 2mm diameter and 10m long electric wire is tightly wrapped with a 1mm thick pl

Question

A 2mm diameter and 10m long electric wire is tightly wrapped with a 1mm thick plastic cover whose thermal conductivity is k = 0.15 W/m.K. Electrical instruments indicate that a current of 10A passes through the wire and there is a voltage drop of 8V along the wire. If the insulated wire is exposed to a medium at T = 30 Degree Celsius, with a heat transfer coefficient of h = 24 W/m².K. Determine:
a) The temperature of the interface of the wire and the plastic cover (assuming steady operation).
b) Determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Explanation / Answer

Given Data

Wire Size - 2 mm diameter and 10m long

Wire Cover - 1 mm thick & Thermal conductivity is k = 0.15 W/m.K.

Current - 10A & Voltage drop - 8V

Medium Temp, T? = 30 Degree Celsius, Medium Heat transfer coefficient h = 24 W/m².K.

To Find Out :

a) Temperature of the interface of the wire and the plastic cover (assuming steady operation).
b) Determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Solution:

Due to electric current passing through the wire , heating of wire core occurs and then this heat will transfer to the plastic cover radially to escape to the surrounding medium

Formula for Heat generated = Voltage X Current = 8 Volt X 10 Ampere = 80 Watts

Heat passes through plastic cover and then to outer medium in SERIES as shown in the diagram

Now we find the thermal resistance at each interface using formulae below

A 2mm diameter and 10m long electric wire is tightly wrapped with a 1mm thick plastic cover whose thermal conductivity is k = 0.15 W/m.K. Electrical instruments indicate that a current of 10A passes through the wire and there is a voltage drop of 8V along the wire. If the insulated wire is exposed to a medium at T? = 30 Degree Celsius, with a heat transfer coefficient of h = 24 W/m².K. Determine:
a) The temperature of the interface of the wire and the plastic cover (assuming steady operation).
b) Determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Given Data

Wire Size - 2 mm diameter and 10m long

Wire Cover - 1 mm thick & Thermal conductivity is k = 0.15 W/m.K.

Current - 10A & Voltage drop - 8V

Medium Temp, T? = 30 Degree Celsius, Medium Heat transfer coefficient h = 24 W/m².K.

To Find Out :

a) Temperature of the interface of the wire and the plastic cover (assuming steady operation).

Solution:

Due to electric current passing through the wire, heating of wire core occurs and then this heat will transfer to the plastic cover radially to escape to the surrounding medium

Formula for Heat generated = Voltage X Current = 8 Volt X 10 Ampere = 80 Watts

Medium T? = 30 C

h = 24 W/m².K.

Plastic Cover

R2 = 2mm

k = 0.15 W/m.K.

Heat passes through plastic cover and then to outer medium in SERIES as shown in the diagram

Wire

R1 = 1mm, L = 10 m

Now we find the thermal resistance at each interface using formulae below

Resistance for plastic = Ln ( R2/R1)

= 0.07358 oC/W

Now we calculate the resistance for heat transfer between plastic cover to medium through convection

= 0.3317 oC/W

For SERIES thermal resistances we find total resistance by

Rtotal = Rplastic + Rmedium = 0.07358 + 0.3317 = 0.4053 oC/W

Now

Heat Transfer Rate = Q =

Hence Tin = Tout + Q x Rtotal

                  = 30 + (80 x 0.4053 oC/W)

                  = 30 + 32.424

              Temperature of interface    = 62.424 oC

To find

b) Determine if doubling the thickness of the plastic cover will increase or decrease this

interface temperature.

Critical Radius of insulation of plastic cover = Rcritical = k/h = 0.15/ 24 = 0.00625 m

                                                                                                                       = 6.25 mm

As this radius is more than current radius of plastic cover we can conclude that

By doubling the thickness new radius of 3 mm will increase the heat transfer until radius reaches 6.25 mm size

Therefore interface temperature will drop down as heat flow is constant

Medium T? = 30 C

h = 24 W/m².K.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.