Question

Aluminum wire, having a diameter of 0.794 mm, is immersed in an oil bath that is at 25°C. Aluminum wire of this size has an electrical resistance of 0.0572W/m. For conditions where an electric current of 100 A is flowing through the wire and the surface coefficient between the wire and oil bath is 550 W/m2.K, determine the steady-state temperature of the wire.

Answer 1

Steady State Temperature is
`442.14^(\circ)C`

Solution:

As per the question:

Diameter, d = 0.794 mm =
`0.794* 10^(- 3) m`

Radius, r =
`(0.794* 10^(- 3))/(2)`

Temperature,
`T_(\infty) = 25^(\circ)C` = 298 K

Electrical resistance of the wire, R = 0.0572
`\Omega/m`

Current, I = 100 A

Surface coefficient, h =
`550 W/m^(2)K`

Now,

To calculate the steady-state temperature:

The value of the thermal coefficient of Aluiminium, from the table of physical props of solid, K = 229
`W/mK`

`B_(i) = (h(r)/(2))/(K)`

`B_(i) = (550(0.794* 10^(- 3))/(4))/(229)`

`B_(i) = 4.767* 10^(- 4)`

`B_(i)` < 0.1

Thus by using lumped capacitance method for the steady steady temperature without radiation:

`T - T_(\infty) = (I^(2)R)/(\pi dh)`

`T = 25^(\circ) + (100^(2)* 0.0572)/(\pi * 0.794* 10^(- 3)* 550) = 442.14^(\circ)C`