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

Answer:

Steady State Temperature is $442.14^{\circ}C$

Solution:

As per the question:

Diameter, d = 0.794 mm = $0.794\times 10^{- 3} m$

Radius, r = $\frac{0.794\times 10^{- 3}}{2}$

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$W/mK$

$B_{i} = \frac{h\frac{r}{2}}{K}$

$B_{i} = \frac{550\frac{0.794\times 10^{- 3}}{4}}{229}$

$B_{i} = 4.767\times 10^{- 4}$

$B_{i}$ < 0.1

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

$T - T_{\infty} = \frac{I^{2}R}{\pi dh}$

$T = 25^{\circ} + \frac{100^{2}\times 0.0572}{\pi \times 0.794\times 10^{- 3}\times 550} = 442.14^{\circ}C$