Question

An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of material having a thermal conductivity of 5 W/m · K. Estimate the temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

Answer 1

Answer:

$T_{1}=94.9^{o}C$

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

Temperature=25°C

The temperature of heater surface follows from the rate equation written as:

$T_{1}=T_{2}+\frac{q}{kS}$

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

$S=\frac{2\pi L}{ln(\frac{4L}{D} )}$

Substitute the given values

$S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m$

The temperature of heater is then:

$T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C$

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

                           $T_{1}=94.9^{o}C$