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.

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:

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

Substitute the given values

The temperature of heater is then:

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

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

the inner core.

good luck: )

klicker