## Electrical Heater Rod

Consider an electrical heater rod that is 5 mm in diameter by 100 mm long, and is installed normal to the surface of a different material that has a 5 W/m∙K thermal conductivity. If the block’s surface temperature is 30°C, and the rod dissipates 50 W, calculate the heater’s temperature.

##
__
__**Hint**

**Hint**

Heat loss can be expressed as:

$$$q=kS(T_{1}-T_{2})$$$

where
$$k$$
is the thermal conductivity,
$$S$$
is the shape factor, and
$$T$$
is the temperature.

##
__
__**Hint 2**

**Hint 2**

Solve for
$$T_1$$
:

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

Heat loss can be expressed as:

$$$q=kS(T_{1}-T_{2})\rightarrow T_{1}=T_{2}+\frac{q}{kS}$$$

Where
$$k$$
is the thermal conductivity,
$$T$$
is the temperature and
$$S$$
can be estimated from the conduction shape factor given in the above table for a “vertical cylinder in a semi-infinite medium”.

$$$S=\frac{2\pi L}{ln(\frac{4L}{D})}=\frac{2\pi(0.1m)}{ln[\frac{4(0.1m)}{0.005m}]}=\frac{0.628m}{4.38}=0.143m$$$

Finally solving for the heater's temperature:

$$$T_{1}=30^{\circ}C+\frac{50W}{5\frac{W}{m\cdot K}(0.143m)}=100^{\circ}C$$$

100°C