Question

You use a horizontal, 4 mm diameter, 100 mm long heater rod to boil water under atmospheric pressure. The rod material is of emissivity 0.5 and maintains its surface temperature at 455˚C during operation. Estimate the power dissipation of the heater rod (W).

Answer 1

Answer:

0.01 W

Explanation:

diameter d of rod = 4 mm

radius r of the rod = d/2 = 4/2 =0.002 m

length L of rod = 100 mm = 0.1 m

temperature of rod = 455 °C

temperature in kelvin = 455 + 273.3 = 728.3 K

emissivity ε of rod = 0.5

external radiative surface area of rod is calculated as

A = $\pi r^{2} L$ = 3.142 x $0.002^{2}$ x 0.1 = 1.26 x $10^{-6}$ m^2

Power dissipiation P = εσA($T^{4}$)

where σ Stefan's constant = 5.67 x $10^{-8}$ W-$m^{2}$-$K^{4}$

P = 0.5 x 5.67 x $10^{-8}$ x  1.26 x $10^{-6}$ x $728.3^{4}$ = 0.01 W