Question

A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of 0.4 cm. Assuming heat to be transferred uniformly from all surfaces, determine
(a) the amount of heat this resistor dissipates during a 24-hour period,
(b) the heat flux, and
(c) the fraction of heat dissipated from the top and bottom surface

Answer 1

Answer:

a. 51.84Kj

b. 2808.99 W/m^2

c. 11.75%

Explanation:

Amount of heat this resistor dissipates during a 24-hour period

= amount of power dissipated * time

= 0.6 * 24 = 14.4 Watt hour

(Note 3.6Watt hour = 1Kj )

=14.4*3.6 = 51.84Kj

Heat flux = amount of power dissipated/ surface area

surface area = area of the two circular end  + area of the curve surface

$=2*\frac{\pi D^{2} }{4} + \pi DL\\=2*\frac{\pi *(\frac{0.4}{100} )^{2} }{4} + \pi *\frac{0.4}{100} *\frac{1.5}{100}$

= 2.136 *10^-4 $m^{2}$

Heat flux =$\frac{0.6}{2.136 * 10^{-4} }$ = 2808.99 $W/m^{2}$

fraction of heat dissipated from the top and bottom surface

$=\frac{\frac{2*\pi D^{2} }{4} }{\frac{2*\pi D^{2}}{4} + \pi DL } \\\\=\\\frac{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} }{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} +\pi *\frac{0.4}{100} *\frac{1.5}{100} } \\\\=\frac{2.51*10^{-5} }{2.136*10^{-4} } \\\\\= 0.1175$

=11.75%

Answer 2

Answer: a. = 3.6Wh

b. = 1179W/m2

c. = 171degree celsius

Explanation: using the formula

Q = Q×dt

Q = 0.15×24

Q = 3.6Wh

Solving for b

Heat flux is calculated using the formular As =2raised to the power of ΠD2 ÷4 + ΠDL

As = 2/4ΠD2 + ΠDL

Where D = 0.003m and L = 0.012m

= 2/4Π(0.003)2 + Π(0.003×0.012)

= 0.000127m2

But q = Q/As

= 0.15/0.000127= 1179W/m2

Solving for c, use the formula

Ts = T + Q/h×As

T=40degree celsius

Q/hAs=0.15/9×0.000127=131 degree celsius.

Ts=40+ 131=171degree Celsius