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
[tex]=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}[/tex]
= 2.136 *10^-4 [tex]m^{2}[/tex]
Heat flux =[tex]\frac{0.6}{2.136 * 10^{-4} }[/tex] = 2808.99 [tex]W/m^{2}[/tex]
fraction of heat dissipated from the top and bottom surface
[tex]=\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[/tex]
=11.75%
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
