Question

A fluorescent light bulb, rated at 100W, is illuminated in air at 25°C and atmospheric pressure. Under these conditions the surface temperature of the glass is 140°C. Determine the rate of heat transfer from the bulb by natural convection. The bulb is cylindrical, having a diameter of 35 mm and a length of 0.8m, and is oriented horizontally.

Answer 1

q = 83.39 W

Step-by-step explanation:

given data

temperature t1 = 25°C

temperature t2 = 140°C

solution

we get here T(f) =
`(25+140)/(2)` = 82.5°C

so here rate of hat transfer is

q = h A ΔT ..........1

q =
`(k)/(D) * Nu * \pi D L * \Delta T` .............2

here we get Nu first

for for 82.5°C

`(\beta\ g)/(v^2)` = 0.625 ×
`10^(8)`
`(m^3K)^(-1)`

so here Ra will be = 0.625 ×
`10^(8)` × 0.035³ × (140-25) × 0.696

Ra = 2.14 ×
`10^(5)`

so Nu will be = [ 0.60 +
`(0.387\ Ra^(1/6))/((1+(0.559/Pr)^(9/16))^(8/27))` ]²

here Pr is 0.696 so solve we get Nu = 9.49

put value in equation 2 we get

q = 0.0304 × 9.49 × π × 0.8 × 115

q = 83.39 W