Answer:
q = 83.39 W
Explanation:
given data
temperature t1 = 25°C
temperature t2 = 140°C
solution
we get here T(f) = [tex]\frac{25+140}{2}[/tex] = 82.5°C
so here rate of hat transfer is
q = h A ΔT ..........1
q = [tex]\frac{k}{D} * Nu * \pi D L * \Delta T[/tex] .............2
here we get Nu first
for for 82.5°C
[tex]\frac{\beta\ g}{v^2}[/tex] = 0.625 ×[tex]10^{8}[/tex] [tex](m^3K)^{-1}[/tex]
so here Ra will be = 0.625 ×[tex]10^{8}[/tex] × 0.035³ × (140-25) × 0.696
Ra = 2.14 × [tex]10^{5}[/tex]
so Nu will be = [ 0.60 + [tex]\frac{0.387\ Ra^{1/6}}{(1+(0.559/Pr)^{9/16})^{8/27}}[/tex] ]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
