Answer:
Q = 125.538 W
Explanation:
Given data:
D = 30 cm
Temperature [tex]T_\infity = 15[/tex] degree celcius
[tex]T_S = 220 + 273 = 473 K[/tex]
Heat coefficient = 12 W/m^2 K
Efficiency 80% = 0.8
[tex]Q = hA(T_S - T_{\infty}) \eta[/tex]
[tex]= 12(\frac{\pi}{4} 0.3^2) (473 - 288) 0.8[/tex]
Q = 125.538 W