Aheated long cylindrical rod is placed in a cross flow of air at 20°c (1 atm) with velocity of 10 m/s. the rod has a diameter of 5 mm and its surface has an emissivity of 0.95. if the surrounding temperature is 20°c and the heat flux dissipated from the rod is 17000 w/m2, determine the surface temperature of the rod. evaluate the air properties at 70°c. the properties of air (1 atm) at 70°c are k

Ts = 413.66 K

Explanation:

given data

temperature = 20°C

velocity = 10 m/s

diameter = 5 mm

surface emissivity = 0.95

surrounding temperature = 20°C

heat flux dissipated = 17000 W/m²

to find out

surface temperature

solution

we know that here properties of air at 70°C

k = 0.02881 W/m.K

v = 1.995 × m²/s

Pr = 0.7177

we find here reynolds no for air flow that is

Re =    

Re =

Re = 2506

now we use churchill and bernstein relation for nusselt no

Nu = = 0.3 +

h = 0.3 +

h = 148.3 W/m².K

so

q conv = h∈(Ts- T∞ )

17000 = 148.3 ( 0.95) ( Ts - (20 + 273 ))

Ts = 413.66 K

answer:

explanation:

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