Question

A heated 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

Answer 1

Answer:

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 ×$10^{-5}$ m²/s

Pr = 0.7177

we find here reynolds no for air flow that is

Re = $\frac{\rho V D }{\mu } = \frac{VD}{v}$    

Re = $\frac{10*0.005}{1.99*10^{-5}}$

Re = 2506

now we use churchill and bernstein relation for nusselt no

Nu = $\frac{hD}{k}$ = 0.3 + $\frac{0.62 Re6{0.5}Pr^{0.33}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}$

h = $\frac{0.02881}{0.005}$0.3 + $\frac{0.62*2506{0.5}0.7177^{0.33}}{[1+(0.4/0.7177)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}$

h = 148.3 W/m².K

so

q conv = h∈(Ts- T∞ )

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

Ts = 413.66 K