Answer to Question #163371 in Electrical Engineering for Mital sosa

Question

1.      Consider a square plate of 0.5 m x 0.5 m in a room at 30 C. One side of the plate is kept at a uniform temperature of 74 C while the other side is insulated. Determine the rate of heat transfer from the plate by free convection under the following three situations: (i) The plate is kept vertical (ii) The plate is kept horizontal with hot surface facing up (iii) The plate is kept horizontal with hot surface facing down.

Answer 1

The properties of air at the film temperature of

$T_f = \large\frac{(T_s + T_{\infin})}{2}$ = $\large\frac{30+74}{2}$ = $52^oC \space$ and 1 atm here

$k= 0.02808 W/m ^o C$

$P_r= 0.7202$

$\nu = 1.896*10^{-5} \frac{m^2}{s}$

$\beta = \large\frac{1}{T_f}$ $=\large\frac{1}{333K}$

(a) Vertical. The characteristic length in this case is the height of the plate, which is L = 0.5m. The Rayleigh number is

$Ra_D = \large\frac{g\beta (T_s - T_{\infin})L^3}{V^2}$$Pr = \large\frac{9.81*\frac{1}{333}(74-30)*0.5^3}{(1.896*10^{-5})^2}$$(0.722) = 3.254*10^{8}$

The natural convection Nusselt number

$Nu = \lbrace 0.825 + \large\frac{0.387Ra_D^{\frac{1}{6}}}{\lbrack1+(\frac{0.492}{Pr})^{\frac9{16}}\rbrack^{\frac{8}{27}}} \rbrace^2$ $=\lbrace 0.825 + \large\frac{0.387(3.254*10^{8})^{\frac{1}{6}}}{\lbrack1+(\frac{0.492}{0.7202})^{\frac9{16}}\rbrack^{\frac{8}{27}}} \rbrace ^2$ $=$ 102.6

$h = \frac{k}{D}Nu = \frac{0.02808 W/m ^oC}{0.5m}(102.6) = 5.76 \frac{W}{m} C$

$A_s = L^2 = (0.5m)^2 = 0.25m^2$

and

$Q = hA_s(T_s - T_{\infin}) = (5.76 \frac{W}{m} C)(0.25m^2)(74-30)^oC = 63.36W$