Respuesta :
Answer:
47.94° C
Explanation:
DATA GIVEN:
length of thin horizontal plate = 16 cm long = 0.16
the width of the thin horizontal plate = 20 cm wide = 0.20 m
temperature of the outside air [tex]T \infty[/tex] = 20° C
Power rating of heating element Q = 20 W
Emissivity of the plate E = 0.9
Surrounding surface temperature [tex]T_{amb}[/tex] = 17° C
Let assume that the surface temperature of the plate is [tex]T_S[/tex] = 40 ° C
then the heat transfer by convection can be expressed by the formula:
[tex]\mathbf{Q_{conv} = hA_s(T_s - T _{\infty})}[/tex]
Also;the heat transfer by radiation is:
[tex]\mathbf{Q_{rad} = E A_2 \sigma (T_s^1 - T^1_{amb})}[/tex]
So; obtaining the properties of Air at 1 atm and film temperature [tex]T_f[/tex]
[tex]T_f = \frac{T_s + T_{\infty}}{2}[/tex]
[tex]T_f = \frac{40+ 20}{2}[/tex]
[tex]T_f = 30^0C[/tex]
Thermal conductivity k = 0.02588 W/m ° C
Prandtl number, Pr = 0.7282
Kinematic viscosity v = 1.608 × 10⁻⁵ m²/s
Volume expansion coefficient ;
[tex]\beta = \frac{1}{T_f}[/tex]
[tex]\beta =\frac{1}{30+273}[/tex]
[tex]\beta = 3.3*10^{-3}K^{-1}[/tex]
The characteristic length [tex]L_c[/tex] of the horizontal plate is as follows:
[tex]L_c = \frac{A_s}{P}[/tex]
[tex]L_c =\frac{L*W}{2(L+W)}[/tex]
[tex]L_c =\frac{0.16*0.20}{2(0.16+0.20)}[/tex]
[tex]L_c =0.044 \ m[/tex]
Rayleigh number Ra = [tex]\frac{g \beta(T_s - T_{\infty})L_c^3 }{v^2}*Pr[/tex]
Ra = [tex]\frac{9.8 *3.3*10^{-3}(40 - 20)0.044^3 }{1.608^2}*0.7282[/tex]
Ra = 155327.931
To determine the heat transfer from the top surface; let's first find the Nusselt number :
[tex]\mathbf{Nu = 0.54Ra^{0.25}}[/tex]
Nu = [tex]\mathbf{0.54(155327.931)^{0.25}}[/tex]
Nu = 10.72
Also, the Heat transfer coefficient is;
[tex]h = \frac{k*Nu}{L_c}[/tex]
[tex]h = \frac{0.02588*10.72}{0.044}[/tex]
[tex]h = 6.3053 \ W/m^2 .K[/tex]
[tex]Q_{top} = hA_s(T_s - T _{\infty})[/tex]
[tex]Q_{top} = 6.3053*0.16*0.20(T_s - 20})[/tex]
[tex]Q_{top} = 0.2017(T_s-20) ------- \ equation (1)[/tex]
Calculating the heat transfer from the bottom surface;
[tex]\mathbf{Nu = 0.27Ra^{0.25}}[/tex]
Nu = [tex]\mathbf{0.27(155327.931)^{0.25}}[/tex]
Nu = 5.36
the Heat transfer coefficient is;
[tex]h = \frac{k*Nu}{L_c}[/tex]
[tex]h = \frac{0.02588*5.36}{0.044}[/tex]
[tex]h = 3.1526 \ W/m^2 .K[/tex]
[tex]Q_{top} = hA_s(T_s - T _{\infty})[/tex]
[tex]Q_{top} =3.1526*0.16*0.20(T_s - 20})[/tex]
[tex]Q_{top} = 0.1008(T_s-20) ------- \ equation (2)[/tex]
The heat transfer by radiation is :
[tex]\mathbf{Q_{rad} =2*s* \sigma A_s(T^4_s-T^4_{amb})}[/tex]
[tex]\mathbf{Q_{rad} =2*0.9* 5.67*10^{-8}*0.16*0.20((T_s+273)^4-(17+273)^4)}[/tex]
[tex]\mathbf{Q_{rad} =3.265*10^{-9}((T_s+273)^4-(290)^4)} ------- \ equation(3)[/tex]
From the steady state condition:
The power rating of the heating element is:
[tex]\mathbf{Q=Q_{rad}+Q_{top}+Q_{bottom}}[/tex]
replacing all the values from the above equations ; we have:
[tex]20 = 3.265*10^{-9}[(T_s+273)^4-(290)^4]+(Ts-20)[0.2017-0.1008][/tex]
[tex]20 = 3.265*10^{-9}[(T_s+273)^4-(290)^4]+(Ts-20)0.3025[/tex]
The surface temperature [tex]T_S[/tex] from solving the above equation = 47.94° C
