Respuesta :
Solution :
Assuming that the wire has an uniform temperature, the equivalent convective heat transfer coefficient is given as :
[tex]$h_T= \epsilon \sigma (T_s+T_{surr})(T_s^2 +T^2_{surr})$[/tex]
[tex]$h_T= 0.20 \times 5.67 \times 10^{-8} (1473+323)(1473^2 +323^2)$[/tex]
[tex]$h_T=46.3 \ W/m^2 .K$[/tex]
The total heat transfer coefficient will be :
[tex]$h_T=(250+46.3) \ W/m^2 .K$[/tex]
[tex]$=296.3 \ W/m^2 .K$[/tex]
Now calculating the maximum volumetric heat generation :
[tex]$\dot {q}_{max}=\frac{2h_t}{r_0}(T_s-T_{\infty})$[/tex]
[tex]$\dot {q}_{max}=\frac{2\times 296.3}{0.0005}(1200-50)$[/tex]
[tex]$= 1.362 \times 10^{9} \ W/m^3$[/tex]
The heat generation inside the wire is given as :
[tex]$\dot{q} = \frac{I^2R}{V}$[/tex]
Here, R is the resistance of the wire
V is the volume of the wire
∴ [tex]$\dot{q} = \frac{I^2\left( \rho \times \frac{L}{A} \right)}{A \times L}$[/tex]
[tex]$=\frac{I^2 \rho}{\left(\frac{\pi}{4}D^2 \right)}$[/tex]
where, ρ is the resistivity.
[tex]$I_{max}= \left(\frac{\dot{q}_{max}}{\rho} \right)^{1/2} \times \frac{\pi}{4}D^2$[/tex]
[tex]$I_{max}= \left(\frac{1.36 \times 10^9}{10^{-6}} \right)^{1/2} \times \frac{3.14}{4}(1 \times 10^{-3})^2$[/tex]
= 28.96 A
Now considering the relation for the current flow through the finite potential difference.
[tex]$E=I_{max} \times R$[/tex]
[tex]$E=I_{max} \times \rho \times \frac{L}{A}$[/tex]
[tex]$L=\frac{AE}{I_{max} \ \rho}$[/tex]
[tex]$L=\frac{\frac{\pi}{4} \times (1 \times 10^{-3})^2 \times 110}{28.96 \times 10^{-6}}$[/tex]
= 2.983 m
Now calculating the power rating of the heater:
[tex]$P= E \times I_{max}$[/tex]
[tex]$P= 110 \times 28.96}$[/tex]
= 3185.6 W
= 3.1856 kW
