Answer:
a). Surface temperature 1051.66 K
b). Insulation temperature 1425.99 K and cable surface temperature 1051.6575 K
c). Thickness 0.0175 m Maximum temperature when thickness is used 590.75 K
Explanation:
[tex]I= 700 A \\D= 5mm = 5x10^{-3}m\\ R=6x10^{-4}\\ T_{f} =30 C= 303 K\\ h= 25 \frac{W}{m^{2}*K }[/tex]
a).
[tex]T_{1}=T_{f} +\frac{Q}{h*\pi*D }[/tex]
Q= heat generated in the ore due to the current
[tex]Q= I^{2} * R[/tex]
[tex]Q= 700^{2} A * 6x10^{-4}[/tex] Ω/m
[tex]Q= 294 \frac{W}{m}[/tex]
[tex]T_{1}=T_{f} +\frac{Q}{h*\pi*D }[/tex] ⇒ [tex]T_{1}=303 K +\frac{294}{25*\pi*5x10^{-3}m }[/tex]
[tex]T_{1} = 1051.66 [/tex] K
b).
[tex]T_{c} = \frac{Q}{\pi *D}*(R_{c} +\frac{1}{h})+T_{f}[/tex]
[tex]R_{c} = 0.02 \frac{m^{2}*K}{W}[/tex]
[tex]T_{c} = \frac{294 \frac{W}{m} }{\pi* 5x10^{-3}m }*(0.02 \frac{m^{2}*K }{W}+\frac{1}{25\frac{W}{m^{2} *K } } ) +303 K[/tex]
[tex]T_{c} = 1425.99 K[/tex]
[tex]T_{i} = Temperature insulation = T_{c} - Q*\frac{R_{c} }{\pi*D*L }[/tex]
[tex]L= 1m[/tex]
[tex]T_{i} = 1425.99 K- 294 \frac{W}{m} *(\frac{0.02 \frac{m^{2*K} }{W} }{\pi *5x10^{-3}m*1m } )[/tex]
[tex]T_{i}= 1051.6575[/tex] K
c).
[tex]t= \frac{d-D}{2} \\d= \frac{2*k}{h} = \frac{2*0.5 \frac{W}{m*K} }{25\frac{W}{m*K} } = 0.04m[/tex]
[tex]t=\frac{0.04m - 5x10^{-3}m }{2} = 0.0175 m[/tex]
[tex]T_{i}= T_{s}-\frac{Q*R_{c}}{\pi*D*L }[/tex]
[tex]T_{s}= Q*R_{t} +T_{f}[/tex]
[tex]R_{t} = \frac{R_{c} }{\pi *D} +\frac{ln\frac{d}{D} }{2*\pi*K } +\frac{1}{h*\pi *d} \\R_{t} = \frac{0.02 \frac{m^{2} *K}{W} }{\pi *5x10^{-3}m } +\frac{ln\frac{0.04m}{0.005m} }{2*\pi*0.5\frac{W}{m*K } } +\frac{1}{25\frac{W}{ m^{2}*K} \pi *0.04m} \\R_{t} = 2.252 \frac{K*m}{W}[/tex]
[tex]T_{s}= 294*2.252 +303= 965.088 [/tex]K
[tex]T_{i}= T_{s}-\frac{Q*R_{c}}{\pi*D*L }[/tex]
[tex]T_{i} = 965.088 k - 294\frac{W}{m} * \frac{0.02 \frac{m^{2}*K }{W} }{\pi* 5x10^{-3}m } =590.75[/tex]K
