Answer:
Heat flow equals 152.287 kW
Explanation:
For a hollow cylinder the resistance to heat flow is given by
[tex]R_{cylinder}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi KL}[/tex]
where,
[tex]r_{2}=[/tex] outer radius
[tex]r_{1}=[/tex] inner radius
[tex]K=[/tex] conductivity of heat
[tex]L=[/tex] length of cylinder
Applying the given values we get
[tex]R_{cylinder}=\frac{ln(\frac{100}{50})}{2\pi \times 60\frac{W}{mK}\times 1}[/tex]
[tex]\therefore R_{cylinder}=0.001838[/tex]
Now the heat flow is given by
[tex]\dot{Q}=\frac{\Delta T}{R_{cylinder}}\\\\\dot{Q}=\frac{400-120}{0.001838}=152.287kW[/tex]
