Answer:
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
k = 50.2
H = k × A × [tex]\frac{[T_{H -}T_{C} ] }{L}[/tex]
Solving for A
A = [tex]\frac{H * L }{k * [ T_{H}- T_{C} ] }[/tex]
A = [tex]\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}[/tex]
A = [tex]\frac{85}{12550}[/tex] = 6.77 ×[tex]10^{-3}[/tex] m²
Now Area of cylinder is :
A = [tex]\frac{\pi }{4}[/tex] d²
solving for d:
d = [tex]\sqrt{\frac{4 * 0.00677 }{\pi } }[/tex]
d = 9.28 cm
