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3 years ago

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A 5 mm dia copper cable is insulated with a material of conductivity of 0.16 W/mK and is exposed to air at 30Â°C with a convecti

on coefficient of 20 W/m^2K. If the surface temperature of the wire can be 120Â°C, determine the insulation thickness for maximum heat flow and the heat dissipated per m length.

1 answer:

Kipish [7]3 years ago

4 0

Answer:

t=5.5 mm

Heat dissipation per unit length = 90.477 W/m

Explanation:

Given that

Diameter d = 5 mm ⇒r = 2.5 mm

Conductivity of insulated material K = 0.16 W/mK

Heat transfer coefficient = 20 $\frac{W}{m^2K}$

When thickness reaches up to critical radius of insulation then heat dissipation will be maximum

We know that critical radius of insulation of wire is given as follow

$r_{c}=\dfrac{K_{insulation}}{h_{surrounding}}$

Now by putting the values

$r_{c}=\dfrac{0.16}{20}$

$r_{c}=8 mm$

So the thickness of insulation

t=8-2.5 mm

t=5.5 mm

As we know that heat transfer due to convection given as follows

Q = hAΔ T

Q=20 x 2 x π x 0.008 x (120-30)

Q = 90.477 W/m

So heat dissipation per unit length = 90.477 W/m

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You can review its deduction on van Wylen 6 Edition, section 8.10.

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