Question

A 1 mm diameter electric wire is covered with 2 mm thick layer of insulation(k= 0.5 W/mk). Air surrounding the wire is at 25 degree Celsius and convective heat transfer coefficient h = 25 W/m2K. The wire temperature is 100 degree Celsius. The maximum value of heat dissipation is

Answer 1

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Answer 2

Answer:

The maximum value of heat dissipation is 50.229 W/m        

Explanation:

Given,

• d₁ = 1 mm = r₁ = 0.5 mm
• r₂ = 0.5 mm + 2 mm = 2.5 mm
• k = 0.5 W/m K
• h = 25 W/m²K
• T₁ = 100°C
• T₀ = 25°C

To find:

The maximum value of heat dissipation

Solution:

The maximum heat dissipation rate:

$\frac{Q_{max} }{T} = \frac{2 \pi ( T_{1} - T_{0}) }{\frac{ln( r_{cr}/r_{1}) }{k}+ \frac{1}{h r_{cr} } }$   ..............................(1)

To find $r_{cr}$:

$r_{cr} = \frac{k}{h} \\r_{cr} = \frac{0.5}{25}\\ r_{cr} = 0.02 m \\r_{cr} = 20 mm$

Radius of electric wire r₁ = 0.5 mm

Radius of thick layer insulation r₂ = 2.5 mm

Wire temperature T₁ = 100°C

The temperature of air surrounding the wire T₀ = 25°C

Heat transfer coefficient h = 25 W/m²K

Substitute the values in equation (1) we get

$\frac{Q_{max} }{T} = \frac{2 \pi ( 100 - 25) }{\frac{ln(20 /0.5) }{0.5}+ \frac{1}{25 (20 *10^{-3}) } }$

       $= \frac{471}{ 7.377 + 2} \\= \frac{471}{9.377} \\= 50.229 W/m$

The maximum value of heat dissipation is 50.229 W/m        

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