Jessie T.

asked • 08/01/22

HELP, please. How can I get to that solution by solving the equation other than by trial and error?

Example 3.38. An electric cable of 12 mm diameter is insulated to increase the current capacity. Due to insulation the current carrying capacity is increased by 15% without increasing cable surface temperature above 70°C. The environmental temperature is 30°C. Assume that the heat transfer coefficient from the bare or insulated cable is 14 W/m2.K. Calculate the conductivity of insulating material.



Considering 2nd and 3rd terms of equation,

we get 1 + ln (11.90k) = 9k or

ln (11.9k) – 9k + 1 = 0

By trial and error Let k = 0.1 W/m.K, then, 1.19 – 0.9 + 1 ≠ 0

Let k = 0.2, then 0.8671 – 1.9422 + 1 = (– 0.075) ≠ 0.031

Let k = 0.2158, satisfy the equation, therefore, Thermal conductivity of insulating material


is 0.2158 W/m.K. Ans.


1 Expert Answer

By:

Ryan C. answered • 08/01/22

Tutor
New to Wyzant

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