3–101E A 0.083-in-diameter electrical wire at 90°F is covered by 0.02-in-thick p

Question

3–101E A 0.083-in-diameter electrical wire at 90°F is covered
by 0.02-in-thick plastic insulation (k 5 0.075 Btu/h·ft·°F). The
wire is exposed to a medium at 50°F, with a combined convection
and radiation heat transfer coefficient of 2.5 Btu/h·ft2·°F.
Determine if the plastic insulation on the wire will increase or
decrease heat transfer from the wire. Answer: It helps

3–102E Repeat Prob. 3–101E, assuming a thermal contact
resistance of 0.001 h·ft2·°F/Btu at the interface of the wire and
the insulation.

Explanation / Answer

>> As, rc = Critical radius

and, as rc = k/h,

where, k = Thermal Conductivity = 0.075 Btu/h.ft.F

and, h = Convection Coefficient = 2.5 Btu/h.ft2.F

=> rc = 0.075/2.5 = 0.03 ft = 0.03*12 = 0.36 in

Now, as t = thickness of insulation = 0.02 in

and, raidus of wire = 0.083/2 = 0.0415 in

So, radius of thickness = 0.02+0.0415 = 0.0615 in

As, r < rc ,

So, Heat Transfer will increase .....ANSWER.......

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.