Solving Exercise: Find Thickness of Rubber Insulation for Copper Wire

AI Thread Summary
The discussion revolves around calculating the thickness of rubber insulation for a copper wire to maintain safe temperature levels. The user attempts to derive the necessary formulas but encounters confusion regarding unit conversions and the correct representation of electrical and thermal resistance. Feedback emphasizes the importance of writing equations symbolically before substituting numerical values and highlights a potential arithmetic error. The conversation also touches on the effects of temperature on wire resistance and the typical insulation thickness for safe current carrying capacity. Ultimately, the user seeks clarity on their approach and calculations while receiving constructive criticism from peers.
Amaelle
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Good day all, I tried to solve the following exercise, I would be glad to get feedback for my attempt ( indeed I was puzzled by the numerical solution I got just need to check it for any mistake:

A copper wire , 2mm in diameter and 100 m in length has an electrical resistance of 0,5 mΩ/m, the wire carries a current of 80 A and is insulated by using a layer of rubber kr=0,15W/mK
determine the thickness of the rubber needed to prevent the maximum temperature in the insulating layer from exceeding 70 °C and the rubber outer surface 40 °C.

My attempt
lets calculate the heat rate q:
q=Relectrical*I2
lets calulate the Resitance of rubber
Rrubber=ln(r2/r1)/(2πLk)

ΔT =Rrubber*qRrubber=ΔT/q
SO
ln(r2/r1)=2πLk*ΔT*q
SO

r2/r1=exp(2πLk*ΔT*q)
which means
r2=r1*exp(2πLk*ΔT*q)


any feed back would be highly appreciated!
thanks!
 
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Are you sure about the current?
 
This is confusing. Try writing the answer in terms of symbols first, then put in the numbers at the very end. I suspect there is an error in unit conversion somewhere. Also, please use symbol R to denote either electrical or thermal resistance but not both.
 
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Chestermiller said:
Are you sure about the current?
thanks for your prompt answer! yes I'm pretty sure about it, but what about my approach is it correct?
 
kuruman said:
This is confusing. Try writing the answer in terms of symbols first, then put in the numbers at the very end. I suspect there is an error in unit conversion somewhere. Also, please use symbol R to denote either electrical or thermal resistance but not both.
thanks a lot for your interest
I tried to arrange the presentation
 
Amaelle said:
thanks for your prompt answer! yes I'm pretty sure about it, but what about my approach is it correct?
In my judgment, if you did the arithmetic right (including units), your approach is correct.

Did they say anything about an external convective heat transfer resistance between the outer surface and the bulk room air?
 
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no nothing ! I have reported the exercice as it is
 
ln(r2/r1)=2πLk*ΔT*q

The q should be in the denominator
 
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Chestermiller said:
ln(r2/r1)=2πLk*ΔT*q

The q should be in the denominator
Yes thank you you are right it was a tipo , but i got still the same awkward numerical results,
Thanks a million for you constructive feedback, you are awsome as always!
 
  • #10
Amaelle said:
Yes thank you you are right it was a tipo , but i got still the same awkward numerical results,
Thanks a million for you constructive feedback, you are awsome as always!
I did not get the same numeral results when I did the calculation.
 
  • #11
Chestermiller said:
I did not get the same numeral results when I did the calculation.
ln(r2/r1)=(2πLk*ΔT)q

q=R*I2=320W
ln(r2/r1)=(2π*100*0,15*30)/320=8,83125
r2=r1exp(8,83125)=6844,83 mm
is that wrong?
 
  • #12
Amaelle said:
ln(r2/r1)=(2πLk*ΔT)q

q=R*I2=320W
ln(r2/r1)=(2π*100*0,15*30)/320=8,83125
r2=r1exp(8,83125)=6844,83 mm
is that wrong?
I must have made a mistake in arithmetic.
 
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  • #13
Chestermiller said:
I must have made a mistake in arithmetic.
oh thanks and i wanted to write ln(r2/r1)=(2πLk*ΔT)/q as you pointed out earlier!
thanks a lot!
 
  • #14
If we shall neglect the temperature drop from surface to air and the resistance increased due temperature it is correct.
Actually, 0.5 ohm will be the resistance of this wire at -2oC and at 70oC it will be 0.6566 ohm.
The surface to air thermal resistance is very high so for 6 mm insulation thickness it will be a minimum wire temperature of 201.15 oC.
That is why Chestermiller ask you if you are sure 80 A is the current.
Usually the thickness of insulation is 0.8 mm and the current carrying capacity [ampacity] will be 32 A
 
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  • #15
Babadag said:
If we shall neglect the temperature drop from surface to air and the resistance increased due temperature it is correct.
Actually, 0.5 ohm will be the resistance of this wire at -2oC and at 70oC it will be 0.6566 ohm.
The surface to air thermal resistance is very high so for 6 mm insulation thickness it will be a minimum wire temperature of 201.15 oC.
That is why Chestermiller ask you if you are sure 80 A is the current.
Usually the thickness of insulation is 0.8 mm and the current carrying capacity [ampacity] will be 32 A
I
Ok then I see :) thanks a lot for your insights!
 

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