- #1
Amaelle
- 310
- 54
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!
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!
Last edited: