Heat loss in a conductor based on Fourier's law

In summary, the admissible current density for a wire with no insulation is Jadm=9.6 A/m, while the admissible current density for a wire with two layers of insulation is Jadm=5.4 A/m.
  • #1
Cassius1n
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Homework Statement


Find the admissible current density Jadm for a wire that has no insulation and also for a wire that has two layers of insulation and compare it to Jadm for the case when the wire has only one layer of insulation.2. The attempt at a solution and equations
In the image I've managed to find Jadm for the case with one layer of insulation, but for the other cases I don't know how to exactly modify the terms in order to obtain the jadm values. Corrections: s=(π*d^2)/4 , S=π*D*l,
WhatsApp Image 2018-06-07 at 17.33.57.jpeg
 

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  • #2
For the uninsulated case you will need to consider the temperature gradient within the wire. This is more complicated because the heat generation is distributed over the cross section. Strictly speaking, the resistance will depend on the temperature, but that makes it really complicated.
And to be exact, you would need to consider the gradient in the wire in the insulated cases too, but it probably becomes insignificant.

For two layers of insulation, I don't see the difficulty. Why isn't that just doubling the thickness?
 
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  • #3
I think the temperature gradient in the cross section of the conductor itself may be neglected so θf is the entire conductor temperature.
Since θs=R*I^2/(K*S)+θa where θa=ambient [air] temperature
then θf=R*I^2*(1/(2*π*λ*l*ln(D/d)+1/(K*S)+θa)
In case of uninsulated conductor R*I^2=K*S*(θf-θa)
R depends on θf [ R=ρo*[1+α*( θf-20)]*lngth/scu
ρo= conductor resistivity at 20oC
α=0.00403 for aluminum and 0.00393 for copper.
K depends on exterior surface type [considering heat evacuation by convection and radiation together].
 
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  • #4
Babadag said:
In case of uninsulated conductor R*I^2=K*S*(θf-θa)
What are K and S here?
 
  • #5
If someone intends to actually calculate the admissible current density -or at least the allowable current [ampacity or current carrying capacity] he has to follow Nehrer and McGrath method or IEC 60287-1-1 and 2-1.
From IEC 60287-2-1 for a cable in open air K[=T4*π*d]=1/h/∆s^(1/4) where h=Z/d^g+E
From Table 2 for a single cable Z=0.21 E=3.94 g=0.6 h=0.21/d^0.6+3.94
∆s=is the excess of cable surface temperature above ambient temperature
S=π*d*length
 
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  • #6
For bare conductor the steady state current could be calculated according to IEEE 738/2006 formula 1(b) :
I=SQRT((qc+qr-qs)/R(T)) where :
qc=Convected heat loss rate per unit length[W/m]
qr=Radiated heat loss rate per unit length[W/m]
qs=Heat gain rate from sun[W/m]
For natural convection [no wind]:
qc = 0.0205*ρf^0.5*D^0.75*(Tc −Ta)^1.25
qr=0.0178*D*ε*{[(Tc+273)/100]^4-[(Ta+273)/100]^4}
qs =α*Qse*sin(θ)*A'
where:
D=Conductor diameter[mm]
pf=Density of air[kg/m^3]
Tc=Conductor temperature[oC]
Ta=Ambient air temperature[oC]
ε=Emissivity (0.23 to 0.91)
α=Solar absorptivity (0.23 to 0.91)
A'=Projected area of conductor per unit length[m^2/m]
θ=Effective angle of incidence of the sun’s rays[degrees]
Qse=Total solar and sky radiated heat flux rate elevation corrected [W/m^2]
Let's say the conductor is a solid copper of 50 mm^2 cross section area [D= 7.8 mm ε=0.5 Tc=80oC no wind, no sun,Ta=40oC]
qc=10.19 W; qr=4,12 W; I=172.9 A; R(80Oc)=0.000478 ohm/m R*I^2=14.3 W/m; S=0.0245 m^2
Then K= 14.3/0.0245=584 W/m^2
 
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  • #7
Thank you for the help!
 

1. What is Fourier's law and how does it relate to heat loss in a conductor?

Fourier's law is a principle in thermodynamics that describes the flow of heat through a material. It states that the rate of heat transfer is directly proportional to the temperature gradient and the cross-sectional area, and inversely proportional to the material's thermal conductivity.

2. How does the temperature gradient affect heat loss in a conductor?

The temperature gradient refers to the difference in temperature between two points in a material. According to Fourier's law, the greater the temperature gradient, the higher the rate of heat transfer and therefore, the greater the heat loss in a conductor.

3. What factors affect the thermal conductivity of a material and how do they impact heat loss?

The thermal conductivity of a material is affected by factors such as its composition, density, and temperature. Materials with higher thermal conductivity, such as metals, will experience greater heat loss compared to materials with lower thermal conductivity, such as insulators.

4. How does the cross-sectional area of a conductor influence heat loss according to Fourier's law?

The cross-sectional area refers to the size of the conductor in the direction of heat flow. According to Fourier's law, a larger cross-sectional area will result in a lower rate of heat transfer and therefore, less heat loss. This is because there is more material to conduct heat, reducing the temperature gradient.

5. Can heat loss in a conductor be prevented or reduced?

Yes, heat loss in a conductor can be reduced by using materials with lower thermal conductivity, increasing the thickness of the conductor, or using insulation to reduce the temperature gradient. In some cases, heat loss can also be prevented by controlling the environment, such as maintaining a constant temperature or using a vacuum to prevent heat transfer through convection or conduction.

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