Finding the power loss in a wire of varying cross-sectional area

[Glenn G]

TL;DR Summary: Finding the power loss in a wire of varying area - my problem is I don't know how to set up the integral

Hopefully you can see in the diagram below that the area of the wire varies linearly with length. I know the equations for resistance and power loss and I can express the resistance of a thin slice but need to integrate over the whole wire to get the full resistance - to find power then it is trivial. I've thought about this for a good while now but can't get it. Please help, this isn't homework (I'm very old and doing this for fun!)

 

[phinds]

I have no answer for you but as an EE I have to comment --- that is one WEIRD "wire"

Just FYI, chicken scratching for problem posts is frowned on here. Best to type it in and Latex is encouraged.

 

[Glenn G]

Ok noted for future

 

[erobz]

TL;DR Summary: Finding the power loss in a wire of varying area - my problem is I don't know how to set up the integral

Hopefully you can see in the diagram below that the area of the wire varies linearly with length. I know the equations for resistance and power loss and I can express the resistance of a thin slice but need to integrate over the whole wire to get the full resistance - to find power then it is trivial. I've thought about this for a good while now but can't get it. Please help, this isn't homework (I'm very old and doing this for fun!)

View attachment 333623

$r$ is a function of $x$, not $dx$.

 

[Steve4Physics]

Hi. Your description says 'the area of the wire varies linearly with length'. But the handwritten work says 'radius increases linearly with length'. The two statements are incompatible!

Also, 'length' is a constant for a given wire. I guess what you really mean is one of:
a) cross-sectional-area (CSA) varies linearly with distance along wire;
or
b) radius varies linearly with distance along wire.

 

[DaveE]

For a conical wire:

the radius $r_1 = \sqrt{\frac{A_1}{\pi}}$ and $r_2 = \sqrt{\frac{A_2}{\pi}}$ so $r(x) = r_1 + x \frac{r_2-r_1}{L}$ where $x \equiv 0$ at $r_1$ and $x \equiv L$ at $r_2$

The resistance of a thin disk at $x$ is $R(x) = \frac{\rho}{\pi r(x)^2}dx$

Then the total resistance is $R = \int_0^L R(x) = \frac{\rho}{\pi} \int_0^L \frac{1}{r(x)^2} \, dx = \frac{\rho}{\pi} \int_0^L \frac{1}{(r_1 + x \frac{r_2-r_1}{L})^2} \, dx = \frac{\rho L}{\pi (r_1-r_2)} \left. \frac{1}{(r_1 + x\frac{r_2-r_1}{L})} \right|_0^L$

$R = \frac{\rho L}{\pi (r_1-r_2)} (\frac{1}{(r_1 + L\frac{r_2-r_1}{L})} -\frac{1}{r_1}) = \frac{\rho L}{\pi (r_1-r_2)} (\frac{1}{r_2 } -\frac{1}{r_1}) = \frac{\rho L}{\pi (r_1-r_2)} (\frac{r_1-r_2}{r_1 r_2 } ) = \frac{\rho L}{\pi r_1 r_2 }$

$R = \frac{\rho L}{\pi \sqrt{\frac{A_1}{\pi}} \sqrt{\frac{A_2}{\pi}} } = \frac{\rho L}{\sqrt{A_1 A_2} }$

 

[Nik_2213]

FWIW, in 'Real World', you would also have change of conductivity with position due temperature rise.
Think 'open', H-shaped fuses...

 

[haruspex]

For a conical wire:

the radius $r_1 = \sqrt{\frac{A_1}{\pi}}$ and $r_2 = \sqrt{\frac{A_2}{\pi}}$ so $r(x) = r_1 + x \frac{r_2-r_1}{L}$ where $x \equiv 0$ at $r_1$ and $x \equiv L$ at $r_2$

The resistance of a thin disk at $x$ is $R(x) = \frac{\rho}{\pi r(x)^2}dx$

Then the total resistance is $R = \int_0^L R(x) = \frac{\rho}{\pi} \int_0^L \frac{1}{r(x)^2} \, dx = \frac{\rho}{\pi} \int_0^L \frac{1}{(r_1 + x \frac{r_2-r_1}{L})^2} \, dx = \frac{\rho L}{\pi (r_1-r_2)} \left. \frac{1}{(r_1 + x\frac{r_2-r_1}{L})} \right|_0^L$

$R = \frac{\rho L}{\pi (r_1-r_2)} (\frac{1}{(r_1 + L\frac{r_2-r_1}{L})} -\frac{1}{r_1}) = \frac{\rho L}{\pi (r_1-r_2)} (\frac{1}{r_2 } -\frac{1}{r_1}) = \frac{\rho L}{\pi (r_1-r_2)} (\frac{r_1-r_2}{r_1 r_2 } ) = \frac{\rho L}{\pi r_1 r_2 }$

$R = \frac{\rho L}{\pi \sqrt{\frac{A_1}{\pi}} \sqrt{\frac{A_2}{\pi}} } = \frac{\rho L}{\sqrt{A_1 A_2} }$

That's fine if $L>>r_2-r_1$. Otherwise there is the complication that current near the surface of the wire has further to travel than that near the core.

 

[DaveE]

That's fine if $L>>r_2-r_1$. Otherwise there is the complication that current near the surface of the wire has further to travel than that near the core.

Yes, there's a subtle assumption that the "disks" are truly just in series; i.e. radial current flow isn't significant.