Doubt about the volume of the air gap in an electrical device

So for the following device: I'm trying to figure out the torque. To do that, I need the area of the gap. In the solution, it says that the gap volume is: $$2gh(r_1+0.5g)\theta$$

But the part of $r_1+0.5g$ is really confusing me. I'm not sure where that comes from. Isn't it supposed to be $2gh(2r_1)\theta$ ?

• The arc at the rotor is of length r1θ; the arc at the stator is of length (r1+g)θ. Average them. – Andrew Morton Apr 24 '18 at 19:20
• The volume of a cuboid is width*height*depth. The volume under consideration can be approximated to 2 cuboids. – Andrew Morton Apr 24 '18 at 19:36
• r1 + g/2 is the half way point of the gap, and gives the best average distance around the arc. – Philip Oakley Apr 24 '18 at 22:15

The volume you are trying to calculate is an area between arcs of the same inner angle, but of different radius. It's called an annular sector: Now, technically, you will need a bit of mathematics to calculate this without any reference but, in the end, the volume is equal to the area of this small region, times the height (which is h in your case).

Practically, according to the shape above, and the picture you posted, you have an area that runs from r to R or, from your shape, it runs from $r_1$ to $r_1 + g$.

You have two coordinates (these are polar coordinates here), the radius $r$ and the angle $θ$ (a parameter), which runs from $0$ to $θ$ (the actual value) in your case. Remember from differential calculus that the area element in polar coordinates is equal to $rdrdθ$ (where $θ$ is in radians). While you 're at it, check this depiction as well, in which $Δr$ represents an infinitesimally small length and $Δθ$ an infinitesimally small angle. Because the elements are infinitesimal, this can be considered approximately equal to a rectangle, so its area (in terms of differential calculus) is the product of its sides: So the area you want to calculate is given by a double integral of the area element over the ranges spanned by the two coordinates in the given area:

$\int_{0}^{θ}(\int_{r_1}^{r_1+g}rdr)dθ$

Because the function in the inner integral does not contain any $θ$, the two integrals can be separated and calculated as multiples:

$\int_{0}^{θ}dθ(\int_{r_1}^{r_1+g}rdr)=\left[θ\right]_{0}^{θ} \left[\frac{r^2}{2}\right]_{r_1}^{r_1+g} =\frac{1}{2}θ \left[r^2\right]_{r_1}^{r_1+g} = \frac{1}{2}θ \left[ \left(r_1+g \right )^2 - \left(r_1\right)^2 \right] = \frac{θ}{2} \left[ r_1^2 + 2r_1g + g^2 - r_1^2 \right] = \frac{θ}{2} \left[ 2r_1g + g^2 \right] = θg \left(r_1 + \frac{g}{2} \right)$

This is only one side of the rotor. Twice that is equal to the area of both sides:

$A = 2g \left(r_1 + \frac{g}{2} \right)θ$

This is the area. If you multiply with the height $h$ (vertical dimension) you get the volume you are looking for:

$V = 2gh\left(r_1 + 0.5g\right)θ$

• +1 and a comment for a well reasoned and clear answer. – KalleMP Apr 24 '18 at 21:46