Hi again,
Willbe:
He he
Trisorion:
Wow, thanks for doing all of that. However, now I'm really confused.
How did you find the best wire diameter without the length or cross sectional area of the core? (or maybe im just missing something) Is the core square or circular in this model?
I am glad that you picked that up. In my haste i did not enter
in all the values for the example core, which includes the window
area and other things. We have to know the window height
and width as well as the core height and width to get the answer
of course.
The window height is the same as the length of the winding.
The window width is the same as the coil 'build'.
The core width is the center width of the core material (the center
width of one lamination for example).
The core height is the core stack height.
Im sorry i didnt make this clear, but i was in a hurry yesterday and
rushed to get that all posted. Since i made a few mistakes, i'll
provide the equations now and all the input data necessary
to do these calculations, and we come out with an interesting
result after all this.
Units: meters, ohms, ohm meters, volts, amps.
The force at the poles:
F=K*I^2*N^2*A/(g+d/um)^2
Area of wire
Awire=pi*(d/2)^2
Area of rectangular window cross section:
Awin=Hwin*Wwin
Circular to square wire area loss factor:
cslf=pi/4
Number of turns that fit into window area:
N=(Awin*cslf)/Awire
The mean wire length:
mwl=2*(Wcore+Wwin)+2*(Hcore+Hwin)
Total wire length:
twl=N*mwl
Wire resistivity:
rho=1.7241e-8*(1+.00393*(Ta)) ohm meters
Total wire resistance:
R=twl*rho/Awire
The force including volts and wire resistance:
F=K2*(V/(R+Rb))^2*N^2
Combining all the equations into one force equation:
F=(Hcore*Hwin^2*Wcore*Wwin^2*K2*V^2)/(d^4*((4*Hwin*rho*Wwin*(2*(Wwin+Wcore)+2*(Hwin+Hcore)))/(d^4*pi)+Rb)^2)
Finding the max:
Code:
dF/dd=(32*Hcore*Hwin^3*rho*Wcore*Wwin^3*(2*(Wwin+Wcore)+2*(Hwin+Hcore))*K2*V^2)/(d^9*pi*((4*Hwin*rho*Wwin*(2*(Wwin+Wcore)+2*(Hwin+Hcore)))/(d^4*pi)+Rb)^3)-(4*Hcore*Hwin^2*Wcore*Wwin^2*K2*V^2)/(d^5*((4*Hwin*rho*Wwin*(2*(Wwin+Wcore)+2*(Hwin+Hcore)))/(d^4*pi)+Rb)^2)
Set that equal to zero to find relative max:
dF/dd=0
Find all solutions to dF/dd=0 ,
and the solution we want is:
d=8^(1/4)*((Hwin*rho*Wwin^2)/(pi*Rb)+(Hwin*rho*Wcore*Wwin)/(pi*Rb)+(Hwin^2*rho*Wwin)/(pi*Rb)+(Hcore*Hwin*rho*Wwin)/(pi*Rb))^(1/4)
Just for reference, at d=optimum:
F=(Hcore*Hwin*pi*Wcore*Wwin*K2*V^2)/(32*rho*Rb*Wwin+32*rho*Rb*Wcore+(32*Hwin+32*Hcore)*rho*Rb)
Now plug in all the values for the equation for d above:
K2=1 (constant for optimization)
V=3 (battery open circuit voltage)
Rb=0.1 (battery internal resistance)
Ta=25 (temperature in degrees C)
Hcore=2/39 (height of core approx 2 inches)
Wcore=1/39 (width of core approx 1 inch)
Hwin=1/39 (height of window approx 1 inch)
Wwin=1/39 (width of window approx 1 inch)
rho=1.7241e-8*(1+.00393*(Ta)) resistivity of wire at ambient temperature Ta
Now graph or use the dF equation to find d, and we get:
d=0.002478 meters (and use a wire gauge chart to find the wire gauge number)
Now from:
Awire=pi*(d/2)^2
N=(Awin*cslf)/Awire
we get:
N=107 turns
and from:
mwl=2*(Wcore+Wwin)+2*(Hcore+Hwin)
twl=N*mwl
R=twl*rho/Awire
Im=V/(R+Rb)
we get:
Im=15 amps
and from:
Asm=pi*(d/2)^2
cmils=Asm/5.067e-10
we get:
cmils=9521
and calculating the current density:
CurrentDensity=cmils/I=635 cmils/amp
so this shouldnt overheat.
Now we take a quick look at what R came out to:
R=twl*rho/Awire
so
R=0.1
Interesting, isnt it?
So we found that the coil resistance R should match the
battery internal resistance.
I'll leave it up to you to prove if we get this result no matter
what the construction is.
My batteries could do 2.4 volts, 20 amps with .06 ohm Rb in series or 1.2 volts, 40 amps with .015 ohm Rb in parallel. 16 amps is not pushing it, even for series. Do I understand this correctly? Are you saying there is no benefit to have more than 16 amps?
Well, it's not quite that simple. We can not simply increase current
because current is dependent on the battery internal resistance,
and because we have optimized for force there should be nothing
we can do to get more force unless we change one of the optimizing
parameters (like the battery internal resistance). But we have to
remember that we are attempting to optimize the wire gauge (via
wire diameter) for a given construction which makes certain things
constant:
Window area
Core area
Battery internal resistance
We have to keep some of these things constant because if we
didnt we could not optimize because after all, if we were able to
increase current indefinitely (without overheating of course) we
would be able to increase the force infinitely even with a core
that saturates. We cant do this though, because for one we
are limited as to what the battery internal resistance is and the
open circuit voltage.
This doesnt mean we cant find out what the effects are when
we increase some parameter, such as core area, but to do
this we would really have to enter in all new parameters and
do a new optimization for wire diameter.
It may also be possible to find other interesting facts such as
maximum force possible for a given construction given all of
the other parameters. Solving for wire diameter and then
inserting that into the equation for force for example provides
for an equation that shows the max force possible for the
best wire diameter, without actually calculating the wire diameter.
This would show the difference between various cores and
window areas because the force for each construction could
be compared.
Since we are talking about optimization processes here, perhaps
you should look into other processes before attempting this one
because after all this construction has a lot of dimensions to
consider. If you dont want to do that then you are just going
to have to keep trying to understand the current process we
are talking about. After that you should be able to factor in
things we havent looked at yet, such as core saturation and
its effect on the optimized wire diameter.
One last note:
A circular cross section for the core is the optimum core shape
because this shape minimizes the wire resistance per turn,
and a square cross section comes in second. Rectangular
cores are common also however, but they wont be as good
as a square cross section. It's easier to get more core area
however with a rectangular cross section because you can
keep adding laminations to make the core height larger
and thus increase the core area.