What quantitiy in the equation cant be higher than 1.8 due to saturation?F=K*I^2*N^2*A/(g+d/um)^2
Thanks MrAl, that was easily one of the best answers I have ever gotten.
Just one more thing. I dont have an osilliscope to test for exact 'hysteresis curves' (I think that is what is called). But lets say my core is average for a piece of steel. How would I calculate when I need a wider core?
What quantitiy in the equation cant be higher than 1.8 due to saturation?
dF/dd=(32*rho*H^3*K2*V^2*W^3*(2*W+2*H+Wwin+Hwin))/(d^9*pi^4*((4*rho*H*W*(2*W+2*H+Wwin+Hwin))
/(d^4*pi^2)+Rb)^3)-(4*H^2*K2*V^2*W^2)/(d^5*pi^2*((4*rho*H*W*(2*W+2*H+Wwin+Hwin))/(d^4*pi^2)+Rb)^2)
What is "window area"?Wwin is the width of the window area,
Hwin is the height of the window area,
What does the first half of this equation mean?R=(4*Awin*mwl*rho)/(d^4*pi^2)
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?We know the current is about 16 amps now
Hi again,
I felt like playing around with the equations a little so i ended up
with an optimization equation.
Combining all the equations we get the equation for force that we will
use for the optimization:
F=(H^2*K2*V^2*W^2)/(d^4*pi^2*((4*rho*H*W*(2*W+2*H+Wwin+Hwin))/(d^4*pi^2)+Rb)^2)
where
H is core center height,
W is core center width,
V is battery open circuit voltage,
d is wire diameter,
rho is resistivity of the wire,
Wwin is the width of the window area,
Hwin is the height of the window area,
Rb is the battery internal resistance,
pi is the constant 3.1415926...
and dimensions are meters, volts, ohm meters, ohms, and later amps when needed.
To optimize all we have to do is look for a relative max.
we set this equal to zeroCode:dF/dd=(32*rho*H^3*K2*V^2*W^3*(2*W+2*H+Wwin+Hwin))/(d^9*pi^4*((4*rho*H*W*(2*W+2*H+Wwin+Hwin)) /(d^4*pi^2)+Rb)^3)-(4*H^2*K2*V^2*W^2)/(d^5*pi^2*((4*rho*H*W*(2*W+2*H+Wwin+Hwin))/(d^4*pi^2)+Rb)^2)
dF/dd=0
and we find four solutions, the one we want is:
d=(sqrt(2)*((2*rho*H*W^2)/Rb+(2*rho*H^2*W)/Rb+(rho*Wwin*H*W)/Rb+(Hwin*rho*H*W)/Rb)^(1/4))/sqrt(pi)
and this leads directly to the wire diameter in meters of about 0.0031
meters, which is roughly a AWG #8 wire gauge.
A simpler way to avoid all the math however is to just plot the equation for
F between wire diameters say 0.0005 meters (AWG=24 approx) to 0.005
(AWG=3 approx) and see where the max occurs graphically. This has been
done in this drawing:
**broken link removed**
Now we must find the current so we can calculate the wire current density
to make sure it doesnt overheat. We will use a 600 cmil/amp criterion for
this test because this is known to work well with transformers which have
the same basic construction.
Calculating the total resistance from:
R=(4*Awin*mwl*rho)/(d^4*pi^2)
we get R=0.0888 approx.
Now we calculate I from
I=V/(0.0888+Rb)
and we get I=15.9 amps and this allows us to check the wire diameter for
heating.
We know the current is about 16 amps now, and since the wire diameter is
0.0033 meters (the actual number 8 gauge wire) we can calculate the
circular mils from:
Asm=pi*(d/2)^2 (d is diameter in meters)
cmils=Asm/5.067e-10
Doing this, we get
Asm=8.55e-6 square meters
cmils=16880
Now that we know the circular mils, we can calculate the current density:
density=cmils/amps
so
density=16880/16=1055,
so the current density is 1055 cmils per amp which is better than
600 cmils per amp so this construction should not overheat.
The other thing we have to do however is check the current draw to make sure
our voltage source can handle the operating level of current. If not, we may
end up decreasing the wire gauge to increase resistance to make it easier
on the source, or we have to think about getting a higher power supply to
drive the coil with.
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?
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)
Well, it's not quite that simple. We can not simply increase currentMy 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?
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