I want to build a solenoid.

Thread Starter

Ziggey

Joined Nov 4, 2015
15
I did the following calculation
L=μAln^2
Values I used for equation
μ= 4π*10^(-7) T/amp m
A= 2π(.493)^2 = 1.527in^2 = .0387 m^2
l = .2032m
n = 258/.2032 T/m = 1269.685 T/m (what is the appropriate unit for n?)
So I have
L = 4π*10^(-7) T/amp m * .0387 m^2 * .2032m * (1269.685)^2 T/m
L = 4.863^-8 (Tm^3)/amp *327578.719 T^2/m
L = .016 T^3m^2/amp
The next calculation I did was
X=2πfL
f = 60 hz
L = .016 H

X= 2π * 60 * .016
X = 6.031
SO finally to find my current I can do
120v (output of my inverter)
6.031 ohms
120/6.031 = 19.897 amps
Does this look about correct?

I found the equations from these sites in case people are wondering.
http://www.allaboutcircuits.com/textbook/alternating-current/chpt-3/ac-inductor-circuits/
http://physics.info/inductance/
 

wayneh

Joined Sep 9, 2010
16,763
Did you tell us your wire gauge and the length of the wire? I can check your coil against my model.

Your result looks plausible but I haven't checked it.
 

MaxHeadRoom

Joined Jul 18, 2013
22,321
Keep in mind that the the current inrush is far higher for an AC solenoid when first energized, for the first half cycle the current is only limited by the resistance of the wire, after this there will be high current (lower inductive reactance) until the solenoid armature actually shifts over, when the inductive reactance is the lowest.
Max.
 

Thread Starter

Ziggey

Joined Nov 4, 2015
15
Did you tell us your wire gauge and the length of the wire? I can check your coil against my model.

Your result looks plausible but I haven't checked it.
Max I'll keep that in mind.
Wayneh
I have 12 gauge wire my core will be 8 In long. That makes 86 turns per layer 3 layers. If I did the math right I got 57 feet ish I dont have my notes in front of me right now...
 

wayneh

Joined Sep 9, 2010
16,763
I get different coil properties. Maybe I don't understand your description. I calculate that a close-wound cylindrical coil, 8" long, 0.493" radius of center cavity, 12ga magnet wire, 3 layers, should require 267 turns and use 82 feet of wire for a total DC resistance of 0.133Ω That could would have an estimated inductance of 280µH.
 

Thread Starter

Ziggey

Joined Nov 4, 2015
15
Did you tell us your wire gauge and the length of the wire? I can check your coil against my model.

Your result looks plausible but I haven't checked it.
Max I'll keep that in mind.
Wayneh
I have 12 gauge wire my core will be 8 In long. That makes 86 turns per layer 3 layers. If I did the math right I got 57 feet ish I dont have my notes in front of me right now...
 
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