Linear Solenoid design parameters

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raju_choppella

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Hi ppl,
I would like your comments on a Linear solenoid that we are developing for application in automotive industry with the following criteria:
1. 12V Dc;
2. Cu wire 42 SWG Class-C;
3. Stroke = 0.5 mm.
4. F = 13 N.
What is your opinion on the Affect due to Resistance factor at 58 Degrees temperature, and the Permeability of Petrol.
Also, would like your help in deciding no. of turns and other parameters that need to be kept in mind.
 
Linear Solenoid

Linear Solenoid

True,
My application is such that I cant help it.

Am having an issue with the Force that the solenoid will provide, my force requirement is 13N, constraints are 42 SWG 50 meters, stroke = 0.3mm,
Am using the relations:
1. B = μ * I / (2*R).
μ = 4π* 10 *
I = Current in the coil.
R = Radius of the coil.
2. F = n*L*I*(μ * I / (2*R)); L= length of the wire, n= number of turns.

Any suggestions??
 
Linear Solenoid design

Linear Solenoid design

BOBIN DESIGN
Date: 26/11/8
Name : Raju Choppella.

Constraints as follows:

WIre Type: 42 SWG, Class-C
Dia: 0.1016 mm
R/L: 2.13Ω/meter.
Length: 46.9 meters (yielding 100 Ω)
Force: 13 N.
V: 12 VDC.

Number of turns:

No. of points of wire along H = H/(0.1016)

No. of points of wire along R = (R-2)/(0.1016)

So,

?n? number of turns = no of points along H * no of points along R.

?n? number of turns = H/(0.1016)* (R-2)/(0.1016)


So,
n = (H) * (R-2)/(0.1016)?

n = (H) * (R-2)*(92.875)-------(1)
We need to add some ?slippage? to counter any errors in the winding. How much? Say 10%?

Outer Radius and Height of the Bobin.

Volume of cylindrical ring = Volume of Wire taken.

π (R? - r?)* H = π*x?*l

So,

R? = √ ((x?*l/H)+r?)

As we know r = 2 mm
x = 0.1016/2 = 0.0508mm
l = 46.9m = 46900mm.

So,
R = √ ((0.0508)?*(46950)/H + (2)?)

R = √ ((121.16/H + (4))---------(2)

Force on a current carrying coil at the center of the coil.

For 1 turn;
B = μ * I / (2*R)

Where;

μ = 4π* 10 *
I = Current in the coil.
R = Radius of the coil.


Force(F) = n*I*L*B (for ?n? turns)-------(3)

Where;
n = number of turns.
I = Current.
L= Length of wire in meters.
B= magnetic Induction.

F = n*L*I*(μ * I / (2*R))

F = μ*L*n*I?/(2R)

Now F = 13 N.

F = (4π* 10 * )* (46.9)*n*I?/(2R)

So,

n*I? = F*2R/( μ*46.9)

n*I? = 13*2*R/(4π* 10 * )* (46.9)

n*I? = 441154.43*R------------(4)

Using equations 1, 2, 3 and 4 we have the following values for a range of bobin?s as below:

H(mm) R(mm) D(mm) n n*I? I? I(Amps)
5 5.313379 10.62676 1604.918117 2344.021 1.461 1.208718
6 4.918672 9.837344 1696.478046 2169.894 1.279 1.130929
7 4.616121 9.232242 1774.056931 2036.422 1.148 1.071448
8 4.3755 8.76 1841.012478 1930.271 1.048 1.023719
9 4.178782 8.36 1899.625892 1843.488 0.9095 0.953677
10 4.014474 8.02 1951.521507 1771.003 0.9704 0.985089
11 3.874861 7.74 1997.898435 1709.412 0.8556 0.924986
12 3.754553 7.5 2039.667621 1656.338 0.8121 0.901166
13 3.649658 7.299315 2077.537437 1610.063 0.775 0.880341
14 3.557286 7.114573 2112.069524 1569.313 0.743 0.861974
15 3.475246 6.950492 2143.716545 1533.12 0.7152 0.845695
 
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