Engineering How to Calculate Reluctance and Estimate Force in a DC Electromagnetic Actuator

[TheRedDevil18]

Homework Statement

Consider the DC electromagnetic actuator shown in Figure 1 which is constructed using Hotrolled

Silicon Steel laminations. All dimensions shown are in millimeters. For this actuator0.5 mm diameter Copper wire is used for the winding, the bobbin has a thickness of 5 mm,

the number of turns is 61 and the current in the coil is 5.25 A. At a particular position of the

plunger, the width of the air gap AG x is 0.125 mm.

Figure 1:

My equivalent circuit diagram:

Questions:(a) Calculate the reluctance of the entire magnetic circuit when the air gap width is

0.125 mm. Neglect fringing in the air gaps but do not neglect the reluctance of

the iron. Work to an accuracy of less or equal to two percent. [2.373x10^5 At/Wb]
(b) Re-calculate the reluctance of the entire magnetic circuit for the condition when

the width of the air gap has been increased to 0.135 mm. Neglect fringing in the

air gaps but do not neglect the reluctance of the iron. Work to an accuracy of

less or equal to two percent. [2.412x10^5 At/Wb]
(c) Estimate the force exerted by the actuator for the condition when the width of
the air gap AGx is 0.125 mm. [-354.6 N]

Homework Equations

The Attempt at a Solution

Hope you guys can read my circuit diagram. It's a bit untidy because I used a mouse to draw it
Rtg refers to "Reluctance of top air gap"
Rcg refers to "Reluctance of centre air gap"
Rbg refers to "Reluctance of bottom air gap"
The rest are the reluctances of the coreF = NI = (61)(5.25) = 320.25 At = Ftotal

Now, because the dimensions are the same for the top and bottom parts of the circuit, hence their reluctances are the same, that is Rgt = Rgb and Rcore(AB) = Rcore(EF). Therefore the flux must divide equally

So the top branch = Total flux/2 = Bottom branch

Therefore Ftotal = Fcore(DC) + Fcg + Fcore(BC) + Ftg + Fcore(AB) + Fcore(AD)

Rcg = Lcg/u*A = (0.125*10^-3)/(4pi*10^-7)(30*10^-3)(35*10^-3)
= 9.47*10^4 At/Wb

Rtg = Ltg/u*A = (0.125*10^-3)/(4pi*10^-7)(15*10^-3)(35*10^-3)
= 1.89*10^5 At/Wb

If you take the ratio of Rtg/Rcg = 2, therefore Rtg = 2*Rcg

Ftg = Flux/2 * Rtg...1)

Fcg = Flux * Rcg....2)

Substituting Rtg = 2*Rcg in equation 1, you get

Ftg = Flux * Rcg, therefore Ftg = Fcg

The next part I'm guessing ill have to estimate the mmf drop across the airgap ?, so after a few tries I estimated a 79.5% drop of the total mmf in the two airgaps

That is 0.795*320.25 = 254.6 At = Fcg+Ftg

But since Ftg = Fcg, therefore Fcg = 127.3 At = Ftg

I now went ahead and calculated B with the formula

Bcg = Fcg*u/Lag = (127.3)(4pi*10^-7)/(0.125*10^-3) = 1.28 T = Btg

Using the BH curve I found H to be 320 At/m

Using the formula F = Hl, I calculated the drops across the core as follows

Fcore(DC) = 320*(50*10^-3) = 16 At
Fcore(BC) = 320*(52.5*10^-3) = 16.8 At
Fcore(AB) = 320*(50*10^-3) = 16 At
Fcore(AD) = 320*(52.5*10^-3) = 16.8 At

Summing up the drops = 254.6 + 16 + 16.8 + 16 + 16.8 = 320.2 At

I therefore get an error of (320.25-320.2/320.25) * 100 = 0.016 %

To get the total reluctance, Flux total = B*(Area of centre limb)
= (1.28)*(30*10^-3)(35*10^-3)
= 1.34*10^-3

Ftotal = Flux total * Total Reluctance
Total Reluctance = Ftotal/Flux Total
= (320.25)/(1.34*10^-3)
= 2.38*10^5 At/Wb

This is my attempt for part a of the question. I need to know if I did it correctly so that I can proceed to the next part. Thanks for any help :)

 

[rude man]

You're not getting responses for one or more of the following reasons:\1. what bobbin?
2. we're supposed to know permeability of "hot rolled silicon steel laminations"?
3. all those numbers give us a headache! I suggest replacing all the numbers on your given diagram by symbols. You can replace them with numbers at the end.
4. what does "Calculate the reluctance of the entire magnetic circuit" mean? There are two "magnetic circuits".

You seem to have the right approach, basically, all things considered.

Hint for (c): virtual work!