Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: electromagnetism

Paul,
I cast around for a usable formula for determining the data for
part b): flux and flux density, each limb.

McGraw-Hill Enc Sci & Technology shows this American
customary style equation in the article "Transformer":

E = f BAN /22500

for
f frequency Hz,
B PEAK flux density in klines/in^2,
A core cross section in in^2,
E rms volts,
N number of turns

You will be accustomed to a rational presentation in SI units, I expect,
but using the equivalent representation, you can see that
knowing f , A, N then you can quickly state B which was required

Considering question c) shows up a weakness of this approach I took
to question b): there is no explicit consideration of the path reluctance,
so increasing the reluctance by a series element of cross section
160 mm^2, length 0.1 mm, relative permeability =1
does not give a sense of the reduction in flux which it causes.
I could always punt - by reworking the reluctance from the figure you
worked, and deducing the reduced flux this way.

However, I suspect you have a more analytical equation to compute
flux or flux density for part b) which would give a more assured
result for c) I imagine that the primary self inductance of the primary,
asked for in part a) is relevant - but we have not presently used it.

If you can suggest this better way, I would be glad to hear of it.

Sincerely

Brian Whatcott

[Paul Giusti wrote:]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
A transformer is constructed from E- and I-shaped cores with coils wound
around the center limb. the middle part of the core has length 25mm and
cross-sectional area 160mm^2, with a mean flux path length of 50mm on each
side.

The relative permeability of the core is 1200. the primary coil has 400
turns, while the secondary has 100 turns.

a) calculate the total reluctance, and the self-inductance of the primary.

b) if 240 V is applied to the primary and a load of 50ohms is connected
across the secondary, determine the flux and flux density in each limb of
the transformer.

c) determine the percentage drop in the flux through the centre limb if a
manufacturing fault leaves a 0.1mm gap between it and the I plate.


At 05:00 PM 4/4/02, Paul Giusti wrote:
...
for part b
since the frequency is not told, i converted the 240 volts in the primary
into 60volts in the secondary(400 turns versus 100 turns)((240/400)*100)
this 60v is terminated by 50 ohms which gives a secondary current of 1.2
Amps(I=v/r)
this is where i got stuck.
any suggestions, if i am right in what i have done.
...
----- Original Message -----

> >firstly, i tried to calculate the total reluctance
>> using S=flux length/mu*A
> > = 5*10^-3/ 1200 * mu0 * 16*10^-3
> > = 207.2 At/Wb
>
> You determined the flux length was 5 mm and the cross section
>area was 16 mm^2 and mu0 was 1.257E-6 N/A^2
>Your handling of scientific notation was unusual.
> You wrote 5*10^-3 for 50 mm instead of 5*10^-2.
>
> But later, when expressing the numbers in decimal fractions, the numbers
> looked better. There you expressed the same length as 0.05



> > i then calculated the self inductance using L=N^2/S
> > =400^2/207.2
> > = 772.2 H
> > this is where i got confused and started e-mailing physics lecturers. i
> > noticed in my notes, for a similar shaped circuit as the one i have got,
> > i have writen beside it " series/parallel circuit, total S=
centre+1/2 side
> > reluctances(assuming symetrical)."
> > i did not understand what this meant and
> > have looked throught books but cannot seem to find an explanation.
> > if that means what i think it does then the total reluctance would be
> > (S=flux length/mu*A) + (2*S=flux length/mu*A)
> > =(2.5^-3/1200*mu0*16*10^-3) + (2*( 5*10^-3/ 1200*mu0*10*10^-3))
> > i am not sure however and genuinely cannot find anything in
> > any books....... after seeing a different question in a
> > book, what i did was :
> >S1= core length/Ur*U0*A = 0.025/(1200 *U0 * 0.16) =103.6 at/wb
> >S2= 2*(flux path lenght / Ur*U0*A) = 2* (0.05/ 1200 *U0 *0.1)
> > =2* 331.57 at/wb = 663.15 at/wb
> >therefore total reluctance = 663.15+766.75 at/wb
> >for S2, i multiplied this by 2 because i treated them as two seperate
> >shapes.
> > you may not be able to see what i mean as you do not have a
> > diagram,
> > which is with the question, and i am not allowed to send attachments.
> > then i worked out the self inductance by using the formula L= N^2/S,
> > however my answer for this depends on the answer for
> > reluctance being correct.
> >paul
>
>
> The central difficulty you expressed was in visualizing reluctance,
> analogous to working out circuit resistance but of a magnetic path. If
> you had a resistor R1, whose ends were connected by two parallel side
> resistors R2 each, you would not have much trouble in visualizing
> the total circuit resistance through the center limb as R1 + R2/2.....


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~




Brian Whatcott
Altus OK Eureka!