Magnetic Circuit: Difference between revisions

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<math>\mathcal{F}=\mathcal{R}_{fg}\Phi_{fg}=1.5915</math>
<math>\mathcal{F}=\mathcal{R}_{fg}\Phi_{fg}=1.5915</math>
<br />
<br />
<br />
<br />
'''Right Arms:'''
'''Right Arms:'''
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<math>\mathcal{R}_{def}=\mathcal{R}_{ghc}=\frac{l_{def}}{\mu A_{def}}=276030</math>
<math>\mathcal{R}_{def}=\mathcal{R}_{ghc}=\frac{l_{def}}{\mu A_{def}}=276030</math>
<br />
<br />
<math>\mathcal{F}_{def}=\mathcal{F}_{ghc} = \mathcal{R}_{def}\Phi_{def}=88.331</math>
<math>\mathcal{F}_{def}=\mathcal{F}_{ghc}=\mathcal{R}_{def}\Phi_{def}=88.331</math>
<br />
<br />
<br />
<br />
'''Center Column:'''
'''Center Column:'''
<br />
<br />
<math>\mathcal{F}_{dc}=\mathcal{F}_{def}+\mathcal{F}_{fg}+ \mathcal{F}_{ghc}=178.25</math>
<math>\mathcal{F}_{dc}=\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=178.25</math>
<br />
<br />
<math>\mathcal{R}_{dc}=\frac{l_{dc}}{\mu A_{dc}}=79577</math>
<math>\mathcal{R}_{dc}=\frac{l_{dc}}{\mu A_{dc}}=79577</math>
<br />
<br />
<math>\Phi_{dc}=\frac{\mathcal{F}_{dc}}{\mathcal{R}_{dc}}=0.0022</math>
<math>\Phi_{dc}=\frac{\mathcal{F}_{dc}}{\mathcal{R}_{dc}}=0.0022</math>
<br />
<br />
<br />
<br />
'''Left Arm:'''
'''Left Arm:'''
<br />
<br />
<math>\Phi_{dabc}=\Phi_{dc}-\Phi_{def}=0.0019</math>
<math>\Phi_{dabc}=\Phi_{dc}-\Phi_{def}=0.0019</math>
<br />
<br />
<math>\mathcal{R}_{dabc}=\frac{l_{dabc}}{\mu A_{dabc}}=2.785\times 10^6</math>
<math>\mathcal{R}_{dabc}=\frac{l_{dabc}}{\mu A_{dabc}}=2.785\times 10^6</math>
<br />
<br />
<math>\mathcal{F}_{dabc}=\mathcal_{dabc}\Phi_{dabc}=5347.6</math>
<math>\mathcal{F}_{dabc}=\mathcal_{dabc}\Phi_{dabc}=5347.6</math>
<br />
<br />
<br />
<br />
'''Conclusions:'''
'''Conclusions:'''
<br />
<br />
 
<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57041</math>
<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{_{dc}+\mathcal{F}_{def} +\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57041</math>
 
<br />
<br />
<center>
<center>
<math>\mathbf{i=\frac{\mathcal{F}_{Total}}{N}=3.57 A}</math>
<math>\mathbf{i=\frac{\mathcal{F}_{Total}}{N}=3.57 A}</math>
</center>
</center>
<br />
<br />
Calculations were performed using the following [[Magnetic Circuit Matlab Script]].
Calculations were performed using the following [[Magnetic Circuit Matlab Script]].

Revision as of 00:19, 11 January 2010

Problem Statement

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Problem 2.16 from Electric Machinery and Transformers, 3rd ed:

A magnetic circuit is given in Figure P2.16. What must be the current in the 1600-turn coil to set up a flux density of 0.1 T in the air-gap? All dimensions are in centimeters. Assume that magnetic flux density varies as

B=[1.5H/(750+H)]

.<ref>Guru and Huseyin, Electric Machinery and Transformers, 3rd ed. (New York: Oxford University Press, 2001), 129.</ref>

Solution

First, we note that the problem statement is incomplete. Assume that the core has a relative permeability of 500. Hence, for all magnetic sections excluding the air gap,

μ=μrμ0=(500)(4π×107)

Also, as recommended in the text, we will neglect fringing.

The lengths and areas of each of the sections to be evaluated are given in the following table.

Table 1: Lengths and Areas for the pertinent secions of the magnetic circuit.
Section fg def ghc dc dabc
Length l (m) 0.01 0.555 0.555 0.48 1.4
Area A (m2) 0.0032 0.0032 0.0032 0.0096 8.0e-4

We must now work backward from the air-gap, since the value of the flux-density is given there. All units are understood to be standard units.
Air Gap:
=lfgμAfg=4973.6
Φfg=BfgAfg=3.2×104
=fgΦfg=1.5915

Right Arms:
Φdef=Φghc=Φfg=3.2×104
def=ghc=ldefμAdef=276030
def=ghc=defΦdef=88.331

Center Column:
dc=def+fg+ghc=178.25
dc=ldcμAdc=79577
Φdc=dcdc=0.0022

Left Arm:
Φdabc=ΦdcΦdef=0.0019
dabc=ldabcμAdabc=2.785×106
Failed to parse (syntax error): {\displaystyle \mathcal{F}_{dabc}=\mathcal_{dabc}\Phi_{dabc}=5347.6}

Conclusions:
Failed to parse (syntax error): {\displaystyle \mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57041}

i=TotalN=3.57A


Calculations were performed using the following Magnetic Circuit Matlab Script.

References

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