Magnetic Circuit: Difference between revisions
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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. |
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. |
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'''Air Gap:''' |
'''Air Gap:''' |
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<math>\mathcal{R}=\frac{l_{fg}}{\mu A_fg} = 4973.6</math> |
<math>\mathcal{R}=\frac{l_{fg}}{\mu A_fg} = 4973.6</math> |
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<math>\Phi_{fg}=B_{fg}A_{fg}=3.2\times10^{-4}</math> |
<math>\Phi_{fg}=B_{fg}A_{fg}=3.2\times10^{-4}</math> |
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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> |
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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> |
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<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> |
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'''Center Column''' |
'''Center Column:''' |
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<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> |
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<math>\mathcal{R}_{dc}=\frac{l_{dc}}{\mu A_{dc}}=79577</math> |
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<math>\Phi_{dc}=\frac{\mathcal{F}_{dc}}{\mathcal{R}_{dc}}=0.0022</math> |
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'''Left Arm:''' |
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<math>\Phi_{dabc}=\Phi_{dc}-\Phi_{def}=0.0019</math> |
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<math>\mathcal{R}_{dabc}=\frac{l_{dabc}}{\mu A_{dabc}}=2.785\times 10^6</math> |
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<math>\mathcal{F}_{dabc}=\mathcal_{dabc}\Phi_{dabc}=5347.6</math> |
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'''Conclusions:''' |
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<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{_{dc}+\mathcal{F}_{def} +\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57041</math> |
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<center> |
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<math>\mathbf{i=\frac{\mathcal{F}_{Total}}{N}=3.57 A}</math> |
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</center> |
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Calculations were performed using the following [[Magnetic Circuit Matlab Script]]. |
Calculations were performed using the following [[Magnetic Circuit Matlab Script]]. |
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Revision as of 23:18, 10 January 2010
Problem Statement
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 .<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,
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.
Section | fg | def | ghc | dc | dabc |
---|---|---|---|---|---|
Length (m) | 0.01 | 0.555 | 0.555 | 0.48 | 1.4 |
Area (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:
Right Arms:
Center Column:
Left Arm:
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}
Calculations were performed using the following Magnetic Circuit Matlab Script.
References
<references />