Magnetic Circuit - Revision history

Kirkbetz: /* Reviewed By */

2010-01-21T19:42:23Z

Reviewed By

← Older revision		Revision as of 12:42, 21 January 2010
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	==Reviewed By==		==Reviewed By==
	Amy Crosby		Amy Crosby

			[[Kirk Betz]]

	==Read By==		==Read By==

	==Comments==		==Comments==

John.hawkins at 18:51, 21 January 2010

2010-01-21T18:51:42Z

← Older revision		Revision as of 11:51, 21 January 2010
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	==Reviewed By==		==Reviewed By==
			Amy Crosby
	==Read By==		==Read By==

	==Comments ~~from reviewers~~==		==Comments==
	~~So I still don't understand how this works but I figured I would just put my comments here and then you can delete them. -Amy~~
	* I believe your length for dabc is incorrect and should be 1.38 unless I am understanding it wrong. if so it will prob change a few numbers :(
	* On your first equation with A sub fg, it looks like the g is too big... just a formatting thing
	* The first equation dealing with the left arm is all small, can you fix that?
	~~Other then that it looks good!~~

	~~(John): Thanks! Fixed them.~~

John.hawkins: Error correction

2010-01-18T18:55:13Z

Error correction

← Older revision		Revision as of 11:55, 18 January 2010
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	! '''Section''' !! fg !! def !! ghc !! dc !! dabc		! '''Section''' !! fg !! def !! ghc !! dc !! dabc
	\|-		\|-
	\| Length <math>l</math> (m)\|\| 0.01 \|\| 0.555\|\| 0.555 \|\| 0.48 \|\| 1.4		\| Length <math>l</math> (m)\|\| 0.01 \|\| 0.555\|\| 0.555 \|\| 0.48 \|\| 1.38
	\|-		\|-
	\| Area <math>A</math> (m<sup>2</sup>) \|\| 0.0032 \|\| 0.0032 \|\| 0.0032 \|\| 0.0096 \|\| 8.0e-4		\| Area <math>A</math> (m<sup>2</sup>) \|\| 0.0032 \|\| 0.0032 \|\| 0.0032 \|\| 0.0096 \|\| 8.0e-4
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	<center>		<center>
	<math>\mathcal{R}_{fg}=\frac{l_{fg}}{\mu ~~A_fg~~} = 4973.6</math>		<math>\mathcal{R}_{fg}=\frac{l_{fg}}{\mu A_{fg}} = 4973.6</math>
	<br />		<br />
	<math>\Phi_{fg}=B_{fg}A_{fg}=3.20\times10^{-4}</math>		<math>\Phi_{fg}=B_{fg}A_{fg}=3.20\times10^{-4}</math>
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	<center>		<center>

	<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}}=2.745\times 10^6</math>
	<br />		<br />
	<math>\mathcal{F}_{dabc}=\mathcal{F}_{dabc}\Phi_{dabc}=~~5347~~.6</math>		<math>\mathcal{F}_{dabc}=\mathcal{F}_{dabc}\Phi_{dabc}=5,271.2</math>
	</center>		</center>

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	<center>		<center>

	<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{F}_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57,~~041~~</math>		<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+

			\mathcal{F}_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+

			\mathcal{F}_{ghc}=5,627.7</math>
	<br />		<br />
	<math>\mathbf{i=\frac{\mathcal{F}_{Total}}{N}=3.57 A}</math>		<math>\mathbf{i=\frac{\mathcal{F}_{Total}}{N}=3.52 A}</math>
	</center>		</center>

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	* The first equation dealing with the left arm is all small, can you fix that?		* The first equation dealing with the left arm is all small, can you fix that?
	Other then that it looks good!		Other then that it looks good!

			(John): Thanks! Fixed them.

Amy.crosby: /* Comments from reviewers */

2010-01-18T07:37:10Z

Comments from reviewers

← Older revision		Revision as of 00:37, 18 January 2010
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	==Comments from reviewers==		==Comments from reviewers==
	So I still don't understand how this works but I figured I would just put my comments here and then you can delete them. -Amy		So I still don't understand how this works but I figured I would just put my comments here and then you can delete them. -Amy
	~~<\n> 1.~~ I believe your length for dabc is incorrect and should be 1.38 unless I am understanding it wrong. if so it will prob change a few numbers :(		* I believe your length for dabc is incorrect and should be 1.38 unless I am understanding it wrong. if so it will prob change a few numbers :(
	~~</n> 2.~~ On your first equation with A sub fg, it looks like the g is too big... just a formatting thing		* On your first equation with A sub fg, it looks like the g is too big... just a formatting thing
	~~/n 3.~~ The first equation dealing with the left arm is all small, can you fix that?		* The first equation dealing with the left arm is all small, can you fix that?
	/n Other then that it looks good!		Other then that it looks good!

Amy.crosby at 07:33, 18 January 2010

2010-01-18T07:33:52Z

← Older revision		Revision as of 00:33, 18 January 2010
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	==Reviewed By==		==Reviewed By==
	==Read By==		==Read By==

			==Comments from reviewers==
			So I still don't understand how this works but I figured I would just put my comments here and then you can delete them. -Amy
			<\n> 1. I believe your length for dabc is incorrect and should be 1.38 unless I am understanding it wrong. if so it will prob change a few numbers :(
			</n> 2. On your first equation with A sub fg, it looks like the g is too big... just a formatting thing
			/n 3. The first equation dealing with the left arm is all small, can you fix that?
			/n Other then that it looks good!

John.hawkins at 02:59, 18 January 2010

2010-01-18T02:59:58Z

← Older revision		Revision as of 19:59, 17 January 2010
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	=Author: John Hawkins=		=Author: John Hawkins=
	==Problem Statement==		==Problem Statement==
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	<references />		<references />

			==Reviewed By==
			==Read By==

John.hawkins at 08:33, 13 January 2010

2010-01-13T08:33:51Z

← Older revision		Revision as of 01:33, 13 January 2010
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	=Author: John Hawkins=		=Author: John Hawkins=
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	<br />		<br />
	'''Air Gap:'''		'''Air Gap:'''

	<br />		<br />

			<center>
	<math>\mathcal{R}_{fg}=\frac{l_{fg}}{\mu A_fg} = 4973.6</math>		<math>\mathcal{R}_{fg}=\frac{l_{fg}}{\mu A_fg} = 4973.6</math>
	<br />		<br />
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	<br />		<br />
	<math>\mathcal{F}_{fg}=\mathcal{R}_{fg}\Phi_{fg}=1.5915</math>		<math>\mathcal{F}_{fg}=\mathcal{R}_{fg}\Phi_{fg}=1.5915</math>
			</center>

	<br />		<br />
	<br />		<br />
	'''Right Arms:'''		'''Right Arms:'''
	<br />		<br />

			<center>
	<math>\Phi_{def}=\Phi_{ghc} = \Phi_{fg} = 3.20\times10^{-4}</math>		<math>\Phi_{def}=\Phi_{ghc} = \Phi_{fg} = 3.20\times10^{-4}</math>
	<br />		<br />
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	<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>
			</center>

	<br />		<br />
	<br />		<br />
	'''Center Column:'''		'''Center Column:'''
	<br />		<br />
			<center>

	<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 />
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	<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>
			</center>

	<br />		<br />
	<br />		<br />
	'''Left Arm:'''		'''Left Arm:'''
	<br />		<br />
			<center>

	<math>\Phi_{dabc}=\Phi_{dc}-\Phi_{def}=0.0019</math>		<math>\Phi_{dabc}=\Phi_{dc}-\Phi_{def}=0.0019</math>
	<br />		<br />
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	<br />		<br />
	<math>\mathcal{F}_{dabc}=\mathcal{F}_{dabc}\Phi_{dabc}=5347.6</math>		<math>\mathcal{F}_{dabc}=\mathcal{F}_{dabc}\Phi_{dabc}=5347.6</math>
			</center>

	<br />		<br />
	<br />		<br />
	'''Conclusions:'''		'''Conclusions:'''
	<br />		<br />
			<center>

	<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{F}_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57,041</math>		<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{F}_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57,041</math>
	<br />		<br />
	<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>

	<br />		<br />

John.hawkins at 07:30, 11 January 2010

2010-01-11T07:30:53Z

← Older revision		Revision as of 00:30, 11 January 2010
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			=Author: John Hawkins=
	==Problem Statement==		==Problem Statement==

John.hawkins at 07:27, 11 January 2010

2010-01-11T07:27:28Z

← Older revision		Revision as of 00:27, 11 January 2010
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	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,		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,

	<center><math>\mu=\mu_r\mu_0=(500)(4\pi\times10^{-7})</math></center>		<center><math>\mu=\mu_r\mu_0=(500)(4\pi\times10^{-7})=6.2832\times10^{-4}</math></center>

	Also, as recommended in the text, we will neglect fringing.		Also, as recommended in the text, we will neglect fringing.
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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. We need only employ the analagous equations to Ohm's Law, KVL, and KCL. All units are standard units.
	<br />		<br />
	'''Air Gap:'''		'''Air Gap:'''
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	<math>\mathcal{R}_{fg}=\frac{l_{fg}}{\mu A_fg} = 4973.6</math>		<math>\mathcal{R}_{fg}=\frac{l_{fg}}{\mu A_fg} = 4973.6</math>
	<br />		<br />
	<math>\Phi_{fg}=B_{fg}A_{fg}=3.2\times10^{-4}</math>		<math>\Phi_{fg}=B_{fg}A_{fg}=3.20\times10^{-4}</math>
	<br />		<br />
	<math>\mathcal{F}_{fg}=\mathcal{R}_{fg}\Phi_{fg}=1.5915</math>		<math>\mathcal{F}_{fg}=\mathcal{R}_{fg}\Phi_{fg}=1.5915</math>
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	'''Right Arms:'''		'''Right Arms:'''
	<br />		<br />
	<math>\Phi_{def}=\Phi_{ghc} = \Phi_{fg} = 3.2\times10^{-4}</math>		<math>\Phi_{def}=\Phi_{ghc} = \Phi_{fg} = 3.20\times10^{-4}</math>
	<br />		<br />
	<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}}=2.7603\times10^5</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>
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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>
	<br />		<br />
	<math>\mathcal{R}_{dc}=\frac{l_{dc}}{\mu A_{dc}}=~~79577~~</math>		<math>\mathcal{R}_{dc}=\frac{l_{dc}}{\mu A_{dc}}=79,577</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>
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	'''Conclusions:'''		'''Conclusions:'''
	<br />		<br />
	<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{F}_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=~~57041~~</math>		<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{F}_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57,041</math>
	<br />		<br />
	~~<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>~~
	<br />		<br />

			Which is the quantity we were looking for.

			<br />

	Calculations were performed using the following [[Magnetic Circuit Matlab Script]].		Calculations were performed using the following [[Magnetic Circuit Matlab Script]].

John.hawkins at 07:22, 11 January 2010

2010-01-11T07:22:03Z

← Older revision		Revision as of 00:22, 11 January 2010
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	'''Air Gap:'''		'''Air Gap:'''
	<br />		<br />
	<math>\mathcal{R}=\frac{l_{fg}}{\mu A_fg} = 4973.6</math>		<math>\mathcal{R}_{fg}=\frac{l_{fg}}{\mu A_fg} = 4973.6</math>
	<br />		<br />
	<math>\Phi_{fg}=B_{fg}A_{fg}=3.2\times10^{-4}</math>		<math>\Phi_{fg}=B_{fg}A_{fg}=3.2\times10^{-4}</math>
	<br />		<br />
	<math>\mathcal{F}=\mathcal{R}_{fg}\Phi_{fg}=1.5915</math>		<math>\mathcal{F}_{fg}=\mathcal{R}_{fg}\Phi_{fg}=1.5915</math>
	<br />		<br />
	<br />		<br />
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	<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{F}_{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{F}_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57041</math>
	<br />		<br />
	<center>		<center>