Reynolds Transport Theorem Derivation - Revision history

Admin at 19:43, 7 November 2010

2010-11-07T19:43:19Z

2010-11-07T19:41:21Z

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 This result can now be used to obtain Reynolds Transport Theorem.
 == References ==
 <references />

2010-11-04T00:45:49Z

← Older revision		Revision as of 00:45, 4 November 2010
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	Reynolds ~~transport theorem~~ can be derived using some mathematical relations. The derivation yields an equation governing the balance of a given quantity in a given control volume.		[[Reynolds Transport Theorem]] can be derived using some mathematical relations. The derivation yields an equation governing the balance of a given quantity in a given control volume.

2010-11-04T00:33:19Z

Euler's Expansion Formula for a Material Volume

← Older revision		Revision as of 00:33, 4 November 2010
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	=== Euler's Expansion Formula for a Material Volume ===		=== Euler's Expansion Formula for a Material Volume ===

	The expression for Euler's expansion formula derived above can be written in a slightly more general way, for a material volume <math>V_{B}^{M}</math>.		The expression for Euler's expansion formula derived above can be written in a slightly more general way, for a material volume <math>V_{B}^{M}</math>, and an initial volume of <math>V_{B,0}</math>.

	If we have a material volume <math>V_{B}^{M}</math> with a spatial location <math>\boldsymbol{x}_{B}(\boldsymbol{\xi},t)</math> and velocity <math>\boldsymbol{v}_{B} = \frac{d \boldsymbol{x}_{B} }{dt}</math>, then the Jacobian is defined as:		If we have a material volume <math>V_{B}^{M}</math> with a spatial location <math>\boldsymbol{x}_{B}(\boldsymbol{\xi},t)</math> and velocity <math>\boldsymbol{v}_{B} = \frac{d \boldsymbol{x}_{B} }{dt}</math>, then the Jacobian is defined as:
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	<math>		<math>
	J = \left\| \frac{ \partial x_{i,B} }{ \partial \xi_j } \right\|		J = \left\| \frac{ \partial x_{i,B} }{ \partial \xi_j } \right\|
			</math>

			such that

			<math>
			V_{B}^{M} = J V_{0,B}
	</math>		</math>

2010-11-03T23:27:18Z

Aris Approach

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	</math>		</math>

	This is also called the Euler Expansion Formula.		This is also called the Euler Expansion Formula.<ref name="Aris">{{Cite book
			\| author=Aris, Rutherford
			\| title=Vectors, Tensors, and the Basic Equations of Fluid Mechanics
			\| year=1962
			\| chapter=4
			\| publisher=Prentice-Hall}}</ref>

	For incompressible fluids, the fluid particle's volume will not change due to compression or dilution, so the Jacobian is zero (that is, the fluid particle volume is always equal to the initial fluid particle volume)		NOTE: For incompressible fluids, the fluid particle's volume will not change due to compression or dilution, so the Jacobian is zero (that is, the fluid particle volume is always equal to the initial fluid particle volume)

	This means that, for a fluid particle,		This means that, for a fluid particle,
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	div (\boldsymbol{v}) = 0		div (\boldsymbol{v}) = 0
	</math>		</math>



	=== Milhas Approach ===		=== Milhas Approach ===

2010-11-03T23:12:56Z

Milhas Approach

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	</math>		</math>

	which is Euler's Expansion Formula.<ref>{{Cite book \| author=Milhaus \| title=Foundations of ~~Radiative~~ Hydrodynamics \| year=~~1900~~ \| publisher=~~Cambridge~~ University Press }}</ref>		which is Euler's Expansion Formula.<ref>{{Cite book
			\| author=Milhaus, Dmitri and Barbara Milhaus
			\| title=Foundations of Radiation Hydrodynamics
			\| year=1984
			\| publisher=Oxford University Press }}</ref>

	=== Euler's Expansion Formula for a Material Volume ===		=== Euler's Expansion Formula for a Material Volume ===

2010-11-03T22:25:51Z

2010-11-03T22:25:27Z

Dilation and the Euler Expansion Formula

2010-11-01T11:01:49Z

← Older revision		Revision as of 11:01, 1 November 2010
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	Reynolds transport theorem can be derived using some mathematical relations. The derivation yields an equation governing the balance of a given quantity in a given control volume.		Reynolds transport theorem can be derived using some mathematical relations. The derivation yields an equation governing the balance of a given quantity in a given control volume.

	{{Fix\|A lot of these B's should be b's - e.g. <math>\nabla \cdot ( \rho b \boldsymbol{v} )</math> versus <math>\nabla \cdot ( \rho B \boldsymbol{v} )</math> - the latter doesn't make sense. Also... There may be some difference between the LHS volume integral and the RHS volume integral. I'm not sure how that will work in the derivation though.}}

2010-11-01T11:00:50Z

Reynolds' Transport Theorem

← Older revision		Revision as of 11:00, 1 November 2010
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	The volume <math>\boldsymbol{x} = \boldsymbol{x}(\boldsymbol{\xi},t)</math> and the Jacobian <math>dV = J dV_0</math> allow us to do this		The volume <math>\boldsymbol{x} = \boldsymbol{x}(\boldsymbol{\xi},t)</math> and the Jacobian <math>dV = J dV_0</math> allow us to do this

	{{Fix\|specify <math>V^{M}_0</math> above... specify it here... replace <math>\mathcal{F}</math> with <math>b</math>...		{{Fix\|specify <math>V^{M}_0</math> above... specify it here... replace <math>\mathcal{F}</math> with <math>b</math>... }}

	So the volume integral can be converted as follows:		So the volume integral can be converted as follows:
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	Next, transforming from <math>V_0</math> back to <math>V</math> (and noting that <math>\frac{dB}{dt} = \iiint_{V^{M}_{B}} \rho b(\mathbf{x},t) dV</math>), it becomes Reynolds Transport Theorem:		Next, transforming from <math>V_0</math> back to <math>V</math> (and noting that <math>\frac{dB}{dt} = \iiint_{V^{M}_{B}} \rho b(\mathbf{x},t) dV</math>), it becomes Reynolds Transport Theorem:

	{{Fix\|where does <math>V(\boldsymbol{x})</math> come into play???}}		{{Fix\|where does <math>V(\boldsymbol{x})</math> come into play? }}

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