Reynolds Transport Theorem Derivation
From charlesreid1
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.
chill out, it's cool - |
This needs to be fixed.
A lot of these B's should be b's - e.g. $ \nabla \cdot ( \rho b \boldsymbol{v} ) $ versus $ \nabla \cdot ( \rho B \boldsymbol{v} ) $ - 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. You can also bug me about this by email: root (at) charlesmartinreid.com |
Nomenclature
Intensive vs. Extensive Properties
For any given amount of fluid, the fluid will have some properties associated with it
Extensive properties $ B $ - change with the mass of the fluid (e.g. concentration of species $ i $)
Intensive properties $ b = \frac{\partial B}{\partial t} $ - are independent of the mass of the fluid (e.g. mass fraction of species $ i $)
Material Volume vs. Control Volume
A material volume (denoted $ V^{M}_{B}(t) $) is a volume of fluid whose boundaries are defined such that there is zero flux of a specified extensive property $ B $ (the typical case is $ B = m $, mass, so that $ b=1 $ and there is zero flux of mass through the material volume's boundaries). The material volume is a function of time, since the boundaries of the volume will change, with the fluid, in time.
A control volume (denoted $ V $) is a fixed and constant volume, whose location is fixed. It is not a function of time.
(See illustrations below)
Fluid particle
A fluid particle is a particle, in the macroscopic (as opposed to the molecular) sense, that consists of a mass of fluid
This mass of fluid has some properties associated with it, an extensive property $ B $ and an intensive property $ b $
The volume of the fluid is a material volume $ V^{M}_{B}(t) $ that changes with respect to time.
The boundaries of a material volume are such that there is zero flux of the property of interest $ B $
That is, for a given fluid particle, with an extensive property $ B $, a material volume $ V^{M}_{B} $ is a volume whose boundaries move such that there is no flux of $ B $ across the boundaries
Derivative Frames of Reference
Frames of References
To begin, the initial position of a fluid particle (not a particle in the molecular sense, but a particle in a macroscopic, continuum sense) may be written as a random variable $ \boldsymbol{\xi} $.
As specified above, a fluid particle volume containing some extensive property $ B $ is contained within a material volume $ V^{M}_{B}(t) $ such that the flux of $ B $ across the boundaries of the material volume are zero. The material volume is a function of time.
The location of this particular fluid particle at a later time is determined, first, by its initial location $ \boldsymbol{\xi} $, and second, by the time that has passed, $ t $.
This later position can be written as $ \boldsymbol{x}(\boldsymbol{\xi},t) $
We will assume that, given a location $ \boldsymbol{x} $, we can backtrack and find the fluid particle's initial location $ \boldsymbol{\xi} $
This means the function $ \boldsymbol{x}(\boldsymbol{\xi},t) $ is assumed to be invertible.
In other words, we assume we can find a function $ \boldsymbol{\xi} = \boldsymbol{\xi}(\boldsymbol{x},t) $ that is continuous and single-valued (i.e. invertible).
The necessary and sufficient condition for invertability is for a non-vanishing Jacobian to exist:
$ J = \frac{ \partial \boldsymbol{x} }{ \partial \boldsymbol{\xi} } $
Now, let's consider some extensive fluid property $ B $. This property field evolves with the state of the fluid, and can be written one of two ways, each with a unique interpretation:
$ B(\boldsymbol{x},t) = B( \boldsymbol{\xi}(\boldsymbol{x},t), t ) $
This way of writing $ B $ can be interpreted as follows: The value of the property $ B $ at the spatial/temporal location $ (\boldsymbol{x},t) $ is the value appropriate to the fluid particle located at $ (\boldsymbol{x},t) $
Alternatively,
$ B(\boldsymbol{\xi},t) = B( \boldsymbol{x}(\boldsymbol{\xi},t), t ) $
This way of writing $ B $ can be interpreted as: The value seen by the particle $ \boldsymbol{\xi} $ at time $ t $ is the value of the property $ B $ at the position the particle occupies at that time $ t $
In keeping with these interpretations, two different temporal derivatives can be written.
Derivatives
Partial Derivative
The partial derivative is denoted by
$ \displaystyle{\frac{\partial B}{\partial t}} $
The partial derivative is the derivative of $ B $ with respect to time, keeping $ \boldsymbol{x} $ constant
This derivative corresponds to the rate of change of $ B $ in a control volume (which is a fixed point in space; see nomenclature above)
Material Derivative
The material derivative is denoted by
$ \displaystyle{\frac{dB}{dt}} $ (alternatively, $ \displaystyle{\frac{DB}{Dt}} $)
The material derivative derivative of B with respect to time, keeping $ \boldsymbol{\xi} $ constant
This derivative corresponds to the rate of change of $ B $ in a material volume (which is a volume whose boundaries are moving with time such that the flux of $ B $ across the boundaries is zero)
Keep in mind that the material derivatives are not partial derivatives because $ \boldsymbol{\xi} $ (the initial particle position) is a constant for a given fluid particle
Position and Velocity
For a material volume $ V^{M}_{B}(t) $, the material volume moves at some rate.
Let the position of the material volume be denoted $ \boldsymbol{x}_{B} $.
Then the velocity of the material volume $ V^{M}_{B} $ is denoted $ \boldsymbol{v}_{B} $
The material derivative of the position of the fluid particle is the velocity:
$ v_{i,B} = \frac{d x_{i,B}}{dt} = \left( \frac{ \partial x_{i,B}(\boldsymbol{\xi},t)}{\partial t}\right)_{\boldsymbol{\xi}} $
In other words, holding $ \boldsymbol{\xi} $ constant (that is, considering a material volume $ V^{M}_{B}(t) $ with the initial position $ \boldsymbol{\xi} $), the rate of change of the fluid particle's current position $ x_{i,B} $ is the velocity of the material volume $ V^{M}_{B} $.
$ \boldsymbol{v}_{B} = \frac{ d \boldsymbol{x}_{B} }{ dt } $
Derivative Relationships
The two derivatives can be related. First, to repeat what these derivatives physically mean:
$ \left( \frac{ \partial B }{ \partial t } \right)_{\boldsymbol{x}} $ : the rate of change of $ B $ at a FIXED POINT $ \boldsymbol{x} $
$ \left( \frac{ dB }{ dt } \right)_{\boldsymbol{\xi}} $ : the rate of change of $ B $ for a FIXED FLUID PARTICLE $ \boldsymbol{\xi} $.
Next, the derivatives of $ B $ with respect to time can be equated at a particular spatial and temporal location, and the chain rule used, to get the relationship between these two derivatives:
$ \begin{array}{rcl} \displaystyle{ \frac{dB}{dt} } &=& \displaystyle{ \frac{\partial}{\partial t} \left( B(\boldsymbol{\xi},t) \right) } \\ &=& \displaystyle { \frac{\partial}{\partial t} \left( B( \boldsymbol{x}( \boldsymbol{\xi},t ), t ) \right) } \\ &=& \displaystyle{ \frac{ \partial B }{\partial x_i} \left( \frac{d x_i}{dt} \right)_{\boldsymbol{\xi}} + \left( \frac{ \partial B }{\partial t} \right)_{x} } \\ &=& \displaystyle{ \frac{ \partial B }{ \partial t } + v_i \frac{ \partial B }{ \partial x_i } } \end{array} $
This can be written conveniently as
$ \frac{ dB }{ dt } = \frac{ \partial B }{ \partial t } + ( v \cdot \nabla ) B $
Practically, in order to obtain the paths of fluid particles from the velocity field, an ordinary differential equation must be solved for each fluid particle:
$ \frac{ d x_i }{ dt } = v_i ( \boldsymbol{x}, t) $
subject to the initial condition $ x_i = \xi_i $ at $ t=0 $.
To solve for acceleration of the particle, another ordinary differential equation must be solved for each fluid particle:
$ \boldsymbol{a} = \frac{ d \boldsymbol{v} }{ dt } = \frac{ \partial \boldsymbol{v} }{ \partial t } + (\boldsymbol{v} \cdot \nabla) \boldsymbol{v} $
in steady flow, the accumulation term goes away, making the acceleration:
$ \boldsymbol{a} = ( \boldsymbol{v} \cdot \nabla ) \boldsymbol{v} $
Dilation and the Euler Expansion Formula
Changing coordinates from $ \boldsymbol{\xi} $ to $ \boldsymbol{x} $ is likely to cause a change in the volume of the fluid particle:
$ dV = \frac{ \partial (x_1, x_2, x_3) }{ \partial (\xi_1, \xi_2, \xi_3) } d \xi_1 d \xi_2 d \xi_3 = J dV_0 $
where $ J $ is the Jacobian.
The Jacobian is equal to:
$ J = \frac{ dV }{ dV_0 } $
This quantity is called the dilation.
The Jacobian is a matrix that looks like this:
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chill out, it's cool - |
This needs to be fixed.
Search Google Books for "Foundations of Radiation Hydrodynamics" (Mihalas and Mihalas), go to page 59, see their derivation of Euler Expansion Formula You can also bug me about this by email: root (at) charlesmartinreid.com |
$ J = \displaystyle{ \left| \begin{array}{ccc} \frac{ \partial x_1 }{ \partial \xi_1 } & \frac{ \partial x_1 }{ \partial \xi_2 } & \frac{ \partial x_1 }{ \partial \xi_3 } \\ \frac{ \partial x_2 }{ \partial \xi_1 } & \frac{ \partial x_2 }{ \partial \xi_2 } & \frac{ \partial x_2 }{ \partial \xi_3 } \\ \frac{ \partial x_3 }{ \partial \xi_1 } & \frac{ \partial x_3 }{ \partial \xi_2 } & \frac{ \partial x_3 }{ \partial \xi_3 } \end{array} \right| } $
The initial volume of the fluid particle is $ d \xi_1 d \xi_2 d \xi_3 = dV_0 $ at $ t=0 $
Motion is continuous, so the volume cannot break up... another way to state that is, $ 0 < J < \infty $ (required so that neither $ J $ nor $ J^{-1} $ vanish, and the mapping from $ \xi $ to $ x $ and vice-versa are continuous and smooth).
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chill out, it's cool - |
This needs to be fixed.
Is the material volume moving at a single velocity? If so, take this into account in the preceding sentence. If not... explain. You can also bug me about this by email: root (at) charlesmartinreid.com |
The natural question to ask is how the volume dilation changes with time - that is, $ \frac{ dJ }{ dt } $
First, the time derivative of the terms in the Jacobian matrix can be simplified:
$ \frac{d}{dt} \left( \frac{\partial x_i}{\partial \xi_j} \right) = \frac{\partial}{\partial \xi_j} \frac{d x_i}{d t} = \frac{ \partial v_i }{ \partial \xi_j } $
where the second step is possible because $ \frac{d}{dt} $ holds $ \boldsymbol{\xi} $ constant.
It was specified above that the velocity is a function of location, $ \boldsymbol{v} = \boldsymbol{v}( x_1, x_2, x_3 ) $. This can be plugged into the relation $ \frac{ \partial v_i }{ \partial \xi_j } $, and the chain rule used, to yield:
$ \frac{ \partial v_i }{ \partial \xi_j } = \frac{ \partial v_i }{ \partial x_1 } \frac{ \partial x_1 }{ \partial \xi_j } + \frac{ \partial v_i }{ \partial x_2 } \frac{ \partial x_2 }{ \partial \xi_j } + \frac{ \partial v_i }{ \partial x_3 } \frac{ \partial x_3 }{ \partial \xi_j } $
which can be generalized to the result:
$ \frac{ d v_i }{ dt } = \frac{ d v_i }{ d x_k } \frac{ d x_k }{ dt } $
This expression gives a way to write the time derivative of the terms in the Jacobian matrix.
|
chill out, it's cool - |
This needs to be fixed.
This is where the Foundation of Radiation Hydrodynamics book's approach would be useful You can also bug me about this by email: root (at) charlesmartinreid.com |
The derivative of the determinant of the Jacobian is the sum of three terms; each term is the Jacobian matrix, with only one row differentiated. Thus,
$ \frac{dJ}{dt} = \displaystyle{ \left| \begin{array}{ccc} \frac{d}{dt} \frac{ \partial x_1 }{ \partial \xi_1 } & \frac{d}{dt} \frac{ \partial x_1 }{ \partial \xi_2 } & \frac{d}{dt} \frac{ \partial x_1 }{ \partial \xi_3 } \\ \frac{ \partial x_2 }{ \partial \xi_1 } & \frac{ \partial x_2 }{ \partial \xi_2 } & \frac{ \partial x_2 }{ \partial \xi_3 } \\ \frac{ \partial x_3 }{ \partial \xi_1 } & \frac{ \partial x_3 }{ \partial \xi_2 } & \frac{ \partial x_3 }{ \partial \xi_3 } \end{array} \right| } + \displaystyle{ \left| \begin{array}{ccc} \frac{ \partial x_1 }{ \partial \xi_1 } & \frac{ \partial x_1 }{ \partial \xi_2 } & \frac{ \partial x_1 }{ \partial \xi_3 } \\ \frac{d}{dt} \frac{ \partial x_2 }{ \partial \xi_1 } & \frac{d}{dt} \frac{ \partial x_2 }{ \partial \xi_2 } & \frac{d}{dt} \frac{ \partial x_2 }{ \partial \xi_3 } \\ \frac{ \partial x_3 }{ \partial \xi_1 } & \frac{ \partial x_3 }{ \partial \xi_2 } & \frac{ \partial x_3 }{ \partial \xi_3 } \end{array} \right| } + \displaystyle{ \left| \begin{array}{ccc} \frac{ \partial x_1 }{ \partial \xi_1 } & \frac{ \partial x_1 }{ \partial \xi_2 } & \frac{ \partial x_1 }{ \partial \xi_3 } \\ \frac{ \partial x_2 }{ \partial \xi_1 } & \frac{ \partial x_2 }{ \partial \xi_2 } & \frac{ \partial x_2 }{ \partial \xi_3 } \\ \frac{d}{dt} \frac{ \partial x_3 }{ \partial \xi_1 } & \frac{d}{dt} \frac{ \partial x_3 }{ \partial \xi_2 } & \frac{d}{dt} \frac{ \partial x_3 }{ \partial \xi_3 } \end{array} \right| } $
Using the identity derived above, it can be shown that in the derivative of the first row, only the terms with k=1 survive, since the coefficients of the other terms have a coefficient that is a determinant of a matrix with two rows the same.
For this reason, the determinant of the first matrix (in the expression for $ \frac{dJ}{dt} $ above) has a value of
$ \frac{ \partial v_1 }{ \partial x_1 } J $
and the others have similar values. This makes the final value of the time-derivative of the Jacobian:
$ \frac{dJ}{dt} = \left( \frac{d v_1}{d x_1} + \frac{ d v_2 }{d x_2} + \frac{ d v_3 }{ d x_3 } \right) J $
which can also be written,
$ \displaystyle{ \frac{1}{J} \frac{dJ}{dt} = div \boldsymbol{v} } $
or,
$ \displaystyle{ \frac{ d \ln{J} }{ dt } = div (\boldsymbol{v}) } $
This is also called the Euler Expansion Formula.
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,
$ div (\boldsymbol{v}) = 0 $
Reynolds' Transport Theorem
All of the above can be put together to derive Reynolds' transport theorem
For any extensive property of a fluid $ B $, there is a corresponding intensive property $ b = \frac{\partial B}{\partial m} $.
A material volume corresponding to a mass of fluid with the property $ b $ can be written as $ V_b(t) $.
The amount of $ b $ is related to the amount of $ B $ by:
$ F(t) = \iiint_{V_b(t)} \rho b(\boldsymbol{x},t) dV $
where $ \rho $ is the fluid density.
The interest is in the material derivative of this extensive property, $ \frac{dF}{dt} $.
The integral is over a varying volume $ V(t) $
So the integral $ \iiint_{V(t)} $ and the derivative $ \frac{d}{dt} $ don't commute.
But if the integration were with respect to a volume in $ \boldsymbol{\xi} $-space, they would
The volume $ \boldsymbol{x} = \boldsymbol{x}(\boldsymbol{\xi},t) $ and the Jacobian $ dV = J dV_0 $ allow us to do this
So the volume integral can be converted as follows:
$ \begin{array}{rcl} \frac{dF}{dt} &=& \frac{d}{dt} \iiint_{V(t)} \rho \mathcal{F}(\boldsymbol{x},t) dV \\ &=& \frac{d}{dt} \iiint_{V_0} \rho \mathcal{F}(\boldsymbol{x}(\boldsymbol{\xi},t)) J dV_0 \\ &=& \iiint_{V_0} \left[ \frac{d \rho \mathcal{F}}{dt} J + \rho \mathcal{F} \frac{dJ}{dt} \right] dV_0 \\ &=& \iiint_{V_0} \left[ \frac{d \rho \mathcal{F}}{dt} + \rho \mathcal{F} \frac{1}{J} \frac{dJ}{dt} \right] J dV_0 \end{array} $
Next, the Euler Expansion formula can be used to simplify the second term in the $ \left[ \right] $ brackets to get:
$ = \iiint_{V_0} \left[ \frac{d \rho \mathcal{F}}{dt} + \rho \mathcal{F} \left( \nabla \cdot \boldsymbol{v}_{\mathcal{F}} \right) \right] J dV_0 $
(where $ \boldsymbol{v}_{\mathcal{F}} $ is the velocity of the intensive property $ \mathcal{F} $ in the fluid)
Next, transforming from $ V_0 $ back to $ V $ (and noting that $ \frac{dF}{dt} = \iiint_{V(t)} \rho \mathcal{F}(\mathbf{x},t) dV $), it becomes Reynolds Transport Theorem:
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$ \frac{dF}{dt} = \left[ \iiint_{V(t)} \frac{d \rho \mathcal{F}}{dt} + \rho \mathcal{F} \left( \nabla \cdot \boldsymbol{v}_{\mathcal{F}} \right) \right] dV $ |
Now the definition of the substantial derivative can be used to put this in different forms
$ \frac{d}{dt} = \frac{ \partial }{ \partial t } + \boldsymbol{v}_{\mathcal{F}} \cdot \nabla $
to get
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$ \frac{dF}{dt} = \iiint_{V(t)} \left[ \frac{ \partial \rho \mathcal{F} }{ \partial t } + \nabla \cdot \left( \rho \mathcal{F} \boldsymbol{v}_{\mathcal{F}} \right) \right] dV $ |
Green's Theorem can be applied to this last divergence term, so that
where $ \boldsymbol{n} $ is the vector normal to the differential surface area $ dS $
This provides another form of Reynolds Transport Theorem:
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$ \frac{dF}{dt} = \iiint_{V(t)} \frac{ \partial \rho \mathcal{F} }{\partial t} dV + \iint_{S(t)} \rho \mathcal{F} \boldsymbol{v}_{\mathcal{F}} \cdot \boldsymbol{n} dS $ |