When normalizing an equation, why are the non-dimensional terms expected to be order 1?












0












$begingroup$


I'm familiar with the procedure of normalization, but I'm unfamiliar with some of the theory involved.



For instance, using the Navier Stokes equations where the density and viscosity can be treated as constant. I have also neglected gravity.



$$rhodot{mathbf{u}}+rholeft(mathbf{u}cdotnablaright)mathbf{u}=-nabla p+munabla^2mathbf{u}$$



Let me argue that for the flow in question, the relevant scaling terms are the free stream velocity $U$, the distance the flow covers $L$, the time it takes the flow to cover this distance $T=L/U$, and finally the pressure difference between the surface and the free stream $p_0-p_infty$. That is ...
$$mathbf{u}=mathbf{u}^*U\
nabla=nabla^* L^{-1}\
t=t^*frac{L}{U}\
p=p^*(p_0-p_infty)$$
where * is to mean non-dimensional terms.



To make this problem simpler, I will also be assuming all three position, and velocity components to scale with $L$ and $U$ respectively; even though this is unlikely to be physical. Considering the $x$ component ...



$$rhodot{u}+rho upartial_x u+rho vpartial_y u+rho wpartial_z u=-partial _x p+mupartial_x^2 u+mupartial_y^2 u+mupartial_z^2 u$$
inserting the scaling terms ...
$$dot{u^*}left(frac{rho U^2}{L}right)+u^*partial_{x^*} u^*left(frac{rho U^2}{L}right)+v^*partial_{y^*} u^*left(frac{rho U^2}{L}right)+ w^*partial_{z^*} u^*left(frac{rho U^2}{L}right)=-partial _{x^*} p^*left(frac{p_0-p_infty}{L}right)+partial_{x^*}^2 u^*left(frac{mu U}{L^2}right)+partial_{y^*}^2 u^*left(frac{mu U}{L^2}right)+partial_{z^*}^2 u^*left(frac{mu U}{L^2}right).$$
Usually I would not stop at this point but, my question is about some of the non-dimensional terms. For instance the term $w^*partial_{z^*}u^*$ has no dimensions and is said to be of order 1. What exactly does this mean?
If I continue with the normalization process ...
$$dot{u^*}+u^*partial_{x^*} u^*+v^*partial_{y^*} u^*+ w^*partial_{z^*} u^*=-partial _{x^*} p^*text{Eu}+left(partial_{x^*}^2 u^*+partial_{y^*}^2 u^*+partial_{z^*}^2 u^*right)frac{1}{text{Re}}$$



where Eu and Re are the non-dimensional numbers the Euler and Reynolds numbers. How does the non-dimensional terms being order 1 contribute to the analysis of experimental data?






field-theory classical-mechanics fluid-dynamics invariance







share|cite|improve this question











$endgroup$












  • $begingroup$
    The purpose of nondimensionalisation is to see the order of each term in the equation. After nondimensionalisation, often people keep only the leading order term (usually order 1) and neglect lower order terms; typically one has a small parameter (let's call this $varepsilon$) from the system and saying that a term is of order 1 means that the dimensionless number in front of the term is some number between 1 to 10.
    $endgroup$
    – Chee Han
    Nov 21 '18 at 21:43










  • $begingroup$
    I guess I'm trying to get at how do you know these non-dimensional terms are order 1?
    $endgroup$
    – WnGatRC456
    Nov 24 '18 at 21:41
















0












$begingroup$


I'm familiar with the procedure of normalization, but I'm unfamiliar with some of the theory involved.



For instance, using the Navier Stokes equations where the density and viscosity can be treated as constant. I have also neglected gravity.



$$rhodot{mathbf{u}}+rholeft(mathbf{u}cdotnablaright)mathbf{u}=-nabla p+munabla^2mathbf{u}$$



Let me argue that for the flow in question, the relevant scaling terms are the free stream velocity $U$, the distance the flow covers $L$, the time it takes the flow to cover this distance $T=L/U$, and finally the pressure difference between the surface and the free stream $p_0-p_infty$. That is ...
$$mathbf{u}=mathbf{u}^*U\
nabla=nabla^* L^{-1}\
t=t^*frac{L}{U}\
p=p^*(p_0-p_infty)$$
where * is to mean non-dimensional terms.



To make this problem simpler, I will also be assuming all three position, and velocity components to scale with $L$ and $U$ respectively; even though this is unlikely to be physical. Considering the $x$ component ...



$$rhodot{u}+rho upartial_x u+rho vpartial_y u+rho wpartial_z u=-partial _x p+mupartial_x^2 u+mupartial_y^2 u+mupartial_z^2 u$$
inserting the scaling terms ...
$$dot{u^*}left(frac{rho U^2}{L}right)+u^*partial_{x^*} u^*left(frac{rho U^2}{L}right)+v^*partial_{y^*} u^*left(frac{rho U^2}{L}right)+ w^*partial_{z^*} u^*left(frac{rho U^2}{L}right)=-partial _{x^*} p^*left(frac{p_0-p_infty}{L}right)+partial_{x^*}^2 u^*left(frac{mu U}{L^2}right)+partial_{y^*}^2 u^*left(frac{mu U}{L^2}right)+partial_{z^*}^2 u^*left(frac{mu U}{L^2}right).$$
Usually I would not stop at this point but, my question is about some of the non-dimensional terms. For instance the term $w^*partial_{z^*}u^*$ has no dimensions and is said to be of order 1. What exactly does this mean?
If I continue with the normalization process ...
$$dot{u^*}+u^*partial_{x^*} u^*+v^*partial_{y^*} u^*+ w^*partial_{z^*} u^*=-partial _{x^*} p^*text{Eu}+left(partial_{x^*}^2 u^*+partial_{y^*}^2 u^*+partial_{z^*}^2 u^*right)frac{1}{text{Re}}$$



where Eu and Re are the non-dimensional numbers the Euler and Reynolds numbers. How does the non-dimensional terms being order 1 contribute to the analysis of experimental data?






field-theory classical-mechanics fluid-dynamics invariance







share|cite|improve this question











$endgroup$












  • $begingroup$
    The purpose of nondimensionalisation is to see the order of each term in the equation. After nondimensionalisation, often people keep only the leading order term (usually order 1) and neglect lower order terms; typically one has a small parameter (let's call this $varepsilon$) from the system and saying that a term is of order 1 means that the dimensionless number in front of the term is some number between 1 to 10.
    $endgroup$
    – Chee Han
    Nov 21 '18 at 21:43










  • $begingroup$
    I guess I'm trying to get at how do you know these non-dimensional terms are order 1?
    $endgroup$
    – WnGatRC456
    Nov 24 '18 at 21:41














0












0








0





$begingroup$


I'm familiar with the procedure of normalization, but I'm unfamiliar with some of the theory involved.



For instance, using the Navier Stokes equations where the density and viscosity can be treated as constant. I have also neglected gravity.



$$rhodot{mathbf{u}}+rholeft(mathbf{u}cdotnablaright)mathbf{u}=-nabla p+munabla^2mathbf{u}$$



Let me argue that for the flow in question, the relevant scaling terms are the free stream velocity $U$, the distance the flow covers $L$, the time it takes the flow to cover this distance $T=L/U$, and finally the pressure difference between the surface and the free stream $p_0-p_infty$. That is ...
$$mathbf{u}=mathbf{u}^*U\
nabla=nabla^* L^{-1}\
t=t^*frac{L}{U}\
p=p^*(p_0-p_infty)$$
where * is to mean non-dimensional terms.



To make this problem simpler, I will also be assuming all three position, and velocity components to scale with $L$ and $U$ respectively; even though this is unlikely to be physical. Considering the $x$ component ...



$$rhodot{u}+rho upartial_x u+rho vpartial_y u+rho wpartial_z u=-partial _x p+mupartial_x^2 u+mupartial_y^2 u+mupartial_z^2 u$$
inserting the scaling terms ...
$$dot{u^*}left(frac{rho U^2}{L}right)+u^*partial_{x^*} u^*left(frac{rho U^2}{L}right)+v^*partial_{y^*} u^*left(frac{rho U^2}{L}right)+ w^*partial_{z^*} u^*left(frac{rho U^2}{L}right)=-partial _{x^*} p^*left(frac{p_0-p_infty}{L}right)+partial_{x^*}^2 u^*left(frac{mu U}{L^2}right)+partial_{y^*}^2 u^*left(frac{mu U}{L^2}right)+partial_{z^*}^2 u^*left(frac{mu U}{L^2}right).$$
Usually I would not stop at this point but, my question is about some of the non-dimensional terms. For instance the term $w^*partial_{z^*}u^*$ has no dimensions and is said to be of order 1. What exactly does this mean?
If I continue with the normalization process ...
$$dot{u^*}+u^*partial_{x^*} u^*+v^*partial_{y^*} u^*+ w^*partial_{z^*} u^*=-partial _{x^*} p^*text{Eu}+left(partial_{x^*}^2 u^*+partial_{y^*}^2 u^*+partial_{z^*}^2 u^*right)frac{1}{text{Re}}$$



where Eu and Re are the non-dimensional numbers the Euler and Reynolds numbers. How does the non-dimensional terms being order 1 contribute to the analysis of experimental data?






field-theory classical-mechanics fluid-dynamics invariance







share|cite|improve this question











$endgroup$




I'm familiar with the procedure of normalization, but I'm unfamiliar with some of the theory involved.



For instance, using the Navier Stokes equations where the density and viscosity can be treated as constant. I have also neglected gravity.



$$rhodot{mathbf{u}}+rholeft(mathbf{u}cdotnablaright)mathbf{u}=-nabla p+munabla^2mathbf{u}$$



Let me argue that for the flow in question, the relevant scaling terms are the free stream velocity $U$, the distance the flow covers $L$, the time it takes the flow to cover this distance $T=L/U$, and finally the pressure difference between the surface and the free stream $p_0-p_infty$. That is ...
$$mathbf{u}=mathbf{u}^*U\
nabla=nabla^* L^{-1}\
t=t^*frac{L}{U}\
p=p^*(p_0-p_infty)$$
where * is to mean non-dimensional terms.



To make this problem simpler, I will also be assuming all three position, and velocity components to scale with $L$ and $U$ respectively; even though this is unlikely to be physical. Considering the $x$ component ...



$$rhodot{u}+rho upartial_x u+rho vpartial_y u+rho wpartial_z u=-partial _x p+mupartial_x^2 u+mupartial_y^2 u+mupartial_z^2 u$$
inserting the scaling terms ...
$$dot{u^*}left(frac{rho U^2}{L}right)+u^*partial_{x^*} u^*left(frac{rho U^2}{L}right)+v^*partial_{y^*} u^*left(frac{rho U^2}{L}right)+ w^*partial_{z^*} u^*left(frac{rho U^2}{L}right)=-partial _{x^*} p^*left(frac{p_0-p_infty}{L}right)+partial_{x^*}^2 u^*left(frac{mu U}{L^2}right)+partial_{y^*}^2 u^*left(frac{mu U}{L^2}right)+partial_{z^*}^2 u^*left(frac{mu U}{L^2}right).$$
Usually I would not stop at this point but, my question is about some of the non-dimensional terms. For instance the term $w^*partial_{z^*}u^*$ has no dimensions and is said to be of order 1. What exactly does this mean?
If I continue with the normalization process ...
$$dot{u^*}+u^*partial_{x^*} u^*+v^*partial_{y^*} u^*+ w^*partial_{z^*} u^*=-partial _{x^*} p^*text{Eu}+left(partial_{x^*}^2 u^*+partial_{y^*}^2 u^*+partial_{z^*}^2 u^*right)frac{1}{text{Re}}$$



where Eu and Re are the non-dimensional numbers the Euler and Reynolds numbers. How does the non-dimensional terms being order 1 contribute to the analysis of experimental data?






field-theory classical-mechanics fluid-dynamics invariance




field-theory classical-mechanics fluid-dynamics invariance






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 21:48







WnGatRC456

















asked Nov 10 '18 at 16:48









WnGatRC456WnGatRC456

10810




10810












  • $begingroup$
    The purpose of nondimensionalisation is to see the order of each term in the equation. After nondimensionalisation, often people keep only the leading order term (usually order 1) and neglect lower order terms; typically one has a small parameter (let's call this $varepsilon$) from the system and saying that a term is of order 1 means that the dimensionless number in front of the term is some number between 1 to 10.
    $endgroup$
    – Chee Han
    Nov 21 '18 at 21:43










  • $begingroup$
    I guess I'm trying to get at how do you know these non-dimensional terms are order 1?
    $endgroup$
    – WnGatRC456
    Nov 24 '18 at 21:41


















  • $begingroup$
    The purpose of nondimensionalisation is to see the order of each term in the equation. After nondimensionalisation, often people keep only the leading order term (usually order 1) and neglect lower order terms; typically one has a small parameter (let's call this $varepsilon$) from the system and saying that a term is of order 1 means that the dimensionless number in front of the term is some number between 1 to 10.
    $endgroup$
    – Chee Han
    Nov 21 '18 at 21:43










  • $begingroup$
    I guess I'm trying to get at how do you know these non-dimensional terms are order 1?
    $endgroup$
    – WnGatRC456
    Nov 24 '18 at 21:41
















$begingroup$
The purpose of nondimensionalisation is to see the order of each term in the equation. After nondimensionalisation, often people keep only the leading order term (usually order 1) and neglect lower order terms; typically one has a small parameter (let's call this $varepsilon$) from the system and saying that a term is of order 1 means that the dimensionless number in front of the term is some number between 1 to 10.
$endgroup$
– Chee Han
Nov 21 '18 at 21:43




$begingroup$
The purpose of nondimensionalisation is to see the order of each term in the equation. After nondimensionalisation, often people keep only the leading order term (usually order 1) and neglect lower order terms; typically one has a small parameter (let's call this $varepsilon$) from the system and saying that a term is of order 1 means that the dimensionless number in front of the term is some number between 1 to 10.
$endgroup$
– Chee Han
Nov 21 '18 at 21:43












$begingroup$
I guess I'm trying to get at how do you know these non-dimensional terms are order 1?
$endgroup$
– WnGatRC456
Nov 24 '18 at 21:41




$begingroup$
I guess I'm trying to get at how do you know these non-dimensional terms are order 1?
$endgroup$
– WnGatRC456
Nov 24 '18 at 21:41










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