Mixing
Introducing Dimensionless Variables into PDE and Performing Change of Variables
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I have the following example in my notes:
Suppose we want to know how long it takes to cool down a long metal cylinder of radius $a$, with thermal diffusivity $alpha$, initial with a uniform temperature $T_i$ , placed in a large bath of water at temperature $T_w$.
Physics tells us that that the temperature, $T$, in the cylinder satisfies the heat equation in cylindrical coordinates. As the initial temperature is uniform we can ignore any angular variation, and as the rod is “long" we can ignore any variation along its length. The problem then becomes one of examining how the temperature changes radially, i.e. we want to find a function $T = T (r , t )$. With all of this the heat equation we want to solve is,
$dfrac partialT partialt = alpha dfrac1r dfracpartialpartialr left(r dfracpartialTpartialr right)$
For simplicity, assume that the water is able to remove all the heat which leaves the rod. This leads to the boundary condition $T(a,t) = T_w$.
As the initial temperature in the rod was uniformly $T_i$ we have,
$T(r, 0) = T_i, 0 le r < a$
We have 4 parameters: $alpha$, $a$, $T_i$ and $T_w$. We can reduce the number of parameters by introducing dimensionless variables. So let
$u = dfracT - T_wT_i - T_w$, $r^* = dfracra$, $t^* = dfractt_0$,
where $t_0$ is some parameter with dimension of time.
Performing the change of variables we find,
$dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$
$dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$
The original PDE can be rewritten as
$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$
What I don't understand is how we went from
$$u = dfracT - T_wT_i - T_w, r^* = dfracra, t^* = dfractt_0$$
to
$$dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$$
$$dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$$
$$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$$
It says that it uses change of variables, but I can't see how this was done, leading to the derivation of these 3 equations.
I would greatly appreciate it if people could please take the time to explain this to me.
pde mathematical-modeling dimensional-analysis
asked Jul 23 at 16:42
The Pointer
2,4752829
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up vote
2
down vote
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I have the following example in my notes:
Suppose we want to know how long it takes to cool down a long metal cylinder of radius $a$, with thermal diffusivity $alpha$, initial with a uniform temperature $T_i$ , placed in a large bath of water at temperature $T_w$.
Physics tells us that that the temperature, $T$, in the cylinder satisfies the heat equation in cylindrical coordinates. As the initial temperature is uniform we can ignore any angular variation, and as the rod is “long" we can ignore any variation along its length. The problem then becomes one of examining how the temperature changes radially, i.e. we want to find a function $T = T (r , t )$. With all of this the heat equation we want to solve is,
$dfrac partialT partialt = alpha dfrac1r dfracpartialpartialr left(r dfracpartialTpartialr right)$
For simplicity, assume that the water is able to remove all the heat which leaves the rod. This leads to the boundary condition $T(a,t) = T_w$.
As the initial temperature in the rod was uniformly $T_i$ we have,
$T(r, 0) = T_i, 0 le r < a$
We have 4 parameters: $alpha$, $a$, $T_i$ and $T_w$. We can reduce the number of parameters by introducing dimensionless variables. So let
$u = dfracT - T_wT_i - T_w$, $r^* = dfracra$, $t^* = dfractt_0$,
where $t_0$ is some parameter with dimension of time.
Performing the change of variables we find,
$dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$
$dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$
The original PDE can be rewritten as
$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$
What I don't understand is how we went from
$$u = dfracT - T_wT_i - T_w, r^* = dfracra, t^* = dfractt_0$$
to
$$dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$$
$$dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$$
$$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$$
It says that it uses change of variables, but I can't see how this was done, leading to the derivation of these 3 equations.
I would greatly appreciate it if people could please take the time to explain this to me.
pde mathematical-modeling dimensional-analysis
asked Jul 23 at 16:42
The Pointer
2,4752829
1
I'll do the first one, you can apply the same approach to get the others. Using $t'$ instead of $t^*$ beginalign fracpartial Tpartial t &= fracpartial Tpartial t' cdot fracd t'dt quad text(chain rule) \ &= fracpartial Tpartial t' cdot frac1t_0 quad (1)endalign Now $$T = (T_i - T_w)u - T_w implies fracpartial Tpartial t' = (T_i - T_w) fracpartial upartial t'$$ and substituting into $(1)$ yields $$fracpartial Tpartial t = frac(T_i - T_w)t_0 fracpartial upartial t'$$
– Mattos
Jul 24 at 1:56
@Mattos I understand now: for change of variables, we need to use the chain rule. Thanks for the clarification.
– The Pointer
Jul 24 at 15:48
Glad I could help.
– Mattos
Jul 24 at 23:57
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the following example in my notes:
Suppose we want to know how long it takes to cool down a long metal cylinder of radius $a$, with thermal diffusivity $alpha$, initial with a uniform temperature $T_i$ , placed in a large bath of water at temperature $T_w$.
Physics tells us that that the temperature, $T$, in the cylinder satisfies the heat equation in cylindrical coordinates. As the initial temperature is uniform we can ignore any angular variation, and as the rod is “long" we can ignore any variation along its length. The problem then becomes one of examining how the temperature changes radially, i.e. we want to find a function $T = T (r , t )$. With all of this the heat equation we want to solve is,
$dfrac partialT partialt = alpha dfrac1r dfracpartialpartialr left(r dfracpartialTpartialr right)$
For simplicity, assume that the water is able to remove all the heat which leaves the rod. This leads to the boundary condition $T(a,t) = T_w$.
As the initial temperature in the rod was uniformly $T_i$ we have,
$T(r, 0) = T_i, 0 le r < a$
We have 4 parameters: $alpha$, $a$, $T_i$ and $T_w$. We can reduce the number of parameters by introducing dimensionless variables. So let
$u = dfracT - T_wT_i - T_w$, $r^* = dfracra$, $t^* = dfractt_0$,
where $t_0$ is some parameter with dimension of time.
Performing the change of variables we find,
$dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$
$dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$
The original PDE can be rewritten as
$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$
What I don't understand is how we went from
$$u = dfracT - T_wT_i - T_w, r^* = dfracra, t^* = dfractt_0$$
to
$$dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$$
$$dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$$
$$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$$
It says that it uses change of variables, but I can't see how this was done, leading to the derivation of these 3 equations.
I would greatly appreciate it if people could please take the time to explain this to me.
pde mathematical-modeling dimensional-analysis
asked Jul 23 at 16:42
The Pointer
2,4752829
I have the following example in my notes:
Suppose we want to know how long it takes to cool down a long metal cylinder of radius $a$, with thermal diffusivity $alpha$, initial with a uniform temperature $T_i$ , placed in a large bath of water at temperature $T_w$.
Physics tells us that that the temperature, $T$, in the cylinder satisfies the heat equation in cylindrical coordinates. As the initial temperature is uniform we can ignore any angular variation, and as the rod is “long" we can ignore any variation along its length. The problem then becomes one of examining how the temperature changes radially, i.e. we want to find a function $T = T (r , t )$. With all of this the heat equation we want to solve is,
$dfrac partialT partialt = alpha dfrac1r dfracpartialpartialr left(r dfracpartialTpartialr right)$
For simplicity, assume that the water is able to remove all the heat which leaves the rod. This leads to the boundary condition $T(a,t) = T_w$.
As the initial temperature in the rod was uniformly $T_i$ we have,
$T(r, 0) = T_i, 0 le r < a$
We have 4 parameters: $alpha$, $a$, $T_i$ and $T_w$. We can reduce the number of parameters by introducing dimensionless variables. So let
$u = dfracT - T_wT_i - T_w$, $r^* = dfracra$, $t^* = dfractt_0$,
where $t_0$ is some parameter with dimension of time.
Performing the change of variables we find,
$dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$
$dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$
The original PDE can be rewritten as
$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$
What I don't understand is how we went from
$$u = dfracT - T_wT_i - T_w, r^* = dfracra, t^* = dfractt_0$$
to
$$dfracpartialTpartialt = dfrac(T_i - T_w)t_0 dfracpartialupartialt^*$$
$$dfracpartialTpartialr = dfrac(T_i - T_w)a dfracpartialupartialr^*$$
$$dfrac1t_0 dfracpartialupartialt^* = dfracalphaa^2 r^* dfracpartialpartialr^* left( r^* dfracpartialupartialr^* right)$$
It says that it uses change of variables, but I can't see how this was done, leading to the derivation of these 3 equations.
I would greatly appreciate it if people could please take the time to explain this to me.
pde mathematical-modeling dimensional-analysis
asked Jul 23 at 16:42
The Pointer
2,4752829
asked Jul 23 at 16:42
The Pointer
2,4752829
asked Jul 23 at 16:42
The Pointer
2,4752829
asked Jul 23 at 16:42
The Pointer
2,4752829
2,4752829
1
I'll do the first one, you can apply the same approach to get the others. Using $t'$ instead of $t^*$ beginalign fracpartial Tpartial t &= fracpartial Tpartial t' cdot fracd t'dt quad text(chain rule) \ &= fracpartial Tpartial t' cdot frac1t_0 quad (1)endalign Now $$T = (T_i - T_w)u - T_w implies fracpartial Tpartial t' = (T_i - T_w) fracpartial upartial t'$$ and substituting into $(1)$ yields $$fracpartial Tpartial t = frac(T_i - T_w)t_0 fracpartial upartial t'$$
– Mattos
Jul 24 at 1:56
@Mattos I understand now: for change of variables, we need to use the chain rule. Thanks for the clarification.
– The Pointer
Jul 24 at 15:48
Glad I could help.
– Mattos
Jul 24 at 23:57
add a comment |Â
1
I'll do the first one, you can apply the same approach to get the others. Using $t'$ instead of $t^*$ beginalign fracpartial Tpartial t &= fracpartial Tpartial t' cdot fracd t'dt quad text(chain rule) \ &= fracpartial Tpartial t' cdot frac1t_0 quad (1)endalign Now $$T = (T_i - T_w)u - T_w implies fracpartial Tpartial t' = (T_i - T_w) fracpartial upartial t'$$ and substituting into $(1)$ yields $$fracpartial Tpartial t = frac(T_i - T_w)t_0 fracpartial upartial t'$$
– Mattos
Jul 24 at 1:56
@Mattos I understand now: for change of variables, we need to use the chain rule. Thanks for the clarification.
– The Pointer
Jul 24 at 15:48
Glad I could help.
– Mattos
Jul 24 at 23:57
1
1
I'll do the first one, you can apply the same approach to get the others. Using $t'$ instead of $t^*$ beginalign fracpartial Tpartial t &= fracpartial Tpartial t' cdot fracd t'dt quad text(chain rule) \ &= fracpartial Tpartial t' cdot frac1t_0 quad (1)endalign Now $$T = (T_i - T_w)u - T_w implies fracpartial Tpartial t' = (T_i - T_w) fracpartial upartial t'$$ and substituting into $(1)$ yields $$fracpartial Tpartial t = frac(T_i - T_w)t_0 fracpartial upartial t'$$
– Mattos
Jul 24 at 1:56
I'll do the first one, you can apply the same approach to get the others. Using $t'$ instead of $t^*$ beginalign fracpartial Tpartial t &= fracpartial Tpartial t' cdot fracd t'dt quad text(chain rule) \ &= fracpartial Tpartial t' cdot frac1t_0 quad (1)endalign Now $$T = (T_i - T_w)u - T_w implies fracpartial Tpartial t' = (T_i - T_w) fracpartial upartial t'$$ and substituting into $(1)$ yields $$fracpartial Tpartial t = frac(T_i - T_w)t_0 fracpartial upartial t'$$
– Mattos
Jul 24 at 1:56
@Mattos I understand now: for change of variables, we need to use the chain rule. Thanks for the clarification.
– The Pointer
Jul 24 at 15:48
@Mattos I understand now: for change of variables, we need to use the chain rule. Thanks for the clarification.
– The Pointer
Jul 24 at 15:48
Glad I could help.
– Mattos
Jul 24 at 23:57
Glad I could help.
– Mattos
Jul 24 at 23:57
add a comment |Â
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I'll do the first one, you can apply the same approach to get the others. Using $t'$ instead of $t^*$ beginalign fracpartial Tpartial t &= fracpartial Tpartial t' cdot fracd t'dt quad text(chain rule) \ &= fracpartial Tpartial t' cdot frac1t_0 quad (1)endalign Now $$T = (T_i - T_w)u - T_w implies fracpartial Tpartial t' = (T_i - T_w) fracpartial upartial t'$$ and substituting into $(1)$ yields $$fracpartial Tpartial t = frac(T_i - T_w)t_0 fracpartial upartial t'$$
– Mattos
Jul 24 at 1:56
@Mattos I understand now: for change of variables, we need to use the chain rule. Thanks for the clarification.
– The Pointer
Jul 24 at 15:48
Glad I could help.
– Mattos
Jul 24 at 23:57