Physical meaning of potential in heat equation
I'm working on the mathematical theory of parabolic equations. The prototype of such equations is heat equation given as follows :
Let $Omega$ be a bounded region of the space and $T>0$ a fixed time. In $Omega_T=(0,T)times Omega$ we consider the following equation
$$u_t =alphaDelta u -a(x)u,$$
$$u(0,x)=f(x),$$
where $f$ is the initial condition, $a$ a bounded potential, $alpha>0$ is a constant, and $Delta$ is the Laplacian. I'm wondering to know the physical meaning of the coefficient $a$ (and may be $alpha$) and it's role in the heat process? Any reference or suggestion would be helpful.
thermodynamics potential diffusion heat-conduction
New contributor
add a comment |
I'm working on the mathematical theory of parabolic equations. The prototype of such equations is heat equation given as follows :
Let $Omega$ be a bounded region of the space and $T>0$ a fixed time. In $Omega_T=(0,T)times Omega$ we consider the following equation
$$u_t =alphaDelta u -a(x)u,$$
$$u(0,x)=f(x),$$
where $f$ is the initial condition, $a$ a bounded potential, $alpha>0$ is a constant, and $Delta$ is the Laplacian. I'm wondering to know the physical meaning of the coefficient $a$ (and may be $alpha$) and it's role in the heat process? Any reference or suggestion would be helpful.
thermodynamics potential diffusion heat-conduction
New contributor
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
– N. Steinle
yesterday
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
– S. Cho
yesterday
Your title asks about a potential? What potential are you talking about?
– Drew
20 hours ago
We call the coefficient $a$ a potential
– S. Cho
9 hours ago
add a comment |
I'm working on the mathematical theory of parabolic equations. The prototype of such equations is heat equation given as follows :
Let $Omega$ be a bounded region of the space and $T>0$ a fixed time. In $Omega_T=(0,T)times Omega$ we consider the following equation
$$u_t =alphaDelta u -a(x)u,$$
$$u(0,x)=f(x),$$
where $f$ is the initial condition, $a$ a bounded potential, $alpha>0$ is a constant, and $Delta$ is the Laplacian. I'm wondering to know the physical meaning of the coefficient $a$ (and may be $alpha$) and it's role in the heat process? Any reference or suggestion would be helpful.
thermodynamics potential diffusion heat-conduction
New contributor
I'm working on the mathematical theory of parabolic equations. The prototype of such equations is heat equation given as follows :
Let $Omega$ be a bounded region of the space and $T>0$ a fixed time. In $Omega_T=(0,T)times Omega$ we consider the following equation
$$u_t =alphaDelta u -a(x)u,$$
$$u(0,x)=f(x),$$
where $f$ is the initial condition, $a$ a bounded potential, $alpha>0$ is a constant, and $Delta$ is the Laplacian. I'm wondering to know the physical meaning of the coefficient $a$ (and may be $alpha$) and it's role in the heat process? Any reference or suggestion would be helpful.
thermodynamics potential diffusion heat-conduction
thermodynamics potential diffusion heat-conduction
New contributor
New contributor
edited 21 hours ago
Qmechanic♦
102k121831156
102k121831156
New contributor
asked yesterday
S. ChoS. Cho
1385
1385
New contributor
New contributor
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
– N. Steinle
yesterday
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
– S. Cho
yesterday
Your title asks about a potential? What potential are you talking about?
– Drew
20 hours ago
We call the coefficient $a$ a potential
– S. Cho
9 hours ago
add a comment |
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
– N. Steinle
yesterday
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
– S. Cho
yesterday
Your title asks about a potential? What potential are you talking about?
– Drew
20 hours ago
We call the coefficient $a$ a potential
– S. Cho
9 hours ago
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
– N. Steinle
yesterday
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
– N. Steinle
yesterday
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
– S. Cho
yesterday
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
– S. Cho
yesterday
Your title asks about a potential? What potential are you talking about?
– Drew
20 hours ago
Your title asks about a potential? What potential are you talking about?
– Drew
20 hours ago
We call the coefficient $a$ a potential
– S. Cho
9 hours ago
We call the coefficient $a$ a potential
– S. Cho
9 hours ago
add a comment |
3 Answers
3
active
oldest
votes
As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$kDelta T(x,t)-h(x)T(x,t)=crhodot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $crho$, then we have
$$alpha Delta u-left(frac{h(x)}{crho}right)u=u_t,$$
with $alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
– S. Cho
9 hours ago
What is $partial_nu$ in this case?
– Chemomechanics
1 hour ago
add a comment |
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $frac{partial u}{partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case, $a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
1
@S.Cho My pleasure! Which wiki article, specifically?
– N. Steinle
yesterday
1
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
– Chester Miller
yesterday
1
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
– t t t t
yesterday
1
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
– N. Steinle
yesterday
1
Oops. You're right. I didn't notice that.
– Chester Miller
yesterday
|
show 5 more comments
In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
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3 Answers
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3 Answers
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active
oldest
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active
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As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$kDelta T(x,t)-h(x)T(x,t)=crhodot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $crho$, then we have
$$alpha Delta u-left(frac{h(x)}{crho}right)u=u_t,$$
with $alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
– S. Cho
9 hours ago
What is $partial_nu$ in this case?
– Chemomechanics
1 hour ago
add a comment |
As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$kDelta T(x,t)-h(x)T(x,t)=crhodot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $crho$, then we have
$$alpha Delta u-left(frac{h(x)}{crho}right)u=u_t,$$
with $alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
– S. Cho
9 hours ago
What is $partial_nu$ in this case?
– Chemomechanics
1 hour ago
add a comment |
As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$kDelta T(x,t)-h(x)T(x,t)=crhodot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $crho$, then we have
$$alpha Delta u-left(frac{h(x)}{crho}right)u=u_t,$$
with $alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$kDelta T(x,t)-h(x)T(x,t)=crhodot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $crho$, then we have
$$alpha Delta u-left(frac{h(x)}{crho}right)u=u_t,$$
with $alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
answered yesterday
ChemomechanicsChemomechanics
4,68131023
4,68131023
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
– S. Cho
9 hours ago
What is $partial_nu$ in this case?
– Chemomechanics
1 hour ago
add a comment |
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
– S. Cho
9 hours ago
What is $partial_nu$ in this case?
– Chemomechanics
1 hour ago
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
– S. Cho
9 hours ago
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
– S. Cho
9 hours ago
What is $partial_nu$ in this case?
– Chemomechanics
1 hour ago
What is $partial_nu$ in this case?
– Chemomechanics
1 hour ago
add a comment |
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $frac{partial u}{partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case, $a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
1
@S.Cho My pleasure! Which wiki article, specifically?
– N. Steinle
yesterday
1
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
– Chester Miller
yesterday
1
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
– t t t t
yesterday
1
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
– N. Steinle
yesterday
1
Oops. You're right. I didn't notice that.
– Chester Miller
yesterday
|
show 5 more comments
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $frac{partial u}{partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case, $a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
1
@S.Cho My pleasure! Which wiki article, specifically?
– N. Steinle
yesterday
1
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
– Chester Miller
yesterday
1
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
– t t t t
yesterday
1
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
– N. Steinle
yesterday
1
Oops. You're right. I didn't notice that.
– Chester Miller
yesterday
|
show 5 more comments
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $frac{partial u}{partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case, $a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $frac{partial u}{partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case, $a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
edited 2 hours ago
answered yesterday
N. SteinleN. Steinle
1,404117
1,404117
1
@S.Cho My pleasure! Which wiki article, specifically?
– N. Steinle
yesterday
1
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
– Chester Miller
yesterday
1
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
– t t t t
yesterday
1
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
– N. Steinle
yesterday
1
Oops. You're right. I didn't notice that.
– Chester Miller
yesterday
|
show 5 more comments
1
@S.Cho My pleasure! Which wiki article, specifically?
– N. Steinle
yesterday
1
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
– Chester Miller
yesterday
1
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
– t t t t
yesterday
1
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
– N. Steinle
yesterday
1
Oops. You're right. I didn't notice that.
– Chester Miller
yesterday
1
1
@S.Cho My pleasure! Which wiki article, specifically?
– N. Steinle
yesterday
@S.Cho My pleasure! Which wiki article, specifically?
– N. Steinle
yesterday
1
1
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
– Chester Miller
yesterday
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
– Chester Miller
yesterday
1
1
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
– t t t t
yesterday
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
– t t t t
yesterday
1
1
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
– N. Steinle
yesterday
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
– N. Steinle
yesterday
1
1
Oops. You're right. I didn't notice that.
– Chester Miller
yesterday
Oops. You're right. I didn't notice that.
– Chester Miller
yesterday
|
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In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
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In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
New contributor
add a comment |
In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
New contributor
In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
New contributor
edited yesterday
New contributor
answered yesterday
Vincent FraticelliVincent Fraticelli
443
443
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New contributor
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add a comment |
S. Cho is a new contributor. Be nice, and check out our Code of Conduct.
S. Cho is a new contributor. Be nice, and check out our Code of Conduct.
S. Cho is a new contributor. Be nice, and check out our Code of Conduct.
S. Cho is a new contributor. Be nice, and check out our Code of Conduct.
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Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
– N. Steinle
yesterday
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
– S. Cho
yesterday
Your title asks about a potential? What potential are you talking about?
– Drew
20 hours ago
We call the coefficient $a$ a potential
– S. Cho
9 hours ago