Mapping PDE Coefficients to a Heat Storage application The Next CEO of Stack OverflowSolving Heat Equation PDEBlack-Scholes PDE to heat equation, nonconstant coefficientsRewriting the heat diffusion equation with temperature dependent diffusion coefficient to include joule heating.What if we change one of Fourier's law of heat conductionPDE Fourier transform(heat equation)2D Heat Transfer Laplacian with Neumann, Robin, and Dirichlet Conditions on a semi-infinite slabApplication of the heat equation, PDESolving Fick's second law with constant surface fluxPDE Cartesian/Polar forms and Numerical Solver issuesHeat equation for a cylinder in cylindrical coordinates

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Mapping PDE Coefficients to a Heat Storage application



The Next CEO of Stack OverflowSolving Heat Equation PDEBlack-Scholes PDE to heat equation, nonconstant coefficientsRewriting the heat diffusion equation with temperature dependent diffusion coefficient to include joule heating.What if we change one of Fourier's law of heat conductionPDE Fourier transform(heat equation)2D Heat Transfer Laplacian with Neumann, Robin, and Dirichlet Conditions on a semi-infinite slabApplication of the heat equation, PDESolving Fick's second law with constant surface fluxPDE Cartesian/Polar forms and Numerical Solver issuesHeat equation for a cylinder in cylindrical coordinates










0












$begingroup$


I am attempting to solve a heat storage problem using Matlab and the PDE Toolbox and could use some assistance with the math portion.



The task is to model a concrete thermal battery with an array of embedded pipes that carry steam from a boiler. The heat from the steam warms the battery so that the heat can be used later when low temperature steam flows through the pipes (while the boiler is ramping up for example). In order to model the entire battery, I figure a representative element needs solving first. In this case, a representative element is one pipe and the concrete directly surrounding it up to the radius of Re.



I'm still learning Matlab, so I looked around online for some examples I might be able to build off of and stumbled upon this example (Nonlinear Heat Transfer in a Thin Plate).



I think this example is similar because the plate can be revolved to match the shape of the concrete element and the bottom edge fixed at a high temperature is similar to the heat from the steam at the center of the element. So the figure showing the temperature of the plate would be modified with the x-axis as the axial (z) coordinate and the y-axis as the radial (r) coordinate.



The part that is giving me a problem (at least for now) is setting up the equation to be in the correct form for the PDE Toolbox. I have to use the specifyCoefficients function to map out values for m, d, c, a, and f in the general PDE:



$$mfrac∂^2u∂t^2 + dfrac∂u∂t - ∇·(c∇u) + au = f$$



The energy equation of the concrete element is:



$$ρc_pfrac∂T∂t = λ(frac∂^2T∂r^2 + frac1rfrac∂T∂r + frac∂^2T∂z^2)$$



I think I have some of the coefficients correct, but I would appreciate input from others.



I believe m = 0



I believe d = concrete_density x concrete_specific_heat x element_thickness



I believe c = concrete_thermal_conductivity x element_thickness



I'm not sure about a or f.



Thanks for the help.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    I am attempting to solve a heat storage problem using Matlab and the PDE Toolbox and could use some assistance with the math portion.



    The task is to model a concrete thermal battery with an array of embedded pipes that carry steam from a boiler. The heat from the steam warms the battery so that the heat can be used later when low temperature steam flows through the pipes (while the boiler is ramping up for example). In order to model the entire battery, I figure a representative element needs solving first. In this case, a representative element is one pipe and the concrete directly surrounding it up to the radius of Re.



    I'm still learning Matlab, so I looked around online for some examples I might be able to build off of and stumbled upon this example (Nonlinear Heat Transfer in a Thin Plate).



    I think this example is similar because the plate can be revolved to match the shape of the concrete element and the bottom edge fixed at a high temperature is similar to the heat from the steam at the center of the element. So the figure showing the temperature of the plate would be modified with the x-axis as the axial (z) coordinate and the y-axis as the radial (r) coordinate.



    The part that is giving me a problem (at least for now) is setting up the equation to be in the correct form for the PDE Toolbox. I have to use the specifyCoefficients function to map out values for m, d, c, a, and f in the general PDE:



    $$mfrac∂^2u∂t^2 + dfrac∂u∂t - ∇·(c∇u) + au = f$$



    The energy equation of the concrete element is:



    $$ρc_pfrac∂T∂t = λ(frac∂^2T∂r^2 + frac1rfrac∂T∂r + frac∂^2T∂z^2)$$



    I think I have some of the coefficients correct, but I would appreciate input from others.



    I believe m = 0



    I believe d = concrete_density x concrete_specific_heat x element_thickness



    I believe c = concrete_thermal_conductivity x element_thickness



    I'm not sure about a or f.



    Thanks for the help.










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      I am attempting to solve a heat storage problem using Matlab and the PDE Toolbox and could use some assistance with the math portion.



      The task is to model a concrete thermal battery with an array of embedded pipes that carry steam from a boiler. The heat from the steam warms the battery so that the heat can be used later when low temperature steam flows through the pipes (while the boiler is ramping up for example). In order to model the entire battery, I figure a representative element needs solving first. In this case, a representative element is one pipe and the concrete directly surrounding it up to the radius of Re.



      I'm still learning Matlab, so I looked around online for some examples I might be able to build off of and stumbled upon this example (Nonlinear Heat Transfer in a Thin Plate).



      I think this example is similar because the plate can be revolved to match the shape of the concrete element and the bottom edge fixed at a high temperature is similar to the heat from the steam at the center of the element. So the figure showing the temperature of the plate would be modified with the x-axis as the axial (z) coordinate and the y-axis as the radial (r) coordinate.



      The part that is giving me a problem (at least for now) is setting up the equation to be in the correct form for the PDE Toolbox. I have to use the specifyCoefficients function to map out values for m, d, c, a, and f in the general PDE:



      $$mfrac∂^2u∂t^2 + dfrac∂u∂t - ∇·(c∇u) + au = f$$



      The energy equation of the concrete element is:



      $$ρc_pfrac∂T∂t = λ(frac∂^2T∂r^2 + frac1rfrac∂T∂r + frac∂^2T∂z^2)$$



      I think I have some of the coefficients correct, but I would appreciate input from others.



      I believe m = 0



      I believe d = concrete_density x concrete_specific_heat x element_thickness



      I believe c = concrete_thermal_conductivity x element_thickness



      I'm not sure about a or f.



      Thanks for the help.










      share|cite|improve this question











      $endgroup$




      I am attempting to solve a heat storage problem using Matlab and the PDE Toolbox and could use some assistance with the math portion.



      The task is to model a concrete thermal battery with an array of embedded pipes that carry steam from a boiler. The heat from the steam warms the battery so that the heat can be used later when low temperature steam flows through the pipes (while the boiler is ramping up for example). In order to model the entire battery, I figure a representative element needs solving first. In this case, a representative element is one pipe and the concrete directly surrounding it up to the radius of Re.



      I'm still learning Matlab, so I looked around online for some examples I might be able to build off of and stumbled upon this example (Nonlinear Heat Transfer in a Thin Plate).



      I think this example is similar because the plate can be revolved to match the shape of the concrete element and the bottom edge fixed at a high temperature is similar to the heat from the steam at the center of the element. So the figure showing the temperature of the plate would be modified with the x-axis as the axial (z) coordinate and the y-axis as the radial (r) coordinate.



      The part that is giving me a problem (at least for now) is setting up the equation to be in the correct form for the PDE Toolbox. I have to use the specifyCoefficients function to map out values for m, d, c, a, and f in the general PDE:



      $$mfrac∂^2u∂t^2 + dfrac∂u∂t - ∇·(c∇u) + au = f$$



      The energy equation of the concrete element is:



      $$ρc_pfrac∂T∂t = λ(frac∂^2T∂r^2 + frac1rfrac∂T∂r + frac∂^2T∂z^2)$$



      I think I have some of the coefficients correct, but I would appreciate input from others.



      I believe m = 0



      I believe d = concrete_density x concrete_specific_heat x element_thickness



      I believe c = concrete_thermal_conductivity x element_thickness



      I'm not sure about a or f.



      Thanks for the help.







      pde matlab






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 19 at 14:44







      Michael

















      asked Mar 18 at 22:39









      MichaelMichael

      11




      11




















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          The two equations you're written are not equivalent and cannot be compared.



          If $c$ is constant than the first equation is



          $$ m fracpartial^2 upartial t^2 + dfracpartial upartial t - cleft(fracpartial^2 upartial r^2 + frac1rfracpartial upartial r + fracpartial^2 upartial z^2right) + au = f $$



          The second equation is



          $$ -lambdafracpartial^2 Tpartial t^2 + rho c_pfracpartial Tpartial t -lambdaleft(frac1rfracpartial Tpartial r + fracpartial^2 Tpartial z^2right) = 0 $$



          The closest you can get is



          beginalign
          m &= -lambda \
          d &= rho c_p \
          c &= lambda \
          a &= 0 \
          f &= 0
          endalign



          But there's still a $fracpartial^2 Tpartial r^2$ missing in the second equation



          If you actually meant to write



          $$ rho c_pfracpartial upartial t - lambdaleft(fracpartial^2 upartial r^2 + frac1rfracpartial upartial r + fracpartial^2 upartial z^2right) = 0 $$



          Then $c=lambda$, $d = rho c_p$ and every other constant is $0$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you! Yes, there was a typo in the second equation. It was supposed to be r, not t.
            $endgroup$
            – Michael
            Mar 19 at 15:04










          • $begingroup$
            Would boundary conditions effect what the other coefficients should be set to?
            $endgroup$
            – Michael
            Mar 19 at 17:50










          • $begingroup$
            No, boundary conditions don't change the equation
            $endgroup$
            – Dylan
            Mar 19 at 17:58











          Your Answer





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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          The two equations you're written are not equivalent and cannot be compared.



          If $c$ is constant than the first equation is



          $$ m fracpartial^2 upartial t^2 + dfracpartial upartial t - cleft(fracpartial^2 upartial r^2 + frac1rfracpartial upartial r + fracpartial^2 upartial z^2right) + au = f $$



          The second equation is



          $$ -lambdafracpartial^2 Tpartial t^2 + rho c_pfracpartial Tpartial t -lambdaleft(frac1rfracpartial Tpartial r + fracpartial^2 Tpartial z^2right) = 0 $$



          The closest you can get is



          beginalign
          m &= -lambda \
          d &= rho c_p \
          c &= lambda \
          a &= 0 \
          f &= 0
          endalign



          But there's still a $fracpartial^2 Tpartial r^2$ missing in the second equation



          If you actually meant to write



          $$ rho c_pfracpartial upartial t - lambdaleft(fracpartial^2 upartial r^2 + frac1rfracpartial upartial r + fracpartial^2 upartial z^2right) = 0 $$



          Then $c=lambda$, $d = rho c_p$ and every other constant is $0$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you! Yes, there was a typo in the second equation. It was supposed to be r, not t.
            $endgroup$
            – Michael
            Mar 19 at 15:04










          • $begingroup$
            Would boundary conditions effect what the other coefficients should be set to?
            $endgroup$
            – Michael
            Mar 19 at 17:50










          • $begingroup$
            No, boundary conditions don't change the equation
            $endgroup$
            – Dylan
            Mar 19 at 17:58















          0












          $begingroup$

          The two equations you're written are not equivalent and cannot be compared.



          If $c$ is constant than the first equation is



          $$ m fracpartial^2 upartial t^2 + dfracpartial upartial t - cleft(fracpartial^2 upartial r^2 + frac1rfracpartial upartial r + fracpartial^2 upartial z^2right) + au = f $$



          The second equation is



          $$ -lambdafracpartial^2 Tpartial t^2 + rho c_pfracpartial Tpartial t -lambdaleft(frac1rfracpartial Tpartial r + fracpartial^2 Tpartial z^2right) = 0 $$



          The closest you can get is



          beginalign
          m &= -lambda \
          d &= rho c_p \
          c &= lambda \
          a &= 0 \
          f &= 0
          endalign



          But there's still a $fracpartial^2 Tpartial r^2$ missing in the second equation



          If you actually meant to write



          $$ rho c_pfracpartial upartial t - lambdaleft(fracpartial^2 upartial r^2 + frac1rfracpartial upartial r + fracpartial^2 upartial z^2right) = 0 $$



          Then $c=lambda$, $d = rho c_p$ and every other constant is $0$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you! Yes, there was a typo in the second equation. It was supposed to be r, not t.
            $endgroup$
            – Michael
            Mar 19 at 15:04










          • $begingroup$
            Would boundary conditions effect what the other coefficients should be set to?
            $endgroup$
            – Michael
            Mar 19 at 17:50










          • $begingroup$
            No, boundary conditions don't change the equation
            $endgroup$
            – Dylan
            Mar 19 at 17:58













          0












          0








          0





          $begingroup$

          The two equations you're written are not equivalent and cannot be compared.



          If $c$ is constant than the first equation is



          $$ m fracpartial^2 upartial t^2 + dfracpartial upartial t - cleft(fracpartial^2 upartial r^2 + frac1rfracpartial upartial r + fracpartial^2 upartial z^2right) + au = f $$



          The second equation is



          $$ -lambdafracpartial^2 Tpartial t^2 + rho c_pfracpartial Tpartial t -lambdaleft(frac1rfracpartial Tpartial r + fracpartial^2 Tpartial z^2right) = 0 $$



          The closest you can get is



          beginalign
          m &= -lambda \
          d &= rho c_p \
          c &= lambda \
          a &= 0 \
          f &= 0
          endalign



          But there's still a $fracpartial^2 Tpartial r^2$ missing in the second equation



          If you actually meant to write



          $$ rho c_pfracpartial upartial t - lambdaleft(fracpartial^2 upartial r^2 + frac1rfracpartial upartial r + fracpartial^2 upartial z^2right) = 0 $$



          Then $c=lambda$, $d = rho c_p$ and every other constant is $0$






          share|cite|improve this answer









          $endgroup$



          The two equations you're written are not equivalent and cannot be compared.



          If $c$ is constant than the first equation is



          $$ m fracpartial^2 upartial t^2 + dfracpartial upartial t - cleft(fracpartial^2 upartial r^2 + frac1rfracpartial upartial r + fracpartial^2 upartial z^2right) + au = f $$



          The second equation is



          $$ -lambdafracpartial^2 Tpartial t^2 + rho c_pfracpartial Tpartial t -lambdaleft(frac1rfracpartial Tpartial r + fracpartial^2 Tpartial z^2right) = 0 $$



          The closest you can get is



          beginalign
          m &= -lambda \
          d &= rho c_p \
          c &= lambda \
          a &= 0 \
          f &= 0
          endalign



          But there's still a $fracpartial^2 Tpartial r^2$ missing in the second equation



          If you actually meant to write



          $$ rho c_pfracpartial upartial t - lambdaleft(fracpartial^2 upartial r^2 + frac1rfracpartial upartial r + fracpartial^2 upartial z^2right) = 0 $$



          Then $c=lambda$, $d = rho c_p$ and every other constant is $0$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 19 at 8:03









          DylanDylan

          14.2k31127




          14.2k31127











          • $begingroup$
            Thank you! Yes, there was a typo in the second equation. It was supposed to be r, not t.
            $endgroup$
            – Michael
            Mar 19 at 15:04










          • $begingroup$
            Would boundary conditions effect what the other coefficients should be set to?
            $endgroup$
            – Michael
            Mar 19 at 17:50










          • $begingroup$
            No, boundary conditions don't change the equation
            $endgroup$
            – Dylan
            Mar 19 at 17:58
















          • $begingroup$
            Thank you! Yes, there was a typo in the second equation. It was supposed to be r, not t.
            $endgroup$
            – Michael
            Mar 19 at 15:04










          • $begingroup$
            Would boundary conditions effect what the other coefficients should be set to?
            $endgroup$
            – Michael
            Mar 19 at 17:50










          • $begingroup$
            No, boundary conditions don't change the equation
            $endgroup$
            – Dylan
            Mar 19 at 17:58















          $begingroup$
          Thank you! Yes, there was a typo in the second equation. It was supposed to be r, not t.
          $endgroup$
          – Michael
          Mar 19 at 15:04




          $begingroup$
          Thank you! Yes, there was a typo in the second equation. It was supposed to be r, not t.
          $endgroup$
          – Michael
          Mar 19 at 15:04












          $begingroup$
          Would boundary conditions effect what the other coefficients should be set to?
          $endgroup$
          – Michael
          Mar 19 at 17:50




          $begingroup$
          Would boundary conditions effect what the other coefficients should be set to?
          $endgroup$
          – Michael
          Mar 19 at 17:50












          $begingroup$
          No, boundary conditions don't change the equation
          $endgroup$
          – Dylan
          Mar 19 at 17:58




          $begingroup$
          No, boundary conditions don't change the equation
          $endgroup$
          – Dylan
          Mar 19 at 17:58

















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