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I would like to plot the temperature profile of a partially filled rotating cylinder. However, I am only familiar with how to plot pressure profiles. Can you please suggest a method for deriving the temperature from the pressure data, assuming an initial temperature of 50 degrees Celsius?

g = 9.8; \[Gamma] = \[Rho]*g; h = 3; r = Sqrt[x^2 + y^2]; 
P[z_] := ((\[Rho]*\[Omega]^2*r^2)/2 - \[Gamma]*z + \[Gamma]*h)/1000;
Column[{
  Text@Style["isobaric surfaces in rotation", 17],
           Show[
     ContourPlot3D[P[z], {x, -1, 1}, {y, -1, 1}, {z, 0, h}, 
     Mesh -> None, ColorFunction -> (Hue[#4/1] &), Contours -> 6, 
     ContourStyle -> [email protected], BoundaryStyle -> None, 
     BoxStyle -> Dashed],
     Plot3D[height, {x, -1, 1}, {y, -1, 1}, 
      PlotStyle -> Opacity[0.7, Blue], Mesh -> False],
      Axes -> False, ImageSize -> {300, 375}]}, Center]

pressure profile

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    $\begingroup$ Please start a notebook with a fresh kernel. Copy your code from this page and see if it works. Anything in blue would mean that its definition is needed. Do you have a (differential) equation relating P[z] to T[z] or something similar? Not that it would make it an on-topic question, but it could facilitate potential respondents. Thanks. $\endgroup$
    – Syed
    Sep 12, 2023 at 4:59
  • $\begingroup$ Do you mean solve heat equation with a given velocity profile? $\endgroup$ Sep 12, 2023 at 5:30
  • $\begingroup$ Yes, I want to solve the heat equation with a given velocity profile. @Trounev $\endgroup$ Sep 12, 2023 at 6:07

1 Answer 1

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For getting an image from the above code define $h, \rho, \omega$ and
r[x_,y_]:=Sqrt[x^2+y^2]

The ideal gas local state equation for a volume element reads

$$p(r,z) \ r^2 \ d\phi\ dr\ dz = \rho(r,z) \ R \ T(r,z) \ r^2 \ d\phi \ dr \ dz$$

The condition of an isolated system in a mechanical potential $V$ with constant temperature (fixed by the force free photon field in equilibrium with temperature relaxation time infinitely shorter than by heat diffusion) to be in a mechanical equlibrium, is, for a small volume dV, that its weight in the gradient field of V is balanced by the directional surface integral of the pressure.

The mechanical force density as a 3-form

$$ \int_V \rho(r,z)\ r^2\ d\phi \ dr \ dz \ (g e_z + r \omega ^2 e_r)$$

has to be equal to the surface integral of pressure over the surface 2-form $dS$

$$ \int_{\partial V} p(r,z) \ dS$$

The integrals can be taken over an flat annulus because of rotational symmetry and therfore independce of $\phi$

$$K=\int_{(r\ \in (R,R+dR),z\ \in (Z,z+dZ))} \rho(r,z)\ r^2 \ dr \ dz (g e_z + r \omega ^2 e_r)=\int_{\partial(r\ in (R,R+dR),z\ in (Z,z+dZ))} p(r,z) dS $$

The four surface integrals yield the gradient of $p$ projected to the radial and vertical directions with the surface scale factors

$$\Delta P=\nabla P(r,z) = \partial_r (r^2 p(r,z)) e_r + \partial_z p(r,z) e_z$$

Up to signs, both forces have to be equal. Replace $\rho$ by $p$ and derive a barometric formula

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