Finite volume methods-2

• Issues:
  • Fluxes with variable properties
    • See S.V. Patankar “Numerical Heat Transfer and Fluid Flow”
  • Grid decoupling

Fluxes with variable properties

• Consider 1-D, steady state heat conduction on a nonuniform grid. $$q_e-q_w = S\Delta x.$$

• Consider $q_e$:

• We need $k_e$. How to evaluate it?

• Try linear interpolation:

  $$\frac{k_E-k_e}{\delta_{e+}} = \frac{k_E-k_P}{\delta_e},$$
  $$k_e = fk_P + (1-f)k_E,$$
  $$f = \frac{\delta_{e+}}{\delta_e}.$$

Unfortunately, this does not work very well.

• The finite volume method is still conservative since $q_e$ is applied to the energy balance for both the P and E cells.
  • But the solution is not physically accurate when applied to systems with variable properties.

Case 1

• Consider the case where cells P and E are different materials.
  • Maybe take E as an insulator
  • Then $k_E = 0$.
  • Expect $q_e = 0$.
  • But linear interpolation to evaluate $q_e$ using $k_e$, as shown above, gives the following: $$k_e = \frac{\delta_{e+}}{\delta_e}k_P,$$ $$q_e = -k_e\frac{T_E-T_P}{\delta_e}\ne0.$$
• We expect a temperature profile like the following:
• Notice that no temperature gradient develops in cell E since it is an insulator. This implies zero heat flux.

Case 2

• Now, take a case with $k_E > k_P$.
• We expect the following temperature profile (assuming uniform properties throughout a given control volume):
• here, $T_I$ is an intermediate temperature at the cell face e.

If we use a uniform temperature gradient throught the region of interest, then consider the heat fluxes that this graph implies

• Denote the heat flux on the left and right sides of e as $q_{e-}$ and $q_{e+}$, respectively.

• Now, since $k_E\ne k_P$ we have $q_{e-}\ne q_{e+}$.

Hence, using (or implying) the same temperature gradient in both cells (materials) gives different heat fluxes in the two cells when using the respective thermal conductivities.

• A linear T profile (same T gradient in both cells) conserves heat flux if the same $k$ is used in both cells.
  • That $k$ could be $k_e$ as evaluated by linear interpolation.
  • But this gives the unphysical behavior of Case 1.
  • And doesn’t imply the physical profile of Case 2.

Solution

Find an evaluation of $k_e$, that is physically consistent with heat fluxes through materials of different properties.

That is, find an evaluation of $k_e$ that allows for different temperature gradients to be implied in cells with different conductivities.

Note, we still want to evaluate the temperature gradient in the usual way, that is $(dT/dx)_e = (T_E-T_P)/\delta_e$.

Construct $k_e$ by equating fluxes in the two cells (so that energy is conserved): * Allow for an intermediate temperature $T_I = T_e$. * Equate the fluxes on the left and right of the interface, and then solve for $T_I$:

• Now equate the flux evaluated over both cells P and E (that is, $q_e$), which is written in terms of $k_e$, and equate this to $q_{e-}$.

• We insert the expression for $T_I$, and then solve for $k_e$:

• This is a harmonic mean instead of an arithmetic mean.
• Use this for finite volume methods with variable $k$, $\mu$, $D$.
• Summary

Results

• Now, for $k_E\rightarrow 0$, $k_e\rightarrow 0$, and $q_e\rightarrow 0$ as it should (unlike the linear interpolation).
• For a case with $k_E\gg k_P$ we have $$k_e = \frac{k_P\delta_e}{\delta_{e-}},$$

• This implies that all the change happens in the P cell, with the E cell offering no resistence. Good! This is consistent with the physical situation.

Grid decoupling

• There is a problem with first derivatives.
• Consider cells W, P, E and a convective flux term where $v$ is velocity and $f$ is some variable.

    |---W---|---P---|---E---|
            w       e

• Integrate this over cell P

  $$(vf)_e-(vf)_w$$

• Then interpolate to find $(vf)_e$ and $(vf)_w$:

  $$ \frac{v_Ef_E+v_Pf_P}{2} - \frac{v_Pf_P + v_Wf_W}{2},$$
  $$ \frac{v_Ef_E-v_Wf_W}{2}.$$

• Note how this totally skips the P cell.

• If we have a $vf$ field as shown below, the first derivative of $vf$ will be zero everywhere and this profile will persist.

    |---5---|---3---|---5---|---3---|---5---|

• This can happen in 2D or 3D:

    |-----|-----|-----|
    |  8  |  3  |  8  |
    |-----|-----|-----|
    |  3  |  5  |  3  |
    |-----|-----|-----|
    |  8  |  3  |  8  |
    |-----|-----|-----|

• Or, suppose we take a smooth solution and add a checkerboard pattern (from noise, or errors, etc.), this patter could persist, or even grow.

• This happens with pressure P in the fluid equations. The $\nabla P$ term gives $P_e-P_w$ hence $(P_E-P_W)/2$.

  • Pressure gradients drive velocity, and we’d like pressure to not skip cells. We don’t want a pressure checkerboard to be preserved or to grow.

• Remedy: Use a staggered grid.

  • Velocity is solved on the orange grid cells (v-cells) and
  • Pressure is solved on the black grid cells (P-cells).
  • When considering the velocity equation we are using the orange grid (v-cells). Then $dP/dx \rightarrow (P_e-P_w)$, but these values are known directly on the black grid (P-cells) and no interpolation is necessary, and no checkerboarding occurs.

• In 2-D, we have P-cells in the center, u-cells shifted to the left, and v-cells shifted down.

• Normally, scalars such as pressure, temperature, species mass fractions, etc. are evaluated using P-cells, and velocity vectors are on u-cells, v-cells, and w-cells.

• This is convenient for discretization, and numerically very stable.

Grids are called either staggered or collocated.