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Two-sided convective heat transfer
Hello All,
This is not a CFD question, but rather a numerical analysis question. I am trying to solve for the temperatures on a plate. The plate is under convective heat transfer from both sides (inside and out). The convection coefficients on both sides are known. However, the convection coefficients are defined with respect to their own adiabatic wall temperatures. That is, the adiabatic wall temperatures for the internal and external convection coefficients are not the same. h = q / (T - T_aw) How can I solve for the plate temperatures? Anybody know an iterative scheme?? Thanks in advance, |

Re: Two-sided convective heat transfer
The issue is not clearly described, nor is the context specified.
The adiabatic wall temperature T_aw is really a fluid property. It is the equilibrium temperature attained by the surface of the wall such that there is no more heat transfer between fluid and wall (normal temperature gradient is zero in the fluid next to the wall). Knowing the bulk fluid temperature and speed on each side, the T_aw of each fluid should be well approximated by the stagnation temperature of each fluid, which can be calculated. I vaguely remember a technical distinction (you can look this up in a heat transfer text) wherein the T_aw may be slightly different from the stagnation temperature (I vaguely remember a recovery factor being involved). If the T_aw on the two sides are different, then conduction must take place through the wall, and thus there is nonzero heat transfer between wall and each fluid, and thus the wall surface temperatures cannot in fact be the adiabatic wall temperatures. I am guessing that the convection coefficients you are given take T_aw to be the driving temperature on the fluid side, rather than its freestream temperature or the average of T_aw and freestream. Next, presumably the other temperature (T) in your h equation is a solid property, being the temperature at the wall surface. (To model the heat transfer between two systems, you need the temperatures of both.) For the steady state, you need to additionally know the wall thickness and the thermal conductivity of its material. For the transient calculation, you need to know the preceding, and also the initial temperature profile of the wall and the specific heat capacity of its material. For the steady state, if you are doing a simple one-dimensional, notebook and pencil sort of calculation, you do not need to iterate. Nor do you need to iterate if you use a CFD code capable of conjugate heat transfer. If you are using two separate codes, a thermal conduction code within the wall, and a CFD one for the fluids, then the physics is artificially decoupled because of the limitations of the codes. In this case, you have to iterate by alternating between the codes and updating the other code's temperature or heat flux boundary conditions each time (many CFD codes allow you to set convection coefficients and wall temperature type boundary conditions, or you may have to do this part with a user-defined subroutine). For the 1D steady state, let the wall surface temperatures be T1 and T2, on the sides corresponding to fluids 1 and 2, and the wall thickness be t. Doing heat transfer balances per unit cross-sectional area, h1(T_aw1 - T1) = k(T1 - T2)/t = h2(T2 - T_aw2) Solve these two equations simultaneously to get T1 and T2. |

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