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October 1, 2004, 08:52 
VISCOUS heat dissipation

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When the Viscous Work Term is not included then it seems that the solver actually does not calculates the dissipated heat separately from velocity gradients, but takes it to be the required amount of energy flow component (above the kinetic and flow work energy components) that insures perfect energy conservation. When the Viscous Work Term is included then we get correct temperature change and heat dissipation even when some walls are moving, because the dissipated energy is calculated separately from velocity gradient for each finite volume as it should be (at least this is my conclusion after some experimentation). To verify whether this is so or not, I have tried to calculate the dissipated heat energy flow in the domain with CEL expressions in CFX post using the basic variables from the results file, and compared it with the value calculated from the temperature change. I have found that the following CEL expression for dissipated heat energy per second: Ph = 0.25*sum((Velocity u.Gradient Y+Velocity u.Gradient Z+Velocity u.Gradient X )^2 *Volume*Dynamic Viscosity)@Assembly gives a value that is in good agreement (deviation below 1%) with the heat dissipation calculated from the temperature change found in the results file as: Phout = massFlowInt((Temperature 298.15 [K])*2400 [J kg^1 K^1])@out Since the Ph dissipated heat energy flow calculated form velocity gradients and viscosity is in good agreement with the same power term calculated from the temperature increase, we could rightly assume that the solver calculates the temperature change for each finite volume using such formula. My dilemma is only about the multiplication factor of 0.25 in the expression for Ph. In the text book about Hydrodynamics of Landau – Lifsic I have found the following formula for calculating the dissipated heat in viscous incompressible flow: E_dot =  0.5*e*Integral((dvi/dxk+dvk/dxi)^2*dV) E_dot is the dissipated energy per second, eis the dynamic viscosity, dVis the infinitesimal volume, and the remaining components in the inner bracket of the integral are the velocity gradient components. This formula is similar to the above CEL expression with the difference that instead of the multiplying constant of 0.5 we use 0.25, and we include all 3 velocity gradient components. My question is whether our CEL expression is correct or not? If not, what would be the correct expression that would be in good agreement with the temperature change calculated by the solver? If it is correct then why should we use 0.25 instead of 0.5 indicated in the textbook, to get agreement with the Phout? Also, why does the formula from the textbook not include all the 3 velocity gradient components? If you have any related comment, or if you know the relevant formula (and or explanation) from other text books, or you also know the answers for the above questions, please tell us your opinion. Any contribution is highly appreciated (even if it does not answers fully the questions), because it can show the way towards the solution. Joseph PS. The test geometry used for such calculations (a hexahedral nozzle 4cm long 1cm wide 1cm high at inlet, 3mm high at outlet, 3 no slip walls and 1 side as symmetry plane, and with 3cm long no slip wall extension of constant cross section at outlet, and 1cm long similar extension at inlet; using glycerol as liquid with inlet velocity 10m/s, 0 Pa static pressure at outlet) and related accuracy issues have been discussed earlier under the heading "ATTENTION! Reliability problems in CFX 5.7". 

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