Q. How does FLOTRAN handle the elevation head of a fluid?A. FLOTRAN subtract the "static" or elevation head from the absolute pressure. In the Navier-Stokes equations, this makes the gravity term look like (Rho-Rhoref)*Grav. This is done for natural circulation problems so that the pressure field at the start, before any density changes have occurred, is uniform. This is good for the numerics of the problem.In order to calculate correct absolute pressures, FLOTRAN includes the static head if gravity is specified. The calculation is rho*grav*height, where height is measured relative to the global coordinate system. Neither that static head nor the absolute pressure is available in the results. This is usually not a significant factor.Now, let us consider an application. Suppose you have a section of vertical pipe filled with water. First of all, we ask ourselves, is the pipe being held out in the atmosphere (where we would expect the pipe to begin to empty at t=0), or is the pipe itself underwater? This latter case is what FLOTRAN calculates if you specify P=0 at both ends. We expect nothing to happen, and numerical roundoff is the only thing that would drive flow.For the former case, one can either run it as a pressure driven problem, where the pressure drop is that calculated by the user: rho*grav*height, or one can simulate the gravity with body force terms: rho*grav applied with the BF command in the appropriate direction.The situation changes when we activate the free surface model where the elevation head IS a part of the FLOTRAN pressure. This is necessary, since the free surface depends on gravity, inertia, and so on. You can look at the pressure field at t=0 in any free surface problem and easily see the elevation head.

 Q. How does FLOTRAN handle the elevation head of a fluid?A. FLOTRAN subtract the "static" or elevation head from the absolute pressure. In the Navier-Stokes equations, this makes the gravity term look like (Rho-Rhoref)*Grav. This is done for natural circulation problems so that the pressure field at the start, before any density changes have occurred, is uniform. This is good for the numerics of the problem.In order to calculate correct absolute pressures, FLOTRAN includes the static head if gravity is specified. The calculation is rho*grav*height, where height is measured relative to the global coordinate system. Neither that static head nor the absolute pressure is available in the results. This is usually not a significant factor.Now, let us consider an application. Suppose you have a section of vertical pipe filled with water. First of all, we ask ourselves, is the pipe being held out in the atmosphere (where we would expect the pipe to begin to empty at t=0), or is the pipe itself underwater? This latter case is what FLOTRAN calculates if you specify P=0 at both ends. We expect nothing to happen, and numerical roundoff is the only thing that would drive flow.For the former case, one can either run it as a pressure driven problem, where the pressure drop is that calculated by the user: rho*grav*height, or one can simulate the gravity with body force terms: rho*grav applied with the BF command in the appropriate direction.The situation changes when we activate the free surface model where the elevation head IS a part of the FLOTRAN pressure. This is necessary, since the free surface depends on gravity, inertia, and so on. You can look at the pressure field at t=0 in any free surface problem and easily see the elevation head.