John C. Chien

July 15, 1998 13:56 
Re: CFD consulting job
Without knowing the exact device or configuration of your problem, it's hard to find a solution. But apparently you have something moving in a pipe ( flow in a pipe over something which is either stationary or rotating). The axial force ( or the tangential force ) normally can be expressed in terms of (=0.5 x fluid density x average fluid velocity x average fluid velocity x frontal area of the paddle x drag coefficient ). Assuming that you are using the same drag coefficient, same area ( the same paddle) , then the new force acting on the paddle will be proportional to ( average fluid velocity x average fluid velocity). Since the average fluid velocity in a 6 inch pipe will be small than that in a 4 inch pipe, with the same flow rate ( flow rate = fluid density x average flow velocity in the pipe x pipe area ), the drag force on the paddle will become smaller for the 6 inch pipe case. There are several things you can do: 1). maintain the average flow velocity, that is add a transition section in front of the paddle to bring the diameter from 6 inch back to 4 inch. In this way, you will get the same flow velocity, same flow rate and the same area and the drag force. 2). you can change the angle of the paddle such that the drag coefficient will be increased ( this depends on the original paddle configuration, whether it's angled relative to the flow direction.) 3). Increase the paddle area, such that ( flow velocity x flow velocity x area ) will be the same for both cases. When you increase the pipe diameter with flow rate fixed, the flow velocity will decrease. As a result, you have to increase the area. ( assuming that area increase does not change the drag coefficient ) Well, on the Internet, it's more or less like the guessing game. Everyone probably will interpret your statement of problem in different ways. This is just my interpretation of your problem, not to be considered as the solution.
