# Flow rate

1. Aug 13, 2008

### Gear2d

1. The problem statement, all variables and given/known data
In an ideal fluid, the pipe length is doubled, while radius is decreased by factor 2 what will have to the velocity and volume that is flowing?

2. Relevant equations

Q=Av = pi*r^2*v

3. The attempt at a solution

From this the volume that will pass, in ideal fluid cases, is constant (same throughout), but the velocity will increase by a factor of 4 since: Q/v = pi*r^2. Is this correct?

Also if I were considering the radius decreasing by a factor of 2, in real fluids, the velocity would increase by it would not increase as much as seen in ideal fluids?

2. Aug 13, 2008

### Topher925

Makes sense to me. Assuming the fluid is incompressible and the volumetric flow rate remains constant. The problem statement doesn't make those assumptions so I will. In the real world, and making those assumptions, there would be a pressure difference across the pipe due to friction affects and an increase of kinetic energy, however the velocity would remain the same as an ideal fluid. This is of course assuming constant volumetric flow rate.