# Closed piping system question!

1. Aug 20, 2013

### Famwoor2

Hello everyone,

Say I have a closed loop comprised of a pump and some piping which connects the inlet of the pump to the outlet of the pump. All of the piping has radius "a," except for a small section, which constricts to radius "b" for a small portion of the line (the inlet and outlet connections are both of radius "a"). Neglecting friction, is there any pressure drop between the inlet and the outlet of the pump if an incompressible fluid were to flow through the system at some flow rate? If so, does Bernoulli's principle in the following fashion account for the magnitude of the pressure drop?

dP = | 1/2*(fluid density)*(flow rate)^2*(1/(pi*a^2)^2-1/(pi*b^2)^2) |

I used (flow rate) = (cross sectional area)*(flow speed) to express the flow speeds, and the fact that the flow rate is the same.

Here it is in TeX form:

$$\Delta P = \frac{1}{2} \rho Q^2 \left( (\frac{1}{\pi a^2})^2 - (\frac{1}{\pi b^2})^2 \right)$$

Thanks for your help and time,
F2

Last edited: Aug 20, 2013
2. Aug 20, 2013

### Staff: Mentor

Without any friction, there is no pressure drop between the two sides of the pump (and in an ideal system you don't need a pump at all to keep it flowing). There is a pressure difference where the cross-section changes, but the same pressure difference (in the opposite direction) occurs where the cross-section changes back.

3. Aug 21, 2013

### gmax137

I think you need to be clear that "no friction" means not only no friction at the pipe wall, but also no friction between the fluid and itself. Otherwise the flow through the restriction will dissipate some energy by creating eddies in the flow, which would appear as heat and noise. The more gradual the approach to and exit from the restriction is, the lower these losses would be.

4. Aug 21, 2013

### Famwoor2

Thanks for the feedback.