# Negative pressure head in a nozzle

1. Apr 8, 2014

### KLM86

Hi,

I am doing some fluid flow calculations for some nozzle designs at work.

For one nozzle in particular, the fluid flows through an inlet pipe, then flows through a short contracted section, then flows into a Helmholtz chamber before entering another contracted section through which it discharges to the atmosphere. The exit diameter (d2) is larger than the diameter of the first contraction upstream of the Helmholtz chamber (d1). Hence U1 > U2. If the nozzle is in a horizontal position, the altitude term of Bernoulli's equation can be neglected. Using gauge pressure, P2=0.

So,

P1 + 0.5*ρ*U12=0.5*ρ*U22

Implying that the term P1 is negative (U1>U2) (?)

I am assuming steady flow occurs through the nozzle and also that the nozzle is designed to produce energy losses that are small enough to be ignored.

This is hypothetical as much as anything as I know that a real nozzle would produce energy losses. However, if the absolute atmospheric pressure was close to zero and U1 was sufficiently greater than U2 to counter any energy loss through the nozzle, then the term P1 could actually be negative in absolute terms?

I'm confused! Someone please tell me where I'm going wrong with this.

Thanks.

Kathryn.

2. Apr 11, 2014

### Baluncore

Welcome to PF.
What is the fluid ?
You assume steady state flow. How do you know it is not an oscillator ?

3. Apr 19, 2014

### yuiop

As I understand it, P1 cannot drop lower than absolute zero pressure, but it can drop lower than than atmospheric pressure, which is why venturi tubes can produce a vacuum. U1 does not increase without bound and as P1 gets close to absolute zero, U1 and U2 both slow down while maintaining the u1*di^2 = u2*d2^2 relationship. If the fluid is a liquid, reduced pressure at the inlet causes the liquid to boil causing bubbles to form and cavitation, increasing the friction and back pressure. Once bubbles form, the liquid is no longer an incompressible fluid.