Nozzle exit loss

The pressure drop of a fluid flowing through a nozzle is equal to 

V = velocity of the fluid, as it flows through the nozzle, in feet per second 

This equation assumes that before the fluid enters the nozzle, its velocity is small, compared to its velocity in the nozzle. The increase in the velocity, or the kinetic energy, of the fluid in the nozzle comes from the pressure of the fluid. This is Bernoulli s equation in action. The energy to accelerate the fluid in the draw-off nozzle comes from the potential energy of the fluid. This is Newton s second law of motion. 

The coefficient used in the equation above (0.34) neglects friction and assumes the process fluid has a low viscosity. For most process nozzles, these are reasonable assumptions. Detailed information on draw-off nozzle coefficients has been published in Crane.1 

The pressure drop of the fluid escaping from the bucket is called nozzle exit loss. Actually, nothing is lost the potential energy, or pressure head of the water in the bucket, is just converted into velocity, or kinetic energy. 

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The nozzle exit loss of the liquid leaving the bottom of the tower. 

The pressure at P1 is now the 1-psi static head, minus the 5-psi nozzle exit loss, or negative 4 psig (or positive 10.7 psia). That is, the pressure at the drain is a substantial partial vacuum, or a negative pressure, meaning that it is below atmospheric pressure (atmospheric pressure at sea level is 14.7 psia). 

This suggests that the pressure in a water drain can get so low, that air could be sucked out of the bathroom and down the drain. Of course, we all see this happen several times a day—typically when we flush a toilet. So much air is drawn into the water drainage piping, that we install vents on our roofs, to release this air. The only requirement, then, for vapors to be drawn into a flowing nozzle is for the nozzle exit loss to be larger than the static head of liquid above the nozzle. 

Increase in velocity through the draw-off nozzle is small, and hence the nozzle exit loss is zero. 

Control valve B is located 56 in below the draw-off nozzle. The pressure drop across the valve is still 2 psi, or 56 in of water. Again, we will neglect nozzle exit loss and friction loss. But in this case, the height of water in the draw-off sump may be zero. But why ... 

Does this mean any frictional losses, due to external piping, at the same elevation as the draw-off nozzle, have to be added to the nozzle exit loss, in determining the liquid level in the sump Yes ... 

I once tried to increase the flow of jet fuel from a crude distillation column by opening the draw-off, flow-control valve. Opening the valve from 30 to 100 percent did not increase the flow of jet fuel at all. This is a sure sign of nozzle exit loss—or cavitation limits. To prove my point, I increased the level of liquid in the draw-off sump from 2 to 4 ft. Since flow is proportional to velocity and head is proportional to (veloc-... 

Figure 16.4 Nozzle exit loss for vapor flow.

The coefficient for nozzle exit loss of 0.34, shown above, assumes that there is no vortex breaker in the nozzle. Vortex breakers are certainly required if nozzle exit velocities exceed 4 ft/s. Vortex breakers are usually not needed if velocities in the draw-off nozzle are less than 2 ft/s. 

If the flow is limited by nozzle exit loss, because the nozzle diameter is small or because the sump is shallow, no process change will increase the flow. However, if the restriction is outside of the vessel because of valves, fittings, or piping on the draw-off hne, located at the same elevation as the draw-off nozzle, then there is an option to marginally increase flow. 

My standard calculation for nozzle exit loss as shown at the start of this chapter is ... 

The 0.34 coefficient shown is the conversion of potential energy to acceleration, which includes a reasonable allowance for turbulence and friction. The theoretical coefficient is about 0.18, which only includes acceleration. In my designs, 1 use 0.34 for nozzle exit losses, which my experience has shown represents what 1 actually observe in the field, plus a small safety factor. 

Account for nozzle entrance and exit losses. (See Fig. 52.4.) The pressure at P2 is lower than Pi due to the acceleration of vapor through the overhead vapor line. That is, the pressure of the vapor is partly converted to velocity. To correct the pressure at P2 to static pressure at Pi, add the nozzle exit loss as follows ... 

Figure 52.4 Pressure at is lower than P, due to nozzle exit loss. Pressure at P is converted to velocity at P. ...

Nozzle exit loss Conversion of pressure into velocity. 

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