I have a system with a pressurized nitrogen that pushes water out of a tank. Bernoulli's equation tells me the exit velocity if I suddenly open the valve, but I'm confused on how to account for a potential nozzle diameter? My intuition tells me if I restrict the flow at the valve, the velocity should increase, yet Bernoulli's equation does not account for this, right? I've calculated the exit velocity to be about 57 [m/s], but what would happen if I add a nozzle that is half the area of the exit pipe? Is it wrong to assume flowrate would be fixed?
I'm sorry if this question seems too simple, but I would appreciate help figuring out what a nozzle would do to the flow.
Best Answer
Bernoulli's Equation is a useful idealisation but one that doesn't take into account any pressure loss (in engineering often referred to as head loss) due to friction in pipes, to bends, constrictions/expansions, valves and other elements commonly encountered in Real Life piping.
It can be adjusted for these losses, as follows:
$$h_1+\frac{p_1}{\rho g}+\frac{v_1^2}{\rho g}+\Delta h_p=h_2+\frac{p_2}{\rho g}+\frac{v_2^2}{\rho g}+ \Delta h_f+\Delta h_m$$
(Ref.)
where $\Delta h_p$ is the pressure contribution from a pump (zero in your case), $\Delta h_f$ is the loss due to pipe friction and $\Delta h_m$ are the so-called 'minor' losses' due to bends, constrictions/expansions, valves and other elements in a pipe system.
You can see how $\Delta h_f$ is calculated in the reference provided above.
For sudden contractions/expansions, $\Delta h_m$ can be calculated as in this wiki entry.
The head losses that make up $\Delta h_m$ are often represented as:
$$\Delta h_i=z_i\frac{v^2}{2g}$$
where $z_i$ is a geometry dependent factor, intrinsic to the specific loss-creating element $i$ and $v$ the flow speed through it.
Then $\Delta h_m=\sum_{i=1}^n \Delta h_i$, for all the loss making elements.
I hope this helps.