## Note: Water Jet Physics.. Equivalent Nozzle Diameters (Dual vs Single)

Question:  What is the diameter of a single nozzle that has equivalent thrust to two smaller diameter nozzles being fed from the same input source (say a skyboard 4 in fire hose)?

I reviewed the physics of water jets in this post https://vamfun.wordpress.com/2013/03/27/water-jet-skyboard-thrust-physics/ for an inventor friend Bob.   He posed the above question to me since he was converting a dual jet concept to a single jet and wanted the same thrust.

I gave the following answer:   Continuity requires that the total mass flow rate exiting  both dual jets must equal the input mass flow rate before the split.   Momentum equations (spoken about in the referenced post) state that the total thrust T for a stationary jet nozzle is

T = Md*Vexit

where Md is the water mass flow rate and Vexit is the water exit velocity.      Vexit must be the same for both the single and dual jets.   Using two jets just splits the mass flow rate into two streams each with mass flow rates of Md/2 but Vexit must remain the same for both cases to generate the identical thrust.    I.e.  the dual nozzle case looks like this:

T = (Md/2)*Vexit + (Md/2)*Vexit   (dual jet case)

= (Md/2 + Md/2)*Vexit

= Md*Vexit   (equivalent  single jet case)

Intuitively we can see that if the exit velocity, Vexit, is the same for both cases then the total areas ejecting water must be the same to get the same mass flow rate.  The area of the single nozzle must be twice the area of one of the dual nozzles.   And since area is proportional to the square of the diameter then the diameter of the single nozzle must be the √2 * the diameter of the smaller nozzle.

How is the diameter calculated given  required thrust T and mass flow rate Md?

First calculate Vexit

Vexit = T/Md

Substituting Vexit into the mass flow rate  equation Md = rho*A*Vexit  where rho is the density of water  and solving for A gives

A = Md/(rho*Vexit) = Md^2/(rho*T)

And finally knowing A we can solve for the diameter:

Diameter =2* √(A/π)

### 2 Responses to Note: Water Jet Physics.. Equivalent Nozzle Diameters (Dual vs Single)

1. Ryan says:

This is an interesting question, and this concept has many factors to consider. In a perfect world, (frictionless) this explanation is correct. If the flow rate is held constant (somehow), and If the sum of the exit nozzle’s areas are the same, the exit velocity will be the same, and therefore the thrust will be the same.

In the real world, dual nozzles will require the water to be split in half somewhere, which will steal a small amount of energy, then the water will have to travel a distance through a smaller diameter tube, which has a much higher friction loss coefficient than larger tubes. The end effect will be that some energy will have been lost in this process, and more energy will need to be applied to the water stream to achieve the same flow rate. In other words, the single nozzle will likely be more efficient.

In practice, unless you can guarantee a constant flow rate, you will probably get slightly more thrust from the single nozzle when the sum of the exit areas are equal, because the exit velocity will likely be higher due to a more efficient water flow path. How much? That’s for another analysis to determine.

2. Davor Eberl says:

Hi,
I noticed your expertise in water jet physics. I would be grateful for your evaluation and honest opinion about my recently patented invention – “water-jet hose manipulation device”. If I put it as simple as possible, my patent is actually a computer controlled and remotely guided flying hose driven by many separately controlled water jets, in general pointing downwards, distributed along the whole length of the hose.
I expect that the range of access and maneuverability of such device could be useful for firefighting as well as in some other technical fields. What do you think?