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)