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 Calculating fluid pressure difference Post Reply Forum
 Posted by: tintin ® 05/17/2004, 16:50:04 Author Profile eMail author Edit Probably very easy for you guys but.. .25inch diameter pipe and I require a jet of water producing a force of app 40lbs. Can anyone help with the method of calculating the required flow rate through the pipe? any assistance much appreciated Tintin

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 Re: Calculating fluid pressure difference Re: Calculating fluid pressure difference -- tintin Post Reply Top of thread Forum
 Posted by: sailorglenn ® 11/12/2006, 14:51:06 Author Profile eMail author Edit a bowthruster in a 40' sailing boat requires about a 4hp electric motor driven 5" prop in the bow.Any idea how to recalculate how to replace this with a waterjet? what orifice size and what pressure of water flow is required?thanks Glenn

 Re: Calculating fluid pressure difference Re: Calculating fluid pressure difference -- tintin Post Reply Top of thread Forum
 Posted by: RandyKimball ® 05/17/2004, 19:11:23 Author Profile eMail author Edit Hello Tintin,Sure but a little more info is in order.Do you want a jet of water out into the air? PSI no longer applies if so, do you want 40 lb per minute? Generally liquid out a nozzle is figured by gallons per minute, or for thrust one could convert it to pounds per minute to move something like a jet boat.Do you want 40 PSI at the opening? If so how long is the .25 dia length of pipe or tube. Pressure is lost at velocity per foot in a container. To get 40 psi at 2 feet from a pump is different than getting 40 psi 50 feet from a pump.Someone may have to jump in and complete the calculations as I will be out of pocket for a while.In the mean time try this area on this site by clicking 'fluids' at the bottom of this message.-randy- The worst suggestion of your lifetime may be the catalyst to the grandest idea of the century, never let suggestions go unsaid nor fail to listen to them. Modified by RandyKimball at Mon, May 17, 2004, 19:19:44

 Re: Calculating fluid pressure difference Re: Re: Calculating fluid pressure difference -- RandyKimball Post Reply Top of thread Forum
 Posted by: tintin ® 05/18/2004, 08:44:07 Author Profile eMail author Edit Thanks for the reply Randy. I'm at the rough sketch stage of designing a water jet powered remote control scale model boat. I'm trying to get an estimate for the pump specs in order to work back to swept volume and rotation speed and from there to motor size, torque and ultimately, electrical power requirements. I hope to keep the internal mass to less than 20lbs and plan to install the hardware in a number of different hulls. This is only a pet project to keep me occupied during recovery from an injury. Any additional advice or suggested reading material/web sites would be much appreciated. ps thanks for the Fluids link Tintin

 Re: Calculating fluid pressure difference Re: Re: Calculating fluid pressure difference -- tintin Post Reply Top of thread Forum
 Posted by: RandyKimball ® 05/18/2004, 19:01:24 Author Profile eMail author Edit I see, (I was on PBR boats in Nam, I was the 'engineer'). You want to move more than 20 lbs of water a minute and use the old saying "equal and opposite reaction". I'm going to give you an example of the real world action. Our PBR's weighed in at 6 tons, loaded. This includded weapons, armor, ammo, fuel, and personel. We had two jet pumps delivering 500 gal per minute (as I recall)[this is 1000 times the weight of a gallon] and we did about 45 to 50 knots (some times 60)after we got on step. Velocity increases as the hull becomes less deep in the water. As the hull climbs the water the resistance drops until the point is reached where the amount of water weight forced through the pump equals the drag in the water depth left, the force required to displace the weight of the water being push aside (6" deep on step), and the air flow dynamics and drag. Knowing at which point too much motor, batteries, and pump weight will prevent you from getting on step is the big question. Additional example; We would load just a little too much ammo (and beer [in case we ran out of drinking water, so don't ask]) and head to shallow water along the beach. Then we would attempt to get on step, often failing on a full load. So we would motor along as we used fuel weight, IF we were in friendly real estate. If we got nervous or shot at we would punch the throtles wide open and toss beer and ammo until she jumped up on step, while returning fire, of coarse. The engines were only going to put out X amount of water weight until velocity increased enough. As the velocity under the hull increases the presure drop into the pump changes and the out put of the pump increases as the hull drag decreases as the pump out put increases.... until .. but you must get on step first. By on step I'm speaking of what a ski does after it breaks the surface and stays on top, same exact scenario.-coarse you know those little jet pumps are for sale, right?Do I know that formula? ... I think not. You need a physics major and a print of the hull design and exact weights, pump out puts per weight, motor torques per weight, battery weights ....... I will tell you, I wish I was doing it!!!! -have a ball, ok?-How much torque is required to move 20 lbs of water a minute is something that can be calculated.-randy- The worst suggestion of your lifetime may be the catalyst to the grandest idea of the century, never let suggestions go unsaid nor fail to listen to them. Modified by RandyKimball at Tue, May 18, 2004, 19:17:14

 Re: Calculating fluid pressure difference Re: Re: Calculating fluid pressure difference -- RandyKimball Post Reply Top of thread Forum
 Posted by: jaybowdler ® 12/16/2005, 18:12:34 Author Profile eMail author Edit Good day wizards of fluid dynamics! I am with the Sacramento Fire Department and am with a group specifying two 28' x 8'6" aluminum jet powered river patrol crafts. Alamar of Oregon is contracted to build the boat.The vessel will include a second pump (500gpm/90psi, 250gpm/150psi) that will supply a water monitor (deck gun) on the bow through a 3" pipe below the deck. With the boat hovering in the current and the deck gun discharging 90 to the side, a rotational force is generated that takes the boat offf station and causes the water to miss the target. This force is defined by fire service equations as follows:Smooth bore tip 1 3/8" Nozzle reaction (NR) = 1.5 x Diameter of nozzle x Nozzle pressure (NP)(pitot pressure in psi).Fog nozzle NR = 0.0505 x flow (gpm) x sq root NP (base pipe pressure psi)I can stand on the bow of a similar vessel with a hand fog nozzle flowing 125gpm at 95 psi with little effect on the boat. If I depress the line such that the water stream now strikes the river surface, then the boat is rotated sharply. My plan is to divert water from the 3" pipe supplying the deck gun through small valved jets below the water line, thus creating a bow thruster to counteract the NR of the deck gun. What I hope you might be able to assist me with is determining the NR of the thruster nozzles. The equations have eliminated the density of hte medium the flow is into. I want to determine the maximum size of thruster dischage diameter needed to stabilize the system, assuming the moment arm of both the thruster and the deck gun are the same.ANy assistance will be greatly appreciated and a solution will be rewarded with lunch upon demand in Sacramento! I can be contacted at 916-216-0305 or jbowdler@sfd.cityofsacramento.org. -Jay- Modified by jaybowdler at Fri, Dec 16, 2005, 18:14:02

 Re: Calculating fluid pressure difference Re: Re: Calculating fluid pressure difference -- jaybowdler Post Reply Top of thread Forum
 Posted by: zekeman ® 12/16/2005, 23:40:18 Author Profile eMail author Edit Firstly, your 1st eq for force should have the D squared, not D since the fundamental equation is NR=2*Area* Pressure= 2*pi*D^2/4*pressure which you could correctly round off to 1.5* d^2*pressure.(Note that D^2 means DxD and the symbol * is the symbol for multiplication).Your second equation is ok, and I calculate that you had a reaction force of about 60 lbs. It seems to me that the thrust force on you would be transmitted to the boat if you are leaning on some part of the structure. So I don't quite understand the difference between pointing the nozzle toward water or air. Please enlighten me since I must be missing something.

 Re: Calculating fluid pressure difference Re: Re: Calculating fluid pressure difference -- zekeman Post Reply Top of thread Forum
 Posted by: jaybowdler ® 01/05/2006, 14:25:24 Author Profile eMail author Edit Hello Wizards of fluid dynamics (Zekeman and istre)!You got me on the math operator symbology. It's been along time. Regarding the difference pointing the nozzle in air as opposed to the water: This is just field observation. Standing on the fore deck of the boat I can flow a hose line 90 deg to the directio of the vessel direction. The operator needs to do little corrections to keepthe bow pointed up stream. Once this system has stabilized, if I then slowly lower the hose stream to where it strikes the water surface near the side of the boat, it upsets the equilibrium significantly. The operator needs to make strong corrections to keep the bow upstream. Observing this I conclude that there is a difference in the nozzle reaction force that is dependant on the viscosity of the medium the flow is discharged into. This is alos seen at a fire where I have been standing on a ladder flowing a hand line into a window and moving the hose stream sideways struck the side of the house near the window and all most flipped the ladder off the wall!All this led me to believe that for a given pressure in a system I could equalize the force of a nozzle flowing into air with a significantly smaller nozzel flowing into water. Possibly these observations are leading me to the wrong conclusions.So I see this scenario at equalibrium:The Nozzle 1 3/8" SBT on the bow flowing into air: NR = 1.5 * d^2 * psi = 1.5 * 1.375^2 * 80 = 226.875 psiThe nozzle below the water line flowing directly opposite the nozzle on the deck. It comes from the same pump source, so it will have the same pressure nearly.NR = 1.5 * d^2 * 80 * den = 226.875 psi, where den is some factor that related the reletive density of air vs water to take into account the observation discussed above.Once den is know, then the second equation can be sovled for d which would be the diameter of the under water nozzle.I hope this make a bit more sense! The key in my mind is the density factor.Thoughts?

 Re: Calculating fluid pressure difference Re: Re: Calculating fluid pressure difference -- zekeman Post Reply Top of thread Forum
 Posted by: zekeman ® 12/17/2005, 00:04:33 Author Profile eMail author Edit I missed your main question so her is my take: In order to fully correct the unbalanced force, a very costly solution is to have the same 3 inch nozzle fed from the same source directly opposite so that the net force on the boat would be zero.i.e. a Tee connection to the pipe. But this would severely reduce the water stream,requiring a doubling of your pump drive.

 Re: Calculating fluid pressure difference Re: Re: Calculating fluid pressure difference -- RandyKimball Post Reply Top of thread Forum
 Posted by: istre ® 05/25/2004, 12:07:42 Author Profile eMail author Edit I'm curious about this one too, because I'm getting some pretty steep pressures for this scenario, and I'm wondering if I'm on track or forgetting something that is causing an error. Starting from the diameter of the nozzle, I get an cross-section of ~0.0491 in^2. I'm also assuming that since this is a model, the pressure of the water at the depth of the nozzle will be negligibly above atmospheric and we can assume that the exit is at 0 psig. Given this condition and the diameter, we should be able to calculate the pressure drop from the pump to the nozzle required to develop the water momentum required to produce a 40lbf driving force.delta P * Area = driving force delta P = Driving force/Area delta P = 40 lbf / (0.0491 in^2) delta P = ~815 psiGoing back to Bernoulli's equation, we haveP1 + (rho/2)*(V1^2) + rho*g*h1 = P2 +(rho/2)*(V2^2) + rho*g*h2Assuming the delta h is negligible and that term can be neglected, and assuming that at the outlet of the pump, V1=0 and at the exit of the nozzle, P2 = 0, we haveP1 = (rho/2)*(V2^2) 815 psi = (62.4 lbm/ft^3)/2*(V2^2)Solving for V2, we get 4170 in/sec and plugging this into flow rate = V * area flow rate = 4170 in/sec * 0.0491 in^2 flow rate = 53.24 gal/minNow, 815 psi just seems extreme and leads to these velocities of 237 or so mph at the exit. I'd just like to know where I took a wrong turn if I did. I realize I made a lot of simplifications that would actually tend to lower the total pressure required from the pump. The one place where I openly question my calculations is in assuming that the pump can be considered separate from the system and that Bernoulli's equation can be applied for the flow from its outlet to the nozzle. If anyone can discount my calculations and let me know why, I'd appreciate it, because if not, it seems like it might be kind of extreme for such a small pump and system to deliver and withstand such high pressures.