how do you determine the rate of flow of water from a pipe tap?
552177
A bucket, a measuring cup, and a stopwatch is all you need.\n\nOpen the tap until it has achieved a steady flow, then put the bucket under it while turning on the stopwatch.\n\nFill the bucket to a desired amount or until you have a fairly long time.\n\nMeasure the water in the bucket and divide it by time. Thats the volume flow rate.\n\nTo get the velocity of the water when it exits the tap, divide the volume flow rate you have computed above with the area of the (circulare) hole.
1063906
You're confusing resistance with pressure. If indeed there is a difference in voltage between A and B then there will be flow.\nThink of a tank filled with water held a few feet above another tank that has a pipe joining the two. Water will flow to the lower container thru the pipe. In this case the pipe has no resistance, which means water flows at maximum possible speed.\nIf both tanks are at the same level, then, sure, there will be no water flow. That's not how the problem was stated, however.
34593
... or to use the hydraulic analogy, current is how much water (electrons) is flowing through the pipe.  Voltage is the water pressure.
1168897
It depends on the pressure pump and the resistance of the pipe.\nSee http://docs.bentley.com/en/HMFlowMaster/FlowMasterHelp-06-02.html
973746
picture a water pipe that is 1/4 inch in size, and a water pipe that is 5 inches in size.\nWhich one can move more water?\n\nnow invision dial up and the small pipe and broadband as the large pipe.\n\nof course the use it to move more stuff.
697169
The equation for power is:\n\nP = W / t\n\nwhere P is power, W is work and t is time. Assuming a perfectly efficient power harnessing system, the work provided is equal to the gravitational work done to pull the water down. \n\nA total of 2400 liters of water (2400 kg) is moved from a height of 20m down to a height of 0m, so:\n\nW = m * g * h\n\nW = (2400 kg)(9.81 m/s²)(20m)\n\nW = 470880 J\n\nThis work is provided over the course of 1 minute (60 second), so the power provided is:\n\nP = 470880 J / 60 s\n\nP = 7848 J/s = 7848 W\n\nFinally, 1 watt = 0.001 340, so 7848 W provides 10.52 hp.\n\nP = 10.52 hp\n\n\nHope that helps! :)
581110
You have to decide the velocity. You need to keep Re<2000. You know the diameter of the pipe, u need to know the coefficient of viscocity. Velocity range is to be calculated by putting all these quantities in the equation for Re, keeping Re between 0 and 2000. Then the velocity is to be controlled in that range.
700759
Let \nQ = volume flow rate in Litre/sec, \nD = diameter of the nozzle in cm and \nV = velocity of water in metre/sec. \n\nThen,  V = (40*Q)/ [pi*(D^2)]\n\nwhere pi = 3.14
41200
If it is a high-volume, low pressure pump, one of the easiest ways is to get a long piece of clear tubing and make a U-tube manometer.  the pressure will cause the water to climb in the tube vented to the atmosphere, while the other end is connected to the pump outlet.  You measure the height of the water column, and can get the head directly in feet of water, a popular unit of measurement in the pump industry.  If it is a high pressure pump, a simple pressure gage from the local plumbing supply or hardware store will read in PSI.  In rough terms, 14.7 psi is approximately equal to 32.2 Feet of water column pressure.
363092
Well I'm not going to do the calculation for you but you can.\n\nDo a web search and you'll find dozens of sites to help you if you don't have a fluid flow handbook.
447990
You could use Bernoulli's equation between two points to solve for the velocity of the fluid in question.  Area of the pipe (or area of the cross section of what the fluid is flowing through) multiplied by the velocity will give you the discharge.  If you aren't familiar with Bernoulli's equation, see the source below.
946895
Let the height be h (inches), radius r.\n\nNow by proportion, h = 5r at every level.\n\nHence volume is given by\n\nV = (1/3)pi*r^2*h which is\n       (5/3)*pi*r^3\n\n"Rate at which water is drained" means find dV/dt\n\ndV/dt = 5*pi*r^2*(dr/dt),\n\nand dr/dt = (1/5)*(dh/dt)\n               = -2/5\n\nYou can finish from there.
946809
x=v(horizontal velocity of water hitting ground)*t( time of travel)\n  = sqrt(2gy) *t; from equation v^2=2gy\nt= sqrt(2h/g); from equation h=.5*g*t^2\nthere you get that equation. I did with kinematics.
483088
this is very easy, just use the pythagorean therom (i might spell it wrong) the formula is: a^2 + b^2 = c^2 (c is the length of the pipe)\n\n15^2 + 8^2 = c^2\n\n225 + 64 = c^2\n\n289 = c^2\n\n17 = c\n\nyou need 17 feet of pipe
198582
From what I recall, the flow rate (GPM) would be the same throughout the pipe.  The equation of continuity for incompressible fluid flow is\n\nA(1)*V(1)=A(2)*V(2)\n\nwhere A is the cross sectional area and V is the velocity of the water (1 and 2 are the before and after positions).  Looking at this, the only thing the different diameter pipes (and thus the CSA) affects is the velocity of the fluid.  When you have a reduction in CSA, the velocity increases (if you partially cover a garden hose, the water shoots out faster), but you're still getting the same amount of fluid out per time unit.\n\nTry this - time how long it takes to fill a cup using your kitchen faucet.  Then, fill the same cup, but cover your faucet like you would a garden hose so that the water shoots out faster, and time that fill.  The times should be the same, plus or minus timing errors.
273013
Go to the link below and it has an online calculator for pump affinity laws.  \n\nThe one you are interested in is:\n\n(pressure at old speed)/(pressure at new speed) =\n\n(old speed/new speed)^2\n\nIncrease the speed (RPM) of the pump 10%, and the discharge pressure goes up 21%    \n( 1.10^2)
580210
Most fluid flow handbooks have this information. The Foxboro company use to have the very best manual for this.You might find one on ebay. Many thermodynamics text books also have this information.