what is the formula for working out how many amps a hydraulic pump will pull at full flow and pressure?
480747
just check out the kilowatt of the pump. then apply the formula kw = 1.713 X V X I X PF.\nwhere V = voltage, PF = power factor, I = Amps.\neg:\na 30 kw pump,at 415 volts and 0.8 PF with three phase supply will draw an amps of\nI = 30000/1.732/415/0.8\nI = 52.17 amps.\n\nIn case of single phase motors just remove the term 1.732
500605
Basically.... it is as simple as (Voltage X (times) amperage = Wattage) example.... \n\na 120volt motor pulling a normal 3 amperes  ....will consume 360 watts of electricity... \n\nThe more amperage the motor requires.... the higher... the horsepower... and also the cost to run or operate it...
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
752166
Let F be the force applied to the pump handle\n\n0.35m x F = 0.04m x 400N\n=> F = 45.7N
252526
none. amps has no weight, its a flow of current. It's like calculating tonnage to speed. How many tons does 100 mph represent?
1169821
Here are some sites. Don't know exactly what you're looking for.\n\nhttp://www.lmnoeng.com/Flow/SmallOrificeGas.htm\nhttp://www.mysolv.com/OrificeCalcIn.asp\nhttp://www.gpengineeringsoft.com/pages/pdtpipedrop.html\nhttp://www.eng-applications.com/orifice.htm\nhttp://www.controlengineering.se/flowcalc/flowcalceng.htm\nhttp://www.pipeflowcalculations.com/
877652
These are GAS LAW problems, so you need to use PV=nRT.  However, for these problems, nRT will be constant so you only need to use P(pressure) and V(volume).  Therefore, for both of these questions, use the equation:\nP1V1 + P2V2 = PtVt   where P1 is the pressure in the first vessel, P2 is the pressure in the second vessel, and Pt is the total pressure of the connected system.\n\nProblem #1\nP1V1 + P2V2 = PtVt  \n(7.0atm)(15.0L) + (P2)(8.4L) = (10.0atm)(23.3L)  \nP2 = 15.4 atm\n\nProblem#2\n(7.2atm)(15.0L) + (4.1atm)(8.3L) = (Pt)(23.3L)  \nPt = 6.1 atm
765689
Yes you are right to subtract the water pressure from the total pressure.\n\nThen use p1v1/t1 = p2v2/t2\n\nSTP is p2v2/t2\n\nSTP = 1 atm and 0C or 273K\n\nSolve for V2.\n\nKeep units the same on both sides!
341744
The easiest example would be to turn on a electrical pump motor. By toggling the switch, it sends electrical current to the motor. This causes the motor to turn, which changes the electrical energy to mechanical energy and then turns the pump, which will then move fluid (Hydraulic Energy).
87578
P = EI (Power equals voltage E times current I)\n\nMany electrical properties can be explained as analogous to water flow.  For example, a transformer "transforms" high voltage and low current into low voltage and high current, keeping the power constant (except for minor losses) at full load.\n\nThis would be equivalent to a high pressure water pump supplying low water flow to the primary winding and taking out high water flow at low pressure from the secondary winding to drive a low pressure water turbine.\n\nYou could represent the transformer core as a hollow square of wood or even thick cardboard.  On one side wrap say ten turns of thin clear plastic tube around the core to represent the primary (first) winding.  On the other side wrap say five turns of larger diameter plastic tube around the core to represent the secondary (second) winding.  Fill the primary tubing with water and red food coloring to represent "hot" high voltage and the secondary tubing with water and blue food coloring to represent "cooler" low voltage.\n\nThe explanation of course is that the high voltage primary winding will have more windings of thin tube (similar to more thinner copper wire windings) because the current will be less than in the secondary winding.\n\nThere will be less windings for the secondary coil using larger tubing because the current will be higher at low pressure.\n\nSupply the secondary tube with a shut off valve (or clip) to represent the fact that there will be no current unless the circuit is open.  In the formula, if current is zero power is also zero (on both sides of the transformer).\n\nYou can represent the pump and turbine with pictures (or drawings) of same but it would not be impossible to rig an actual pump and turbine, just not necessary to "see" what is happening.\n\nThe turbine can use up more and more power until the maximum power of the primary winding is reached but not more.  The magnet of the core will be "saturated at full power."\n\nThe rest is explaining magnetic flux in the core and how it is produced by the primary winding and cuts through the wires of the secondary to produce voltage and current if there is a load.  Each winding of a coil adds to the effect.  A transformer with equal primary and secondary windings would have the same voltage and current under load.  I believe that the phase would shift 180 degrees because of lead/lag.   Good luck.
762826
Hydraulic machines work on Pascals principle.\nAccording to it the pressure exerted on a confined fluid is tranmitted equally and undiminshed in all directions.\nSo we can apply a small force on a large area and that pressure is transmitted throughout the fluid.\nThis pressure can be used to lift relatively large load(or force) by making it act on a small area of cross-section on the platform of the machine.\nPascals Principle is : P1=P2\nor\nF1/A1=F2/A2
504417
When selecting a pump for an application, the best design is a pump that will provide both the required flow rate and discharge pressure.\n\nOn occasion in order to provide sufficient suction head for the pump a booster pump is installed in the suction. I have seen this on occasion for both centrifugal and positive displacement pumps.\n\nThis arrangement is usually avoided if at all possible because of the problems with operating the two pumps at the same time. Failure of one can cause problems for the other. When it is done there should be safety devices to stop both pumps if one fails.\n\nFor existing facilities there are times when operations may change and more flow or higher pressures are required than the existing pumping system can provide. This is the other time when a pump might be placed in series with another pump. \n\nIf there is a need for greater flow rates, a second pump in parallel can be used. If they are centrifugal pumps, one must be careful to allow for system pressure drop. If one pump is capable of 100 gpm and the other is capable of 50 gpm the net flow may not be 150 gpm. This is because the over all system pressure drop will go up with the higher flow rate and the existing pump will actually pump less liquid than before so the 100 gpm pump may now only pump 90 gpm.
844584
sl = 240 kPa (35 psi) (equivalent to 4.6 m or 15 ft) \nhd = (-)80 kPa [(-)11.5 psi] (equivalent to the injector being 1.5 m or 5 ft lower than the pump) \ndp = 69 kPa (10 psi) \nfl = CQ2L, where \nC = friction loss coefficient for the following pipe diameters: \n0.00114 for 100 mm (4 in) hose \n0.000455 for 125 mm (5 in) hose \n0.000284 for 150 mm (6 in) hose \n0.000142 for 150 mm (6 in) aluminum pipe \nQ = flow rate in litres/second \nL = hose length in metres
735996
D-W Formula for,\n\nHead loss due to friction in piping network\n\n=(friction factor*length*velocity^2) divided by (2*g*inside diameter of pipe)\n\nSo, from above formulae and my practical experience ,head loss can not be greater than pipe length in a fairly designed piping network.
1105154
pressure is: \n p(density)gh / gc =y(specific weight)h\nflow of liquid flowing inside the pipeline is:\n use the bernoulli equation:\n [p1/p(density)g  +  v1^2/2g  +  z1]  =  [p2/p(density)g  +  v2^2/2g  +  z2  +  hf]\n\nnote: p(density) is rho and y(specific weight) ir gamma.
827251
V = πr²h\nA (top & bottom) = πr²\nEnd Area = 2πr²\nLateral Area = 2πrh\nCost of tube = 2πrhC, C being cost per unit area\nCost of ends = 2πr²C*1.2\nTotal cost, P = 2πC(1.2r² + rh)\nh = V/πr², so\nP = 2πC(1.2r² + rV/πr²)\nP = 2πC(1.2r² + V/πr)\ndP/dr = 2πC(2.4r - V/πr²) = 0 for max or min, so\n2.4r = V/πr²\nr³ = V/2.4π\nr³ = 5/2.4π\nr = 0.872"\nh = 5/π0.872²\nh = 2.093"\n\nCheck:\nC(2.4*π0.872² + 10/0.872) = 17.201050C\nC(2.4*π0.87² + 10/0.87) = 17.201444C (+0.000394)\nC(2.4*π0.874² + 10/0.874) = 17.201137C (+0.000087)
103361
The above are correct but I think this is more readable:\n\nY = F*L /(dL*A)\n\nY=Young's modulus in units of pressure (F/A)\nF=force\nA=cross-section area\nL=length\ndL=change in length