how proportional hydraulic work?
437002
no book but with a proportional valve, two or more different actions can be controlled with the same system, even if the flow of the devices are different.
482206
Is exponentially proportionally to the load.
313557
leftknight is about right, but it uses hydraulic fluid, not water. These presses range from small hand-pumped laboratory units to huge monsters with thousands of tons force. They are used in metal forming and making bricks and structural concrete pieces.
926824
Mechanical jacks are classic levers (fulcrums).  You apply a lesser force through a longer distance on the input in order to achieve a greater force over a lesser distance on the output.\n\nHydraulic jacks are based on the property of non-compressibility of liquids and uniform pressure throughout a static liquid.  You apply a given force to a small area on the input (the pump) which transfers that pressure to a larger area (the piston) thus amplifying the force (but again, reducing the distance).
400812
I think you meant to say which ratios are proportional.  To find that answer treat each ratio like a fraction and reduce them to their simplest form.  When in simplest form if they are equal then they are proportional.  Another method would involve cross multiplication (diagonal multiplication) in your first two proportions you would multiply 1x5 and 2x10.  If your answers are equal then the ratios are proportional.
139840
The formula for either is P=rho*g*h, where rho is the density of the material (mass per unit volume) and g is acceleration of gravity, and h is the depth of the material at which you are measuring the pressure. \n\nPressure is a force per unit area. If force are unbalanced, then one gets acceleration of something i.e., something moves, by Newton's second law, F=ma.
60954
It depends on usage. You know that there are two main things in pump, Head and Discharge, and we have a rating curve for these two parameter. If you want to send water to higher elevation, you can send less water (Decrease Discharge) and vise-versa.\n\nin normal case and normal usage, we have some pumps around 2 lit/sec for a very low high to around 2000 lit/sec. Pumps that are used in wells is around 30 to 50 lit/sec.\nBut I advice you to chose apropriate pump by rating curve.\nOne of the largest company in pump industries is KSB. You can find more information in their sites. Below:
24462
Yes, it does. Pressure = Force/Area. So area is inversely proportional to pressure. So pressure does decrease with increase in area.
963983
There are many ways to do this problem.  The best way to probably handle density, and perhaps mass even, is using Archimedes principle of buoyancy.  You can find a detailed discussion on www.wikipedia.com and search for Archimedes or buoyancy.  The main idea is that "the weight of the displaced fluid is directly proportional to the volume of the displaced fluid".  \n\ngood luck!
1066290
First, you need the volume. The formula yields: V = pi rh/3 = 20000pi/3 cubic feet. Since it weighs 50 pounds per cubic foot, the ratio 50 lbs/ft.^3 = 1. So multiplying this to our volume equation, V = 20000pi/3 ft.^3 = 20000pi/3 ft.^3 * 50lbs/ft.^3 = 100000pi/3 lbs. The work function is (25-x)100000pi/3 = W(x). Integrate this result from x=0 to x=25, and you get the answer.
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).
461412
1) assuming it's inversely proportional then you have multiplied by m+2/m so it's dm/(m+2)\n2)(m+2)d/m
596792
Jet engines work on Newton's 3rd law. Throw sometihng back with a certain force and you will be propelled by that same force.\n\nNow, by continuity eqn, Av = A'v'\n\nIf you throw something out of a small area at a high velocity, the rate of flow in the opposite direction is then proportional.
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! :)
942928
All substances, gases, solids, and liquids, including water, are compressible. Some \nsubstances just require more pressure per change in volume to compress than others. The \nnumerical value of the compressibility depends both on the temperature and on the pressure \ntoo. That is how hard you have to squeeze at one pressure to obtain a certain fractional \ndecrease in the volume is different than at some other pressure. The same for two different \ntemperatures. Under most "normal" lab conditions however solids and liquids are usually \nconsidered to be incompressible.\n\nTechnically speaking, water can be compressed, but very little even at high pressures.  For \npractical design purposes, water is considered an incompressible fluid, that is, its density \ndoes not change with pressure.  The reason anything is compressible is due to how close the \natoms are packed together.  If you think about air, it is highly compressible because there \nis considerable spacing between the atoms, so it is relatively easy to force the atoms closer \ntogether.  When you have a liquid, the atoms are much closer together and considerable \npressure is required to make them any closer.  Solids will compress some as well, but it \ntakes significant pressure.  Temperature also plays a factor in density (and thus \ncompressibility), just ask your chemistry teacher about the ideal gas law.\nI am not sure what your book was using a "water compression system", but hydraulic systems \nare used almost everywhere to transmit power across a system.  What a hydraulic system does \nis put a liquid (usual an oil of some type) under pressure, route that pressurized liquid \nthrough pipes or hoses, and releases the pressure at a place where work needs to be done, \nusually through a motor or cylinder that transfers the pressure into work.  You might want \nto research on hydraulic systems (hydraulic lifts, hydraulic elevators, etc.) to get a better \nunderstanding on how energy can be transferred around in a hydraulic system.
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!
936671
The lever gives you a 5:1 advantage, this means that what ever you get for the hydraulic advantage, you will divide the answer by 5 because of the added mechanical advantage of the lever.\n\nThink of it like this: Without the lever, the hydraulic advantage is 1.3/.3 or 4.33:1\nThe pressure to balance on hydraulic port #1 is 500/4.33 or 115.38 lbs. and the lever makes the total needed to balance 23.08 lbs. or rounded to approx. 23 lbs.