What is the correct charge pressure for a 1 gallon bladder accumulator with a 5300 PSIG pump inlet pressure.?
413210
hi chuck, \n\n5300 lbs of pressure?  yikes!  hate to be working near that thing if it should blow!\n\nanyway the point of a bladder accumulator is to expand/contract thereby absorbing transient fluctuations in the pressure, right?  so i would expect you would want it to be somewhere near the middle of its range - too empty and it won't be full enough to compensate for pressure dips, and too full and it won't be able to expand to handle pressure spikes.  so somewhere near halfway i.e. 60% sounds reasonable to me.  \n\nthis is one of those things where a calculation of the pressure depends on a LOT of weird variables like the elastic constant of the bladder, the shape of it etc.  your best bet to get maximum damping is probably to just do an empirical trial and error sort of thing.\n\n200 Hz is well within the audible range of the human ear.  G3 on the piano (find middle A, go down an octave plus one white key - that's low G) is supposed to be tuned to 196 Hz, so your 200 cps hum should sound almost like low G.  depending on the amplitude of the pulses you could probably just listen to the pipes, or use a stethoscope if it's too quiet, and then vary the charge pressure looking for a minimum in the hum.\n\ngood luck!
40789
At Mustang Racing Technologies we run a Novi 1200 supercharger on our 2005 at 8.5 psi.  We have no problems and have had it higher.  With current boost it is putting out around 400 horses.  If you are planning on more than 5 psi you need some serious computer remapping and fuel help.  CPR fuel rails, injectors, and an SCT tune are suggested.  Don't worry about the intake.
297405
Depends on your application.\n\nPump water from a shallow (less than 20ft)or deep (greater than 20ft) well?\nHow much volume(gpm) and pressure(psi) are you looking for?\nBased on standard house current at 110VAC or off of a car battery(12V) and how long do you plan to have this pump running (few minutes at a time or hr's or day's on end)?\n\nYour question can be simple or a major under taking. \n\nA simple pump, using an old washmachine pump and motor you can come up with a basic pump. This will work on a shallow application, meaning the pump can't draw water from a resavoir being very deep, maybe a few feet. \nA sump pump is also a very shallow well type, works from a jet principle, but can't deliver a lot of pressure 10-15psi for a short period of time.\n\nWater wells which are deep, > 30ft, usually have the pump submerged into the water which then pushes the water out vs drawing the water up.\nSome of the older pumps were of a piston design, inefficient compared todays jet pump standards.\n\nOtherwise an electric moter and a mechanical pump.
710153
Usually psi (pounds per square inch).\n\nAloha
832830
Your head loss is going to be the the total weight of the water columns (height X area X density) between the inlet and the outlet.\n\nThe height of the current point at freefall is the head required by the pump beyond the inlet pressure supply to provide the required pressure at the outlet. It is the pressure expressed as weight of the water column divided by the outlet area.\n\nE-mail me more specifics.
131400
I believe it was around 300 psi.
1085954
I think the above person is correct is the physics, elementary as it may be. However there is no engineering insight present what so ever.  Despite the engr smarts remark. Though he did say "a little" engr smarts. An overstatement.  4 atmospheres is very very common (14.7psi/atm  x 4atm =58 psi) so pressure obviously is not going to be an issue. The issue is a challenge. Without the help of a processor, how do you open the supply and exhaust valves 8.33 times per second? 2 supply and 2 exhaust valves must actuate 8.33 times per second. That’s pretty quick. A PLC could do it of course. But it costs money, as stated. Perhaps you could use a cam system where each cam opens each respective valve at the appropriate time. Then all you have to control is the speed of the camshaft. The 4/2 valve ought to do the trick. As long as it can handle the duty cycle.
338000
I only had time to do the first one, btu the answer is \ne-2.  Check your email for the work.
227723
Regardless of what size capacitor you choose, there will be a small ripple on the output voltage, which of course gets bigger as the load increases.  You need to determine what is an acceptable ripple voltage for your load to calculate out a value.\n\nIf you want a totally smooth DC output, you'll need to first rectify and filter, then use a voltage regulator with a voltage set point that's a few volts below the voltage output from the capacitor.  Follow that by a smaller capacitor after the regulator to help catch any remaining transients and response after regulation.\n\nSomething to note:   Your capacitor's voltage rating should be much higher than the actual working voltage you will subject it to.  Eletrolytics like to blow up (literally) if you run them past their maximum rating, even for a short time.   Also remember this, the AC voltage that you measure at the output of the transformer (or the output of the rectifier, unfiltered) is the RMS voltage, not the peak voltage of the AC waveform.  Once you add the filter cap, your DC voltage will appear to be higher because the capacitor is charging to the peak voltage, which is basically the RMS voltage multiplied by 1.414   Keep this in mind as well when selecting the working voltage rating for your capacitor.
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!
963774
Use the fact that PV/T = a constant, we find the constant by pluging in the values we have, T = 20C, but we must convert that to K first so 20C = (273 + 20) = 293K.  And 1 atm should be converted to physicall units of Nt/m^2 for an easier solution.  1 Atm = 101325 Nt/m^2.    So our constant is (101325)(4.5)/293 = 1.336 x 10^7 so now you have one equation as\n\nPV/T = 1.336 x 10^7 setting T = 260C or 533K gives\n1)  PV = 7.121 x 10^10\n\nThe next equation that we need is the force on the cylinder end given the pressure, which would be\n\n2)  .01P = F\n\nThe spring will stretch a distance 3)  d = F / (2.00 x 10^3)  \n\nThe volume changes by  4)  V = 4.5 + d * .010   \n\nThis gives you 4 equations which you can now solve to find the solution for V, d, P and F
581128
In order to achieve 600 ppm, \n600/120 pound= 5 pound product should be added to 1000 gal water.\n∴ In 10 gallon water, 5 pound * 10/1000 = 0.05 pound of product should be added. This is to be added in 30 seconds.\n\nHence, the flow rate of product is: 0.05/(1/2) = 0.1 lb/minute = 0.1 *60*24 lb/day = 144 lb/day * 1 gal/10 lb = 14.4 gal/day \n  \n\n⇒ %output = (14.4 gal/day)/(22 gal / day) = 65.45.. ←Answer
396771
go here and check on accu check products on the left hand side. \nhttp://www.accu-chek.com/us/rewrite/home/en_US\n\ncheck on downloads on the left side and see if your meter is in there. Good luck.
744792
Pressure at the bottom = p1 = 12.7 * 0.0136 * 9.8 *10000 + 101300= 118226.56 Pa\nT1 at the bottom = 273+33=306 K\nT2 at surface = 273+33.6=306.6 K\nV1=0.115 cu.cm.\np1*v1/T1 =p2*v2/T2\n118226.56 * 0.115 /306 =101300*v2/306.6\nwe get v2 = 0.1345 cu.cm
105685
1 bar =  1x10^5 Pa and 1 atm = 101325 Pa\ntherefore 1 bar = 0.987 atm\n\n_______\nOK, got it.  Just for future reference, you can clarify a question using the "add details" button in the stick note in the top right rather than posting it as the answer to a different question :)\n\nWith a signficant increase in height, water pressure will change signficantly.  Basically, that extra 33-foot column of water weighs quite a bit and is pushing down on things under it.\n\nThe density of water is 1 g/cm^3 = 0.001 kg/cm^3.  33 feet = approx. 1,000 cm.  A 1,000 cm high column of water with one cm^2 cross-secional area would weigh about 1kg, adding 1kg/cm^2.  1kg/cm^2 = 1.097 bar.  So being three stories up would add just about 1 bar of water pressure.\n\nI'm not sure how these water tank assemblies work, but it seems to me that if you put the tank three stories up and didn't pressurize it you would have about 1 bar of water pressure at the bottom.  If you pressurized the tank so you had 1 bar of water pressure at the top, you'd have 2 bars down where your dishwasher is.
1135012
Look at fig.1 on the link below. This is a very useful chart which will give you a thorough knowledge about the test pressure. This website will also give you a good conception about the test procedures. \n\nFor one line answer: \nTest pressure is 1.5 times or 150% of the MAWP (Maximum Allowable Working Pressure) and not the MOP (Maximum Operating Pressure). \n\nWell, I have completed a year-long project on pressure vessel just a month ago. So I could provide you this link so easily. Let me know whether the answer was helpful.
908061
Water always loses pressure going up, regardless of the steepness of the incline. The rate is about 15 psi per vertical foot.