A fail-open valve receives 7.8 psig from a controller. What is the valve opening in percentage?
820885
Ultimately the valve should be open 60%. This is a fail open valve which means it is fully open at 0psi instrument air supply. An instrument air signal is 3-15psi range, meaning in the case of a fail open valve that 3psi is valve just starting to close, 6psi valve is closed 25%, 9psi closed 50%, 12psi closed 75%, and 15psi fully closed. your application is 7.8 psi supply which is 4.8psi above the start of the range (3psi) putting the valve 40% closed and in turn making it 60% open. \n\n    3psi  /      6psi   /       9psi   /     12psi    /    15psi\n    0%   /       25%  /       50%  /      75%     /    100%\n\nThere may be some speculators to this question because the adverse condition is the fail open valve, if the question was formed to apply to a fail closed valve the answer would be 40%open and 60% closed.
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.
791045
The volumetric flow rate is proportional to the pressure at the bottom (which is rho*gh, or density times gravity times fluid height) and to the area of the orifice (which is pi*r^2, where r is the radius, 1 mm = 0.001 m).  There is a valve constant involved, which you can assume is equal to 1 if you're not given additional information.
380471
pneumatic operated butterfly valves with 3-15 psi control using a 4-20ma I to P converter based on a voice coil type machanics, don't try one of the 4-20 ma electronic I to P converters, Radio frequency, Variable frequency drives, Changes in grounding, make the electronic I to P converters less accurate and reliable. Add a 4-20 loop to the position control and your world will be perfect.
665350
As the volume is constant,\n(P/T)initial = (P/T) final\n1/(25+273.15) = P/(50+273.15)\nP = 1.084 atm\n\n(Taking the pressure of the initial gas was 1 atm.
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB
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)
365444
0.500L x 0.250moles/L x 58.44g/mol = 7.31g\n(a)
445032
Click on "Add details," post the problem, and, as SMeghead said, someone might be able to help.\n\n(EDIT: You didn't provide too much information. From what you gave, about all we can say is that 12.5% of the solute - 1% from 8% - evaporated. If you need to know how many grams, milliliters, cubic centimeters, etc. evaporated, we'll need more information.)
826679
Let's assume that the flask is upside down in a water bath.\nAt 100 C, it contains some water (volume = x), and a volume of hot gas (VH).  When cooled to 0 C, the volume of the gas decreases, and some water (214 mL) is pulled into the flask. The volume of the cool gas is now VC. \n\nSome relationships to get you started:\n\nVH - x = 300 mL\nVH - VC = 214 mL\n\nCharles' Law\n\nVC / TC  =   VH / TH\n\nor\n\nTH /TC    =    VH / VC
711363
The volume will be 549.5 mL.  Let me know if you want to see the work.   Use this equation PV/T=P*V*/T*.    P,V, and T are just the pressure, volume, and temperature for the initial system (they are usually represented as P1,V1, and T1 but I left out the numbers which should be subscripts for simpliofication because yahoo does not have many text options.)  P*,V*, AND T* are the pressure, volume, and temperature after the system has been changed.  You could use any units pretty much for the pressure and volume as long as you keep the units consistent.  However always use Kelvin for the temperature.  Kelvin = 273 + Celcius Temp.
747804
There are multiple units usable.  Try pascals, psi, torr, atm
1019372
try this one\n\nValve Selection Handbook, Fifth Edition : Engineering Fundamentals for Selecting the Right Valve Design for Every Industrial Flow Application (Hardcover)\nby Peter Smith, R. W. Zappe \n\nhttp://www.amazon.com/gp/product/0750677171/sr=8-2/qid=1149712777/ref=sr_1_2/102-7470698-5004951?%5Fencoding=UTF8
708158
well aside from that you said the vapor pressure is 18.6 and put 18.9 in your calculations, it looks great! Besides, even with that difference, it doesn't change the answer at all, its still .877 atm with both values.
41783
If you mean a fluid seal, like a hydraulic seal, then its 60 to 80 percent. I had to look this up recently, and found the info in a Parker Catalog. I was milling the groove for a seal.
652894
the amount of pressure exerted on an object
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.