How to calculate the volume of water in a 1 foot section of 10" pipe with a flow rate of 200gpm?
450433
well, if it is steady flow and incompressible, then the flow rate doesnt matter (cuz equal volume of water flows in as the amount that flows out so no change in volume). so the diameter of the pipe is 10" so the radius is 5" and you can find the area by doing A=pi r^2. and then 1 foot is 12 inches so then multiply that by 12 to get the volume in um inch^3. so its 300pi"^3. hope that helps.
1051954
All you need to do is know the density of water- not too difficult to find! THe psi necessary to support a column then must match the weight of that column. So, if a column is 1inch in area and Hinches high, if you know D the density of water per cubic inch in pounds, the height of the column in inches times this value is the psi needed.\n\nPressure psi=H*D\n\nThis also works for metric, just use cms and kgs as the base sizes.
261582
Of course, you would have to convert the English to Metric, but...\n\nObtain the volume (in cubic feet) of the water (i.e. Length times width times height)\nConvert cubic feet to gallons (i.e. 7.48 gallons/cubic foot)\nObtain the weight of the water (i.e. 8.34 pounds/gallon)
498449
It depends.  If the pipe is vertical and long enough you won't EVER push it all the way through.  Once the pipe is full enough, it will take 500 PSI at the base just to keep it in there!\n\nIt also depends on the length of the pipe (you could assume here whatever length of 1 inch pipe it takes to contain 50 cu inch of liquid, or 50/(PI*.5^2) = 200/pi inches) but more importantly on the viscousity of the liquid (water, which you can look up).  There is also frictional drag by the pipe, but I bet you are supposed to ignore that. \n\nFlow rate = pi*P*r^4/(8 * v*l)  where v is vicousity, P is the pressure difference between the ends of the pipe, l is the length of the pipe and r is the radius.  Putting in that length expression and dividing the volume by the flow rate to get the time, you end up with \n\ntime=pi^2*P*r^6/(8*v*vol^2)\n\nand vol=50 cu inches in this case\n\nThis makes the assumption that the length of pipe in question is s.t. it always contains 50 cu inches of water at any point in time,  and you just want to see how long it takes to push 50 cu inches out as you add more in the back end.  If you change the volume and pipe length as a function of time (the pipe is still the same starting length and starts full but no new water is added as the 50 cu inches moves down the pipe) you will have to solve a simple differential equation. Be sure to put down the appropriate expressions for how mass and volume of water are a function of length!
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
386344
1067 gallons/minute, approx.\n\nSee reference
704981
V = LWH\nV = (200cm)(200cm)(1cm)\nV = 40000 ml\nV = 40 Liters
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!
1011234
united states gallons per minute
601391
Depends on the height.\n\nV=pi*r^2*h\n\n1 ft^3 = 7.48 gallons (U.S.)
498327
Multiply the number of cubic feet by 7.481 to get the number of gallons (U.S.). \n\nThis formula applies to "liquid gallons", which are the kind commonly used today.
1177171
multiply by 100 and divide by the other amount of water
361907
If the tank is a cylinder, the Volume is πr²h.  If you get the volume in cubic feet, multiply by 7.5 to get gallons.
753895
PV=nRT\nn=PV/RT\n\nR=10.7316 ft³ · psi · °R-1 · lbmol-1\nP=14.7 psi\nT=530 R, 540 R\nV=100ft^3\n@70F=530R\nn=14.7*100/(10.7316*530)=.2585\nmol wt=28.8\nm=28.8*.2584=7.44 lb\n@ 80f=540R\nn=14.7*100/(10.7316*540)=.2537\nm=.2537*28.8=7.31 lb
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.
898274
Yes, you can do it, but it will require iterations.  \n\n1) Convert all units to standard engineering units (feet, pounds, seconds).\n\n2) Set up the mechanical energy equation:\n(p2 - p1)/gamma = z2 - z1 + (f*L*v^2)/(2*D*g)\nwhere p2 -p1 is the pressure across the pipe (20 psi, but convert to lb/ft^2), z2 - z1 is the rise in the pipe (if any), gamma is the specific weight of water (62.4 lb/ft^3), L is the length of the pipe (50 ft), D is the I.D. of the pipe (ft), and g is the acceleration of gravity.  \n\n3) Now you have a problem: v is the velocity of the water, and f is the friction factor, which depends on v.  So, you guess a value for v, and calculate f from it.  You look up f on a chart (it's called a "Moody chart").  But you'll need the Reynolds number and relative roughness of the pipe surface to do that.  Then you try another v, find f, etc.  Or, you can try this:\n\nhttp://www.fluidmech.net/jscalc/vmd03.htm\n\n4) Just keep plugging in different velocities until the head loss (right side of the equation) equals the left side.  Then convert to volumetric flow rate.\n\nHey, I get 14.2 ft/s, 0.0774 ft^3/s = 34.7 gpm
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.