given the gpm and pipe diameter can the pressure head be determined?
481625
Yes, it is possible to find pressure drop per unit length from  some more information about density and viscosity of the fluid and roughness of the pipe. Any book on fluid dynamics gives the procedure.\n\nTheformula is:\nPr head = 2f(v^2/g)(N/D)\nwhere \nf= friction factor (generally found from a chart)\nv= average velocity of fluid \ng= conversion factor (having magnitude equal to std accln due to gravity)\nN= Equivaalent length\nD= diameter of pipe.\nYou have to follow some standard system to work out this problem (see standard reference)
223337
Without knowing exactly what you need to know it's difficult, but I found this website that might help.\n\nhttp://www.kbr30.dial.pipex.com/educ19.shtml
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.
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/
222690
This all depends on the diameter of the second pipe, if you reduce the diameter then the pressure in the second pipe will increase. i.e. psi. to make sure you got the same flow rate then you will need to know the orginal flow rate. and the flow rate of the second pipe. then you can get it balanced out, by reformulating the psi equation/ the flow rate for both pipes.
525014
Let's make this very simple.\n\nTo measure the volume of a cylinder, you find the diameter of the interior portion of the cylinder.  Take that number and divide by two...that's your radius (r).  Now measure the height of the interior of your cylinder...that's your height (h).\n\nNow, take the radius (r) and the height (h) and plug them into the formula below:\n\nV = π*(r)*(r)*(h)\n\n(π equals roughly 3.1516)\n\nNow if your answer is in cubic inches, multiply V by .0036 for British gallons or .00433 for US gallons (yes, they are different).  If V is in cubic centimeters, multiply by .00022 (British) or .00026 (US).\n\nI don't know how to make it any clearer...good luck.
483088
this is very easy, just use the pythagorean therom (i might spell it wrong) the formula is: a^2 + b^2 = c^2 (c is the length of the pipe)\n\n15^2 + 8^2 = c^2\n\n225 + 64 = c^2\n\n289 = c^2\n\n17 = c\n\nyou need 17 feet of pipe
938684
If you desire the external surface area of the pipe,\n\nA = pi*d*L\n\nMake sure d & L are in the same units and your answer will be in those units squared.\n\nIf you are looking for the internal cross sectional area:\n\nAc = (pi*d^2)/4
657436
I assume the pump performance curve is unavailable, pity. You know what the current rpms, pressure and power is. I would apply the pump affinity laws in order to determine what the increase in rpm(flow) and power it will take to make the pressure you require. It may be  easier getting a bigger motor for the existing pump, than going through the time and effort of a single new pump, pumps in series, piping reconfiguration etc, all of which will work.
943088
Depends on the head type...unless you need to be extremely exact assuming 1/2 the vessel diameter is pretty good(i.e. a semi-elliptical or 2:1 head).\nKeep in mind though that pressure vessel heads have some straight-side to them when they are manufactured and the way I understand it is this amount is determined by the equipment manufacturer.
953550
see this website :\nhttp://www.hydraulicspneumatics.com/200/FPE/Accessories/Article/True/6443/Accessories
423881
true - high sustained pressure and a signifigant distance to any pressure disturbance or variation
237589
There are two aspects to your question.  The area of the cross-section through a round valve stem is Pi x r**2 (Pi times the radius squared.)  Working with a magnified drawing of the stem with keyway, you could determine the percent of area of the cross-section removed by the keyway and calculate an effective new diameter."  This could be used to determine how much meteal is needed to make the stem, but not how the stem will perform.\n\nThe use of a keyway not only eliminates some metal from the stem but changes how the stem is loaded while being torqued.  That depends on the keyway design and also how the keyway is machined.  Sharp corners in the keyway may cause severe stress risers, multiplying local stress and leading to failure if overloaded.  You may need to calculate allowed torque on stems with and without keyways to determine equivalent effective diameters from a performance point of view.
278523
you need to figure out the inside and outside diameter of each pipe example 1 7/8 inside dia, will fit over 1 7/8 outside dia, hope this helps good luck
1031813
use the formula pi*r^2*h\nwhere pi=1.34\n r=radius (in inches)\n  h=height of cylinder (inches)
745326
The volume of a cylinder is the area of the base (a circle) times\nthe height of the cylinder.  The pipe represents a cylinder.\nThe area of a circle is PI times the radius squared (PI X r^2)\n\ndiameter of pipe = 24 inches implies the radius = 12 inches implies radius = 1 foot.  Length of pipe is height of cylinder\nnew value of PI = 3.1416 exactly (?)\n\nVol of Cylinder = Area of Base                      X Height of Cylinder \nVol of Cylinder = Area of Circle                     X Height of Cylinder \nVol                   = PI(3.1416) X radius(1 ft.)^2 X height(8 ft.)\nVol                   = (3.1416 sq.ft.)                     X height(8 ft.)\nVolume            =  25.1328 cu.ft. exactly
711880
The Cos of the pressure angle is equal to the Base Diamater over the pitch diamater of either gear in the gear set.  So since;\n\nCos(φ)=Dbase/Dpitch \n\nTherefore\n\nφ=cos^-1(Dbase/Dpitch )