what are the components that made up the total head of a location along the pipe?
176188
Unable to determine your question.\nMore details are needed to answer your question!
347099
I have used the Bernoulli equation to calculate pressure drop through piping and to size pumps (determine required head pressure).  You can't get much more real than that.
788282
Hardware store.\n\nCopper pipe\nsteel bolt\nbrass bolt\naluminium knife (or fork)\n\nHacksaw them to approx same size ?
121975
The machine consists of a screw inside a hollow pipe. The lower end of the device is put in the water, and the screw is then turned (usually by a windmill or by animal or human labor). As the bottom end of the tube turns, it scoops up an amount of water. This puddle of water will slide up in the spiral tube as the shaft is turned, until finally it falls out from the top of the spiral tube and feeds the irrigation system.
1116662
The pitch diam. at the end of 1" pipe is 1.213" and a 1" pipe reamer tapers from 1.103" to 1.212", so the reamer should be inserted all the way, or close to it.
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
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
1111034
Perhaps the link will help a little.  If you bend a pipe or metal bar, some of the metal will be in tension (stretched) and some will be in compression (squeezed).  There is a neutral plane between the two conditions that is neither stretched nor squeezed.  Often the plane passes through the centroid of the pipe (or bar).  This is easy to find if the cross-section is regular (round, square, etc.) but more difficult to find if the area is irregular or of variable properties such as density.  The centroid is similar to the center of gravity for the shape.  Hope that helps a little.
845325
A red tube used for cleaning pipes.\n\nCoach
247204
simply they are pipes which convey graphic informations \nmore pipelines the better graphic it has
719561
it's where you place a T intersection (like for where the shower faucets go) with a riser above it to stop "water hammer" by giving the system some room for the air to pressure to go to.
1105154
pressure is: \n p(density)gh / gc =y(specific weight)h\nflow of liquid flowing inside the pipeline is:\n use the bernoulli equation:\n [p1/p(density)g  +  v1^2/2g  +  z1]  =  [p2/p(density)g  +  v2^2/2g  +  z2  +  hf]\n\nnote: p(density) is rho and y(specific weight) ir gamma.
696672
I have a drawing that shows the various types of shell and tube exchangers. I believe their functions are self evident if you see the cross sectins.\n\nSend me you email address and I'll send the drawing.
825100
The sewer can be the outter edge of the circle...the Pub should be the Inner circle :)
129691
See my answer to your next question.  For phase lead, C is in series and R in shunt.
237112
the cylinder shape and the contents.
995793
Division 1, Division 2 and Division 3.\n\nKidding, i think the base, handle and the eyepiece/nose piece.\nNot sure though.