how can I predict pressure loss in hoses?
108206
I think you can use Bernoulli's equation to predict the difference in pressure from one area of the hose to the other and, therefore, the pressure drop:\n\nP = (flow rate * 8 * viscosity * length of hose) / (pi * hose radius^4)\n\nNot sure if it works though...typically used for fluids...
114490
u may want to try Handbook of Hydraulic Fluid Technology\nMechanical Engineering (Marcel Dekker, Inc.) by George E. Totten ISBN: 0824760220
249398
I suppose you are reffering to the small digital scales. They have a device whos electrical resistance changes under pressure.
437096
Try this...\nhttp://www.tirepressuremonitor.com/\n\nor this........\nhttp://www.hunter.com/pub/undercar/techlink/
1005042
For the correct reading, read off the base of the meniscus. \n\nSurface tension between the liquid and the cylinder's inner circumference will raise the reading and give an erroneous reading.
1062484
if you are asking an instrument to measure a parameter outside of its design range, then it is not possible to state with accuracy how reliable the measured value is.  Your best bet is to look at the equations for flow through an orifice and see if the pressure drop is linear, or not, or what the relation is.  You might be able to extrapolate a resaonable approximation of the true flow if it is not too far away from the 3000 value.  If the meter was reliable below 3000, the manufacturer would have stated so.
877652
These are GAS LAW problems, so you need to use PV=nRT.  However, for these problems, nRT will be constant so you only need to use P(pressure) and V(volume).  Therefore, for both of these questions, use the equation:\nP1V1 + P2V2 = PtVt   where P1 is the pressure in the first vessel, P2 is the pressure in the second vessel, and Pt is the total pressure of the connected system.\n\nProblem #1\nP1V1 + P2V2 = PtVt  \n(7.0atm)(15.0L) + (P2)(8.4L) = (10.0atm)(23.3L)  \nP2 = 15.4 atm\n\nProblem#2\n(7.2atm)(15.0L) + (4.1atm)(8.3L) = (Pt)(23.3L)  \nPt = 6.1 atm
963866
Cover one end of your garden hose with clear plastic (like Saran Wrap) and wrap rubber bands around it to hold the plastic in place.  Then remove your garden hose from the faucet and cover the faucet end of the garden hose with plastic wrap and use rubber bands to hold that in place as well.  Lay down the garden hose over a relatively flat area of the ground where you wish to determine level elevation.  Then look at each end for air bubbles.  (Make sure the garden hose isn't draped over a hill, or this procedure won't help.)\n\nIf you find air bubbles at one end of the hose, it means that end is at a higher elevation than the other.  Move the bubble end of the garden hose, without moving the other end until the bubbles leave the plastic.  Then check the other end of the garden hose.  If you find bubbles, move that end only until the bubbles leave the plastic.  \n\nRepeat this procedure until you don't see bubbles at either end.  That is when the ground is level along the length of the garden hose.
1034025
The pressure rating versus temperature for these classes of flange ratings is independent of the material being contained.\n\nThe best way to get a table or chart of the relationship between pressure and temperature is to contact a pipe, valve and fitting supplier in your yellow pages. If you can't find one, try the online yellow pages for Houston, Tulsa, Oklahoma City, Dallas or New Orleans.\n\nThe actual ratings are part of copyrighted documents published by ANSI and usually aren't available for free.
748177
Shear stress, wind loads, dead loads, compression...I think.
967963
Either air pressure, or fluid pressure on a fluid present in the cylinder.
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
302322
I won't do your homework for you. But I will tell you this:\n\n\n1a&b. These should be common sense. Go outside and turn on the garden hose and tell your teacher what you've observed.\n2a. Look up BERNOULLI'S LAW as it pertains to fluid pressure through a tube. There is a formula that related pressure to hose diameter and volume.\nb. The answer is: high pressure is bad for vessels weakened by atherosclerosis.
845517
One way is to use the Clausius-Clayperon Equation.  See the link below and use the boiling point of water for one of the pressure and temperature pairs.  \n\nP.S.  It's equation 22 in the linked document.
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.
622897
You'll have to look up ASTM standards for exfiltration or pressure testing for manholes (structures) and for pipes.  Pricing usually comes with installation & cost of materials.  Pricing can depend on depth of pipe, material of pipe, and size of pipe.
794091
I'd suggest you first try the site listed below where you can find a downloadable spreadsheet that predicts velocity. You can deduce the author's equations from it. It's actually a time-stepping simulation rather than analytical equations. However, a bit of theory wouldn't hurt either. First, to keep it simple, let's assume the pressure is constant for T, the lesser of the time the valve is open and the time the projectile remains in the barrel. This is actually not a good assumption since many spudguns have pressure vessels that are smaller than or not much bigger than the barrel volume. Anyway, during T the spud is subjected to a force F=P*barrel area = P*pi*d^2/4. The net force accelerating the spud is F - friction force, so its acceleration A=(F-Ffriction)/M and its velocity v=A*T. I don't know the practice on valving. It may be that open valve time is always less than barrel transit time, but if not, you can compute the distance S that the spud would travel during the open valve interval T; S=0.5*A*T^2. If S>l, then T is the time spent in the barrel, which is SQRT(2*l/A), and you should recompute v. \nAnother assumption would be that each shot entirely empties the pressure tank. In this case we simplify by disregarding pipe friction, thermodynamic losses (like failure to fully expand due to cooling) and any other losses you can think of, and assume the potential energy of the compressed air is converted to kinetic energy. The potential energy PE = 0.5*P*vol (where vol is tank volume = L*pi*D^2/4). The kinetic energy of the spud is 0.5*M*v^2. Thus v=SQRT(P*vol/M).\nAbout other questions, 1 psi = 6894.75729 Pascals, and speed of sound =340.29 m/sec.\nNote that with what you have given in your question, you have a lot of units conversion to do.\nEDIT: See friction stuff added above. The spreadsheet includes friction effects.