Can i run 10 pounds per minute of steam through a 1" line with 10 pounds pressure drop? how do i know?
363092
Well I'm not going to do the calculation for you but you can.\n\nDo a web search and you'll find dozens of sites to help you if you don't have a fluid flow handbook.
338800
Maybe about 10 pints.
711392
If its just proof-of-concept, could you use the pressure of a tank of liquid CO2 to pressurize your vessel with CO2?? Add some liquid CO2  (or even just some dry ice!!) to your vessel from the tank and warm it up to get the desired pressure?\n\nOr possible collaborate with somebody who does CO2 cleaning research?? You might be able to get some preliminary data cheaper by renting their equipment than building your own.
611212
One litre of water can be contained in a cube 10cm on each side.\nSo if you have, say, a 10cm square-section piece of roof guttering and you have water flowing along it (filling it to the brim), then every 10 centimetres the water moves means a whole litre has passed any given point.  If the water is flowing at 1 metre per second, then that's 10 lots of 10 centimetres and the flow rate is therefore 10 litres per second.\n\nRivers aren't nearly as simple, because they are not nice neat square troughs and besides, there are other factors such as turbulence, viscosity etc to take into account.\nBut the general principle is the same; try to measure the area of the cross-section of the flow and multiply it by the speed of flow to get the volume of water passing through.
264933
try substracting atmospherical pressure, and then you have the in-gage pressure.
380084
The answer is Pressure\n\n10 pts plz! im only level 1!
1136394
I would go by weight personally.  Timing it requires you to have consistent pressure which is hard to do especially if you have a high volume day.  This works best if you supply the containers to eliminate complexity of zeroing a scale, or the error of having differently weighted containers.  You'll need a platform that sends a varying electrical signal based on load, you could rig a potentiometer to change resistance based on the position of the platform, as the springs compress and the platform sinks the resistance decreases, now run a voltage through this potentiometer and a couple of resistors to get a variable voltage then run this into an op amp that turns on and off a relay that controls a servo that opens and closes a valve.  Adjust the resistances until you get it switching off when you want it to. I suggest ball parking it based on the weight of a gallon of water+container, then adjusting it up since there will be a small amount of momentum from the water falling from the faucet.  Setting up the Op Amp circuit isn't too tough I suggest getting a text book on them.  I suggest http://www.amazon.com/gp/product/0130149918/ref=pd_sxp_grid_i_2_2/002-1800445-2936816?%5Fencoding=UTF8\nIt's accessible to someone with little background in electronics, and is loaded with example circuits that you can virtually copy out of the book for any application.  It is also very robust considering it's simplicity.\n\nAs far as making it coin operated, use a locking mechanism that uses coins as it's key, then have that activate the power supply running to the op amp circuitry, and valve control, add another op amp circuit that turns off the power after the valve is closed.\n\nIf you're dead set on a timer, use the coin locking mechanism, but have it turn a dial timer, when the timer goes off it sends an electrical signal to an op amp that activates a relay and turns off the valve.
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!
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.
693386
I believe it developed on time as a practical ecomonic choice or industry standard.\n\nYou would want to use a very low pressure if possible but still it needs to be high enough to actuate instuments and control devices.
155279
By measuring the change in pressure.
1070639
Something is wrong with this picture... You are getting 0 psi out while putting 30 psi in?  There has to be some pressure coming out.... otherwise the CFM is going to be zero too.  I realize there is going to be some line loss - but your scenario does not account for a loss of that magnitude.  \n\nI'd check that gauge on the output side... maybe switch the in and out and see if the readings change. \n\nCheers!
619720
Well, the volume of water (in cubic inches) within a foot-long piece of pipe will be:\n\nV = 3.14159*r^2*(12in)\n\n(where r is the inner radius of the pipe in inches, 1" and 2" respectively)\n\nThen convert that volume from cubic inches into liquid gallons using the site listed below.\n\nedit: That sounds like an interesting idea.  You are heating the pipes via a solar panel?  In any case, good luck on saving energy bill money.
1077336
there are a few diff ways for this.  the way i do it is with a rotating lazer level. you can rent them at most rental places, another way is with a transit/builders level but if you dont know how...you will be lost cause you wont know how to read it. or the cheap and easy way.   get some clear plastic hose hold the ends up and fill it with water. create a "benchmark". use a post or somthing else you can tie it to and walk away without spilling. on the other end make a leg for you to rest it on roughley as high as the post. now as you move around you can see if it isflat by marking the level in the tube cause water will seek its own level. i prob left out some things in the explanition but im sure you will pick it up.....it was good enough for the romans and it can work for you.
266904
i don't know i don't think it's possible\n\ni tried 1010-10=1000\n\nso u c cant just do it with one straight line
226848
You have some very good answers here but I can add a little more.  Consider a pipe to be elastic (true!) but imagine that it is made of rubber.  Consider water to be nearly incompressible (a cubic inch of water at the bottom of the 7 mile deep Mariana trench still measures nearly a cubic inch).  Water can possess energy in two ways (neglecting heat); pressure energy or velocity energy and can easily convert back and forth.  When you extinguish velocity of a slug (or stream) of water you produce pressure and the pressure when relaxed can produce velocity.  This can go on and on until the energy is dissipated for example by friction.\n\nWhen a stream of water is flowing along an elastic pipe (think rubber) and you suddenly trip shut a guillotine valve (very fast acting) the velocity of the column of water is instantly extinguished and converted to pressure which starts as a shock wave stretching the rubber pipe toward the upstream end perhaps until it reaches the vessel supplying the water.  But the stream is then reversed as the rubber pipe relaxes squeezing out the water toward the vessel and continues until the rubber pipe is under a partial vacuum and shrinks.  In an ideal system (no friction), pressure waves would continue the travel back and forth along the pipe converting pressure to velocity and back to presure as the pipe expands and contracts under the changing pressure.  They call it water hammer because it actually can sound like someone hitting the pipe with a hammer.  Actually the pipe expands suddenly sending off a sound wave through the air.\n\nWater hammer can be avoided several ways.  For example if the pipe also contains air bubbles (accidently or on purpose) the bubbles will absorb any hydraulic shock wave preventing the elastic pipe from expanding and the bubbles re-expand slowly, slowing down the entire procedure.  The plumbing leading to many sinks in homes has a small "empty" vertical section near the sink to hold an air bubble which avoids a water hammer if the sink faucet is shut too fast.  It is sometimes necessary to drain the system with valves open to refill the air trap with air if it gets waterlogged.\n\nWater hammer can be avoided by shutting valves slowly rather than instantly.  Water hammer can occur in a pipe filled with steam because steam is easily compressed compared to an air bubble and a slug of water can be accelerated along a pipe by condensing steam until it "strikes" the end of the run and converts velocity to over-pressure.  I once got a frantic call at home (3 AM?) from a power plant under construction.  While testing a large pipe run for the first time under steam, the long say 12-inch pipe began to jump around threatening to break its pipe hangers and making a terrific racket.  Almost without waking up I told them to crack open any lowpoint valves or fittings to drain off all condensation and the pipe went back to sleep too.  In the finished plant there were automatic condensation drain valves to avoid water hammer which could damage the steam turbines too, etc.
2591
Your problem can not be solved accurately unless you also state the length and roughness of the pipe.  Picture two pipes, one ten feet long and the other the length of a football field.  Clearly the longer pipe will greatly resist flow.  Also a smooth copper pipe will resist flow less than a rusty rough one.  The Crane company published a book that can be used to estimate flow through pipes based on lab tests.  Good luck.