A fluid traveling through a pipe of radius 4in. is moving @ 400 ft/sec. Constricted to a radius of 2in. what?
536856
as you know that\na1v1=a2v2 \nso \npi r1 square v1= pi r2 square v2\nv2= 1600 ft/sec\n\ni hope it helps
704981
V = LWH\nV = (200cm)(200cm)(1cm)\nV = 40000 ml\nV = 40 Liters
696976
Let v1 = speed of the plug relative to a dot\nLet v2 = speed of the infinite pipe relative to a dot\nLet phi = angle of the pipe relative to some plane\nspeed of the pipe along the plane is v2 * Cos(phi)\nphi = 0 if we select a plane along the axis of the pipe\nTherefore the vector describing the speed of the pipe along the plane is v2*Cos(0) = v2\nThe plug speed relative to the plane is also found to be v1\nAn infinite pipe will act like a line that is very far away\nThe plug will act like a dot moving along that line if it is very far away too\nAssume that our origin is very far away from the pipe - if the pipe is infinite then space has to be very big and the assumption is correct for all 3-dimensional space.\nThe speed of the plug relative to the pipe is v1-v2\nv2 = 0 since if it was greater than 0 then it would have taken an infinite amount of energy to move it since it would have an infinite mass and no energy would have been left over to make the plug.\nSince the ends of the pipe are infinitely far away then the pipe must have come into existence after the plug\nThe guy who got the pipe wrapped around the plug would never have been able to do it if the plug was moving\nTherefore v1 = 0\nIf v1=0 and v2=0 then v1=v2\nSo in fact the plug is stopped relative to the pipe.
708432
Poiseuille's Law: dV/dt = (πR^4/8η)(∆P/L)\n\nWe're given\n\ndV = 490 cm^3 or 0.00049 m^3\ndt = 35 min or 2100 sec\nL = 2.5 cm or 0.025 m\n\nη - viscosity of water is 8.90 × 10-4 Pa•s or 8.90 × 10-3 dyne-sec/cm2at about 25°C.\n\nWe also have ∆p = ρg∆h\n\nρ is the density of pure water (1000 kg/m^3)\ng is the acceleration due to gravity (9.81 m/s^2)\n∆h is the height of the fluid column (we are given 0.8 m)\n\nHence, ∆p = (1000)(9.81)(0.8) = 7848 N/m^2\n\nNow we solve for D, the diameter of the needle, which is = 2R\n\nD = 2R = 2[(8ηLdV)/(π∆Pdt)]^-4\n\n= 2 [(8(8.90×10^-4)(0.025)(0.00049)) / ((3.147)(7848)(2100))]^(1/4)\n\n= 2(2.025 x 10^-4) = 4.05 x 10^-4 m or 0.405 mm
484307
in air= 340m/s\nin water=1500m/s\nin solid=4500m/s
887467
Im a little rusty on the thermodynamics, but lets see if this helps.\nV/T (initial)=V/T (final) since no gas is lost or added  Raising the temperature on a container that is free to expand does not change the pressure, so you can find the change in volume with temperature.  Since V=A*h, you can find h.\n\nNow, the problem is that the vessel is not entirely free to expand, and acts on a spring, with force equal to F=k*h.  This force is related to the pressure created by the expanding vessel as simply F/A.\n\nAll values are given, so I think you are set.
1078399
i'm not very sure if this is what you're looking for, but for a closed system with constant temperature, pV=p'V'. since volume, V = pi x (diameter/2)^2 x length, with constant length of the pipe, we have V = k(diameter)^2, where k is constant. hence, pV=p'V' is rewritten as p(diameter)^2 = p'(diameter')^2, i.e. p'/p =(80/250)^2. the new pressure is 0.1024 the original pressure.
581110
You have to decide the velocity. You need to keep Re<2000. You know the diameter of the pipe, u need to know the coefficient of viscocity. Velocity range is to be calculated by putting all these quantities in the equation for Re, keeping Re between 0 and 2000. Then the velocity is to be controlled in that range.
742316
Torqu equals force applied times radius, so the equation would be T=Fr    or    T=mar    so, T=(4)(9.8)(6), or 235.2 n/m\n\nMoment of inertia is not needed to solve this problem unless it later asks for another different answer.
12926
First, find the area of the circle that is the end of the pipe.  Pi * r squared = 3.1416 * (.25 / 2) = 0.7854 square inches.\n\nThen multiply by the length of the pipe.  BUT, since you want cubic inches, and the pipe is 750 FEET long, convert feet to inches (750 * 12).\n\nSo the answer is 750 * 12 * .7854 = 7068.6 cubic inches\n\nThe prior answer above assumed 750 INCH LONG pipe not 750 FOOT long pipe (750')
82505
First, you need to find the wavelength of the frequency.\nSpeed=Wavelegth*frequency\nWavelength= 342 m/s / 440 1/s =.78 meters\n\nThe wave traveling into the tube will hit its first resonance frequency one quarter of the way along the wave.  This is the first spot where that wave travels into the tube at such an angle that when it bounces off the water, it can escape the tube again; that is what creates the "resonance" or the louder vibrating sound.  So the first resonance would be @ .78m/4=.19m ; the second @ .19*2=.39m ; and the third at .19*3=.58 meters.
800864
Dear Little Giggl:  I think the answer to your question is VELOCITY, but it is NOT viscosity - which is the RESISTANCE property in liquid, that prevents quick flowing.\n\nFlow dynamics is said to be measured according to density and pressure which help determine velocity measurement of liquid.\n\nThank you for the question. . .it helped me to learn something.
1022690
1502 pipe is rate to 15000 psi, for the rate you can take the size of the pipe times itself then double that amount, so that a 2" pipe will be 2X2X2=8, this gives you the barrels per minute ( 42 gal in an oil field barrel). When you go above this point you get what is called turbulent flow, and any thing above 8 bpm on the 2" pipe can and will begin to wash or erode the inner walls of the pipe. Now with that said, pressure and time will affect how much eroding you will see, the higher the pressure and longer you pump fluids, sand, and gasses through the pipe the more washing you will get. Now if you are pumping into a formation, as in fracturing a formation, also called fraccing a well, there are many points as to what rate you will be able to get. You can add\nchemicals to the pumping medium to lower the pressure, but by rule of thumb all frac companies refer back to the formula I showed you. A 2" pipe is 8 bpm, a 3" pipe is 18 bpm and a 4" pipe is 32 bpm. Any mount above that will be destroying the pipe.
838461
The angle formed by an object at a given external point \nfor 1/2 in pipe at 250 Ft:\nRight triangle is (draw a picture)  hypotinuse is 250X12 = 3000" \nSide opposite angle is 1/4"\ntherefore: sin A = .25/3000\nA = .00477.......
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.
22742
4 inches. the radius is half of the diameter.
787117
If the cross-secional area of the aorta is 2.54 cm^2, and the blood flows at 48.9 cm/s, then in each second, 2.54*48.9 CUBIC centimeters of blood is flowing through the aorta.  All you have to do is multply.\n\nFor the second part, you use the answer from the first part.\n\n48.9*2.54 cubic centimeters per second is flowing through a bunch of capillaries with a combined cross-sectional area of 2720 cm^2.  Therefore the speed must be 48.9*2.43/2720 cm/second.  It's always a good idea to check units: 48.9*2.43 cm^2/sec/2720 cm^2 comes out in units of cm/second.\n\nSame concept as "wide rivers run slowly"