How do you calculate the flow rate of the water in a river.Like how many gallons per minute pass a given point
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.
859177
Volumetric flow = velocity x cross-section area = 43cm/s x 2.0 cm2 = 86 cm3/s\n\nMass flow = Vol flow x density = 86 cm3/s x 1.0 g/cm3 = 86 g/s\n\nCapillary Velocity = Vol Flow / capillary cross-section = 86 cm3/s   /  3.1x10^3 cm2 = 0.028 cm/s
966711
The volume of the water in the cup at any moment is:\n\nV = (1/3) pi [ (Rw)^2 + (1)Rw + (1)^2 ] \n   = (1/3) pi [ (Rw)^2 + Rw + 1 ] h\n\nRw = Radius of upper part of water\nh = height of water\n\nRw / h = 2 / 6\n\n6 Rw = 2 h\n\nRw = 1/3 h\n\nSubstitute in to the volume equation\n\nV = (1/3) pi [ (1/9) h^2 + (1/3) h + 1] h\n = (1/3) pi [ (1/9) h^3 + (1/3) h^2 + h ]\n\ndV / dt  = (1/3) pi [ (1/3) h^2 + (2/3) h + 1] (dh / dt)\n\nSolve for dh /dt  and plug in the values h=4 and dV/dt = 3
810640
See picture: http://mudandmuck.com/str2/BOAT.JPG\nShe must travel √(x^2 +4) miles by boat, and 6-x miles walking.\nTherefore, the time t is the sum of the distances divided by the rates, or\nt = √(x^2 +4) / 2 + (6-x)/5\n\nTake the derivitive of t with respect to x and set = 0 to find the minimum time.\n\nt = (1/2)(x^2 + 4)^(1/2) - x/5 + 6/5\ndt/dx = (1/2)(1/2)[(x^2+4)^(-1/2)](2x) - 1/5\n= x/[2√(x^2+4)] - 1/5 = 0\n5x = 2√(x^2+4)\n25x^2 = 4x^2 + 16\n21x^2 = 16\nx^2 = 16/21\nx = √(16/21) = 4√21/21 ≈ .87287156 miles from the point on shore nearest the boat.\n\n{Substituting this value back into the equation to find time, we see that the quickest time is ≈2.116515139 hours.}
987046
http://www.watertiger.net/RO/how_ro_works.htm\nyou find it here
1097124
Aplly the following formula:\n\nG = w * A, where:\n\nG - flow rate [m^3/s]\nw - speed of fluid [m/s]\nA - transversal section area of the pipe [m^2]
252385
Wow! At last I found a good question & that too with a sensible asker who is wiling to do the research part. Here is my answer just for you-\n\nIn case of a flood, the flow of river is most of the times "turbulent". Do a Google search on "turbulence"+"fluids". Also search for "Mechanics of fluids".\n\nBut, the force in case of turbulent motion is pretty complicated so, I'll explain it in case of a slower speed of flood when the flow is "streamline" i.e. when the flow is not irregular.\n\nWhile flowing, the magnitudes of the velocities of streamliness are inversely proportional to the spacing between two streamlines in a plane. So, the narrower the hole is, more speedily the water will pass from it & thus, would exert more force.\n\nNow, try and do a Google search on these topics-\n1- Bernoulli's Principle\n\n2- Torricelli's Law of Efflux\n\nThose laws will help you with your report a lot.
177213
I don't think there is a minimum.  If it is a river it is implied that the water flows.  Ecosystems develop depending on what conditions exist.
1088769
P = d * Q * g * H * efficiency\nWhere,\np= power in KW\nd= density of fluid used\nQ= discharge\nH = head\nefficiency= given for a turbine
772330
You just need a few basics to estimate things.\n\n1.  6000 watts is 6 BTU/sec.\n2. 1 BTU heats 1 lb of water 1 degree F\n3. 1 gallon of water weighs 8.3 lbs.\n4. A nominal natural convection heat transfer coefficient is 1 BTU/(ft^2 * hr * deg F)\n\nIf you barrel is well insulated, just use the above (except 4) and you will see that 25 gallons will heat at 0.027 deg F per second.  \n\nIf you arent insulated, you need to use the heat transfer to see what portion of the 6000 watts is lost as the temperature rises.  You can readily do this in general form and the result should be an exponentially asymptotic temperature.
890224
Velocities are vectors, so you´ll have to proceed according to vector rules for addition/subtraction.  Draw a horizontal line, with an arrowhead pointing to the left. This represents the (unknown) velocity due West. Now draw another line, vertical, this time.  Put an arrowhead in the lower end of this line.  This stands for the 50 kph wind speed to the South.  These two lines should make a right (90°) angle.  The lines haven' t to be drawn to scale.\n\nFinally, draw an oblique line connecting both remaining ends.  This is the aircraft speed; to indicate direction, put an arrowhead  at its leftmost end.\n\nNow, you have a triangle of which two sides are known.  This, and the fact it is a right triangle, suffices to solve the entire triangle. The angle you seek is that formed between the horizontal and oblique lines.  Let θ denote this angle.  From elementary trigonomety, sin θ = opposite side/hypotenuse = 50/205 = 0.2439... All you have to do now is to find the angle whose sine is 0.2439...  This is 14.1169...°.  From the figure you have drawn, this angle lies N of W.\n\nThere are several ways to figure out the speed to ground.  You may use Pythagoras' Theorem; or, having found the opposite angle, you may compute it as 205 × cos 14.117° = 198.809...\n\nJust follow the same procedure to solve the other problem.  You'll have to decide which direction is "river across" and which is "downstream"; I´ll suggest "right" and "down", respectively.  Just plug in the proper figures.
303745
54 mi in 3 hours is a speed of 18mph (54/3=18),\n54 mi in 3.6 hours is a speed of 15mph (54/3.6=15),\nTHe current is half the difference between the two - it helps as much as it stops you....\nTherefore current is 1.5mph ((18-15)/2=1.5) and the boatspeed in calm water is 16.5mph (15+1.5 or 18-1.5).\n\nThe equations are, with boat rate=x and current=c;\nx+c=18      (1)\nx-c=15      (2)\n(1) ==> c=18-x\ninsert c into (2) ==> x-18+x=15\n==>2x=33 ==>x=33/2=16.5\nInsert x into (1) or (2) to get c=1.5   \n\nHope it helps,\nFredrik, MSc
791045
The volumetric flow rate is proportional to the pressure at the bottom (which is rho*gh, or density times gravity times fluid height) and to the area of the orifice (which is pi*r^2, where r is the radius, 1 mm = 0.001 m).  There is a valve constant involved, which you can assume is equal to 1 if you're not given additional information.
60733
simply use a hydrometer.
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.
799401
Darcy's Formula is the general equation for flow of fluids. I was going to list it here, but the notation would be too difficult to express in this format.\n\nFind a copy of the book Flow of Fluids Through Valves, Fittings, and Pipe, Technical paper No. 410 by Crane Valves North America
381756
You need to use Boyle's Law to calculate this.  Pressure is directly related to volume.  PV(inside) = PV(outside)  If the pressure of your tank is 2500PSI and holds 50L and atmospheric pressure is 14.7PSI we can calculate the volume once all the gas is released.  Since we can not get the last 14.7PSI out of the tank due to atmospheric pressure we have to subtract that from our tank pressure.  So we have...\n(2485.3PSI)(50L) = (14.7PSI)(X)  solving for X we would have a total volume of 169.07L.  Divide this by the flow rate and you can determine how long it should last.