Formula (in metric) for calulating volume of water in a drum?
963728
Area of base of drum (1/2 of diameter of drum in metric units squared * pi) multiplied by height of water in drum in metric units = volume of water in drum in cubic metric units.\n\nSo if the drum were 100cm wide, and the water height was 200cm, the volume would be:\n\n(100/2)^2 * pi * 200 = 2500 * pi * 200 = 500000 * pi = 1,570,796 cm^3 (cubic centimeters or cc)
984991
Density = mass / volume\nTo find the volume of an irregular object submerge it in water the amount the water rises is the volume of the object.
162851
Put the stones in a container partly filled with water. Mark the before and after levels. Then measure the volume of the amount of water it takes to fill up from the low line to the high mark. This
168920
This is ultrasonic, not electromagnetic, so C is irrelevant.  The equation is:\nF=V/L where V=sound velocity in water (~1475 m/s), F=frequency in Hz, L=wavelength (0.0003 m)\nResult is you need 4.9 MHz or higher.
185076
Get the dimensions of the rectangular tank into the same units, then get the volume. the unit could be in^3, cm^3, m^3, ft^3, mi^3, mm^3, km^3, etc. Have them be converted to cm^3 using the ff. conversion units:\n\n1 mi = 5280 ft\n1 ft = 12 in\n1 in = 2.54 cm\n1 m = 10 dm = 100 cm = 1000mm = .1 dam = .01 hm = .001 km\n\nNow, 1 cm^3 = 1 mL. Have it converted into liters using 1000 mL = 1 L. To get gallons, use the ff. conversion unit:\n3.8 L = 1 gal\n\nHope this helps ^_^
22880
There's a formula for volumes of gasses:\n\nPV = nRT.\n\nIf I remember right, P = pressure, V = volume, n = number of moles, T = temperature, and R = molar density, or something molar that is specific to the gas.
562654
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
486266
the formula for volume of a cylinder is V = h * pi * r^2.\nAlternatively, the surface area is A = h * pi * r * 2.\n\nSo, if you have either the surface area or the volume, you can calculate for the height.\n\nH = Volume / pi * r^2\nH = Area / pi * r * 2
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]
525014
Let's make this very simple.\n\nTo measure the volume of a cylinder, you find the diameter of the interior portion of the cylinder.  Take that number and divide by two...that's your radius (r).  Now measure the height of the interior of your cylinder...that's your height (h).\n\nNow, take the radius (r) and the height (h) and plug them into the formula below:\n\nV = π*(r)*(r)*(h)\n\n(π equals roughly 3.1516)\n\nNow if your answer is in cubic inches, multiply V by .0036 for British gallons or .00433 for US gallons (yes, they are different).  If V is in cubic centimeters, multiply by .00022 (British) or .00026 (US).\n\nI don't know how to make it any clearer...good luck.
1051954
All you need to do is know the density of water- not too difficult to find! THe psi necessary to support a column then must match the weight of that column. So, if a column is 1inch in area and Hinches high, if you know D the density of water per cubic inch in pounds, the height of the column in inches times this value is the psi needed.\n\nPressure psi=H*D\n\nThis also works for metric, just use cms and kgs as the base sizes.
119253
several ways to measure volume.\nfor a liquid you can use a graduated cylinder and measure the volume in milliliters, liters or some similar unit.\nfor a reqular shaped solid you can calculate the volume by measuring the dimensions.\nfor a strange shaped solid you can submerge it in water and measure the difference before and after. Archimedes figured that one out about 2000 yrs ago.
918429
use the formulas your given\nto find the volume of the cylinder\nthen pretend the semi hemi on  the top is a sphere find the volume and multiply by 1/2\n\nadd two totals to find volume
418026
Volume of Cylinder is pi(r-squared)times height.\nDiameter is 10 inches, so raius is 5 inches\n\npi(25)(12)=942 cubic inches\n\nUsing the volume conversion in the link below, we find this to be roughly 4.08 gallons
700759
Let \nQ = volume flow rate in Litre/sec, \nD = diameter of the nozzle in cm and \nV = velocity of water in metre/sec. \n\nThen,  V = (40*Q)/ [pi*(D^2)]\n\nwhere pi = 3.14
502554
Most of the world uses the Metric system, so in order to have a container with a general amount of material, a 55 gallon drum was set because it equals roughly a 200 liter drum (208 liters to be exact because a gallon is 3.785 liters). Big enough to hold plenty of material yet movable by one person.
580754
i haven't done this in a while.\n\nThe volume of the reservoir is equal to (1/2)*(height)*("base")*length.  the reservoir is a rectangular prism with the "base" being the distance across the surface of the water.  We are given the shape of the full volume so we can scale down to the level of the water to find what the volume in the reservoir at its current level.\n\nif you think about the cross-section of the volume, and use half of that forming a right triangle, we can determine the ratio of the surface (the "base") to the height (water depth).  In this case, it is (40/2)/(20*sqrt(3)) or 1/sqrt(3).  so ("base"/2)/(height) = 1/sqrt(3)  and then "base" = (2*height)/sqrt(3).\n\nsubstitute that into the volume equation so that\nvolume = [(height^2)*length]/sqrt(3) or literally,\nV = [(h^2)*L]/sqrt(3).  differentiate with respect to time,\n(dV/dt) = [2*h*(dh/dt)*L]/sqrt(3) and substitute the numbers in:\ndV/dt = 200 - 50 = 150\nh = 10*sqrt(3)\nL = 450\nand solve for dh/dt which is the rate of change of the water level in yds/hour.  these are odd units, you might want to change to inches/hour.\n\ni did it this way and got 1/60 yds/hr or 0.6 inches/hr.\n\nbut again it has been a long time....