formula for measuring water in tanks?
611169
Base area x height of water.
350090
Use the formula PV=rnT\n\np= pressure in atm\nV= volume in L or dm^3\nr=number of moles\nn= constant: 0.0821 atm x L/ moles x K (This constant is used to balance out units)\nt= temperature in K\n\nSo for your problem:\n2.3 atm x 120 L = r x (0.0821 atm x L/ moles x K) x 340 K\n\nthe constant should allow you to be left with moles as your unit\nusing algebra, r is left, which also stands for the number of moles (meaning the problem coordinates with your question)\n\nusing this setup, i got 9.89 moles
765689
Yes you are right to subtract the water pressure from the total pressure.\n\nThen use p1v1/t1 = p2v2/t2\n\nSTP is p2v2/t2\n\nSTP = 1 atm and 0C or 273K\n\nSolve for V2.\n\nKeep units the same on both sides!
593980
1 gallon is 3.78 liters.\nin +4 C, 1 g water = 1 ml water\n3780g water we have.\n3780/18= 210 moles\nUsing avagadro number you can calculate\n1264,2*10^23 water molecules
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.
470447
You need to know the specific gravity of the weight.  (Specific gravity is weight per unit of volume, typically expressed in grams per milliliter).  The specific gravity of water is generally accepted as 1 (meaning that it weighs 1 gram per milliliter).  The equation to determine volume of water displaced would be:\n\n[10g/(S.G. of the weight)] X 1 = Vol. of water displaced\n\nNotes:\nS.G. is short for Specific Gravity\n\nThe 1 in the equation is the S.G. of water.\n\nIn actuality, the S.G. of water varies with temperature, so you may need to adjust the S.G. of water to the correct value if the problem gives you that information.\n\nThe answer will be volume of water in grams.
220777
use the following formula .. \n\ntotatal surface area = \n[(H/D)*4(pie)(0.5D*0.5D)] + [ (H-t)/(D-2t)*4(pie)*(0.5D-t)(0.5D-t)] + [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]\n\nthis is considering D = tank diametere is from outer surface and not the cavity diameter \n\n\nouter surface area of cover = (H/D)*4*(pie)(0.5D*0.5D)\ninner surface area of cover = [(H-t)/(D-2t)*4*(pie)*(0.5D-t)(0.5D-t)] \n\nthe base line are = [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]
217873
Measure concentration. Use a dummy (with water only) measuring that with a LabPro probe. Then compare that to a certain volume of a variable soft drink. I did this in high school, hope it works!
1162775
if u evr seen a water tank they r usually round\nhttp://www2.warnerbros.com/main/homepage/homepage.html\n\nvolume = sqr ft x ht\nsqr ft  (of a circle) = pi x half the diameter(radius) squared\n\nvol = pi x (1/2x38) ^2x 100\nvol=3.1415926535897932384626433832795 x (1/2 x 38)^2 x 100\nvol=113,411.49479459153590850142613639 cubic feet
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
552177
A bucket, a measuring cup, and a stopwatch is all you need.\n\nOpen the tap until it has achieved a steady flow, then put the bucket under it while turning on the stopwatch.\n\nFill the bucket to a desired amount or until you have a fairly long time.\n\nMeasure the water in the bucket and divide it by time. Thats the volume flow rate.\n\nTo get the velocity of the water when it exits the tap, divide the volume flow rate you have computed above with the area of the (circulare) hole.
1105154
pressure is: \n p(density)gh / gc =y(specific weight)h\nflow of liquid flowing inside the pipeline is:\n use the bernoulli equation:\n [p1/p(density)g  +  v1^2/2g  +  z1]  =  [p2/p(density)g  +  v2^2/2g  +  z2  +  hf]\n\nnote: p(density) is rho and y(specific weight) ir gamma.
804606
For cubic measure:  L x W x H = CI / 231 = G\nL = Length\nW = Width\nH = Height\nCI = Cubic Inches (Volume)\nG = Gallons\nMeasure the Length, Width, and Height in Inches.\n231 Cubic Inches to one Gallon.\nConvert Volume to Cubic Inches\nLength x Width x Height  = Cubic Inches  / 231 = Gallons\n\nFor Right cylinder measure: Pi x r'2 x H = CI / 231 = G\nPi = 3.14159\nr = radius (half of diameter) '2 = squared\nH = Height (or length)\nCI = Cubic Inches (Volume)\nConvert Volume to Cubic Inches. \nPi x r squared x H = CI / 231 = gallons
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 ^_^
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.
1071773
PV=nRT\n\nV=nRT/P\n\nR=.08206\nT=373 K\nP= 1 atm.\n\nYou will have to figure out n, the number of moles in one gallon of water.  The answer will be in liters, convert it to gallons
410052
Gallons of water. It depends on how dense the material is that you are weighing.  Gallons are a measure of volume and tons are a measure of weight.