what is the diam. of a 550 gal oil tank (round) not oval?
601391
Depends on the height.\n\nV=pi*r^2*h\n\n1 ft^3 = 7.48 gallons (U.S.)
856204
1. 285000 X 7.481 = 2132085 gal\n\n2.  I'm almost positive that this one is written incorrectly.\n\n3.  Use FOIL\n5x X 5x = 25x^2\n5x X 3 = 15x\n5x X 3 = 15x\n3 X 3 = 9\n25x^2 + 30x + 9\n\n4. w = 12/5x\nw = 900/y^2
361907
If the tank is a cylinder, the Volume is πr²h.  If you get the volume in cubic feet, multiply by 7.5 to get gallons.
374994
The density of diesel fuel varies from one refined lot to the next. The density of diesel fuel ranges from 82%-95% that of water.\n\nSince the weight of water is 8 lbs/gallon, it would stand to reason that a gallon of diesel fuel would be between 6.5 lbs and 7.6 lbs.
313123
Yikes.  I have been off tanks since Oct 2000.  I cannot remember how many gallons of FRH the replentisher holds.  The replentisher resivor is located above the gun in the turret.  And you fill the transmission with plain old motor oil.
322779
Set the volume of rain from the roof (12m x 3m x .1m) equal to the volume of the tank in the form of pi x (2m)^2 x H, and solve for H.\n\nI got 0.2865m.\n\n*Edit: the person below is correct, if the dimensions given are only half the area draining into the tank, twice as much water would drain to the tank, making h = 0.573m
1136335
http://www.ioga.com/Special/crudeoil_Hist.htm\nThat should help.
1161603
40000 pounds...8 lbs. per gal.
916491
i highly recommend the company USTANKCO  see their web page at www.ustankco.com they make frac tanks 400 barrel  500 barrel frac masters
268773
My recommendation is steel, high pressure tank, and 80-100 cu ft.\n\nSteel is negatively buoyant when empty.  This allows you to wear less weight when compared to aluminum tanks.  Avoid neutrally buoyant aluminum tanks.  These are just aluminum tanks with an extra 5#s added to it.\n\nHigh pressure because you get more capacity for the size.\n\n80-100 cu ft since these are the most versatile and will most likely match your buddy.  While 80 cu ft hp tanks are great single tanks, if you plan on doubles, then 100 hp tanks are better since these will trim out better for most people.
172572
42 us gallons.\nthis is the standard used by the petoleum industry, think about it you cant put 55 gallons of oil in a 55 gallon container once it heats up and expands you have problems eh.
644639
You could measure the volume of fuel in the tank by weighing the tank accurately as it is filled with fuel. The weight of a gallon of fuel is known accurately. Home heating oil tanks utilize crude floats to measure the height of the oil level in the tank, but more sophisticated level indicators are available for industrial uses.\nSearch http://www.ThomasNet.com for that application.\nhttp://www.thomasnet.com/nsearch.html?cov=NA&which=prod&what=tank+level+sensors&navsec=search&heading=96118542
653606
44000*0.00012*56 = 26.4\n\n=44026.5Gal\n\nBe aware that actually the thermal coefficient by Volume (B) stricly speaking varies with temperature
84469
Deriving this is one of my favorite derivations.  I only will use exact legal values.\n\n500 mL = how many english fluid ounces.  We'll start with the gallon, 128 fluid ounces.\n\nLegally, there are 231 cubic inches per gallon.\n\n1 gal = 231 in^3\n\nNow take the cube root of both sides, remembering the law of exponents, (a*b)^c = a^c * b^c. A cube root is an exponent of (1/3). Dimensional analysis is the theory that you can manipulate the [gal] and the [in] like they were algebraic unknowns.\n\n1 cube_root_gallon = cube_root(231) * inches ` ` ` ` ` (1)\n\nNow a cube root of a gallon is the length of a side of a cube with volume one gallon. Lets convert this inches to centimeters. Legally\n\n` ` 2.54 cm = 1 inch\n\nmultiply this equation by 1/inch both sides to get\n\n` ` 2.54 cm/inch = inch/inch\n\nnotice that the right side is equivalent to 1\n\n` ` 2.54 cm/inch = 1\n\nNot multiply this equivalent to 1 on the right hand side of equation (1) to get\n\n1 cube_root_gallon = cube_root(231) * inches * 2.54 cm/inch\n\nand cancel the "inch" dimension on the numerator and demonenator right hand side\n\n1 cube_root_gallon = cube_root(231) * 2.54 cm\n\nnow cube both sides to get gallons again, remembering the law of exponents, that is each factor on the right is cubed as well\n\n\n1 gallon = 231 * (2.54)^3 cm^3\n\nNow working on the left hand side, 4 quarts = 1 gallon, so by a similar argument 1 = 4 quarts/gallon and multiplying this to the left hand side of the equation (since multiplying by 1 doesn't change the value)\n\n1 gallon*4 quarts/gallon = 231 * (2.54)^3 cm^3\n\ncancelling "gallon"\n\n4 quarts = 231 * (2.54)^3 cm^3\n\nand 2 pints = 1 quart, so 2 pints/quart = 1, multiplying and cancelling\n\n4 * 2 pints = 231 * (2.54)^3 cm^3\n\nand a pint is legally 16 fluid ounces, 1 pint = 16 fluid oz, thus 1 = 16 fluid oz / pint; multiplying this value of 1 on the left hand side won't change anything, then cancelling "pints"\n\n4 * 2 * 16 fluid oz = 231 * (2.54)^3 cm^3\n\nnow multiply both sides by 500/(231 * (2.54)^3)\n\n4 * 2 * 16 * 500 / (231 * 2.54^3 ) fluid oz = 500 cm^3\n\nsince 1 mL is exactly equivalent to 1 cm^3, our answer is on the left-hand-side.  via good calculator\n\n500 mL = 16.907011350921498584313594983 fluid oz.
214638
I am assume stand temperature and pressures at the surface of the earth.\n\nThe volume of the liquid is: \n1 inch x 1 acre \n= 1 in. x 43560 sq. ft.\n= 1 in. x 208.71 ft x 208.71 ft (assume square acre)\n= 1 in. x 2504.52 in x 2504.52 in\n= 6272620.4 cubic inches \n= 27154.2 Gallons (assumung US Gallon)
679893
liters gallon quart ounce\ncalories kilocalories degrees btu
1032297
1 Gallon = 8.345404 Pounds