procedure for designing a underground rcc water tank
132818
It's basically a swimming pool with a slab over, though you have the opportunity of using stub walls or columns to reduce the slab spans. If the tank is small the sizes will be determined by practical rather than structural considerations. I'm hoping to build such a tank later this year as part of extentions to my house.
367333
to get the volume, you would need to use the formul V=dwl (d-depth, w-width, l-length).\n\nbefore you can plug in the numbers, you have to make sure your numbers' units are the same. KHDBdcm.\n\n6km=6000m, 2km=2000m, and .4km=400m\n\nnow, 1 cubic meter equals 1000 L, so you want to do this in m.\n\nV=(6000)(2000)(400)\nV=4,800,000,000 m^3\n  =(4.8)10^12L\n\n--\n\nnow youve got 6,000,000 people who use 420 L of water each day.\n\ndivide how much water youve got by how much water each person uses. \n\n=1.142857 x 10^10 \n\nthats how many people you can serve.\n\n\nthen divide that by 6million, and you find you can serve them for 1904.8 days.
536892
Put in a sprinkler system controlled from inside.
380089
this is a great site. i hope it will help. this is where i understood it. \n\nhttp://www.candu.org/candu_reactors.html
309465
Self-Contained Underwater Breathing Apparatus
216979
You need to integrate part of a circle.  Imagine this cylindrical tank centered at (0,-4) in order to accomodate the 8" depth.  First, you want to find your X values of integration.  Imagine that the liquid in the tank (which is upside down now) makes a horizontal line.  The distance from the center to the end of one of these lines is the radius, or 12".  The distance vertically is 4", or the radius minus the depth.  Therefore, you can find the horizontal distance by using right triangle properties. The answer is the square root of 128, each direction.  So you will be integrating from -(128)^(1/2) to 128^(1/2).\n\nNext, you must set up the integral.  The equation for this circle would be X^2 + (Y+4)^2 = 144.  Solving for Y gives us\n\nY = (144-X^2)^(1/2) - 4\n\nIntegrating this would give us\n\n[1/2(144-x^2) + 72arcsin(x/12)-4x] from the negative square root of 128 to the positive square root of 128.\n\nevaluating this gives us approximately 102.75 in^2\n\nMultiply this area by the length of the tank to get the volume...\n\n102.75 in^2 * 68 in = 6886.89 in^3\n\n1 in^3 = .004329 gal, so...\n\n3437.36 in^3 * .004329 gal/in^3 =30.24 gal.\n\nAs far as I can see, that seems logical.\n\nNote: I had the wrong integral formula to begin, so this has been edited.
182882
Here's some ideas\n\n(1)\nMeasure the hardness and ion content of your local water supply.  You can perform precipitation experiments, measure solid residue after evaporation, and more.\n\n(2)\nSeparate the protein solids and fat from milk.  I think you can do this by adding salt to the milk.  (I've never done it, may be too easy.)\n\n(3)\nDo a project on the danger of confined spaces.  There is a really cool visual experiment that goes along with this.  If you mix vinegar and baking soda in a tall glass you will get a glass full of carbon dioxide.  Since the carbon dioxide is heavier than air, it will stay in the glass for a while (invisibly so).  You can then make a "pouring action" and pour the CO2 into a second glass, although there is no visible flow of gas.  You also need a 3rd glass that has a small, lit candle at the bottom of the glass.  You then make the same pouring action with the second glass into the third and when the CO2 displaces the oxygen and nitrogen in the third glass the candle will get snuffed out.  You can even do this experiment at home for fun!\n\n(4)\nMeasure the pH of common household materials: foods, cleaners, etc.
185240
This answer will depend on the density of the water that is used.  According to Wikipedia the density of Water is \n1000 kg/m^3.  The equation for density is as follows:\n\nDensity=Mass/Volume\n\nMultiply both sides by volume gives us the following equation\nMass=Density*Volume.\n\nGiven: \nVolume=1 cubic foot\n\nDensity of Water=1000 kg/m^3 \nYou most likely want how many lbs of water you need; thus we need to convert the Density of water from kg/m^3 to \nlbs/ft^3\n\nThus using an online conversion:\n1 kilogram/cubic meter = 0.062 427 962 pound/cubic foot\n1000kg/m^3= 62.43 lb/ft^3\n\nThus finally\nMass=Density*Volume\nMass=62.43 * 1\nMass=62.43 lbs approximate!\n\nNow since your designing it you want to design over 25%\nso therefore design for 78.04 lbs.  Worry about the pressure on the sides more when designing this system.
663099
C- combustion chamber
526471
Take a differential height dh at height h from the vertex.  The differential volume is dV = π*r^2*dh, where r is the radius of the tank at level h.  r@27ft is 10ft, r at 0ft is 0ft, so r = (10/27)*h.  Thus\n\ndV = π*2.7^2*h^2*dh.  Divide both sides by dt to get\n\ndV/dt = π*2.7^2*h^2*dh/dt.\n\nYou are given dV/dt, that is the volumetric flow rate into the tank = 10cfm   dh/dt is the rate of depth increase:\n\ndh/dt = dV/dt / (π*2.7^2*h^2)
418507
I can not answer your question, but if you contact some distributors or manufacturers of tanks they may be able to help you.\n\nGood luck!
67007
Large storage tanks are usually cylindrical and vertical for all three reasons you mention; economics, space and technical.\n\nUsing calculus (maxima and minima) it is possible to design a tank that contains the most volume for the least tank metal.  This also applies to canned foods.  Also, cylinders are very strong and can not bulge compared to the flat sides of a box-shaped tank.  Also, the pressure in the tank is greatest at the bottom of the tank and the cylindrical plates there can be thicker and stronger than at the top saving metal and tank weight.  Of course, the bottom plate of the tank could bulge except it rests on a concrete floor.\n\nAlthough a cylindrical tank has a smaller "foot print" than a square tank, it can rise to almost any height required to provide the needed volume, taking good advantage of the 3D space available.\n\nOn the technical side, a cylindrical tank is stronger when empty than a box-shaped tank and easier to transport to the needed location without damage.  It is easier to design good welds for a tank that will not constantly flex in operation as the tank is filled and drained and cylindrical tanks flex less than box-shaped tanks.  For onsite construction, it is probably easier to weld curved metal plates into cylinders than to make a huge flat plate from a patchworks of smaller plates.  Cylendrical tanks are less likely to buckle than box-shaped tanks due to ground tremmors.
157507
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.
87555
1 liter is a cubic decimeter. There are 1000 dm3 per m3, 264.17205 gallons per m3 and 3.7854118 liters per gallon.\n\nVolume can be calculated by the equation:\narea*height. So if you have a nice symmetrical tank just try to calculate its bottom (or top) area and then multiply with the height.\n\nArea is calculated with either:\nside*lenght of the tank for squares or for circles:\nradius^2*pi\n\nIf your tank is neither cylindrical nor cubical you may have your work cut out for you. Determine which shape it resembles the most.\nA sphere:\nV=(4*pi*r^3)/3\n\nA prism:\nV=B*h\nwhere h is height and B is the bottomarea of the parallell trapeze:\nB=(h*(a+b))/2\n\nFor a pyramid:\nV=(B*h)/3\n\nFor a straight circular cone:\nV=(pi*r^2*h)/3\n\nIf all else fails just fill that sucker with water (if it´s not a monstersized tank). You need to check it for leaks anyway. Weigh it empty and subtract that from the weight of the tank full. 1 liter (dm3)= 1 kg.
586323
Depends on what you mean by types.\n\nDo you mean the type of material used in constuction or the type of activity the building is intended for.\n\nMost large building use a chilled water system where they use chillers to cool water down to about 42 deg F then pump the water throught the building to air handler units which blow air across the coils filled with the water.  This heats the water, but cools the air.  As the air is circulated throughout the area the water is returned to the chiller to be recooled and the process starts all over again.\n\nHowever, an important point is capacity.  That will be determined by both what the intended use of the building is and what sort of material is used.
593165
aquifers, undergound streams, etc. anything under ground
340591
I have a drawing that shows the various types of shell and tube exchangers. I believe their functions are self evident if you see the cross sectins.\n\nSend me you email address and I'll send the drawing.