weight of a steel pipe measuring 4' in dia. by 22' long by 1/2" think?
452629
The weight of steel you can find through density and volume.The density /specific weight of steel C 35 --0.0784 N/cc.\n\nCalculate the volume of the steel component and then multiply it with the density given above. Remember to take care of unit you are following.
278523
you need to figure out the inside and outside diameter of each pipe example 1 7/8 inside dia, will fit over 1 7/8 outside dia, hope this helps good luck
898274
Yes, you can do it, but it will require iterations.  \n\n1) Convert all units to standard engineering units (feet, pounds, seconds).\n\n2) Set up the mechanical energy equation:\n(p2 - p1)/gamma = z2 - z1 + (f*L*v^2)/(2*D*g)\nwhere p2 -p1 is the pressure across the pipe (20 psi, but convert to lb/ft^2), z2 - z1 is the rise in the pipe (if any), gamma is the specific weight of water (62.4 lb/ft^3), L is the length of the pipe (50 ft), D is the I.D. of the pipe (ft), and g is the acceleration of gravity.  \n\n3) Now you have a problem: v is the velocity of the water, and f is the friction factor, which depends on v.  So, you guess a value for v, and calculate f from it.  You look up f on a chart (it's called a "Moody chart").  But you'll need the Reynolds number and relative roughness of the pipe surface to do that.  Then you try another v, find f, etc.  Or, you can try this:\n\nhttp://www.fluidmech.net/jscalc/vmd03.htm\n\n4) Just keep plugging in different velocities until the head loss (right side of the equation) equals the left side.  Then convert to volumetric flow rate.\n\nHey, I get 14.2 ft/s, 0.0774 ft^3/s = 34.7 gpm
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!
1078399
i'm not very sure if this is what you're looking for, but for a closed system with constant temperature, pV=p'V'. since volume, V = pi x (diameter/2)^2 x length, with constant length of the pipe, we have V = k(diameter)^2, where k is constant. hence, pV=p'V' is rewritten as p(diameter)^2 = p'(diameter')^2, i.e. p'/p =(80/250)^2. the new pressure is 0.1024 the original pressure.
662092
Let x be a steel bar and y be the box.\n\n25x + y = 79\n16x + y = 61\n\n9 steel bars was taken away so\n\n9x = 79 - 61\n9x = 18\nx = 2\n\nNow plug 2 for x in one of the above equations\n\n16(2) + y = 61\n\nSolve for y\n\ny = 61-32\ny = 29.\n\nBox weighs 29 pounds.
100213
Diameter will be 1 units\n\nCircumference =piD\nif Circumference =pi\nthen D=1
9559
You're talking about weight of steel tubing in lbs. per foot, which is\n10.6802   X  Wall Thickness X  ( Diameter - Wall Thickness ),\nand for Round Tubing it is 10.6802 x T x (OD - T), where OD is Outside Diameter. The websites I listed below just offer the formula, with no explanation for the constant, but I'll try to give it some explanation.\n\nThe constant must be related to the properties of the material in question, steel in our case, which is made of carbon and iron. The concentration of these depends on the application and it causes the weight to vary, consequently, the constant will vary also.\n\nThe 10.6802 must be the constant of weight (if we have to put a name on it) for commercial piping steel, and its units should be lbs/ft, so that you will get pounds in your result. I’m not a mechanics or materials engineer so I’m just putting the basics of material’s science to work here.\n\nI recommend you look in The Science & Engineering of Materials, by Donald R. Askeland, Pradeep P. Phulé, or any other materials book.
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
732742
awnser :length =18feet\n\np=2L+2W\n60=2L+24\n-24     -24\n36=2L\ndivide by 2\n18=L
910691
You need to work out the cross sections - which is bigger\n\ncross section is the area of the circle\n\n= pi times the radius squared\n\nwhat is bigger 2 pipes\n\n2 x 3.142 x 0.001^2\n\nor\n\n1 pipe\n\n3.142 x 0.002^2\n\nI get \n2 pipes = 6.284E-06\n1 pipes = 1.257E-05 therefore approx twice as much
114607
Is 4ft the radius or the diameter? (diameter is twice the radius).  Use the formula 2 x 22/7 x r  OR 22/7 x d (both of which are the same since 2r=d)
709877
1 inch = 25.4 millimeters.\n\nHere's a calculator\nhttp://education.yahoo.com/reference/weights_and_measures/index?p=length
781568
yes it can be done. The 203 x 133 are I hope in millimetres.This would be 8 inch x 5 1/4 inch if you measure the beam it should be somewhere close.The weight of 30kg would depend upon the thickness of the metal in the different parts of the beam.A steel stockist could give you the thickness of the different weights per metre so you could compare the millimetre thickness given to the inch measure. To convert millimetre to inch  divide the number of millimetres by 25.4 if you know any engineers they should be able to assist.Not being at work I do not have the information to hand.If your local library has "KEMPS"  year book in its reference section that could help.Kellys directory would give you steel stockists, if you cannot find one locally.The info is probably on the internet somewhere.Can investigate further for you if you cannot resolve.
923963
Traffic timing for this much weight is also considerable. A\n   building code shows that you might need 15 or more for\n   this area so titanium is probably out.  The core value of\n   bronzed steel is ok for the average busy building.  Size again\n   must be measurables ranging from 2.7 to 3.7 made in 20 foot\n   can be made in 35 precise.
820893
that depends on what kind of steel.  Not every steel is made the same way or with the same density so it varies.  I can tell you that a W14X56 is a 14" wide flange beam and weighs 56 lb/ft
1106941
Scale isn't a measurement, it's a ratio.  So if your model at 1/32 scale is 2 inches long, the original item is 2*32 or 64 inches long, which is 5 feet 4 inches.\n\nDo the same math for 1/64.\n\nThis comes in handy on maps also, where the scale can be 1 inch to 1 mile, 100 miles, or more.  You can measure on the map how may inches and then just multiply.