how pressure and bending interacted in dented pipe?
490087
Flow in a straight pipe will have a certain pressure drop depending upon the pipe material and flow etc. When the flow in a pipe goes through a bend there will be a drop in the static pressure. Dented pipe is likewise a flow disturbance and will result in some static pressure drop.
302337
You can bend the flow of water from the hose by swinging it on the air.\nIn fact you cannot bend its properties called H2O. \nYou cannot bend it either using your hand or vice grip.\n\nWater is not compressible either.\n\nHydrokinesis on only a fiction it could not be true.
1117570
If I remember right you can try dropping it into boiling water (be careful) .. it will heat up the air inside, expanding it and possibly fixing the dent ... don't leave it in long and did I mention BE CAREFUL .. if you are not an adult please don't try it until you get some help !!
49028
Yes the discharge coefficent must change as with the increase in the diameter the viscous forceacting in the center of the pipe will be reduced resulting in a higher discharge and a higher discarge coefficent
794147
You might work up some Q&A while showing this old demo of the effect of atm pressure...thereby adding some visual interest to the subject.\n\nAdd a few oz of water to a square-sided, metal, one-gallon can having a tightly fitted cap...bring to a vigorous boil (cap now off, please :-))..) and continue boiling until all air is expelled...screw cap in place and allow to cool...they will see the can collapse.\n\nWork up questions about the vacuum thus created (by the condensation of the water vapor) allowing atm pressure to exert its force.
984213
http://en.wikipedia.org/wiki/Pressure_gradient_force\nhttp://en.wikipedia.org/wiki/Temperature_gradient\nHope this helps, good luck!
384170
These won't affect the integrety of the pipe. They are from the machine which extruded the metal into the shape of a pipe.\n\nThe rating characteristics: PSI, material make up, corrosion protection added, and ID are all met.\n\nA Pipe is made to a standard which must be assured by the manufacturer for the specific application intended. Testing on a periodic basis with a certificate of compliance is usually issued.
302322
I won't do your homework for you. But I will tell you this:\n\n\n1a&b. These should be common sense. Go outside and turn on the garden hose and tell your teacher what you've observed.\n2a. Look up BERNOULLI'S LAW as it pertains to fluid pressure through a tube. There is a formula that related pressure to hose diameter and volume.\nb. The answer is: high pressure is bad for vessels weakened by atherosclerosis.
825980
Elastic materials spring back to their original shape when they are stressed. Inelastic materials simply bend and stay bent, which is called plastic deformation.
83027
Here's a link that might help you:  http://www.mebs.co.uk/deflection_fields.htm
170252
Imagine that you have a rod say 10 feet long and one square inch in cross-section suspended from a tall ceiling with a loading hook on the bottom.\n\nIf you place say a 100 pound weight on the hook you load the entire length of rod.  At every cross-section of the rod (a horizontal cut) the "stress" will be 100 PSI (pounds per square inch).\n\nThe load is taken up by distorting the molecules within the rod.  If the rod is rubber it might stretch say one foot total.  This is the strain.  The strain would be 1 ft per 10 ft or 1/10 or 0.1 where the units cancel out.\n\nIf the rubber rod continues to stretch with time, there is a time componment which is called creep.  There is a feedback component to stress/load/strain in that more load increases stress and more strain or molecular distortion occurs to take up the added load.\n\nUsually the stress/strain/load properties are applied to metals such as a steel rod and depend on distortion (strain) of entire metal crystals within the rod.  The strain is almost impossible to see and special strain gages must be used.  Also steel can support very heavy loads and exhibit creep only near their load limits.  If the load limit is exceeded, permanent deformation will occur or even necking (reduction of diameter) and ultimately rupture.\n\nThe stress/strain characteristics of metals are determined on special tensile test machines and the data are used to design configurations far more complex than a rod.
940353
The engineering graph is plotted using force vs. elongation.  Stress is then calculated using force divided by cross sectional area, which is a constant.  This cannot simulate true stress because it doesn't account for "necking."\n\nNecking occurs when the sample is about to break.  Usually you can see with your own eyes that the sample begins to thin out before it breaks, thus reducing the cross sectional area (which was held constant in engineering graph).  When necking starts to happen, even the force applied may be smaller, the stress does not decrease, since Stress = Force / Area.
281995
Is it elasticity? Try to solve the problem with some formula below:\nYoung's Modulus(Y)=Stress/Strain\n=(F/A)/(dl/L) \nwhere, F=force applied, \nA=cross sectional area, \ndl=leangth increase and L=total lenght.\nBulk Modulus(B)\n=dP/(dV/V) \nwhere P=pressure and V=volume.
3469
Ground yourself by touching a metal pipe.
570511
both.  the friction, as well as creating its own heat, slows the airflow, which aggrivates the compression along the leading edges.\n\nI'm pretty sure.  don't put this on a physics test and then blame me.
967963
Either air pressure, or fluid pressure on a fluid present in the cylinder.
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.