pressure balance in offshore wells?
370097
depends on the hydrosataic pressure of the aquifer relative to the elivation or the water bearing strata
527053
well the oil is under pressure, so when we drill all we are doing is releasing the pressure. Then when the pressure is gone there is still some oil left. sometimes to get out the oil we have to pump water down to force the oil up.\nSo it is not like you are left with any empty space if that is what you are thinking.\n\nthere are no consequences that I am aware of.
943088
Depends on the head type...unless you need to be extremely exact assuming 1/2 the vessel diameter is pretty good(i.e. a semi-elliptical or 2:1 head).\nKeep in mind though that pressure vessel heads have some straight-side to them when they are manufactured and the way I understand it is this amount is determined by the equipment manufacturer.
278765
Typical pressures for brackish water range from 225 to 375 lbf/in² (1.6 to 2.6 MPa). In the case of seawater, they range from 800 to 1,180 lbf/in² (6 to 8 MPa).
165642
power and lubrication
1097483
P/T=P/T\n1/(20+273)=P/(600+273)\nP=30 atm\n\nThe new pressure would be 30 atms, but this is considering that the volume remains constant.
814078
Pressure and volume are inverses of each other and therefore you can't in the first place say they are at constant pressure unless they were in the same container (When they do mix).. so when they are separated, you can't have the same pressure at different volumes. Though your experiment shows the achievement of equilibrium of high to low pressure, as the bigger one gets bigger (right?)...\n\nAnyways, finding an inexpensive experiment (on the web) is really difficult.  Dalton's law can be used as you probably know by collecting water over gases (which requires some lab equipment and some chemicals not readily available so to say) and also has an application in scuba diving.  I'll ask someone for you that I know... Can you give me 'till Monday or Tuesday to get a good answer for this or when does your presentation have to be done?
425733
***Important Note***\nYou should consult a petroleum engineer who is familiar with the specifics of your work if you require this information for actual field work.  The sources below are fine if you are looking for general information not to be used in the field.\n\nYou need to be more specific about the information you need.  There are several 2 7/8" drill pipes available and the weight varies.  If you need to know the weight of drill pipe that has become stuck while drilling, you have to take buoyancy into account.  Try the first link for the weight and the second link for the discussion about buoyancy.
92719
Let's consider a deep-sea diver. The pressure pushing on the diver is less at shallow depths than at greater depths. This is because at shallow depths, there is a shorter column on water above the diver than there would be at greater depths. But the pressure the diver feels is not just felt from above! The force is transmitted through the fluid and pushes the diver from all directions, even from below. \n\nPressure within the Earth can be thought of the same way that we think of water pressure on a diver. Under the surface of the ocean, you are under pressure from all directions because of the weight of the column of water above you. Within the Earth, an object is under pressure from all directions due to the weight of the column of rock above the object. This pressure is often called confining pressure, because it is the push from all directions that tends to confine material to the smallest possible volume (space).
60954
It depends on usage. You know that there are two main things in pump, Head and Discharge, and we have a rating curve for these two parameter. If you want to send water to higher elevation, you can send less water (Decrease Discharge) and vise-versa.\n\nin normal case and normal usage, we have some pumps around 2 lit/sec for a very low high to around 2000 lit/sec. Pumps that are used in wells is around 30 to 50 lit/sec.\nBut I advice you to chose apropriate pump by rating curve.\nOne of the largest company in pump industries is KSB. You can find more information in their sites. Below:
110015
At least 3000 psi.  Recommend at least 180 degree water also.
561732
water displacement and boyles law of bouyancy
1167987
You should read the following papers published by Gus Archie who worked for Shell Oil Company. Archie was credited with coining the term petrophysics. \n\nHis papers that describe the empirical relationship between resistivity and porosity:\n\nThis is the key paper by Archie:\n\nArchie, G.E., 1942, The electrical resistivity log as an aid in determining some reservoir characteristics, Journal of Petroleum Technology, V 5, pp 54-62    (you may find this as Transactions of American Institution of Mining, Metallugical, and Petroleum Engineers or just as Trans. AIME. )\n\nOthers:\nArchie, G.E., 1947, Electrical Resistivity an aid in Core-Analysis Interpretation, Bulletin of the American Association of Petroleum Geologists, Vol 31, No 2, pp 350-366\n********************************************\nHis paper that describes the relationships between pore distribution and porosity/permeability, and fluid distribution:\n\nG.E. Archie, 1950, Introduction to Petrophysics of Reservoir Rocks, Bulletin of the American Association of Petroleum Geologists, Vol 34, No 5, pp 943-961\n\n***********************************\nArchie's paper for carbonates was the first attempt to relate rock fabrics to petrophysical properties in carbonates, which is of limited usefulness today. If you are specifically interested in carbonates I would recommend reading the book "Carbonate Reservoir Characterization" by F. Jerry Lucia, 1999, that describes the Lucia transform methods for carbonates, or read some of Lucia's other publications.\n\nG.E.Archie, 1952, Classification of Carbonate Reservoir Rocks and Petrophysical Considerations, Bulletin of the American Association of Petroleum Geologists, Vol 36, No 2, pp 278-298\n**********************************************\n\nArchie's work can be summarized in the following relationship:\n\nR sample ~ Ro = F Rw\n\nR is resistivity, F is formation factor\n\nArchie eventually admited that a relationship between formation factor and permeability did not exist, but one did seem to exist for porosity. It can be expressed as a power law of the form:\n F ~ 1/phi^m \n\n(m is the cementation exponent)\n\nUnder conditions of constant saturation, the lower the porosity, the higher the apparent resistivity will be. \n\nHere is a list of key petrophysical papers:\nhttp://www.petrophysics.net/papers.htm\n\nThe AAPG Bulletin papers can be downloaded from \nhttp://www.aapg.org/ if you are a member. If not, most university libraries carry this publication. A major university library should have the Journal of Petroleum Technology paper as well.
1140416
you need to employ th ebournoulli equation and the fanninf (or darcy ) friction factor equation ... \n\nThe Darcy-Weisbach Equation for pressure loss due to friction (in the SI metric system) is:\n\nΔP = 4f(L/D)(ρ v2)/(2 gc)\n\nwhere: \nΔP = pressure loss in Pascals \nf = Fanning friction factor (dimensionless) from Moody's diagram \nL = pipe length in meters \nD = pipe diameter in meters \nρ = fluid density in kilograms per cubic meter \nv = fluid velocity in meters per second \ngc = gravitational conversion constant = 1 (kg-m)/(N-s2) in the SI metric system\n\nAnd for head loss due to friction, the equation is:\n\nΔh = f(L/D)(v2)/(2 g)\n\nwhere: \nΔh = head loss in meters \nf = friction factor (dimensionless) \nL = pipe length in meters \nD = pipe diameter in meters \nv = fluid velocity in meters per second \ng = local gravitational acceleration of 9.807 m/s2
963360
A possible reason this is incorrect is when you plugged your numbers into PV=nRT, you only had 1 mmHg while the problem reads 17.5 mmHg (2333.14 Pa). Also, you are to subtract this from the barometric pressure to get your actual pressure, not multiply (according to the link mei gave). I don't remember doing problems exactly like this, but it appears otherwise you have set it up correctly. The only other thing I can suggest is to double check your units on R. You didn't list them.
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.
967963
Either air pressure, or fluid pressure on a fluid present in the cylinder.