formula for duration of flow of medical oxygen cylinder?
480385
OXYGEN CYLINDER ENDURANCE CALCULATION FORMULA\nduration of flow in minutes = (gauge pressure - safe residual pressure) x constant\nflow rate in liters per minute\nsafe residual pressure: 200\nconstants: D cylinder 0.16\nE cylinder 0.28\nM cylinder 1.56\nH cylinder 3.14\nExample\nto calculate the duration flow in minutes for a full M cylinder being used at 15 LPM:\nduration of flow in minutes = (3000 - 200) x 1.56 = 291.2 minutes = 4 hours 50 minutes\n15LPM\nExample\nIn order to figure out whether an ambulance's oxygen cylinder has the minimum O2 required for a trip, use the\nformula above.\nminimum required gauge pressure = duration of trip in minutes x LPM + 200\nconstant for the size of cylinder\n· Minnedosa to Winnipeg = 2.5 hours = 150 minutes\n· cylinder in use: M cylinder constant is 1.56\n· oxygen flow rate for the patient = 15 LPM\n· minimum required gauge pressure = 150 min x 15 LPM + 200 = 1642 PSI\n1.56
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
358957
Volume = pi * r^2 * h\n\nr is 1/2 the radius of the drilled hole = (D'/2)\nh is equivalent to the diameter of the original cylinder = D
917169
L = μ(0) μ(r) N^2 A/l\n\nwhere\n \nμ(0) is the permeability of free space (4π × 10-7 henrys per metre) \nμ(r) is the relative permeability of the core (dimensionless) \nN is the number of turns. \nA is the cross sectional area of the coil in square metres. \nl is the length of the coil in metres\n\nHere we assume that μ(r) = 1, since the solenoid doesn't enclose anything like a metal core. Then\n\nL = [μ(0) N^2 (πR^2)] / l, where R is the radius of the solenoid\n\n= [(4π × 10-^7)(2920)^2(π(0.005)^2)] / (0.1)\n\n= 8.44 × 10-^3 henries
602334
PV = nRT\n\nV = 3 L\nn = g CO2/MW CO2 + g O2/MW O2\nT = 273 K\nR = 0.082057 L-atm / mol-K\n\nsolve for P
787117
If the cross-secional area of the aorta is 2.54 cm^2, and the blood flows at 48.9 cm/s, then in each second, 2.54*48.9 CUBIC centimeters of blood is flowing through the aorta.  All you have to do is multply.\n\nFor the second part, you use the answer from the first part.\n\n48.9*2.54 cubic centimeters per second is flowing through a bunch of capillaries with a combined cross-sectional area of 2720 cm^2.  Therefore the speed must be 48.9*2.43/2720 cm/second.  It's always a good idea to check units: 48.9*2.43 cm^2/sec/2720 cm^2 comes out in units of cm/second.\n\nSame concept as "wide rivers run slowly"
477933
Check out the link below:\n\nhttp://www.respiratorygroup.com/products/high_pressure/med_o2_spec.aspx
1127531
Surely you know the foumla for Re?\n\nRe=(density*velocity*length)/viscosity\n\nRe=(1000*1*25*10^-3)/1014*10^-3\n\nRe=24.655 (3dp)\n\nI think you've got yourself a laminar flow.
975044
Every time you exhale, you breathe out most of the air you breathed in. While the air is in your lungs, oxygen travels from air to blood and CO2 from blood to air, in each case moving naturally to where it is less concentrated.\n\nHow long it takes the average oxygen molecule from being absorbed to being breathed out again as part of a carbon dioxide molecule is a really good question - I have no idea if anyone has even calculated it, but we can try.\n\nMost people pass out if their blood oxygen level drops below about 80% - and most people can hold their breath for maybe 4-5 minutes (the record is almost 9 minutes by Houdini, but that's someone trained and a bit freakish). That implies you use 20% of the oxygen in, say, 5 minutes - so it would take 25 minutes to use it all, if that were physically possible.\n\nBTW, I'm prettty sure it's 8-9 pints of blood. The normal O2 reading in arterial blood is around 10-13 kPa, but I'd have to admit I don't know how to convert that to a mass (or free volume) of oxygen.\n----\nOoh, here you go: arterial blood contains about 20ml of oxygen per 100cc. Your blood volume is about 5 litres, so that would mean your body contains rather less than a litre of oxygen (because not all of your blood is fully oxygenated).
589422
For a column of vibrating air with one closed end and one open end the equation is:\n\nlambda = 4L\n\nWhere L is the length of the tube and lambda is the wavelength of the standing sound wave.  Given the fact you know the frequency, and the speeed of sound, I think you can figure out the rest.
532639
I believe in a generalized fashion the equation for flow rate is:\n\nFlowrate = K*(cv)*((p)^.5)\n\nK = a variable to adjust for various fluids
994673
I am not sure of the final answer you are looking for.  If you are looking for the number of molecules transferred after a certain time t, the answer is simply n=-DAt(dc/dz).  This of course assumes that dc/dz remains relatively constant.  If not you will need a dc/dz as a function of t.
940323
Reynold's number if the ratio of intertial force to viscous forces:\n\nRe = rho * v * L / mu\n\nrho: fluid density\nv: fluid velocity\nL: characteristic length\nmu: fluid viscosity\n\nNow you are given a scenario where the model is 1/30 the size of its prototye.  That means L2=L1 / 30.  Velocity is the same and the fluid air is the same, which means mu is the same.  We know that density will be different because we have to change pressure (rho2 not equal to rho1).  Here's what the problem looks like:\n\nRe = rho1 * v * L1  / mu = rho2 *v*L2 / mu\n\nWe know that L2 = L1/30 so let's plug that in:\n\nRe = rho1 *v *L1/ mu = rho2 * v * (L1/30)/mu\n\nRearranging:\n\nrho2/rho1 = 30\n\nNow use the perfect gas law.  Let's manipulate it to get a form that is right for this problem.\n\nPV = NRT = m (R/MW) T\nP = rho * (R/MW)*T\nrho=(P/T)*(MW/R)\n\nWrite the last eqn above as ratio rho2/rho1, with constant T and MW (molec. weight)\n\nrho2/rho1 = P2/P1\n\nTherefore, P2/P1 =30.  P2 is 30 times P1.
234957
Check out this site: http://www.questconsult.com/~jrm/enthpres.html\n\nUse the component id for oxygen (28) and mole fraction (1.00), press submit form. Makes a nice Enthalpy-Pressure diagram.
970312
If you can figure out how to calculate the number of grams of O2 it takes, then you just need to us Pv = nRT to get the volume. I forget the equation for combustion though...need a combustion/engine textbook.
136796
heart rate\nblood pressure\namount of Carbon dioxide entering the lungs and the amount od oxygen.
359943
Would need to know the orifice diameter that is flowing air from this 150 psi source.........and if you are asking cu ft per minute.......\n\nThe rate of flow won't change whether there was 6 or 60 cylinders, but more overall delivered cu ft when the cascade was depleted.