27,68089 inch of water = 1 psi 51,715 mm of mercury = 1 psi; 742 col. Hg = 14,347 psi 742 col. Hg = 0,515 inch of water
Simple equation: ECD, ppg = [ (annular pressure, loss, psi ) ÷ (0.052 x TVD, ft) ] + (mud weight, in use, ppg)
limit is 86% of 93 psi = 86/100 x 93 psi ≈ 79.98 psi As the psi is measured to the nearest 0.1 psi, the safe limit is a measured 79.9 psi.
That's going to depend on many things you haven't mentioned, including how deep/thick you need it to be.
3113 is the symmetrical number - it reads the same backwards as it does forwards.
1419 add 3311 = 4730
LCM(3311, 55) = 16555
This depends on what your target pressure is. If the height of the pipe is H in feet, the pressure at the bottom of the pipe from the column of water is 0.036127292 * H * 12 where 0.036127292 is the density of water in lbs/in^3. Thus, if your initial water pressure is I, the pressure at H feet will be I - 0.036127292 * H * 12 So you will have Height | Pressure +--------- 0 ft | 40 psi 10 ft | 35 psi 20 ft | 31 psi 30 ft | 27 psi 40 ft | 23 psi 50 ft | 18 psi 60 ft | 14 psi 70 ft | 10 psi 80 ft | 5 psi 90 ft | 1 psi 100 ft | 0 psi One will be required above 90 ft. You'll probably want one above 40 ft. Derivations: eq means equals. This wiki eats equals signs. D eq diameter of the pipe in inches, H eqheight in feet D / 2 eq R Area (A)(in^2) eq pi * R ^ 2 Volume (V)(in^3) eq A * H * 12 eq pi * R ^ 2 * H * 12 Water weight (W)(lbs) eq V * 0.036127292 eq pi * R ^ 2 * H * 12 * 0.036127292 Pressure (P)(psi) eq W / A eq V * 0.036127292 eq pi * R ^ 2 * H * 12 * 0.036127292 / pi * R ^ 2 eq H * 12 * 0.036127292
Um... there are 1 psi (lb/in2) = 144 psf (lbf/ft2) , if by (lb ft) you meant pounnds per square foot. then there is 1/144 psi in 1 (lbf/ft2)
The pressure is only dependent on the height of water above the measuring point. 1 psi corresponds to a height of 2.3 ft, so 8 ft = 3.5 psi The pressure is only dependent on the height of water above the measuring point. 1 psi corresponds to a height of 2.3 ft, so 8 ft = 3.5 psi
About 415psi at 900ft
The Solution involves the use of two basic pressure formulas, namely V= square root of 2hg where: V= discharge velocity of nozzle or muzzle (ft/sec) g= acceleration due to gravity (32 ft/sec/sec) h= velocity head (ft) An P = 0.433 h where: P- pressure head (psi) h=head(ft) Take 75 psi P= 75 psi = 0.433h h=173.2 ft V= square root of 2(32)(173.2)=105.3 ft/sec
super
10 feet x 0.433 psi/ft = 4.33 psi at the base of the cylinder.
20 ft of head can be easily converted to psi by using the specific weight of water: 62.4 lb/ft3 and converting the units to inches. 20ft * 62.4lb/ft3 / 144in2/ft2 = 8.67 psi
P = ( d H2O ) ( h H2O ) ( g / gc )P = ( 62.36 lbm/ft^3 ) ( 60 in ) ( 1 ft/ 12 in ) ( 32.17 ft/s^2 / 32.17 lbm - ft / lbf - s^2 )P = 311.8 psf
'Hydrostatic Pressure' is the Term used for 'the force exerted by a body of fluid at rest. The pressure increases with increase in depth.There are two ways to Calculate water (clean water) pressure at any depth (both yields almost same results):1. The Hydrostatic pressure of water is 0.433 Psi/ft (Pounds per square inch Per feet). So at 5000 feet, the pressure is: 0.433 Psi/ft. * 5000 ft = 2165 Psianother way to go about it is:2. Water pressure increases at 14.7 psi every 34 feet depth. Thus Pressure at 5000 ft will be: (5000 ft / 34 ft) * 14.7 psi = 2162 Psi.