This question is misguided. You don't need pressure, you need energy.
Energy = work done = mass of water x g (acceleration due to gravity) x height raised.
Pressure = force/area over which it is applied (e.g. the cross section of the water pipe.)
So the pressure needed is given by
P = mgh/area
Be sure your value of g is in the right units for your mass and height measurements.
Differential air pressure will be shown by raising the water level in, for example, a U shaped tube, where differentail pressure is applied between the two arms. Yes, to move the water, energy is used. BTW, energy equals force (weight) times distance, but you are not lifting all the water by one foot.
Because of the weight of the water, the water pressure increases at a rate of 2.31 psi per foot of depth.
Yes, because the density of oil is less than water. (The pressure would be less with oil.)
Depends on the size of the foot - a child's food will take far less pressure to crush than an adult man's foot.
Each foot of elevation change is equal to 0.433 PSI of water pressure. Elevating the water tank 10 feet would add 4.43 psi.
The pressure of water increases about 0.445 psi per foot of depth. If we "zero" our meter so we have "no" pressure at the surface (ignoring the normal 14.7 psi of air pressure at sea level), at 18 feet we will have 0.445 psi/ft times 18 feet, which is 8.01 psi, or right at about 8 psi.
Water pressure at a depth of about 44 feet is about 20psi
Raise your leg and your foot
Because of the weight of the water, the water pressure increases at a rate of 2.31 psi per foot of depth.
Every one feet the pressure raises by 24.9 psi.
Assuming you mean a 14 foot by 20 foot pool, you need the following amount of water to raise it 1 inch:261.818 US liquid gallons; OR218.009 imperial (UK) gallons; OR991.089 liters
0.43197 psi
5300 gallons
If it is fresh water, and the surface is at sea level, then the pressure at the surface is 14.69 psi. As you submerge, then the pressure from the weight of the water above you is added to the air pressure above the water. For each foot that you descend, the water pressure will increase by 0.4331 psi, so at 328 feet deep, the water pressure is 142.0568 psi. Add the 14.69 psi air pressure to get 156.7468 psi.
A pressure foot is used to measure air pressure inside a tank. This means that every square foot of the tank, there is a pound of pressure.
By elevating it higher as every foot higher will increase the static pressure almost a half a pound
The pressure increase is dependent on density of the water. Pure water at 60F has about 27.78 inches of water column (INWC or INWG) per psi or approximately 0.43197 psi/ft. This is from memory; but it should be close. Fresh water: 0.43 psi per foot Sea water: 0.44 psi per foot. So, for each additional 10 feet of depth, figure about 4.3 to 4.4 psi increase in pressure. You can calculate this yourself by using the fact that fresh water weighs about 62.4 pounds per cubic foot (pcf) and sea water weighs about 64 pcf. Divide those numbers by 144 (the "footprint" of one cubic foot, 12 x 12) and there you go. It's interesting to note that this pressure is independent of volume or expanse. i.e. the water pressure behind a fresh water dam at 100 feet deep is about 43.3 psi regardless of whether the dam's reservoir is 25 miles long or 10 feet long. Depth and density are the only relevant parameters needed to determine pressure. Dive pressure, however, would be the water pressure of 43.3 plus the air pressure above the water. So the net pressure on your ears & body would be 43.3 plus 14.7 (one atmosphere)totalling 58 psi,or about 4 atmosphers. That's four times our normal experience. Worthy of careful consideration. Regarding diving - internal pressure inside ones body EQUALS the atmospheric pressure (14.7 psi). [Otherwise we would be squashed by the athmospere.] Therefore, the net result is still dependent on depth ONLY - in example given it would be around 2.95 atmospheres.
mean and dirty a one foot column of water will produce 1/2 psig head