# Water Pressure on a Door

*Most recent answer: 10/22/2007*

Q:

I am designing a flood defence system for external doors of properties and need to know what water pressure will be exurted on the flood defence system infront of the door. The size of the area is 100cms wide x 140cms high.
Thanks

- MARK HUGHES (age 22)

University Of Wales Institute Cardiff, UK

- MARK HUGHES (age 22)

University Of Wales Institute Cardiff, UK

A:

No problem, Mark!

The pressure in a fluid such as water depends on the depth as long as the water is not moving. Your flood defense system will need to have additional strength if you expect waves to come crashing into it. I’ll also assume for this calculation that your defense surface is vertical.

In a stationary fluid, the pressure is P=rho*g*h, where rho is the density of the fluid, g is Newton’s gravitational constant, and h is the depth from the surface. If you really want to be correct, we should also add on one atmosphere’s worth of pressure due to all the air on top of the water pressing down on it. The reason that’s not important usually is that the same air pressure pushes on the back side of the flood defense wall and the net force is just from the water.

I’ll do this in SI units. The density of water rho is 1000 kg per cubic meter. g is 9.81 m/s**2, and h is any number from zero to 1.4 meters, depending on how far down the side we are. Call the width (1 meter) W and the total height H (1.4 meters). The force on a piece of your defense door is the area of the piece times the pressure where the piece is. We’ll think of little strips of height dh, width W, and depth h. The area of each strip is W*dh, and so the force on the strip is rho*g*h*W*dh. If we add all these up, we integrate over h, and get 0.5*rho*g*H*H*W (the same as finding the area of a triangle in this case). Putting in the numbers, that’s about 9614 Newtons, or about 2160 pounds.

Tom

The pressure in a fluid such as water depends on the depth as long as the water is not moving. Your flood defense system will need to have additional strength if you expect waves to come crashing into it. I’ll also assume for this calculation that your defense surface is vertical.

In a stationary fluid, the pressure is P=rho*g*h, where rho is the density of the fluid, g is Newton’s gravitational constant, and h is the depth from the surface. If you really want to be correct, we should also add on one atmosphere’s worth of pressure due to all the air on top of the water pressing down on it. The reason that’s not important usually is that the same air pressure pushes on the back side of the flood defense wall and the net force is just from the water.

I’ll do this in SI units. The density of water rho is 1000 kg per cubic meter. g is 9.81 m/s**2, and h is any number from zero to 1.4 meters, depending on how far down the side we are. Call the width (1 meter) W and the total height H (1.4 meters). The force on a piece of your defense door is the area of the piece times the pressure where the piece is. We’ll think of little strips of height dh, width W, and depth h. The area of each strip is W*dh, and so the force on the strip is rho*g*h*W*dh. If we add all these up, we integrate over h, and get 0.5*rho*g*H*H*W (the same as finding the area of a triangle in this case). Putting in the numbers, that’s about 9614 Newtons, or about 2160 pounds.

Tom

*(published on 10/22/2007)*