Calculating Pressure: Understanding the Relationship Between Volume and Force

[Idea04]

I'm kind of confused on the value to determine pressure. I'm told that 2.31 feet equals 1 psi of pressure. But the formula F/A force divided by area brings up a different number. So I've got 6 feet hight of water with a 1 inch diamter area. So what I did is use LxWxH/1728 x 10 to get value in gallons (imperial gallons) and multiplied by 10.0200098 pounds to get the weight. Then divided by .785398163 square inches. So 4.175004083lbs / .785398163 = 5.315780301 psi. But according to the value above is 2.31 feet to 1 psi is 2.594402597 psi for 6 feet. What is wrong here?

 

[sdekivit]

according to my table: $1 \psi = 6.89 \cdot 10^{3} Pa = 1 lbf \cdot in^{-2}$

Note the pressure is always a force on a surface and has units in the order of $N\cdot m^{-2}$

 

[Andrew Mason]

I'm kind of confused on the value to determine pressure. I'm told that 2.31 feet equals 1 psi of pressure. But the formula F/A force divided by area brings up a different number. So I've got 6 feet hight of water with a 1 inch diamter area. So what I did is use LxWxH/1728 x 10 to get value in gallons (imperial gallons) and multiplied by 10.0200098 pounds to get the weight. Then divided by .785398163 square inches. So 4.175004083lbs / .785398163 = 5.315780301 psi. But according to the value above is 2.31 feet to 1 psi is 2.594402597 psi for 6 feet. What is wrong here?

Why use imperial units?

The volume of a 6 foot, 1 inch diameter column is 72 x .785 = 56.5 in.^3. = 926 cm^3 = .926 kg. = 2.04 lb. So the pressure is 2.04/.785 = 2.60 lb/in^2.

AM

 

[Hawknc]

Well...yeah, the volume of a cylinder isn't LxWxH for starters. ;)