Value to determine pressure

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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?
 
according to my table: [itex] 1 \psi = 6.89 \cdot 10^{3} Pa = 1 lbf \cdot in^{-2} [/itex]

Note the pressure is always a force on a surface and has units in the order of [itex] N\cdot m^{-2}[/itex]
 
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Andrew Mason

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Idea04 said:
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
 
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Well...yeah, the volume of a cylinder isn't LxWxH for starters. ;)
 

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