Fluid pressure problem and units confusion

AI Thread Summary
The discussion centers on calculating fluid pressure at a depth of 10 meters in a lake, emphasizing the use of appropriate units. The correct pressure at this depth is derived using the equation P = P0 + ρgh, resulting in a pressure of 1.97 atm when using Pascals. Participants express confusion over unit conversions, noting discrepancies when applying different units like kPa and atm. There is a consensus that the equation is most straightforward when using Pascals, leading to questions about how to adapt it for other units. Clarifying these unit preferences is essential for accurate calculations in fluid pressure problems.
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Homework Statement



Exercise Find the pressure at a depth of 10 m below the
surface of a lake if the pressure at the surface is 1 atm.
(Answer: With P0 = 1atm = 101kPa, ρ=103kg/m^{3}, and g= 9.81 N/kg, we have P = P0 + ρgh = 1.97 atm. The pressure at a depth of 10 m is nearly twice that at the surface.)

Homework Equations



P = P0 + ρgh

1 atm = 101.325 kPa = 101325 Pa

ρ = 10^{3} kg/m^{3}

The Attempt at a Solution



Doing what they did:

P = 101 kPa + 10^{3}*9.81 N/kg*10 m = 98,201 kPa = 972 atm.

Hmm... Trying it with atm:

P = 1 atm + 10^{3}*9.81 N/kg*10 m = 98,101 atm.

Trying it with Pa:

P = 101,325 Pa + 10^{3}*9.81 N/kg*10 m = 199,425 Pa = 1.97 atm.

So I was only able to get the answer they did when using Pa. I think they neglected to mention that pivotal part when solving the problem. I have questions two:

1). Why does the equation have a strong preference for Pa?
2). How would I have to restructure the equation to use kPa, atm, or mmHg?
 
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1 Pa = 1 N/m.

1kPa = 103 N/m2

103*9.81 N/kg*10 m = 9.81×104 Pa = 98.1 kPa
 
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