Basic Hydrostatic Equation for manometers

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The basic hydrostatic equation can be simplified for manometers as follows
[tex]P_B - P_A= - \rho*g(z_B-z_A) [/tex]
I am having trouble figuring out what the order in which I should put the values of the pressure and elevations. Which is the higher elevation and which is the higher pressure? The order matters because the pressure can be below atmospheric. I can't seem to figure this out like I do with mechanics which pretty intuitive.
 

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  • #2
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The basic hydrostatic equation can be simplified for manometers as follows
[tex]P_B - P_A= - \rho*g(z_B-z_A) [/tex]
I am having trouble figuring out what the order in which I should put the values of the pressure and elevations. Which is the higher elevation and which is the higher pressure? The order matters because the pressure can be below atmospheric. I can't seem to figure this out like I do with mechanics which pretty intuitive.
I don't think it matters. Give us a specific example where it matters.
 
  • #3
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well if the final answer I get for pressure is negative that means that the pressure is lower than atmospheric pressure. Same if I try to find the the elevation; if z is negative that would mean that the elevation is below reference. Sadly I don't have an example to give at the moment because I haven't started any questions yet
 
  • #4
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well if the final answer I get for pressure is negative that means that the pressure is lower than atmospheric pressure. Same if I try to find the the elevation; if z is negative that would mean that the elevation is below reference. Sadly I don't have an example to give at the moment because I haven't started any questions yet
If the pressure is negative, that just means that the elevation is above the surface. But the equations doesn't apply above the surface. I recommend that you invent some sample problems that you think will cause you conceptual problems, and then actually run the calculations. I will be glad to help with explanations, if necessary.

Chet
 
  • #5
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Here is a problem I took out of the book.
A is at 1.5 kPa, density of gasoline is 680 kg/m3 and glycerine is 1260 kg/m3.
I am solving this as follows for gasoline. z here is the height of B
1.5 kPa -0 kPa = 9.81 m/s2 * 680 kg/m3 (1.5m-z)
My original question is
on the left hand side should it be
1.5 kPa -0 kPa or 0 kPa - 1.5 kPa
and on the right hand side should it be
1.5m - z or z- 1.5m
Also I am doing something wrong here can you point out the mistake
 

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  • #6
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Here's how I would do it.

Pressure at bottom of tank (z = 0) = 1.5 kPa + 9.8 (680)(1.5) + 9.8(1270)(1)
Pressure at bottom working down from arm B = 0 + 9.8(680)(z-1)+9.8(1270)(1)

So,

1.5 kPa + 9.8 (680)(1.5)=0 + 9.8(680)(z-1)

So,

1.5 kPa - 0 = -9.8(680)(2.5-z)

So this is in perfect agreement with your general equation. But, I personally would never do the problem using your general equation. I would do it the way I did above.

Do you next want to try the C arm?
 
  • #7
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I get what you are saying but at the same time I have trouble explaining it. Anyways the answer to arm C would be
1.5 kPa + 9.8 (680)(1.5) + 9.8(1260)(1) = 9.8(1260)(1) + 9.8(1260)(z-1)
what I have trouble understanding and explaining is the term 9.8(1260)(z-1). What is the physical meaning behind this term?
 
  • #8
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I get what you are saying but at the same time I have trouble explaining it. Anyways the answer to arm C would be
1.5 kPa + 9.8 (680)(1.5) + 9.8(1260)(1) = 9.8(1260)(1) + 9.8(1260)(z-1)
what I have trouble understanding and explaining is the term 9.8(1260)(z-1). What is the physical meaning behind this term?
What a great question!!! The term 9.8(1260)(z-1) is the pressure difference between the free surface and the level z = 1 in tube C, and the sum of the terms 9.8(1260)(1) + 9.8(1260)(z-1) =9.8(1260)z is the pressure difference between the free surface in tube C and the baseline z = 0. I'm guessing now that you might be saying to yourself the following:

"How can we get the pressure drop in the tube C without even knowing its diameter or its curved shape. From a free body diagram, shouldn't the pressure difference in tube C just be the weight of the fluid in tube C divided by the area of the tube?"

The answer is no. There is an additional force in the free body diagram. This is the force exerted by the tube wall on the fluid, which has a component in the upward direction. This upward force cancels part of the weight of the fluid so that the remainder for the pressure is just ρgΔz, irrespective of the shape of the tube or its cross sectional area. One can prove this by integrating the vertical component of the pressure exerted by the tube wall over the surface area of the tube wall.

I hope that this makes sense.

Chet
 
  • #9
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Yep this makes sense. Thank you for the help.
 

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