What is the peizometric head difference between A and B?

[Padmal]

Homework Statement

Water flows in an inclined pipeline in an upward direction from A to B. A differential U-tube manometer is connected

between A and B and indicates a mercury level difference of 30 cm. The difference in elevation between A and B is 2 m.
What is the peizometric head difference between A and B?
1. 2.52 m
2. 4.52 m
3. 3.78 m
4. 5.22 m

Homework Equations

Pressure Head = Pressure / Specific weight

The Attempt at a Solution

My Calculation
P_a = P_b + 2 + (30/100)x13.6
P_a - P_b = 6.08 m (Which is not an answer)

 

[haruspex]

All I know about piezometric head is what I just read online, at several references.
As I read it, it purely relates to height difference, so the answer should be 2m. It should be unrelated to the flow. The actual pressure difference (4.08m head) is greater, presumably, because of drag in the pipe. You seem to have added the two, which strikes me as unreasonable. Taking the difference is more logical, but still doesn't match any definition I can find, nor any of the proposed answers.
What definition have you been given?

 

[Padmal]

I think we can graphically represent the question as above.
So if we calculate equalizing the pressure differences noted by the blue line,
Z₂w + P_A = P_B + Z₁w + 13.6hw (w is the specific weight of water)
But Z₁ = Z₂ - h + 2 m

P_A - P_B = 5.78 m (Which is again not the answer) :(

 

[haruspex]

I think we can graphically represent the question as above.
So if we calculate equalizing the pressure differences noted by the blue line,
Z₂w + P_A = P_B + Z₁w + 13.6hw (w is the specific weight of water)
But Z₁ = Z₂ - h + 2 m

P_A - P_B = 5.78 m (Which is again not the answer) :(

OK, so the 'mercury pressure difference' is actually a 'mercury minus water pressure difference'? That explains how the decimal part can become either .78 or .22.
But what definition of piezometric head are you using? You seem to be taking it to mean simply the pressure difference (as a height of water). I've not been able to find one that would make the answer different from 2m.

 

[Padmal]

OK, so the 'mercury pressure difference' is actually a 'mercury minus water pressure difference'? That explains how the decimal part can become either .78 or .22.
But what definition of piezometric head are you using? You seem to be taking it to mean simply the pressure difference (as a height of water). I've not been able to find one that would make the answer different from 2m.

Sir, thank you very much for the answer! One of my friends got 5.22 m and my definition of piezometric head is the height of water. How did you get the 5.22 instead of 5.78?

 

[haruspex]

my definition of piezometric head is the height of water

I understand that it is a pressure expressed as a height of water, but what height (or what pressure)?
According to all references I've found, piezometric head is the same as hydraulic head.

I finally found this diagram: http://en.wikipedia.org/wiki/Hydraulic_head#/media/File:Relation_between_heads_flowing.svg.
This indicates that the elevation head (z) is simply the geophysical height difference (z_A-z_B=-2m of water here) and the pressure head is the gauge pressure difference between A and B, expressed as a height of water ($\phi$). The piezometric/hydraulic head (h) is the elevation head plus the pressure head, $h = \phi+z$. That very equation appears at http://en.wikipedia.org/wiki/Hydraulic_head#Components_of_hydraulic_head.

According to your diagram of the manometer, the gauge pressure difference will be (13.6-1)hg (because height h of the manometer has Hg on one side and water on the other), so $\phi_A-\phi_B= 12.6*0.3m=3.78m$. Thus $h_A-h_B =\phi_A-\phi_B+z_A-z_B=3.78m+(-2m)=1.78m$.
This makes sense intuitively. As you can see in the diagram at the link, the hydraulic head is effectively the backpressure from the water flow (drag). In the present case the water is flowing up. The pressure required to make it do so (3.78m head) is the pressure required to overcome the height difference (2m head) plus the drag. So the drag is 1.78m head.

Of course, all this depends on that linked diagram being a correct representation of these pressures. I haven't found any other link which expresses it in a way I consider unambiguous.

I didn't get 5.22m by any means. I mentioned .22 because, having calculated 3.78 for gauge pressure difference, if a later part of the calculation required you to subtract that from a larger whole number of metres you would end up with .22 as the decimal part.

 

[Padmal]

According to your diagram of the manometer, the gauge pressure difference will be (13.6-1)hg (because height h of the manometer has Hg on one side and water on the other), so $\phi_A-\phi_B= 12.6*0.3m=3.78m$. Thus $h_A-h_B =\phi_A-\phi_B+z_A-z_B=3.78m+(-2m)=1.78m$.
This makes sense intuitively. As you can see in the diagram at the link, the hydraulic head is effectively the backpressure from the water flow (drag). In the present case the water is flowing up. The pressure required to make it do so (3.78m head) is the pressure required to overcome the height difference (2m head) plus the drag. So the drag is 1.78m head.

As 3.78 is one of the choices given, I think that is the correct answer. $\phi_A-\phi_B= 12.6*0.3m=3.78m$ could be the correct peizometric head!

 

[haruspex]

As 3.78 is one of the choices given, I think that is the correct answer. $\phi_A-\phi_B= 12.6*0.3m=3.78m$ could be the correct peizometric head!

It could equally be one of the wrong answers the question setter anticipates.
(Btw, it's piezo-, not peizo-.)