# Pressure calculations in a manometer

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1. May 27, 2016

### Faiq

1. The problem statement, all variables and given/known data
I have attached a picture of the question.

2. Relevant equations
P = pgh

3. The attempt at a solution
I can calculate pressures in a barometer using the simple rule that at same height the pressure exerted by tubes is same. But the same rule seems not to imply in manometers. Please explain how to work out the respective equations for each section.

2. May 27, 2016

### tommyxu3

The law you mentioned may still work here!
By the law, you can get two relations, respectively between x and P, y and P.

3. May 27, 2016

### Merlin3189

I wouldn't have thought you were expected to use P = ρgh here, though obviously you can. There are several equations you could build and then solve.
This just looks like a simple question to test understanding of manometers and you can answer more or less by inspection. Each option has two clues and each clue rules out half the options.

4. May 27, 2016

### tommyxu3

I think this may be applied to get their relations. Why not?

5. May 27, 2016

### Merlin3189

Yes, you can use it. I just thought that this is not a first question on manometers, rather it is a slightly complex situation to test understanding.

But there is an obvious connection between all the measurements and between the measurements and the pressure P, based on what manometers do - measure pressure. When you use a manometer, you don't normally do density calculations, you just read off the pressure.

In this question, all liquids are mercury, all gases are air, all height measurements are in mm and all pressures are given in mm of mercury. There is no need to translate anything into Pascals or kilograms or anything else. Just look at the pressures.

6. May 27, 2016

### tommyxu3

Ahh...that's what I mean.
However, I think when you directly calculate the pressure in mercury, you have, to be honest, applied the relation between pressure and the liquid's height.

7. May 27, 2016

### Merlin3189

I suppose you could say that, but you didn't need to know the density of mercury nor the value of g, just that pressure is proportional to height and that, therefore, we often specify pressures simply in terms of height of a specific well known liquid.

I see this as an exercise in understanding manometers, not in understanding or using P = ρgh. That equation may be used to develop the theory of manometers, but the general principle that you arrive at, is simply that the difference in pressure is proportional to the difference in height. You can, but don't need to, go back to first principles for each question, but IMO that obscures the simple relationship here.

If OP is still around and wants to proceed in the way he suggests, all he needs to do is to calculate P for each given value of x. That eliminates two options.
Then calculate P from each given value of y . That eliminates two options.
Only one option is left that can produce the correct P from both values.

So I would draw the table with two extra columns:
x ; Px ; y ; Py ; Given value of P
Where Px or Py means the pressure calculated from the given value of x or y.

8. May 27, 2016

### Faiq

Thanks for all the insights
I have worked out its solution using these two equations. If possible please verify them
Assuming pg to be constant and equal to 1
Atm + x = Pressure P
Atm + 50 - y = Pressure P

I still have one question though.
Why are we taking the whole value of Pressure P for both equations? I mean shouldn't it be divided by 2?
( I have worked out the solution by dividing the manometer in two parts and then treated both parts as a u-tube barometer)

9. May 27, 2016

### Staff: Mentor

These equations are correct. Why do you feel that each equation for P should be divided by 2?

10. May 27, 2016

### Merlin3189

That's right.

P is P. It is the pressure of the air in the middle. It is the same on the left part and the right part. It does not get split or shared, so there is no need to halve it nor anything else. That is what pressure is like: it acts equally in all directions and, at the same depth, it is equal throughout a body of fluid. (Here we neglect the depth of air in the middle, as its effect is negligible.)

ρg is indeed constant. For it to be 1, it would need to be in units of "mm of Hg", which is fine, but I think it would be better just to say you were working with pressures in mm of Hg.

So which answer did you choose?

11. May 27, 2016

### Faiq

I chose B

12. May 27, 2016

### Faiq

Btw the answer will not remain same right if the value of gravitational acceleration was not 9.81 since the atmospheric pressure would be different?

13. May 27, 2016

### Merlin3189

I chose B too.

It would never have occurred to me that g had anything to do with this question (still doesn't.) And if g changed I think you'd have better things to worry about than this.

If g changed, I guess atmospheric pressure would change too ( though I suspect it mightn't be a very simple relationship. And atmospheric pressure changes daily despite g being constant.) But of course the pressure of 1mm of Hg would also change with g. So maybe these would balance out. But the air in the middle has a pressure which does not depend on g. Would you want it to be at the old 780 mm of Hg pressure (104 kPa), or the new 780 mm of Hg pressure (104 * gnew/gold kPa)?

The air in the middle at P is an enclosed mass of air, so assuming constant temperature, its pressure changes with its volume as x and y change.
On a day when atmospheric pressure was equivalent to 780 mm Hg, C could have been the answer to this question. But since today atm pressure is at 760 mm Hg, if atmospheric pressure increased tomorrow to 780 mm Hg, this B situation would not change into C. When atmospheric pressure increases, x must reduce, P must increase, and the difference between P and atmospheric pressure must reduce.

14. May 28, 2016

### Faiq

Okay thank you for helping