Finding Gauge Pressure in this Manometer

[atpeef]

Homework Statement

Finding Gauge Pressure

Relevant Equations

Pgage = Pabs - Patm

Hi everyone. I would like to ask how do you find the gauge pressure at point 4.

I know the formula, which is:

Pgage = Pabs - Patm

But I don't know which one is the Pabs.

This is my attempt:

P4 = (600 kg/m^3 × 9.8 m/s^2 × 0.05 m) + Ptank

This is all I can do.

Please professors help me on this.

Tell me please the clue and hints as much as possible.

 

[Steve4Physics]

Hi @atpeef and welcometo PF.

You may find it helpful to look through the forum rules so you know what to do and what to expect here: https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

Here's a question for you: is there anywhere in the system that is at exactly the same pressure as point 4? Struck-through as misleading.

 

[haruspex]

But I don't know which one is the Pabs.

You ask that as though Pabs is the pressure at some particular point; it isn't. At every point you can express the pressure either relative to atmospheric pressure (that is its "gauge" pressure) or as its absolute pressure P_abs, i.e. relative to absolute zero pressure. The difference, of course, is atmospheric pressure, P_atm.

P4 = (600 kg/m^3 × 9.8 m/s^2 × 0.05 m) + Ptank

Since you don't know Ptank, that is not going to help. Start at a point where the pressure is known.

 

[nasu]

What is (are) the fluid(s) in the system? Why different colours?
Oh, is "S" the relative density of the fluid?

 

[atpeef]

What is (are) the fluid(s) in the system? Why different colours?
Oh, is "S" the relative density of the fluid?

I think S mean a Specific Gravity (or SG) which is SG = ρ/ρwater.

So

S = 0.6 mean ρ = 600 kg/m^3
S = 0.8 mean ρ = 800 kg/m^3
S = 1.0 mean ρ = 1000 kg/m^3

I think they are in different colors just to illustrate a difference between one fluid to another.

 

[atpeef]

How do I type an equation here (perhaps in Latex format)? I tried with $any equation$ and $$any equation$$ and click the preview icon but it doesn't work.

 

[haruspex]

How do I type an equation here (perhaps in Latex format)? I tried with $any equation$ and $$any equation$$ and click the preview icon but it doesn't work.

Under the left hand end of the text entry area you should see the button "LaTeX Guide".

 

[erobz]

How do I type an equation here (perhaps in Latex format)? I tried with $any equation$ and $$any equation$$ and click the preview icon but it doesn't work.

There is a “bug”. You can’t see latex in preview until latex already exists in the thread.

 

[Lnewqban]

Welcome, @atpeef !

The whole mass of liquid inside the gauge is under the atmospheric pressure (P1) shown at the open end.
The weight of the columns of liquid (hydrostatic pressure) only adds up to that value for each section located under the liquid surface.

The lower each section of tube is below the open surface, the greater the total pressure is within the liquid.
Therefore, it would have a maximum value at the bottom of the U-shape.
Section 4 is in between that point and the open surface.

Please, see:
https://en.m.wikipedia.org/wiki/Hydrostatics

https://en.m.wikipedia.org/wiki/Atmospheric_pressure

https://en.m.wikipedia.org/wiki/Pascal's_law

 

[atpeef]

In order to find the $P_{gage}$ at point 4, do I have to find the value of the $P_{tank}$ first?

If yes, this is what I got:

$P_{tank} + \left( 600 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.05 \ m \right) + \left( 800 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.15 \ m \right) = \left( 1000 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.1 \ m \right) + \left( 600 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.2 \ m \right) + P_{1}$

$P_{tank} + 294 \ Pa + 1,176 \ Pa = 980 \ Pa + 1,176 \ Pa + 101 \ kPa$

$P_{tank} + 1,470 \ Pa = 2,156 \ Pa + 101 \ kPa$

$P_{tank} = 2,156 \ Pa - 1,470 \ Pa + 101 \ kPa$

$P_{tank} = 686 \ Pa + 101 \ kPa$

$P_{tank} = 0.686 \ kPa + 101 \ kPa$

$P_{tank} = 101.686 \ kPa$

What is the next step?

Is it available another way to find the value of the $P_{gage}$ at point 4 without have to find the value of the $P_{tank}$ first? If yes, how to do it step-by-step?

 

[Lnewqban]

If you would imaginarily remove the pipe and tank and liquid above cross-section 4, how much pressure you would need to apply on that surface to keep the liquid columns as balanced as the diagram shows?

 

[Steve4Physics]

In order to find the $P_{gage}$ at point 4, do I have to find the value of the $P_{tank}$ first?

No. (Also, it's worth noting that the the tank-pressure is not a single value. The pressure at the top of the tank is less than the pressure at the bottom, though this is not an issue here.)

If yes, this is what I got:

$P_{tank} + \left( 600 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.05 \ m \right) + \left( 800 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.15 \ m \right) = \left( 1000 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.1 \ m \right) + \left( 600 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.2 \ m \right) + P_{1}$

This doesn't help. But note, it gives the pressure at point 3, not point 4.

Is it available another way to find the value of the $P_{gage}$ at point 4 without have to find the value of the $P_{tank}$ first? If yes, how to do it step-by-step?

The rules here mean that we won't give you a step-by-step answer. But see if this helps.

What are the the pressure-differences between:
a) point 1 and point 2?
b) point 2 and point 3?
c) point 3 and point 4?

Does knowing these values help you answer the question?

[Minor edits.]

 

[atpeef]

If you would imaginarily remove the pipe and tank and liquid above cross-section 4, how much pressure you would need to apply on that surface to keep the liquid columns as balanced as the diagram shows?

I don't know. Please explain.

No. (Also, it's worth noting that the the tank-pressure is not a single value. The pressure at the top of the tank is less than the pressure at the bottom, though this is not an issue here.)This doesn't help. But note, it gives the pressure at point 3, not point 4.The rules here mean that we won't give you a step-by-step answer. But see if this helps.

What are the the pressure-differences between:
a) point 1 and point 2?
b) point 2 and point 3?
c) point 3 and point 4?

Does knowing these values help you answer the question?

[Minor edits.]

The pressure difference between point 1 and point 2 is:

$P_{2} - P_{1} = \left( 600 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.2 \ m \right) = 1,176 \ Pa$

The pressure difference between point 2 and point 3 is:

I don't know, please tell me how to find this pressure.

The pressure difference between point 3 and point 4 is:

I don't know, please tell me how to find this pressure.

Please help me slowly.

 

[nasu]

Why can't you use the same metod for the other pressure differences? You did it for the difference between 1 and 2.

 

[Steve4Physics]

The pressure difference between point 1 and point 2 is:

$P_{2} - P_{1} = \left( 600 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.2 \ m \right) = 1,176 \ Pa$

Yes, good.

The pressure difference between point 2 and point 3 is:

I don't know, please tell me how to find this pressure.

The pressure difference ($\Delta P$) between any 2 points (height difference $\Delta h$) in a continuous 'section' of a single liquid (density $\rho$) is given by: $\Delta P = \rho g \Delta h$.

This pressure difference is not affected by what's outside, or by the shape of, the 'section’ of liquid.

Look at the Post #1 diagram. Between points 2 and 3 there is a continuous section of a single liquid with density 1000kg/m³. Point 3 is 10cm lower than point 2.

Can you now work out the pressure difference between points 2 and 3.?

Can you now work out the pressure difference between points 3 and 4?

And then, can you use the three pressure differences you have found to answer the question?

 

[atpeef]

$P_{3} = P_{5} = \left( 1000 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.1 \ m \right) + P_{2}$

So, the pressure difference between point 2 and point 3 is:

$P_{3} - P_{2} = \left( 1000 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.1 \ m \right) = 980 \ Pa$

Is this correct?

 

[haruspex]

View attachment 337767

$P_{3} = P_{5} = \left( 1000 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.1 \ m \right) + P_{2}$

So, the pressure difference between point 2 and point 3 is:

$P_{3} - P_{2} = \left( 1000 \ \frac{kg}{m^3} × 9.8 \ \frac{m}{s^2} × 0.1 \ m \right) = 980 \ Pa$

Is this correct?

Yes.

 

[atpeef]

OK, now how do I find the pressure difference between point 3 and point 4?
Is it including a $P_{tank}$?

 

[nasu]

Read (again?) post 15, the one by @Steve4Physics.

 

[atpeef]

I don't understand after read the post #15.
Is it including/consist of $P_{tank}$?
Please anwer this question.
If the answer is yes, then I will write it.

 

[nasu]

Yes, good.The pressure difference ($\Delta P$) between any 2 points (height difference $\Delta h$) in a continuous 'section' of a single liquid (density $\rho$) is given by: $\Delta P = \rho g \Delta h$.

This pressure difference is not affected by what's outside

What is confusing in the above? It is quoted from post 15.

 

[atpeef]

$P_{3} = \left( 800 \ \frac{kg}{m^3} + 9.8 \ \frac{m}{s^2} + 0.15 \ m \right) + P_{4}$

So, the pressure difference between point 3 and point 4 is:

$P_{3} - P_{4} = \left( 800 \ \frac{kg}{m^3} + 9.8 \ \frac{m}{s^2} + 0.15 \ m \right) = 1,176 \ Pa$

Is this correct?

 

[nasu]

It looks OK to me.

 

[atpeef]

OK, now what is the conclusion of all of these pressure differences? What it has to do with the $P_{gauge}$ at point 4?

 

[nasu]

Can't you find the pressure at point 4 after you have calculated all these differences?

 

[Steve4Physics]

OK, now what is the conclusion of all of these pressure differences? What it has to do with the $P_{gauge}$ at point 4?

@atpeef, can you answer this?

A diver is going to explore a winding flooded tunnel.

At the start point, A, before jumping in, his pressure gauge reads 10 units (atmospheric pressure).

He jumps in and eventually reaches point B. From A to B, the pressure increased by 5 units.

He reaches point C. From B to C, the pressure increased by 10 units.

He reaches point D. From C to D, the pressure decreased by 3 units.

Questions;

a) What is the absolute pressure at D?

b) What is the gauge pressure at D?

 

[atpeef]

Can't you find the pressure at point 4 after you have calculated all these differences?

Are you saying that by combining all of these 3 pressure differences, I may find the value of $P_{4}$?

 

[Steve4Physics]

Are you saying that by combining all of these 3 pressure differences, I may find the value of $P_{4}$?

Yes - assuming you combine them correctly of course. Please answer the questions in Post #26 so we can check your understanding.

 

[atpeef]

@atpeef, can you answer this?

A diver is going to explore a winding flooded tunnel.

At the start point, A, before jumping in, his pressure gauge reads 10 units (atmospheric pressure).

He jumps in and eventually reaches point B. From A to B, the pressure increased by 5 units.

He reaches point C. From B to C, the pressure increased by 10 units.

He reaches point D. From C to D, the pressure decreased by 3 units.

Questions;

a) What is the absolute pressure at D?

b) What is the gauge pressure at D?

a) Is it 22 units?
b) Is it 22 units - $P_{atm}$?

 

[Steve4Physics]

a) Is it 22 units?

Yes, that's the absolute pressure.

b) Is it 22 units - $P_{atm}$?

Yes, that's the gauge pressure. Note that you are told $P_{atm}$ = 10 units so the gauge pressure is 22-10=12 units.

Can you complete your original question now?

 

[Steve4Physics]

In addition to Post #30 ...

$P_{3} = \left( 800 \ \frac{kg}{m^3} + 9.8 \ \frac{m}{s^2} + 0.15 \ m \right) + P_{4}$

So, the pressure difference between point 3 and point 4 is:

$P_{3} - P_{4} = \left( 800 \ \frac{kg}{m^3} + 9.8 \ \frac{m}{s^2} + 0.15 \ m \right) = 1,176 \ Pa$

Is this correct?

The final value is correct but there are some errors in what you have typed. You have typed additions ($+$) where you should have multiplications ($\times$).

 

[Lnewqban]

I don't know. Please explain.

If you would imaginarily remove the pipe and tank and liquid above cross-section 4, how the remaining liquid columns would naturally move to find a new balance?
How much effort would be needed to avoid that movement?