Thermodynamics - Double U Tube Manometer with 2 Closed Ends

In summary, Thermodynamics -- Double U Tube Manometer with 2 Closed Ends: - Fresh water and sea water in parallel horizontal pipelines are connected to each other by a double U tube manometer. - The density of sea water is 1035 kg/m^3 while the density of air is 1.2 kg/m^3. - Standard temp and pressure are in use. - A pressure difference between the pipelines is sought. - The air column can be ignored in the analysis. - If the oil column is replaced with sea water, what is the pressure difference in kPa between the two pipelines?
  • #1
ynotbathe
3
0
Thermodynamics -- Double U Tube Manometer with 2 Closed Ends

Homework Statement


GIVEN:
Fresh water and sea water in parallel horizontal pipelines are connected to each other by a double U tube manometer.

Density of sea water is 1035 kg/m^3
Density of air is 1.2 kg/m^3
Standard temp and pressure

A link to the relevant diagram is here:
https://engineering.purdue.edu/ME200/Fall%202009/SpecialProblems/SpecialProblems3-5.pdf"

FIND:
a) Pressure difference between the pipelines
b) Can the air column be ignored in your analysis?
c) If the air is replaced with oil of specific gravity 0.72, then what is the pressure difference in kPa between the two pipelines? Can the oil column be neglected?

Homework Equations


P-Patm = rho * g * h of liquid

The Attempt at a Solution


The problem I am running into is that each end of the manometer is closed so I'm not sure how to find an initial pressure. If one end were open to atmospheric pressure I think I could find the pressure difference and work my way back, if that makes sense. However since both ends are closed "bulbs" of sea and fresh water, I'm not really sure where to start.

Right now I have labeled the point where the fresh water meets the mercury as point a, the point in the mercury at that same level as point b, the point where the air meets the sea water as point d, and the point at that same level as point c. I am thinking that I can then use the rho*g*l equation somehow, I'm just unclear as to how since Patm isn't evident. Pressure at point a and point be should be equal though, and pressure at point c and point d should be equal.

I am thinking that there is some sort of equivalency I can use or something... I have the entire manometer hookup as my system but perhaps that is where I am going wrong.

I am sorry I don't have much worked out, if anyone can even give me some hints as to a starting point I would appreciate it greatly.

Thanks!
 
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  • #2


Hi ynotbathe, welcome to PF. You don't need an absolute pressure, because you're only asked to find a pressure difference. Go ahead and assign a pressure [itex]p_0[/itex] to any of the points you mentioned; it's going to disappear in the final analysis. Although you've given one version of the [itex]\rho g h[/itex] equation, there's a more general version that doesn't necessitate using [itex]P_\mathrm{atm}[/itex]. Does this help?
 
  • #3


could you explain how you went about doing this problem? I am a little lost
 
  • #4


Mapes said:
Hi ynotbathe, welcome to PF. You don't need an absolute pressure, because you're only asked to find a pressure difference. Go ahead and assign a pressure [itex]p_0[/itex] to any of the points you mentioned; it's going to disappear in the final analysis. Although you've given one version of the [itex]\rho g h[/itex] equation, there's a more general version that doesn't necessitate using [itex]P_\mathrm{atm}[/itex]. Does this help?

This is what I did, but I'm pretty sure it's wrong:

Pa = Pb = arbitrary constant = rho*g*h = 13570 kg/m^3 (density of mercury) * 9.8 m/s^2 (gravitational constant) * .1 m (difference in height) = 13298.6 Pa

Pc = Pd = arbitrary constant = rho*g*h = 1.2 kg/m^3 (density of air) * 9.8 m/s^2 (gravitational constant) * .4 m (difference in height) = 4.707 Pa

In addition to not thinking these numbers are right, I'm not really sure where to go from here. Any hints or suggestions to what I'm doing wrong?

Thanks!
 
  • #5


The [itex]\rho g h[/itex] equation is for finding the difference in pressure between two points. Lacking an absolute reference, as is the case here, you can only find a pressure relative to an arbitrary point. So pick an arbitrary point -- say, one of the two pipes -- and work your way to the other pipe.
 
  • #6


u can try this method. Start with one of the enclosed bulb (eg; P1) add the "rho"gh terms when u go down the tube and subtract the terms when you go up the tube until you reach the other enclosed bulb. The effect of air air column is usually neglected in most problem as the density is small and so, the "rho"gh term for the air column is assumed to be zero.Try this link, http://www.fuzzyturtle.net/TDEC/tdec202/hw01.pdf"
 
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Related to Thermodynamics - Double U Tube Manometer with 2 Closed Ends

1. What is a double U tube manometer with 2 closed ends?

A double U tube manometer with 2 closed ends is a device used to measure the difference in pressure between two points in a system. It consists of two U-shaped tubes connected at the bottom with a closed top. The two sides of the U tubes are filled with different fluids, typically water and mercury, and the difference in height between the two columns of fluid represents the pressure difference in the system.

2. How does a double U tube manometer work?

A double U tube manometer works based on the principle of hydrostatic pressure. The difference in height between the two columns of fluid is directly proportional to the difference in pressure between the two points in the system. The fluid in the U tubes will move to balance out the pressure difference, creating an equal but opposite difference in height between the two columns.

3. What is the significance of having 2 closed ends in a double U tube manometer?

The 2 closed ends in a double U tube manometer ensure that the pressure at both ends of the system being measured is the same, eliminating the need to account for atmospheric pressure. This allows for a more accurate measurement of the pressure difference between the two points in the system.

4. How is a double U tube manometer used in thermodynamics?

In thermodynamics, a double U tube manometer is used to measure pressure differences in various systems. It can be used to measure the pressure difference across a valve, in a heat exchanger, or in a gas flow system. It is a simple and effective tool for measuring and monitoring pressure in thermodynamic processes.

5. What are the limitations of using a double U tube manometer?

The main limitation of using a double U tube manometer is that it can only measure pressure differences, not absolute pressure. This means that it cannot provide an accurate measurement of the actual pressure in a system, only the difference in pressure between two points. Additionally, the use of different fluids in the U tubes may cause inaccuracies due to differences in density and surface tension.

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