Mercury U Tube Problem: Solving for Height with Fluid Physics

[twotaileddemon]

I really honestly believe that I worked this out correctly, and explained my logic.. but I just wanted to have this checked to be on the safe side. Thanks a lot for helping, I appreciate the time everyone puts into reading this ^^;;

Homework Statement

Mercury is poured into a U tube in which the cross sectional area of the right limb is three units larger than the left one. The level of mercury in the narrow limb is 30 cm from the upper end of the tube. How much will the mercury level rise in the right limb when the left limb is filled to the top with water?

Homework Equations

(p_w)(g)(h_w)(A_1) = (p_m)(g)(h_m)(A_2)
Where m = mercury, w = water, p = density, g = 9.8 m/s^2, h = height, and A = area
(h_m) = (p_w)(h_w)(A_1) / (p_m)(A_2)
30 cm = .30 m
Density of water = 1000 kg/m^3
Density of mercury = 13,600 kg/m^3

The Attempt at a Solution

(h_m) = (1000 kg/m^3)(x)(.3m) / (13,600 kg/m^3)(3x) = .007m
The x's cancel out.. unit left remaining is just m on top because you want height, which is in meters.
Logically speaking, since the limb on the right is 3x as large the height of the mercury will not be as great as .30 m on the other side because the area is larger. So .007 m sounds like a reasonable number. Also considering the density of mercury is much larger than that of water, it makes sense that not a very large height at all is needed to hold all the mercury.

 

[OlderDan]

What would you say about the pressures on the two sides of the tube at the level of the bottom of the water column in the smaller side. Are they the same? If not, how are they different?

 

[billiards]

I got 0.007537688 metres. Or 0.008m I guess, I worked it out by saying the weight of the displaced water equalled the weight of the displaced mercury. I'm assuming by upper end of the table you meant top of the tube?

 

[twotaileddemon]

Oh, I made a typo.. it should say tube, not table.. sorry!

What would you say about the pressures on the two sides of the tube at the level of the bottom of the water column in the smaller side. Are they the same? If not, how are they different?

At the bottom of the water column on the smaller side..
Well if the whole left is eventually filled with water, and mercury is pushed to the right, I would assume that there is some sort of difference or at least interaction in pressure between the water and mercury

 

[twotaileddemon]

Oh, and I didn't take into account 1.013 x 10^5 atm of pressure because both sides should experience it..

 

[OlderDan]

Oh, I made a typo.. it should say tube, not table.. sorry!

At the bottom of the water column on the smaller side..
Well if the whole left is eventually filled with water, and mercury is pushed to the right, I would assume that there is some sort of difference or at least interaction in pressure between the water and mercury

Where does the following equation you posted earlier come from?

2. Homework Equations
(p_w)(g)(h_w)(A_1) = (p_m)(g)(h_m)(A_2)

This equation does not give you equal pressures, and finding a level where the pressures are equal is essential in this problem. Equilibrium is obtained when the pressure on both sides of the tube is equal at some level where both sides contain mercury. The highest point where that is true is at the top of the mercury on the water side. The additionl pressure from the water column on the one side must equal the additional pressure from the mercury above this level on the other side. I cannot see where you have equated those pressures in your analysis.

 

[OlderDan]

Oh, and I didn't take into account 1.013 x 10^5 atm of pressure because both sides should experience it..

That's correct. You do not need to consider the atmospheric pressure.

 

[twotaileddemon]

I believe the equation I got from earlier was related to Pascal's principle... I got it directly from my class notes.

I thought about using Bernoulli's equation..
P_1 + .5pv_1^2 + pgy_1 = P2 + .5pv_2^2 + pgy_2 but bebcause I don't have a v (velocity) to use then I couldn't solve it.
I know that a_1v_1 = a_2v_2, so perhaps I could use that to solve the for a missing velocity, though that would still give me an unknown variable..

not to mention I don't have P_1 or P_2, so that's even more unknown variables.

 

[twotaileddemon]

Equilibrium is obtained when the pressure on both sides of the tube is equal at some level where both sides contain mercury. The highest point where that is true is at the top of the mercury on the water side.

I understand up to this point.

The additional pressure from the water column on the one side must equal the additional pressure from the mercury above this level on the other side. I cannot see where you have equated those pressures in your analysis.

Why is this true? Mercury and water should have different pressures?

 

[OlderDan]

I believe the equation I got from earlier was related to Pascal's principle... I got it directly from my class notes.

I thought about using Bernoulli's equation..
P_1 + .5pv_1^2 + pgy_1 = P2 + .5pv_2^2 + pgy_2 but bebcause I don't have a v (velocity) to use then I couldn't solve it.
I know that a_1v_1 = a_2v_2, so perhaps I could use that to solve the for a missing velocity, though that would still give me an unknown variable..

not to mention I don't have P_1 or P_2, so that's even more unknown variables.

The equation you wrote is equating two forces being applied to pistions of different areas. I don't know the context of your class notes, but what you need in this problem is to equate two pressures, not two forces. You start out with the top of the mercury at the same level in both sides of the tube, and you have 30cm of space above the mercury in the narrow side. Then you pour in some water until the narrow side is filled. As you do this, the mercury level in the narrow side drops by some distance we can call d, and the mercury on the wider side moves up a distance we can call D.

In terms of d and D:
How high is the column of water?
How much higher is the mercury in the wider side than the mercury in the narrow side?
What is the relationship between D and d?

The pressure at the top of the mercury in the narrow side must be the same as at a point at the same level in the wider side. (This is consistent with Bernoulli's equation, but since nothing is moving you don't need to pursue Bernoulli's equation to do the problem.) In the narrow side, the pressure at this level is due to the column of water above it (and the atmosphere). In the wider side, the pressure at this level is due to the column of mercury above it (and the atmosphere). You have to equate the pressure due to the mercury column on one side with the pressure due to the water column on the other side.

If you answer the questions above, you will have the heights of these two columns in terms of the information given in the problem and the variables d and D. Which of these is the answer to the question being asked in the problem? Equate the pressures, relate these variables, and solve the problem.

 

[twotaileddemon]

argh.. I had all these equations written out neatly using bold to emphasize my text according to your guidance with quotes but my internet window randomly shut off on me and I lost all the text! I'm so upset...

I said that, in terms of pressure, P_1 + p_wgd = P_2 + p_mgD. They should be proportional, in that as one height decreases the other should increase, and if one pressure decreases the other pressure should increase
The column of water, or d, therefore, must be d = (P_2 - P_1 + pmgD)/(p_wg)
The mercury on the wider side , after water is added, should be similar..
D = (P_1 - P_2 + p_wgd)/(p_mg)
As D increases, d should decrease and vice versa.

In terms of pressure, P_1 (the pressure on the left) = P_2 + p_mgD - p_wgd and P_2 = P_1 + p_wgd - p_mgD

Am I going about this right before I continue?

 

[twotaileddemon]

And congrats on your award ^^ You well deserve it

 

[OlderDan]

argh.. I had all these equations written out neatly using bold to emphasize my text according to your guidance with quotes but my internet window randomly shut off on me and I lost all the text! I'm so upset...

I said that, in terms of pressure, P_1 + p_wgd = P_2 + p_mgD. They should be proportional, in that as one height decreases the other should increase, and if one pressure decreases the other pressure should increase
The column of water, or d, therefore, must be d = (P_2 - P_1 + pmgD)/(p_wg)
The mercury on the wider side , after water is added, should be similar..
D = (P_1 - P_2 + p_wgd)/(p_mg)
As D increases, d should decrease and vice versa.

In terms of pressure, P_1 (the pressure on the left) = P_2 + p_mgD - p_wgd and P_2 = P_1 + p_wgd - p_mgD

Am I going about this right before I continue?

Make a careful side by side sketch of your tube before and after the water is poured in. Before the water is added, the mercury levels are the same on both sides because the only thing above the mercury is the atmosphere. The narrow side has 30cm of empty tube above the mercury.

When the water is poured in, the pressure at one level is equal on both sides of the tube only at levels that have mercury on both sides. Extend the line from the no-water sketch across the with-water sketch. The mercury level in the wider side has moved above the line by D and the mercury level in the narrow side has moved below the line by d. Draw a horizontal line at the top of the mercury on the narrow side. The height of the water column above this line is not 30cm because the mercury level has dropped. What is the height? In the wider side, the height of the mercury column above this line is not D because D is the distance from the line in the left sketch, not the new line in the right sketch.

You have to get these distances right before you can solve the problem. There is a simple relationship between D and d, but you are not seeing it. The sketch should help.

Thanks for the congrats. Glad I can be of some help.

 

[twotaileddemon]

Okay.. I think the drawing should help

"The height of the water column above this line is not 30 cm because the mercury level has dropped. What is this height?"

The height should be d - 30 cm.

"the wider side, the height of the mercury column above this line is not D because D is the distance from the line in the left sketch, not the new line in the right sketch."
The height should be 30 cm + D

Are these correct now? I drew the diagram and extended the lines and the two equations I just typed seem to be consistant. Now should I compare the pgh's on each side?

 

[OlderDan]

Okay.. I think the drawing should help

"The height of the water column above this line is not 30 cm because the mercury level has dropped. What is this height?"

The height should be d - 30 cm.

"the wider side, the height of the mercury column above this line is not D because D is the distance from the line in the left sketch, not the new line in the right sketch."
The height should be 30 cm + D

Are these correct now? I drew the diagram and extended the lines and the two equations I just typed seem to be consistant. Now should I compare the pgh's on each side?

Sorry, but no they are not right. The water column is more than 30cm high when the narrow side is filled, and D has no direct connection to the 30cm. D is whatever it has to be to contain the mercury that was pushed out of the narrow side into the wider side.

 

[twotaileddemon]

I'll try again then..
seeing as how water in the left will be lower than before, that means its height must have increased, in that case it's d + 30
Therefore, when the water on the right rises, it should be (1/3)(d+30) because its area is 3x as large. I know you said that 30 cm has no direct connection to D, but the d + 30 is the height increased on the left side. If it increases that much, D must increase .333 times that much.

 

[OlderDan]

I'll try again then..
seeing as how water in the left will be lower than before, that means its height must have increased, in that case it's d + 30
Therefore, when the water on the right rises, it should be (1/3)(d+30) because its area is 3x as large. I know you said that 30 cm has no direct connection to D, but the d + 30 is the height increased on the left side. If it increases that much, D must increase .333 times that much.

You are getting close. d + 30cm is correct for the narrow side, but the distance the mercury level moved in the narrow side is d, not d + 30. So D is?

 

[twotaileddemon]

You mean in the latter part of the sentence the mercury level moved in the wider side is d, not d+30? Let me think why.. both sides originally started at the same place.. if the mercury level goes down d distance (d+30 from original) D has no "30" to start with.. oh okay, now it's clear. In that case, it's only d like you said. D is therefore d/3.

 

[OlderDan]

You mean in the latter part of the sentence the mercury level moved in the wider side is d, not d+30? Let me think why.. both sides originally started at the same place.. if the mercury level goes down d distance (d+30 from original) D has no "30" to start with.. oh okay, now it's clear. In that case, it's only d like you said. D is therefore d/3.

OK good. Now you know the height of the water above the "new line" at the bottom of the water level is d + 30cm, and you know D = d/3. At the level of this new line the pressure is the same on both sides. The pressure produced by the d + 30cm of water must be the same as the pressure pruduced by the mercury above this line in the sider side. What is the height of the mercury in the wider side relative to this line?

 

[twotaileddemon]

On the right side, the pressure is pgD/3..so D must be 3P/pg ?

 

[OlderDan]

On the right side, the pressure is pgD/3..so D must be 3P/pg ?

You can calculate the pressure at the bottom of the water on the narrow side in terms of d and the density of water because you know the water column is d + 30cm high. On the wider side, you can calculate the pressure at the same level in terms of the density of mercury and the height of the mercury, but you must first express the height of the mercury relative to the level of the bottom of water in terms of d and D, or since you know D = d/3, just in terms of d. You have not done that yet, and you can't finish the problem until you do.

 

[billiards]

I looked back at your first attempt and I can tell you where you went wrong. The equation you gave works but you used the wrong value for h_w.

h_w=0.3+h_m

you just used 0.3 before because you forgot to account for the lowering of the mercury. I plugged in this value into your formula and got exactly the same answer as I got before, when I used a more intuitive method. I.e. I said the weight of the displaced mercury was equal to the weight of the column of water.

 

[twotaileddemon]

Oh okay.. because D = d/3, if you have d, then finding D becomes really easy
In that case, I know the density of water is 1000 kg/m^3 and therefore (1000)(9.8)(d+30) = P_l, but we don't know exactly what P is.
(1000)(9.8)(d + .3) = P_l
9800d + 2940 = P_l
13600(9.8)(d/3) = P_r
44427d = P_r

Pr = P_l
44427d = 9800d + 2940
34627d = 2940
d = .085 m
D = .085/3 = .028 m?

 

[twotaileddemon]

I looked back at your first attempt and I can tell you where you went wrong. The equation you gave works but you used the wrong value for h_w.

h_w=0.3+h_m

you just used 0.3 before because you forgot to account for the lowering of the mercury. I plugged in this value into your formula and got exactly the same answer as I got before, when I used a more intuitive method. I.e. I said the weight of the displaced mercury was equal to the weight of the column of water.

Okay, that makes sense.

 

[twotaileddemon]

Physics is always so fun to learn for some reason.. and just seeing everyone guiding me makes me want to do the same after I learn enough physics to do so. I really look forward to the day when I can answer questions in this forum with certainty in my answer.. hopefully I'll learn tons in college next year so I can return the kindness of everyone here ^^

 

[OlderDan]

Oh okay.. because D = d/3, if you have d, then finding D becomes really easy
In that case, I know the density of water is 1000 kg/m^3 and therefore (1000)(9.8)(d+30) = P_l, but we don't know exactly what P is.
(1000)(9.8)(d + .3) = P_l
9800d + 2940 = P_l
13600(9.8)(d/3) = P_r
44427d = P_r

Pr = P_l
44427d = 9800d + 2940
34627d = 2940
d = .085 m
D = .085/3 = .028 m?

You still do not have the correct height for the mercury. See if you can figure out what it should be before the picture shows up.

 

[twotaileddemon]

A .pdf file? You didn't have to go to such an extent! I really appreciate it honestly, you are a very kind person.

But before I look at it, I will try to see what I did wrong.
I know on the left that (p_w)(g)(.3m + d) will provide the level of mercury on the left. This seems valid. So I must have done something wrong on the right.
I had said (p_m)(g)(d/3) will give the area the mercury will increase from its original position. The statement in itself is true, but perhaps that is not the correct way to express the pressure on that side. Though it may give the correct height increased, I don't think it gives the complete solution for the pressure. I just need to think about how to account for that. pgh is the correct equation, so it must have something to do with the height because I'm positive p_m(g) is used. In that case, the only other thing I could see doing would be to add the distance the mercury traveled on the left to the right to account for the distance between the two pressures completely. In that case, (p_m)(g)(d/3 + d) = P_2, and d/3 = D when I solve for d when setting this and the P_1 equation equal to each other.

Okay.. I'll post and then open the .pdf now. Thanks again

 

[twotaileddemon]

! The diagram is so elaborate. I'm so sorry that I made you spend so much time on it ;_;..

But I understand the equation now.. I knew the right forumla originally, I just needed to know the distance, and now that I've been shown how I think I will be able to apply it to other problems.

Thanks xinfinity :)