# Solving U-Tube Fluid Problem: h1 Calculation

• mirandasatterley
In summary, the problem involves a U-tube filled with mercury and water is poured into one side. The length of the water column in the right arm is 20 cm. To find the height of the mercury in the left arm, an equation is used and it is suggested to work from a reference level where the pressure is the same on both sides. The equation is P = 1atm + ρ(Hg)*g*h(Hg) = 1atm + ρ(H2O)*g*h(H2O) and h(H2O) is known. The final ingredient needed is to relate the h(Hg) found to the original level before the water was added.
mirandasatterley
The problem is;
“Mercury is poured into a U-tube. The left arm of the tube has cross-sectional area A1 of 10.0cm^2, and the right arm has a cross-sectional area A2 of 5.0cm^2. 100 grams of water are then poured into the right arm. (a) Determine the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6g/cm^3, what distance h does the mercury rise in the left arm?

For part (a) of this problem;
m(water) = d(water) V(water)
m(water) = d(water) A2 h(water)
h(water) = 20 cm

For part (b);
We use the point where the water and mercury meet in the right arm and the point horizontal to that in the left arm. And we get an equation;
P1 = P(atm) + d(Hg) g (h1 +h2), where hi is the height on the diagram from the horizontal line to the top of the mercury and h2 is the height from the horizontal line where the water and mercury meet to the horizontal line on the diagram.

P2 = P(atm) + d(water) g h (water)
P1 = P2
d(Hg) g (h1 + h2) = d(water) g h(water)
d(Hg) (h1 + h2) = d(water) h(water)
(h1 + h2) = (d(water) h(water))/ d(Hg)
= 0.0147 m,
But I only want to know h1. Any ideas how I could approach this?

#### Attachments

• utube pic.doc
57 KB · Views: 442
It may be awhile before your diagram shows, and I am having some difficulty interpreting the words. I don't see why you need an h1 and h2. If the reference level is where the water meets the mercury, and you know the height of the water column, all you need is the height of the mercury above the reference level on the other side.

We were supposed to use a point where the horizontal reference line is lined up with the mercury on both sides of the u-tube. On the right side, this point is where the water meet the mercury, which is lower than the initial point where the mercury began to rise. So there is an h1 from where the mercury initially began to the height it got pushed up and h2 is from where the water meets the mercury on the left side to the initial position of the mercury.

mirandasatterley said:
We were supposed to use a point where the horizontal reference line is lined up with the mercury on both sides of the u-tube. On the right side, this point is where the water meet the mercury, which is lower than the initial point where the mercury began to rise. So there is an h1 from where the mercury initially began to the height it got pushed up and h2 is from where the water meets the mercury on the left side to the initial position of the mercury.
OK. I can see your diagram now. It is very blurry and I cannot read the labels but I get the idea. You still should consider working from a reference height where you know the pressure on both sides is the same, and that would be a point in the mercury or at the mercury water boundary. You can relate all of your h values to this level and express your answer in the terms required.

Sorry about the blurry picture, i couldn't find a better one.

I'm trying to use;
(h1 +h2) = 1.47cm
and write another equation using all of the heights but when i sub them into solve for h1, the h1's cancel out.

mirandasatterley said:
Sorry about the blurry picture, i couldn't find a better one.

I'm trying to use;
(h1 +h2) = 1.47cm
and write another equation using all of the heights but when i sub them into solve for h1, the h1's cancel out.
I don't think I can help you with your h1 and h2. I'm not convinced you have the conditions set up correctly in the first place. What I can tell you is that the pressure on both sides is the same at the same level only if you are in the mercury on both sides. If you draw a line across from the mercury water boundary to the other side, then the weight of the 100 grams of water on the right divided by the right-side area has to equal the weight of the mercury on the left above this level divided by the left-side area. This will let you find the height of mercury above this level on the left. You can figure out whatever h values you need fom there.

Okay, so;
m(water)/A(right) = m(Hg)/A(left)
m(Hg) = (m(water)A(left))/A(right)
If i sub in my values,
m(Hg) = 200g

Then are you suggesting that I do,
d(Hg)V(Hg) = 200g
d(Hg) A h(Hg) = 200g
And then to slove for h(Hg), and then find the h I need?

If this is right, is the A that I use the cross-sectional area on the right side?

mirandasatterley said:
Okay, so;
m(water)/A(right) = m(Hg)/A(left)
m(Hg) = (m(water)A(left))/A(right)
If i sub in my values,
m(Hg) = 200g

Then are you suggesting that I do,
d(Hg)V(Hg) = 200g
d(Hg) A h(Hg) = 200g
And then to slove for h(Hg), and then find the h I need?

If this is right, is the A that I use the cross-sectional area on the right side?
Ah(Hg) is the volume of the 200g of mercury on the side that contains only mercury. So which A is it?

You may soon start to recognize that all this calculation involving areas and volumes leads you back to the conclusion that the pressure at this reference level is

P = 1atm + ρ(Hg)*g*h(Hg) = 1atm + ρ(H2O)*g*h(H2O)

You already know h(H2O) = 20cm.

There is a final ingredient needed to relate the h(Hg) you will find to the original level before the water was added. What becomes of any mercury that is pushed out of the tube on the right?

OlderDan said:
There is a final ingredient needed to relate the h(Hg) you will find to the original level before the water was added. What becomes of any mercury that is pushed out of the tube on the right?

The mercury remains in the tube, it is just pushed higher up the tube

mirandasatterley said:
The mercury remains in the tube, it is just pushed higher up the tube
So if the mercury surface on the right side goes down 3cm, what happens to the mercury surface on the left?

OlderDan said:
So if the mercury surface on the right side goes down 3cm, what happens to the mercury surface on the left?

The left side of the tube is larger than the right side

Ok,,

Ok, so I am doing this question as well, and I got all the way to where this thread ends, found the height of the mercury column on the side with only mercury from the reference level where the water and mercury meet, done. However, I am stuck on how to get to the next part, i understand you must relate the volume of mercury pushed out of the right side tube to the volume of mercury on the left tube, i.e. V(right)Hg = V(left)Hg or h(right)(5.00cm^2) = h(left)(10.0cm^2) so h(left) = h(right)/2 right? but how do you find out how much mercury is pushed out of the tube?

Help Anyone?

amcca064 said:
Ok, so I am doing this question as well, and I got all the way to where this thread ends, found the height of the mercury column on the side with only mercury from the reference level where the water and mercury meet, done. However, I am stuck on how to get to the next part, i understand you must relate the volume of mercury pushed out of the right side tube to the volume of mercury on the left tube, i.e. V(right)Hg = V(left)Hg or h(right)(5.00cm^2) = h(left)(10.0cm^2) so h(left) = h(right)/2 right? but how do you find out how much mercury is pushed out of the tube?
If you have in fact found the height of the mercury in the left tube that is above the level where the water and mercury meet in the right tube, then you can use your statement above to find the level where the mercury was on both sides before the water was added. The volume of water below that level on the right is the volume of mercury that was pushed out of the right tube into the left tube. So what is the relationship between the height of the water below that level on the right and the height of the mercury that is above that level on the left?

ok so the total height of the mercury on the left is the height of mercury above the "equilibrium" (without water) line plus the height of water below that line, H(total) = H(Hg) + H(H20) and since they are of the same volume the height of the water is twice that of the height of the mercury, so H(total) = H(Hg) + 2(H(Hg)) so H(total)/3 = H(Hg) right?wow, I think I actually understand now! haha Its amazing how simple it seems once you get it. Thanks!

## What is the purpose of solving U-Tube fluid problems?

The purpose of solving U-Tube fluid problems is to calculate the height difference between two fluids in a U-shaped tube, also known as the "manometer" height. This measurement can be used to determine pressure differences, fluid flow rates, and other important variables in various fluid systems.

## What is the formula for calculating the "manometer" height?

The formula for calculating the "manometer" height (h1) is: h1 = (ρ1-ρ2)gh/ρ2, where ρ1 and ρ2 are the densities of the two fluids, g is the gravitational constant, and h is the height difference between the two fluids in the U-tube.

## What are the units for the "manometer" height?

The units for the "manometer" height (h1) are typically in meters (m) or centimeters (cm), as it is a measurement of height. However, it is important to ensure that the units of ρ1 and ρ2 are consistent with the units of h1 in order to get an accurate calculation.

## Can the "manometer" height be negative?

Yes, the "manometer" height (h1) can be negative if the density of the first fluid (ρ1) is less than the density of the second fluid (ρ2). This indicates that the first fluid is lighter and therefore rises above the second fluid in the U-tube. However, in most cases, the "manometer" height is positive as the first fluid is typically denser.

## What are some common sources of error when calculating the "manometer" height?

Some common sources of error when calculating the "manometer" height include inaccurate measurement of the height difference (h) between the two fluids, using incorrect units for the densities (ρ1 and ρ2), and not taking into account the effects of atmospheric pressure. It is important to carefully measure and double-check all variables in the formula to ensure an accurate calculation.

• Introductory Physics Homework Help
Replies
2
Views
5K
• Introductory Physics Homework Help
Replies
18
Views
5K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
17
Views
2K