Pressures in a U Tube along a line - different liquids

[brotherbobby]

Homework Statement

Shown in the diagram below is a U-Tube with water on the left side and oil on the right. The system is in equilibrium. The horizontal line $\mathbf{AB}$ is drawn such that it passes through water through the left tube and through oil in the right tube. Are the pressures the same along this line in the right and left tubes?

Relevant Equations

Absolute pressure at a point of depth $h$ due to a liquid is $P_A = P_G + P_{\text{atm}}$, where $P_G = \rho_L gh$ is the gauge pressure due to the liquid of density $\rho_L$ and $P_{\text{atm}}$ is the atmospheric pressure.

Yes, pressure is same along the line for both the tubes since the liquids are in equlibrium. (I ignore the slightly longer air column in the left compared to the right).

Is my answer correct?

 

[Merlin3189]

Think about lines at other positions. Eg PQ, RS, TU

 

[Chestermiller]

No. The pressure at B is higher than at A if the density of the oil is less than the density of the water. Can you figure out why?

 

[brotherbobby]

Think about lines at other positions. Eg PQ, RS, TU
View attachment 257169

Thank you for the trouble. Let me respond to you first and then I'd reply to Chestermiller's post ( Chestermiller's ) above later.

For PQ : Pressure is higher at Q than at P : $\boxed{\mathbf{P_Q > P_P}}$. This is because the pressure due to the column of oil above it. $P_{\text{atm}}$ is the same for both tubes. (Of course, I am igoring the slight amount of extra pressure over P due to the air column. Even if I didn't, the answer would stand because pressure due to oil is more than pressure due to the same height of air, given higher density of oil).

For RS : Same as before, pressure is more at S than at R : $\boxed{\mathbf{P_S > P_R}}$. Same justification too. Oil has higher density than air.

For TU : Pressure at T is the same as the pressure at U : $\boxed{\mathbf{P_T = P_U}}$. I do not know why this is, except that I am told (in books) that pressure at the same height in a liquid is the same. I can agree to this if we have only one liquid. But in this problem, there is oil above U while only water above T. I do not understand how can the pressures be the same, only that they are as per what books say.

 

[brotherbobby]

No. The pressure at B is higher than at A if the density of the oil is less than the density of the water. Can you figure out why?

Let me paste the picture first. I do not know (please see my response to Merlin above), but I am told from textbooks that the pressure at two different points along a horizontal line (same height) is the same for a given liquid . I do not know know why this is if there are other liquids "on top" of the liquid in question. However, assuming that to be true, we have (see picture alongside), $\mathbf{P_T = P_U}$.

Now let's go "up" from TU to AB. We encounter the same colume lengths of oil and water. But since the density of water is greater than the density of oil ($\rho_{\text{oil}} < \rho_{\text{water}}$), the fall in pressure is greater when we arrive at A than we do at B. Hence, as you had asked me, the pressure at P is greater at B than at A. $\mathbf{P_B > P_A}$.

I assume that answers my (initial) question, so many thanks. But as to why pressures at T and U has to be same is still in question. Can you clarify that?

 

[kuruman]

But as to why pressures at T and U has to be same is still in question. Can you clarify that?

Imagine a horizontal line VW near the bottom of the U so that its entirety is in the blue fluid. The pressure is the same all along that line. Now raise that line a bit so that it is still below line TU but only some of it is in the blue fluid. Is the pressure the same or different at each side of the line? Keep raising the line and keep asking yourself that question. What happens when line VW coincides with line TU? What happens when it goes above line TU?

 

[BvU]

But as to why pressures at T and U has to be same is still in question

Familiar with$$\Delta p = \rho g\Delta h\quad ?$$and with communicating vessels ? (wiki)

 

[Lnewqban]

...
For TU : Pressure at T is the same as the pressure at U : $\boxed{\mathbf{P_T = P_U}}$. I do not know why this is, except that I am told (in books) that pressure at the same height in a liquid is the same. I can agree to this if we have only one liquid. But in this problem, there is oil above U while only water above T. I do not understand how can the pressures be the same, only that they are as per what books say.

...
I assume that answers my (initial) question, so many thanks. But as to why pressures at T and U has to be same is still in question. Can you clarify that?

It is important that you understand what is physically happening.
It is all about the effect of gravity acceleration over the masses of both fluids in your problem (in other words, about the existing weight of fluids above each point within the columns of liquids), combined with the volume each fluid occupies.
Let's forget for a moment about the abstract concepts of pressures and densities and let's make an equivalent system, also in perfect balance, considering things that you can easily see and feel.

Let's use 19 cubes of identical dimensions (1 cm x 1 cm x 1 cm).
13 of those cubes weight 20 Newtons each.
6 of those cubes weight 10 Newton each.

Next, let's place two towers of cubes on a scale balance in equilibrium.
Left tower is formed by 8 of the 20-N cubes.
Right tower is formed by 6 of the 10-N cubes stacked on top of 5 of the 20-N cubes.

Those two towers should weight the same.
If you could, create a diagram like described.
Then, you will clearly see how much weight (or pressure) the interphase of each pair of contiguos cubes supports.
If you connect interphases under similar load, you could also see the vertical pressure variation, or the variation in pressure as a function of elevation for each fluid of your problem.

 

[Chestermiller]

For a fluid in static equilibrium, the pressure varies vertically and horizontally according to the equations:
$$\frac{\partial p}{\partial z}=-\rho g$$
$$\frac{\partial p}{\partial x}=0$$where z is the elevation in the direction opposite to gravity and x is the horizontal direction. The reason that the pressure cannot vary with x is that there is no force acting on the fluid in the horizontal direction. If there were a pressure gradient in the x direction, then you would have to have accelerating flow in that direction, but this would mean that the fluid is not in static equilibrium.
So, if you start at T and integrate the pressure variation around the curved channel towards U, the pressure will first increase as you move downward, but then, after passing through the bottom of the curved tubing,, the pressure will begin increasing. And, when you finally arrive at the elevation of point U, the pressure will be back to the value it had at T.

 

[brotherbobby]

Imagine a horizontal line VW near the bottom of the U so that its entirety is in the blue fluid. The pressure is the same all along that line. Now raise that line a bit so that it is still below line TU but only some of it is in the blue fluid. Is the pressure the same or different at each side of the line? Keep raising the line and keep asking yourself that question. What happens when line VW coincides with line TU? What happens when it goes above line TU?

I will respond to Kuruman's quote first, but thank you all for your questions and hints.

Let me copy and paste the figure first, adding the few lines that you mentioned.

Line VW :

Yes the pressure is the same along the line VW. $\boxed{\mathbf{P_V = P_W}}$. This is because if the pressures were not the same then we would have had a flow of water from the higher to the lower pressure, which would violate equilibrium.

Line V' W' :

We move further up along the line to the (dotted) V'W'. This line doesn't entirely lie in the water - some of it is out of the system in air. However, in response to my own question, I now realize that the pressures at both points V' and W' are the same : $\boxed{\mathbf{P_{V'} = P_{W'}}}$. This can be explained by keeping Pascal's principle in mind. The pressures at V' and W' act in all directions, so let's take their values acting "straight down". If the pressures had not been the same, there would have been motion of fluid (water) from the point of higher to lower pressure, which would jeopardise equilibrium.

Line TU :

Line TU is the highest point of water, right up to the water-oil interface. The pressures at T and U are the same : $\boxed{\mathbf{P_T = P_U}}$. The equality holds to this line, but no further. For instance, for reasons I have mentioned earlier in this thread : $\boxed{\mathbf{P_A < P_B}}$. The argument is the same as for the line V'W' above - Pascal's principle.

If my reasoning is correct, it answers my doubt for the critical line V'W'.

We can assume correctness for the statement : The pressure along a horizontal line is the same long as the line moves through the same liquid, irrespective of whether the line passes through the same or different vessels containing the liquid.

 

[brotherbobby]

Familiar with$$\Delta p = \rho g\Delta h\quad ?$$and with communicating vessels ? (wiki)

Yes I was aware of the general notion of communicating vessels, but your link on it (the first one) was illuminating. It taught me, among other things, that the pressure at the bottom of a vessel is independent of the weight of liquid in the vessel, and dependent instead on the height of the liquid. A bit of thinking has helped me draw the non-intuitive diagram that I copy and paste for you on the right.

Both vessels A and B have the same height and the same cross sectional area at their base. They are filled entirely with water. Clearly, the pressure at their bases are equal : $\mathbf{P_A = P_B}$. The base areas are equal, so the force due to each liquid is also the same : $\mathbf{F_A = F_B}$.

Imagine each vessel is carried by a boy, lifting up. We understand that the weight of water in the two vessels are not the same : $\mathbf{w_A < w_B}$. Hence, if $F'$ is the lifting force applied by each boy, can we say : $\color{red}{\mathbf{F'_A = F'_B\;?}}$

 

[kuruman]

We can assume correctness for the statement : The pressure along a horizontal line is the same long as the line moves through the same liquid, irrespective of whether the line passes through the same or different vessels containing the liquid.

That is exactly what I hoped you would conclude. Once you go above the "same fluid line" TU, the pressure at each side changes at a different rate with respect to height because the densities are different. I think you got it now.

 

[BvU]

can we say : F′A=F′B?

Absolutely not ! Compare sheets 9 and 10 in the link. In the first, holding up the piston requires 150 kgf, both left and right.

In the second, the bottom of the container is holding up the liquid column, and whoever holds up the whole thing carries the total mass, which is small on the left and much too much on the right.

I regret using the link: the author makes a complete mess of units and values !

 

[Chestermiller]

Yes I was aware of the general notion of communicating vessels, but your link on it (the first one) was illuminating. It taught me, among other things, that the pressure at the bottom of a vessel is independent of the weight of liquid in the vessel, and dependent instead on the height of the liquid. A bit of thinking has helped me draw the non-intuitive diagram that I copy and paste for you on the right.

View attachment 257279Both vessels A and B have the same height and the same cross sectional area at their base. They are filled entirely with water. Clearly, the pressure at their bases are equal : $\mathbf{P_A = P_B}$. The base areas are equal, so the force due to each liquid is also the same : $\mathbf{F_A = F_B}$.

Imagine each vessel is carried by a boy, lifting up. We understand that the weight of water in the two vessels are not the same : $\mathbf{w_A < w_B}$. Hence, if $F'$ is the lifting force applied by each boy, can we say : $\color{red}{\mathbf{F'_A = F'_B\;?}}$

In the primed figure (left), is the fluid exerting pressure on the lid? If so, what is holding the lid down?

 

[Lnewqban]

Line TU :

Line TU is the highest point of water, right up to the water-oil interface. The pressures at T and U are the same : $\boxed{\mathbf{P_T = P_U}}$. The equality holds to this line, but no further. For instance, for reasons I have mentioned earlier in this thread : $\boxed{\mathbf{P_A < P_B}}$. The argument is the same as for the line V'W' above - Pascal's principle.

Could you represent four or five lines joining points of similar static pressure for both sides (water and oil) above the line T-U?
Do you believe that the water column A-R weights as much as the full oil column?

 

[Merlin3189]

 

[kuruman]

Just to complete the picture (and beat the topic to death) here is a plot of pressure vs depth for the two sides. The density ratio is 2:1.

 

[brotherbobby]

Absolutely not ! Compare sheets 9 and 10 in the link. In the first, holding up the piston requires 150 kgf, both left and right.

In the second, the bottom of the container is holding up the liquid column, and whoever holds up the whole thing carries the total mass, which is small on the left and much too much on the right.

I regret using the link: the author makes a complete mess of units and values !

I understand the difference between a piston (which is movable) and the rigid bottom of a container.

However, if the force due to liquid pressures are the same ($F_A = F_B$) because pressures are the same, why should for forces exerted by someone to hold the piston (or rigid bottom) be different?

In other words, as per the diagram alongside, why should $\mathbf{F'_A \neq F'_B}$?

(I realize the obvious - clearly, anyone trying to carry the two vessels would find the first easier than the second, owing to its lower weight. However, given that both liquids apply equal forces on their bases, I cannot reconcile between the two things).

 

[BvU]

why should for forces exerted by someone to hold the piston (or rigid bottom) be different?

If the bottom is rigid, fixed, attached, welded to the sides, the force needed is the force to carry the container including the liquid. The container bottom exerts the offsetting force for the water column pressure.

If the bottom is a piston, the force needed is the force to keep the piston in place. The container is kept in postion by other means and doesn't move

In the latter case, for A, when you do move the piston, the change of the liquid column height change will be enormous, and with it the force needed to hold the piston.

 

[Chestermiller]

I understand the difference between a piston (which is movable) and the rigid bottom of a container.
View attachment 257321However, if the force due to liquid pressures are the same ($F_A = F_B$) because pressures are the same, why should for forces exerted by someone to hold the piston (or rigid bottom) be different?

In other words, as per the diagram alongside, why should $\mathbf{F'_A \neq F'_B}$?

(I realize the obvious - clearly, anyone trying to carry the two vessels would find the first easier than the second, owing to its lower weight. However, given that both liquids apply equal forces on their bases, I cannot reconcile between the two things).

In post #14, I asked you some leading questions to help you resolve this issue. You have failed to respond to my questions. Not very respectful of a Physics Forums mentor! If you are unable to answer them, I will answer them for you.

 

[kuruman]

It might be informative to watch video 1 and then video 2. They are well done.

 

[brotherbobby]

In post #14, I asked you some leading questions to help you resolve this issue. You have failed to respond to my questions. Not very respectful of a Physics Forums mentor! If you are unable to answer them, I will answer them for you.

I am sorry - I thought the issue was resolved. Let me draw the diagram again and get back to replying to your question.

You had asked - "In the primed figure (left), is the fluid exerting pressure on the lid? If so, what is holding the lid down? "

My answer : Yes, the fluid is applying pressure on both lids, A and B. I think you meant "what is holding the lid *up*"? The applied (external) forces acting up, $F'_A\; \text{or}\; F'_B$, is what is holding the lid up in both cases.

As per my doubt, we know that $F_A = F_B$ (because $P_A = P_B$ and the base areas of both vessels are equal). Does that therefore mean that $F'_A = F'_B$? Clearly, and intuitively, no, because the person holding both vessels would obviously have to apply a higher lifting force on B than on A since $w_B > w_A$.

How do we reconcile to the fact that the base pressures are equal ? ($F_A = F_B$)

The answer lied in post 19 above. The bottom surfaces are rigid, not movable pistons. Hence the forces (effort) needed to carry the vessels is equal to the weight of the liquids, ignoring the light masses of the vessels. However, a movable piston fitted at the bottom of the vessels would move identically in both cases owing to the equality of liquid forces. But even in that case, other arrangements needed to hold the liquids (meaning their own weights) must be made to keep the liquids in place.

Please tell me what you may have to when you have the time.

(A private note : I am sorry for the huge delay in reply. I am a physics tutor myself and exams are on. I would rarely have the time to come on here for the next two weeks. But I will try to do so and spend a good 10 to 15 minutes answering your responses to my queries. Thanking you all for the trouble).

 

[Chestermiller]

The pressure acting on the lower lid of A (situated just a small distance above the bottom of A) is much higher than the pressure on the top of B. So that lower lid has to be held down in some way. It is held down by tension in the short side-wall in A connecting the lid to the bottom. This tension in the side-wall also acts on the base of A (isolated as a free body). So there are 3 forces (not 2) acting on the base of A:

1. The upward force $F'_A$ (the weight of the fluid and container)
2. The downward pressure force of the fluid $F_A$
3. The upward tension in the sidewall (equal to $F_A-F'_A$)