EK physics 1001 question 577

[justhanging]

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The question comes with a drawing that unfortunately I can't get on here. Its a fluid question showing one long tube and 3 small pipes running vertical than asks for the relative heights of water in each pipe.

The answer says that the height is related by 1atm + 0.5*rho*v^2

This comes from Bernoulli equation but am not really seeing it.

 

[gkdubey]

I really wish someone would post the answer to your question here. I have NO idea how to solve this problem.

 

[sps27]

The pictures are not very clear but manageable. I too am not sure about the answer they have given. I was thinking more in the lines of c>a>b, but that is not an option. It seems they have neglected the difference in height completely, even though the question explicitly states measuring from the floor.

 

[Schenker]

EDIT: Well, I goofed. Sorry about that! Fixed and simplified.

(Before starting, I'm going to assume that you're solid with the concept of Q = AV. The flow speed at a any point in the pipe depends only on the diameter of the pipe at that point. In other words, water flows at the same speed at A and C, since the diameter is the same, and slower at B, since the diameter is larger. Also, I'm going to assume familiarity with the fact that pressure drops at higher flow rates.)

Let h_a, h_b, and h_c represent the height of each respective water column as measured from the floor. For simplicity, I'm going to treat the distance from the floor to the horizontal line that passes through the center of the pipe at A and B (but that passes below C) as being 0.

First, I will show that h_b > h_a. Next, I will show that h_a = h_c.

So lets look at the center of the pipe at A. Call the pressure at that point P_a. Look at the column of water above A that is at a height of h_a. The pressure, P_a, at a depth of h_a, is

🙂 P_a = (rho)(g)(h_a).

Similarly,

🙂 P_b = (rho)(g)(h_b).

We know that P_a < P_b from the fact that the flow speed at A is greater than at B. By the last two smiley-formatted equations, P_a < P_b implies that (rho)(g)(h_a) < (rho)(g)(h_b). This means that h_a < h_b.


Now, to show that h_a = h_c, we will need to let P_c denote the pressure at a point that is NOT the center of the pipe at C. Choose a point on the horizontal line that connects the centers of A and B but is in the skinny section to the right of the fat section B. Call this point pointC. Let P_c be the pressure at pointC.

Measured from the top of column C, pointC is at a depth of h_c, so

🙂 P_c = (rho)(g)(h_c).

The center of the pipe at A and pointC are at the same height, and the pipe is of the same diameter at each of these two points. Thus, P_a = P_c. From before, we had that

🙂 P_a = (rho)(g)(h_a).

Setting these last two smiley-formatted equations equal to each other yields

🙂 (rho)(g)(h_a) = (rho)(g)(h_c)

which shows that h_a = h_c.

 

[sps27]

EDIT: Well, I goofed. Sorry about that! Fixed and simplified.

(Before starting, I'm going to assume that you're solid with the concept of Q = AV. The flow speed at a any point in the pipe depends only on the diameter of the pipe at that point. In other words, water flows at the same speed at A and C, since the diameter is the same, and slower at B, since the diameter is larger. Also, I'm going to assume familiarity with the fact that pressure drops at higher flow rates.)

Let h_a, h_b, and h_c represent the height of each respective water column as measured from the floor. For simplicity, I'm going to treat the distance from the floor to the horizontal line that passes through the center of the pipe at A and B (but that passes below C) as being 0.

First, I will show that h_b > h_a. Next, I will show that h_a = h_c.

So lets look at the center of the pipe at A. Call the pressure at that point P_a. Look at the column of water above A that is at a height of h_a. The pressure, P_a, at a depth of h_a, is

🙂 P_a = (rho)(g)(h_a).

Similarly,

🙂 P_b = (rho)(g)(h_b).

We know that P_a < P_b from the fact that the flow speed at A is greater than at B. By the last two smiley-formatted equations, P_a < P_b implies that (rho)(g)(h_a) < (rho)(g)(h_b). This means that h_a < h_b.


Now, to show that h_a = h_c, we will need to let P_c denote the pressure at a point that is NOT the center of the pipe at C. Choose a point on the horizontal line that connects the centers of A and B but is in the skinny section to the right of the fat section B. Call this point pointC. Let P_c be the pressure at pointC.

Measured from the top of column C, pointC is at a depth of h_c, so

🙂 P_c = (rho)(g)(h_c).

The center of the pipe at A and pointC are at the same height, and the pipe is of the same diameter at each of these two points. Thus, P_a = P_c. From before, we had that

🙂 P_a = (rho)(g)(h_a).

Setting these last two smiley-formatted equations equal to each other yields

🙂 (rho)(g)(h_a) = (rho)(g)(h_c)

which shows that h_a = h_c.

Thanks for the response. I understand that h_b will be greater than h_a. Higher cross sectional area at b, hence lower velocity, hence higher pressure. Moreover they are at the same height so the (rho)(g)(h) term is same for both 'a' or 'b'. But 'c' is at a height above 'a' or 'b' and so gravitational potential energy for fluid at 'c' will be higher than either 'a' or 'b'.

I follow your method up till 'a' and 'b' but I don't understand moving 'c' to be at the same level as 'a' or 'b'. I am not sure we can do that. The point 'c' is given to be above 'a' or 'b' and I think they are telling us that fluid at 'c' is at a higher gravitational potential energy. The area of cross section at 'c' being same as 'a', it means the velocity is the same so the pressure term due to velocity is the same. But we still have to take into account the height difference, I think.

 

[Schenker]

I follow your method up till 'a' and 'b' but I don't understand moving 'c' to be at the same level as 'a' or 'b'. I am not sure we can do that. The point 'c' is given to be above 'a' or 'b' and I think they are telling us that fluid at 'c' is at a higher gravitational potential energy. The area of cross section at 'c' being same as 'a', it means the velocity is the same so the pressure term due to velocity is the same. But we still have to take into account the height difference, I think.

I'm pretending that the vertical section of the pipe and the center of C are part of column C.

In fact, if you bent the pipe so that sections A, B, and C were level, the height of the water column above C as measured from the floor would not change.

If you are unconvinced:

You are right that, while the pressure term due to velocity is the same at A and C, the height difference means that A is at a higher pressure than C. This means that column C will be shorter off the pipe than column A, but column C is higher off the ground than column A by exactly that amount.

Let's do some math to confirm. Let h_ac be the difference between the heights of the pipe at A and C. Then the difference in pressures at A and C is (rho)(g)(h_ac). In other words,

(1) P_a = P_C+ (rho)(g)(h_ac)

where P_a is the pressure at A and P_C is the pressure at C. (P_C is not the same as P_c from my previous post, because they're pressures at different points.)

We have another equation for each P_a and P_C, namely,

(2) P_a = (rho)(g)(h_a)
(3) P_C = (rho)(g)(h_C)

where h_a is the height of column A as measured from the pipe at A and h_C is the height of column C as measured from the pipe at C.

As in my previous post, I will treat the vertical distance from the floor to the pipe at A as being 0, so h_a is also the height of column A as measured from the floor, but h_C is not. As in my previous post, let h_c be the height of column C from the floor. Note that

(4) h_c = h_ac + h_C.

Now, plug in (2) and (3) into (1). We get

(5) (rho)(g)(h_a) = (rho)(g)(h_C) + (rho)(g)(h_ac)

which reduces to

(6) h_a = h_C + h_ac

when you divide out (rho)(g). Finally, (4) and (6) imply that

(7) h_a = h_c.

In other words, column A and column C are the same height when measured from the floor.

 

[sps27]

I'm pretending that the vertical section of the pipe and the center of C are part of column C.

In fact, if you bent the pipe so that sections A, B, and C were level, the height of the water column above C as measured from the floor would not change.

If you are unconvinced:

You are right that, while the pressure term due to velocity is the same at A and C, the height difference means that A is at a higher pressure than C. This means that column C will be shorter off the pipe than column A, but column C is higher off the ground than column A by exactly that amount.

Let's do some math to confirm. Let h_ac be the difference between the heights of the pipe at A and C. Then the difference in pressures at A and C is (rho)(g)(h_ac). In other words,

(1) P_a = P_C+ (rho)(g)(h_ac)

where P_a is the pressure at A and P_C is the pressure at C. (P_C is not the same as P_c from my previous post, because they're pressures at different points.)

We have another equation for each P_a and P_C, namely,

(2) P_a = (rho)(g)(h_a)
(3) P_C = (rho)(g)(h_C)

where h_a is the height of column A as measured from the pipe at A and h_C is the height of column C as measured from the pipe at C.

As in my previous post, I will treat the vertical distance from the floor to the pipe at A as being 0, so h_a is also the height of column A as measured from the floor, but h_C is not. As in my previous post, let h_c be the height of column C from the floor. Note that

(4) h_c = h_ac + h_C.

Now, plug in (2) and (3) into (1). We get

(5) (rho)(g)(h_a) = (rho)(g)(h_C) + (rho)(g)(h_ac)

which reduces to

(6) h_a = h_C + h_ac

when you divide out (rho)(g). Finally, (4) and (6) imply that

(7) h_a = h_c.

In other words, column A and column C are the same height when measured from the floor.

Thanks bud! I do appreciate your painstaking analysis to explain this. I understand your last post much better. Thank you!