Question about venturimeter formula

[songoku]

TL;DR Summary

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The level of this learning material is high - school level

From my note, there are two equations that can be used related to venturimeter:

1. $P+\rho gh + \frac{1}{2} \rho v^2 = \text{constant}$

2. $P_1 + \frac{1}{2} \rho {v_1}^2 = P_2 + \frac{1}{2} \rho {v_2}^2$

I want to ask about second equation, how can we get that equation? My teacher said it is also from Bernoulli equation and we just cross out the $\frac{1}{2} \rho v^2$ part but I do not understand why. Don't the liquid have different velocity ($v_1$ and $v_2$) so why we can cross that part?

Thanks

 

[etotheipi]

Let $h_1 = h_2$!

 

[kuruman]

The second equation is really the first equation in the case of horizontal flow.
The first equation gives $$P_1+\rho gh_1 + \frac{1}{2} \rho v_1^2 = P_2+\rho gh_2 + \frac{1}{2} \rho v_2^2$$. Then set $h_1=h_2$ as @etotheipi suggested.

 

[songoku]

I am really sorry, I mistyped the second equation

The second equation should be: $P_1 + \rho gh_1 = P_2 + \rho gh_2$

That is the reason why I wrote the teacher crossed out the $\frac{1}{2} \rho v^2$ part and I don't understand why it can be done so.

I apologize once again

Thanks

 

[etotheipi]

Let $v_1$ = $v_2$! nevermind

 

[songoku]

Let $v_1$ = $v_2$!

Why? From flow rate continuity equation: $A_1 . v_1 = A_2 . v_2$ and $A_1 \neq A_2$ so why we can set $v_1 = v_2$?

Thanks

 

[kuruman]

Let $v_1$ = $v_2$!

That would mean $A_1=A_2$ which is not the case here.

Why? From flow rate continuity equation: $A_1 . v_1 = A_2 . v_2$ and $A_1 \neq A_2$ so why we can set $v_1 = v_2$?

Thanks

We can't. Either you misunderstood what the teacher said, or the teacher was incorrect.

 

[etotheipi]

That would mean $A_1=A_2$ which is not the case here.

Wait, where are we getting that from? I haven't seen any areas mentioned. If $A_1 = A_2$, then $v_1 = v_2$.

 

[kuruman]

Wait, where are we getting that from? I haven't seen any areas mentioned. If $A_1 = A_2$, then $v_1 = v_2$.

We get that from the figure posted by OP.

 

[etotheipi]

We get that from the figure posted by OP.

Whoops, missed that. In that case, teacher is wrong.

 

[songoku]

We can't. Either you misunderstood what the teacher said, or the teacher was incorrect.

The teacher said something like: the second equation is for the vertical pipes where the heights (or depth, I am not sure) of the liquid are different (one is $h_1$ and the other is $h_2$)

 

[kuruman]

Whoops, missed that. In that case, teacher is wrong.

Perhaps teacher canceled the wrong terms thinking that the speeds are the same since the flow is horizontal and there is no change in gravitational potential energy per unit volume. If teacher is correct, then it is also true that $h_1=h_2$ which also means that $P_1=P_2$ which also means the $1=2$, i.e. we are at the same section of the pipe. That's the trivial solution.

 

[etotheipi]

I think teacher makes a huge mess of this problem.

It should be noted by the OP that the $h_1$ and $h_2$ drawn in the figure are not the same as the $h_1$ and $h_2$ that appear in the Bernoulli equation. The heights drawn in the figure only serve to represent the pressure difference. The most simplifying you can do is$$\frac{1}{2} \rho v_1^2 + p_1 = \frac{1}{2}\rho v_2^2 + p_2$$

 

[kuruman]

I am really sorry, I mistyped the second equation

The second equation should be: $P_1 + \rho gh_1 = P_2 + \rho gh_2$

That is the reason why I wrote the teacher crossed out the $\frac{1}{2} \rho v^2$ part and I don't understand why it can be done so.

I apologize once again

Thanks

If $h_1$ and $h_2$ are as shown in the figure, then it seems to me that $P_1=\rho g h_1$ and likewise for index $2$ so that the Bernoulli equation for the two parts of the horizontal tube should be
$$\frac{1}{2}\rho v_1^2 + \rho gh_1 = \frac{1}{2}\rho v_2^2 + \rho gh_2.$$

 

[songoku]

It should be noted by the OP that the $h_1$ and $h_2$ drawn in the figure are not the same as the $h_1$ and $h_2$ that appear in the Bernoulli equation.

Yes, they are not the same

If $h_1$ and $h_2$ are as shown in the figure

Yes, they are the height / depth of liquid in small vertical tubes measured from "horizontal line" that makes both "$h_1$" and "$h_2$" in Bernoulli equation have same value

Perhaps teacher canceled the wrong terms thinking that the speeds are the same since the flow is horizontal and there is no change in gravitational potential energy per unit volume. If teacher is correct, then it is also true that $h_1=h_2$ which also means that $P_1=P_2$ which also means the $1=2$, i.e. we are at the same section of the pipe. That's the trivial solution.

I am not sure if this helps. There is derivation of formula of speed $v_1$ in my note:

1. At horizontal pipe, the height will be the same so the equation can be reduced to:
$P_1 + \frac{1}{2} \rho {v_1}^2 = P_2 + \frac{1}{2} \rho {v_2}^2$

2. At vertical pipe, $P_1 + \rho gh_1 = P_2 + \rho gh_2$ where $h_1$ and $h_2$ are the depth of the liquid in the two vertical pipes (based on the picture, $h_1 > h_2$)

3. Flow rate continuity equation: $A_1 . v_1 = A_2 . v_2 \rightarrow v_2 = \frac{A_1}{A_2}.v_1$

From part (1):
$P_1 - P_2 = \frac{1}{2} \rho ({v_2}^2 - {v_1}^2)$
$P_1 - P_2 = \frac{1}{2} \rho \left[ \left( \frac{A_1}{A_2} \right)^2 .{v_1}^2 - {v_1}^2 \right]$

From part (2):
$P_1 - P_2 = \rho g (h_2 - h_1)$

$P_1 - P_2 = \rho g \Delta h$ where $\Delta h$ is the difference between vertical height of liquid in vertical pipes

Equating those two equations:
$\frac{1}{2} \rho \left[ \left( \frac{A_1}{A_2} \right)^2 .{v_1}^2 - {v_1}^2 \right] = \rho g \Delta h$

After simplifying, we get:
$v_1=\sqrt {\frac{2g \Delta h}{\left( \frac{A_1}{A_2} \right)^2 -1}}$

So based on the derivation, this equation can be obtained by using equation from part (2)

Thanks

 

[kuruman]

Are we abandoning the horizontal venturimeter questione you posted in #1?

When you say "vertical pipe" what is the complete picture? Fluid enters the pipe and then what? Is the fluid in free fall going down or pushed in from the bottom at some pressure? What do $A_1$ and $A_2$ in post #15.3 represent? If a steady stream of an incompressible fluid is in free fall, as it accelerates its cross-sectional area decreases (see figure) because the speed increases.

It seems to me that this thread started from an explanation of an equation related to horizontal flow and has been transformed to a question about vertical flow. Can you please clarify what you wish to understand?

 

[songoku]

Are we abandoning the horizontal venturimeter questione you posted in #1?

No we aren't. My question is related to that picture

When you say "vertical pipe" what is the complete picture? Fluid enters the pipe and then what? Is the fluid in free fall going down or pushed in from the bottom at some pressure?

I post new picture to describe the explanation as best as I can

I think (1) and (2) is the position of $P_1$ and $P_2$

What do $A_1$ and $A_2$ in post #15.3 represent?

$A_1$ is the area of larger horizontal pipe and $A_2$ is the area of smaller horizontal pipe

Thanks

 

[kuruman]

No we aren't. My question is related to that pictureI post new picture to describe the explanation as best as I can
View attachment 271748
I think (1) and (2) is the position of $P_1$ and $P_2$$A_1$ is the area of larger horizontal pipe and $A_2$ is the area of smaller horizontal pipe

Thanks

Then what I said in post #14 applies. If you know the input speed $v_1$, you can use the equation shown therein to solve for $v_2$. Is that what you needed to do?

 

[songoku]

Then what I said in post #14 applies. If you know the input speed $v_1$, you can use the equation shown therein to solve for $v_2$. Is that what you needed to do?

No, I just want to know how to get the second equation. My teacher did like this:

At vertical pipe:
$P_1 + \rho gh_1 + \frac{1}{2} \rho {v_1}^2 = P_2 + \rho gh_2 + \frac{1}{2} \rho {v_2}^2$

Then he crossed out $\frac{1}{2} \rho {v_1}^2$ and $\frac{1}{2} \rho {v_2}^2$ and ended up with $P_1 + \rho gh_1 = P_2 + \rho gh_2$. I want to know why we can cross these 2 terms out

Thanks

 

[kuruman]

No, I just want to know how to get the second equation. My teacher did like this:

At vertical pipe:
$P_1 + \rho gh_1 + \frac{1}{2} \rho {v_1}^2 = P_2 + \rho gh_2 + \frac{1}{2} \rho {v_2}^2$

Then he crossed out $\frac{1}{2} \rho {v_1}^2$ and $\frac{1}{2} \rho {v_2}^2$ and ended up with $P_1 + \rho gh_1 = P_2 + \rho gh_2$. I want to know why we can cross these 2 terms out

Thanks

I don't think "we" can cross these terms out. We went over that in posts #7 - #10. If your teacher did it, you should ask him/her why it was done.

 

[songoku]

I don't think "we" can cross these terms out. We went over that in posts #7 - #10. If your teacher did it, you should ask him/her why it was done.

Ok so for now I can say that the derivation in post #15 is not valid?

Thanks

 

[kuruman]

Ok so for now I can say that the derivation in post #15 is not valid?
Thanks

The derivation in #15 is almost correct if your task is to find $v_1$ in terms of the geometry and the height difference in the vertical columns. The problem with your derivation is this
You say in #15(1) $P_1 + \frac{1}{2} \rho {v_1}^2 = P_2 + \frac{1}{2} \rho {v_2}^2$.
You know that $v_2>v_1$ because $A_2<A_1$. It follows that, for the equality to hold, it must be true that $P_2<P_1.$

Then you say in #15(2) $P_1 + \rho gh_1 = P_2 + \rho gh_2$. Note that from the figure $h_2<h_1$ and we just concluded that $P_2<P_1$. This second equality cannot be correct because the two smaller terms are on the same side. If you must use the Bernoulli equation for a static vertical column, you must write it at two points in the same column. Say we pick column 1, and choose as point A the surface of the fluid that is at atmospheric pressure and point B at the axis of the tube. Then, starting from the general Bernoulli equation,$$P_A + \rho gh_A + \frac{1}{2} \rho {v_A}^2 = P_B + \rho gh_B + \frac{1}{2} \rho {v_B}^2$$Here is what your teacher did, I think. The fluid in column 1 is not moving vertically, therefore $v_1=v_2=0$. The model here is that you have a static vertical column of fluid standing on a horizontally moving tube of fluid. The gauge pressure at $A$ is $P_A=0$ and $P_B=P_1$; the heights are $h_A=h_1$ and $h_B=0.$ WIth all this, the equation becomes $$0 + \rho gh_1 +0= P_1 + 0 + 0~\Rightarrow~P_1=\rho gh_1.$$You do the same with a second Bernoulli equation for the other vertical columns and get $P_2=\rho g h_2$. Finally you write a third Bernoulli equation for the horizontal fluid that is moving and connect points 1 and 2. That third equation is what I put down in #14 which you have to solve for $v_1$.

 

[songoku]

Thank you very much etotheipi and kuruman