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Negative velocity squared ?

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>A Venturi tube without manometer is given with initial area $A$ and area in its throat as $a$, initial pressure as $P_1 = 3 \ \ \mathrm{atm}$ and $A = 5a$, Find velocity of water as it enter the tube.

Figure

by equation of continuity :-

$$Av_1 = av_2 \implies 5av_1 = av_2 \implies 5v_1 = v_2 \tag{1}$$

Since $\Delta F = 0$
Thus, $$P \propto \frac1A \implies P_1 A = P_2a \implies P_1\times 5a = P_2a \implies 5P_1 = P_2 \tag{2}$$

By Bernoulli's principle :-

$$\Delta P + \rho g\Delta h + \frac12\rho\Delta v^2 = 0$$
$$(P_2 - P_1) + \frac12\rho(v_2^2 - v_1^2) = 0$$

From 1 and 2,

$$(5P_1 - P_1) + \frac12\rho(25v_1^2 - v_1^2) = 0$$
$$(P_1) + \rho(3v_1^2) = 0$$

$$\sqrt{-\frac{P_1} {3\rho }} = v_1$$

Now this is impossible and I am a idiot to get to this result but even after staring at this for, I can't get spot the flaw. Surely some of you can do that.

I am grateful for any help .

asked Jan 22, 2017 in Physics Problems by Motu (250 points)
edited May 11, 2018 by sammy gerbil
$P_1>P_2$. An increase in velocity results in a decrease in pressure. See http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html.
The question does not make much sense. You only have 2 equations (the Bernoulli Equation and the Continuity Equation), so you can only calculate 2 unknowns. The trouble is that you have 3 unknowns : $P_2, v_1, v_2$.

Your use of $P \propto \frac{1}{A}$ is incorrect.
So it should be $P \propto A$, but does not it conflicts with the definition of pressure $(p=F/A)$ ??
What is $F$? How do you know that $F$ is constant? ie that $\Delta F=0$?
$F$ is the force, I thought the force is constant because the force is only applied from one side of the tube, so it must be constant. right ?
What exactly is this force acting on? And what happens to it when the tube gets narrower at the restriction? If this is the only force acting here, why doesn't the water (or the tube) accelerate? It is not obvious what forces are acting in this situation, nor why $F$ should be constant at different parts of the tube. It is straightforward how to apply $P=F/A$ in a static situation - eg connected cylinders with pistons; then $P$ is constant but $F$ is not. It is not obvious to me why $F$ should be constant in this case.

As in my 1st comment here, if $P=F/A$ can be applied here and $F$ is constant then we should have $P_2 \gt P_1$ because $A_2 \lt A_1$. This contradicts Bernoulli's Equation which predicts that $P_1 \gt P_2$.
So it should be $P\propto A$, $P_1/A=P_2/a => P_1/5a=P_2/a=> P_1=5P_2$. right ?
Why should it be $P \propto A$? Where are you getting that from? Why not $P \propto A^2$?

No, as I said in my 2nd comment, the question does not make sense to me. There is not enough information to solve it.
Yes your comment seems correct to me. I have no answer why it should be $P \propto A$ and $P \propto A^2$, I never thought that. You have a strong understanding in Physics, wish you were my teacher. I am sorry for my stupid comment.

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