Movement of the water level in the barrel - calculate its velocity, etc.

[Lotto]

Homework Statement

The barrel of a cylindrical shape containing the $A_1$ cross-section has a discharge hole in the bottom containing the $A_2$ cross-section. $A_1$ is much bigger than $A_2$. Water's coming out through the opening. At time $t_0 = 0$, the level has a height of $h_0$ above the bottom of the barrel.

Calculate the level's velocity as a function of time, its acceleration and the time when the barrel is empty.

Relevant Equations

$A_1 \frac{\mathrm d h}{\mathrm d t}=-A_2\sqrt{2hg}$

Here is only my solution:

$A_1 \frac{\mathrm d h}{\mathrm d t}=-A_2\sqrt{2hg}$,

so by integrating we get

$h(t)=\left(\sqrt{h_0}-\frac{A_2}{2A_1}\sqrt{2g} t\right)^2.$

Setting $h(T)=0$ we get

$T=\frac{A_1}{A_2}\sqrt{\frac{2h_0}{g}}.$

By doing the first time derivative of $h$ we get

$v(t)=\frac{{A_2}^2}{{A_1}^2}gt-\frac{A_2}{A_1}\sqrt{2h_0 g}.$

$v(t)$ has negative value because it points downward.

By doing the first time derivative of $v$ we get

$a(t)=\frac{{A_2}^2}{{A_1}^2}g.$

It is correct? Is there an "easier" way to solve it?

 

[erobz]

Homework Statement: The barrel of a cylindrical shape containing the $A_1$ cross-section has a discharge hole in the bottom containing the $A_2$ cross-section. $A_1$ is much bigger than $A_2$. Water's coming out through the opening. At time $t_0 = 0$, the level has a height of $h_0$ above the bottom of the barrel.

Calculate the level's velocity as a function of time, its acceleration and the time when the barrel is empty.
Relevant Equations: $A_1 \frac{\mathrm d h}{\mathrm d t}=-A_2\sqrt{2hg}$

Here is only my solution:

$A_1 \frac{\mathrm d h}{\mathrm d t}=-A_2\sqrt{2hg}$,

so by integrating we get

$h(t)=\left(\sqrt{h_0}-\frac{A_2}{2A_1}\sqrt{2g} t\right)^2.$

Setting $h(T)=0$ we get

$T=\frac{A_1}{A_2}\sqrt{\frac{2h_0}{g}}.$

By doing the first time derivative of $h$ we get

$v(t)=\frac{{A_2}^2}{{A_1}^2}gt-\frac{A_2}{A_1}\sqrt{2h_0 g}.$

$v(t)$ has negative value because it points downward.

By doing the first time derivative of $v$ we get

$a(t)=\frac{{A_2}^2}{{A_1}^2}g.$

It is correct? Is there an "easier" way to solve it?

Procedurally it looks ok to me. I haven't checked the math in every part to final result(but they look reasonable). The criterion $A_1 \gg A_2$ neglects acceleration of the fluid body inside the barrel. Also, in reality at $t = 0 , v = 0$ which is not reflected in the simplification. As for a simpler method, not that I know of.

 

[kuruman]

$h(t)=\left(\sqrt{h_0}-\frac{A_2}{2A_1}\sqrt{2g} t\right)^2.$

By doing the first time derivative of $h$ we get

$v(t)=\frac{{A_2}^2}{{A_1}^2}gt-\frac{A_2}{A_1}\sqrt{2h_0 g}.$

The time derivative doesn't look right. Remember the chain rule.

 

[Lotto]

The time derivative doesn't look right. Remember the chain rule.

I think it is right. I calculated it again and got the same result.

 

[kuruman]

I think it is right. I calculated it again and got the same result.

Yes, it is right. I initially misread $t$ as being under the radical, but that would make the expression dimensionally incorrect. Sorry about the confusion.

 

[haruspex]

The Bernoulli equation you started with is for steady flow, so no acceleration of the flow over time Yet you find there is an acceleration. Also, it is evident that the downward acceleration cannot be constant. It will decline as the level drops, eventually being negative.

This is, in fact, a very difficult problem. See posts #17, #19 at https://www.physicsforums.com/threads/velocity-of-efflux-out-of-a-water-tank.868030/#post-5455954. It is unlikely there is a closed form for the motion as a function of time.

 

[erobz]

The Bernoulli equation you started with is for steady flow, so no acceleration.

Perhaps just a clarification for the OP: "convective" acceleration is fine in Bernoulli's i.e. flow can accelerate along a streamline depending on pipe geometry. "Steady flow" in this case means the flow velocity at any fixed position along a streamline is not changing in time. The latter is the criterion @haruspex is saying it fails.

 

[haruspex]

Perhaps just a clarification for the OP: "convective" acceleration is fine in Bernoulli's i.e. flow can accelerate along a streamline depending on pipe geometry. "Steady flow" in this case means the flow velocity at any fixed position along a streamline is not changing in time. The latter is the criterion @haruspex is saying it fails.

Thanks, I've clarified it.

 

[kuruman]

I have a question about this. I understand that the steady flow Bernoulli equation is not applicable here. The link provided by @haruspex in @Chestermiller's post #6 has the equation $$g h(t)=\frac{v_x^2(t)}{2}\left[1-\left(\frac{a}{A}\right)^2\right]+h(t)\frac{a}{A}\frac{dv_x(t)}{dt}\tag{9}.$$ The first term is what gets from the steady flow Bernoulli equation. In this problem we are told that $A_1<<A_2$ which translates to $a<<A$. Why can we not ignore the second term relative to the first in Equation (9) and proceed with the solution? In other words, use the approximation that the steady flow form of Bernoulli's equation is applicable when the area of the exit orifice is much smaller area than the tank's area. It might be a simplifying approximation to a complex situation, but so is ignoring air resistance in projectile motion.

 

[erobz]

I have a question about this. I understand that the steady flow Bernoulli equation is not applicable here. The link provided by @haruspex in @Chestermiller's post #6 has the equation $$g h(t)=\frac{v_x^2(t)}{2}\left[1-\left(\frac{a}{A}\right)^2\right]+h(t)\frac{a}{A}\frac{dv_x(t)}{dt}\tag{9}.$$ The first term is what gets from the steady flow Bernoulli equation. In this problem we are told that $A_1<<A_2$ which translates to $a<<A$. Why can we not ignore the second term relative to the first in Equation (9) and proceed with the solution? In other words, use the approximation that the steady flow form of Bernoulli's equation is applicable when the area of the exit orifice is much smaller area than the tank's area. It might be a simplifying approximation to a complex situation, but so is ignoring air resistance in projectile motion.

I think you can ignore it under the assumption $A_1 \gg A_2$ ( $A_1$ is the barrel cross section, $A_2$ the hole ) just as you say. Maybe I didn't state that clearly when I brought it up, but my only objection was just to make note that the time varying flow was not strictly valid. I felt it was somewhat "hand wavy" to use that model for this type of problem without mentioning it.

 

[haruspex]

I have a question about this. I understand that the steady flow Bernoulli equation is not applicable here. The link provided by @haruspex in @Chestermiller's post #6 has the equation $$g h(t)=\frac{v_x^2(t)}{2}\left[1-\left(\frac{a}{A}\right)^2\right]+h(t)\frac{a}{A}\frac{dv_x(t)}{dt}\tag{9}.$$ The first term is what gets from the steady flow Bernoulli equation. In this problem we are told that $A_1<<A_2$ which translates to $a<<A$. Why can we not ignore the second term relative to the first in Equation (9) and proceed with the solution? In other words, use the approximation that the steady flow form of Bernoulli's equation is applicable when the area of the exit orifice is much smaller area than the tank's area. It might be a simplifying approximation to a complex situation, but so is ignoring air resistance in projectile motion.

True, but as I noted in post #6, the acceleration cannot be constant. There must an initial downward acceleration, but later it is upward. The equations at the link I posted shed no light on this; when the velocity is zero the starting equation collapses to 0=0. If I model the equation obtained for the velocity, at $h=h_0$ it produces zero, so nothing moves.

Consider a thin element of water right at the exit when the hole appears. It is suddenly subject to a non infinitesimal pressure difference. This impulse should produce an 'instantaneous' velocity of $\sqrt{gh_0}$. Incompressibility implies the surface of the tank water instantaneously acquires a downward velocity of $\frac aA\sqrt{gh_0}$. It would seem that a full understanding would involve e.g. the speed of sound in water.
But that kick-start is enough. Modelling the equations from there, for a<<A, the tankwater acceleration immediately settles down to $\frac{a^2}{A^2}g$ upwards, as found in post #1.

There is another change in behaviour when the tank is nearly empty. The model says the upward acceleration starts to increase, even for a<<A. This adds roughly $\frac 12\sqrt{\frac {h_0}g}$ to the time to empty. However, the model would be inaccurate there anyway since we would now have horizontal flows in the tank converging on the exit.

 

[erobz]

I've solved this numerically for a bottle rocket thrust not too long ago. In that problem the heavily transient phase is over quite rapidly, but then again, the entire ejection is over quite rapidly too.

I can't set $P_o$ to 1 atm, because of my gas expanding to vacuum (relative to atmospheric) due to the sealed top, but if I change $D_b$ to 200 cm so that it 2 orders of magnitude than the nozzle diameter, the graphs are approaching each other very rapidly. How significant it is surely depends on the actual dimensions of the hole and barrel which we know nothing about.