Fluid Dynamics: Waterflow out of a Tank

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The discussion focuses on fluid dynamics related to a cylindrical tank with a hole at the bottom, addressing three main questions about the relationship between water flow speeds and heights. The first question explores the relationship between the speed of water flowing out of the hole and the speed at which the water level drops, with the user deriving a formula involving conservation of mass. The second question pertains to the time dependency of the water height, leading to the conclusion that the speed of the water level drop can be expressed as the derivative of height with respect to time. The third question involves calculating the distance the water travels horizontally before hitting the ground, treated as a projectile motion problem. The user is seeking guidance on how to further develop these relationships and calculations.
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Hi, this is some questions about fluid dynamics(mostly). There are three somewhat connected questions here, I will try to organize it as best as I can.

Homework Statement


A sylindrical tank filled with water is standing on a table, the tank has a small hole at the side of the tank at the very bottom(not underneath). The tank is placed on the table so that the water coming out of the hole will drop directly to the floor(it won't touch the table).
Variables:
Hight of table(hight from the hole in the tank to the ground) = H
Hight of waterlevel inside the tank = h
(inside)Radius of hole at the bottom of the tank = r
(inside)Radius of the tank = R
Speed at which the waterlevel inside the tank drops = V
Speed of the water flowing out of the hole = v
Distance on the floor, from the floor directly beneath the hole to where the water hits the floor = L.
I've added a sketch of the variables.

mekoblig2.jpg

The airpressure is constant everywhere, g is gravity and works in the negative y direction, and the density of the water is constant.

Q1: What is the relation between the speed v and the speed V?
Q2: The speed V is related to the timedependancy of hight h by some differentiate. What is the relation?
Q3: What is the length L as a function of hight h?

Homework Equations


- Bernoullis equation for incompressible liquid(see attempt at solution).

The Attempt at a Solution



A1: The relation can be given by \rho t v \pi r^2 = \rho t V \pi R^2 = constant
Wich means vr^2 = VR^2 = constant. I don't know what more i need to do in order to show the relation, is this enough?
A2: I don't know how to solve this.
A3: I used Bernoullis equation to get a model for v: v = \sqrt{2gh}. So now I have the speed v out of the hole(i think), but don't know how to get from there. It is supposed to be a function of hight h, so I think I'm not supposed to mix time into it, in which case it would be easier.

I'm mostly interested in a small push in the right direction, and any help would be much appreciated=)

Edit: I forgot to say the mass is conserved!
 
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Q2: The speed V is related to the timedependancy of hight h by some differentiate. What is the relation?

V is the speed the water level drops. Look at its units. What derivative in variable h has the same units and, of course, defines the same thing.

Q3: What is the length L as a function of hight h?
This is a typical projectile being shot off a cliff problem.
 
LawrenceC said:
Q2: The speed V is related to the timedependancy of hight h by some differentiate. What is the relation?

V is the speed the water level drops. Look at its units. What derivative in variable h has the same units and, of course, defines the same thing.

Q3: What is the length L as a function of hight h?
This is a typical projectile being shot off a cliff problem.

Hmm, okey, so given that my A1 is correct, I would think I could solve Q2 like this:

Velocity is the derivative of position with respect to time, so if h is the displacement:

\frac{dh}{dt} = V(t)

If I am correct in A1, then vr^2 = VR^2 which means \frac{vr^2}{R^2} = V(t) which again means; \frac{dh}{dt} = \frac{vr^2}{R^2}. Is this correct?

And for Q2: I haven't really been doing any cannonball off cliffs problems, but I would solve it as:

v(t) = \sqrt{2gh} \Rightarrow position = t\sqrt{2gh} which represents the position of the water in x-direction as a function of time. And:
acc = -g \Rightarrow Velocity = -gt \Rightarrow position = -\frac{1}{2}gt^2 which represents the position of the water in y-direction, giving me:

position = x(t) = \sqrt{2gh} \boldsymbol{i} - \frac{1}{2}gt^2 \boldsymbol{j} Wich I guess would be nice for a graph. But how do I go from there? Would it be possible to solve it as a quadratic equation (?) :

x(t) = -gt^2 + \sqrt{2gh}t + H, and pick the positive root?
 
Q2: The speed V is related to the timedependancy of hight h by some differentiate. What is the relation?

Unless I am misunderstanding the question, I would simply answer by saying V = dh/dt.

For the water passing through the air part, first determine how long it takes the water to reach the ground. The basic assumption is that the horizontal velocity is constant. The portions of these problems are solved separately. You already have written the correct formula to determine the time for it to reach the ground - y-direction computation for t. Solve for t and use it to determine range L.
 
LawrenceC said:
Q2: The speed V is related to the timedependancy of hight h by some differentiate. What is the relation?

Unless I am misunderstanding the question, I would simply answer by saying V = dh/dt.

For the water passing through the air part, first determine how long it takes the water to reach the ground. The basic assumption is that the horizontal velocity is constant. The portions of these problems are solved separately. You already have written the correct formula to determine the time for it to reach the ground - y-direction computation for t. Solve for t and use it to determine range L.

I'm going to work a bit more on it tomorrow=) Thanks for all the help!
 
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