# Waterflow out of a tank

by Noorac
Tags: tank, waterflow
 P: 13 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. 1. The problem statement, all variables and given/known data 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 wich 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. 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? 2. Relevant equations - Bernoullis equation for incompressible liquid(see attempt at solution). 3. 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 wich 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!
 P: 1,195 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.
P: 13
 Quote by LawrenceC 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 $vr^2 = VR^2$ wich means $\frac{vr^2}{R^2} = V(t)$ wich 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:

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

 P: 1,195 Waterflow out of a tank 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.
P: 13
 Quote by LawrenceC 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 gonna work a bit more on it tomorrow=) Thanks for all the help!

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