# Differential Equations Problems: Modeling, 2nd Order Linear w/ NonConstant Coeff.

• Jun 6th 2010, 08:37 PM
justaguy
Differential Equations Problems: Modeling, 2nd Order Linear w/ NonConstant Coeff.
Hey,

I was wondering if you guys could help me out with a couple of differential equations problems that have been driving me crazy.

In the first, I'm told that there is a hemispherical tank of radius four feet filled with water, and that at time zero, a circular hole of diameter one inch is opened at the bottom of the tank. I'm asked to find the time it takes the tank to drain.

I have no idea how to even model this one. I can't figure out the physics involved. I'm sure it must be something simple that I'm missing, but I don't know what. I've only taken the introductory physics series, so the only equations that I know which relate to fluid flow are Bernoulli's equation and the equation of continuity. I thought maybe I could use the fact that the volume rate of flow is equal to the area of the opening times the velocity of the fluid passing through the opening, but that still leaves the problem of finding the velocity of the fluid passing through the opening. Obviously, Torricelli's Theorem doesn't apply since the velocity of the water at the top of the tank will approach the velocity of the water passing through the opening as the water level falls. And even if it did apply, I would only be able to solve for the velocity in terms of the water level, which is itself an unknown function of time.

The second problem is finding the general solution to a second order linear ODE with nonconstant coefficients. The equation is $x^2 y'' - (x - 327/400) y = 0$. This doesn't fit any general forms for solvable second order linear ODEs that I can find. Hopefully, there's a method that can be applied to this equation that I've missed, and one of you knows it.

• Jun 7th 2010, 03:18 AM
Ackbeet
Bernoulli All the Way
I think your intuition about Bernoulli is right. Recall Bernoulli's equation:

$\frac{v^{2}}{2}+gz+\frac{p}{\rho}=\text{constant},$

where:

$v$ is the fluid flow speed at a point on the streamline,
$g$ is the acceleration due to gravity,
$z$ is the elevation of the point (the one that has velocity $v$) above a reference plane, with the positive $z$ direction upwards
$p$ is the pressure at the point, and
$\rho$ is the density of the fluid at all points in the fluid.

I just got this equation from the wiki. You should definitely double-check this against a textbook.

So, how could you get a handle on each of these variables? Going down the list:
$v$ might be something of an intermediate target variable (a variable for which you wish to solve). If you knew $v$, you could, as you say, find the volume rate of flow, which could be integrated, I think, to find the amount of fluid left in the tank at any particular time. So $v$ would be nice to know. Let's treat that as the target variable for now, then. $g=9.8\;[\text{m}/\text{s}^{2}]$. I think you'd be wise to choose $z$ to be at the orifice, the circular hole drilled in the bottom; that is, after all, the point at which you'd like to know $v$. So $z$ isn't going to change.
Now the pressure is going to change depending on how much water is above the circular hole, right? Perhaps you remember doing force calculations of various kinds of tanks in Calculus II? That's a fairly standard problem, because it illustrates the Calc II method very nicely. Obviously, as the water drains out, you'll have less and less water above the hole, and the pressure should get less and less. (How do you relate force to pressure? The usual equation is $p=\frac{F}{A}$, where $A$ is the area over which the force is applied.) So that's what changes with time. The last parameter, $\rho$ will, I think, be constant, since you're dealing with liquid water, which is mostly incompressible. You can look up the density of water in the units you choose to use.

So hopefully, this starts to get more clear: you've got Bernoulli's equation, which can relate velocity to pressure. What about the constant on the RHS? I'd just plug in known values of all variables for one instant in time and find out what the constant is. Make sense?
• Jun 7th 2010, 05:27 AM
justaguy
Ackbeet,

Thank you for the reply. It really does seem like this problem should be solved using Bernoulli's equation. I tried to follow your advice, and this is where I am currently:

Starting with Bernoulli's (The equation you pulled from Wiki is correct):

$v^2/2+gz+P/d=constant$

I'm solving for the velocity at the orifice, so $z=0$:

$v^2/2+P/d=constant$

In order to solve for the constant on the RHS, I use the initial state of the system, i.e., $v=0$ when $P=dgh=(1.94 slug/ft^3)(32.2 ft/sec^2)(4 ft)=249.9 lb/ft^2$. Therefore,

$P/d=constant=128.8ft^2/sec^2$

Substituting back into Bernoulli's:

$v(t)^2/2+P(t)/d=128.8$

Solving for v(t):

$v(t)=Sqrt[257.6-1.03P(t)]$

The problem is, the velocity at the orifice (and thus $dV/dt$) varies with time, and to use Bernoulli's equation to solve for the velocity as a function of time, I would need the pressure as a function of time.

(If there are any errors in my thinking or calculations, it's because it's five-thirty in the morning and I haven't slept yet. I can be obsessive.)
• Jun 7th 2010, 05:59 AM
Ackbeet
Hmm. You can use English units if you like - I find them generally a bit cumbersome.

Several comments: 1. You might consider a different choice of origin. While the form of Bernoulli's equation that we've both quoted assumes $z$ is positive up, it makes no assumption regarding choice of origin. Whenever I see anything spherical, I usually find that the origin is best chosen to be the center of the sphere.

2. I don't think your initial condition is correct. Think about it: the hole has just been cut in the bottom of the tank. That's the situation you're dealing with; the tank is assumed not to change geometry until after the water has finished pouring out. The initial velocity of the water out of the tank is not going to be zero. I would use the final condition, actually, to solve for the constant. Then you know the velocity is zero, because there's no water left.

3. You can get the pressure from Calculus II techniques. You know the geometry of the tank, which means you can compute the force at any point on the tank. Once you know the force, divide by the area of the hole at the bottom to get the pressure. That will give you pressure as a function of the height of the water.

See where this leads.
• Jun 7th 2010, 01:46 PM
justaguy
Yeah, I don't like working in FPS either, but it's what the problem was given in, so I figured I'd just stick with it.

I tried working with the origin at the top of the hemisphere, and it simplified the equations a little, but it didn't really resolve anything. And you're absolutely right about my initial condition, of course the velocity isn't zero at time zero (that was pretty dumb of me, but what can I say, it was late). As for the pressure, we don't even really need to use calculus. The fluid pressure is equal to the density of the water times the acceleration due to gravity times the height, so right there we have water pressure as a function of height. One mistake I think we might be making is including the water pressure term twice. Bernoulli's equation already takes into account the water pressure with the term dgh, or, if we divide through by the density, gh. I believe the pressure term in the equation ought to be just the external pressure to which the orifice is exposed, i.e., atmospheric pressure. Having done all this, I'm still confronted with the same problem. Whenever I solve my equations, I'm left with a function in terms of geometry (the height), but I need a function in terms of time. The problem becomes figuring out how the height changes in time, which is obviously physically determined, but I just can't figure out by what law/principle. This has really got me stumped.
• Jun 7th 2010, 02:08 PM
Ackbeet
I think you may be right about the pressure in Bernoulli's equation - I'm not an expert on it.

With regard to getting the height as a function of time: is it not true that $\frac{dV}{dt}=v\cdot A$, where $V$ is the volume of fluid passing through the orifice as a function of time, $v$ is the velocity of the fluid going through the orifice, and $A$ is the area of the orifice? Now, you can find $V(h)$, correct? It'll be something you use Calc II techniques for, I imagine. So, what you really have then is $\frac{dV(h(t))}{dt}=v\cdot A=\frac{dV}{dh}\,\frac{dh}{dt}$. Then you can rearrange and integrate to find $h(t)$.
• Jun 7th 2010, 02:14 PM
justaguy
Yeah, that seems promising. Let me play around with it and see if it works out.
• Jun 7th 2010, 03:00 PM
justaguy
I think I've got it! The volume flow rate is, as you said, the product of area and velocity. We can relatively easily solve for the area of the surface of the fluid as a function of the height (defining our coordinate system at the bottom of the hemisphere--I think it's easier to work with Bernoulli's equation this way): $A(surface) = Pi*h(h-8)$. The velocity of the surface is simply $dh/dt$. This is equal to the negative (since the height is decreasing) of the area of the opening, $Pi/576$, times the velocity of the fluid through the opening. To solve for the velocity of the opening, I used a modified form of Bernoulli's:

$P(1)+(1/2)dv(1)^2+dgh(1)=P(2)+(1/2)dv(2)^2+dgh(2)$. (The (1) and (2) are supposed to represent subscripts.)

Since $h(1) = h, h(2) = 0, v(1) = dh/dt$, and $P(1) = P(2) = P(atm)$ (according to the assumption I made in the last post--I'm pretty confident that it's right),

$v(2) = Sqrt[(dh/dt)^2+2gh]$.

Finally, we have to multiply the RHS by the contraction coefficient, 0.6 for water, to take into account the effects of the surface tension. Thus,

$Pi*h(8-h)(dh/dt)=-(0.6Pi/576)Sqrt[(dh/dt)^2+2gh]$.

Well, you only had to kick me in the right direction about eight times, but I think that this is it.

I really appreciate your help. Let me know what you think.
• Jun 7th 2010, 03:32 PM
justaguy
Regarding the other problem I posted, it was apparently copied down wrong. It was actually to be a first order linear ODE, and so solving it just became a matter of applying the integrating factor. However, there is apparently a method to solve the problem as written, so if any of you guys know it, I would still be curious to see how it's done.