Fluid flow problem

**process91** · September 27th 2011, 10:24 AM

A tank with vertical sides has a square cross-section of area 4 ft squared. Water is leaving the tank through an orifice of area 5/3 inches squared. Water also flows into the tank at the rate of 100 cubic inches per second. Show that the water level approaches the value (25/24)^2 ft above the orifice.

Rate of discharge of volume through the orifice is $4.8A_0 \sqrt{y}$ cubic feet per second, where $A_0$ = size of orifice in square feet

$\dfrac{dV}{dt}=100/12^3-4.8A_0 \sqrt{y}$

Also, $V(y)=\int_0^y 4 du$ so
$\dfrac{dV}{dy}=4$

By the chain rule, $\dfrac{dV}{dt}=\dfrac{dV}{dy} \dfrac{dy}{dt}$ so
$100/12^3-4.8A_0 \sqrt{y} = 4 \dfrac{dy}{dt}$

The problem right before this was the same except water was not being added at all, and that was an easily solvable differential equation. I am stuck on this, and when I got the answer from WolframAlpha it did not look encouraging that I was on the right track, so I think I have set up the problem incorrectly.

**alexmahone** · September 27th 2011, 10:42 AM

Originally Posted by process91

A tank with vertical sides has a square cross-section of area 4 ft squared. Water is leaving the tank through an orifice of area 5/3 inches squared. Water also flows into the tank at the rate of 100 cubic inches per second. Show that the water level approaches the value (25/24)^2 ft above the orifice.

Rate of discharge of volume through the orifice is $4.8A_0 \sqrt{y}$ cubic feet per second, where $A_0$ = size of orifice in square feet

$\dfrac{dV}{dt}=100/12^3-4.8A_0 \sqrt{y}$

Also, $V(y)=\int_0^y 4 du$ so
$\dfrac{dV}{dy}=4$

By the chain rule, $\dfrac{dV}{dt}=\dfrac{dV}{dy} \dfrac{dy}{dt}$ so
$100/12^3-4.8A_0 \sqrt{y} = 4 \dfrac{dy}{dt}$

The problem right before this was the same except water was not being added at all, and that was an easily solvable differential equation. I am stuck on this, and when I got the answer from WolframAlpha it did not look encouraging that I was on the right track, so I think I have set up the problem incorrectly.

By Torricelli's law, the velocity of water through the orifice $=\sqrt{2gy}=\sqrt{2*32*y}=8\sqrt{y}$

So, the volume of water flowing out of the orifice per unit time $=\frac{1}{12^2}*\frac{5}{3}*8\sqrt{y}=\frac{5\sqrt {y}}{54}\ ft^3/s$

**process91** · September 27th 2011, 10:47 AM

In this problem, we are given that the fluid viscosity/opening make it such that the coefficient of the general version of Torricelli's law yields that the velocity of the stream is $4.8A_0\sqrt{y}$ .

Actually, I think I may have gotten it. Instead of trying to solve my differential equation, I just examine it at y=(25/24)^2, and see that it is zero. Examining the second derivative, we find that it is always negative so this is a maximum of y. For y less than this, the derivative is positive and above this value the derivative is negative. Hence no matter what value of y the problem starts with, as t increases it will approach (25/24)^2.

Does that seem valid?

**alexmahone** · September 27th 2011, 11:03 AM

Originally Posted by process91

In this problem, we are given that the fluid viscosity/opening make it such that the coefficient of the general version of Torricelli's law yields that the velocity of the stream is $4.8A_0\sqrt{y}$ .

Actually, I think I may have gotten it. Instead of trying to solve my differential equation, I just examine it at y=(25/24)^2, and see that it is zero. Examining the second derivative, we find that it is always negative so this is a maximum of y. For y less than this, the derivative is positive and above this value the derivative is negative. Hence no matter what value of y the problem starts with, as t increases it will approach (25/24)^2.

Does that seem valid?

It may be valid but it's cheating because the question could just as easily have asked you to find the limiting water level.

Why don't you post your attempt at solving the DE?

**process91** · September 27th 2011, 11:09 AM

If asked to find the limiting water level I could set dy/dt to zero and solve for y, then verify that values above and below that y are positive/negative respectively.

I have no attempt at solving that DE, no methods gone over thus far in the book seem applicable. This is in Apostol Calculus Vol I., section 8.28 #19. For first-order differential equations he covered homogeneous and separable options, as well as Bernoulli's equation. If there is a way to solve this, and my way is indeed cheating, please let me know because I do want to do this properly.

I know, in theory, that I would solve the DE for y in terms of t, then take the limit as t increases without bound. WolframAlpha's solution to the DE does not seem to illuminate any solution to the problem, however.

**alexmahone** · September 27th 2011, 11:12 AM

Originally Posted by process91

If asked to find the limiting water level I could set dy/dt to zero and solve for y, then verify that values above and below that y are positive/negative respectively.

I have no attempt at solving that DE, no methods gone over thus far in the book seem applicable. This is in Apostol Calculus Vol I., section 8.28 #19. For first-order differential equations he covered homogeneous and separable options, as well as Bernoulli's equation. If there is a way to solve this, and my way is indeed cheating, please let me know because I do want to do this properly.

I know, in theory, that I would solve the DE for y in terms of t, then take the limit as t increases without bound. WolframAlpha's solution to the DE does not seem to illuminate any solution to the problem, however.

This is a separable DE.

**process91** · September 27th 2011, 11:15 AM

The DE which incorporates both the water being discharged and the water being added is

$100/12^3-4.8A_0 \sqrt{y} = 4 \dfrac{dy}{dt}$

This doesn't appear separable to me, can you point me in the right direction?

**alexmahone** · September 27th 2011, 11:19 AM

Originally Posted by process91

The DE which incorporates both the water being discharged and the water being added is

$100/12^3-4.8A_0 \sqrt{y} = 4 \dfrac{dy}{dt}$

This doesn't appear separable to me, can you point me in the right direction?

$4\frac{dy}{dt}=\frac{100}{12^3}-4.8*\frac{1}{12^2}*\frac{5}{3}\sqrt{y}$

$4\frac{dy}{dt}=\frac{100}{12^3}-\frac{8}{12^2}\sqrt{y}$

$12^3*4\frac{dy}{dt}=100-96\sqrt{y}$

$1728\frac{dy}{dt}=25-24\sqrt{y}$

$1728\left(\frac{dy}{25-24\sqrt{y}}\right)=dt$

**process91** · September 27th 2011, 11:31 AM

Thanks, don't know what my mental block was on that.

Anyway, after integrating I'm getting something which implicitly defines y, but I have no way to figure out what happens as t increases without bound.

$24\sqrt{y}+25\ln(24\sqrt{y}-25)=-\frac{t}{6} + 25 + C$

Obviously as t increases without bound I see that both sides of this equality must decrease without bound, and the only way for the one on the left to do that is for y=(24/25)^2, but this still feels a little like cheating. Any more mathematically standard way to handle it that I'm overlooking?

**alexmahone** · September 27th 2011, 11:36 AM

Originally Posted by process91

Thanks, don't know what my mental block was on that.

Anyway, after integrating I'm getting something which implicitly defines y, but I have no way to figure out what happens as t increases without bound.

$24\sqrt{y}+25\ln(24\sqrt{y}-25)=-\frac{t}{6} + 25 + C$

Obviously as t increases without bound I see that both sides of this equality must decrease without bound, and the only way for the one on the left to do that is for y=(24/25)^2, but this still feels a little like cheating. Any more mathematically standard way to handle it that I'm overlooking?

$24\sqrt{y}+25\ln|24\sqrt{y}-25|\to -\infty$

$\implies \ln|24\sqrt{y}-25|\to -\infty$

$\implies 24\sqrt{y}-25\to 0$

$\implies y\to (24/25)^2$

(Note that you can find $C$ by using the initial condition $y(0)=y_0$ .)

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