Height to time equation: water cylinder with pinhole

[Jonathan Densil]

Hi guys,

I was just doing an experiment with relating the weight flow rate of water out of cylinder with a pin hole at the bottom. The equation that I put together was: $$\dot{W}=\rho g C_d A \sqrt{2gh}$$ ##\dot{W}## is the weight flow rate, ##\rho## is the density of water, ##g## is the acceleration due to gravity, ##C_d## is the coefficient of discharge of the orifice, ##A## is the area of the orifice, ##h## is the height of the water above the hole. The problem that I am facing is that the weight flow rate changes with time along with the height of the water. Thus, I thought that if I can find an equation for the height as it relates to time, I can plug it into the equation above, and my equation would be complete. I also need to see how you got to whatever equation makes sense. I am looking for something like: ##h = kt##. As in k can be any function including any variables that make sense. I believe this requires derivatives and integrals so don't be afraid to include and explain them.

Thank you very much for your help.
Jonathan

 

[Chestermiller]

If ##A_x## is the cross sectional area of the tank and h(t) is the height of liquid in the tank, what is the weight of liquid in the tank at any time?

 

[Jonathan Densil]

The weight of the water in the bottle changes in relation to time. At 11.06 seconds, it is 10.51481546 N, at 46.18 seconds, it is 5.507255163 N. The ##A## in the equation above is the area of the hole. I was hoping to include the ##A_x## in the height-time equation. This is what Wikipedia gave:

Now,

where A and a are the cross sections of container and tube respectively, dh is the height of liquid in container corresponding to dx in the tube which decreases in same time dt.

I guess this makes sense if you are solving for time. I want to solve for height. First of all, I don't know where the ##\frac{Adh}{a\sqrt(2gh)}## came from. Second of all, I don't understand how they integrated ##\int\frac{Adh}{a\sqrt(2gh)}##, did they move the dh in the numerator out to become ##\int\frac{A}{a\sqrt(2gh)}dh##? If what they have done is right, then can you please rearrange and solve the derivative and integral in terms of height and show steps, and if it is not right then can you please propose and explain your solution Thank you for your help,

Jonathan

 

[Chestermiller]

Well, I guess I'll have to answer my own question.

$$W=\rho g A_xh$$So,$$\frac{dW}{dt}=\rho g A_x\frac{dh}{dt}=-\rho g C_DA\sqrt{2gh}$$
This gives me:$$\frac{dh}{dt}=-\frac{A}{A_x}C_D\sqrt{2gh}$$
Do you know how to solve this equation for h as a function of t, subject to the initial condition h = h₀ at t = 0?

 

[Jonathan Densil]

Thank you very much for your help, I really do appreciate it but, if I am talking complete nonsense it is because I am in Grade 11 and teaching myself calculus on my own time. So...please bear with me. I'm not sure how to deal with the initial conditions that you set but, I think what you can do is multiply both sides by dt and take the integral of both sides, exposing the height on the left.
$$dh=-\frac{A}{A_x}dtC_D\sqrt{2gh}$$
$$\int dh=\int-\frac{A}{A_x}dtC_D\sqrt{2gh}$$ This would give...
$$h= ...something\ t\ something...$$
I also wanted to get rid of the h under the root sign, and just have h = some function and variables and t.
I know that this isn't solved or probably anywhere close to what you wanted but this, this is as far as I have gone, after which I don't know how to proceed. If I am at least on the right track, please say so and if I am not, please correct me. Thank you again for your patience.

Jonathan

 

[Chestermiller]

Suppose I rewrote the equation as $$h^{-1/2}dh=-\frac{A}{A_x}C_D\sqrt{2g}dt$$
Could you integrate each side of the equation then, at least as an indefinite integral (i.e., with an arbitrary constant)?

 

[Jonathan Densil]

Would this be:
$$\int \frac{dh}{\sqrt{h}} =\int -\frac{A}{A_x}C_D\sqrt{2g}dt$$ By factoring constants:
$$\int \frac{dh}{\sqrt{h}} = -\frac{A}{A_x}C_D\sqrt{2g}\int dt$$ Solving:
$$ 2\sqrt{h} = -\frac{A}{A_x}C_Dt\sqrt{2g}$$ Solving for h:
$$ h = \frac{2gt^2A^2C_D^2}{4A_x^2}$$
Did I do this correctly?

 

[Chestermiller]

Would this be:
$$\int \frac{dh}{\sqrt{h}} =\int -\frac{A}{A_x}C_D\sqrt{2g}dt$$ By factoring constants:
$$\int \frac{dh}{\sqrt{h}} = -\frac{A}{A_x}C_D\sqrt{2g}\int dt$$ Solving:
$$ 2\sqrt{h} = -\frac{A}{A_x}C_Dt\sqrt{2g}$$ Solving for h:
$$ h = \frac{2gt^2A^2C_D^2}{4A_x^2}$$
Did I do this correctly?

Not quite. You left out the constant of integration:
$$ 2\sqrt{h} = -\frac{A}{A_x}C_Dt\sqrt{2g}+C$$
To determine the constant of integration, you need to substitute the initial condition into this equation. What do you get for C? When you substitute this back into the equation for arbitrary h, what do you get then?

 

[Jonathan Densil]

When ##h = h_0##, ##t = 0##, and ##h_0 = 0.14## m, I get ##C = 2\sqrt{0.14}## or ##\frac{\sqrt{14}}{5}## or ##0.7483314774##

 

[Chestermiller]

When ##h = h_0##, ##t = 0##, and ##h_0 = 0.14## m, I get ##C = 2\sqrt{0.14}## or ##\frac{\sqrt{14}}{5}## or ##0.7483314774##

This is correct, but I has hoping that you would substitute ##C=2\sqrt{h_0}## algebraically into the equation for h, and then solve for h. Alternately, you could leave the equation in terms of ##\sqrt {h}##. This would show that ##\sqrt {h}## is a linear function of t, with an initial value of ##\sqrt {h_0}##. What would you get if you plotted your data on a graph of ##\sqrt {h}## versus t? What would be the slope and what would be the intercept? What would you get if you plotted ##\sqrt{W}## vs t?

 

[Jonathan Densil]

With time on the x-axis and the ##\sqrt{h}## on the y-axis, I get the slope to be ##-0.002625## and the intercept is ##0.3670##. The correlation is -0.9998.

 

[Chestermiller]

With time on the x-axis and the ##\sqrt{h}## on the y-axis, I get the slope to be ##-0.002625## and the intercept is ##0.3670##. The correlation is -0.9998.

View attachment 96238

Excellent. Now the slope should be equal to some combination of the parameters in your model. Algebraically, from your final equation, what is the slope equal to? If you know the pinhole area, you should be able to use the slope of your graph to determine the discharge coefficient.

 

[Jonathan Densil]

um... I got ##9014.33573##
##A = 0.0000001781393481## ##m^2## (area of the pin hole)
##A_x=0.0094717086## ##m^2## (cross sectional area of the cylinder)
##g = 9.81## ##ms^-2##
##C=\frac{\sqrt{14}}{5}##
##m = -0.002625## (slope of the graph)
$$C_D=\frac{m-C}{-\frac{A}{A_x}\sqrt{2g}}$$
I don't think that is the right answer because when I looked up the coefficient of discharge for a hole like mine, I got 0.98... is there anything that I did wrong?

 

[Chestermiller]

um... I got ##9014.33573##
##A = 0.0000001781393481## ##m^2## (area of the pin hole)
##A_x=0.0094717086## ##m^2## (cross sectional area of the cylinder)
##g = 9.81## ##ms^-2##
##C=\frac{\sqrt{14}}{5}##
##m = -0.002625## (slope of the graph)
$$C_D=\frac{m-C}{-\frac{A}{A_x}\sqrt{2g}}$$
I don't think that is the right answer because when I looked up the coefficient of discharge for a hole like mine, I got 0.98... is there anything that I did wrong?

The C shouldn't be in there.

 

[Jonathan Densil]

Without the C I get 31.50999077, it's still way off.

 

[Chestermiller]

Without the C I get 31.50999077, it's still way off.

How accurate is the pin hole area?

 

[Jonathan Densil]

I used a ##\frac{3}{16}## inch diameter drillbit.

 

[Chestermiller]

I used a ##\frac{3}{16}## inch diameter drillbit.

For a 3/16" hole, 1 get 100 x your area.

 

[Jonathan Densil]

True, you are right, the coefficient that I get is 0.315, thank you very much.

 

[Chestermiller]

True, you are right, the coefficient that I get is 0.315, thank you very much.

Not so fast. That's still pretty far off. Is there a factor of ##\pi## missing somewhere?

 

[Jonathan Densil]

Well the cross sectional area of the tank is:
$$A = \frac{C^2}{4\pi}$$
$$A = \frac{0.345^2}{4\pi}$$
$$A = 0.0094717086\ m^2$$

The area of the hole:
$$A = \frac{\pi d^2}{4}$$
$$A = \frac{\pi 0.0047625^2}{4}$$
$$A = 0.00001781393481\ m^2$$

I'm not sure where the extra ##\pi## disappeared...

 

[Chestermiller]

Well the cross sectional area of the tank is:
$$A = \frac{C^2}{4\pi}$$
$$A = \frac{0.345^2}{4\pi}$$
$$A = 0.0094717086\ m^2$$

The area of the hole:
$$A = \frac{\pi d^2}{4}$$
$$A = \frac{\pi 0.0047625^2}{4}$$
$$A = 0.00001781393481\ m^2$$

I'm not sure where the extra ##\pi## disappeared...

So the tank is 11 cm in diameter. That's about 4". Does that sound about right?

 

[Jonathan Densil]

Yah, that sounds about right. I used a 2 Litre pop bottle. It has straight sides, a conical top, and the weird bit at the bottom. I just used drilled the hole above the weird bit at the bottom and measured from 14 cm above the hole, just under the conical top. So essentially a cylinder.The hole is pretty much a circle since I drilled it slowly and steadily with a sharp drill bit. Unless I was in a university lab with university equipment I don't think I could have gotten more accurate.

 

[Chestermiller]

You're missing a factor of 2 in your equation for the slope. It should be :

$$C_D=\frac{2m}{-\frac{A}{A_x}\sqrt{2g}}$$

Can you see where the factor of 2 comes from?

 

[Jonathan Densil]

I'm guessing it's because I'm plotting time against the square root of the height... but it still only gives me 0.63 if the 2 is negative. Does 0.63 as the coefficient of discharge make sense because the hole that I made isn't exactly perfect?

 

[Chestermiller]

I'm guessing it's because I'm plotting time against the square root of the height... but it still only gives me 0.63 if the 2 is negative. Does 0.63 as the coefficient of discharge make sense because the hole that I made isn't exactly perfect?

The 2 isn't negative; the slope is negative. It seems like the value of 0.63 is, for whatever reason, the correct Cd for your system.

As far as the factor of 2 is concerned, rewrite your equation in terms of ##\sqrt{h}## rather than ##2\sqrt{h}##, and back out what the slope should be.

 

[Jonathan Densil]

Oh ya, I forgot that I had it written in ##2\sqrt{h}##. Now that I know all of the constants in my equation, how do I deal with the integral constant, should I replace it with the value of the constant (##\frac{\sqrt{14}}{5}##) or make ##C =2 \sqrt{h_0}## and then plug it back into the equation? If I use the latter method, I would have: $$2(\sqrt{h}-\sqrt{h_0}) = -\frac{A}{A_x}C_Dt \sqrt{2g}$$ If that was right, I wanted to have on ##h## on the left side because it would be easier to plug into my first equation at the beginning of this thread.

 

[Jonathan Densil]

Actually (this is exciting) the height is really the change in height:
$$\dot{W} = \rho C_D A \sqrt{2g\Delta h}$$ and the equation above is:
$$\Delta h= -\frac{2gt^2A^2C_D^2}{4A_x^2}$$
Thus,
$$\dot{W} = \frac{\rho C_D^2 A^2 gt}{A_x}$$
Would this be correct?

 

[Chestermiller]

Oh ya, I forgot that I had it written in ##2\sqrt{h}##. Now that I know all of the constants in my equation, how do I deal with the integral constant, should I replace it with the value of the constant (##\frac{\sqrt{14}}{5}##) or make ##C =2 \sqrt{h_0}## and then plug it back into the equation? If I use the latter method, I would have: $$2(\sqrt{h}-\sqrt{h_0}) = -\frac{A}{A_x}C_Dt \sqrt{2g}$$ If that was right, I wanted to have on ##h## on the left side because it would be easier to plug into my first equation at the beginning of this thread.

Yes, but you can't plot the data and get a straight line when you do that. Anyhow, I'm sure you know how to solve this equation for h algebraically.

Chet

 

[Chestermiller]

Actually (this is exciting) the height is really the change in height:
$$\dot{W} = \rho C_D A \sqrt{2g\Delta h}$$ and the equation above is:
$$\Delta h= -\frac{2gt^2A^2C_D^2}{4A_x^2}$$
Thus,
$$\dot{W} = \frac{\rho C_D^2 A^2 gt}{A_x}$$
Would this be correct?

None of this is correct.

 

[Jonathan Densil]

I'm sorry for taking so much of your time, I really appreciate what you've done for me, but what would I have to do to finish my equation where I can replace the height in my first equation so that the weight flow rate is in terms of time?

 

[Chestermiller]

I'm sorry for taking so much of your time, I really appreciate what you've done for me, but what would I have to do to finish my equation where I can replace the height in my first equation so that the weight flow rate is in terms of time?

I don't understand why you want that, but, if it is important to you, the first step is to solve the equation in post #28 algebraically for h as a function of t.

 

[Jonathan Densil]

I am confused as to which equation you want me to solve, post # 28 has 3.

 

[Chestermiller]

I am confused as to which equation you want me to solve, post # 28 has 3.

I meant post #27. Typo.

 

[Jonathan Densil]

After I have ##\sqrt{h}-\sqrt{h_0} = -\frac{AC_Dt\sqrt{2g}}{2A_x}##, I'm not sure how to isolate for ##h## on its own...sorry, could you please help me on that please?