Height to time equation: water cylinder with pinhole

[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?

 

[Chestermiller]

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?

$\sqrt{h}=\sqrt{h_0} -\frac{AC_Dt\sqrt{2g}}{2A_x}$

$${h}=h_0\left(1 -\frac{AC_D\sqrt{2g/h_0}}{2A_x}t\right)^2$$

 

[Jonathan Densil]

Oh, I thought you didn't want the height to be on the right side at all. Wouldn't it be :
$$h = h_0\left(1-\frac{AC_D\sqrt{2g}}{2A_xh_0}\right)$$

 

[Chestermiller]

Oh, I thought you didn't want the height to be on the right side at all. Wouldn't it be :
$$h = h_0\left(1-\frac{AC_D\sqrt{2g}}{2A_xh_0}\right)$$

This equation is not correct. The algebra leading to it is not correct. You can't do this kind of thing if you have not adequately mastered algebra.

The only height on the right side of my equation is the initial height (which is a constant independent of time). So that just goes along with the other constants multiplying the t.

 

[Jonathan Densil]

sorry,

$$ h = h_0\left(1-\frac{AC_D\sqrt{2g}}{2A_xh_0}\right)^2$$

 

[Chestermiller]

sorry,

$$ h = h_0\left(1-\frac{AC_D\sqrt{2g}}{2A_xh_0}\right)^2$$

That should be a $\sqrt{h_0}$ in the denominator, and the second term should have a t.

 

[Jonathan Densil]

Right that makes sense, I'm not sure why I mind-blanked there. I forgot I was factoring $\sqrt{h_0}$ and not $h_0$. So now I can plug it into the first equation to get: $$\dot{W} =\rho g C_DA\sqrt{2gh_0}-\frac{t\rho g^2 C_D^2 A^2}{A_x}$$
$$ \dot{W} = \rho g C_DA\left(\sqrt{2gh_0}-\frac{tgC_DA}{A_x}\right)$$
Is this correct?

 

[Chestermiller]

Right that makes sense, I'm not sure why I mind-blanked there. I forgot I was factoring $\sqrt{h_0}$ and not $h_0$. So now I can plug it into the first equation to get: $$\dot{W} =\rho g C_DA\sqrt{2gh_0}-\frac{t\rho g^2 C_D^2 A^2}{A_x}$$
$$ \dot{W} = \rho g C_DA\left(\sqrt{2gh_0}-\frac{tgC_DA}{A_x}\right)$$
Is this correct?

I think so, but not the first version.

 

[Jonathan Densil]

The second equation was just a factored version of the first one... is there any way of verifying that this equation makes sense? Here is the graph for weight to time:

 

[Chestermiller]

The second equation was just a factored version of the first one... is there any way of verifying that this equation makes sense? Here is the graph for weight to time:
View attachment 96255

This is not the rate at which the weight is leaving, right? It is the weight itself vs time, right? If $\sqrt{h}$ vs time is a straight line, and W is proportional to h, then, if you plot $\sqrt{W}$ vs time, that should be a straight line also. Plot it up and see.

 

[Jonathan Densil]

You were right, it is a straight line.

 

[Jonathan Densil]

I am using a software from Vernier called Logger Pro, I used it to take the numerical derivative of the weight vales and plotted it against time. If my formula above is correct, then $\dot{W} \propto t$. This is what it came up with:

However, the derivative values that it came up with were all much smaller than what I calculated by inputting time values into my formula. For example, at 8.52 seconds, I got the flow rate to be -0.907... but the software gave me -0.164 and at 38.5 seconds I got -0.3197... but the software got -0.128. Why is this the case, is it because my equation is wrong or is it because the software had no definite formula to get the derivative of and just used data points?

 

[Chestermiller]

I am using a software from Vernier called Logger Pro, I used it to take the numerical derivative of the weight vales and plotted it against time. If my formula above is correct, then $\dot{W} \propto t$. This is what it came up with:
View attachment 96258
However, the derivative values that it came up with were all much smaller than what I calculated by inputting time values into my formula. For example, at 8.52 seconds, I got the flow rate to be -0.907... but the software gave me -0.164 and at 38.5 seconds I got -0.3197... but the software got -0.128. Why is this the case, is it because my equation is wrong or is it because the software had no definite formula to get the derivative of and just used data points?

I'm having trouble following exactly what you did.

 

[Jonathan Densil]

Logger Pro did a numerical derivation of the weight values up in post #43. Those are the weight instances as a function of time. For example at time A the weight was x Newtons and at time B, the weight was y Newtons. If you plot the weight against time, the slope at any point in time would be the flow rate at the instance in time. I made Logger Pro take a numerical derivation of the weight values as a function of time to get the weight flow rate which I then plotted in post #46. Since I couldn't figure out the equation that relates force and time from the graph, I took a numerical derivation instead of an analytical derivation. My problem is, when I plugged in values for time into my equation, I got a different answer for the weight flow rate than what Logger Pro got. So my question is why is this the case? Is my equation wrong or is Logger Pro wrong? And is there any more surefire way of getting the weight flow rate from my data?

 

[Chestermiller]

Logger Pro did a numerical derivation of the weight values up in post #43. Those are the weight instances as a function of time. For example at time A the weight was x Newtons and at time B, the weight was y Newtons. If you plot the weight against time, the slope at any point in time would be the flow rate at the instance in time. I made Logger Pro take a numerical derivation of the weight values as a function of time to get the weight flow rate which I then plotted in post #46. Since I couldn't figure out the equation that relates force and time from the graph, I took a numerical derivation instead of an analytical derivation. My problem is, when I plugged in values for time into my equation, I got a different answer for the weight flow rate than what Logger Pro got. So my question is why is this the case? Is my equation wrong or is Logger Pro wrong? And is there any more surefire way of getting the weight flow rate from my data?

The derivatives that you calculated by hand seem way out of line with the graph of weight vs time. Did you remember to divide by $\Delta t$?

 

[Jonathan Densil]

Why would I need to divide by time. I just plugged in the parameters and values into my equation above. If I did divide by time I would get $-0.10655...$ and ##-0.008305...# respectively for 8.52 seconds and 38.5 seconds.

 

[Jonathan Densil]

I didn't calculate the derivative by hand, I just made Logger Pro do a numerical derivative from the data given with respect to time. I tried to verify it by using my formula and plugging in values.

 

[Jonathan Densil]

In any case, what does a negative flow rate mean? Do you just take the absolute value of it because clearly the slope of the tangent of the graph in post #43 is negative.

 

[Chestermiller]

I didn't calculate the derivative by hand, I just made Logger Pro do a numerical derivative from the data given with respect to time. I tried to verify it by using my formula and plugging in values.

The finite difference approximation to the dW/dt is $\Delta W/\Delta t$. I didn't say to divide by time. I said to divide by the time interval between the point A and point B. The graph of W dot vs time made by Logger Pro looks consistent with the graph of W vs time to me.

 

[Chestermiller]

In any case, what does a negative flow rate mean? Do you just take the absolute value of it because clearly the slope of the tangent of the graph in post #43 is negative.

Sure. The weight is decreasing with time, right?

 

[Jonathan Densil]

Yah, the weight is decreasing, but how do I divide my formula by the time interval? Can you show me please an example with my data of how to get the same answer that Logger Pro is getting?

 

[Chestermiller]

Yah, the weight is decreasing, but how do I divide my formula by the time interval? Can you show me please an example with my data of how to get the same answer that Logger Pro is getting?

Give me the weights and times at two points, A and B.

 

[Jonathan Densil]

Point A: 9.131339746N at 19.58 seconds
Point B: 3.281800887 N at 67.6 seconds

 

[Chestermiller]

Point A: 9.131339746N at 19.58 seconds
Point B: 3.281800887 N at 67.6 seconds

$$\frac{\Delta W}{\Delta t}=\frac{3.2818-9.1313}{67.6-19.58}=-0.122$$
This is over a rather large time interval, so we need to associate this value of dW/dt with the average value of time over the time interval (19.58+67.6)/2=43.6 sec. See where the point (43.6, -0.122) would fall on the Logger Pro graph

 

[Jonathan Densil]

Ohhhh... so in what cases would I use my formula?

 

[Chestermiller]

Ohhhh... so in what cases would I use my formula?

I don't understand. What is "your formula?"