Height to time equation: water cylinder with pinhole

[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?"

 

[Jonathan Densil]

The one that we've been trying to figure out:
$$\dot{W} = \rho g C_D A \left(\sqrt{2gh_0}-\frac{t g C_D A}{A_x}\right)$$
When would I have to use this?

 

[Jonathan Densil]

Shouldn't that also have given -1.22 if I plug in 43.6 seconds or 48.02 seconds (I don't know which one to use, the average or the $\Delta t$)?

 

[Chestermiller]

The one that we've been trying to figure out:
$$\dot{W} = \rho g C_D A \left(\sqrt{2gh_0}-\frac{t g C_D A}{A_x}\right)$$
When would I have to use this?

Now that you know the value of Cd, you can plot this equation up and see what you get. Plot a graph of this equation so that we can compare it with the Logger Pro graph for dW/dt.

 

[Chestermiller]

Shouldn't that also have given -1.22 if I plug in 43.6 seconds or 48.02 seconds (I don't know which one to use, the average or the $\Delta t$)?

You calculate the derivative from the difference in the times, but you plot the derivative at the average of the times. It is just a coincidence that the difference and the average are about the same value for this particular case.

 

[Jonathan Densil]

The weight flow rate against time. Sorry for the lack of labels, I did it in Mathematica and don't know how to add labels

 

[Chestermiller]

View attachment 96264

The weight flow rate against time. Sorry for the lack of labels, I did it in Mathematica and don't know how to add labels

Excellent. This is almost the exact negative of dW/dt calculated by Logger Pro from the raw data.

So you can see that, once you have determined Cd, you don't need the data any more.

 

[Jonathan Densil]

Thank you very, very much, I really appreciate your help. The equation works and very happy that I understand how I got it. I really am very thankful for your help. I struggled to make sense of it all for an entire week with my math and physics teachers at school. Thanks again,

Kind regards,
Jonathan

 

[Jonathan Densil]

Hello Chestmiller,

I just have a quick question about uncertainties. I took the derivative of the weight to get weight flow rate. So, if I have the uncertainty for the weight, how do I find the uncertainty of the weight flow rate.

 

[Chestermiller]

Hello Chestmiller,

I just have a quick question about uncertainties. I took the derivative of the weight to get weight flow rate. So, if I have the uncertainty for the weight, how do I find the uncertainty of the weight flow rate.

I don't know how to do this. Maybe, if you submit it to the mathematics forums (specifying very precisely what is involved mathematically), you can get some help there. Mention that you have an analytic expression for the rate of weight loss vs time, and experimental data on weight as a function of time.

 

[Jonathan Densil]

Do you know any good, trustworthy forums as this one?

 

[Chestermiller]

My inclination would be Physics Forums, Science Education, Homework and Coursework Questions, Calculus and Beyond

You have an analytic expression for the derivative of a function based on a fit to experimental data on the function itself. You know the uncertainty of the fit to the experimental data, and you would like to determine the uncertainty in the corresponding analytic expression for the derivative.