Error Propagation - Reconciling Two Approaches

  • #1
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Hi,
I am trying to find the error propagated by calculating the sum of a set of mass flow rates collected over the same length of time. The sum of mass flow rates can be calculated with two approaches, since the collection time is the same for all of them. Approach (1) is adding up all of the individual mass flow rates, and Approach (2) is adding up all of the masses and then dividing by the collection time. However, when I use standard error propagation formulas (http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm) to derive an expression for the error in the sum of mass flow rates, I get two different answers depending on which of the two above approaches I use. I have attached my work. Can someone please help me figure out where math has gone wrong, or my conceptualization?
Thank you
 

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  • #2
BvU
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Hi,

Can you make clear what you have measured ?
It is not clear at all if you have taken one measurement of t or many
The respective roles of ##m##, ##\dot m## and ##M## is not clear either
 
  • #3
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Hi,

Can you make clear what you have measured ?
It is not clear at all if you have taken one measurement of t or many
The respective roles of ##m##, ##\dot m## and ##M## is not clear either
I have calculated the mass, ##m_{i}##, of liquid exiting the outlet of stream ##i## for ##n## number of streams over a measured time period of ##t##, by measuring the mass of liquid + container, ##M_{1,i}##, and subtracting from that the measurement for the mass of the container, ##M_{2,i}##. ##M_{1,i}## and ##M_{2,i}## both have the same measurement error (the error of the scale), which is the same for all ##n## streams. Thus, the error in ##m_{i}## is the same for all streams. The liquid from each stream was collected simultaneously in separate containers, therefore, there is only one ##t## measurement for all of them. The mass flow rate, ##\dot{m}_{i}##, is calculated for each stream ##i## by dividing the mass of stream ##i##, ##m_{i}##, by the collection time, ##t##.

So do you think that may be why the results are different? Is it because one approach assumes that there is a different ##t## measurement for each stream (several ##t_{i}##'s), and one doesn't? So I should use the approach where you sum all of the masses, and then divide by ##t##?
 
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  • #4
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Just want to rewrite all my work using LaTeX and clarify a few things...

Objective: Show that if [itex]\sum^n_{i=1}\dot{m}_i=\frac{\sum^n_{i=1}m_{i}}{t}[/itex] then [itex]\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}=\delta{\left(\frac{\sum^n_{i=1}m_{i}}{t}\right)}[/itex]

Define: [itex]M_{1,i}\ \mbox{and}\ M_{2,i}\ \mbox{are measured masses}\\\ \\m_{i}=M_{1,i}-M_{2,i}\ \mbox{for all}\ i\\\ \\\dot{m_{i}}=\frac{m_{i}}{t}\ \mbox{for all}\ i[/itex]

Given: [itex]t=const.\ \mbox{(i.e., all masses were collected over the same time period)}\\\delta{M_{1,i}}=\delta{M_{2,i}}=\delta{M}=const.\ \mbox{for all}\ i[/itex]

Proof: [itex]\sum^n_{i=1}\dot{m}_i\stackrel{?}{=}\frac{\sum^n_{i=1}m_{i}}{t}\\\ \\\dot{m}_{1}+\dot{m}_{2}+...+\dot{m}_{n}\stackrel{?}{=}\frac{\sum^n_{i=1}m_{i}}{t}\\\ \\\frac{m_{1}}{t}+\frac{m_{2}}{t}+...+\frac{m_{n}}{t}\stackrel{?}{=}\frac{\sum^n_{i=1}m_{i}}{t}\\\ \\\frac{1}{t}\left(m_{1}+m_{2}+...+m_{n}\right)\stackrel{?}{=}\frac{\sum^n_{i=1}m_{i}}{t}\\\ \\\frac{\sum^n_{i=1}m_{i}}{t}=\frac{\sum^n_{i=1}m_{i}}{t}[/itex]

Error Analysis: [itex]\delta{m_{i}}=\sqrt{\left(\delta{M_{1,i}}\right)^2+\left(\delta{M_{2,i}}\right)^2}=\sqrt{2\left(\delta{M}\right)^2}=\sqrt{2}\delta{M}=\delta{m}=const.\ \mbox{for all}\ i\\\ \\\delta{\left(\frac{\sum^n_{i=1}m_{i}}{t}\right)}=\frac{\sum^n_{i=1}m_{i}}{t}\sqrt{\left(\frac{\delta{\left(\sum^n_{i=1}m_{i}\right)}}{\sum^n_{i=1}m_{i}}\right)^2+\left(\frac{\delta{t}}{t}\right)^2}\\\ \\\delta{\left(\sum^n_{i=1}m_{i}\right)}=\sqrt{\sum^n_{i=1}\left(\delta{m_{i}}\right)^2}=\sqrt{\sum^n_{i=1}\left(\delta{m}\right)^2}=\sqrt{n\left(\delta{m}\right)^2}=\sqrt{2n}\delta{M}\\\ \\\delta{\left(\frac{\sum^n_{i=1}m_{i}}{t}\right)}=\frac{\sum^n_{i=1}m_{i}}{t}\sqrt{\left(\frac{\sqrt{2n}\delta{M}}{\sum^n_{i=1}m_{i}}\right)^2+\left(\frac{\delta{t}}{t}\right)^2}\\\ \\\delta{\left(\frac{\sum^n_{i=1}m_{i}}{t}\right)}=\frac{\sum^n_{i=1}m_{i}}{t}\sqrt{2n\left(\frac{\delta{M}}{\sum^n_{i=1}m_{i}}\right)^2+\left(\frac{\delta{t}}{t}\right)^2}\ \ \ \ \ \ \ \ \ \ \mbox{*(EQN. 1)*}\\\ \\\ \\\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}=\sqrt{\sum^n_{i=1}\left(\delta{\dot{m}_{i}}\right)^2}\\\ \\\ \ \ \ \ \ \ \ \ \ \delta{\dot{m}_{i}}=\dot{m}_{i}\sqrt{\left(\frac{\delta{m_{i}}}{m_{i}}\right)^2+\left(\frac{\delta{t}}{t}\right)^2}\\\ \\\ \ \ \ \ \ \ \ \ \ \delta{\dot{m}_{i}}=\dot{m}_{i}\sqrt{2\left(\frac{\delta{M}}{m_{i}}\right)^2+\left(\frac{\delta{t}}{t}\right)^2}\\\ \\\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}=\sqrt{\sum^n_{i=1}\dot{m}_{i}^2\left[2\left(\frac{\delta{M}}{m_{i}}\right)^2+\left(\frac{\delta{t}}{t}\right)^2\right]}\\\ \\\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}=\sqrt{\sum^n_{i=1}\left(\frac{m_{i}}{t}\right)^2\left[2\left(\frac{\delta{M}}{m_{i}}\right)^2+\left(\frac{\delta{t}}{t}\right)^2\right]}\\\ \\\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}=\sqrt{\frac{1}{t^2}\sum^n_{i=1}m_{i}^2\left[2\left(\frac{\delta{M}}{m_{i}}\right)^2+\left(\frac{\delta{t}}{t}\right)^2\right]}\\\ \\\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}=\frac{1}{t}\sqrt{\sum^n_{i=1}\left[2\left(\delta{M}\right)^2+m_{i}^2\left(\frac{\delta{t}}{t}\right)^2\right]}\\\ \\\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}=\frac{1}{t}\sqrt{\sum^n_{i=1}2\left(\delta{M}\right)^2+\sum^n_{i=1}m_{i}^2\left(\frac{\delta{t}}{t}\right)^2}\\\ \\\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}=\frac{1}{t}\sqrt{2n\left(\delta{M}\right)^2+\sum^n_{i=1}m_{i}^2\left(\frac{\delta{t}}{t}\right)^2}\\\ \\\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}=\frac{\sum^n_{i=1}m_{i}}{t}\sqrt{2n\left(\frac{\delta{M}}{\sum^n_{i=1}m_{i}}\right)^2+\frac{\sum^n_{i=1}m_{i}^2}{\left(\sum^n_{i=1}m_{i}\right)^2}\left(\frac{\delta{t}}{t}\right)^2}\ \ \ \ \ \ \ \ \ \ \mbox{*(EQN. 2)*}[/itex]

(EQN. 2) is not equivalent to (EQN. 1), because the second term in the radical is multiplied by the extra [itex]\frac{\sum^n_{i=1}m_{i}^2}{\left(\sum^n_{i=1}m_{i}\right)^2}[/itex] term. This would suggest that [itex]\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}\neq\delta{\left(\frac{\sum^n_{i=1}m_{i}}{t}\right)}[/itex]. However, if [itex]\sum^n_{i=1}\dot{m}_i=\frac{\sum^n_{i=1}m_{i}}{t}[/itex] then why shouldn't [itex]\delta{\left(\sum^n_{i=1}\dot{m}_{i}\right)}=\delta{\left(\frac{\sum^n_{i=1}m_{i}}{t}\right)}[/itex]? Where have I gone wrong?
 
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  • #5
BvU
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My compliments for your extensive typesetting work ! I didn't dare ask, because it's a huge task indeed.
So, to make sure:

you calculate the total mass flow from N measurements of ##M_{1,i}## and 1 measurement of ##M_{2,i}## and one single time measurement t ?

To me that's difficult to imagine: how can you have one container and N simultaneous measurements ?
But let me accept that that is possible, then the errors in the ##m_i## contain
  • one contribution from ##M_{1,i}## that you can assume to be distributed according to a probability distribution (Gaussian, if the resolution is smaller than the standard deviation).
  • The other contribution is NOT distributed, but exactly the same for each measurement ##i##.

The thing to do is then first do ##\sum M_{2,i}## , let that be ##M_{2,{\rm\, tot}}##.
Internal error on ##M_{2,{\rm\, tot}}## (i.e. based on estimated measurement errors, as opposed to external error, based on observed standard deviation) is indeed ##\sqrt N \delta M##.

Next step is to subtract the one single ##M_{1,i}## measurement N times:$$ m_{\rm\, tot} = \sum M_{2,i} - N M_{1,i} $$with internal error $$
\left ( \delta m_{\rm\, tot} \right )^2 = N \left (\delta M \right )^2 +N^2 \left (\delta M \right )^2
$$

Last step is divide by t, a simple quadratic addition of relative errors in ##m_{\rm\, tot}## and ##t##.

Depending on how big N is you can forget N versus N^2 -- and you probably will be sorry for not weighing the pile of N containers together instead of only one ...

Is this what you did and is what I proposed clear ?

I must admit that I did not dig completely through your derivation of eq 2 because I think it goes off the rails at the point you introduce the ##\delta m_i## and subsequently treat them as independent.

--
 
  • #6
94
1
My compliments for your extensive typesetting work ! I didn't dare ask, because it's a huge task indeed.
So, to make sure:

you calculate the total mass flow from N measurements of ##M_{1,i}## and 1 measurement of ##M_{2,i}## and one single time measurement t ?

To me that's difficult to imagine: how can you have one container and N simultaneous measurements ?
But let me accept that that is possible, then the errors in the ##m_i## contain
  • one contribution from ##M_{1,i}## that you can assume to be distributed according to a probability distribution (Gaussian, if the resolution is smaller than the standard deviation).
  • The other contribution is NOT distributed, but exactly the same for each measurement ##i##.

The thing to do is then first do ##\sum M_{2,i}## , let that be ##M_{2,{\rm\, tot}}##.
Internal error on ##M_{2,{\rm\, tot}}## (i.e. based on estimated measurement errors, as opposed to external error, based on observed standard deviation) is indeed ##\sqrt N \delta M##.

Next step is to subtract the one single ##M_{1,i}## measurement N times:$$ m_{\rm\, tot} = \sum M_{2,i} - N M_{1,i} $$with internal error $$
\left ( \delta m_{\rm\, tot} \right )^2 = N \left (\delta M \right )^2 +N^2 \left (\delta M \right )^2
$$

Last step is divide by t, a simple quadratic addition of relative errors in ##m_{\rm\, tot}## and ##t##.

Depending on how big N is you can forget N versus N^2 -- and you probably will be sorry for not weighing the pile of N containers together instead of only one ...

Is this what you did and is what I proposed clear ?

I must admit that I did not dig completely through your derivation of eq 2 because I think it goes off the rails at the point you introduce the ##\delta m_i## and subsequently treat them as independent.

--
Thanks for the reply, and for the compliment! It was certainly painful to type all of it out...Also, no, there are ##N## streams and ##N## separate containers with which to collect liquid from each stream. So for each stream ##i##, the mass of liquid, ##m_{i}##, is calculated by subtracting the mass of the container for that stream, ##M_{2,i}##, from the mass of liquid + container for that stream, ##M_{1,i}##. Liquid is collected from all streams simultaneously for the same amount of time. Sorry, I should have explicitly stated this.
 

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