# Mass Flow Rate Uncertainty: A simple problem gone horribly, horribly wrong!

The Mass Flow Rate in a water flow system determined by collecting the discharge over a timed interval is 0.2 kg/s. The scales used can be read to the nearest 0.05kg and stop watch is accurate to 0.2 s. Estimate the precision with which the flow rate can be calculated for time intervals of a.) 10s and b.) 60s

Mdot= mass flow rate

Ok so I found the change in mass by using the flow-rate and given time: m = (Mdot)(change in time) so m = (0.2kg/s)(10s) = 2kg

Then i found the uncertainty in the time and mass:

Ut = 0.2s/10s = 0.02s
Um= 0.05kg/2kg

Then I took uncertainty of the mass flow rate, which came from a bunch of partials like this (d is delta):

UMdot = +/- {[(m/Mdot)(dMdot/delta m)(Um)^2 + (t/Mdot)(dMdot/dt)(Ut)^2]}^(1/2)

Then

UMdot = +/- {[((1)( +/- 0.025)^2 + ((-1)(0.02)^2]}^1/2

=0.032 = 3.2%

BUT the answer is exactly half of that, 1.6%

Where the hell did I go wrong?

I can figure out b.) no problem once I figure out why the initial problem isn't working.

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Filip Larsen
Gold Member
If the scale can be read to nearest 0.05 kg, what is then the uncertainty in the mass? (hint: it is not ±0.05 kg as that would mean there had to be 0.1 kg between each mark). Same goes for your clock reading.

If the scale can be read to nearest 0.05 kg, what is then the uncertainty in the mass? (hint: it is not ±0.05 kg as that would mean there had to be 0.1 kg between each mark). Same goes for your clock reading.
so half of the smallest increment? Making it ±0.025 kg.

Filip Larsen
Gold Member
so half of the smallest increment? Making it ±0.025 kg.
Indeed. The assumption is that when you read the weight off the scale you select the nearest mass mark and that means you will select a mark no further away from the true reading than half of the distance between the marks. Or, in other words, the true reading will (with some high probability) lie in the range ± 0.025 from the mark.