Uncertainty - Reynolds number and the friction factor

[Joon]

Homework Statement

Correctly present the table of information. The values in the table are deliberately in a wrong format.

The calculated Re values have been analysed to have an uncertainty of ± 0.4% and the calculated f values an uncertainty of ± 0.1%.

Homework Equations

The Attempt at a Solution

From the table attached, the first Reynolds number is 94988.9752496553.
I need to fix the number of decimal places (reduce or simplify).[/B]

What I have tried is calculating the uncertainty for this number, 94988.9752496553 x 0.004 = 379.955901.
Since the number of decimal places must be the same for a value and its uncertainty, should I change the Reynolds number to 6 decimal places? Please help.

It is similar for the friction factor f, the values are up to 10 decimal places and I need to simplify the values in the table.

Thank you.

 

[BvU]

Hi,

should I change the Reynolds number to 6 decimal places?

No.

You are given a single digit relative error, so you can not trust more than one digit of the error in Re. 379.955901 is preposterous. One digit means e.g. the first one is ( 95.0 $\pm$ 0.4 ) * 10³ .
For the last one, you would get ( 35.9 $\pm$ 0.1 ) * 10³ .

 

[BvU]

But I don't get why you said +-0.4 or +-0.1 when they are percentage uncertainty

0.4% of 95.0 is 0.4
0.4% of 35.9 is 0.1

 

[Joon]

Thanks, I get it.

Should I say (9.50 +- 0.04 ) x 10^4 or (95.0 +-0.4) x 10^3?
I know both are correct but is the first one better?

 

[BvU]

I can't see a preference for one or the other.

 

[Joon]

Thanks a lot for your replies, I very much appreciate it!

 

[Joon]

Sorry but I have one more question.

For the friction factors, f, how many significant figures would be the most ideal?
e.g. for the first one, if 3 s.f are taken into account, it becomes (1.96 +- 0.00) x 10^(-2) and 0.00 uncertainties for all the rest. If 4 s.f are taken into account then all the uncertainties become (+- 0.002) x 10^(-2).

To consider uncertainties, would it be better to choose the latter option?

 

[BvU]

You have 0.1% as the given uncertainty in the f values. That is 1 in 1000 -- that is four figures!

(equally, with 0.1% relative uncertainty, 0.5 in 500 requires you to report 500.0 $\pm$ 0.5 )

Stating an uncertainty of 0.00 isn't helpful

So (1960 $\pm$ 2) $\times$ 10^-5
But I would personally prefer to report (19.60 $\pm$ 0.02) $\times$ 10^-3
(engineers do that often: pick powers of 1000).
(even though I'm not an engineer)
Perhaps people who do a lot of programming would like (1.960 $\pm$ 0.002) $\times$ 10^-2
(1 digit before the decimal point) Excel, e.g., does it like that for format 'scientific'-- but excel isn't the law.

De gustibus non disputandum

 

[Joon]

Thanks, I get it. To make sure if I understood 100%, I'll write how I approached to the problem, could you check my steps please?

For the Reynolds numbers, 0.4% is 1 in 250, which is 1/250. Following the rules of uncertainty calculations, Reynolds number (with decimals up to 10) / 250 (3 significant figures) should end up with a value of 3 s.f.

Similarly, for the f values, it is 1 in 1000, which means f values(up to 10 decimal places) / 1000(4 s.f.) should end up with a value of 4 s.f.
Am I correct?

There's something I cannot understand though, following the rules of uncertainty calculation, shouldn't it be the same for multiplication and division? (ending up with a value that has the same number of s.f. as the factor that has the least number of s.f.?) E.g. 191049820 * 0.004 should end up with a value of 1 s.f. as 0.004 has 1 s.f.

 

[BvU]

shouldn't it be the same for multiplication and division

I think so, yes.
The 4 as single signifcant figure (without further information) means the factor is 3.5 to 4.5 times 10^-3, so the product is between 67 and 86 times 10⁴. Best you can do is report is 8 $\times$ 10⁴.

Similarly, suppose you get 191049820 / 0.004 then the quotient is (4.8 -- between 5.5 and 4.2) times 10¹⁰ and stating 4.8 10¹⁰ would artificially reduce the perceived accuracy by an unjustified factor of 10. Therefore: 5 10¹⁰

$\$

 

[Joon]

Thank you. I have successfully completed my task, thanks again for your help.
Have a nice day!