Question

A fluid moves through a tube of length 1 meter and radius r=0.006±0.00025 meters under a pressure p=4⋅105±2000 pascals, at a rate v=0.375⋅10−9 m3 per unit time. Use differentials to estimate the maximum error in the viscosity η given by
η=π8pr4v

Answer 1

Answer:

Maximum error for viscosity is 17.14%

Step-by-step explanation:

We know that everything is changing with respect to the time, "r" is changing with respect to the time, and also "p" just "v" will not change with the time according to the information given, so we can find the implicit derivative with respect to the time, and since

$n = (\frac{\pi}{8}) (\frac{pr^4}{v})$

The implicit derivative with respect to the time would be

$\frac{dn}{dt} = \frac{\pi}{8} ( \frac{r^4}{v} \frac{dp}{dt} + \frac{4pr^3}{v}\frac{dr}{dt} )$

If we multiply everything by    dt   we get  

$dn = \frac{\pi}{8} ( \frac{r^4}{v} dp + \frac{4pr^3}{v} dr})$

Remember that  the error is given by   $\frac{dn}{n}$   therefore doing some algebra we get that

$\frac{dn}{n} = 4 \frac{dr}{r} + \frac{dp}{p}$

Since,    r = 0.006   ,   dr = 0.00025 ,  p = 4*105   ,   dp = 2000  we get that

$\frac{dn}{n} = 0.1714$

Which means that  the maximum error for viscosity is 17.14%.