Question

A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate v=0.5⋅10−9 m3 per unit time. Use differentials to estimate the maximum error in the viscosity η given by

Answer 1

Answer:

The  maximum error is  $\Delta \eta = 2032.9$

Explanation:

From the question we are told that

     The length  is  $l = 1\ m$

      The radius is  $r = 0.002 \pm 0.0002 \ m$

        The pressure is  $P = 4 *10^{5} \ \pm 1750$

        The  rate  is  $v = 0.5*10^{-9} \ m^3 /t$

       The viscosity is  $\eta = \frac{\pi}{8} * \frac{P * r^4}{v}$

The error in the viscosity is mathematically represented  as

       $\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v$

   Where  $\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}$

and         $\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}$

and          $\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}$

So  

             $\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]$

substituting values

            $\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]$

  $\Delta \eta = \frac{\pi}{8} [56 + 5120 ]$

   $\Delta \eta = 647 \pi$

    $\Delta \eta = 2032.9$