Answer:
The maximum error is [tex]\Delta \eta = 2032.9[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 1\ m[/tex]
The radius is [tex]r = 0.002 \pm 0.0002 \ m[/tex]
The pressure is [tex]P = 4 *10^{5} \ \pm 1750[/tex]
The rate is [tex]v = 0.5*10^{-9} \ m^3 /t[/tex]
The viscosity is [tex]\eta = \frac{\pi}{8} * \frac{P * r^4}{v}[/tex]
The error in the viscosity is mathematically represented as
[tex]\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v[/tex]
Where [tex]\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}[/tex]
and [tex]\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}[/tex]
and [tex]\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}[/tex]
So
[tex]\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][/tex]
substituting values
[tex]\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 ][/tex]
[tex]\Delta \eta = \frac{\pi}{8} [56 + 5120 ][/tex]
[tex]\Delta \eta = 647 \pi[/tex]
[tex]\Delta \eta = 2032.9[/tex]
