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

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

Step-by-step 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 = (\pi)/(8) * (P * r^4)/(v)`

The error in the viscosity is mathematically represented as

`\Delta \eta = | (\delta \eta)/(\delta P)| * \Delta P + |(\delta \eta)/(\delta r) |* \Delta r + |(\delta \eta)/(\delta v) |* \Delta v`

Where
`(\delta \eta )/(\delta P) = (\pi)/(8) * (r^4)/(v)`

and
`(\delta \eta )/(\delta r) = (\pi)/(8) * (4* Pr^3)/(v)`

and
`(\delta \eta )/(\delta v) = - (\pi)/(8) * (Pr^4)/(v^2)`

So

`\Delta \eta = (\pi)/(8) [ |(r^4)/(v) | * \Delta P + | (4 * P * r^3)/(v) |* \Delta r + |-(P* r^4)/(v^2) |* \Delta v]`

substituting values

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

`\Delta \eta = (\pi)/(8) [56 + 5120 ]`

`\Delta \eta = 647 \pi`

`\Delta \eta = 2032.9`