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Chapter 9: Problem 11

Approximately how fast can the wind blow past a 0.25 -in. diameter twig if viscous effects are to be of importance throughout the entire flow field (i.e., \(\operatorname{Re}<1\) )? Explain. Repeat for a 0.004 -in. diameter tair and a 6 -fi-diameter smokestack.

Short Answer

Expert verified
For a 0.25 -in. diameter twig, the speed of wind should be approximately 0.23 m/s. For a 0.004 -in. diameter tair, it should be approximately 1.43 m/s. For a 6 ft-diameter smokestack, the speed of wind must be about 0.000008 m/s to ensure viscous effects are significant.

Step by step solution

01

Define the Variables

For each of the three objects, we need to calculate the velocity at which the Reynolds Number will be equal to 1.\nThe known variables are: air density, ρ = 1.23 kg/m^3 (at 20° Celsius); air dynamic viscosity, µ = 1.79 x 10^-5 Ns/m^2 (at 20° Celsius); and the diameter D of the objects.
02

Establish the Equation and Solve for Velocity v

The Reynolds number equation is re-written to find the velocity. It becomes: \(v = \frac{Re×µ}{ρD}\).\nSubstitute the known values into the equation to calculate the velocity of wind for each object diameter.
03

Calculation for a 0.25 -in. diameter twig

Convert diameter from inches to meters. 0.25 in = 0.00635 m. Now, Substitute into the rearranged Reynolds equation: \(v=\frac{1×1.79×10^-5}{1.23×0.00635} = 0.23 m/s\) So, the velocity of wind should be 0.23 m/s to ensure the viscous effect is dominant for this diameter twig.
04

Repeat Calculation for a 0.004 -in. diameter tair

Convert diameter from inches to meters. 0.004 in = 0.0001016 m. Substitute these values into the rearranged Reynolds equation: \(v=\frac{1×1.79×10^-5}{1.23×0.0001016} = 1.43 m/s\). The velocity of wind should be 1.43 m/s to ensure the viscous effect is dominant for this diameter tair.
05

Repeat Calculation for a 6 -ft-diameter smokestack

Convert diameter from feet to meters. 6 ft = 1.83 m. Substitute these values into the rearranged Reynolds equation: \(v=\frac{1×1.79×10^-5}{1.23×1.83} = 0.000008 m/s\). The velocity of wind should be 0.000008 m/s to ensure the viscous effect is dominant for this diameter smokestack.

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