Chapter 10, Problem 54P

To determine

To calculate: The diameter of long air duct.

Answer to Problem 54P

The diameter is $0.11\text{m}$ .

Explanation of Solution

Given:

Coefficient of viscosity, $\eta =0.2\text{Pa·s}$

Length, $l=21\text{m}$

Gauge pressure, $P=0.71\times {10}^{-3}\text{atm}$

Room’s dimension, $\text{9m×12m×4m}$

Formula used:

According to the Poiseuille's equation,

  $Q=\frac{\pi r^4\left(P_2-P_1\right)}{8\eta l}$

Where, $Q$ is the volume discharge flow

  $\left(P_2-P_1\right)$ , Pressure difference between any two ends of object

  $\eta$ Viscosity of fluid,

  $r$ is the inner radius,

  $l$ is the length of the pipe,

Calculation:

Substitute the values in the given equation, and then the result will be,

  $\begin{array}{l}Q=\frac{\pi r^4\left(P_2-P_1\right)}{8\eta l}\Rightarrow r={\left(\frac{8Q\eta l}{\pi \Delta P}\right)}^{1/4}\\ r={\left(\frac{8\times \frac{\left(9\times 12\times 4{\text{m}}^3\right)}{600s}\times 0.018\times {10}^{-3}\times 21}{3.14\times 0.71\times {10}^{-3}\text{atm}\times 1.013\times {10}^5\frac{\text{Pa}}{\text{atm}}}\right)}^{1/4}={\left(\frac{2.17728}{225.1694}\right)}^{1/4}={\left(5.801712\times {10}^{-3}\right)}^{1/4}\\ r={\left(96.6952\times {10}^{-4}\right)}^{1/4}\text{m}=0.055\text{m}\end{array}$

Now the diameter is,

  $d=2r=2\times 0.055\text{m}=0.11\text{m}$

Conclusion:

The diameter is $0.11\text{m}$ .