Question

In an underground mine, an airflow of \(30\) m\(^3\) s\(^{-1}\) is delivered through a circular opening having a diameter of \(5\) m and a length of \(500\) m. Assuming that there is no change in the surface characteristics, the diameter of the opening, in \(m\), required to double the quantity of airflow at the same pressure loss is . (Rounded off to two decimal places)

Hint:

Use Atkinson's equation for airway pressure loss and note how it depends on the diameter of a circular opening through the perimeter and area.

Answer 1

Correct Answer: 6.6

This problem is really about how the resistance of a mine airway changes with its size. Instead of substituting straight into Atkinson's equation, we work through the airway resistance concept, which mine ventilation engineers use directly.

The airway resistance $R$ is defined so that the pressure loss is $H = RQ^2$, where

\[ R = k\frac{PL}{A^3} \]

Here $k$ is the friction factor, which stays fixed because "no change in surface characteristics" means the airway wall roughness is the same. $L$ is the airway length, which also stays fixed at 500 m.

For a circular airway of diameter $D$, the perimeter is $P = \pi D$ and the area is $A = \pi D^2/4$. Putting these into the resistance formula:

\[ R = k\frac{\pi D \cdot L}{(\pi D^2/4)^3} = \frac{64kL}{\pi^2}\cdot\frac{1}{D^5} \]

So resistance falls sharply as the diameter grows, specifically as $R \propto 1/D^5$.

Now use the condition that pressure loss stays the same while airflow doubles. Since $H = RQ^2$ is fixed:

\[ R_1 Q_1^2 = R_2 Q_2^2 \implies \frac{R_2}{R_1} = \left(\frac{Q_1}{Q_2}\right)^2 = \left(\frac{30}{60}\right)^2 = \frac{1}{4} \]

Since $R \propto 1/D^5$, the ratio of resistances converts to a ratio of diameters:

\[ \frac{R_2}{R_1} = \left(\frac{D_1}{D_2}\right)^5 = \frac{1}{4} \implies \left(\frac{D_2}{D_1}\right)^5 = 4 \]\[ D_2 = D_1 \times 4^{1/5} = 5 \times 1.31951 = 6.5975 \text{ m} \]

Let's summarize:

• Airway resistance depends on diameter as $R \propto 1/D^5$ for a circular opening.
• Doubling the airflow at the same pressure loss means the resistance must drop to a quarter.
• Solving for the new diameter gives $D_2 = 6.60$ m, about 1.32 times the original diameter.

So the required diameter of the opening is 6.60 m.