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

Step 1: Recall Atkinson's equation for airway pressure loss.
The frictional pressure loss H in a mine airway is given by Atkinson's equation:
\[ H = k \frac{P L Q^2}{A^3} \]
where k is the friction factor of the airway surface, P is the perimeter of the cross-section, L is the length of the airway, Q is the quantity of airflow, and A is the cross-sectional area.

Step 2: Write P and A in terms of the diameter for a circular airway.
For a circular opening of diameter D,
\[ P = \pi D, \qquad A = \frac{\pi D^2}{4} \]
Substituting these into Atkinson's equation gives
\[ H = k \frac{(\pi D) L Q^2}{(\pi D^2/4)^3} = \frac{64\,k L Q^2}{\pi^2 D^5} \]
So for a fixed k and L, the pressure loss depends on the airflow and diameter as \(H \propto Q^2/D^5\).

Step 3: Apply the condition of equal pressure loss.
The surface characteristics (so k) and the length L stay the same, and the pressure loss H must stay the same while the airflow doubles from \(Q_1 = 30\) to \(Q_2 = 60\) m\(^3\)/s.
\[ \frac{Q_1^2}{D_1^5} = \frac{Q_2^2}{D_2^5} \]
\[ D_2^5 = D_1^5 \left(\frac{Q_2}{Q_1}\right)^2 \]

Step 4: Substitute the numbers and solve for D2.
\[ D_2^5 = 5^5 \times \left(\frac{60}{30}\right)^2 = 3125 \times 4 = 12500 \]
\[ D_2 = 12500^{1/5} \]
\[ D_2 = 6.5975 \text{ m} \]

Final Answer:
\[ \boxed{D_2 = 6.60 \text{ m}} \]