Question

A main mechanical ventilator installed for an underground mine develops a pressure of 25 mm wg. A natural ventilation pressure (NVP) of 15 mm wg acting in the mine aids the ventilator, and a total 500 \( \text{m}^3\,\text{min}^{-1} \) of air is circulated in the mine. Considering the same NVP aiding the ventilator, the pressure required, in mm wg, to be generated by the ventilator to circulate 1000 \( \text{m}^3\,\text{min}^{-1} \) of air, is . (answer in integer)

Hint:

Total pressure overcoming a fixed set of airways follows a square law with airflow, and NVP simply adds to the ventilator pressure when it is aiding.

Answer 1

Correct Answer: 145

Step 1: Set up the total pressure at the first duty point.
When an NVP aids the ventilator, the two pressures add together to push air through the mine. So the total pressure driving the airflow at the first condition is \( P_1 = 25 + 15 = 40 \) mm wg, moving \( Q_1 = 500\ \text{m}^3\,\text{min}^{-1} \) of air.

Step 2: Apply the mine resistance law.
For a fixed set of airways the resistance \( R \) does not change, and total pressure varies with the square of the airflow, \( P = RQ^2 \). This is the mine characteristic curve. Since \( R \) is the same at both duty points, \( \dfrac{P_2}{P_1} = \left( \dfrac{Q_2}{Q_1} \right)^2 \).

Step 3: Find the total pressure needed at 1000 m3/min.
Here \( Q_2 = 1000\ \text{m}^3\,\text{min}^{-1} \), so \( \dfrac{Q_2}{Q_1} = 2 \). Then \( P_2 = P_1 \times 2^2 = 40 \times 4 = 160 \) mm wg. This 160 mm wg is the total pressure the airways demand at the higher flow.

Step 4: Take out the NVP contribution.
The question says the same NVP is still aiding the ventilator, so it still supplies 15 mm wg of the total. The ventilator only has to make up the rest: \( P_{fan,2} = 160 - 15 = 145 \) mm wg.

Final Answer:
The ventilator must generate 145 mm wg to circulate 1000 m3/min of air through the mine. \[ \boxed{145\ \text{mm wg}} \]