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: Understanding the Question:
A fan and a natural ventilation pressure (NVP) push air through the same set of mine airways together. We know the fan pressure and NVP at one airflow, and we need the new fan pressure at double that airflow, with the NVP staying the same.

Step 2: Key Formula or Approach:
Airway resistance stays fixed when only the quantity of air changes, so pressure lost to resistance grows with the square of the quantity, $P \propto Q^2$. Because the NVP is aiding the fan, the total driving pressure at any flow is $P_{total} = P_{fan} + NVP$, and it is this total pressure that obeys the square law, not the fan pressure alone.

Step 3: Detailed Explanation:
At the first flow, $Q_1 = 500\ \text{m}^3/\text{min}$, the total pressure is $P_{total,1} = 25 + 15 = 40$ mm wg.
The new flow $Q_2 = 1000\ \text{m}^3/\text{min}$ is exactly double $Q_1$, so the flow ratio is 2.
Squaring this ratio gives the new total pressure: $P_{total,2} = 40 \times 2^2 = 40 \times 4 = 160$ mm wg.
The NVP has not changed, so it still contributes 15 mm wg toward this 160 mm wg total. Whatever is left is what the fan itself must generate: $P_{fan,2} = 160 - 15 = 145$ mm wg.

Step 4: Final Answer:
The ventilator has to develop 145 mm wg to move 1000 m3/min of air, with the 15 mm wg NVP still helping it along.