Question

In a mine ventilation system, the resistance of an airway is such that when air quantity is \(10~\mathrm{m^3/s}\), the pressure drop is \(100~\mathrm{Pa}\). What will be the pressure drop when the air quantity is increased to \(20~\mathrm{m^3/s}\)?

• A. \(400~\mathrm{Pa}\)
• B. \(300~\mathrm{Pa}\)
• C. \(200~\mathrm{Pa}\)
• D. \(100~\mathrm{Pa}\)

Hint:

For mine airways, \[ \boxed{P\propto Q^2.} \] If air quantity doubles, the pressure drop becomes four times.

Answer 1

The Correct Option is A

Step 1: Use Atkinson's equation. Pressure drop in an airway is given by \[ P \propto Q^2, \] where \(P\) is the pressure drop and \(Q\) is the quantity of air.

Step 2: Calculate the new pressure drop. Given, \[ P_1=100~\mathrm{Pa}, \qquad Q_1=10~\mathrm{m^3/s}, \qquad Q_2=20~\mathrm{m^3/s}. \] Using \[ \frac{P_2}{P_1} = \left(\frac{Q_2}{Q_1}\right)^2, \] \[ P_2 = 100\left(\frac{20}{10}\right)^2 = 100\times4 = 400~\mathrm{Pa}. \] Hence, \[ \boxed{400~\mathrm{Pa}} \] is the correct answer. Therefore, \[ \boxed{(A)} \] is the correct answer.