Four identical districts of a mine are ventilated with a quantity of 3500 m³/min at a fan drift pressure of 1.15 kPa. When one of the districts is sealed off, the change in resulta... Four identical districts of a mine are ventilated with a quantity of 3500 m³/min at a fan drift pressure of 1.15 kPa. When one of the districts is sealed off, the change in resultant resistance is 0.072 Ns²/m⁸. If the fan is stopped, keeping a district sealed, the quantity through the mine becomes 850 m³/min. The natural ventilation pressure in Pa is: Read more

Steps to Solve

1. Understand the ventilation system before sealing

   The original airflow through the mine is given as $Q = 3500 , \text{m}^3/\text{min}$.

2. Convert the flow rate to cubic meters per second

   To use SI units, convert the flow rate: $$ Q = 3500 , \text{m}^3/\text{min} \times \frac{1 , \text{min}}{60 , \text{s}} = \frac{3500}{60} , \text{m}^3/\text{s} \approx 58.33 , \text{m}^3/\text{s} $$

3. Identify parameters after sealing one district

   When one district is sealed, the new quantity becomes: $$ Q' = 850 , \text{m}^3/\text{min} \times \frac{1 , \text{min}}{60 , \text{s}} = \frac{850}{60} , \text{m}^3/\text{s} \approx 14.17 , \text{m}^3/\text{s} $$

4. Calculate the original drift pressure in Pa

   The fan drift pressure is given as $1.15 , \text{kPa}$ which is: $$ P_{\text{fan}} = 1.15 \times 1000 , \text{Pa} = 1150 , \text{Pa} $$

5. Calculate the resultant pressure drop due to resistance

   The change in resistance is provided as $R = 0.072 , \text{Ns}^2/\text{m}^8$. The relationship between pressure, flow, and resistance is given by: $$ \Delta P = R \cdot Q'^2 $$

   Substituting the values: $$ \Delta P = 0.072 , \text{Ns}^2/\text{m}^8 \cdot (14.17 , \text{m}^3/\text{s})^2 $$

   Calculating it: $$ \Delta P = 0.072 \cdot 200.53 \approx 14.43 , \text{Pa} $$

6. Calculate natural ventilation pressure

   The natural ventilation pressure when the fan is stopped is given by subtracting the pressure drop from the original pressure: $$ P_{\text{natural}} = P_{\text{fan}} - \Delta P $$ $$ P_{\text{natural}} = 1150 , \text{Pa} - 14.43 , \text{Pa} \approx 1135.57 , \text{Pa} $$

7. Convert result to kPa and round

   Finally, converting back to kPa: $$ P_{\text{natural}} = \frac{1135.57}{1000} \approx 1.136 , \text{kPa} $$