Question

In a longwall panel, air flows at a velocity of 1.2 m/s through a 900 m long gate road of 2.5 m height and 3 m width. The coefficient of friction is 0.022 Ns$^2$m$^{-4}$. The frictional pressure drop, in Pa, between two ends of the gate road is _______ (rounded off to 3 decimal places).

Hint:

For calculating pressure drop due to friction in air flow through ducts, use the Darcy-Weisbach equation with the appropriate values for friction factor, length, and hydraulic diameter.

Answer 1

Step 1: Use the Darcy-Weisbach equation for pressure drop due to friction. \[ \Delta P = \frac{4fL\rho v^2}{D} \] Where: \(\Delta P\) = pressure drop (Pa) 
\(f\) = friction factor (Ns$^2$m$^{-4}$) 
\(L\) = length of the gate road (900 m) 
\(\rho\) = density of air (approximately 1.225 kg/m$^3$ at 0°C) 
\(v\) = velocity of airflow (1.2 m/s) 
\(D\) = hydraulic diameter (calculated from the cross-sectional area of the gate road) 
Step 2: Calculate the hydraulic diameter \(D\) for the rectangular cross-section. The area \(A\) of the gate road is: \[ A = 2.5 \times 3 = 7.5 \, {m}^2 \] The perimeter \(P\) of the gate road is: \[ P = 2 \times (2.5 + 3) = 11 \, {m} \] Now, calculate the hydraulic diameter \(D\): \[ D = \frac{4A}{P} = \frac{4 \times 7.5}{11} = 2.727 \, {m} \] Step 3: Plug in values into the Darcy-Weisbach equation. \[ \Delta P = \frac{4 \times 0.022 \times 900 \times 1.225 \times (1.2)^2}{2.727} \approx 45 \, {Pa} \] Answer: The frictional pressure drop is 45 Pa.