Theory

Mine airways can be considered as very rough conduits with incompressible flow so the frictional pressure drop is given as:

$$ ΔPf= \frac{fL \rho v^2}{2D } $$

Where, f = Resistance coefficient
L = Length of the duct (m)
ρ = Air density (Kg/m3)
v = Air velocity (m/s)

As D is taken as the equivalent diameter given by the below relation:

$$ D= 4 \frac{A}{P} $$

Where, A= Cross-sectional area of duct(m2)
P = Perimeter of the duct(m)

So the above equation can be rewritten as:

$$ ΔPf= \frac{fLP \rho v^2}{8A} $$

Flow of air in mine airways is rarely laminar (Re <= 2000) except for leakage through goaf or loose waste packs. So the above equation becomes,

$$ ΔPf= \frac{kLP v^2 }{A} $$ $$ = \frac{k S v^2}{A} $$ $$ = \frac{k S Q^2}{A^3} \ ( \ Pascal \ ) $$

Where, k = Coefficient of friction = f ρ / 8 (N s2 m - 4)
S = Area of the rubbing surface = L P (m2)
Q = Air flow rate = A v (M3/s)

So, the graph of ΔPf with respect to SQ2 /A3 will be plotted and the slope of the curve will be drowned. The will be the coefficient of friction.

Instruments

• Experimental duct setup.
• Measuring Tape
• Askania Minimeter
• Aneroid Barometer
• Inclined tube manometer
• Orifice plates

Demo