Mine Ventilation

Auxiliary Ventilation

Ventilating blind headings with a duct and fan — duct resistance, leakage and the air power delivered to the face.

PART 1

Topic Breakdown & Traps

The Engineering Principle

A blind heading (development end) has no through-flow, so it is ventilated by a fan-and-duct system. The duct itself imposes a resistance

P = R Q^{2}

, and inevitable leakage through joints means the quantity reaching the face is less than the quantity the fan delivers — the ratio is the duct (delivery) efficiency. To guarantee a required face quantity you must oversize the fan flow accordingly.

The Core Formula Matrix

Duct resistance pressure:

P = R Q^{2}

(

R

= duct resistance,

Q

= quantity in the duct).

Delivery efficiency:

η_{d} = \frac{Q _{f a ce}}{Q _{f an}} \Rightarrow Q_{f an} = \frac{Q _{f a ce}}{η _{d}}

.

Air power of the auxiliary fan:

P_{a} = P Q_{f an}

The ‘IIT Traps’

⚠
Leakage means fan flow > face flow. Size the fan for $Q_{f a ce} / η_{d}$ , not just the face requirement.
⚠
**Duct pressure follows the square law $R Q^{2}$ .** Doubling the duct quantity quadruples its pressure loss.
⚠
Use the fan (duct) quantity for fan air power, not the face quantity.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ

A ventilation duct has resistance

R = 2 N\cdots^{2} / m^{8}

and passes

3 m^{3} / s

. The pressure loss in the duct is:

Q2MEDIUM2 Marks · NAT

A heading requires

4 m^{3} / s

at the face. With a duct delivery efficiency of

80%

, the fan must deliver ______ m³/s. (Round off to two decimal places.)

m³/s

Q3HARD2 Marks · NAT

The auxiliary fan delivers

5 m^{3} / s

through a duct of resistance

R = 3 N\cdots^{2} / m^{8}

. The air power of the fan is ______ W. (Round off to the nearest whole number.)