Mine ventilation is one of those GATE Mining topics where a student who has practised a handful of formulas can pick up marks quickly, while a student who has only read the theory loses them just as quickly. The questions are usually numerical, the maths is simple, and the traps are almost always about units, density and the difference between series and parallel.
This post collects the formulas you actually need for the Mine Ventilation part of the GATE MN syllabus, explains each one in plain language, shows worked examples with the arithmetic written out, and ends with six practice questions. If you are new to the syllabus, read our GATE Mining Engineering syllabus guide first, and browse the other mine ventilation and GATE exam articles on this blog.
Before you use these: constants and conventions differ slightly between textbooks, for example the value of k, the orifice constant and the standard air density. In an exam, use the values given in the question, and if none are given, state the standard values you assume (air density 1.2 kg/m³). Check your own textbook for the version your syllabus follows.
Quick formula sheet
| What you want | Formula | Units |
|---|---|---|
| Air quantity | Q = A × V | Q in m³/s, A in m², V in m/s |
| Frictional pressure drop (Atkinson) | p = k × O × L × V² ÷ A | p in Pa (N/m²) |
| Airway resistance | R = k × O × L ÷ A³ | Ns²/m⁸ |
| Square law | p = R × Q² | p in Pa |
| Series | R = R₁ + R₂ + R₃ ...; Q is the same | |
| Parallel | 1/√R = 1/√R₁ + 1/√R₂ + ...; p is the same | |
| Equivalent orifice | A_e = 1.19 × Q ÷ √p = 1.19 ÷ √R | A_e in m² |
| Velocity pressure | p_v = ½ × ρ × V² | p_v in Pa |
| Natural ventilation pressure | NVP = g × H × (ρ_down − ρ_up) | Pa |
| Air power | P_air = Q × p | W (if Q in m³/s, p in Pa) |
| Fan efficiency | η = air power ÷ shaft power | |
| Fan laws (same fan and system) | Q ∝ N, p ∝ N², Power ∝ N³ | N = speed (rpm) |
| Gas dilution | Q = q ÷ (C_allowed − C_intake) | C as fractions |
| Sensible heat carried by air | H = ρ × Q × c_p × ΔT | kW (c_p in kJ/kg·K) |
| Water gauge to pascal | 1 mm w.g. = 9.81 Pa |
1. Air quantity and velocity
Q is the volume of air passing a section every second, A is the cross-sectional area and V is the mean velocity. This is the equation you will use in nearly every problem. Watch the units: many questions give Q in m³/min, and you must convert (1 m³/s = 60 m³/min) before using the pressure formulas.
Pitot tube: the velocity pressure is p_v = ½ρV², so V = √(2p_v/ρ). With ρ = 1.2 kg/m³, V = 1.29 × √p_v. For example, if p_v = 15 Pa, then V = √(2 × 15 ÷ 1.2) = √25 = 5 m/s.
2. Atkinson's equation and airway resistance
When air flows through a roadway, friction against the walls causes a pressure loss. Atkinson's equation gives it:
Here k is the coefficient of friction (in Ns²/m⁴, standard-density value), O is the perimeter, L is the length, V is the velocity and A is the area. Grouping the constants gives the resistance and the square law:
Density correction: k is given for standard air (1.2 kg/m³). For other densities, k = k_std × (ρ ÷ 1.2).
Worked example 1
A roadway is 4 m wide and 3 m high (A = 12 m², O = 14 m), 800 m long, with k = 0.012 Ns²/m⁴ and Q = 30 m³/s.
- R = 0.012 × 14 × 800 ÷ 12³ = 134.4 ÷ 1,728 = 0.0778 Ns²/m⁸
- p = R Q² = 0.0778 × 30² = 0.0778 × 900 = 70 Pa
- Check with velocity: V = 30 ÷ 12 = 2.5 m/s, and p = 0.012 × 14 × 800 × 2.5² ÷ 12 = 840 ÷ 12 = 70 Pa. ✓
- Equivalent orifice: A_e = 1.19 ÷ √0.0778 = 1.19 ÷ 0.279 ≈ 4.27 m²
3. Series and parallel airways
Series (one after another)
- The same quantity of air flows through each airway: Q is the same.
- The pressure drops add up: p_total = p₁ + p₂ + ...
- The resistances add up: R_total = R₁ + R₂ + ...
Parallel (splits)
- The same pressure drop acts across each branch: p is the same.
- The quantities add up: Q_total = Q₁ + Q₂ + ...
- The resistances combine with square roots: 1/√R_total = 1/√R₁ + 1/√R₂ + ...
- The share of air in a branch: Q_i ÷ Q_total = (1/√R_i) ÷ Σ(1/√R_j)
Equivalent orifice version: in parallel, A_e,total = A₁ + A₂ + ... (areas simply add). In series, 1/A_e² = 1/A₁² + 1/A₂² + ...
Worked example 2: two splits
A total of 40 m³/s splits into two parallel airways with R₁ = 0.4 and R₂ = 0.9 Ns²/m⁸.
- 1/√R₁ = 1/0.6325 = 1.5811 and 1/√R₂ = 1/0.9487 = 1.0541, so the sum = 2.6352
- Q₁ = 40 × 1.5811 ÷ 2.6352 = 24 m³/s
- Q₂ = 40 × 1.0541 ÷ 2.6352 = 16 m³/s
- Pressure drop = R₁Q₁² = 0.4 × 24² = 0.4 × 576 = 230.4 Pa. Check with the other branch: 0.9 × 16² = 0.9 × 256 = 230.4 Pa. ✓
- Combined resistance = (1/2.6352)² = 0.144, and 0.144 × 40² = 230.4 Pa. ✓
4. Equivalent orifice
The equivalent orifice of a mine is the area of a sharp-edged hole that would pass the same air quantity under the same pressure difference. It is a convenient way to describe how "tight" or "open" a mine is. Starting from the orifice equation Q = C_d × A × √(2p/ρ) with C_d = 0.65 and ρ = 1.2 kg/m³:
Use it with the mine's total pressure and total quantity. A smaller A_e means a harder-to-ventilate mine, and a larger A_e means an easier one. Textbooks give different classification bands, so use the one your syllabus follows.
5. Natural ventilation pressure
Air in the downcast shaft is cooler and denser than air in the upcast, and this difference drives a natural flow:
H is the vertical depth and ρ is the average density of the air column in each shaft.
Worked example 3
H = 400 m, ρ_down = 1.20 kg/m³ and ρ_up = 1.16 kg/m³.
NVP = 9.81 × 400 × (1.20 − 1.16) = 9.81 × 400 × 0.04 = 157 Pa. It helps the main fan in winter, and it can oppose or even reverse the flow in summer if the temperature relationship changes.
6. Fans: power, efficiency and fan laws
With Q in m³/s and p in Pa, the answer is in watts. Fan total pressure = fan static pressure + velocity pressure at the fan outlet. The fan operates where its characteristic curve meets the mine characteristic p = RQ².
Fan laws (same fan, same system, same density)
- Quantity: Q₂ / Q₁ = N₂ / N₁
- Pressure: p₂ / p₁ = (N₂ / N₁)²
- Power: P₂ / P₁ = (N₂ / N₁)³
- If the density changes, pressure and power scale directly with density.
Fans in series add their pressures at the same quantity. Fans in parallel add their quantities at the same pressure.
Worked example 4: air power and shaft power
In example 2 the mine passes 40 m³/s at 230.4 Pa.
- Air power = 40 × 230.4 = 9,216 W ≈ 9.2 kW
- If the fan efficiency is 70 percent, shaft power = 9,216 ÷ 0.7 ≈ 13,166 W ≈ 13.2 kW
Worked example 5: increasing the fan speed by 20 percent
N₂/N₁ = 1.2.
- Q = 40 × 1.2 = 48 m³/s
- p = 230.4 × 1.2² = 230.4 × 1.44 = 331.8 Pa
- Air power = 9.216 × 1.2³ = 9.216 × 1.728 = 15.9 kW
Notice how a 20 percent increase in speed raises the power by about 73 percent. Speed is expensive.
7. Quantity needed to dilute gases
q is the gas emission rate in m³/s, and the concentrations are expressed as fractions (1 percent = 0.01).
Worked example 6
Methane emission is 0.4 m³/s and the allowed concentration (assumed here as 1 percent for illustration) is 0.01.
- With clean intake air: Q = 0.4 ÷ 0.01 = 40 m³/s
- If the intake air already has 0.1 percent: Q = 0.4 ÷ (0.01 − 0.001) = 0.4 ÷ 0.009 = 44.4 m³/s
The statutory limits come from the regulations in force, not from this example. For the gases themselves, read Mine Gases and Firedamp Explained and Black Damp in Coal Mines.
8. Heat and humidity
For heat removal by the air stream, the sensible heat carried is:
Worked example 7
Q = 40 m³/s, ρ = 1.2 kg/m³, so the mass flow = 48 kg/s. If the air warms by 3 °C:
H = 48 × 1.005 × 3 = 144.7 kW of sensible heat.
Other formulas you may meet:
- Specific humidity: W = 0.622 × p_v ÷ (P − p_v), in kg of water vapour per kg of dry air (p_v is the partial pressure of vapour and P is the total pressure).
- Relative humidity: RH = p_v ÷ p_s, where p_s is the saturation pressure at the same temperature.
- Autocompression: air going down a shaft heats up as it is compressed, by roughly 1 °C per 100 m of depth for dry air. This is one reason deep mines become hot.
9. Ventilation networks: Kirchhoff's laws and Hardy Cross
- First law (junctions): the air quantities entering a junction equal those leaving: ΣQ = 0.
- Second law (meshes): around any closed loop, the pressure drops add up to zero (counting fans and natural ventilation pressure as sources): Σp = 0.
For complex networks, the Hardy Cross method corrects an assumed flow distribution step by step. For each loop, the correction is:
Add ΔQ to the flow in every branch of the loop (with signs), then repeat until ΔQ is negligibly small. In an exam you will normally be asked to do one iteration.
Common mistakes in GATE ventilation numericals
- Adding resistances in parallel. They combine using square roots, never directly.
- Forgetting to convert units: m³/min to m³/s, mm w.g. to Pa, and cm² to m².
- Using the wrong power of A. Resistance has A³ in the denominator, not A².
- Ignoring density. If air density is not 1.2, correct k, and remember that fan pressure and power scale with density.
- Mixing up the sign of NVP. It is helpful when the downcast air is denser than the upcast air, and the other way round it opposes the fan.
- Rounding too early. In splits, keep four decimal places in the square roots.
- Using a formula from memory without checking the units. A quick unit check catches most slips.
Practice questions (tap to check)
- A. 12 Pa
- B. 24 Pa
- C. 48 Pa
- D. 96 Pa
Show Answer
- A. 25 Pa
- B. 50 Pa
- C. 100 Pa
- D. 5 Pa
Show Answer
- A. 0.5
- B. 0.25
- C. 0.125
- D. 0.0625
Show Answer
- A. 10 percent
- B. 21 percent
- C. 33.1 percent
- D. 46.4 percent
Show Answer
- A. 0.6 m²
- B. 1.2 m²
- C. 2.38 m²
- D. 4.76 m²
Show Answer
- A. 59 Pa
- B. 118 Pa
- C. 236 Pa
- D. 353 Pa
Show Answer
How to study this topic
- Copy the formula sheet onto one page and keep it where you study.
- Solve one numerical from each section every day for a week. Time yourself.
- Always check your answer by computing it a second way (for example with V and with Q, as in worked example 1).
- Learn the concepts behind the numbers: why mines have splits, why booster fans are used and what natural ventilation does.
- Then practise previous year questions by topic. Our mine ventilation label has more material.
Frequently asked questions
Is mine ventilation important for GATE Mining?
Yes. It is part of the section on surface environment, mine ventilation and underground hazards, and it usually brings numerical questions that can be solved quickly once the formulas are clear.
Which formulas are asked most often?
Atkinson's equation, p = RQ², series and parallel combinations, equivalent orifice, fan laws and air power, and natural ventilation pressure.
Do I need to memorise the value of k?
Usually it is given in the question. If it is not, note the value you assume.
What is the standard air density?
1.2 kg/m³ is the value used in most mine ventilation problems.
Disclaimer: this article is for exam preparation and general education. Formulas, constants and conventions vary slightly by textbook. Statutory limits and requirements come from the regulations in force, so always check the latest official text.
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