Mine Ventilation - Questions from DGMS, GATE and PSU Exams

An Air Sample taken from old waterlogged workings showed the following composition. CO₂ -3.0 percent, O₂ - 16.2 percent, N₂ - 79.5 percent, CH₄ - 1.3 percent. (The composition of normal air by volume is O₂- 20.93 percent, N₂ - 79.04 percent & CO₂ - 0.03 percent). Find the Black damp content of the sample. (DGMS FCMC 2021)

Solution

CO₂ eq = CO₂ wrt O₂

= (CO_2,std/O_2,std)O_2,std 

= (0.03/20.93) x 16.2 = 0.023

N₂ eq. = N₂ wrt O₂ 

=(N_2,std/O_2,std) x O_2,ret = (79.04/20.93) x 16.2

= 61.12

Excess N₂ = 79.09 – 61.12 = 17.97

Excess CO₂ = 3 – 0.023 = 2.977

Blackdamp = 17.97 + 2.977 = 20.94%

These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

A sealed-off area air sample consists of 16.0% O₂, 3.0% CO₂, and the rest is N₂. Assume that the standard composition of atmospheric air is 21.0% O₂ and 79.0% N₂. The percentage of black damp in the air sample is ––– (round off to 2 decimal places). (GATE 2020)

 

Solution

CO₂ eq. = CO₂ wrt O₂

= (CO_2,std/O_2,std)O_2,ret     

= (0/21.0)O_2,ret = 0

Because CO₂ = 0%; O₂ = 21% in standard air.

N₂ eq. = N₂ wrt O₂

= (N_2,std/O_2,std)O_2,ret = (79/21) x 16 = 60.19

Because N₂ in standard air = 79%

Excess N₂ = 79 – 60.19 = 18.81

Excess CO₂ = 3.00 – 0.00 = 3.00

Black damp = 18.81 + 3.00 = 21.81

The analysis of a sample of air from old workings is reported to be O₂ – 19.3%, CO₂ – 0.4%, N₂ = 79.8%, CH₄ – 0.5%. Find the percentage of air and black damp in the sample, as well as composition of black damp. (AMIE Summer 2016)

Solution

Standard composition is not given.  These values can be taken as CO₂ – 0.03, O₂ – 20.93 and N₂ – 79.04 respectively (if not given).

CO₂ eq. = CO₂ wrt O₂

= (CO_2,std/O_2,std)O_2,ret

= (0.03/20.93) x 19.3

= 0.028

N₂ eq. = N₂ wrt O₂   

= (N_2,std/O_2,std)O_2,ret

= (79.04/20.93) x 19.3

 = 72.88

Excess N₂ = 79.04 – 72.88 = 6.16

Excess CO₂ = 0.4 – 0.028 = 0.372

Blackdamp = 6.16 + 0.372 = 6.532%

Atmosphere air  = O₂,ret + N₂ eq. + Eq CO₂ eq. 

= 19.3 + 72.88 + 0.028 = 92.208%

These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

In a sealed off area, the analysis report shows the following gas percentage O₂ - 19.2, CO - 0.008, CO₂ - 2, and N₂ - 78, the Graham's ratio is 一 (DGMS FCMC 2015).

Solution

Given that

O₂ – 19.2

CO – 0.008

N₂ – 78

Graham ratio = CO/O₂ absorbed

= 100CO/(0.265 x N₂ – O₂)

Putting values

Graham ratio = 100 x 0.008/(0.265 x 78 – 19.2)

= 0.54

An air sample inside a sealed off area indicates 2% of CO, 6% of O₂, 9% CO₂ & 83% N₂. The Graham's ratio expressed in % will be — (DGMS SCMC 2015)

Solution

O₂ – 6%

CO – 2%

N₂ – 83%

Graham ratio = CO/O₂ absorbed

= 100CO/(0.265 x N2 – O2)

Putting values

= 100 x 2/(0.265 x 83 – 6)

= 12.50

These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

An air sample taken from a return airway yields the following analysis: N₂ = 79.22 %, O₂ = 20.5 %, CO = 18 ppm. Find the Graham's ratio. (DGMS SCMC 2021)

Solution

Given data

N₂ = 79.22 %

O₂ = 20.5 %

CO = 18 ppm

= 18 x 10^–4%

(Because 1 ppm = 10^–4 %)

Graham ratio = CO/O₂ absorbed

= 100CO/(0.265 x N₂ – O₂)

= 100 x 18 x 10^–4/(0.265 x 79.22 – 20.5)

= 0.36

The point ‘A’ as shown in the Coward flammability diagram represents the gas composition of a sealed-off area of a coal mine. The volume of the sealed-off area is, 10000 m³. Inert gas is proposed to be injected into the sealed-off area so that the gas composition comes below the LEL (lower explosibility limit). The minimum volume of the inert gas required (at the same pressure as that of the sealed-off area), in m³ is_______. (round off to one decimal place). (GATE 2022)

Solution

V_{excess inert} = V[(C_methane/C_{methane, cr}) – 1]

= 10,000 x [(7/5.8) – 1]

= 2068 m³

Here C_methane,cr is the value of methane at nose which is taken as 5.8%. That of oxygen is taken as 12.1%.

These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

A mine worker inhales normal air, whereas the exhaled air contains 16.65% O₂ and 3.83% CO₂. The respiratory quotient of breathing for the worker is _____. (GATE 2013)

Solution

RQ = CO₂ exhaled/O₂ consumed

= 3.83/(20.93 - 16.65)

= 0.89

Intake air containing 0.2% methane enters a section of an underground mine where emission of methane is 0.05 m³/s. Assuming that the threshold limit value of methane is 1.25%, the minimum quantity of fresh air required in m³/s is —. (GATE 2015)

Solution

We know that

Q_v(C_i/100) + q_g = (Q_v + q_g).(C_r,max/100)

Given that

Q_v  =  ?

C_i = 0.2%

C_r,max = 1.25%q_g = 0.05 m³/s

Putting these values

Q_v(0.2/100) + 0.05 = (Q_v + 0.05)(1.25/100)  
∴ 0.2Q_v + 5 = (Q_v + 0.05) x 1.25 

∴ Q_v = 4.9375/1.05 = 4.70 m³/s

These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

In a longwall face, % of methane at intake and return gates indicate 0.1 % and 0.4 %, air quantity flowing at the face is 1800 m³/min, production from the face is 1500 TPD, methane emission rate in m³/tonne of coal output is equal to —. (DGMS FCMC 2021)

Solution

Given that

Q_v = 1800 m³/min = 30 m³/s

C_i = 0.1%

C_r,max = 0.4%

Production capacity = 1500 tpd

q_g = ?

Using

Q_v = [100q_g/(C_r,max - C_i)] - q_g

Putting values 

30 = [100q_g/(0.4 - 0.1] - q_g

30 = 333.33q_g – q_g 

∴ q_g =  0.09 m³/s per 1500 t

∴ q_g = 7766 m³/day per 1500 t

= 7766/1500 m³/day per tonne

= 5.184 m³/day per tonne

These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

An underground coal mine panel produces 520 tonnes/day deploying 220, 200 and 192 persons in three shifts. As per CMR 1957, the minimum quantity of air in m³/min to be delivered at the last ventilation connection of the panel is —. (GATE 2010)

Solution

As per CMR 153-2-(i)(2017), in every ventilating district not less than 6 m³ per minute of air per person employed in the district on the largest shift

The minimum quantity of air in m³/min to be delivered at the last ventilation connection will be decided based on maximum number of persons per shift. Here, in this case, it is 220.

∴ Minimum quantity of air = 6 x 220 = 1320 m³/min.

A rectangular mine airway of 2.0 m width and 2.5 m height has a bend with deflection of 𝜋/4 radian. If the radius of curvature of the bend is 4.0 m, the shock factor of the bend is (round off to three decimals)

(a) 0.014*

(b) 0.024

(c) 0.051

(d) 0.071 (GATE 2024)

Solution: 

Shock factor for a bend is

X = [0.25/R²√a] (2i/π)²

where R =radius ratio = r/W where r is radius of curvature and W is width.

a = aspect ratio = H/W where H is height.

i = angle of deflection in radian

Given values are:

W = 2 m

H = 2.5 m

r = 4 m

i = π/4 radian

Now, R = r/W = 4/2 = 2

a = H/W = 2.5/2 = 1.25

Hence, X = [0.25/(2²)√1.25][(2 x π/4)/π]²

= 0.014

These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

Polluted air with particulate matters of diameter 50 μm enter with a horizontal velocity of 1.0 m/s at a height of 0.5 m from the bottom of a dry settling chamber. The density of the particle is 2000 kg/m³ and dynamic viscosity of the air is 1.8 × 10⁻⁵ kg/m-s. Assume streamline flow and the density of air is negligible as compared to particles and uniform horizontal velocity of 1.0 m/s of gas and particles within the chamber. Considering particle settling follows Stoke’s law, the minimum length in m, of the chamber required for settling of the particle at its bottom, is ______. (round off up to 2 decimals). (GATE 2024)

Solution: V_t = ρ_pD_pg/18μ

= 2000 x (50 x 10^-6)² x 9.8/(18 x 1.8 x 10^-5)

= 0.15 m/s

Now, Time of fall = L/V_t

= 0.5 m/0.15 m/s = 3.33 s

Velocity = length of chamber needed for settling/time of fall

1 m/s = length/3.33

Hence, length of chamber required for settling = 3.33 m

A rectangular face of 2.0 m x 2.5 m dimension is blasted with 20 kg explosive in a 1000 m long drive. One kilogram of explosive produces 2200 cm³ of nitrous fumes. The face is ventilated with a duct, located 10.0 m away from the face, to dilute the fumes. The quantity of air, in m³/s to be circulated for reducing the concentration of nitrous fumes to 5 ppm within a period of 5 minutes, is ____. (round off up to 2 decimals)

[Use the relation, t = 2.303(V_m/Q)log(q/V_mc) + (V - V_m)/Q, where t = time, V_m = volume of the tunnel over which the mixing of the gases produced at the face, and air delivered by the fan occurs, 𝑐 = concentration at time, t, 𝑉 = volume of tunnel, 𝑄 = quantity of air flow, 𝑞 = total volume of nitrous fumes produced]. (GATE 2024)

Solution

V = 2 x 2.5 x 1000 = 5000 m³

q = 20 x 2200 cm³ = 44000 cm³ = 0.044 m³

c = 5 p.p.m. = 0.0005%

V_m = 10 x 2 x 2.5 = 50 m³

Now using given equation

5 = [2.303 x (50/Q)log(0.044/50 x 0.0005)] + [(5000 - 50)/Q]

Which gives Q = 905.76 m³/min = 15.096 m³/s 

These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.