Mine Ventilation - Questions from DGMS 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

= (CO2,std/O2,std)O2,std 

= (0.03/20.93) x 16.2 = 0.023

N2 eq. = N2 wrt O2 

=(N2,std/O2,std) x O2,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 and PSU) by amiestudycircle.com.  WhatsApp channel: Miners Adda 

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

= (CO2,std/O2,std)O2,ret
= (0.03/20.93) x 19.3

= 0.028

N2 eq. = N2 wrt O2   

= (N2,std/O2,std)O2,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 and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required. WhatsApp channel: Miners Adda

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 and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required. WhatsApp channel: Miners Adda

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

Vexcess inert = V[(Cmethane/Cmethane, cr) – 1]

= 10,000 x [(7/5.8) – 1]

= 2068 m3

Here Cmethane,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 and PSU) by amiestudycircle.com.  WhatsApp channel: Miners Adda

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

Qv = 1800 m3/min = 30 m3/s

Ci = 0.1%

Cr,max = 0.4%

Production capacity = 1500 tpd

qg = ?

Using

Qv = [100qg/(Cr,max - Ci)] - qg

Putting values 

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

30 = 333.33qg – qg 

∴ qg =  0.09 m3/s per 1500 t

∴ qg = 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 and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

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 and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

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: Vt = ρpDpg/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/Vt

= 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

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