assignment mine ventilation


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Page 1: Assignment Mine Ventilation
Page 2: Assignment Mine Ventilation

Q.1>What are the physiological effects of high wet bulb temperature in mines?

ANS>At high wet-bulb temperatures the rate of cooling gets a result,the body temperature rises.

TABLE-1.Rise in body temperature with wet-bulb temperature

The rise in temperature of the body varies from person to person and depends on the degree of acclimatization. A moderate rise in the body temperature of the order of 1.4k is not harmful,but when the body temperature rises above 312k and /or the heart rate exceeds 140 beats per minuteheat intolerance,that may ultimately lead to heat stroke ,appears.

Haldane puts the tolerable limit at 304.3 k for partially acclimatized men at rest in still saturated air.

Caplan reported that with the prevalent high dry-bulb temperatures in the KOLAR-GOLD fields of the order of 316-322 k,cases of heat strokes are frequent at wet-bulb temperature 307.5-308.5 kbut were low at a temperature below 305.5 k .

Wet-bulb temperatureIn k

Rise in body temperature In k

Up to 302.15 0.11-0.66

302.65-304.85 0.33-0.77

305.35-307.65 0.66-1.55

Above 307.65 1.44-1.90

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Q.2>Write in detail the different air conditioning systems that are used in mines.

ANS>The methods of improving the cooling power of mine air include Increasing the quantity of ventilating air Circulating dried air Cooling or refrigeration of circulating air


The increased quantity of air not only dilutes heat produced in the mine but also produces a higher air velocity which improves the cooling power of mine air.

There is a minimum quantity of air required for a mine for supporting men ,flame lamps Taking a min. of air velocity of 0.5 m s -1 in any part of the mine and considering

provision of comfortable environment for hard work(metabolic rate 270 W m-2),the maximum allowed wet –bulb temperature in any part of mine will be300.5 k .

It would be advisable to design the ventilation system on the basis of this wet-bulb temperature permissible law is higher ,i.e. 303.5 k in india .

However ,the increase in volume has its limitations. In every hot and humid mines where an increased air velocity produces little bodily comfort ,an extra volume of air will be of no help.

Moreover ,mine airways should be large enough to take extra quantity of air without causing excessive frictional pressure loss and excessive air velocity .

High pressure loss increases the power costs of ventilation and high air velocity raises dust in the roadways. hence both of these are undesirable .


We know a fairly high dry-bulb temperature can be tolerable if the relative humidity or the wet –bulb temperature is low .that is why in deep mines where the air temperature is high , maintaining the air dry helps a great deal in improving the working conditions.

There is no economical process of drying the air as such as refrigeration . Drying of mine airs by passing it through calcium desiccators or magnesium chloride or

silica gel becomes very costly . and the advantage gained by drying is greatly compensated by the heat produced by absorption which raises the temperature of mine air.

Spraying of fuel oil over the surface of mine air ways has been found to considerably reduce evaporation from the surface .


REFRIGERATION of mine air is necessary when its temperature becomes excessive so that no further increase in the quantity of air would improve environmental conditions

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.refrigeration plants are usually designed to produce tolerable environmental conditions throughout the year at the working places in the mines .

Normally , the air is cooled and dehumidified so that it is saturated at 275 to 278 k . It is then conducted to the workings as such or after mixing with a stream of uncooled air

so as to obtain the desired face temperature . Hence , a refrigeration plant should be designed to have a capacity , sufficient for cooling

the farthest face under the worst surface temperature conditions as may occur in the summer .

Calculation of cooling load of a refrigerator as follows : QC = q +Q ɖ ( H1 - H2 ) KW

Where Qc =total required cooling load ,q =total heat added to the mine air



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Q 3>What is the amount of heat extracted when 1 cubic meter of air at 310 k d.b. /301 k w.b. is cooled to 305 k d.b. ? what is the wet bulb temperature? What would be the amount of heat extracted if the air were cooled further to 278 k saturated ? what would be the mass of water vapour condensed in the process?

ANS >from psychometric chart ,at 310 k d.b./301 k w.b. ,enthalpy is 88.5 kJ /kg

When mine air is cooled to 305 k d.b. ,the wet bulb temperature will be 299.8 k and the enthalpy would be 83 kJ /kg.

The heat extracted from the mine air will be =change in enthalpy

=88.5 - 83 kJ/kg

= 5.5 kJ/kg

Now, the mine air is further cooled to 278 k saturated

At 310 k d.b. and 301 k w.b.,vapour pressure can be calculated as

Ew.b. = 0.6105 ҽ ( 17.27 t/237.3 +t) (1)

E =Ew.b. - 0.000644* P *(d.b.t. – w.b.t ) (2)

Vapour pressure E = 3.183 k Pa

Mixing ratio m =622 E /P - E (3)

= 18 .42 g /kg of dry air

Cpa1 (specific heat capacity of dry air )for the temp . range 273.15 k to 310 k

= 995.68 + 0.029 (273.15 +d.b.t /2 ) (4)

= 1287.255 J /kg.k

Cpv1 (specific heat of water vapour )for the range 273.15 k to 310 k

= 1553.7 + 0.645 ( 273.15 +d.b.t /2) +35169/ (273.15 +d.b.t )/2 (5)

= 1863.92 J/kg.k

L (latent heat of vapourization) at 273.15 k =2.5004 *10 6 J /kg

Therefore,enthalpy of air at 310 k d.b. /301 k w.b.

H = Cpa1(d.b.t -273.15) +0.001 m Cpv1(d.b.t – 273.15 ) +0.001 m L

=94.74 kJ/kg of dry air

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At temperature 278 k saturated ,vapour pressure can be calculated using eqn (1)

E1 = 0.878 kPa

Mixing ratio M1 (using eqn (3) ) = 4.974 g/kg of dry air

Amount of liquefied vapour = m – M1



Cpa2(specific heat of dry air ) using eqn (4)

=1003.67 J/kg.k

Cpv2(specific heat of water vapour) using eqn 5

=1859.06 J/kg.k

Cpw(specific heat of liquefied water vapour )

=4820.5 – 2.18 T J/kg.k

=4219.75 J/kg.k

Enthalpy of saturated air is H1 =1.00367 *4.85 +0.001 *4.974*1.85906*4.85 +0.001*4.974*2500.4+0.001*13.44*4.21975*4.85

=17.61 kJ/kg

The difference in enthalpy per 1 kg of air is =94.74 -17.61 =77.13 kJ/kg

The density of dry air 310 k/301k = (P-E)1000/287.1*d.b.t

=1.207 kg per cubic meter

1 m3 of air weighs 1.207 kg

Therefore total heat extracted from the mine air is 1.207 *77.13

=93.14 kJ

Mass of water vapour condensed is 93.14 = M *4.187 *(290.8 -278)

M =1.737 kg of water vapour


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