Mining Engineering Lecture Insights: Equipment and Ventilation
Delve into the world of mining engineering with this lecture covering equipment and natural ventilation in shaft systems. Explore concepts like air pressure differentials, water gauge measurement, air current horsepower calculation, and the relationship between air velocity and pressure.
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Tishk International University Engineering Faculty Petroleum & Mining Eng. Department MINING ENGINEERING PTR-324 Lecture 8 3dr. Grade- Spring Semester 2021-2022 Instructor: Professor, Dr. Hamed M. Jassim
8th. Lecture: Other Equipment & Aids Natural Ventilation:
The following equation may hold, in connection with the figure above: H2 = H1 + HA The weight of air column in the main Down-Cast shaft ( B ) is greater than the weight of air column in the Up Cast shaft ( A ) leading to initiate an air flow shaft ( B ) to shaft ( A ). Water Gauge ( WG ): The air pressure difference between the openings in the mine entries and the openings in the exits is measured according to the difference in the water gauge (WG) of the two openings in inches. This is accomplished by using a tube of shape ( U ), by putting one of its ends in the entry and the other end in the exit ( or one end is put in the DC shaft and the other end in the UC shaft ). WG ( in inches ) x 5.2 = lb / ft2 To convert the pressure difference from Water Gauge in inches to pressure units in ( lb/ft2 ) by using the followingrelation:
Air Current Horse Power (Hp): This is the power of the fan which pushes the needed quantity of air in the ventilation process. It also represents, generally, the rate of doing work with respect to time or the force which is needed to overcome the wasted energy in the air currents due to friction and length of ventilation distance, etc. This is calculated according to the following relationship: HP = P x Q / 0.55 The Relation Between Air Velocity ( V ) and Water Gauge Pressure ( WG ): Since the length of air head ( h ) is calculated from: h = P / w Where, P = air pressure in ( lb / ft2) w = air density in ( lb / ft3)
Since we can calculate the air velocity from the relation: V = 2 g h And by substitution, we get: V = 2 g P / w = 2 g x 5.2 WG / w Whereby, V will be in the unit of ( ft / s ), Or it could be in: V = 1097.5 WG / w Whereby, V will be in ( ft / Min ) We can also find the air pressure in ( WG ) using the following equation: WG = 0.000,000,831 V2w
