2Ventilation Fundamentals Air Quantity: Air quantity is the product of the air velocity times the cross-sectional area of the airway. Q = AV.Velocity: V is the rate of airflow in linear feet per minute.Area: A is the cross-sectional area of the entry or duct through which the air flows, expressed in square feet.Perimeter: O is the linear distance in feet of the airway rubbing surface at right angles to the direction of the airstream.Water Gage: This is a common instrument used in mine ventilation for measuring differential pressures in inches of water.
3Ventilation Fundamentals Static Pressure (SP): This is pressure, either negative or positive, exerted in all directions, measured in inches of water gage.Velocity Pressure (VP): This is pressure exerted by the kinetic energy of air movement, measured in inches of water gage.Total Pressure (TP): This is the algebraic sum of the static pressure and velocity pressure, either negative or positive.High velocity air measurement: Using a Pitot Tube to measure velocity pressure in inches of water, air velocity can be determined using the following equation:V (in fpm) = 4000
6Air MeasurementsFor measuring velocities from 120 to 2000 feet per minute, ordinary commercial types of medium-velocity vane anemometers are practical, convenient and accurate.The vane anemometer is a small windmill geared to a mechanical counter through a small clutch, which is engaged for recording revolutions.Pictured at the right is a mine supervisor determining the air velocity by traversing an entry with a vane anemometer.
7Fundamentals of Airflow The principles of airflow are:Airflow in a mine is induced by pressure differences between intake and exhaust openings.The pressure difference is caused by imposing some form of pressure at one point or a series of points in the ventilating system.The pressure created must be great enough to overcome frictional resistance and shock losses.Passageways, both intakes and returns, must be provided to conduct the airflow.Air always flows from a point of higher to lower pressure.Airflow follows a square-law relationship between volumes and pressures. In other words, twice the volume requires four times the pressure.Mine-ventilating pressures, with respect to atmospheric pressures, may be either positive (blowing) or negative (exhausting).The pressure drop for each split leaving from a common point and returning to a common point will be the same regardless of the air quantity flowing in each split.
8Fundamentals of Airflow Pressure LossesPressure losses are divided into two separate groups:Friction pressure losses caused by the resistance of the walls on the airstream.Shock pressure losses caused by abrupt changes in the velocity of air movement.The common method of measuring ventilating pressures producing circulation is equivalent inches of water gage. One inch of water equals a pressure of 5.2 lbs per sq. ft.For general and easy application, pressure loss in inches of water (H) is:H = RQ2where: R is the resistance factor of the airway or mine, andQ is the quantity of flow, expressed in units of 100,000 cfm.
9Fundamentals of Airflow Pressure LossesH = RQ2 can also be written as:H = KLOQ2 / 5.2A3where: K is the friction factor which can be provided by tables in mine ventilation texts,L is the length of the airway in feet,O is the perimeter of the airway in feet,V is the velocity in feet per minute, andA is the cross-sectional area of the airway in square feet
10Fan Systems and Requirements Fans induce airflow in underground mines. Mechanical ventilation is governed by the general Fan Laws:Air quantity varies directly as fan speed; in other words, twice thevolume requires twice the speed.Induced pressure varies directly as the fan speed squared; in otherwords, twice the fan speed develops four times the pressure.The fan’s input horsepower varies directly as the fan speed cubed;in other words, twice the volume requires eight times the power.The mechanical efficiency of the fan is independent of the fanspeed.
11Fan Systems and Requirements The performance of a fan in a ventilating system is determined by its characteristic curve (a matter of design controlled by the manufacturer) and the mine resistance.The resistance of the mine is a matter of layout and maintenance of the ventilating network and is controlled by the mine operator.Brake horsepower (HPb):HPb = [(H)(Q)] / [(6350) (Efan)]where: Efan is the fan efficiency,expressed as a decimal.
12Fan Systems and Requirements The amount of airflow induced in a mine will depend on the fan characteristic and mine resistance. See the figure at the right.Mine fans are available for most conditions of mine resistance and desired volume relationships. Modern fans are built with variable pitch blades that permit a wide range of application for the single fan.On the next page, you will see a graph of the intersection of a mine characteristic curve and a fan curve. The point of intersection is called the operating point. Notice how the operating point changes when the mine resistance is reduced.
14Ventilation Plan Requirements Requirements for ventilation plans for underground coal mines are specified in 30 CFR 75. Subpart D deals with ventilation, while Subpart E deals with combustible materials and rock dusting.Requirements for ventilation plans for underground metal and nonmetal mines are specified in 30 CFR 57. Subpart G deals with ventilation, while Subpart T deals with safety standards for methane.
15Review Questions (Answers on the next slide) A 1500-ft long slope with a cross-sectional area of 150 sq. ft passes 270,000 cfm. What is the head loss for the slope if R equals 0.32?1.96 inches of water gage2.12 inches of water gage2.33 inches of water gage2.55 inches of water gageThe slope in the previous problem is 10 ft high and 15 ft wide. Determine the value for K.10.815.457.674.9If, by design, the maximum velocity of air in an entry five feet high and twenty feet wide is 600 fpm, what is the maximum quantity that the entry can handle?50000 cfm60000 cfm75000 cfm90000 cfm
16Answers to the Review Questions A 1500-ft long slope with a cross-sectional area of 150 sq. ft passes 270,000 cfm. What is the head loss for the slope if R equals 0.32?c inches of water gageThe slope in the previous problem is 10 ft high and 15 ft wide. Determine the value for K.d. 74.9If, by design, the maximum velocity of air in an entry five feet high and twenty feet wide is 600 fpm, what is the maximum quantity that the entry can handle?b cfm
17ReferencesTextBise, Christopher, 2003, Mining Engineering Analysis, Society for Mining, Metallurgy, and Exploration, Inc.Kingery, Donald S., 1960, “Introduction to Mine Ventilating Principles and Practices,” U. S. Bureau of Mines Bulletin 589.Mine Safety and Health Administration, 30 CFR 57, and 30 CFR 75.