Ventilation Program Day II Advanced Math & Problem Solving Presented by

Review of Formula Terms a = sectional area of airway measured in square feet (ft.2)   Rectangle or square………height x width = area Trapezoid ………. top width + bottom width x height = area 2 Circle………………………..¶ x r2 = area Note: ¶ = 3.1416

Perimeter o = perimeter of airway measured in linear feet   Rectangle or Square…………… Top width +bottom width + side 1 + side 2 Circle………………………………¶ x diameter Please go to work sheet and do questions 1 thru 5

Velocity v = velocity of air current measured in feet per minute (fpm)   Smoke tube………………………distance decimal time Anemometer…………………. Magnehelic……………………V.P. = 4003 X  i (Velocity Pressure)

Magnehelic The magnehelic is a differential pressure gage which can measure positive, negative, or differential pressure to within + 2 % accuracy. It is generally used when a high speed anemometer is not available.

Pitot Tube The Pitot tube (named after Henri Pitot in 1732) measures a fluid velocity by converting the kinetic energy of the flow into potential energy. The conversion takes place at the stagnation point, located at the Pitot tube entrance (see the schematic). A pressure higher than the free-stream (i.e. dynamic) pressure results from the kinematic to potential conversion. This "static" pressure is measured by comparing it to the flow's dynamic pressure with a differential manometer.

Taking an air reading using a Magnehelic and a Pitot tube: When high velocity air movement will damage the anemometer.   Take magnehelic reading(inches of water), then use formula;  4003 x  i. = V.P. or ventilation pressure , which is in fpm Ventilation Tubing Air Flow Pitot Tube Magnehelic 1..2..3..4..

Q A V q = quantity of air, in cubic feet per minute (cfm)   Quantity of air(cfm)......................................... q = a x v Velocity of air................................................…v = q a Area(when velocity and quantity are known).......................................... ………..a = q v Q A V Please go to work sheet and do questions 6 thru 11

Perimeters - Trapezoid: o = Top Width + Bottom Width + Side 1 + Side 2 This formula is used to find the angle side(Z) of a right triangle. The height(Y) is given and the top and bottom portion of the trapezoid are given. To find X, subtract the top width from the bottom width, then divide by 2 Use Pythagoras Theorem: Z = X2 + Y2  Z Y = height X Complete by finding the perimeter by adding the top + the bottom + the right side + the left side.

Perimeters - Trapezoid Determine the perimeter of an entry 18 feet across the top, 19 feet across the bottom, and 6 feet high. Solution: X = 19 ft. - 18 ft. 2 X = 1.0 ft. X = .5 ft. Z =  X2 + Y2 Z =  (.5 ft.)2 + (6 ft.)2 Z =  (.25 ft.) + (36 ft.) Z =  (36.25 ft.) Z = 6.02 ft. o = Top Width + Bottom Width + Side 1 + Side 2 o = 18 ft. + 19 ft. + 6.02 ft. + 6.02 ft. o = 49.04 feet 18 ft. 6 ft. 19 ft.

Perimeters - Trapezoid Determine the perimeter of an entry 20 feet across the top, 23 feet across the bottom, and 5 feet 6 inches high. Solution: X = 20ft. - 23ft. 2 X = 3 ft. X = 1.5 ft. Z =  X2 + Y2 Z =  (1.5 ft.)2 + (5.5 ft.)2 Z =  (2.25 ft.) + (30.25 ft.) Z =  (32.5 ft.) Z = 5.7 ft. o = Top Width + Bottom Width + Side 1 + Side 2 o = 20 ft. + 23 ft. + 5.7 ft. + 5.7 ft. o = 54.4 feet 20’ 5’6” 23’

Perimeters - Trapezoid Solution: X =17ft. - 20ft. 2 X = 3 ft. X = 1.5 ft. Z =  X2 + Y2 Z =  (1.5 ft.)2 + (4 ft.)2 Z =  (2.25 ft.) + (16.0 ft.) Z =  (18.25 ft.) Z = 4.27 ft. o = Top Width + Bottom Width + Side 1 + Side 2 o = 20 ft. + 17 ft. + 4.27 ft. + 4.27 ft. o = 45.54 feet Determine the perimeter of an entry 17 feet across top, 20 feet across bottom, and 4 feet high. 17 ft. 4 ft. 20 ft.

Please go to work sheet and do questions 12 Solve this Problem: A mast is 50 feet high, the anchor pin is 30 feet away. How much wire rope is needed to secure the top of the mast to the anchor pin? Solution: First, identify that a right angle exists, then use Pythagoras Theorem Z =  X2 + Y2 Z =  (30 ft.)2 + (50 ft.)2 Z =  (900 ft.) + (2500 ft.) Z =  (3400 ft.) Z = 58.3 ft. 50ft. Please go to work sheet and do questions 12 30 ft.

Please go to question # 13 in the work sheet Using the U-tube To mine Outside 4 3 2 1 Add negative side and positive side for mine’s water gauge Please go to question # 13 in the work sheet

Formula Equations Atmospheric Air Pressure (Barometric pressure-Mercury) 1 inch Hg = 876 feet in air column Subtract the top Hg barometric reading from the Bottom Hg barometric reading Then multiply by 876 Top Reading Bottom Reading

Atmospheric Air Pressure What is the depth of the air shaft, if the Barometer reads 29.75 inches at top of the shaft and 30.95 inches a the bottom? Barometric Difference = Barometric Reading (Bottom) -Barometric Reading (Top) 1 (mercury) inch = 876 feet in (Barometric Pressure) air column Solution: Barometric Difference = Barometric Reading (Bottom) - Barometric Reading (Top) 30.95 - 29.75 = 1.2 inches 1.2 inches x 876 = 1,051.2 feet

Atmospheric Air Pressure What is the depth of the air shaft, if the Barometer reads 29.35 inches at top of the shaft and 29.65 inches a the bottom? Barometric Difference = Barometric Reading (Bottom) -Barometric Reading (Top) 1 (mercury) inch = 876 feet in (Barometric Pressure) air column Solution: Barometric Difference = Barometric Reading (Bottom) - Barometric Reading (Top) 29.65 - 29.35 = 0.3 inches 0.3inches x 876 = 262.8 feet Please go to question # 14 in the work sheet

Please go to work sheet and do question 15 Water (gallons)…………………………………….1 cubic foot = 7.46 gallons Water (weight)………………………………………1 cubic foot = 62.4 lbs Please go to work sheet and do question 15

Rubbing Surface (ft2) s = lo Rubbing Surface = Length x Perimeter S Formula Equations Rubbing Surface (ft2) s = lo Rubbing Surface = Length x Perimeter S L O

Practice Problem - Rubbing Surface An entry is 10 feet high and 22 feet wide with a total length of 2,000 ft. What is the rubbing surface? s = lo o = Top Width + Bottom Width + Side 1 + Side 2 Solution: o = W1+W2 +S1+S2 o = 10’+22’+10’+22’ o = 64 ft. s=lo s = 2,000 ft x 64 ft. s = 128,000 sq. ft.

Practice Problem - Rubbing Surface An entry is 12 feet high and 18 feet 6 inches wide with a length of 1,500 feet. What is the rubbing surface? s = lo o = Top Width + Bottom Width + Side1 + Side2 Solution: o = W1+W2 +S1+S2 o = 12’+18.5’+12’+18.5’ o = 61.0 ft. s=lo s = 1,500 ft x 61.0 ft. s = 91,500 sq. ft.

Practice Problem - Rubbing Surface An entry is 5 feet high and 19 feet wide and 1,750 feet long. What is the rubbing surface? s=lo o = Top Width + Bottom Width + Side 1 + Side 2 Solution: o = W1+W2 +S1+S2 o = 5’+19’+5’+19’ o = 48.0 ft. s=lo s = 1,750 ft x 48.0 ft. s = 84,000 sq. ft.

Practice Problem Rubbing Surface ; Trapezoid Solution: X = Bottom Width - Top Width 2 X = 22’ - 18’ X = 4’ X = 2’ Z =  (X2 + Y2) Z =  (22+102) Z =  (4+100) Z =  (104) Z = 10.19 ft. o = Top + Bottom+ Side1+Side2 o = 18’+(2+18+2)+10.19’+10.19’ o = 60.38 ft. S = lo s = 3,000’ x 60.38 ft. s = 181,140 sq. ft. An entry measures 18 feet across the top and 22 feet across the bottom and 10 feet high with a length of 3,000 feet. What is the rubbing surface? o = Top Width + Bottom Width + Side 1 + Side 2 Pythagoras’s Theorem: Z =  (X2 + Y2) X = Bottom Width - Top Width 2 s = lo 18’ Z Y 10’ X 22’

Practice Problem - Rubbing Surface ; Circle What is the rubbing surface of a circular shaft 3,500 feet long with a diameter of 18 feet? o = ¶ x Diameter (¶ = 3.1416) s = lo Solution: o = ¶ x Diameter o = 3.1416 x 18’ o = 56.5488 ft. s = lo s = 3,500’ x 56.5488’ s = 197,920.8 sq.ft.

Practice Problem - Rubbing Surface ; Circle What is the rubbing surface of a circular shaft 2,500 feet long with a diameter of 15 feet 6 inches? o = ¶ x Diameter (¶ = 3.1416) s = lo Solution: o = ¶ x Diameter o = 3.1416 x 15.5’ o = 48.6948 ft. s = lo s = 2,500’ x 48.698’ s = 121,737.0 sq.ft. Please go to work sheet and do question 16 & 17

Formulas for Methane Evaluation Quantity of Gas – CH4/cfm QG Quantity of Return Air - cfm QR Percent of Gas %G Quantity of Intake Air - cfm Qr

Formulas for Methane Evaluation METHANOMETER CONVERSION: .5% of Methane = .005 (2 decimal places) 1.0% of Methane = .01 For Quantity of Methane in a 24 hour period: QG (cfm) X 60 (minutes) X 24 (hours)

Formulas for Methane Evaluation The formula to find the quantity of gas (CFM) when the percent of gas and the quantity of return air are known: QG = QR X %G The formula to find the Percent of Gas when the quantity of gas and the Quantity of return air are known: %G = _QG_ QR The formula to find the quantity of return air when the quantity of gas and the percent of gas are known: QR = _QG_ %G Algebraic Circle QG QR %G

Methane Evaluation A return airway has a quantity of 11,000 CFM, which has 0.4% gas. What is the quantity of gas? QG = QR X %G Solution: QG = QR X %G QG = 11,000 CFM x .004 QG = 44 CFM CH4

Methane Evaluation A return airway has a quantity of 32,000 CFM, which has 0.1% gas. What is the quantity of gas? QG = QR X %G Solution: QG = QR X %G QG = 32,000 CFM x .001 QG = 32 CFM CH4

Methane Evaluation A return airway has a quantity of 17,500 CFM, which has 2.0% gas. What is the quantity of gas? QG = QR X %G Solution: QG = QR X %G QG = 17,500 CFM x .02 QG = 350 CFM CH4

(convert to percentage) Methane Evaluation A return airway has a quantity of 12,500 CFM, with 110 CFM/CH4. What is the percentage of gas? %G = _QG_ QR Solution: %G = _QG_ QR %G = 110 CFM 12,500 CFM %G = 0.0088 (convert to percentage) .88 % CH4 (round off) .9 % CH4

(convert to percentage) Methane Evaluation Solution: A = ¶ x R2 A = 3.1416 x 8.52 A = 3.1416 x 72.25 A = 226.98 sq. ft. Q = AV Q = 226.98 ft2 x 180 fpm Q = 40,856 CFM %G = _QG_ QR %G = 75 CFM 40,856 CFM %G = 0.0018 (convert to percentage) .18 % CH4 (.2 % CH4) A return leg of an air shaft has a diameter of 17 feet, with a velocity of 180 fpm, and a quantity of gas of 75 CFM/CH4. What is the percentage of gas? %G = _QG_ QR A = ¶ x R2 Q = AV

Methane Evaluation QR = _QG_ %G QR = _QG_ QR = 120 CFM/CH4 %G .02 The quantity of gas in the return airway was 120 CFM/CH4 with 2.0 % CH4. What was the quantity? QR = _QG_ %G Solution: QR = _QG_ %G QR = 120 CFM/CH4 .02 QR = 6,000 CFM

Please go to work sheet and do question 18 thru 21 Methane Evaluation The quantity of gas in the return airway was 95 CFM/CH4 with .5 % CH4. What was the quantity? QR = _QG_ %G Solution: QR = _QG_ %G QR = 95 CFM/CH4 .005 QR = 19,000 CFM Please go to work sheet and do question 18 thru 21

24 hour Methane Evaluation Example: A Methanometer reading of 1.0% in the return. The Anemometer reading was 200,000 cfm. Solution: QG (cfm) X 60 (minutes) X 24 (hours) QG = .01 X 60 (minutes) X 24 (hours)

Please go to work sheet and do question 22 & 23 Methane Evaluation Solution: A = HW A = 10’ x 20’ A = 200 ft2 Q = AV Q = 200 ft2 x 150 fpm Q = 30,000 CFM QG = QR X %G QG = 30,000 CFM x .01 QG = 300 CFM/CH4 QG (CFM) x 60 (minutes) x 24(hours) 30 x 60 x 24 432,000/CH4/24 hour A mine entry measured 10’ high and 20’ wide and the anemometer reading was 150 fpm, the methane reading was 1.0 %. What is quantity of gas liberated in a 24 hour period? A = HW Q = AV QG = QR X %G QG (CFM) x 60 (minutes) x 24 (hours) Please go to work sheet and do question 22 & 23

Formulas for Methane Evaluation The formula to find the quantity of return air when the quantity of gas and quantity of intake air are known: QR = Qr + QG

Formulas for Methane Evaluation The formula to find the amount of air to add to reduce the percent of gas in an air current: Air to add = QG - QR new % G To find total volume of air, do not subtract the return air

Methane Evaluation – CH4 … Air to Add The quantity of return air was 10,500 cfm and found to contain 2.3 % CH4. How much extra air is needed to reduce the methane content to 1.5 %. QG = QR x %G Air to add = QG - QR new % G Solution: QG = QR x %G QG = 10,500 cfm x .023 QG = 241.5 CFM/CH4 Air to add = QG - QR new % G Air to add = 241.5 cfm/ch4 - 10,500 cfm .015 Air to add = 16,100 - 10,500 cfm Air to add = 5,600 cfm

Methane Evaluation - CH4 Air to Add The quantity of return air was 14,500 cfm and found to contain 3.4 % CH4. What is the total volume needed to reduce the methane content to 2.0 %. QG = QR x %G Air to add = QG - QR new % G Solution: QG = QR x %G QG = 14,500 cfm x .034 QG = 493 CFM/CH4 Air to add = QG - QR new % G Air to add = 493 cfm/ch4 - (14,500 cfm) .02 new % G Total Volume = 24,650 cfm Please go to work sheet and do question 24

Formula Equations: Equivalent Orifice (ft2) E.O. = .0004 X Q (new air reading) I This is formula for calculating Regulators

Equal Orifice  E.O. = .0004 x Q (new)  I E.O. = .0004 x 18,000 cfm If the new section requires 18, 000 cfm & the water gauge is 1.2 inches, what is the size of the regulator need to be? E.O. = .0004 x Q (new)  I E.O. = .0004 x 18,000 cfm  1.2 in. E.O. = 7.2 1.09 E.O. = 6.6 sq.ft.

E.O. = .0004 x 17,500 cfm  2.8 in. E.O. = 7.0 1.67 E.O. = 4.19 sq.ft. Equal Orifice  If a new section requires 17,500 cfm, the water gauge is 2.8 inches, what is the size of the regulator? E.O. = .0004 x Q (new)  I E.O. = .0004 x 17,500 cfm  2.8 in. E.O. = 7.0 1.67 E.O. = 4.19 sq.ft. Please go to question # 25 thru 28 in the work sheet

Formula Equations: Horsepower h = __u___ 33,000 Horsepower = Units of Power  33,000 (One horsepower equals 33,000 units of power or it can move 33,000 pounds one foot vertically in one minute, 330 pounds 100 feet vertically in one minute, or 33 pounds 1,000 feet vertically in one minute.)

k = coefficient of friction (The Resistance Of One Square Foot Of Rubbing Surface of an entry To An Air Current With A Velocity Of One Foot Per Minute) {.00000002} Mine entry Air Pressure Air Velocity Mine Entry

Horsepower  The entry is 3,000 feet long, it is 5 feet high, 20 feet wide. How much horse- power is required to move 350 fpm of air? h = __u___ 33,000 u = ksv3 k = .00000002 s = lo v3 = Solution: V3 = (350)3 V3 = 42,875,000 fpm o = S1+S2+top+bottom o = 5’+20’+5’+20’ o = 50 ft s = lo s = 3,000 ft x 50 ft s = 150,000 sq. ft. next slide

Horsepower  (cont.) 37 Solution: (cont.) u = ksv3 u = .00000002 x 150,000 sq. ft. x 42,875,000 fpm u = 128,625 foot-pounds per minute h = __u___ 33,000 h = 128,625 foot-pounds per minute h = 3.897 Horsepower Please go to question # 29 & 30 in the work sheet

Fan Chart Exercise Please go to question # 31 thru 35 in the work sheet

Sling Psychrometer To operate — saturate the wick of the wet bulb thermometer in clean water and whirl the sling psychrometer until the temperature stops dropping. Read the two thermometers. Place wet bulb temperature over the dry bulb temperature scale on the slide rule — the arrow will then point directly to the accurate relative humidity. . Range on thermometers is 20 to 110°F.

Partial Relative Humidity Difference Between Dry Bulb and Wet Bulb Temperatures Relative Humidity None 100% 0.5° 96% 1.0° 93% 1.5° 89% 9.0° 44% 9.5° 42% 14.5° 19% 15.0° 17% 18.0° 5% Partial Relative Humidity   Temperature conversion Fahrenheit to Centigrade……………………Co = 32 - Fo temp. x .555 Centigrade to Fahrenheit…………………….Fo = Co x 1.8 + 32 Please go to question # 36 thru 40 in the work sheet