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Ventilation Program Day II

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

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Ventilation Program Day II

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  1. Ventilation ProgramDay II Advanced Math & Problem Solving Presented by

  2. 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

  3. 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

  4. 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)

  5. 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.

  6. 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.

  7. 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..

  8. 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

  9. 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.

  10. 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. 2 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 Perimeters - Trapezoid 18 ft. 6 ft. 19 ft.

  11. 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. 2 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 Perimeters - Trapezoid 20’ 5’6” 23’

  12. Determine the perimeter of an entry 17 feet across top, 20 feet across bottom, and 4 feet high. Solution: X =17ft. - 20ft. 2 X = 3 ft. 2 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 Perimeters - Trapezoid 17 ft. 4 ft. 20 ft.

  13. 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. Solve this Problem: 50ft. Please go to work sheet and do questions 12 30 ft.

  14. Using the U-tube To mine Outside 4 3 2 1 0 1 2 3 4 Add negative side and positive side for mine’s water gauge Please go to question # 13 in the work sheet

  15. 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

  16. 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

  17. 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 Atmospheric Air Pressure Please go to question # 14 in the work sheet

  18. 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

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

  20. 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 + Side2 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

  21. 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

  22. 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

  23. 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 Solution: X = Bottom Width - Top Width 2 X = 22’ - 18’ 2 X = 4’ 2 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. Practice Problem Rubbing Surface ; Trapezoid 18’ Z Y 10’ X 22’

  24. 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

  25. 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. Practice Problem - Rubbing Surface ; Circle Please go to work sheet and do question 16 & 17

  26. 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

  27. 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)

  28. 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 Formulas for Methane Evaluation Algebraic Circle QG QR %G

  29. 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

  30. 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

  31. 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 Methane Evaluation

  32. 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 Methane Evaluation

  33. 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 = ¶ xR2 Q = AV Solution: A = ¶ xR2 A = 3.1416 x 8.52 A = 3.1416 x 72.25 A = 226.98 sq. ft. Q = AV Q = 226.98 ft2 x180 fpm Q = 40,856 CFM %G = _QG_ QR %G = 75 CFM 40,856 CFM %G = 0.0018 (convert to percentage) .18 % CH4 (.2 % CH4) Methane Evaluation

  34. 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 Methane Evaluation

  35. 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 Methane Evaluation Please go to work sheet and do question 18 thru 21

  36. 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) 24 hour Methane Evaluation

  37. 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) 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 Methane Evaluation Please go to work sheet and do question 22 & 23

  38. 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

  39. 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

  40. 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

  41. 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 Methane Evaluation - CH4 Air to Add Please go to work sheet and do question 24

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

  43. Equal Orifice  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.

  44. 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

  45. 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.)

  46. 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

  47. 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 

  48. 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 33,000 h = 3.897 Horsepower Please go to question # 29 & 30 in the work sheet

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

  50. 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.

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