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## Ventilation 1 - Program

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**Ventilation 1 - Program**Presented by Training Staff Bureau of Deep Mine Safety Basic Math & Problem Solving**Review of Formula Terms**• a = sectional area of airway, in square feet (ft.2) • l = length of airway, in feet (ft.) • o = perimeter of airway, in feet (ft.) • s = rubbing surface, in square feet (ft2) • v = velocity of air current, in feet per minute (fpm) • q = quantity of air, in cubic feet per minute (cfm)**COMMON AREA FORMULAS**• Rectangular or Square Dimension: Area = Height X Width Note: Please remember to convert inches into the decimal equivalent of one foot - inches divided by 12**Determine the area of a mine entry that is 19 feet wide and**7 feet high: Solution: A = W x H A = 19 x 7’ A = 133 sq. ft. Practice Problems – Area ; Rectangle 7’ 19’**Determine the area of a mine entry that is 18 feet wide and**5 feet, 6 inches high: Solution: A = W x H A = 5.5’ x 18’ A = 99 sq. ft. Practice Problems – Area ; Rectangle 5’6’’ 18’**Determine the area of a mine entry that is 17 feet 3 inches**wide and 6 feet 9 inches high: Solution: A = W x H A = 17.25’ x 6.75’ A = 116.44 sq. ft. Practice Problems 6’9’’ 17’3’’**COMMON AREA FORMULAS**• Trapezoid: Area = Top Width + Bottom WidthX Height 2**Determine the area of a mine entry that is 6 foot high, and**18 feet wide across the top, and is 19 feet wide across the bottom. Solution: Area = Top Width + Bottom WidthX Height 2 A = 18’ + 19’ x 6’ 2 A = 37’ x 6’ 2 A = 18.5’ x 6’ A = 111.00 sq. ft. Practice Problems – Area ; Trapezoid 18’ 6’ 19’**Determine the area of a mine entry that is 5 foot high, and**20 feet wide across the top, and is 22 feet wide across the bottom. Solution: Area = Top Width + Bottom WidthX Height 2 A = 20’ + 22’ x 5’ 2 A = 42’ x 5’ 2 A = 21’ x 5’ A = 105 sq. ft. Practice Problems – Area ; Trapezoid 20’ 5’ 22’**Determine the area of a mine entry that is 4 foot 6 inches**high, and 17 feet wide across the top, and is 20 feet wide across the bottom. Solution: Area = Top Width + Bottom WidthX Height 2 A = 17’ + 20’ x 4.5’ 2 A = 37’ x 4.5’ 2 A = 18.5’ x 4.5’ A = 83.25 sq. ft. Practice Problems 17’ 4’6’’ 20’**COMMON AREA FORMULAS - Circle**• Circular: A = ¶ x D2 4 or A = ¶ xR2 Please use the following For Pi……… ¶ = 3.1416 diameter radius**Determine the area of a circle that has an diameter of 20**feet 9inches. Solution: A = ¶ xR2 R = 20.75 = 10.375 2 A = 3.1416 x 10.3752 A = 3.1416 x 107.640 A = 338.16 sq. ft. Practice Problems –Area ; Circle R**Determine the area of a circular air shaft with a diameter**of 20 feet Solution: A = ¶ xR2 R = 20 = 10 2 A = 3.1416 x 102 A = 3.1416 x 100 A = 314.16 sq. ft. Area - Circle 20”**Determine the area of a circle that has an diameter of 17**feet. Solution: A = ¶ x r2 R = 17 = 8.5 2 A = 3.1416 x 8.52 A = 3.1416 x 72.25 A = 226.98 sq. ft. Practice Problems 17’**Perimeters**• Square or Rectangle o = Top Width + Bottom Width + Side 1 + Side 2 Remember, perimeter measured in linear feet**Determine the perimeter of an entry 7 feet high and 22 feet**wide. Solution: o = Top Width + Bottom Width + Side 1 + Side 2 o = 22’ + 22’ + 7’ + 7’ o = 58 feet Practice Problem – Perimeter ; Rectangle 7 ft. 22 ft.**Determine the perimeter of an entry 6 feet 6 inches high and**20 feet 3 inches wide. Solution: o = Top Width + Bottom Width + Side 1 + Side 2 o = 6.5’ + 6.5’ + 20.25’ + 20.25’ o = 53.5 feet Practice Problem – Perimeter ; Rectangle 6ft.6in. 20ft.3in.**Perimeters - Circle**o = ¶ x Diameter ¶ = 3.1416 Diameter**Determine the perimeter of a circular air shaft with a**diameter of 17 feet, 6 inches. Solution: o = ¶ x Diameter o = 3.1416 x 17.5 ft. o = 54.978 ft. Perimeter - Circle 17’6”**Determine the perimeter of a circular air shaft with a**diameter of 20 feet Solution: o = ¶ x Diameter o = 3.1416 x 20 ft. o = 62.83 ft. Perimeter - Circle 20”**Determine the perimeter of a circular air shaft with a**radius of 9 feet. Solution: D = 2 x r D = 2 x 9 ft. D = 18 ft. ¶ = 3.1416 o = ¶ x Diameter o = 3.1416 x 18.0 ft. o = 56.548 ft. Perimeter - Circle 9’**Formula Equations**Algebraic Circle • Quantity of Air (cfm) Q = AV Quantity = Area X Velocity • Velocity of air (fpm) V = _ Q_ A Velocity = Quantity Area • Area (when velocity and quantity a known) A = _Q_ V Area = Quantity Velocity Q A V**Find the quantity of air passing thru an entry 17 feet 6**inches wide and 9 feet high, with 180 fpm registered on the anemometer. A = WH Q = AV Solution: A = WH A = 17.5’ x 9’ A = 157.5 sq. ft. Q = AV Q = (157.5 sq.ft.)(180 fpm) Q = 28,350 CFM Practice Problem - Quantity**Find the quantity of air passing thru and entry 18 feet wide**and 6 feet 6 inches high, with 110 fpm registered on the anemometer. A = WH Q = AV Solution: A = WH A = 18’ x 6.5’ A = 117 sq. ft. Q = AV Q = (117 sq.ft.)(110 fpm) Q = 12,870 CFM Practice Problem - Quantity**What is the velocity in a entry 10 feet high and 22 feet**wide, with a quantity of 11,380 CFM? A = WH V = _Q_ A Solution: A = WH A = 22 ft. x 10 ft. A = 220 sq. ft. V = _Q_ A V = 11,380 CFM 220 sq.ft. V = 51.72 fpm Practice Problem - Velocity**An entry has 12,500 CFM of air with a velocity of 150 fpm.**What is the area of the entry? A = _Q_ V Solution: A = _Q_ V A = 12,500 CFM 150 fpm A = 83.33 sq. ft. Practice Problem - Area