Ventilation 1 - Program Basic Math & Problem Solving 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)

Rectangular or Square Dimension: 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

Practice Problems – Area ; Rectangle 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. 7’ 19’

Practice Problems – Area ; Rectangle 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. 5’6’’ 18’

Practice Problems Solution: 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. 6’9’’ 17’3’’

Trapezoid: COMMON AREA FORMULAS Area = Top Width + Bottom Width X Height 2

Practice Problems – Area ; Trapezoid 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 Width X Height 2 A = 18’ + 19’ x 6’ A = 37’ x 6’ A = 18.5’ x 6’ A = 111.00 sq. ft. 18’ 6’ 19’

Practice Problems – Area ; Trapezoid 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 Width X Height 2 A = 20’ + 22’ x 5’ A = 42’ x 5’ A = 21’ x 5’ A = 105 sq. ft. 20’ 5’ 22’

Practice Problems Solution: 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 Width X Height 2 A = 17’ + 20’ x 4.5’ A = 37’ x 4.5’ A = 18.5’ x 4.5’ A = 83.25 sq. ft. 17’ 4’6’’ 20’

COMMON AREA FORMULAS - Circle Circular: A = ¶ x D2 4 or A = ¶ x R2 Please use the following For Pi……… ¶ = 3.1416 diameter radius

Practice Problems –Area ; Circle Determine the area of a circle that has an diameter of 20 feet 9inches. Solution: A = ¶ x R2 R = 20.75 = 10.375 2 A = 3.1416 x 10.3752 A = 3.1416 x 107.640 A = 338.16 sq. ft. R

Area - Circle Determine the area of a circular air shaft with a diameter of 20 feet Solution: A = ¶ x R2 R = 20 = 10 2 A = 3.1416 x 102 A = 3.1416 x 100 A = 314.16 sq. ft. 20”

Practice Problems Solution: A = ¶ x r2 A = 3.1416 x 8.52 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. 17’

Square or Rectangle Perimeters o = Top Width + Bottom Width + Side 1 + Side 2 Remember, perimeter measured in linear feet

Practice Problem – Perimeter ; Rectangle 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 7 ft. 22 ft.

Practice Problem – Perimeter ; Rectangle 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 6ft.6in. 20ft.3in.

Perimeters - Circle o = ¶ x Diameter ¶ = 3.1416 Diameter

Solution: Perimeter - Circle 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. 17’6”

Solution: Perimeter - Circle 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. 20”

Solution: Perimeter - Circle 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. 9’

Formula Equations 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 Algebraic Circle Q A V

Practice Problem - Quantity 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 - Velocity 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 - Area 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.