ventilation 1 - program presented by training staff bureau of deep mine safety basic math &...
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Ventilation 1 - ProgramVentilation 1 - Program
Presented by Training StaffBureau of Deep Mine Safety
Basic Math & Problem Solving
Review of Formula TermsReview 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 FORMULASCOMMON 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 ; RectanglePractice Problems – Area ; Rectangle
Determine the area of a mine entry that is 19 feet wide and 7 feet high:
7’
19’
Solution:A = W x HA = 19 x 7’A = 133 sq. ft.
Practice Problems – Area ; RectanglePractice Problems – Area ; Rectangle
Determine the area of a mine entry that is 18 feet wide and 5 feet, 6 inches high:
18’
5’6’’
Solution:A = W x HA = 5.5’ x 18’A = 99 sq. ft.
Practice Problems
3 Determine the area of a mine entry that is 17 feet 3 inches wide and 6 feet 9 inches high:
3 Solution:A = W x HA = 17.25’ x 6.75’A = 116.44 sq. ft.
6’9’’
17’3’’
COMMON AREA FORMULASCOMMON AREA FORMULAS
Trapezoid:
Area = Top Width + Bottom Width X Height
2
18’
19’
6’
Practice Problems – Area ; Trapezoid 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’ 2A = 37’ x 6’ 2A = 18.5’ x 6’A = 111.00 sq. ft.
Practice Problems – Area ; Trapezoid 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’ 2A = 42’ x 5’ 2A = 21’ x 5’A = 105 sq. ft.
5’
20’
22’
Practice Problems
6 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.
6 Solution:Area = Top Width + Bottom Width X
Height 2
A = 17’ + 20’ x 4.5’ 2A = 37’ x 4.5’ 2A = 18.5’ x 4.5’A = 83.25 sq. ft.
4’6’’
17’
20’
COMMON AREA FORMULAS - CircleCOMMON AREA FORMULAS - Circle
Circular:A = ¶ x D2
4 or
A = ¶ x R2
Please use the following For Pi………
¶ = 3.1416
radius
diameter
Practice Problems –Area ; CirclePractice 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 - CircleArea - 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
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 72.25A = 226.98 sq. ft.
17’
PerimetersPerimeters
Square or Rectangleo = Top Width + Bottom Width + Side 1 + Side
2
Remember, perimeter measured in linear feet
Practice Problem – Perimeter ; RectanglePractice 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 ; RectanglePractice 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 - CirclePerimeters - Circle
o = ¶ x Diameter
¶ = 3.1416
Diameter
Perimeter - CirclePerimeter - Circle
Determine the perimeter of a circular air shaft with a diameter of 17 feet, 6 inches.
Solution:
o = ¶ x Diametero = 3.1416 x 17.5 ft.o = 54.978 ft.
17’6”
Perimeter - CirclePerimeter - Circle
Determine the perimeter of a circular air shaft with a diameter of 20 feet
Solution:
o = ¶ x Diametero = 3.1416 x 20 ft.o = 62.83 ft.
20”
Perimeter - CirclePerimeter - 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 Diametero = 3.1416 x 18.0 ft.o = 56.548 ft.
9’
Formula EquationsFormula Equations
Quantity of Air (cfm)Q = AVQuantity = Area X Velocity
Velocity of air (fpm)V = _ Q_ AVelocity = Quantity Area
Area (when velocity and quantity a known)A = _Q_ V Area = Quantity Velocity
Q
A V
Algebraic Circle
Practice Problem - QuantityPractice 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 = WHQ = AV
Solution:A = WHA = 17.5’ x 9’A = 157.5 sq. ft.
Q = AVQ = (157.5 sq.ft.)(180
fpm)Q = 28,350 CFM
Practice Problem - QuantityPractice 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 = WHQ = AV
Solution:A = WHA = 18’ x 6.5’A = 117 sq. ft.
Q = AVQ = (117 sq.ft.)(110
fpm)Q = 12,870 CFM
Practice Problem - VelocityPractice 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 = WHA = 22 ft. x 10 ft.A = 220 sq. ft.
V = _Q_ AV = 11,380 CFM 220 sq.ft.V = 51.72 fpm
Practice Problem - AreaPractice 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 fpmA = 83.33 sq. ft.