Transcript Ventilation - Math - the Mining Quiz List

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
2
 Circle………………………..¶ x r2 = area
Note: ¶ = 3.1416
x
height = area
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.
Ventilation Tubing
Air Flow
Pitot Tube
Magnehelic
Take magnehelic reading(inches of water), then use formula;
4003 x  i.
= V.P. or ventilation pressure , which is in fpm
 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
Please go to work
sheet and do
questions 6 thru
Algebraic Circle
11
Q
A
V
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.
18 ft.
6 ft.
19 ft.
Solution:
X = 19 ft. - 18 ft.
2
X = 1.0 ft.
2
X = .5 ft.
Z
Z
Z
Z
Z
o
o
=  X2 + Y2
=  (.5 ft.)2 + (6 ft.)2
=  (.25 ft.) + (36 ft.)
=  (36.25 ft.)
= 6.02 ft.
= Top Width + Bottom Width + Side 1 + Side 2
= 18 ft. + 19 ft. + 6.02 ft. + 6.02 ft.
o = 49.04 feet
Perimeters - Trapezoid
 Determine the perimeter
of an entry 20 feet across
the top, 23 feet across
the bottom, and 5 feet 6
inches high.
20’
5’6”
23’
 Solution:
X = 20ft. - 23ft.
2
X = 3 ft.
2
X = 1.5 ft.
Z
Z
Z
Z
Z
=
=
=
=
=
 X2 + Y2
 (1.5 ft.)2 + (5.5 ft.)2
 (2.25 ft.) + (30.25 ft.)
 (32.5 ft.)
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
 Determine the
perimeter of an entry
17 feet across top, 20
feet across bottom,
and 4 feet high.
17 ft.
4 ft.
20 ft.
 Solution:
X =17ft. - 20ft.
2
X = 3 ft.
2
X = 1.5 ft.
Z
Z
Z
Z
Z
=
=
=
=
=
 X2 + Y2
 (1.5 ft.)2 + (4 ft.)2
 (2.25 ft.) + (16.0 ft.)
 (18.25 ft.)
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
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.
50
ft.
Please go to work sheet and do
questions 12
30 ft.
Using the U-tube
Outside
4
To mine
3
2
1
0
1
Add negative side and
positive side for mine’s
water gauge
2
3
4
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
Top Reading
Subtract the top Hg barometric reading from the
Bottom Hg barometric reading
Then multiply by 876
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
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
Formula Equations
Rubbing Surface (ft2)
s = lo
Rubbing Surface = Length x Perimeter
S
Algebraic Circle
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:
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
X
10’
Y
22’
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 ; 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
1.0% of Methane = .01
(2 decimal places)
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
Algebraic Circle
The formula to find the Percent of Gas when
the quantity of gas and the Quantity of
return air are known:
QG
%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
%G
QR
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
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
Methane Evaluation
Solution:
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
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)
Methane Evaluation
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
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)
Methane Evaluation
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
Please go to work
sheet and do
question 22 & 23
432,000/CH4/24 hour
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 %.
Solution:
QG = QR x %G
Air to add =
Air to add =
QG
- QR
new % G
QG = QR x %G
QG = 10,500 cfm x .023
QG = 241.5 CFM/CH4
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 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 %.
Air to add =
Solution:
Q G = Q R x %G
QG = 14,500 cfm x .034
QG = 493 CFM/CH4
Air to add =
QG = QR x %G
QG
- QR
new % G
CH4 Air
QG
- QR
new % G
Air to add = 493 cfm/ch4 - (14,500
cfm)
.02 new % G
Total Volume =
Please go to work sheet and do question 24
24,650 cfm
Formula Equations:
Equivalent Orifice (ft2)
E.O. = .0004 X Q (new air reading)
I
This is formula for calculating Regulators
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.
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
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
33,000
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%
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