Ventilation Program Day II

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 squareheight 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 tubedistance
• 
  decimal time
•  
• Anemometer.
•  
•  
• MagnehelicV.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
Perimeters - Trapezoid

• 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

18 ft.
6 ft.
19 ft.
11
Perimeters - Trapezoid

• 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

20
56
23
12
Perimeters - Trapezoid

• 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

• Determine the perimeter of an entry 17 feet
  across top, 20 feet across bottom, and 4 feet
  high.

17 ft.
4 ft.
20 ft.
13
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.

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 mines
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
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

17
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
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
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 W1W2 S1S2
• o 10221022
• o 64 ft.
• slo
• s 2,000 ft x 64 ft.
• s 128,000 sq. ft.

21
Practice Problem - Rubbing Surface

• Solution
• o W1W2 S1S2
• o 1218.51218.5
• o 61.0 ft.
• slo
• s 1,500 ft x 61.0 ft.
• s 91,500 sq. ft.

• 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

22
Practice Problem - Rubbing Surface

• An entry is 5 feet high and 19 feet wide and
  1,750 feet long. What is the rubbing surface?
• slo
• o Top Width Bottom Width Side 1 Side 2

• Solution
• o W1W2 S1S2
• o 519519
• o 48.0 ft.
• slo
• s 1,750 ft x 48.0 ft.
• s 84,000 sq. ft.

23
Practice Problem Rubbing Surface Trapezoid

• Solution
• X Bottom Width - Top Width
• 2
• X 22 - 18
• 2
• X 4
• 2
• X 2
• Z ? (X2 Y2)
• Z ? (22102)
• Z ? (4100)
• Z ? (104)
• Z 10.19 ft.
• o Top Bottom Side1Side2
• o 18(2182)10.1910.19
• o 60.38 ft.

• 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
• Pythagorass Theorem Z ? (X2 Y2)
• X Bottom Width - Top Width
• 2
• s lo

18
Z
10
Y
X
22
24
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.

25
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
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
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
• 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
  

Algebraic Circle
QG
QR
G
29
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

30
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

31
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

32
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

33
Methane Evaluation

• Solution
• 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)

• 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

34
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

35
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
36
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)
  

37
Methane Evaluation

• 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

• 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)

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
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 .
• 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

41
Methane Evaluation - CH4 Air 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 .
• 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

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
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 S1S2topbottom
• o 520520
• o 50 ft
• s lo
• s 3,000 ft x 50 ft
• s 150,000 sq. ft.
• next slide

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 110F.

51
Partial Relative Humidity

  Temperature conversion Fahrenheit to
CentigradeCo 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