Ventilation Program Day II Advanced Math & Problem Solving Presented by

Ventilation ProgramVentilation ProgramDay IIDay II

Advanced Math & Problem Solving

Presented by

Review of Formula TermsReview of Formula Terms

a = sectional area of airway measured in square feet (ft.a = sectional area of airway measured in square feet (ft.22)) Rectangle or squareRectangle or square………height x width = area………height x width = area TrapezoidTrapezoid ………. ………. top width + bottom width top width + bottom width x height = area x height = area 22 CircleCircle………………………..¶ x r………………………..¶ x r22 = area = area

Note: ¶ = 3.1416Note: ¶ = 3.1416

PerimeterPerimeter

o = perimeter of airway measured in o = perimeter of airway measured in linear feetlinear feet

Rectangle or SquareRectangle or Square………………………… Top width +bottom width + side 1 + side 2 Top width +bottom width + side 1 + side 2

Circle………………………………¶ x diameterCircle………………………………¶ x diameter

Please go to work sheet and do questions 1 thru 5

v = velocity of air current measured in feet per minute (fpm)v = velocity of air current measured in feet per minute (fpm)

Smoke tube………………………Smoke tube………………………distance distance

decimal timedecimal time

Anemometer………………….Anemometer………………….

Magnehelic……………………V.P. = 4003 X Magnehelic……………………V.P. = 4003 X ii

(Velocity Pressure)(Velocity Pressure)

VelocityVelocity

MagnehelicThe magnehelic is a The magnehelic is a differential pressure differential pressure gage which can measure gage which can measure positive, negative, or positive, negative, or differential pressure to differential pressure to within within ++ 2 % accuracy. 2 % accuracy.

It is generally used It is generally used when a high speed when a high speed anemometer is not anemometer is not available.available.

Pitot Tube•The Pitot tube (named after The Pitot tube (named after Henri Pitot in 1732) Henri Pitot in 1732) measures a fluid velocity by measures a fluid velocity by converting the kinetic energy converting the kinetic energy of the flow into potential of the flow into potential energy. The conversion takes energy. The conversion takes place at the place at the stagnation pointstagnation point, , located at the Pitot tube located at the Pitot tube entrance (see the entrance (see the schematic). schematic).

•A pressure higher than the A pressure higher than the free-stream (i.e. dynamic) free-stream (i.e. dynamic) pressure results from the pressure results from the kinematic to potential kinematic to potential conversion. This "static" conversion. This "static" pressure is measured by pressure is measured by comparing it to the flow's comparing it to the flow's dynamic pressure with a dynamic pressure with a differential manometer. differential manometer.

Taking an air reading using a Magnehelic and a Pitot tube: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;Take magnehelic reading(inches of water), then use formula;

4003 x 4003 x i. = V.P. or ventilation pressure , which is in fpm i. = V.P. or ventilation pressure , which is in fpm

Magnehelic

Ventilation TubingVentilation Tubing

Air Flow

Pitot TubePitot Tube

q = quantity of air, in cubic feet per minute (cfm)q = quantity of air, in cubic feet per minute (cfm)

Quantity of air(cfm).........................................Quantity of air(cfm)......................................... q = a x v q = a x v

Velocity of air................................................…Velocity of air................................................…v = v = qq

a a Area(when velocity and quantityArea(when velocity and quantity

are known)..........................................are known).......................................... ………..a = ………..a = q q

vv Q

A V Algebraic Circle

Please go to work sheet and do questions 6 thru 11

Perimeters - Trapezoid:Perimeters - Trapezoid:

o = Top Width + Bottom Width + Side 1 + Side 2o = Top Width + Bottom Width + Side 1 + Side 2This formula is used to find the angle side(Z) of a right triangle. The height(Y) is given 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.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 2To find X, subtract the top width from the bottom width, then divide by 2

Use Pythagoras Theorem:Use Pythagoras Theorem: Z = XZ = X22 + Y + Y22

Y = height

Z

X

Complete by finding the perimeter by adding the top + the bottom + Complete by finding the perimeter by adding the top + the bottom + the right side + the left side.the right side + the left side.

Perimeters - TrapezoidPerimeters - 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. 2X = 1.0 ft. 2X = .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 2o = 18 ft. + 19 ft. + 6.02 ft. + 6.02 ft.

o = 49.04 feet

18 ft.

19 ft.

6 ft.


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. 2X = 3 ft. 2X = 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’

23’

5’6”


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

Solution:

X =17ft. - 20ft. 2X = 3 ft. 2X = 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 2o = 20 ft. + 17 ft. + 4.27 ft. + 4.27 ft.o = 45.54 feet

17 ft.

4 ft.

20 ft.

Solve this Problem:Solve this Problem:

A mast is 50 feet high, the A mast is 50 feet high, the anchor pin is 30 feet away. anchor pin is 30 feet away. How much wire rope is How much wire rope is needed to secure the top needed to secure the top of the mast to the anchor of the mast to the anchor pin?pin?

Solution:Solution:

First, identify that a right angle First, identify that a right angle exists, then use Pythagoras exists, then use Pythagoras TheoremTheorem

Z = Z = X X22 + Y + Y2 2

Z = Z = (30 ft.) (30 ft.)22 + (50 ft.) + (50 ft.)22

Z = Z = (900 ft.) + (2500 ft.) (900 ft.) + (2500 ft.)

Z = Z = (3400 ft.) (3400 ft.)

Z = 58.3 ft.Z = 58.3 ft.50ft.

30 ft.

Please go to work sheet and do questions 12

4

3

2

1

0

1

2

3

4

To mineOutside

Add negative side and positive side for mine’s water gauge

Using the U-tubeUsing the U-tube

Please go to question # 13 in the work sheet

Formula EquationsFormula Equations

Atmospheric Air Pressure (Barometric pressure-Mercury)

1 inch Hg = 876 feet in air column

Subtract the top Hg barometric reading from theBottom Hg barometric reading

Then multiply by 876

Top Reading

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

feetPlease 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 EquationsFormula Equations

Rubbing Surface (ft2)s = loRubbing Surface = Length x Perimeter

S

L O

Algebraic Circle

Practice Problem - Rubbing SurfacePractice 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


Solution:o = W1+W2 +S1+S2

o = 10’+22’+10’+22’o = 64 ft.

s=los = 2,000 ft x 64 ft.s = 128,000 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

Solution:o = W1+W2 +S1+S2

o = 12’+18.5’+12’+18.5’

o = 61.0 ft.

s=los = 1,500 ft x 61.0

ft.s = 91,500 sq. ft.


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=los = 1,750 ft x 48.0 ft.s = 84,000 sq. ft.

Practice Problem Practice Problem Rubbing Surface ; TrapezoidRubbing Surface ; Trapezoid

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?


Pythagoras’s Theorem: Z = (X2 + Y2)

X = Bottom Width - Top Width 2 s = lo

Solution:X = Bottom Width - Top Width 2X = 22’ - 18’ 2X = 4’ 2X = 2’

Z = (X2 + Y2) Z = (22+102)Z = (4+100) Z = (104)Z = 10.19 ft.

o = Top + Bottom+ Side1+Side2o =

18’+(2+18+2)+10.19’+10.19’o = 60.38 ft.

S = los = 3,000’ x 60.38 ft.s = 181,140 sq. ft.

18’

22’

10’Z

X

Y

Practice Problem - Rubbing Surface ; CirclePractice 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 Diametero = 3.1416 x 18’o = 56.5488 ft.

s = los = 3,500’ x 56.5488’s = 197,920.8 sq.ft.

Practice Problem - Rubbing Surface ; CirclePractice 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 Diametero = 3.1416 x 15.5’o = 48.6948 ft.

s = los = 2,500’ x 48.698’s = 121,737.0 sq.ft.

Please go to work sheet and do question 16 & 17

Formulas for Methane EvaluationFormulas 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


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)


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

QG

%GQR

Algebraic Circle

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


QG = QR X %G


QG = 32,000 CFM x .001

QG = 32 CFM CH4

Methane Evaluation


QG = QR X %G


QG = 17,500 CFM x .02QG = 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

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

Solution:A = ¶ x R2

A = 3.1416 x 8.52

A = 3.1416 x 72.25A = 226.98 sq. ft.

Q = AV

Q = 226.98 ft2 x 180 fpmQ = 40,856 CFM

%G = _QG_

QR

%G = 75 CFM40,856 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

.02QR = 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

.005QR = 19,000 CFM

Please go to work sheet and do question 18 thru 21

24 hour Methane Evaluation

Example:Example:

A Methanometer reading of 1.0% in the A Methanometer reading of 1.0% in the return. The Anemometer reading was return. The Anemometer reading was 200,000 cfm.200,000 cfm.

Solution:Solution:QQGG (cfm) X 60 (minutes) X 24 (hours) (cfm) X 60 (minutes) X 24 (hours)

QQGG = .01 X 60 (minutes) X 24 (hours) = .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 = HWA = 10’ x 20’A = 200 ft2

Q = AVQ = 200 ft2 x 150 fpmQ = 30,000 CFM

QG = QR X %G

QG = 30,000 CFM x .01QG = 300 CFM/CH4

QG (CFM) x 60 (minutes) x 24(hours)

30 x 60 x 24

432,000/CH4/24 hourPlease go to work sheet and do question 22 & 23

Formulas for Methane Formulas for Methane EvaluationEvaluation

The formula to find the quantity of The formula to find the quantity of return air when the quantity of gas return air when the quantity of gas and quantity of intake air are known:and quantity of intake air are known:

QQRR = Q = Qr r + Q+ QGG

Formulas for Methane Evaluation

The formula to find the amount of air to addair 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 %.

QG = QR x %G


new % G

Solution:QG = QR x %G

QG = 10,500 cfm x .023

QG = 241.5 CFM/CH4


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 - 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


new % G

Solution:QG = QR x %G

QG = 14,500 cfm x .034

QG = 493 CFM/CH4


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

Formula Equations:

Equivalent Orifice (ftEquivalent Orifice (ft22))E.O. = E.O. = .0004 X Q (new air reading).0004 X Q (new air reading)

II

This is formula for calculating This is formula for calculating RegulatorsRegulators

Equal Orifice If the new section requires 18, 000 cfm & If the new section requires 18, 000 cfm &

the water gauge is 1.2 inches, what is the the water gauge is 1.2 inches, what is the size of the regulator need to be?size of the regulator need to be?

E.O. = E.O. = .0004 x Q (new).0004 x Q (new) II

E.O. = E.O. = .0004 x 18,000 cfm.0004 x 18,000 cfm 1.2 in.1.2 in.

E.O. = E.O. = 7.27.2 1.091.09

E.O. = 6.6 sq.ft.E.O. = 6.6 sq.ft.

Equal Orifice If a new section requires 17,500 cfm, the If a new section requires 17,500 cfm, the

water gauge is 2.8 inches, what is the water gauge is 2.8 inches, what is the size of the regulator?size of the regulator?

E.O. = E.O. = .0004 x Q (new).0004 x Q (new) II

E.O. = E.O. = .0004 x 17,500 cfm.0004 x 17,500 cfm 2.8 in. 2.8 in.

E.O. = E.O. = 7.07.0 1.671.67

E.O. = 4.19 sq.ft.E.O. = 4.19 sq.ft.

Please go to question # 25 thru 28 in the work sheet

Formula Equations:

Horsepowerh = __u___ 33,000Horsepower = 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.)

Mine entry

k = coefficient of friction

(The Resistance Of One Square Foot Of Rubbing Surface of (The Resistance Of One Square Foot Of Rubbing Surface of an entry To An Air Current With A Velocity Of One Foot Per an entry To An Air Current With A Velocity Of One Foot Per Minute)Minute) {.00000002}

Mine EntryMine Entry

Horsepower

The entry is 3,000 feet long, it is 5 feet high, 20 feet wide. How much horsepower is required to move 350 fpm of air?

h = __u___ 33,000u = ksv3

k = .00000002 s = lov3 =

Solution:V3 = (350)3

V3 = 42,875,000 fpm

o = S1+S2+top+bottom

o = 5’+20’+5’+20’o = 50 ft

s = los = 3,000 ft x 50 fts = 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___ 33,000h = 128,625 foot-pounds per minute 33,000h = 3.897 Horsepower

Please go to question # 29 & 30 in the work sheet


Fan Chart Exercise

Sling Psychrometer

To operate — saturate the wick of the wet bulb thermometer in clean water To operate — saturate the wick of the wet bulb thermometer in clean water and whirl the sling psychrometer until the temperature stops dropping. and whirl the sling psychrometer until the temperature stops dropping.

Read the two thermometers. Place wet bulb temperature over the dry bulb Read the two thermometers. Place wet bulb temperature over the dry bulb temperature scale on the slide rule — the arrow will then point directly temperature scale on the slide rule — the arrow will then point directly to the accurate relative humidity. . Range on thermometers is 20 to to the accurate relative humidity. . Range on thermometers is 20 to 110°F.110°F.

Partial Relative Humidity

Difference BetweenDry Bulb and Wet BulbTemperatures

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 conversionTemperature conversion

Fahrenheit to Centigrade……………………CFahrenheit to Centigrade……………………Coo = 32 - F = 32 - Foo temp. x .555 temp. x .555

Centigrade to Fahrenheit…………………….FCentigrade to Fahrenheit…………………….Foo = C = Coo x 1.8 + 32 x 1.8 + 32


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Ventilation Program Day II Advanced Math & Problem Solving Presented by