Basic Vent Math

8/17/2019 Basic Vent Math

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entilation 1 - Programentilation 1 - Program

Presented by Training Staff

Bureau of Deep Mine Safety

Basic Math & Problem

Solving


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evie! of "orm ula Term s

evie! of "orm ula Term s

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)


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#$MM$% &'& "$M()&S #$MM$% &'& "$M()&S

Rectangular or Square Dimension

Area = Height X Width

!ote "lease remember to convert inches into the #ecimalequivalent of one foot $ inches #ivi#e# by %2


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Practice Problems * &rea + ectangle Practice Problems * &rea + ectangle

Determine the area of amine entry that is %&feet wi#e an# ' feet

high

7’

19’

Solution

A = W x H

= %& '*A = 133 sq. ft.


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Practice Problems * &rea + ectangle Practice Problems * &rea + ectangle

Determine the area of amine entry that is %+feet wi#e an# feet, -

inches high

18’

5’6’’

Solution

A = W x H

= .* %+*A = 99 sq. ft.


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Practice Problems

Determine the area ofa mine entry that is%' feet inches wi#e

an# - feet & incheshigh

Solution

A = W x H

= %'.2* -.'* = %%-.// sq. ft.

6’9’’

17’3’’


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#$MM$% &'& "$M()&S #$MM$% &'& "$M()&S

0rape1oi#

Area = Top Width + Bottom Width X Height

2


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18’

19’

6’

Practice Problems * &rea + Trape,oidPractice Problems * &rea + Trape,oid

Determine the area ofa mine entry that is- foot high, an# %+

feet wi#e across thetop, an# is %& feetwi#e across thebottom.

Solution Area = Top Width + Bottom Width X Height

2

= %+* %&* -* 2

= '* -*

2

= %+.* -*

A = 111. sq. ft.


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Practice Problems * &rea + Trape,oidPractice Problems * &rea + Trape,oid

Determine the area ofa mine entry that is foot high, an# 23

feet wi#e across thetop, an# is 22 feetwi#e across thebottom.

Solution Area = Top Width + Bottom Width X

Height

2

= 23* 22* *

2

= /2* *

2 = 2%* *

A = 1! sq. ft.

5’

20’

22’


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Practice Problems

- Determine the area ofa mine entry that is /foot - inches high,

an# %' feet wi#eacross the top, an# is23 feet wi#e acrossthe bottom.

- SolutionArea = Top Width + Bottom Width X

Height

2

= %'* 23* /.*

2

= '* /.*

2 = %+.* /.*

A = "3.2! sq. ft.

4’6’’

17’

20’


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#$MM$% &'& "$M()&S - #ircle #$MM$% &'& "$M()&S - #ircle

4ircularA = # x $2

%

or

A = # x &2

'lease (se the follo)i*gor 'i,,,

# = 3.1%1-

radius

diameter


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Practice Problems *&rea + #ircle Practice Problems *&rea + #ircle

Determine the area ofa circle that has an#iameter of 23 feet

&inches.

Solution

A = ¶ x R2

& = 2.! = %3.' 2

= .%/%- %3.'2

= .%/%- %3'.-/3

A = 33".1- sq. ft.

R


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&rea - #ircle &rea - #ircle

Determine the area ofa circular air shaftwith a #iameter of 23

feet

Solution

A = ¶ x R2

& = 2 = %3 2

= .%/%- %32

= .%/%- %33

A = 31%.1- sq. ft.

20”


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Practice Problems

Determine the area ofa circle that has an#iameter of %' feet.

Solution

A = ¶ x r 2

R = %' = +. 2

= .%/%- +.2

= .%/%- '2.2

= 22-.&+ sq. ft.

17’


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Perimeters Perimeters

Square or Rectangleo = 0op 5i#th 6ottom 5i#th Si#e % Si#e 2

Remember, perimeter measure# in linear feet


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Practice Problem * Perimeter + ectangle Practice Problem * Perimeter + ectangle

Determine theperimeter of anentry ' feet highan# 22 feet wi#e.

Solution o = Top Width + Bottom Width + /ide 1 +

/ide 2

o = 22* 22* '* '* o = !" feet

7 ft.

22 ft.


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Practice Problem * Perimeter + ectangle Practice Problem * Perimeter + ectangle

Determine theperimeter of anentry - feet -inches high an#23 feet incheswi#e.

Solution

o = Top Width + Bottom Width + /ide 1 +/ide 2

o = -.* -.* 23.2* 23.2*

o = !3.! feet

6ft.6in.

20ft.3in.


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Perimeters - #ircle Perimeters - #ircle

o = 7 Diameter

7 = .%/%-

Diameter


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Perimeter - #ircle Perimeter - #ircle

Determine theperimeter of a circularair shaft with a

#iameter of %' feet, -inches.

Solution

o = 7 Diameter

o = .%/%- %'. ft.

o = !%.9" ft.

17’6”


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Determine theperimeter of a circularair shaft with a

#iameter of 23 feet

Solution

o = 7 Diameter

o = .%/%- 23 ft.

o = -2."3 ft.

20”


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Determine theperimeter of a circularair shaft with a ra#ius

of & feet.

Solution D = 2 r

D = 2 & ft.

D = %+ ft.

7 = .%/%-

o = 7 Diameter

o = .%/%- %+.3 ft.

o = !-.!%" ft.

9’


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"ormula 'uations "ormula 'uations

8uantity of ir (cfm)

8 = 9

8uantity = rea : 9elocity

9elocity of air (fpm)

9 = ; 8;

9elocity = 8uantity ÷ rea

rea (when velocity an# quantity a


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Practice Problem - .uantity Practice Problem - .uantity

in# the quantity of airpassing thru an entry%' feet - inches wi#e

an# & feet high, with%+3 fpm registere# onthe anemometer.

= 5>

8 = 9

/ol(tio*0

A = WH

= %'.* &*

= %'. sq. ft.

8 = 9

8 = (%'. sq.ft.)(%+3fpm)

= 2"3! 4


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Practice Problem - .uantity Practice Problem - .uantity

in# the quantity ofair passing thru an#entry %+ feet wi#ean# - feet - incheshigh, with %%3 fpmregistere# on theanemometer.

= 5>

8 = 9

Solution

= 5>

= %+* -.*

= %%' sq. ft.

8 = 9

8 = (%%' sq.ft.)(%%3fpm)

= 12" 4


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Practice Problem - Velocity Practice Problem - Velocity

5hat is the velocityin a entry %3 feethigh an# 22 feet

wi#e, with aquantity of %%,+34?@

= 5>

9 = ;8;

Solution

= 5>

= 22 ft. %3 ft.

= 223 sq. ft.

9 = ;8;

9 = %%,+3 4?

223 sq.ft.

5 = !1.2 fpm


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Practice Problem - &rea Practice Problem - &rea

n entry has %2,334? of air with avelocity of %3 fpm.

5hat is the area ofthe entry@

= ;8;

9

Solution

= ;8;

9 = %2,33 4?

%3 fpm

A = "3.33 sq. ft.

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Basic Vent Math