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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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Perimeter - #ircle Perimeter - #ircle
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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Perimeter - #ircle Perimeter - #ircle
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.