air flow calculations 15-05-09

7/30/2019 Air Flow Calculations 15-05-09

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Air Flow CalculationsSize of pipes (Blow Room)

Cross-sectional area of a pipe = R2 = ( D/2 ) 2 = 42D (i)

Cross-sectional area of main Pipe = Area of sub pipes

( Dm/2 )2 x = ( D1/2 )2 x + ( D2/2 )2 x + ( D3/2 )2 x + Main Supply Pipe

Dm2 x = 1 ( D12+D22 + D32 + .)4 4 Dm Sub Pipes

Dm = D12 + D22 + D32+ (ii) D1Where

Dm = Dia of main Pipe D2

D1 = Dia of pipe no.1

D2 = Dia of pipe no. 2 D3D3 = Dia of pipe no. 3Section Q1, Q2 and Q3Basic Air Flow Equation:

Q = AV OR V =A

QOR A =

V

Q(iii)

Q = Air flow rate (volume) in cfmA = Area of opening in ft2

V = Air speed in fpm (ft per mint)

Rate of air flow (Volume of air) = Cross-Sectional Area x Velocity or (Linear Speed)Rate of air flow (cfm) = Cross-Sectional Area in ft2 x Linear speed in ft/ mint (fpm).

Section Q4 and Q5Q. No: 1 A main pipe in Blow-Room have to feed three branch pipes, their respective dias are 12", 10"

and 8". Find the dia of main supply pipe.Solution:

D1 = 12"D2 = 10"D3 = 8"

Dm = D12 + D22 + D32

= ( 12 )2 + ( 10 )2 + ( 8 ) 2

= ( 144 ) + ( 100 ) + ( 64 )= 308

= 17.55 inches


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Q. No. 2. A main stream pipe feeds 4 branches with the following dia 6", 8", 10" & 12" respectivelythe capacity of the main is only 90% to that of the branches, what should be the dia of mastream pipe?

Solution:

D1 = 6"D2 = 8"D3 = 10"D4 = 12"

Dm = 36 + 64 + 100 + 144

= 344 = 18.54 inches

Above dia is at 100% capacity, so at 90% capacity

Dm(90%) = 18.54 x 90/100= 16.69 inches

Q. No. 3. In a ventilated system, a rectangular suction duct trunk of 8" x 12" is to be replaced by acircular duct of 25 % greater capacity, what will be the dia of circular duct?

Solution:

Area of Cross-Section of a Rectangular Duct = 8 x 12= 96 sq inch.

Increased area of duct by 25% = 96 x 125/100

= 120 sq inch

Area = D24

120 = D2 4D2 = 120 x 4

D2 = 152.72

D = 152.72

D = 12.35 inches


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Q. No.4 The tower of a Blow Room is 10 ft sq in cross-section discharging into the base pipes withthe following particular.

1 Pipe of 12" dia having linear air speed of 3000 ft/min.1 pipe of 10" dia having linear air speed of 2000 ft/min.5 pipes each of 12" dia having linear speed of 1500 ft/min.

Calculate the velocity of air flow in ft/min in the tower.

Solution:

Q = A V

Volume of air in tower = Sum of volume of air in branch pipes.

V1 = Volume of air in first pipe =4

D2 x linear speed in ft/mint.

=4

x (12/12)2 x 3000 = 2357.14 ft3/mint.

V2 = Volume of air in 2nd pipe =

4

(10/12)2 x 2000

= 1091.26 ft3/mint.

V3 = Volume of 5 pipes = (4

D2 x 1500) x 5

= (4

(12/12)2 x 1500) x 5

= 5892.85 ft3

/mint.Total volume = Sum of volume of air in individual pipes

= V1 + V2 + V3= 2357.14 + 1091.26 + 5892.85= 9341.25 ft3 /min

Volume of air in tower = 9341.25 ft3/mint.Area of cross-section of tower = 10 x 10 = 100 sq ft

Linear speed of air = Volume

Area= 9341.25 = 93.4125 ft/mint100


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Q. No. 5 Areturn air grille has a free net area of 1728 sq. inchand the velocity is measured with ananeno meter at 500 ft/min (fpm). Find the air volume (cfm).

Solution:

Q= AV

1728 inch2 =12 ft2,

144Air Volume = 12 x 500 = 6000 cfm

Air Ventilation by Air Change Method:

Q0 (cfm) =(mint)60

Nxft 03

or N0 = cfm/changeQ0

(iv)

= 60NV

=changepermints

V

Q0 = Out side air quantity (cfm).

N0 = No. of out door air changes per hr.

Ft3 = Cubic contents of conditioned space. (Volume of conditioned space)

Q. No. 6 A room is 20 x 20 x 10 ft. Find the required cfm of out side air (ventilation rate) based ontwo air changes per hour.

Solution:

Q0= 20 x 20 x 10 x 260

= 4000 x 260

= 133.3 cfm


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No. of Air Changes Per Hour:

Q. No. 7 A dry cleaning establishment is 400 ft long, 15 ft wide and 10ft high.Find the cfm based on an air change every 3minutes.

Solution:

Q0 (cfm) = changepermintsV

=3

ft60,0003

= 20,000 cfm

Q. No. 8 A room measuring 30 x 20 yds & 10 yds high is ventilated by 2 pipes of 15" dia each. I% of space in the room is taken up by the machines & it is desired to change the air twevery hour (30 minutes per change). Calculate linear air speed in ft/min in the pipes.

Solution:

Volume of room = 30 x 20 x 10

= 6000 yds3= 6000 x (3)3 ft3 = 6000 x 27 = 162000 ft3

= 162000 ft3

10 % decreased volume = 162000 x 90/100

= 145,800 ft3

Q= changepermintV

= 30800,145

= 4860 cfm

Area of Cross-Section of 2 pipes = 4

(15/12)2

x 2 = 2.455 ft2

As we know that

Volume = Area of Cross-Section x Linear speed

Hence Linear speed =Area

Volume=

455.2

4860= 1979.63 ft/minute

air flow calculations 15-05-09

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