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Ventilation > deck_3354128 (1) > Flashcards

deck_3354128 (1) Flashcards

1
Q

Velocity from velocity pressure

A

4005(VP).5

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2
Q

V in FPM

A

4005(VP).5

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3
Q

Velocity in FPM for non- normal conditions

A

4005[VP/df].5

- DF from table 5.10 air density correction factor

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4
Q

What is velocity if VP is 2.7iw, temp is 100degrees F and BP is 31”hg

A

Table 5-10 air density correction factor @ 31mmhg, 100F the Df=.98

V=4005[2.7/.98].5

6648 fpm

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5
Q

Reynolds for Turbulent vs laminar flow

A

129.2 Q/d

d= in ft

Q=cfm

>4000= Turbulent

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6
Q

With a duct that is 1’ in diameter what is reynolds number for flow of:

1) 10 CFM
2) 100 CFM

A

1) 129.2 Q/D @10cfm
129. 2*10/1ft = 1292 -
2) 129.2*100/1 = 12,920 - >4000 so turbulent

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7
Q

Measurements to account for duct turbulance

Vavg

VPavg

A

Vavg= .9 Vcl

VPavg= .81 VPcl

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8
Q

Flanges and baffles

A
  • Confine and restrict flow contour lines at a hood face
  • Flanges reduce required Q by 25
  • Flange width >= Squareroot of surface area of the hood
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9
Q

Design Requirements- Flow dist in hoods

A

Rectangular and round hoods

1) Make hood plenum as long as possible
2) ensure take off angle is
3) use internal baffles as needed
4) use multiple take offs for wide hoods
5) use a slotted face

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10
Q

Slot Hood Design Guidelines

A

Vslot= 2000 fpm

Vplenum=

  1. Use multiple take offs for slot width >10ft
  2. Use internal baffles/ slitter vanes if neede
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11
Q

Hood Entry Loss

A
  1. some loss due to vena contracts as air enters the face
  2. Tapered hoods, measure TP(&SPh ) at 1 Duct diameter past take off
  3. Plain and flanged hooods- measure 3D from hood face
  4. Two ways to determine He

Hood Entry Coefficent Ce

Hood Entry loss Coefficient Fd

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12
Q

Determine VP

A

(V/4005)2

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13
Q

he Hood Entry Loss

A

he= |SP|- VPavg

VPavg= .81(VPcl)

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14
Q

Hood Entry Coefficient

A

Ce Qactual/Qideal

Ce = (VP/SPh).5

Ce= (VP/ |sp| ) .5

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15
Q

Hood Entry Loss Coefficient, Fd

A
  1. Ce=
  2. Fd= 1-Ce2 / Ce2
  3. he= FdVP

SPh= (Fd*VP)+VP

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16
Q

Hood Entry Coefficient (Ce)

A

Ce= (VP/|sp|).5

Where VP= .81(VPcl)

17
Q

Hood Entry Loss Coefficient (Fd)

A

Fd= 1-Ce2/Ce2

  • Find VP, find |sp|, then computer Ce
18
Q

Determining Sph from Fd

A

Sph= (Fd* VP)+VP

Fd from IVM fig 5-15

VP=.81(VPcl)

19
Q

Hood Entry Loss from Fd

A

He= Fd* VP

Fd = figure 5-15 IVM

20
Q

A hood Sp is -2iw. Centerline duct velocity pressure is 1.2iw

What is

  1. he
  2. Ce
  3. Fd
A

VP= .81(Vpcl) .81*1.2=.97iw

  1. he= |sp|- .97; 2.0-.97= 1.03 iw
  2. Ce= (VP/|sp|).5 = .7
  3. Fd= 1-Ce2/Ce2= 1-.72/.72 = 1.0
21
Q

A round flanged faced hood is attached toa 10” diamter duct through a 90 degree takeoff at a designed flow rare of 800 cfm. What is expected Sph?

A
  1. Find V - V=Q/A= 1467 cfm
  2. Look up Fd from fig 5-15. Fd=.15
  3. SPh= (Fd+1)(V/4005)2

= .15 iw

22
Q

Using Ce or Fd with measurement of Sph to mesure Q

A

Q=4005* Aduct* Ce* (Sqrt SPh)

Q=4005* Aduct * (sqrt SPh/Fd+1)

23
Q

A tapered conical hood enters a 5” duct via a 100º take off. The hood Sp is 1.9 iw

What is flow rate through hood

A
  1. Table 5-15 of IVM for Entry Loss Fd= .18
  2. Q=4005* Aduct* (sqrt SPh/Fd+1)

692.94 CFM

24
Q

Losses in compound hood

For a compound hood w/ Vs= 2000 fpm

and Vd= 3500 fpm. Whats the total Sph for a rectangular to round 90º take off

A

FsVPs+FdVPd+VPmax

( VPmax the larger of VPs or VPd)

Fs= 1.78 , Fd.25 (from 5-15 table)

SPh= (1.78)(2000/4005)2+ (.25)(3500/4005)2+ (3500/4005)2

25
Q

For a compound hood, estimate flow rate from SPh

If slot is 8’x2” and Duct diamter is 1ft and SPh= 1.38”w, what is flow rate?

Rect to round is 90º, Vs=2000; Vd= 3500 fpm

A

Q=4005*Aduct*(sqrt Sph/1+f)

F= Fd+Fs(Ad/As)2

  1. from table 5-15, Fs=1.78; Fd=.25
  2. As= 8’*2’/12”= 1.33 ft2
  3. Ad= π(12/12)2 /4= .78ft3

F=.88

26
Q

Generic ranges of duct velocity based on size of contaminant

A

1000-2000- Gas, vapor, smoke

2000-2500- Fine aerosols

2500-3000- Low dens. aerosols; cotton lint, flour, powders

3000-3500- dry dusts and powders

3500-4000- grinding, buffing, granite dust

4000-4500- heavy dusts

>4500 large wet dusts, moist cement

27
Q

Determining duct size

A hood captures welding fume with hood flow range of 275 cfm. What is the right duct size

A

Dmax=(4Q/π*Vmin).5

Vmin= min req velocity

2000-2500 fpm required

.418 ft ~5 inches

28
Q

Hood design Flow Chart

A
  1. Determine null point and capture velocity
  2. Determine hood geometry
  3. Determine hood flow rate (devalle)
  4. Size slots, plenum, takeoffs
  5. Determine duct transport velocity
  6. Determinehood entry loss and hood Sph
29
Q

Slot Design

A

Vslot= 2000fpm

W/L=

Aft=Q/Vslot -Q from devalle,

W=Aft/Length (ft)

Flange width greater than sqrt of hood surface area

30
Q

Plenum Design

For a 12’ long hood, what should be plenum depth if CFM is 2925

A

Plenum velocity= 1000 fpm

W=Q/V*L

2925//1000*12’

W= .24 ft= 3”

Plenum should be 12’ wide and 3” deep

31
Q

Takee offs from Plenum

A

should be 90º or less

do a triangle, bisect triangle to a right angle. detemine angles and feet along base to determine height of triangle

32
Q

Duct size

A
  1. A=Q/V
  2. D=(4A/π).5
  3. Adjust for closest duct size
  4. Final Vd = Q/Aduct size
33
Q

VP from V

A

Vp=(V/4005)2

34
Q

After plenum and duct determined on compound hood, refigure Sph

A

1) compound hood- he=FsVPs+FdVPd

Fs and Fd in table 5-15

2) Sph= VPmax+he

VPmax determined from either VPs or VPd- which ever one is larger

35
Q

Friction losses in round duct work

A straight run of galvanized sheet duct 6” in diameter 100 ft long and carriers 500 cfm

A

Hf=a*Vb/Qc = See Loeffler equation constants in table

1) Use Q to determine V and VP (2546 fpm, VP=.40iw)
2) Hf= .0307 V .533/Q.612 = .0447
3) hf = HfL(VP)= .0447* 100’* .40 iw= 1.8iw

36
Q

Elbow Loss Coefficient

What is fittting loss in a 60º 3 peice elbow with a duct diamter of 8” centerline radius of 12” with a flow rate of 500 cfm

A
  1. R/D= 12”/8”= 1.5
  2. look up R/D and number of peices
  3. 5 r/d = 3 peice= .34 (assumes a 90º)
  4. Find VP from Q = .13 iw

V=Q/A; VP=(V/4005)2

  1. F=Ø/90º (for non-90º elbows)
  2. h=F*VP = .34* (60/90)*.13
37
Q

Static pressure regaines for expansions

LEV moving air at 1000cfm through slot hood into a 10” duct. Duct expands to a diameter of 20” through a 20º taper. What is SP after expansion if it was prior to expansion at -1.5iw?

A

1) find V and VP for both duct sizes

V1=Q/A1= 1833fpm, VP1= .21iw

V2=Q/A2= 458fpm, VP2= .013iw

2) Go to expansion matrix table. Find taper degree on left and ratio of the expansion on top

20º taper+ D2/D1= 20”/10”= 2:1 = R=.48

3) Sp2=Sp1 + R (VP1-VP2) = -1.41
4) Delta SP= -1.41-(-1.5) = .09iw

38
Q

Tapered Contractions in duct work

Air flowing in a 20” duct at 1000 cfm. The duct contracts to a diameter of 10” if SP before contraction was -1.5iw, what is SP after contraction?

A

1) find V1 (458), V2 (1833) and VP1 (.013), VP2 (.21)
2) Go to contraction matrix. Find ratio for abrupt or tapered

A2/A1=102/202 = .25- K~.44

3) SP2=SP1+VP1-VP2-K*VP2
- 1.5+ .013-.21-.44(.21)= -1.79 iw
4) Delta SP= -1.79-(-1.5)= -.29 iw

39
Q

Determine if air flow has accellerated or decreased with a branch entry.

You have branch 2 meeting after branch one and going into main duct

A

Determine V, VP and SP of all three ducts

1) VPr=Q1/Q3VP1+Q2/Q3VP2
2) VP3>VPr than there has been acceleartion and losses

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