deck_3354128 (1)

Question 1

Velocity from velocity pressure

Answer 1

4005(VP)^.5

Question 2

V in FPM

Answer 2

4005(VP)^.5

Question 3

Velocity in FPM for non- normal conditions

Answer 3

4005[VP/df]^.5

^- DF from table 5.10 air density correction factor

Question 4

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

Answer 4

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

V=4005[2.7/.98]^.5

6648 fpm

Question 5

Reynolds for Turbulent vs laminar flow

Answer 5

129.2 Q/d

d= in ft

Q=cfm

>4000= Turbulent

Question 6

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

1) 10 CFM
2) 100 CFM

Answer 6

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

Question 7

Measurements to account for duct turbulance

Vavg

VPavg

Answer 7

V_avg= .9 V_cl

VP_avg= .81 VP_cl

Question 8

Flanges and baffles

Answer 8

• 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

Question 9

Design Requirements- Flow dist in hoods

Answer 9

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

Question 10

Slot Hood Design Guidelines

Answer 10

V_slot= 2000 fpm

V_plenum=

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

Question 11

Hood Entry Loss

Answer 11

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

Hood Entry Coefficent C_e

Hood Entry loss Coefficient F_d

Question 12

Determine VP

Answer 12

(V/4005)²

Question 13

h_e Hood Entry Loss

Answer 13

h_e= |SP|- VP_avg

VP_avg= .81(VP_cl)

Question 14

Hood Entry Coefficient

Answer 14

C_e Q_actual/Q_ideal

C_e = (VP/SPh)^.5

C_e= (VP/ |sp| ) ^.5

Question 15

Hood Entry Loss Coefficient, F_d

Answer 15

1. Ce=
2. F_d= 1-C_e² / C_e²
3. h_e= F_dVP

SP_h= (Fd*VP)+VP

Question 16

Hood Entry Coefficient (Ce)

Answer 16

Ce= (VP/|sp|)^.5

Where VP= .81(VP_cl)

Question 17

Hood Entry Loss Coefficient (Fd)

Answer 17

Fd= 1-Ce²/Ce²

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

Question 18

Determining Sp_h from F_d

Answer 18

Sp_h= (Fd* VP)+VP

Fd from IVM fig 5-15

VP=.81(VP_cl)

Question 19

Hood Entry Loss from Fd

Answer 19

H_e= F_d* VP

Fd = figure 5-15 IVM

Question 20

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

What is

1. h_e
2. C_e
3. F_d

Answer 20

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

1. h_e= |sp|- .97; 2.0-.97= 1.03 iw
2. C_e= (VP/|sp|)^{.5 =} .7
3. F_d= 1-C_e²/C_e²= 1-.7²/.7² = 1.0

Question 21

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 Sp_h?

Answer 21

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

= .15 iw

Question 22

Using Ce or Fd with measurement of Sph to mesure Q

Answer 22

Q=4005* A_duct* C_e* (Sqrt SP_h)

Q=4005* A_duct * (sqrt SP_h/F_d+1)

Question 23

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

Answer 23

1. Table 5-15 of IVM for Entry Loss F_d= .18
2. Q=4005* A_duct* (sqrt SP_h/F_d+1)

692.94 CFM

Question 24

Losses in compound hood

For a compound hood w/ V_s= 2000 fpm

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

Answer 24

F_sVP_s+F_dVP_d+VP_max

( VP_max the larger of VPs or VPd)

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

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

Question 25

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

Answer 25

Q=4005*A_duct*(sqrt Sp_h/1+f)

F= F_d+F_s(A_d/A_s)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

Question 26

Generic ranges of duct velocity based on size of contaminant

Answer 26

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

Question 27

Determining duct size

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

Answer 27

Dmax=(4Q/π*V_min)^.5

Vmin= min req velocity

2000-2500 fpm required

.418 ft ~5 inches

Question 28

Hood design Flow Chart

Answer 28

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

Question 29

Slot Design

Answer 29

V_slot= 2000_fpm

W/L=

A_ft=Q/V_slot -Q from devalle,

W=A_ft/Length (ft)

Flange width greater than sqrt of hood surface area

Question 30

Plenum Design

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

Answer 30

Plenum velocity= 1000 fpm

W=Q/V*L

2925//1000*12’

W= .24 ft= 3”

Plenum should be 12’ wide and 3” deep

Question 31

Takee offs from Plenum

Answer 31

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

Question 32

Duct size

Answer 32

1. A=Q/V
2. D=(4A/π)^.5
3. Adjust for closest duct size
4. Final V_d = Q/A_{duct size}

Question 33

VP from V

Answer 33

Vp=(V/4005)²

Question 34

After plenum and duct determined on compound hood, refigure Sph

Answer 34

1) compound hood- h_e=F_sVP_s+F_dVP_d

Fs and Fd in table 5-15

2) Sph= VP_max+h_e

VP_max determined from either VP_s or VP_d- which ever one is larger

Question 35

Friction losses in round duct work

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

Answer 35

H_f=a*V^b/Q^c = 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) h_f = H_fL(VP)= .0447* 100’* .40 iw= 1.8iw

Question 36

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

Answer 36

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

4. F=Ø/90º (for non-90º elbows)
5. h=F*VP = .34* (60/90)*.13

Question 37

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?

Answer 37

1) find V and VP for both duct sizes

V₁=Q/A₁= 1833_fpm, VP₁= .21_iw

V₂=Q/A₂= 458_fpm, VP₂= .013_iw

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) = .09_iw

Question 38

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?

Answer 38

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

Question 39

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

Answer 39

Determine V, VP and SP of all three ducts

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