Slot Hood Design

An exterior hood may be an opening on a welding table or slots on the side of a tank. The exterior hood should be located in the path of the emission if transferring larger particulates such as sand. There are four main types of exterior hoods: Canopy: A one- or two-sided overhead hood that receives upward airflow from hot air or gas. Hood “losses” Minimum duct design velocities. Slotted hoods Slots are used for proper distribution of air across a large area of contamination dispersion.

1. Slot Hood Designs
2. Slot Hood Design Images
3. Slot Hood Exhaust Design
4. Slot Hood Design Guide
5. Slot Hood Design

Sizing exhaust hoods - air volume flow and capture velocities - online exhaust hood calculator

Exhaust hoods are essential in kitchens, laboratories and industrial application for removing

• fumes
• mists
• vapors
• aerosols
• particulates
• hazardous substances
• polluting contaminants

before they 'escapes' to from the surrounding air.

In general for an exhaust hood to be efficient

• the height - y - should not exceed 1.20 m (4 ft)
• the distance - x - should not be less than 1/3 y
• the capture velocity - v₁ - should not be less than 0.15 - 0.20 m/s (30 - 40 ft/min)

Note! Potential hazardous and polluting applications requires special solutions. Always check local regulations before design.

Capture velocity - v₁ - for an exhaust hood can be estimated with the empirical equation

v₁ = q / 2 y² c (1)

where

v₁ = capturing velocity (m/s)

q = air volume flow (m³/s)

y = distance between table and exhaust hood (m)

c = circumference of the hood (m)

(1) can be modified to calculate air volume flow

q = 2 v₁ y² c (1a)

Example - Kitchen Exhaust Hood

Required air volume flow for a exhaust hood with circumference 3 m located 1.2 m above a stove can be calculated as

q = 2 (0.2 m/s) (1.2 m²)² (3 m)

= 1.7 m³/s

Note! The units don't match since the equation is empirical (a result of experiments).

Exhaust Hood Calculator

The air flow volume in the exhaust hood can be calculated below

Exhaust Hood with internal Plate

The efficiency of an exhaust hood can be improved by adding an internal plate.

The required air volume for an exhaust hood with an internal plate can in general be reduced to approximately 80% compared to an exhaust hood without a plate.

Exhaust Hood with Side Walls

The exhaust hood efficiency can be further improved by adding side walls.

Related Topics

• Ventilation - Systems for ventilation and air handling - air change rates, ducts and pressure drops, charts and diagrams and more

Related Documents

• Convective Air Flow from a Single Heat Source - Calculate the vertical air flow and air velocity generated by a single heat source
• Evacuation of Air - minimum Capture Velocities to avoid Pollution products transferred to the Room - Capture velocity to avoid pollution products from galvanic baths, spray painting boxes and more contaminating the surrounding room and environment
• Exhaust Outlets - Capturing Air Velocities - Capturing air velocity in front of an exhaust outlet - online exhaust outlet velocitycalculator

Tag Search

• en: exhaust hoods velocity air volume
• es: campanas de extracción de volumen de aire de velocidad
• de: Dunstabzugshauben Geschwindigkeitsluftvolumen

EXERCISE 1

WHICH LOSSES WOULD YOU EXPECT TO BE THE MOST SIGNIFICANT IN A LOCAL EXHAUST SYSTEM?

Most significant loss in a Local Exhaust System:

Hood Entry Loss

Other Losses:

Friction Losses

Branch Losses

EXERCISE 2

SEE FIGURE BELOW. WHAT INDIVIDUAL LOSSES CAN YOU IDENTIFY?

Losses:

Hood

Elbow

Friction

Entry

Entry Loss to Fan

Stack

Slot Hood Designs

EXERCISE 3

WHAT IS THE FRICTION LOSS (IN INCHES OF WATER) OF STANDARD AIR FLOWING THROUGH A 5 INCH DUCT AT V = 4,000 FPM?

Solution:

From I.V. Manual, Table 5-5

At D = 5' and Duct Velocity = 4,000 FPM : Vp=1' wg

Duct Friction Loss = 0.054*Vp per foot = 0.054' wg per foot

Friction Loss: 0.054' wg per foot

EXERCISE 4

WHAT IS THE FRICTION LOSS IN TERMS OF VELOCITY PRESSURE (VP)?

Solution:

Again, from Appendix A - Chart 5 or I.V. Manual

Friction Loss: 0.054' wg per foot

EXERCISE 5

FOR THE FOLLOWING GRINDING WHEEL HOOD SYSTEM, CALCULATE (a) AIR VOLUME, (b) DUCT VELOCITY AND (c) PERCENT LOSS IN VELOCITY HEAD.

Given: Ce = 0.78
SPh = 2.50 inches water
DUCT DIAMETER = 5 inches
DUCT AREA (A) = 0.136 sq. ft.

Solution:

(a) Air Volume (Q) = V x A = 4005ÖVP x A = 4005 x Ce ÖSPh x A

= 4005 x 0.78 x Ö2.50 x 0.136 = 672 CFM

(b) Duct Velocity (V) = V = Q / A = 672 / 0.136 = 4950 FP

(c) Percent loss in Velocity head:

VP = (V/4005)² = (4950 / 4005)² = 1.52' WG

Static Pressure (SPh) = VP + he or 2.50 = 1.52 + he;

Hood entry loss (he) = 2.50 - 1.52 = 0.98' WG

Hood entry loss factor (Fh) = (he / VP) = (0.98 / 1.52) = 0.645

Percent loss in Velocity head = Fh x 100 = 0.645 x 100 = 64.5%

EXERCISE 6

IN A SIMPLE HOOD SHOWN BELOW, THE HOOD STATIC PRESSURE IS EQUAL TO THE VELOCITY PRESSURE IN THE DUCT PLUS THE HOOD ENTRY LOSS. IF THE FACE VELOCITY (Vf) IS 250 FPM AND DUCT VELOCITY (Vd) IS 3000 FPM, CALCULATE HOOD STATIC PRESSURE (SPh).

Solution:

Face Velocity (Vf) = Q / A face = 250 FPM

Duct Velocity (Vd) = Q / A duct = 3000 FPM

VPd = (Vd / 4005 )² = 0.56 ' WG

Fh (From Chart 18) = 0.25

SPh = Hood entry loss of transition (Hed = Fh x VPd) + VPd

= (0.25 x 0.56 ) + o.56

= 0.70 ' WG

EXERCISE 7

FILL IN THE BLANKS:

PROBLEM NO:	SHAPE OFDUCT	DIN.	ASQ. FT.	VFPM	QCFM	TPIN. WG	SPIN. WG	VPIN.WG
A	SQUARE.	12*	4005	-2.00
B	CIRCLE.	2.182	3000	-2.50
C	CIRCLE.	14	3000	-0.86
D	CIRCLE.	393	-0.70	0.25
E	CIRCLE.	30	2000	-3.64
F	CIRCLE.	4005	2184	-5.20

*-Length of the sideNOTE: ALL PROBLEMS ARE ON THE UPSTREAM SIDE OF THE FAN . FLOW CONDITIONS ARE AT STP.Solution:

Problem A:

Given: L = 12 inch, SP = - 2.00 inch Q = 4005 CFM

A = 1’ x 1’ = 1 Sq. Ft.

Q = V / A or 4005 = V / 1’ x 1’ or V = 4005 FPM

From Appendix A - Chart 5 or I.V. Manual, Table 5.7B

At V = 4005 FPM, VP = 1 inch wg

TP = SP + VP

= - 2.00 + 1.00 = -1.00 inch wg

Problem B:

Given: Area (A) = 2.182 Sq. Ft., V = 3000 FPM SP = -2.50 inch wg

Area = p D² / 4 or 2.182 = 3.14 D² / 4; D² = (2.182 x 4) / 3.14

D = 20 inch

From Appendix A - Chart 5 or I.V. Manual, Table 5-7B

At D = 20 inch and V = 3000 FPM, Q = 6000 CFM and VP = 0.57

TP = SP + VP

TP = -2.50 + 0.57 = -1.93 inch wg

EXERCISE 8

FILL IN THE BLANKS

PROBLEM NO.	DINCH	QCFM	VFPM	VPIN. WG	FLF	FL/F
A	6	2000
B	140	400
C	20	7300
D	1600	0.02

FLF: Friction Loss Factor FL/F: Friction Loss/Foot

NOTE: ALL FLOW IN ROUND GALVANIZED DUCTS AT STP.

Solution:

Problem A:

Given: D = 6 inch, V = 2000 FPM

From Appendix A - Chart 5 or I.V. Manual, Tables 5-5 and 5-7B

At D = 6 inch and V = 2000 FPM

Q = 350 CFM, FLF = 0.45 VP/Foot , VP = 0.25 inch WG and

Friction Loss = 0.0456*0.25= 0.0114 inch WG/ft

Problem B:

D = 8', VP = 0.01 inch WG,

Friction Loss Factor = 0.0321 Vp/ft

Friction Loss = 0.000321 inch WG/ft

Problem C:

Q = 15926.4CFM, VP = 3.2 inch WG,

Friction Loss Factor = 0.0096 VP/ft

Friction Loss = 0.31inch WG/ft

Problem D:

Q = 1600 CFM, V = 566 FPM, D= 23 inch

Friction Loss Factor = 0.0088 VP/ft

Friction Loss = 0.00018inch WG/ft

EXERCISE 9

WHAT IS THE ACTUAL LOSS IN INCHES WG OF AIR FLOWING THROUGH A THREE-PIECE, 60° AND 45° ELBOW (R/D = 1.25) AT Vd = 3,500 FPM.

Solution:

Given: Three piece, 60° Elbow, R/D = 1.25 Vd = 3,500 FPM

From Appendix A - Chart 13 or I.V. Manual, Table 5-13

Elbow Loss Factor for angles < 90° , Kf = (f / 90° ) x K₉₀

Loss Coefficient = (0.42+0.34)/2 = 0.38

Therefore, Kf = (60 / 90) x 0.42 = 0.25 inch WG

From Appendix A - Chart 5 or I.V. Manual, Table 5-7B

At Duct Velocity Vd = 3,500 FPM, Velocity Pressure VP = 0.765

Elbow Loss = ( K x VP )

Elbow Loss= 0.25 x 0.76 = 0.19 inch WG

For 45° , three-piece elbow:

Elbow Loss Factor = (45 / 90) x 0.38 = 0.14 inch WG

Elbow Loss= 0.14 x 0.76 = 0.11 inch WG

EXERCISE 10

WHAT IS THE ACTUAL LOSS IN INCHES OF WG FOR A BRANCH ENTRY OF 30° WHERE THE ENTRY VELOCITY IS 4,500 FPM?

Solution:

Given: V = 4,500 FPM

From Appendix A - Chart 5, VP = 1.26 Inch WG

(I.V Manual, Table 5-B)

From Appendix A - Chart 14, At 30° branch entry, K = 0.18

(I.V Manual, Table 5-14)

Entry Loss (SPL) = 0.18 x 1.26 = 0.23 inch WG

SPL = 0.23 inch WG

EXERCISE 11

WHAT IS THE ESTIMATED FLOW RATE THROUGH A HOOD IF Ce = 0.82, SPh = -2.0 INCHES WG, AND THE DUCT DIAMETER IS 12 INCHES?

Solution:

Coefficient of Entry (Ce) = e(VP / SPh)

Or SPh = VP / Ce² Or VP = Ce² x SPh

VP = (0.82)² x 2.0 = 0.67 x 2.0 = 1.34 inch WG

From Appendix A - Chart 7, At VP = 1.34, V = 4,636 FPM

From Appendix A - Chart 5, At V = 4,636 FPM, Q = 3.750 CFM

EXERCISE 12FIND Fn, He, Ce FOR THE FOLLOWING HOODS:

A BELL MOUTH HOOD, IF THE DUCT VELOCITY = 3,000 FPM.

Solution:

From Appendix A - Chart 11A, Coefficient of Entry (Ce) = 0.98 and

Loss Factor (K) = 0.04

From Appendix A - Chart 7, At V = 3,000 FPM, VP = 0.56 inch WG

Ce = v ( VP / SPh ), SPh = (VP / Ce² ) = 0.56 / 0.98²

SPh = (0.56 / 0.96) = 0.58 inch WG

Hood Entry Loss (He) = SPh - VP = 0.58 - 0.56 = 0.02 inch WG

A ROUND TAPERED HOOD, INCLUDED ANGLE OF 90-DEGREES AND DUCT VELOCITY V= 2,300 FPM.

Solution:

From Appendix A - Chart 11 B, Fn = 0.15

(I.V. Manual-Fig 5-12)

From Appendix A - Chart 7, At V = 2,300 FPM, VP = 0.33 inch WG

He =Fn * Vp =0.15*0.33 = 0.05' WG

SPh = He + Vp = 0.05 + 0.33 = 0.38 inch WG

Ce = Sqrt(Vp/ SPh ) = 0.93

A ROUND PLAIN DUCT HOOD, IF DUCT VELOCITY = 3,750 FPM.

Solution:

From Appendix A - Chart 11 A, Ce = 0.72, Fn = 0.93

(I.V. Manual-Fig 3-16)

From Appendix A - Chart 7, At V = 3,750 FPM, VP = 0.875 inch WG

(I.V. Manual, Table 5-7B)

SPh = (0.875 / 0.72² ) = (0.875 / 0.52) = 1.68 inch WG

He = SPh - VP = 1.68 - 0.875 = 0.805 inch WG

IF AT A LATER DATE, THE HOOD STATIC PRESSURE FOR EACH OF THE ABOVE HOODS IS FOUND TO BE, SPh = -2.00 INCH WG, AND D = 12 INCH, WHAT IS THE ESTIMATED FLOW RATE (Q) IN EACH SYSTEM?

Solution:

Given: 19(a): D = 12', Ce = 0.98,

SPh = - 2.00 inch WG,

Area (A) = p r² = (3.14 x 36) / 144 = 0.78 Sq. Ft.

Q = 4005 Ce x eSPh *A = 4005 x 0.98 x v 2.00*0.78

Q = 4,360 CFM

EXERCISE 13

A FLANGED DUCT HOOD IS TO BE USED TO CONTROL WELDING FUMES AND GASES. THE HOOD CAN BE MOVED WITHIN NINE INCHES OF THE WELDING POINT. THE PLANT MANAGER WOULD LIKE TO MAINTAIN A CAPTURE VELOCITY AT THE WELDING POINT OF 150 FPM. WHAT Q WOULD YOU SUGGEST?

Solution:

From Appendix A - Chart 11C (Welding Hood, Portable…)

Q = 2 p X² Vc = 2 x 3.14 x (9/12)² x 150

Q = 530 CFM

EXERCISE 14

WHAT CAPTURE VELOCITIES ARE RECOMMENDED FOR THE FOLLOWING OPERATIONS?

Solution:

ARC WELDING 100 FPM

GRINDING WHEEL WITH A GOOD ENCLOSURE 200 - 500 FP

LAB FUME HOOD 100 FPM

EXERCISE 15

FOR THE BOOTH SHOWN, FIND THE TARGET AIRFLOW (Qt), GIVE THAT ITS FACE VELOCITY (Vf) = 100 FPM.

Solution:

Area = 6' x 8' = 48 Sq. ft.

Qt = Vf x A = (100 ft/min) x (48 Sq. ft.)
Q = 4800 CFM

EXERCISE 16
FIND Qt, GIVEN THAT Vf = 125 FPM
Solution:
Area (A) = 5' x 3' = 15 Sq. ft.

Qt = Vf x A = (125 ft./min.) x (15 Sq. ft.)
Qt = 1875 CFM

EXERCISE 17

FIND Q, GIVEN THAT: DUCT VELOCITY (Vd) = 2510 FPM, DUCT DIAMETER (D) = 24'
Solution:

Area = p D² / 4 = 3.14 x 24 x 24 /4 Sq. Inch

= 3.14 x 24 x 24 / 4 x 12 x 12= 3.14 Sq. ft.
Q = (Vd) x A = (2510 ft/min.) x (3.14 Sq. Ft.)= 7885 CFM
Q = 7885 CFM

EXERCISE 18

FIND Q GIVEN THAT: DUCT DIAMETER (D) = 6', AND DUCT VELOCITY (Vd) = 3550 FPM
Solution:

Area (A) = p D² / 4 = 3.14 x 6 x 6 / 4 x 12 x 12= 0.1963 Sq. Ft.

Q = (Vd )x A = (3550 ft./min.) x (0.1963 Sq. Ft.)= 697 CFM
Q = 697 CFM

EXERCISE 19

FIND THE AIRFLOW IN THE MAIN DUCT:
Solution:

Q = V x A = (2890 ft./min.) x (Area of 10' duct)

= (2890) x (p D² / 4)

= 2890 x 3.14 x 10 x 10 / 4 x 12 x 12

= 2890 x 0.5454 = 1576 CFM

Q = 1576 CFM

EXERCISE 20

FIND DUCT VELOCITY (Vd) GIVEN THAT: Q = 7885 CFM AND DUCT DIAMETER = 24'

Solution:

Area (A) = 3.14 x 24 x 24 / (4 x 12 x 12)

= 3.14 Sq. Ft.

Duct Velocity (V) = Q / A = 7885 / 3.14

= 2510 ft./min.
V = 2510 FPM

EXERCISE 21

FIND DUCT VELOCITY (Vd) GIVEN THAT Q = 697 CFM AND D = 6'
Solution:

Area = 3.14 x 6 x 6 / (4 x 12 x 12) = 0.1963 Sq. Ft.

V = Q / A = 697 / 0.1963

= 3551 FPM
V = 3551 FPM

EXERCISE 22

CHOOSE A DIAMETER SUCH THAT Vd ³ 3500 FPM AND THEN CALCULATE THE ACTUAL DUCT VELOCITY.
Solution:

Area (At) = Qt /Vt = (697 CFM) / (3500 FPM) = 0.1991 Sq. Ft.

Also, Area = 3.14 x D² / 4 x 12 x 12 = 0.1991

D² = 0.1991 x 4 x 12 x 12 / 3.14

D = 24 x Ö (0.1991 / 3.14) = 6.04'

Since we must choose either a 6' or 7' diameter duct, then duct velocity (Vd) will not exactly equal to Vt. Let us calculate the duct velocity for both:

V = Q / A

For 6' diameter duct

V = 697 / 3.14 x 6 x 6 / 4 x 12 x 12 = 697 x 0.1963 = 3551 FPM

For 7' diameter duct

V = 697 / 3.14 x 7 x 7 / 4 x 12 x 12 = 697 x 0.2673 = 2608 FPM

Slot Hood Design Images

If Q = Qt, to ensure that Vd > Vt, choose the smaller diameter.

(Alternate Solution: Let V = Vt, Q = Vt x Area of larger duct

Q = 3500 x 3.14 x 7 x 7 / 4 x 12 x 12 = 935 CFM)

EXERCISE 22

FOR THE SYSTEM SKETCHED BELOW, CHOOSE THE DUCT DIAMETERS AND COMPUTE THE DUCT VELOCITIES FOR BRANCHES 'a' AND 'b'.
Solution:

For Branch 'a':

Area (A) = Qa / Va = 1000 / 2500 = 0.40 Sqft.

Also, Area = p D² / 4 = 3.14 x D² / 4 x 12 x 12 S q. ft

Duct Diameter (Da) = Ö (0.40 x 4 x 12 x 12 / 3.14) = 8.56'
Choose Da = 8'

Area (A) = 3.14 (8 /4 x 12 x 12)² = 0.3491 Sq. Ft.

Va = Qa / Aa = 1000 /0.3491 = 2865 FPM

For Branch 'b':

Area (A) = Qb / Vb = 1000 / 4000 = 0.25 Sq. Ft.

Duct Diameter (Db) = 4 x 12 x 12 Ö (0.25 / 3.14) = 6.77'
Choose D_b = 6'
Area (A) = 3.14 (6' / 4 x 12 x 12)² = 0.1963 Sq. Ft.

Vb = Qb / Ab = 1000 / 0.1963 = 5094 FPM

EXERCISE 23

CHOOSE THE DUCT DIAMETER AND COMPUTE THE DUCT VELOCITY FOR THE MAIN DUCT, GIVEN THAT Vt = 2500 FPM.
Solution:

For the main duct:

Vt = Highest transport velocity upstream = 2500 FPM

Qm = Qa + Qb = 1200 + 805 = 2005 CFM

(we are interested in the actual Q, not Qt)

Area (A) = Qm / Vt = 2005 / 2500 = 0.802 Sq. Ft.

Diameter (Dt) = 4 x 12 x12 x Ö (0.802 / 3.14) = 12.13'
Choose Dm = 12'

Area = 3.14 (12'/24')² = 0.7854 Sq. Ft.

Vm = Qm / A = 2005 / 0.7854 = 2553 FPM

EXERCISE 24

FOR THE SYSTEM SKETCHED BELOW, WHAT IS THE AVERAGE FACE VELOCITY?
Solution:

Assume that the density is constant in the duct, the Continuity Equation is

: A1 V1 = A2 V2 Q1 = Q2V2 = A1 V1 / A2= [p (8)² / 4] x 1000 / [p (4)² /4]

Slot Hood Exhaust Design

= 4000 FPM
V2 = 4000 FPM

EXERCISE 25

FOR THE SYSTEM SKETCHED BELOW, WHAT IS THE AVERAGE FACE VELOCITY?
Solution:

Slot Hood Design Guide

From the Continuity Equation, V1 A1 = V2 A2

V2 = 2000 x (3.14 x 4 x 4 / 4) / 6 x 12

Slot Hood Design

= 349 FPM
V2 = 349 FPM