Slot Hood Design
Feb 17, 2017 It recommends hood construction, location, fire protection, specialty hoods, identification, inspection, testing and maintenance and exhaust. SEFA is the Scientific Equipment & Furniture Association. Its publication “Laboratory Fume Hoods Recommended Practices” covers design requirements of hoods, face velocities and testing. Slot hood designs can vary widely. Although variable slot widths offer flexibility, they are subject to tampering. Fixed slots and unobstructed (no internal baffling) plenums are generally the most reliable designs. Plenum velocities are often designed at 1/2 the slot velocity.
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 |
| Friction Losses |
| Branch Losses |
SEE FIGURE BELOW. WHAT INDIVIDUAL LOSSES CAN YOU IDENTIFY?
Losses:| Hood |
| Elbow |
| Friction |
| Entry |
| Entry Loss to Fan |
| Stack |
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
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
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%
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
FILL IN THE BLANKS:
| SQUARE. | 12* | 4005 | -2.00 | |||||
| CIRCLE. | 2.182 | 3000 | -2.50 | |||||
| CIRCLE. | 14 | 3000 | -0.86 | |||||
| CIRCLE. | 393 | -0.70 | 0.25 | |||||
| CIRCLE. | 30 | 2000 | -3.64 | |||||
| CIRCLE. | 4005 | 2184 | -5.20 |
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
FILL IN THE BLANKS
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
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 |
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= 0.14 x 0.76 = 0.11 inch WG
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
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
| A BELL MOUTH HOOD, IF THE DUCT VELOCITY = 3,000 FPM. |
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. |
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. |
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? |
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
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
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 |
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
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
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
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
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
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
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
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
Slot Hood Design
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
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)
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 Db = 6'
Area (A) = 3.14 (6' / 4 x 12 x 12)² = 0.1963 Sq. Ft.
Vb = Qb / Ab = 1000 / 0.1963 = 5094 FPM
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
Fume Hood Design
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]
= 4000 FPM
V2 = 4000 FPM
FOR THE SYSTEM SKETCHED BELOW, WHAT IS THE AVERAGE FACE VELOCITY?
Solution:
From the Continuity Equation, V1 A1 = V2 A2
V2 = 2000 x (3.14 x 4 x 4 / 4) / 6 x 12
Slot Hood Exhaust Design
= 349 FPM
V2 = 349 FPM
Slot Hood Design Images
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