Slot Hood Design

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Hood Design

Problem:

Slotted Fume Hoods are ideal as a capture mechanism for fumes produced by welding, grinding, buffing, etc. Unlike a standard fume hood, a slotted fume hood is more energy efficient and cost effective by creating a more restrictive and faster airflow pattern from a smaller fan.

• Practical Slot Hood Design for Acid Mist Control of a Battery Charging Operation Marie L. Fitzgerald Douglas Battery Manufacturing Company, 500 Battery Drive, Winston-Salem, North Carolina, 27117, USA & Jonathan F. Hale Jackson-Hale Environmental Technologies (JHET) Inc., P.O. Box 1839, Clemmons, North Carolina, 27012, USA.
• Hood 'losses' Minimum duct design velocities. Slotted hoods Slots are used for proper distribution of air across a large area of contamination dispersion.

Solution:

C_e = (VP_d / SP_h)^1/2

where C_e is coefficient of entry,

VP_d is duct velocity pressure,

SP_h is hood static pressure.

SP_h = h_e + VP_dwhere h_e is the hood entry loss.

For plain opening hood,

VP_d = SP_h * Ce² = 3' * 0.82² = 2.017' wg

h_e= 3' – 2.017' = 0.983' wg

Percent loss in velocity head = 0.983 / 2.017 = 48.74 %

For flanged opening hood,

VP_d = SP_h * Ce² = 3' * 0.76² = 1.733' wg

h_e= 3' – 1.733' = 1.267' wg

Percent loss in velocity head = 1.267 / 1.733 = 73.11 %

For cone hood,

VP_d = SP_h * Ce² = 3' * 0.93² = 2.595' wg

h_e= 3' – 2.595' = 0.405' wg

Percent loss in velocity head = 0.405 / 2.595 = 15.61 %

Slot Hood Design Images

With an increase of C_e, hood entry loss and % loss in velocity decreases.

Slot Hood Designs

Problem:

Find the static pressure of a hood if the velocity pressure is 4.5 'H₂O and the coefficient of entry is 0.82.

Solution:

C_e = (VP / SP)^1/2

where C_e is coefficient of entry,

VP is velocity pressure,

SP is static pressure.

SP = VP / C_e² = 4.5 / 0.82² = 6.7' of water

Problem:

A 25' wide hood needs a slot velocity of 200 fpm with a 40 cfs volume. What size of slot opening is required?

Solution:

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Required area of slot = Q / V = (40 cfs * 60) / (200 fpm) = 12 ft²

Size of slot opening = 12 ft² / (25' / 12) = 5.76 ft = 69.12'

Problem:

Assume the hood static pressure in a 8' duct is 3.0' H₂O and the hood manufacturer indicates that Ce is 0.89. What is the estimated flow rate?

Solution:

Hood entry coefficient, C_e = √ (VP_duct / SP_hood)

So, VP_duct = C_e ² * SP_hood = 0.89² * 3 = 2.376 'wg

Mgm online casino nj. V_duct = 4005 * √VP_duct = 6173.42 fpm

Q = V_duct * A = 6173.42 * [π * (8/12)² / 4] = 2154.93 cfm

Problem:

Assume a hood is exhausting 3,000 cfm of air and the hood static pressure was measured at 2.17' w.g. Three months later the hood static pressure is 5.0 ' H₂O. Assume continued standard conditions (55^oF & 29.92' H₂O). Calculate, by how much the air flow through the duct has been reduced?

Solution:

For standard air, Q = 4005 A C_e √ SP_hood

For the same hood, Q is proportional to √ SP_hood

So, Q₂ = Q₁ (√ SP_{hood, 2} / √ SP_{hood, 1}) = 3000 (√ 5 / √ 2.17) = 4553.83 cfm

Increase in air flow rate = (4553.83 – 3000) = 1553.83 cfm

% Increase in air flow rate = (4553.83 – 3000) / 3000 = 51.8 %

Problem:

Compare the flow rate for a lateral hood located at the edge of an open surface tank (4ft*3ft) with the rate for a canopy hood located 3 ft above the tank. The tank contains a solution of ammonium phosphate in water at 250^oF. Ammonia gas is released from the tank. State your assumptions.

Problem:

Slotted Fume Hoods are ideal as a capture mechanism for fumes produced by welding, grinding, buffing, etc. Unlike a standard fume hood, a slotted fume hood is more energy efficient and cost effective by creating a more restrictive and faster airflow pattern from a smaller fan.

• Practical Slot Hood Design for Acid Mist Control of a Battery Charging Operation Marie L. Fitzgerald Douglas Battery Manufacturing Company, 500 Battery Drive, Winston-Salem, North Carolina, 27117, USA & Jonathan F. Hale Jackson-Hale Environmental Technologies (JHET) Inc., P.O. Box 1839, Clemmons, North Carolina, 27012, USA.
• Hood 'losses' Minimum duct design velocities. Slotted hoods Slots are used for proper distribution of air across a large area of contamination dispersion.

Solution:

C_e = (VP_d / SP_h)^1/2

where C_e is coefficient of entry,

VP_d is duct velocity pressure,

SP_h is hood static pressure.

SP_h = h_e + VP_dwhere h_e is the hood entry loss.

For plain opening hood,

VP_d = SP_h * Ce² = 3' * 0.82² = 2.017' wg

h_e= 3' – 2.017' = 0.983' wg

Percent loss in velocity head = 0.983 / 2.017 = 48.74 %

For flanged opening hood,

VP_d = SP_h * Ce² = 3' * 0.76² = 1.733' wg

h_e= 3' – 1.733' = 1.267' wg

Percent loss in velocity head = 1.267 / 1.733 = 73.11 %

For cone hood,

VP_d = SP_h * Ce² = 3' * 0.93² = 2.595' wg

h_e= 3' – 2.595' = 0.405' wg

Percent loss in velocity head = 0.405 / 2.595 = 15.61 %

Slot Hood Design Images

With an increase of C_e, hood entry loss and % loss in velocity decreases.

Slot Hood Designs

Problem:

Find the static pressure of a hood if the velocity pressure is 4.5 'H₂O and the coefficient of entry is 0.82.

Solution:

C_e = (VP / SP)^1/2

where C_e is coefficient of entry,

VP is velocity pressure,

SP is static pressure.

SP = VP / C_e² = 4.5 / 0.82² = 6.7' of water

Problem:

A 25' wide hood needs a slot velocity of 200 fpm with a 40 cfs volume. What size of slot opening is required?

Solution:

Come visit the Palm Beach Kennel Club's poker room, one of Florida's largest and only poker room with two World Series of Poker stops. FEBRUARY TOURNAMENT SCHEDULE. Palm Beach Kennel Club is committed to ensuring that this website is accessible to everyone. We are currently improving our website's ADA Compliance for the experience of all visitors. If you have any questions or suggestions regarding the accessibility of this site, please contact us at info@pbkennelclub.com. 5400 Bradenton Rd. Sarasota, FL 34234 MAP. 941-355-7744 ext 1054. Mon and Tues: 10am- 12am. Wednesday: 10am-2am. Thursday: 10am-12am. Fri-Sat 10am- 4am. Sunday Noon - Midnight. Kennel club poker tournament schedule. A schedule of Palm Beach Kennel Club poker tournaments, including time, buy-in, blind structure, and more. You also can find Palm Beach Kennel Club phone number, address, and website info.

Required area of slot = Q / V = (40 cfs * 60) / (200 fpm) = 12 ft²

Size of slot opening = 12 ft² / (25' / 12) = 5.76 ft = 69.12'

Problem:

Assume the hood static pressure in a 8' duct is 3.0' H₂O and the hood manufacturer indicates that Ce is 0.89. What is the estimated flow rate?

Solution:

Hood entry coefficient, C_e = √ (VP_duct / SP_hood)

So, VP_duct = C_e ² * SP_hood = 0.89² * 3 = 2.376 'wg

Mgm online casino nj. V_duct = 4005 * √VP_duct = 6173.42 fpm

Q = V_duct * A = 6173.42 * [π * (8/12)² / 4] = 2154.93 cfm

Problem:

Assume a hood is exhausting 3,000 cfm of air and the hood static pressure was measured at 2.17' w.g. Three months later the hood static pressure is 5.0 ' H₂O. Assume continued standard conditions (55^oF & 29.92' H₂O). Calculate, by how much the air flow through the duct has been reduced?

Solution:

For standard air, Q = 4005 A C_e √ SP_hood

For the same hood, Q is proportional to √ SP_hood

So, Q₂ = Q₁ (√ SP_{hood, 2} / √ SP_{hood, 1}) = 3000 (√ 5 / √ 2.17) = 4553.83 cfm

Increase in air flow rate = (4553.83 – 3000) = 1553.83 cfm

% Increase in air flow rate = (4553.83 – 3000) / 3000 = 51.8 %

Problem:

Compare the flow rate for a lateral hood located at the edge of an open surface tank (4ft*3ft) with the rate for a canopy hood located 3 ft above the tank. The tank contains a solution of ammonium phosphate in water at 250^oF. Ammonia gas is released from the tank. State your assumptions.

Solution:

Lateral Hood:

Q = V (10 X² + A)

where, Q = Air Flow (cfm), V = Centerline Velocity at X distance from hood (fpm)

X = Distance of source from edge along axis (ft), A = Area of hood opening (ft²)

Assumptions: Capture Velocity = 75 fpm, Hood Dimensions = 4 ft by 3 ft (same as tank)

V = 75 fpm, X = 0 (hood at edge of tank), A = 4 * 3 = 12 ft²

Q = 75 * 12 = 900cfm

Slot Hood Design Pictures

Rectangular Canopy Hood:

Height of hood above tank = 3 ftLow Canopy Hood required.

Slot Hood Design

Q = 6.2 * b^1.33 * Δt^0.42 * L

where, Q = Air Flow (cfm), b = Width of hood (ft),

Δt = Difference in source and ambient temperature (F), L = Length of hood (ft)

Assumptions: Hood Dimensions = 4 ft by 3 ft (same as tank)

b = 3 ft, L = 4 ft, Δt = (250 – 70) = 180 F

Q = 6.2 * 3^1.33 * 180^0.42 * 4 = 946.75 cfm