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. '68-'69 Camaro S/S Hood Item No. CRC-099: $7.50: 14. Mini Rectangle Pro Stock Hood Scoop Item No. CRC-102: $5.50: 15. Aero IV Pro Stock Hood Scoop Item No. CRC-103: $6.00: 16. Stock Vega Hood Item No. CRC-104: $7.00: 17. Short Old Style Pro Hood Scoop Item No. CRC-106: $5.50: 18. Cowl Induction Hood Scoop Item No. CRC-107: $5.00: 19.
Hood Design
Problem:
Consider the following hoods:Plain opening hood with Ce = 0.82 |
Flanged opening hood with Ce = 0.76 |
Cone hood with Ce = 0.93 |
Solution:
Ce = (VPd / SPh)1/2
where Ce is coefficient of entry,
VPd is duct velocity pressure,
SPh is hood static pressure.
SPh = he + VPdwhere he is the hood entry loss.
For plain opening hood,
VPd = SPh * Ce² = 3” * 0.82² = 2.017” wg
he= 3” – 2.017” = 0.983” wg
Percent loss in velocity head = 0.983 / 2.017 = 48.74 %
For flanged opening hood,
VPd = SPh * Ce² = 3” * 0.76² = 1.733” wg
he= 3” – 1.733” = 1.267” wg
Percent loss in velocity head = 1.267 / 1.733 = 73.11 %
For cone hood,
VPd = SPh * Ce² = 3” * 0.93² = 2.595” wg
he= 3” – 2.595” = 0.405” wg
Percent loss in velocity head = 0.405 / 2.595 = 15.61 %
With an increase of Ce, hood entry loss and % loss in velocity decreases.
Cone hood is the better of the three given hoods with only 15% loss in velocity head.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:
Ce = (VP / SP)1/2
where Ce is coefficient of entry,
VP is velocity pressure,
SP is static pressure.
SP = VP / Ce² = 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:
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, Ce = √ (VPduct / SPhood)
So, VPduct = Ce ² * SPhood = 0.89² * 3 = 2.376 “wg
Vduct = 4005 * √VPduct = 6173.42 fpm
Q = Vduct * 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 (55oF & 29.92' H2O). Calculate, by how much the air flow through the duct has been reduced?
Solution:
For standard air, Q = 4005 A Ce √ SPhood
For the same hood, Q is proportional to √ SPhood
So, Q2 = Q1 (√ SPhood, 2 / √ SPhood, 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 250oF. 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
Rectangular Canopy Hood:
Height of hood above tank = 3 ftLow Canopy Hood required.
Q = 6.2 * b1.33 * Δt0.42 * 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 * 31.33 * 1800.42 * 4 = 946.75 cfm
The primary goal of an industrial ventilation system hood is to capture and transfer environmental contaminants. A hood’s size and shape is designed specific to its end application but is typically classified within the enclosing hood or exterior hood category.
An enclosing hood will completely or partially surround the point where contaminants are generated. An enclosing hood is typically preferred but may not be practical due to potential interference with employee workstations.
Exterior hoods are placed next to the point where contaminants are generated without creating an enclosure. 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:
A specific velocity is required, depending on the type of contaminant being captured. To achieve the required velocity, carefully consider the hood’s shape, size and location.
An industrial ventilation system hood is one of the most important components of an individual’s workstation. A worker will be more likely to use the hood and the ventilation system properly if ergonomic elements are considered. Among these considerations are:
IVI’s engineering and design team has years of experience designing and sizing industrial ventilation systems. IVI can assist in the design of a new system or the redesign of an existing system.