Hood Design
Problem:
Consider the following hoods:
Plain opening hood with C_{e} = 0.82 | |
Flanged opening hood with C_{e} = 0.76 | |
Cone hood with C_{e} = 0.93 |
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_{d} where 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 %
With an increase of C_{e},
hood entry loss and % loss in velocity decreases.
Problem:
Find the static pressure of a hood if the velocity pressure is 4.5 "H_{2}O 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_{e}² = 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" H_{2}O 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
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 " H_{2}O. Assume continued standard conditions (55^{o}F & 29.92" H_{2}O). 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, Q_{2} = Q_{1} (√ 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^{o}F. 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 = 900 cfm
Rectangular Canopy Hood:
Height of hood above tank = 3 ft
Low Canopy Hood required.
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