Hood Design

 

Problem:

Consider the following hoods:
bullet Plain opening hood with Ce = 0.82
bullet Flanged opening hood with Ce = 0.76
bullet Cone hood with Ce = 0.93
For each hood, SPh is 3" of water and the duct diameter is 6".
Compute the hood entry loss and percent loss in velocity head for the above case. Comment on your results.

Solution:

Ce = (VPd / SPh)1/2  

where Ce is coefficient of entry,

           VPd is duct velocity pressure,

           SPh is hood static pressure.

SPh = he + VPd     where 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 = 900 cfm

Rectangular Canopy Hood:

Height of hood above tank = 3 ft            Low 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