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Author: Subject: Calculating Pressure Drop in a Fume Hood System
Magpie
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Calculating Pressure Drop in a Fume Hood System

1. Introduction
Here is a short tutorial on how to calculate the pressure drop in a fume hood circular duct system. This pressure drop value is needed to properly size the blower.

Pressure drop is commonly referred to as “static pressure” (SP) in blower vendor catalogs. The units are “inches of water” or “cm of water.” This is actually a “head loss” which is equivalent to a pressure loss, or pressure drop.

There are two forms of pressure loss: “major losses” and “minor losses.” In this case the term “minor loss” is actually a misnomer as it constitutes most of the pressure loss in a typical fume hood system.

2. “Major Losses”
These are the pressure losses due to the straight runs of duct. To determine this you must assume an air flowrate, Q, in cubic feet per min (cfm). Other units may be used. Then you must know the duct diameter in feet, D. Again, other units may be used. Now you can calculate the average velocity, V, of the air in feet per minute, fpm:

V = Q/A = Q/[(πD^2)/4], where A is the cross-sectional area of the duct

Now, the easy part: Go to the Engineering Toolbox: http://www.engineeringtoolbox.com/ductwork-friction-loss-d_1...

Here you can pick off the friction loss per 100 feet of duct in units of inches of water. To convert to cm of water multiply by 2.54.

3. “Minor Losses”
Minor losses are those losses (pressure drops) that are attributable to the non-duct components (fittings) of the ducting system. Typical air duct system fittings are elbows, tees, expansions, reducers, screens, and louvers.

Minor losses are proportional to the energy term V^2/(2g) where V is the average velocity in feet per second and g is the acceleration of gravity, 32.2 feet/s^2.

Minor loss = KV^2/(2g) in feet of fluid (air)

where K is a coefficient that varies for each ducting fixture. One source of K’s is the Engineering Toolbox: http://www.engineeringtoolbox.com/minor-loss-air-ducts-fitti...

You will note that here the coefficients are referred to by the Greek letter zeta, ζ. This is the same as K.

Listed below are K values from Streeter (ref 1):

Standard 90° elbow ……………………………..……0.9
Medium sweep 90° elbow……………………………..0.75
Long sweep 90° elbow………………………………...0.60

Minor losses from other sources:

90° sharp bend..............................................1.3
(Engineering Toolbox)

90° bend, rounded, (r/d >1)...............................0.25
(Engineering Toolbox)

10” x 16” louver (70% free area)………………………………….(see example below)
10” x 14” insect screen (7 mesh, 1/8” apertures, 77% open area)…(see example below)
fume hood …………………………………………………………(see example below)

The sum of the minor losses must be converted to inches of water from the units of feet of air. This is done using the respective densities of air and water. The conversion factor is therefore = (0.075 lb/ft^3)/(62.3 lb/ft^3) = 0.0012 at sea level and room temperature.

4. Total Pressure Drop
Sum the major and minor losses in “inches of water.”

Example (my own fume hood system):

Given:

fluid: air at 70°F and 1 atm pressure

1. hood opening face velocity = 1.0 ft/s = 60ft/min
2. hood opening = 44.25” x 27.5”
3. duct diameter (ID) = 8”, smooth, circular
4. length of duct = 15 feet
5. 2ea 11.25° smooth bends, r/d = 1.5
6. 1ea 22.5° smooth bend, r/d = 1.6
7. 1ea 90° smooth bend, r/d = 2.6
8. 1ea 10” x 16” louvered outlet, 70% open area
9. 1ea 7 mesh (1/8”) insect screen, 10” x 14”, 77% open area
10. 1ea 90° sharp bend at outlet plenum

Required: pressure drop in inches of water

Analysis:

For the hood opening: area, A= (44.5”)(27.5”)/144 = 8.5 ft^2

Required air flow = Q = AV = (8.5ft^2)(1.0 ft/s) = 8.5 ft^3/s = 510 cfm

Velocity of air in the duct = Q/A = Q/[(πD^2)/4] = (8.5ft^3/s)/[π{(0.667 ft)^2}/4] = 24.3 fps = 1460 fpm

From the Engineering Toolbox:

Major loss = 0.42” H2O/100 ft; therefore for 15 feet this = 0.06” H2O

For the minor losses:

K1 = 0.05 for the 11.25° smooth bend (a guess); 2(0.05) = 0.1

K2 = 0.25 for the 90° smooth bend (Engineering Toolbox)

K3 = 0.05 for the 22.5° smooth bend (a guess)

K4= 1.3 for the 90° sharp bend at the outlet plenum
(Engineering Toolbox)

For the 10” x 16“ (70% open area) louver:

Open area = (0.7)(10” x 16”)/144 = 0.77 ft^2
Velocity = Q/A = (510 ft^3/min)/(0.77 ft^2) = 662 ft/min

Using the R.L. Craig Co. Airflow Resistance graph at:

http://www.rlcraigco.com/pdf/louvers-selection-design-consid...

Pressure drop = 0.06” H2O

For the 10” x 14” insect screen (7 mesh):

Use the equation in Figure 1 in the paper by Valera et al (ref 2):

www.inia.es/gcontrec/pub/273-279-(08_05)-Aerodynamic_1165240539140.pdf

The pressure drop (in Pa) = 1.3248 u^2 + 3.1999u – 0.8748

Where u = the velocity in m/s.

Screen open area = (0.77)(10” x 14”)/144 = 0.75 ft^2

u = (510 ft^3/min)/(0.75 ft^2) = 680 ft/min
= (680 ft/min)(0.3048m/ft)(min/60s) = 3.45 m/s

Pressure drop = 1.3248 (3.45)^2 + 3.1999(3.45) – 0.8748 = 25.8 Pa
= (25.8 Pa)(1” H2O/248.7Pa) = 0.10”H2O

For the hood itself:

According to the Kewaunee website the hood loss itself for a 48” hood with an 8.1 ft^2 opening and a face velocity of 100 fpm is 0.25” H2O.

Therefore assuming a hood loss value of 0.20” H2O for a face velocity of 60 ft/min should be conservative.

The sum of the K’s = K1 + K2 + K3 + K4 = 0.1 + 0.25 + 0.05 +1.3 = 1.7

Therefore these minor losses = (1.7)[(24.3 ft/s)^2]/[(2)(32.2 ft/s^2)] = 15.6 feet of fluid (air)

= (15.6)(0.0012) = 0.019 ft of water = 0.22” H2O

5. Results
The total head loss then, is = 0.06” (major losses) + 0.06” (louver) + 0.10” (insect screen) + 0.20” (hood) + 0.22” (K’s) = 0.64” H2O

6. Discussion
A hood face velocity of 1.0 ft/s is, IMO, a minimum for safety. Kewaunee states that 1.67 ft/s is the minimum recommended.

The determination of the “major losses” is likely accurate within 5%. The accuracies of the “minor loss” values, however, are more difficult to quantify. They might be off by 25% or more.

The Handbook of the American Society of Heating, Refrigerating and Air-Conditioning Engineers (ASHRAE) would probably be the best source of minor loss correlations and data but I do not have access to this publication. If anyone else can provide more accurate estimates for the minor losses please do. My purpose here was to provide a relatively simple method for the calculations without sacrificing too much rigor. I didn’t want to get into calculating Reynolds numbers, friction factors, etc.

7. Recommendation
Estimate the losses as accurately as you can then buy a blower with a 25% overcapacity.

8. References
1. Fluid Mechanics, by V. L. Streeter, 3rd ed, (1962), p. 224.
2. “Aerodynamic Analysis of Several Insect-Proof Screens Used in Greenhouses,” by D. L. Valera et al, Spanish Journal of Agricultural Research (2006) 4 (4), p. 277.

[Edited on 14-7-2016 by Magpie]

[Edited on 15-7-2016 by Magpie]

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blogfast25
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Nice work, Magpie.

j_sum1
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Thanks Magpie.
I will be building a fume hood some time in the next several months. But I will need a rather long duct to vent the fumes to a suitable location. And due to the geometry of the house I will probably require a fairly narrow diameter with some bends. I was considering having a two fan system -- entry and exit blower. Whatever I decide, these calculations will be useful.

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 Sciencemadness Discussion Board » Special topics » Prepublication » Calculating Pressure Drop in a Fume Hood System Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication   » References Non-chemistry   » Forum Matters   » Legal and Societal Issues   » Whimsy   » Detritus   » The Moderators' Lounge