1. Industrial Ventilation vs. IAQ
Heating
Ventilation
Air
Conditioning

2. Industrial Ventilation vs. IAQ24

3. Industrial Ventilation vs. IAQ

4. Industrial Ventilation vs. IAQ

5. Routes of Entry
Inhalation
Ingestion
Absorption
Injection

6. Control Options Process change Substitution Isolation Ventilation
Administrative control
Personal protective equipment

7. Problem Characterization
AIRFLOW
EMISSION SOURCE
EMISSION SOURCE BEHAVIOR
Where are the sources?
Which emission sources contribute to exposure?
What is the relative contribution of each source to exposure?
Characterize each contributor (chemical composition, temperature, rate of emission, direction of emission, initial emission velocity, continuous or intermittent, time intervals of emission)
AIR BEHAVIOR
How does the air move?
Characterize the air (air temperature, mixing potential, supply and return flow, air changes per hour, wind speed and direction, effects of weather and season)
WORKER BEHAVIOR
How do workers interact with the emission sources?
Characterize employee involvement (worker location, work practices, worker education and training, cooperation)
NOTE: Due to airflow, two workers in the same general vicinity may have different exposures.
BURTON 2-1

8. Burton Ex. 2-1 GROUP EXERCISE
Study the figure on page 2-4 and discuss potential control measures that you might use to correct the problem.
BURTON 2-4

9. THE BEHAVIOR OF AIR

10. The Atmosphere Reaches 50 miles into space.
Pressure = 14.7 pounds per square inch.

11. Composition of Air

12. Pressure Measurement
Vacuum
Atmospheric Pressure
14.7 psia

13. Pressure Measurement 14.7 psia = 407in. Water 14.7 psia =
29.92 in. Mercury (Hg.)

14. How Do We Make Air Move ?

15. Pressure Differences in air pressure cause movement.
The pressure inside a ventilation duct is measured relative to the atmospheric pressure.

16. Pressure Differential Causes Movement
FLOW
LOW
HIGH
FAN
BURTON 3-6

17. Negative Pressure = Less Than Atmospheric

18. Positive Pressure = Greater Than Atmospheric

19. Pressure Relationships

20. Pressure Terms
Static Pressure
Velocity Pressure
Total Pressure

21. Static Pressure Flow SP Static pressure (SP) is
exerted in all directions.
The pressure inside a ventilation duct is measured relative to the atmospheric pressure.
It is measured perpendicular to air flow.

22. Velocity Pressure Flow SP VP Velocity Pressure (VP) is
kinetic (moving pressure) resulting from air flow.
It is measured parallel to the direction of the flow.

23. Total Pressure
Flow SP VP TP
Total pressure (TP) is the algebraic sum of the VP and SP.
It is measured parallel to the direction of the flow.

24. Pressure Upstream and Downstream of the Fan
TP SP VP
Up-stream
Down-stream
BURTON 3-8

25. What is use of the term “Velocity Pressure” ?
Determine the air flow.
To design the system.
V = 4005(VP)1/2

26. What is use of the term “Static Pressure” ?
Accelerate the air.
Overcome resistance to friction.
Friction losses are associated with hood entries, duct entries, dampers, air cleaners, and elbows.

27. Static Pressure and Velocity Pressure are Mutually Convertible
When air is accelerated, the static pressure is converted to velocity pressure. =
Think of static pressure as a “potential pressure”.
When air is decelerated, the velocity pressure can be transformed back into static pressure.

28. Conservation of Mass Mass in = Mass out.
Air speeds up when the duct area is smaller.
Q = VA
Q = Cubic Feet Per Minute
V = Velocity
A = Area

29. Dilution Ventilation YES NO non-hazardous
gas, vapor, respirable particle
uniform time emission
emissions not close to people
moderate climate NO
toxic material
large particulate
emission varies widely over time
large, point source emissions
people in vicinity
severe climate
irritation or complaints
Emissions mix with the workroom are then are diluted to acceptable levels.
Large amounts of heat are lost.
Fresh air>worker>emission source>out
BURTON 4-1

30. Volume Vapor Flow Rate BURTON 4-3 See Burton 3-17
At standard conditions, one pound-molecular weight (lb-mole) of a material will evaporate to fill 387 cubic feet of space.
q is the emission rate.
Finding the pounds evaporated is the hardest thing to do.
q = (387 * lbs. evaporated)/ (MW * t * d)
q = volume of vapor flow rate, acfm
MW = molecular weight
t = time interval over which the liquid
evaporates, minutes
d = density correction factor, unitless
BURTON 4-3

31. Estimating Dilution Air Volume
Mixing factors are in Chart 20 on Burton Appendix Page 32.
Mixing factors of 2.0 to 3.0 are typical.
q = (387 * lbs. evaporated)/ (MW * t * d) - from previous equation.
Dilution air should flow in the direction from clean to dirty.
Cover the material on Burton 4-6 related to good design and operation practices.
Qd = [q * 106 * K mixing]/Ca
Qd - dilution air volume flow
rate
q volume flow rate of
vapor emitted
Ca - acceptable exposure
concentration
K mixing - mixing factor
BURTON 4-5

32. Poor Dilution

33. Good Dilution

34. Example 4-1
What is q, the volume flow rate of vapor formed, if 0.5 gallons of toluene are evaporated uniformly over an 8-hr. shift? What volume flow rate Qd is required for dilution to 10 ppm, if Kmixing = 2 ? (Assume STP; d = 1.0)
What is the average face velocity of air in a room 10ft. * 8ft. * 40ft for these conditions?
Q should be Qd
This example done by instructor. The calculation is performed in the next slide.
BURTON 4-6

35. Strategy Ex. 4-2 Step 1: Calculate the volume flow rate of the vapor
emitted q.
q = (387 * lbs. evaporated)/ (MW * t * d)
Note:lbs. Evaporated = gal. * 8.31 * SG
Step 2. Calculate the dilution air volume flow rate
Qd.
Qd = q * 106 * K mixing
Ca (ppm)
Step 3: Calculate the face velocity.
V face = Qd/A
Step 4: Calculate the air changes/ hour.
N = (Qd * 60)/Vr
Given Information: 5 gallons of perchloroethylene, SG perchloroethylene = 1.63, MW = 165.8
Want to dilute to 25 ppm, Kmixing = 2, time = 480 min.., density = 1.
Strategy: To calculate the dilution air volume flow rate Qd, you first need to calculate q, the volume flow rate of vapor emitted. With this information we can calculate Qd. With this answer in hand, we will then calculate air changes/hour.
Step 1: Calculate the volume flow rate of the vapor emitted q.
q = (387 * lbs. evaporated)/ (MW * t * d)
Note:lbs. Evaporated = gal. * 8.31 * SG
q = (387 * 5 gal * 8.31 lbs.gal * 1.63)/(165.8 *480 * 1.0) = 0.3 scfm
Step 2. Calculate the dilution air volume flow rate Qd.
Qd = q * 106 * K mixing
Ca (ppm)
Qd = (0.33 * 106 * 2)/ 25 = 26,400 scfm
Step 3: Calculate the face velocity
V face = Qd/A = 26,400/500 = about 53 fpm
Step 4: Calculate the air changes/ hour
N = (Qd * 60)/Vr = (26,400 * 60)/30,000 = about 53 air changes/ hour

36. Purge and Buildup
Purge and buildup - predict contaminant buildup or purge rate.
Steady state -equilibrium maintained.
Instructor does example 4-5 on page 4-11.
Note: Have the class correct the Qd notation error in their book.
t = - Kmixing *( Vr/Qd) * ln { ( (q * 106)/ Qd) - C2) }
( (q * 106)/ Qd) - C1)
q is the generation rate in cubic feet per minute
C2 is the concentration in the air of the emitted substance in ppm, after time t, in minutes
Note: If the concentration is given in mg/cubic meter convert using formula:
ppm * MW = mg/cubic meter * 24.1
ln is the natural log
C1 is the initial concentration in ppm
C2 is the final concentration in ppm
Qd is the dilution volume flow rate
Vr is the room volume in cubic feet
K is the mixing factor
BURTON 4-9

37. Example 4-5
An automobile garage was severely contaminated with carbon monoxide.
How long will it take to purge the garage?
STP Kmixing = 1.5
C1 = 10,000 ppm (initial conc.)
C2 = 25 ppm (final conc.)
q = 0 scfm
Qd = 3,000 scfm
t = - Kmixing *( Vr/Qd) * ln { ( (q * 106)/ Qd) - C2) }
( (q * 106)/ Qd) - C1)
t = *( 11,500/3000) * ln { ( (0 * 106)/ 3000) - 25) }
( (0 * 106)/ 3000) -10,00)
t = -1.5 * ln(-25/-10,000) = * ln = 34 minutes
BURTON 4-11

38. Chapter 11 - Makeup Air Balance
Exhausted air must be replaced.
Negative pressure without makeup air.
BURTON 11-1

39. Make up Air
Fresh air supplied into the breathing zone of the associate.

40. Overcoming Negative Static Pressure
Changes in static pressure involving radial (squirrel cage) fans cause a small change in the volumetric flow rate.
Changes in static pressure involving axial (propeller) fans cause a large change in the volumetric flow rate.
A radial fan is commonly known as a squirrel cage fan.
An axial fan is a propeller fan.
Propeller fans are sensitive to static pressure. They can handle one-half inch of static pressure.
Referring to illustration in book. A radial fan will have less volumetric flow loss per unit change in static pressure.
One-quarter inch of static pressure in a room will create a 26-lb force on a door. Crack drafts will approach 2600fpm. Such drafts may disrupt capture velocities in adjacent hoods.
BURTON 11-2

41. INDUSTRIAL VENTILATION 2-4
Good Makeup Air
RULE OF THUMB
Provide 2-3 cfm/sq. ft.
Explain thermal effects.
INDUSTRIAL VENTILATION 2-4

42. INDUSTRIAL VENTILATION 2-4
Bad Makeup Air
RULE OF THUMB
Provide 2-3 cfm/sq.. ft.
Explain thermal effects.
INDUSTRIAL VENTILATION 2-4

43. Reentrainment
BURTON 11-9

44. Reentrainment BURTON 11-9
Bad design can result in exhaust air being recycled.
BURTON 11-9

45. Avoiding Reentrainment 10-50-3000 RULE
Place stacks at least fifty feet from air intakes, ten feet above adjacent roof lines and/or air intakes if within fifty feet, and to provide a stack exit velocity of 3000 fpm.
Avoid rain caps.
BURTON 14-5

46. Recirculation of Exhaust Air
Good for non-toxic particulate control.
Can recover 40-60% of heat energy.
Read criteria on 12-1 and 12-2.
Not a preferred method. OK for wood dust (except carcinogens like red cedar).
Not OK for solvents. (Formaldehyde does not adhere well to carbon)
Can recover % of heat energy.
BURTON 12-1

47. Types of Ventilation Systems
Process exhaust exhausts major process offgasses, such as the hot process gas from a furnace.
Local exhaust is dedicated to employee protection.
BURTON 5-1

48. Why Choose Local Ventilation?
No other controls
Containment
Employee in vicinity
Emissions vary with time
Sources large and few
Fixed source
Codes
BURTON 5-2

49. Exercise 5-3
Form your group and try exercise Compare the operation to the parameters listed below:
No other controls available
Hazardous contaminant
Employee in immediate vicinity
Emissions vary with time
Emission sources large and few
Fixed emission source
Codes & standards
This exercise is performed by the class.
BURTON 5-3

50. Components of a Local Exhaust System
Hoods are discussed in Chapter 6.
Piping is covered in Chapter 7.
Fans are discussed in Chapter 8.
BURTON 5-4

51. Static Pressure Review
Static pressure is the “potential energy” of the system. It is converted into the “kinetic energy” of velocity, the heat energy of friction losses, and so forth.
BURTON 5-5

52. Energy Conservation BURTON 5-6 TYPES OF LOSSES Duct friction losses.
Duct losses in elbows, contractions, expansions, and orifices.
Entry losses in branch entries, cleaner entries.
Hood entry losses due to turbulence and the vena contracta.
System effect losses at the fan.
Special fitting losses such as blast gates, valves, air cleaners, etc..
Other - stack losses …
Optional: Class exercise 5-6.
BURTON 5-6

53. Basic Air Flow Equations
Q = V * A
TP = SP + VP
V = 4005(VP/d)0.5
ADD: SP(loss) = K(loss) * VP * d
BURTON 5-7

54. Static Pressure Loss Static Pressure Loss = Kloss * VP * d BURTON 5-8
SPloss = Kloss * VP * d
Sploss = loss of static pressure
Kloss = loss factor, determined experimentally
VP = the average velocity pressure in the duct
d = density correction factor
The formula for velocity from velocity pressure assuming density of 1.
V = 4005(VP)0.5
Solving for VP
VP = (V/4005)2
SPloss = Kloss * (V/4005)2
The figure 4005 incorporates conversion factors.
When a small quantity of air flows through a large hood, entry losses are small. When air flows at low velocity down a duct, friction losses are low. But as the quantity of air increases, or as the velocity of the air in the duct increases, the losses increase by the square of the increase.
BURTON 5-8

55. Elbow Loss
Air moving through elbows spends static pressure because of:
directional change
friction
shock losses
turbulent mixing
air bunching up
SP(loss) = K(elbow )* VP * d
Air bunches up in elbows so elbows sometimes designed 2 gauges lower so they wear longer
BURTON 5-9

56. Elbow Loss Ex. 5-8
What is the elbow loss factor K(elbow) where the elbow radius of curvature is R/D = 2.0 in a smooth transition elbow.
Done by instructor. Use Chart 13. Appendix page 25.
K=0.13
BURTON 5-9

57. Elbow Loss Exercise 5-9
What is the actual loss in inches of water of air flowing through a 60-degree, 3-piece elbow at V = 3440 fpm? R/D = 1.5, STP, d=1.
Work with R/D ratio of 1.5.
Done by the class.
Use Chart #13. Appendix page 25.
BURTON 5-10

58. Elbow Loss Exercise 5-9 SPloss = K * VP * d
Use Chart 13, Appendix pg. 25 for information on a 90-degree 3- piece elbow with R/D = 1.5
Let K = (angle/90) * K 90
VP = (V/4005)2
Done by class
K = 0.34 for a 90-degree 3-piece elbow with R/D = 1.5.
Let K = (60/90) * K90 = 0.67 * 0.34 = 0.23
V = 3440ft/sec.
VP = (3440/4005)2 = 0.74 in w.g.
SP loss = K * VP * d = 0.23 * 0.74 * 1 = 0.17 in w.g.

59. Friction Loss as a Function of Duct Length
Friction Loss = K * VP * L * R * d
K is a value taken from Chart #5,
appendix page 9
VP is duct velocity pressure, in w.g.
L is the length of the duct in feet
d is the density correction factor
R is roughness correction factor
Done by instructor.
BURTON 5-11

60. Exercise 5-10
What is the friction loss for a length of galvanized duct with the following parameters? D = 8in., Q = 1000scfm, L = 43 ft. R = 1.
K = from Chart #5, appendix page 9
Area of 8 in dust = from chart also Pi * (4/12)2
Q = VA, V = Q/A = 1000/ = 2865 fpm
VP = (V/4005)2 = (2865/4005)2 = 0.51 in w.g.
Friction Loss = K * VP * L * R * d = * *43 * 1 * 1 = 0.68 in w.g.
Useful equations:
Q = VA, V = Q/A
VP = (V/4005)2
Friction Loss = K * VP * L * R * d

61. Tee Losses
Use Chart 14. Appendix page 26.
BURTON 5-11

62. Tee Losses Ex. 5-12
What is the estimated static pressure loss in inches of water for a branch entry of 30 degrees where the branch entry velocity is 4500 fpm?
Done by class.
Use Chart 14, appendix page 26.
K = 0.18
VP = (4500/4005)2 = 1.26 in w.g.
SPloss = 0.18 * 1.26 = 0.23 in w.g.
(Use modified ACGIH method)
Use Chart 14, appendix page 26
SPloss = K * VP
VP = (V/4005)2
BURTON 5-12

63. Converting Static Pressure To Velocity Pressure
At the hood, all of the available static pressure is converted to velocity pressure and hood entry loss.
SPh = VP + he
At the hood, all of the available static pressure is converted to velocity pressure and hood entry losses. It takes one VP plus hood entry losses to accelerate the air.
BURTON 6-2

64. Measuring Hood Static Pressure
Measure hood static pressure 4-6 duct diameters downstream from the hood.
4-6 D
BURTON 6-2

65. Hood Entry Losses
The hood entry loss is the sum total of all losses from the hood face to the point of measurement in the duct.
SP(loss) = K * VP * d
he = K * VP * d
Substituting into the hood static equation, we have:
SPh = VP + he = VP + (K*VP*d)
= VP (1 + K * d)
BURTON 6-2

66. Example 6-1
What is the hood static pressure when the duct velocity pressure is VP = 1.10 in. w.g. and the hood entry loss is
he = 1.00 in w.g.
SPh = VP + he
SPh =
= in w.g.
Remind the class that the static pressure is negative upstream of the fan.
BURTON 6-3

67. Vena Contracta
The greatest loss normally occurs at the entrance to the duct, due to the vena contracta formed in the throat of the duct.
The vena contracta is smaller when the hood has a flange.
BURTON 6-3

68. Ce = Q(actual)/Q(ideal)
Hood Efficiency
A hood’s efficiency can be described by the ratio of actual to ideal flow. This ratio is called the Coefficient of Entry, Ce.
Ce = Q(actual)/Q(ideal)
The closer the ratio is to 1, the more efficient the hood.
BURTON 6-4

69. Hood Static Pressure and Entry Losses Example 6-5
The average velocity in a duct serving a hood is V = 2000 fpm. The loss factor for the hood has been obtained from the manufacturer as Khood = What are the he and SPh? (Assume STP, d = 1)
Done by instructor. Explain to the class that since your measurement is taken upstream of the fan, that this was the reason that there is a negative sign in the equation for SPh.
From Chart 7, appendix pg. 12, VP = 0.25in. w.g.
he = K*VP*d = 2.2*0.25*1 = 0.55 in. w.g.
SPh = -(he + VP) = -( ) = -0.80
in w.g.
BURTON 6-5

70. Hand Grinding Table Example 6-6
Assume that a special hand grinding table hood has been built and the following data have been measured:
SPh = in w.g., V = 4000fpm, and the duct diameter is 18 in. (Assume STP, d=1)
Done by instructor.
Ce = (VP/SPh) 0.5 is from Burton 6-4.
I assume that this illustration would demonstrate to the class how the parameters are derived.
VP = (V/4005)2 = (4000/4005)2 = in w.g.
Unit conversion: 18in * 1ft/12in = 1.5 ft.
Area =  D2/4 = 3.14 * (1.5)2/4 = sq. ft.
Q = VA = 4000fpm* sq. ft. = 7,068 scfm
he = SPh - VP = = 1.0 in. w.g.
Ce = (VP/SPh) 0.5= (1/2.5)0.5 = 0.63
K = he/VP = 1.50/1.00 = 1.50
BURTON 6-6

71. Types of Hoods Receiving Capturing Enclosing BURTON 6-10
Discuss design approaches on pages 6-9 and 6-10.
BURTON 6-10

72. Hood Types
SLOTTED HOOD

73. Hood Types
ENCLOSED HOOD

74. Hood Types
ENCLOSING HOOD

75. Hood Types
CAPTURING HOOD

76. Grinding Wheel Hood Example Example 6-9
Determine the volume flow rate, transport velocity, duct diameter, loss factor K, Ce, he, and SPh, for a grinding wheel hood, wheel diameter = 13in. (low surface speed), straight take off [sto], STP)
Step 1. Proceed to Chart 11C in appendix page 18.
K=0.65, Ce = 0.78, Vtrans. = 4500fpm,
Q = 35Wd - 50 = [(35*13)-50] = 405 cfm
Step 2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed. Line up
nomograph with V = 4500 and Q = 405
D = 4 in. = sq. ft.
Step 3. Find the actual velocity.
Vactual = Q/A = 405/ = 4639 fpm
D = 4in, Vactual = 4639fpm
Step 4. Calculate the velocity pressure in the duct for a velocity of 4639 fpm using Chart #7 in
appendix page 12.
VP = 1.34 in w.g.
Alternate way to calculate the VP =(V/4005)2 = (4639/4005)2 = 1.34 in w.g.
Step 5.Calculate hood entry loss.
he = K*VP = 0.65*1.34 = 0.87 in w.g.
VP = 1.34 in w.g, he = 0.87
Step 5. Calculate the hood static pressure.
(formula different than used in the book)
SPh = VP + he = =
-2.21 in w.g.
(minus sign added because hood is upstream of the fan)
BURTON 6-12

77. EXERCISE 6-10 USEFUL FORMULAS
Q = V * A
V = 4005(VP)1/2
VP = (V/4005)2
he = K * VP
SPh = VP + he
In exercise 6-10 a, they list a picture of a buffing hood instead of a grinding wheel hood.
A, B, and C - Students
D and E Instructor
BURTON 6-12 AND 6-13

78. Exercise 6-10a
Where appropriate, determine the volume flow rate, transport velocity, duct diameter, loss factor K, Ce, he, and SPh for a grinding wheel hood with a wheel diameter of 14 in. (low surface speed, tapered takeoff [tto]. Note: the picture in the book is for a buffing hood.
In exercise 6-10 a, they list a picture of a buffing hood instead of a grinding wheel hood.
A, B, and C - Students
D and E Instructor
1. Use Chart 11C, appendix pg. 18 to find Q, Vtrans., K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
Q = (35*14)-50 = 440 cfm
K=0.40, Ce=0.85, Vtrans = 4500 fpm
D = 4 in. = sq. ft.
Vactual = Q/A = 440/ = 5040 fpm
VP = (Vactual/4005)2 = (5040/4005)2 = 1.58 in. w.g.
he = K * VP = 0.40 * VP = 0.40 * 1.58 = in w.g.
SPh = VP + he = = 2.21
BURTON 6-12

79. Exercise 6-10a Strategy
1. Use Chart 11C, appendix pg. 18 to find Q, Vtrans., K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
In exercise 6-10 a, they list a picture of a buffing hood instead of a grinding wheel hood.
A, B, and C - Students
D and E Instructor
1. Use Chart 11C, appendix pg. 18 to find Q, Vtrans., K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
Q = (35*14)-50 = 440 cfm
K=0.40, Ce=0.85, Vtrans = 4500 fpm
D = 4 in. = sq. ft.
Vactual = Q/A = 440/ = 5040 fpm
VP = (Vactual/4005)2 = (5040/4005)2 = 1.58 in. w.g.
he = K * VP = 0.40 * VP = 0.40 * 1.58 = in w.g.
SPh = VP + he = = 2.21

80. Exercise 6-10b
Where appropriate, determine the volume flow rate, transport velocity, duct diameter, loss factor K, Ce, he, and SPh for a hand grinding table 10 feet long by 2 feet wide.
A, B, and C - Students
D and E Instructor
1. Use Chart 11C, appendix pg. 18 to find Q, Vtrans., K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
Q = 250 * 20 = 5000 cfm
K=0.50, Ce=0.82, Vtrans = 3500 fpm
D = 16 in. = sq. ft.
Vactual = Q/A = 5000/1.396 = 3580 fpm
VP = (Vactual/4005)2 = (3580/4005)2 = 0.80 in. w.g.
he = K * VP = 0.50 * 0.80 = 0.40 in w.g.
SPh = VP + he = = 1.2
BURTON 6-13

81. Exercise 6-10b Strategy
1. Use Chart 11C, appendix pg. 18 to find Q, Vtrans., K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
A, B, and C - Students
D and E Instructor
1. Use Chart 11C, appendix pg. 18 to find Q, Vtrans., K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
Q = 250 * 20 = 5000 cfm
K=0.50, Ce=0.82, Vtrans = 3500 fpm
D = 16 in. = sq. ft.
Vactual = Q/A = 5000/1.396 = 3580 fpm
VP = (Vactual/4005)2 = (3580/4005)2 = 0.80 in. w.g.
he = K * VP = 0.50 * 0.80 = 0.40 in w.g.
SPh = VP + he = = 1.2

82. Exercise 6-10c
Where appropriate, determine the volume flow rate, transport velocity, duct diameter, loss factor K, Ce, he, and SPh for a band saw used to cut wood that has a blade width of 1 inch.
A, B, and C - Students
D and E Instructor
1. Use Chart 11E, appendix pg. 20 to find Q, Vtrans., K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
Q = (250 * 1 in.)+ 300 = 550 cfm
K=1.75, Ce=0.60, Vtrans = 3500 fpm
D = 5 1/2 in. = sq. ft.
Vactual = Q/A = 3333 = fpm
VP = (Vactual/4005)2 = (3333/4005)2 = in. w.g.
he = K * VP = 1.75*0.69 = in w.g.
SPh = VP + he = = 1.91 in w.g.
BURTON 6-13

83. Exercise 6-10c Strategy
1. Use Chart 11E, appendix pg. 20 to find Q, Vtrans., K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
A, B, and C - Students
D and E Instructor
1. Use Chart 11E, appendix pg. 20 to find Q, Vtrans., K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
Q = (250 * 1 in.)+ 300 = 550 cfm
K=1.75, Ce=0.60, Vtrans = 3500 fpm
D = 5 1/2 in. = sq. ft.
Vactual = Q/A = 3333 = fpm
VP = (Vactual/4005)2 = (3333/4005)2 = in. w.g.
he = K * VP = 1.75*0.69 = in w.g.
SPh = VP + he = = 1.91 in w.g.

84. Exercise 6-10d
Where appropriate, determine the volume flow rate, transport velocity, duct diameter, loss factor K, Ce, he, and SPh for a bell-mouthed hood used for welding. X=10 in., Vc = 100 fpm, Vtrans = 3000 fpm.
A, B, and C - Students
D and E Instructor
1. Use Chart 11A, appendix pg. 16 to find Q, K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
Q = 3X2Vc = 3*3.14*(10/12)2 * 100 = 650 cfm
K= 0.04, Ce=0.98, Vtrans = 3000 fpm
D = 6 in. = sq. ft.
Vactual = Q/A = 650/ = 3310 fpm
VP = (Vactual/4005)2 = (3310/4005)2 = in. w.g.
he = K * VP = 0.04*0.68 = in w.g.
SPh = VP + he = = 0.71 in w.g.
BURTON 6-13

85. Exercise 6-10d Strategy
1. Use Chart 11A, appendix pg. 16 to find Q, K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
A, B, and C - Students
D and E Instructor
1. Use Chart 11A, appendix pg. 16 to find Q, K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
Q = 3X2Vc = 3*3.14*(10/12)2 * 100 = 650 cfm
K= 0.04, Ce=0.98, Vtrans = 3000 fpm
D = 6 in. = sq. ft.
Vactual = Q/A = 650/ = 3310 fpm
VP = (Vactual/4005)2 = (3310/4005)2 = in. w.g.
he = K * VP = 0.04*0.68 = in w.g.
SPh = VP + he = = 0.71 in w.g.

86. Exercise 6-10e
Where appropriate, determine the volume flow rate, transport velocity, duct diameter, loss factor K, Ce, he, and SPh for a canopy hood used for a hot-liquid open surfaced tank. P = 16 ft., X = 3 ft., Vcontrol = 125 fpm, Vtrans = 2000fpm.
A, B, and C - Students
D and E Instructor
1. Use Chart 11B, appendix pg. 17 to find Q, K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
Q = 1.4PXVcontrol = 1.4*16*3*125 = 8400 cfm
K= 0.25, Ce=0.89, Vtrans = 2000 fpm
D = 28 in. = sq. ft.
Vactual = Q/A = 8400/4.276 = 1965 fpm
VP = (Vactual/4005)2 = (1965/4005)2 = in. w.g.
he = K * VP = 0.25*0.24 = in w.g.
SPh = VP + he = = 0.30 in w.g.
BURTON 6-13

87. Exercise 6-10e Strategy
1. Use Chart 11B, appendix pg. 17 to find Q, K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
A, B, and C - Students
D and E Instructor
1. Use Chart 11B, appendix pg. 17 to find Q, K, and Ce.
2. Use Chart 5A in appendix pg. 9 to find the diameter of the pipe needed and it’s area.
3. Calculate Vactual = Q/A
4. VP = (Vactual/4005)2
5. he = K * VP
6. SPh = VP + he
Q = 1.4PXVcontrol = 1.4*16*3*125 = 8400 cfm
K= 0.25, Ce=0.89, Vtrans = 2000 fpm
D = 28 in. = sq. ft.
Vactual = Q/A = 8400/4.276 = 1965 fpm
VP = (Vactual/4005)2 = (1965/4005)2 = in. w.g.
he = K * VP = 0.25*0.24 = in w.g.
SPh = VP + he = = 0.30 in w.g.

88. Factors Influencing Hood Performance
Competition
Mixing
Work practices
Competition: Competing air currents greater than 100 fpm may effect the hood’s effectiveness. Blow lightly on your hand so you barely feel air movement = 100fpm.
Mixing:
Cooling fans
Ventilation system outlets
People walking by (a person walks about 300 fpm)
Mobile equipment
Obstructions
Work Practices: Employee stands between paint sprayer and hood system of paint booth.
BURTON 6-17

89. Canopy Hoods Use only for hot processes with rising air.
Estimate initial and terminal velocities of rising air stream.
The volume of air exhausted from the hood must exceed the volume of air arriving at the hood face.
Warm rising air expands as it rises. Make the cross-sectional area of the hood face 125% larger than the plume of hot air.
Avoid canopy hoods if an employee must work over the source.
BURTON 6-19

90. Chapter 7 Selection and Design of Ductwork
Air does not flow smoothly in industrial ductwork. Because of it’s high velocity ( fpm) it is either totally turbulent, or approaching it. This turbulence is desired to assist in transporting particles through the ductwork.
Chart No. 9, appendix pg. 14, provides recommended transport velocities for different operations and particle sizes.
BURTON 7-1

91. Exercise 7-2
Standard air (d=1) moves through an 8 in. galvanized duct system at 4000 fpm. Estimate VP, find the loss factors K from the Charts, and then estimate static pressure loss for each component in each branch. (Note: treat the branch entry as two 45-degree entries and use the ACGIH value for K on Chart 14.)
Done by instructor. Correct errors.
BURTON 7-4

92. Exercise 7-2a, Flanged Hood
Done by instructor. Correct errors.
D = 8in. Vave. = 4000 fpm
Find VP, Loss factors K, and estimate static pressure losses.
a. Flanged hood - From Chart 11A, appendix pg. 16, K = 0.50.
VP = (Vave./4005)2 = (Vave./4005)2 =
1 in. w.g.
he = K * VP * d = 0.5*1*1 =
0.5 in. w.g.
BURTON 7-4

93. Exercise 7-2b, Plain Duct Hood
Done by instructor. Correct errors.
D = 8in. Vave. = 4000 fpm
Find VP, Loss factors K, and estimate static pressure losses.
b. Plain duct hood - From Chart 11A, appendix pg. 16, K = 0.93.
VP = (Vave./4005)2 = (Vave./4005)2 =
1 in. w.g.
he = K * VP * d = 0.93*1*1 =
0.93 in. w.g.
BURTON 7-4

94. Exercise 7-2c, Elbow, 3-piece
Done by instructor. Correct error, answer should be K = SPloss = 0.42
D = 8in. Vave. = 4000 fpm
Find VP, Loss factors K, and estimate static pressure losses.
c. Elbow, 3-piece, R/D =1, 90-degree From Chart 13, appendix pg. 25, K = 0.42
VP = (Vave./4005)2 = (Vave./4005)2 =
1 in. w.g.
SP loss = K * VP * d = 0.42*1*1 =
0.42 in. w.g.
BURTON 7-4

95. Exercise 7-2d, Elbow, 5-piece
Done by instructor. Correct error, answer should be K = SPloss = 0.24
D = 8in. Vave. = 4000 fpm
Find VP, Loss factors K, and estimate static pressure losses.
d. Elbow, 5-piece, R/D =1.5, 90-degree From Chart 13, appendix pg. 25, K = 0.24
VP = (Vave./4005)2 = (Vave./4005)2 =
1 in. w.g.
SP loss = K * VP * d = 0.42*1*1 =
0.24 in. w.g.
BURTON 7-4

96. Exercise7-2e, Elbow, 4-piece
Done by instructor. Correct error, answer should be K = SPloss = 0.12
D = 8in. Vave. = 4000 fpm
Find VP, Loss factors K, and estimate static pressure losses.
e. Elbow, 4-piece, R/D =2.0, 45-degree From Chart 13, appendix pg. 25, K = 0.24/2 = 0.12
VP = (Vave./4005)2 = (Vave./4005)2 =
1 in. w.g.
SP loss = K * VP * d = 0.42*1*1 =
0.12 in. w.g.
BURTON 7-4

97. Exercise 7-2f, Branch Entry
Done by instructor. Correct error, answer should be K = 0.28 (each branch). SPloss = 0.56
D = 8in. Vave. = 4000 fpm
Find VP, Loss factors K, and estimate static pressure losses.
f. Branch entry, 90-degree included angle. From Chart 14, appendix pg. 26, K = 0.28 (each branch)
VP = (Vave./4005)2 = (Vave./4005)2 =
1 in. w.g.
SP loss = K *2* VP * d = 0.28*2*1 =
0.56 in. w.g.
BURTON 7-4

98. Exercise 7-2g, 50 ft. of Duct BURTON 7-4
Done by instructor. Correct error, answer should be K = 0.03 SPloss = 0.03
D = 8in. Vave. = 4000 fpm
Find VP, Loss factors K, and estimate static pressure losses.
f. Friction loss, L = 50 ft galv. duct. From Chart 5A, appendix pg. 9, K = 0.03 (per foot)
VP = (Vave./4005)2 = (Vave./4005)2 =
1 in. w.g.
SP loss = K * VP * L * R * d = 0.03*1*50*1*1 = 0.03 in. w.g.
BURTON 7-4

99. Roughness Example 7-1
Standard air is flowing in 40 feet of a 24 in. concrete pipe at the 4000 fpm. What is the correction factor, R? The loss factor K?
From Chart 10, appendix pg. 15, R = 2.1
From Chart 5A, appendix pg 9, K = per foot
Friction loss = K*VP*L*R*d
Friction loss = *1*40*2.1*1 = 0.63 in w.g.
BURTON 7-5

100. Duct Shapes
Use round duct whenever possible, it resists collapsing, provides better aerosol transport conditions, and may be less expensive.
For rectangular and square duct, one formula for the circular equivalent is:
Dc = 1.3*[(ab)5/(a+b)2]0.125
Dc = circular equivalent, in inches
a is the length of one side in inches and b is the length of the other side in inches
BURTON 7-6

101. Pressure Diagrams BURTON 7-11
Every ventilation system is self balancing. Static pressure at any point in the system is directly related to the requirements for air flow at that point. If we increase the static pressure at the fan, the system will readjust itself to restore balance. If static pressure demand is reduced at one point in the system, the whole system will redistribute the additional static pressure to achieve a new balance.
For example, at the hood the static pressure is precisely the amount necessary to get air through the hood and flowing down the duct at actual duct velocity.
The static pressure at the fan is the amount required to move air through the hood, down the duct, through the elbows, etc., and bring it at duct velocity to the fan entry.
BURTON 7-11

102. Chapter 8 Fan Selection and Operation
AXIAL FANS
propeller fans
CENTRIFUGAL FANS
radial fans
forward inclined
backward inclined
The backward-inclined wheel is often used when the air stream is relatively clean.
The foreward-inclined wheel is rarely used in industrial systems. It is not very rugged and tends to clog or abrade away.
BURTON 8-2

103. Fan Total Pressure
The fan total pressure (FTP) represents all energy requirements for moving air through the ventilation system.
The fan total pressure is often referred to as the fan total static pressure drop.
FTP = TP outlet - TP inlet
FTP = SPout - VP out - SPin - VP in
FTP = SPout - SPin
BURTON 8-3

104. Exercise 8-1
Find the Fan Total Pressure given that the SPin = -5.0 in w.g, SPout = in w.g.
VPin = VPout = 1.0 in. w.g.
FTP = SPout - SPin =
(-5.0) = 5.4 in w.g.
Done by instructor.
Ftp = SPout - SPin
BURTON 8-3

105. Exercise 8-2 Fan Static Pressure
The fan static pressure out of the fan is defined as the fan total pressure minus the average velocity pressure out of the fan.
FSP = Fan TP - VPout
Done by instructor.
FSP = SPout - SPin - VPin
Fan static pressure represents the system losses, I.e., the amount of static pressure converted to useless heat or noise.
FTP = SPout - VP out - SPin - VP in -VPout
FSP = SPout - SPin - VP in
SPin = -5 in. w.g., SPout = in. w.g.
VPin = VPout = 1.0 in. w.g.
FSP = (-5.0) - 1 = 4.4 in. w.g.
BURTON 8-4

106. SOP and Fan Curves
To develop a system curve, the fan should be turned at different rpms and the flow and the absolute values of the static pressures at the fan are plotted.
BURTON 8-5

107. Developing Fan Curves
BURTON 8-6

108. SOP on Steep Part of Curve
BURTON 8-7

109. Example 8-1
Choose an appropriate fan for a system operating point of Q = 10,000 scfm and FTP = 1.5 in. w.g.
Done by instructor.
first chart , rpm = 550 (no.730BL fan)
second chart, none possible.
third chart, rpm = 556
BURTON 8-8

110. Exercise 8-3
Find a fan and appropriate rpm for a fan exhausting 15,000 cfm at a fan TP = 2.0 in. w.g.
Done by instructor.
805BL
600rpm
Can use 890
BURTON 8-8

111. Exercise8-4
Find a suitable fan and the appropriate rpm for a ventilation system exhausting 480 cfm at a fan TP = 13.8 in. w.g.
Done by instructor.
17.5 in wheel at 3500 rpm
BURTON 8-8

112. Commercial Fan Curves
BURTON 8-9

113. Commercial Fan Curves
BURTON 8-10

114. Commercial Fan Curves
BURTON 8-11

115. System Effect Losses BURTON 8-12
System effect loss occurs when air does not enter and leave the fan wheel in the smooth, uniform, balanced manner found during ideal testing.
BURTON 8-12

116. Six-and-Three Rule BURTON 8-13
Six diameters of straight duct into a fan, and three diameters of straight duct out of a fan before any elbows.
BURTON 8-13

117. Air Horsepower
Air horsepower refers to the minimum amount of power to move a volume of air against the fan total pressure. It represents the power to get the air through the duct system.
ahp = ( FTP * Q * d)/6356
Air horsepower refers to the minimum amount or power to move a volume of air against the fan total pressure. It represents the power to get the air through the duct system.
ahp air horsepower
FTP fan total pressure (in w.g.)
d density correction factor
BURTON 8-14

118. Brake Horsepower
Brake horsepower refers to the actual power required to operate the fan so that it fulfills the job of moving the specified cfm against the FTP. It takes into account fan inefficiencies, i.e. losses in the fan.
bhp = ahp/ME
Brake horsepower refers to the actual power required to operate the fan so that it fulfills its job of moving the specified cfm against the specified FTP.
bhp brake horsepower
ME mechanical efficiency
ME = 0.60 will probably work well enough for a rough estimate of a radial fan.
BURTON 8-15

119. Shaft Horsepower
Shaft horsepower is bhp plus any power required for drive losses, bearing losses, and pulley losses between the fan and the shaft of the motor.
shp = bhp * Kdl
Shaft horsepower is bhp plus any power required for drive losses, bearing losses, and pulley losses between the fan and the shaft of the motor.
Kdl is approx for pulley drives on larger motors, 1.30 for pulley drives on motors < 2hp, and 1.05 for direct drive.
BURTON 8-15

120. Rated Horsepower
Rated horsepower is the nameplate horsepower on the motor.
Rated horsepower is the nameplate horsepower on the motor.
Horsepower: 1, 1.5, 2,3,5,7.5,10,15,20,25
BURTON 8-15

121. Example 8-4
What is the required power for the system and what rated power motor would you use?
FTP = 5.0 in. w.g. ,
Q = scfm
ME = 0.60, Kdl = 1.10, d = 1,
f = 6356
Done by instructor.
shp = (FTP*Q*Kdl*d)/6356*ME
shp = (5*12000*1.10*1)/6356*0.60 =
17.3 hp
rhp>1.33shp>17.3*1.33>23 (use 25 hp motor)
BURTON 8-16

122. Exercise 8-7
Estimate the ahp, bhp, shp, and the rated power motor you would choose for the following system.
Fan TP = 10.0 in. w.g.,
Q = 5000 scfm
Kdl = 1.15, STP(d=1),
f = 6356, ME = 0.65
ahp = ( FTP * Q * d)/6356
bhp = ahp/ME
shp = bhp * Kdl
Fan TP = 10.0 in. w.g., Q = 5000 scfm
Kdl = 1.15, STP(d=1), f = 6356
ahp = ( FTP * Q * d)/6356 = (10*5000*1)/6356 = 8 hp
bhp = ahp/ME = 8/0.65 = 12.3 bhp
shp = bhp * Kdl = 12.3 * 1.15 = 14.1
rhp = 1.33 *shp = 1.33*14.1 = 20hp
BURTON 8-17

123. Fan Laws
BURTON 8-19

124. Local Exhaust Ventilation Design
BURTON 9-1

125. Plenum Design
BURTON 9-3

126. BALANCING
Balancing during the design phase means adjusting losses in duct runs leading to a junction that the predicted loss in each run is essentially equal.
BURTON 9-4

127. Example 9-2
Design an local exhaust system based on the criteria listed in the example.
Done by instructor.
BURTON 9-5

128. Done by instructor.

129. Example 9-3
Design a local exhaust system based upon the criteria listed on this page.
BURTON 9-11