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Module 5: Flowcharts and Ventilation Systems - Review Exercises

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Instructions:

Work these problems on a sheet of paper and check your answers against those provided below.

#1

Identify each of the following flowchart symbols.

1. Nozzle
   
2. Emission point
   
3. Solid or liquid stream
   
4. Valve

1. Fan
   
2. Fugitive emission source
   
3. Solid or liquid stream
   
4. Pump

1. Major equipment
   
2. Damper
   
3. Hood
   
4. Stack

1. Pump
   
2. Hood
   
3. Gas stream
   
4. Fan

1. Nozzle
   
2. Emission point
   
3. Solid or liquid stream
   
4. Valve

Answer: i. c. Solid or liquid stream

Answer: ii. d. Pump

Answer: iii. a. Major equipment

Answer: iv. d. Fan

Answer: v. b. Emission point

To review material, see Module 5 lesson on Flowcharts (Symbols).

#2

Which static pressure reading appears to be illogical according to the flowchart?

1. Duct A
   
2. Duct B
   
3. Duct D
   
4. All static pressure readings given appear to be logical.

Answer: d. All static pressure readings given appear to be logical.

The gas stream decreases in static pressure as it approaches the fan inlet. The static pressure of the gas stream rises after the fan.

To review material, see Module 5 lesson on Flowcharts (Diagrams).

#3

Calculate the static pressure at the inlet to the centrifugal fan. Exclude frictional losses of ducts and entry losses. See Figure 2 (Example Flowchart).

1. 1.2 in. W.C.
   
2. -1.2 in. W.C.
   
3. -9.2 in. W.C.
   
4. -13.2 in. W.C.

Answer: c. -9.2 in. W.C.

Solution:

To review material, see Module 5 lesson on Flowcharts (Diagrams).

#4

The temperature in Duct A was checked by plant personnel and determined to be correct. Which of the other temperature readings appears to be illogical according to the flowchart? See Figure 2 (Example Flowchart).

1. Duct B
   
2. Duct C
   
3. Duct D
   
4. All temperature readings appear to be logical.

Answer: a. Duct B

The temperatures should decrease as the gas moves through the system because no significant source of heat is added to the gas. There is no reason for the gas stream in Duct B to be hotter than the gas stream in Duct A. Otherwise the temperature trend appears to be logical. Note that the gas temperature sometimes rises slightly after the gas stream is compressed in a fan due to the Joule-Thompson effect.

To review material, see Module 5 lesson on Flowcharts (Diagrams).

#5

The hood capture efficiency is 92% and the wet scrubber control system has a collection efficiency of 95%. If the process served by this system is generating 140 lb_m of pollutant per hour, calculate the fugitive emissions and the stack emissions.

1. Fugitive emissions = 1.50 lb_m/hr and Stack emissions = 0.80 lb_m/hr
   
2. Fugitive emissions = 11.20 lb_m/hr and Stack emissions = 6.44 lb_m/hr
   
3. Fugitive emissions = 14.0 lb_m/hr and Stack emissions = 3.54 lb_m/hr
   
4. Fugitive emissions = 20.50 lb_m/hr and Stack emissions = 9.80 lb_m/hr

Answer: b. Fugitive emissions = 11.20 lb_m/hr and Stack emissions = 6.44 lb_m/hr

Solution:

1. Calculate the fugitive emissions.
   

2. Calculate the captured emissions.
   

3. Calculate the stack emissions.
   

   To review material, see Module 5 lesson on Hoods (Introduction).

#6

A decline in hood capture efficiency can affect:

1. Fugitive emissions.
   
2. Stack emissions.
   
3. Hood static pressure.
   
4. a and b, only
   
5. a, b, and c

Answer: d. a and b, only

Maintaining high hood capture efficiencies is very important because small declines in the efficiency rate can cause significant increases in fugitive emissions. If the system includes a stack, a decline in the hood capture efficiency would affect stack emissions because less contaminated air would be drawn into the air pollution control system and exhausted through the stack.

Changes in the hood static pressure can affect the hood capture efficiency but changes in hood capture efficiencies do not cause the hood static pressure to change. For example, if the hood is moved further away from the source, the hood capture efficiency drops but the static pressure remains the same (assuming no other changes were made to the system).

To review material, see Module 5 lessons on Hoods (Introduction) and Hoods (Monitoring Performance).

#7

Hood capture efficiency can be improved by:

1. Adding baffles or flanges.
   
2. Lowering the airflow rate.
   
3. Increasing the distance between the hood face and the contaminant source.
   
4. Increasing cross drafts around the hood entrance.

Answer: a. Adding baffles or flanges.

Hood capture efficiency can be improved by (1) adding baffles or flanges to help block the movement of clean air into the hood and reduce cross drafts (2) decreasing the distance between the hood face and the contaminant source, and (3) ensuring that the airflow rate is sufficient to maintain the required hood capture velocity.

To review material, see Module 5 lesson on Hoods (Operating Principles).

#8

What is the required volumetric flow rate for a flanged hood with dimensions of 10 in. by 16 in.? The distance between the hood face and the contaminant source is 9 in. and the hood capture velocity is 350 fpm.

1. 1,768 ACFM
   
2. 2,260 ACFM
   
3. 2,357 ACFM
   
4. 43,477 ACFM

Answer: a. 1,768 ACFM

Solution:

Where:

1. Calculate the area of the hood opening.
   

2. Calculate the volumetric flow rate, Q, required to attain the recommended capture velocity of 350 fpm at a distance of 9 in. from the hood.
   

   To review material, see Module 5 lesson on Hoods (Operating Principles).

#9

A hood with a hood coefficient of entry (Fd) of 0.49 has a hood static pressure of -1.16 in. W.C. What is the gas flow rate through a 1.5-foot diameter duct from the hood? Use EPA standard conditions for temperature and pressure.

1. 4,165 ACFM
   
2. 4,304 ACFM
   
3. 6,247 ACFM
   
4. 8,015 ACFM

Answer: c. 6,247 ACFM

Solution:

Step 1. Calculate the velocity pressure (VP) using the following equation.

1. Calculate the hood entry loss (h_e) using the following equation.

   

2. Calculate the velocity pressure.

   

Step 2. Calculate the gas velocity at standard conditions using the following equation. The density of air at EPA standard conditions is 0.075 lb_m/ft³.

Step 3. Calculate the gas flow rate.

To review material, see Module 2 lesson on Velocity and Module 5 lesson on Hoods (Monitoring Performance).

#10

A centrifugal fan is moving 1,000 ACFM of air at a temperature of 450°F and a fan inlet pressure of -15 in. W.C. What would the actual airflow rate be if the gas temperature decreased to 68°F, the inlet pressure remained unchanged, and the fan rotational speed remained the same?

1. The airflow rate will increase to 1,580 ACFM.
   
2. The airflow rate will decrease to 580 ACFM.
   
3. The airflow rate will remain at 1000 ACFM.
   
4. The airflow rate will increase to 1,723 ACFM.

Answer: c. The airflow rate will remain at 1000 ACFM.

According to the first fan law shown below, fans move a constant volume of air as long as the fan speed remains the same.

To review material, see Module 5 lesson on Fans (Operating Principles).

#11

A centrifugal fan is operating with a motor current of 120 amps. The gas density entering the fan during normal operation is 0.045 lb_m/ft³. Estimate the motor current at standard conditions when the gas density is approximately 0.075 lb_m/ft³.

1. 500 amps
   
2. 72 amps
   
3. 159 amps
   
4. 200 amps

Answer: d. 200 amps

Solution:

To review material, see Module 5 lesson on Fans (Performance).

#12

Use the fan laws to determine the new fan speed, brake horsepower, and static pressure when the volumetric flow rate increases from 20,000 to 22,500 ACFM. The current fan operating conditions are provided below.

i. What is the new fan speed?

1. 350 rpm
   
2. 400 rpm
   
3. 450 rpm
   
4. 490 rpm

ii. What is the new brake horsepower?

1. 9.5 BHP
   
2. 10.2 BHP
   
3. 15.1 BHP
   
4. 17.1 BHP

iii. What is the new static pressure?

1. -2.8 in. W.C.
   
2. -3.2 in. W.C.
   
3. -3.9 in. W.C.
   
4. -5.7 in. W.C.

Answer: i. c. 450 rpm

Solution:

Answer: ii. d. 17.1 BHP

Solution:

Answer: iii. b. -3.2 in. W.C.

Solution:

To review material, see Module 5 lessons on Fans (Operating Principles) and Fans (Performance).

#13

How can changes in the fan motor current be used to evaluate system performance?

1. Can indicate shifts in the system characteristic curve.
   
2. Can indicate changes in the gas flow rate.
   
3. Can indicate changes in the gas temperature.
   
4. All of the above.

Answer: d. All of the above.

Changes in the fan motor current can indicate shifts in the system characteristic curve, changes in the gas flow rate, and/or changes in gas temperature. For example, the fan motor current would increase in the following situations:

• When the system resistance decreases (e.g. a decrease in the pressure drop across a fabric filter due to overcleaning the bags)
  
• When the airflow rate increases
  
• When the gas temperature decreases (higher gas density)
  

To review material, see Module 5 lesson on Fans (Performance).

#14

A system consists of the following components (in order): hood, fabric filter, centrifugal fan, and stack. The fabric filter static pressure drop has increased from 4.5 in. W.C. to 6. in. W.C.

i. Assuming no other changes have occurred to the system, what will happen to the hood static pressure?

1. It will decrease (become less negative and closer to zero).
   
2. It will increase (become more negative).
   
3. It will remain unchanged.

ii. What will happen to the hood capture efficiency?

1. It will increase.
   
2. It will decrease.
   
3. It will remain unchanged.

Answer: i. a. It will decrease (become less negative and closer to zero).

The increased pressure drop across the fabric filter indicates that the system resistance has increased (increased blockage of airflow through the fabric filters). The Figure below shows a shift in the system characteristic curve to the left (increased resistance), which is accompanied by a decrease in the airflow rate.

With increased system resistance, less airflow will enter the hood and the hood static pressure will decrease as illustrated in the Figure below.

Answer: ii. b. It will decrease.

A decrease in the hood static pressure below the value recommended for a particular system will result in a lower hood capture efficiency.

To review material, see Module 5 lesson on Fans (Operating Principles) and Hoods (Monitoring Performance).

#15

Suppose a hole has developed in the ductwork between the hood and the fan.

i. What would happen to the fan's baseline operating point and the volumetric flow rate?

1. The fan's operating point would shift to the right and the volumetric flow rate would increase.
   
2. The fan's operating point would shift to the left and the volumetric flow rate would decrease.
   
3. The fan's operating point would shift to the left and the volumetric flow rate would increase.
   
4. There would be no effect on the fan's baseline operating point.

ii. What would happen to the hood static pressure and the hood capture efficiency?

1. The hood static pressure would increase and the hood capture efficiency would decrease.
   
2. The hood static pressure would increase and the hood capture efficiency would increase.
   
3. The hood static pressure would decrease and the hood capture efficiency would decrease.
   
4. The hood static pressure would decrease and the hood capture efficiency would increase.

Answer: i. a. The fan's operating point would shift to the right and the volumetric flow rate would increase.

The industrial ventilation system will draw air from the path of least resistance. The hole in the ductwork will have less resistance than the ductwork, bends, and hoods that are upstream from the hole. Thus a portion of the air will enter the hole and the amount of air entering the hood will be reduced. The decrease in the system static pressure (due to the fact that some air is bypassing the equipment upstream of the hole) will cause the system characteristic curve to shift down and to the right and increase the gas flow rate. The gas flow rate will increase due to the air infiltration.

Answer: ii. c. The hood static pressure would decrease and the hood capture efficiency would decrease.

With a lower airflow rate entering the hood due to the short circuiting effect produced by the hole, the hood static pressure decreases (see Figure 4 Relationship Between Hood Static Pressure and Flow Rate Entering Hood).

To review material, see Module 5 lesson on Fans (Operating Principles) and Hoods (Monitoring Performance).

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