Appendix A - Northern Arizona Universityitepsrv1.itep.nau.edu/itep_course_downloads/DAI resources/APTI... · 1.0 - 1/20/2005 A-1 Appendix A Practice Problem Answers and Explanations

1.0 - 1/20/2005 A-1

Appendix A Practice Problem Answers and Explanations

Lesson 2 1. Give the number of significant digits for each of the following:

a. 3.7 = 2

b. 2.06 = 3

c. 17.41 = 4

d. 0.114 = 3

e. 0.00134 = 3

f. 12000.0 = 6

g. 12000 = 2

h. 1200.001 = 7

2. Give the most accurate/precise number for the following calculations:

a. 1.50 + 2.317 = 3.82

b. 1.50 - 2.317 = -0.82

c. 1500 x 3.94 = 5900

d. 1500 ÷ 3.94 = 380

e. 1.500 ÷ 3.94 = 0.381

Appendix A ______________________________________________________________________________________________________________________

A-2 1.0 - 1/20/2005

Lesson 3 Solve the following problems. Carry the answers to a maximum of three decimal places.

1. 2 8 5+ + = 15 14. ( )8 5 2÷ + = 1.143

2. 8 5 2+ + = 15 15. 2 8 5− ÷ = 0.4

3. 2 8 5× × = 80 16. ( )2 8 5− ÷ = -1.2

4. 8 5 2× × = 80 17. 8 5 2− ÷ = 5.5

5. 2 8 5− − = -11 18. ( )8 5 2− ÷ = 1.5

6. 8 5 2− − = 1 19. 2 8 5 8+ ÷ × = 14.8

7. 2 8 5÷ ÷ = 0.05 20. ( )2 8 5 8+ ÷ × = 28.8

8. 8 5 2÷ ÷ = 0.8 21. ( )2 8 5 8− ÷ + = 6.8

9. 2 8 5+ ÷ = 3.6 22. ( )8 5 2× + = 56

10. ( )2 8 5+ ÷ = 2 23. ( )( )2 8 5 8+ ÷ × = 16

11. 8 5 2+ ÷ = 10.5 24. ( )2 8 5 8− ÷ + = 1.385

12. ( )8 5 2+ ÷ = 6.5 25. ( ) ( )2 8 5 8− ÷ + = 0.462

13. 8 5 2÷ + = 3.6

Problems 1 through 8 illustrate the use of the fifth rule of order. The calculations were performed working from left to right.

Algebraic orders 3 and 4 are used in solving problems 9, 11, 13, 14, 15, 17, and 19. First, the multiplication or division was calculated and then the addition and subtraction was completed.

Problems 10, 12, 16, 18 and 20 through 25 require the application of rules 1, 3, 4 and 5.

Lesson 4 1. Answer: c = a - b

a = b + c Solve for c Reverse the equation a b= + + =b c to read c a .

Get b to the right side by subtracting b from both sides. b c b a b c a b/ + − / = − → = −

2. Answer: d abc

= −

a bc

d= + Solve for d.

Switch sides.bc

d = a→ +

Answers and Explanations for Practice Problems _______________________________________________________________________________________________________________________

1.0 - 1/20/2005 A-3

Subtract bc

from both sides. bc

d bc

abc

d abc

→ + − = −

→ = −

3. Answer: b ac cd or b = c(a - d)= −

abc

d= + Solve for b.

Switch sides. bc

+ d = a →

Subtract d from both sides. b

c + d - d= a - d

bc

a d

→

→ = −

( )Multipy both sides by c. bc

c = a - d c

b ac dc

→/

⎛⎝⎜

⎞⎠⎟× / ×

→ = −

4. Answer: c ba d

=−

abc

d= + Solve for c.

Swap sides. bc

d a→ + =

Write the equation as an inverse. cb

1d

1a

→ + =

Subtract 1d

from both sides. cb

1d

1d

1a

1d

cb

1a d

→ + − = −

→ =−

( ) ( )Multiply both sides by b. cb

b 1a d

b c ba d

→// =

−→ =

−

5. Answer: PP VV12 2

1

=

P V P V1 1 2 2= Boyle’s Law: Solve for P1.

Divide both sides by V . P VV

P VV

P P VV1

1 1

1

2 2

11

2 2

1→

//

= → =

Appendix A ______________________________________________________________________________________________________________________

A-4 1.0 - 1/20/2005

6. Answer: VP VP21 1

2

=

PP VV12 2

1= Boyle’s Law: Solve for V2.

( ) ( )Reverse the equation and multiply by V . P VV

V P V

P V P V

12 2

11 1 1

2 2 1 1

→/

/ =

→ =

Divide both sides by P . P VP

P VP

V P VP2

2 2

2

1 1

22

1 1

2→

//

= → =

7. Answer: TV TV11 2

2

=

VT

VT

1

1

2

2

= Charles’ Law: Solve for T1.

Write the inverse of the equation. VT

VT

TV

= TV

1

1

2

2

1

1

2

2→ = →

( ) ( )Multiply by V . TV

V TV

V T T VV1

1

11

2

21 1

2 1

2→

// = → =

8. Answer: TT VV21 2

1=

VT

VT

1

1

2

2

= Charles’ Law: Solve for T2.

Invert the equation and swap sides. VT

VT

TV

= TV

1

1

2

2

2

2

1

1→ = →

Multiply by . TV

V = TV

V T2

22

1

12 2

2V T VV21

1→

// → =

Lesson 5 1. Rewrite the following values using scientific notation:

a. 0.00020589 = 2.06 x 10-4 b. 160007423 = 1.60 x 108

2 Using the base 10, find the logarithm (rounded to the nearest thousandth) for the following:

a. log 9.68 = 0.9858 = 0.986 b. log 1.4 = 0.1461 = 0.146 c. log 135.96 = 2.13341 = 2.133 d. log 0.00143 = -2.844663 = -2.845


1.0 - 1/20/2005 A-5

3. Using the base 10, find the inverse log (antilog) (rounded to the nearest thousandth) for the following:

a. log x = 3.263 → x = 1832.314 or 103.263 = 1832.314 b. log x = -1.002 → x = 0.100 or 10-1.002 = 0.100 c. log x = 0.7792 → x = 6.015 or 100.7792 = 6.015 d. log x = 0.0256 → x = 1.061 or 100.0256 = 1.061

4. Answer: 3.61 x 10-12 ergs For light having a wavelength of 550 nm, find the energy of the photon. The wavelength of light can be expressed in units of nanometers (nm) (10-9 m). The energy of a photon (∆E ) is related to the light wavelength by the expression:

Where: h Planck' s constant erg = frequency of the light wave

=

Where: c = speed of light msec

= wavelength of light

∆E h

c

=

= × −

×

−

υ

υ6 62 10

3 0 10

27

8

. sec

.l

l

∆E h hc

6.62 10 ergm

secm

27= = = × −×

×=−

−νl

sec3 10550 10

8

9

6.623

55010 ergs .0361 10 ergs 3.61 10 ergs27 8 9 10 12× × = × = ×− + + − −0

5. Answer: 24.8% Using the following formula, find the opacity (Op) when the optical density (D) is 0.124. The answer is to be presented as a percentage.

D log 11 Op

=−

0.124 log 11 Op

=−

inv log 0.124 inv log log 11 Op

=−

⎛⎝⎜

⎞⎠⎟

( )1.330 11 Op

1.330 1 Op 1 =−

→ − = →

1.330 -1.330 Op = 1 1.330 Op 1.330 1 → = − →

Op 1.330 -1 0.330 = 0.2481 = 24.8%= =1330 1330. .

Appendix A ______________________________________________________________________________________________________________________

A-6 1.0 - 1/20/2005

Lesson 6 1. Answer: 61 m

Convert 200 ft into meters. There are 0.3048 meters per foot.

200 ft .3048 mft 60.96 m 61 m× = =0

2. Answer: 0.15 in. Hg Convert 2.0 in. of water into inches of mercury (to the closest 0.01 in.).

There are 13.6 in. H2O/in. Hg. 2.0 in. H O

13.6 in. H Oin. Hg

.1471 in. Hg .15 in Hg2

2= =0 0

3. Answer: 357,210 lb/hr Convert 45 kg/sec to pounds per hour.

There are 2.205 pounds per kilogram. 45 kg/sec x 2.205 lb/kg = 99.225 lb/sec x 3600 sec/hr = 357,210 lb/hr

4. Answer: 56,548 ft2

We must know the maximum effective filter surface for the filter bags in a baghouse. The filters are in the shape of cylinders 12 in. in diameter by 30 in. in height, as shown in Figure 6-3. The bottom of the cylinder is closed with a wooden plug and the filtered air exits through the top. There are six compartments in the baghouse and each compartment contains 1200 filter bags. To the nearest square foot, what is the maximum effective filter area in the baghouse?

Figure 6-3.

We are interested in finding the total surface area of the bag filters. The surface of a cylinder (excluding the ends) is found by using the formula a = πdh and applying the area of one filter to the total number of filters in the baghouse. The area of one filter is the surface area of a cylinder and is expressed as: a = πdh


1.0 - 1/20/2005 A-7

The surface area of all of the filters can be expressed: a = πdh x 6 x 1200 = 3.1416 x 12 in. x 30 in. x 6 x 1200 = 8143008.158 in.2 The problem asked for an answer in square feet. 1 ft2 = 144 in.2

Thus: 8143008.158 in.2 /144 = 56548.66776 ft2 = 56548 ft2 We could have converted first and: a = πdh x 6 x 1200 = 3.1416 x 1 ft x 2.5 ft x 6 x 1200 = 56548 ft2 Note that since we are looking for maximum effective area, we cannot round up to 56549.

5. Answer: 10,800 lb Under a baghouse are three pyramid hoppers. The particulate matter being collected has a density of 25 lb/ft3. The hoppers (Figure 6-4) have base dimensions of 6 ft per side and a height of 12 ft. How many pounds of particulate matter can be stored collectively in the three hoppers?

Figure 6-4.

The volume of a pyramid is a 13 b h2=

( )a 3 1 / 3 b h b htotal2 2= =

a 36 ft 12 ft 432 fttotal2 3= × =

Total particulate matter:

432 ft 25 lbft 10800 lb3

3× =

Appendix A ______________________________________________________________________________________________________________________

A-8 1.0 - 1/20/2005

6. Answer: 18 pints Three hoppers and the baghouse must be painted. The baghouse has a surface area of 208 ft2. The dimensions for one panel (side) of a hopper are provided in Figure 6-5. A gallon of the proper paint will cover 300 ft2. The paint may be acquired in 1-pint increments. How many pints of paint must be secured?

Figure 6-5.

Surface to be painted (one side of a hopper):

A 6 ft 12.7 ft2

38.1 ft2=×

=

Surface of all hoppers plus the baghouse:

= × + =38.1 ft 12 208 ft 665.2 ft2 2 2

Coverage per pint = 300 ftgal

8 ptgal

37.5 ftpt

2 2

=

665.2 ft37.5 ft

pt17.74 pt 18 pt

2

2 = =

7. Answer: 10.0 days

This problem refers to the “Intro Problem” in Lesson 6. Each of the three hoppers are independent units and may be emptied only one at a time. Assume that it takes eight hours to open, empty, clean, and return each hopper to use. Assume that the distribution into the three hoppers is equal and that while one hopper is being emptied, shutters in the baghouse evenly distribute the incoming particulate matter into the two remaining hoppers. Particulate matter weighs 2.5 lb/ft3 and is being collected at the rate of 5.2 lb/hr, 24 hr/day. How often must the emptying operation be initiated? Allow a 12.5% safety factor in the computation and provide an answer to the closest one-third of a day.

This problem required a little more problem-solving than math. Although the amounts of particulate matter are important in the computation, time is the primary element of interest.


1.0 - 1/20/2005 A-9

How long does it take to fill the hoppers? There is room for 604.0 ft3 of particulate matter in the hoppers. The particulate matter weighs 2.5 lb/ ft3. The total weight of the particulate matter when the hoppers are full is:

604.0 ft 2.5 lb ft 1510.0 lb3 3× =

The hoppers are filling at the rate of 5.2 lb/hr. Therefore, to fill the hoppers (assuming they were empty) it will take: 1510.0 lb 5.2 lb / hr 290.38 hr=

This part of the problem could also have been completed using the capacity of just one hopper and one-third of the fill amount per hour.

With no safety factor, when must emptying operations begin? It will take eight hours to empty and clean the first hopper, so emptying must begin before the hoppers are full.

8 hr 5.2 lbhr 41.6 lb× =

Emptying must begin while there is still 41.6 lb of capacity remaining, or 20.8 lb of capacity in the two remaining hoppers. Thus, hopper capacity minus eight hours of collection time determines when emptying must begin, in terms of weight. 1510.0 lb 41.6 lb 1468.4 lb− = With the particulate matter collecting at the rate of 5.2 ft3 per hour, dividing the weight (when cleaning must start) by the collection rate will yield the time until emptying must begin.

1468.4 lb5.2 lb

hr282.38 hr=

282.38 hr24 hr 11.7660 days 11 days2

3= =

With no safety factor included, emptying must begin every11.0 days.

Insert a safety factor A safety factor of 12.5% is to be used in the final determination.

282.38 hr 0.125 35.2975 hr282.38 hr - 35.2975 hr = 247.0825 hr247.0825 hr 24 hr =10.295 days

To the nearest 13 day = 10 days

× =

Note that rounding in this case is toward the safety side. Although 10.295 would round to 10.3 and is slightly closer to 10 1

3 days, rounding should be toward the safer side.

Appendix A ______________________________________________________________________________________________________________________

A-10 1.0 - 1/20/2005

Lesson 7 1. Answer: 105.4 ft

A cable must be attached to the outside bottom of a smokestack, go up the outside of the stack, cross the top wall, and extend 3 ft ±1/10 ft down into the inside of the stack. The smokestack is 100 ft tall and the top wall (rim) is 2 ft thick. The inside wall of the smokestack is perpendicular to the ground. The outside wall angle at the base of the stack is 95°. How long must the cable be (to the nearest tenth of a foot)?

Figure 7-14. Figure 7-14a.

The length of the outside wall (c) in Figure 7-14a forms the hypotenuse of a triangle with angle A = 180° - 95° = 85°

sin A ac

c asin A

100 ftsin 85

100.3820 ft= → = = =o

100.3820 ft 2 ft 3 ft 105.3820 ft 105.4 ft+ + = =

2. Answer: 1.27 km A pipeline, constructed with straight sections of pipe must go around an obstruction. Two sections of pipe form legs of 4.675 km and 3.756 km (Figure 7-15). The angle of intersection is 115.85°. How much additional pipe was needed to get from point A to point B using the longer route?

Figure 7-15.


1.0 - 1/20/2005 A-11

It appears that there is no right angle, and since we are given sides a and b and angle C, the Law of Cosines will have to be used.

c a b 2ab cosC2 2 2= + −

c (4.675) (3.756) 2 (4.675) (3.756) cos 115.85 51.27545 km2 2 2 2= + − =o

c 51.27545 km 7.161 km2= =

Now add the distance actually traversed (sides a + b) and subtract the shortest distance (side c).

4.675 km 3.756 km 7.161 km 1.270 km+ − =

3. Answer: 1.263 mi A plume from a 75 ft high smokestack is rising at an angle of 26.55° with respect to the ground (horizontal). Assume the plume is moving horizontally at a rate of 10 mph and that the ascent is at a steady rate. How high, in miles (to the third decimal), will the plume have reached after 15 minutes?

Figure 7-16. Figure 7-16a.

The plume is moving along the ground at a rate of 10 mph. In 15 minutes, the plume will move one-fourth of the 10 miles, or 2.5 miles. Solving for the tangent of A:

( )tan A ab

a b tan A 2.5 mi tan 25.55 1.195 mi= → = = =o

But 1.195 mi is not the correct answer. The plume was already 75 ft above the ground when first observed.

75 ft5280 ft

mi.0142 mi 1.195 mi .014 mi 1.209 mi= → + =0 0

4. Answer: 89.62° A bracket for a filter is set such that when viewed from point A (Figure 7-17), the bracket drops 2.00 in. for each 25 ft of length. What is the angle of decline (or what is the angle at A)?

Appendix A ______________________________________________________________________________________________________________________

A-12 1.0 - 1/20/2005

Figure 7-17.

Using point A, the length of the side opposite and the side adjacent are given. Therefore, the tangent can again be used.

tan A ab

25 ft 12 in.ft

2 in.150= =

×=

A inv tan 150 89.618 89.62= = =o o o

5. Answer: ramp length = 25.35 ft; distance to lip = 25.11 ft In designing a new sampling site, a service door is to be located 2.65 ft above ground level. A steel ramp is to be installed at an angle of 6° of incline. A reinforced vertical lip will be cast in concrete to retain the lower end of the ramp (Figure 7-18). To the nearest one-hundredth of a foot, how long must the ramp be, and how far from the building must the vertical lip be cast?

Figure 7-18.

Ramp length

sin A ac

c asin A

2.65 ftsin 6

25.3519 25.35 ft= → = = = =o

Distance to lip

cos A bc

b c cosA 25.35 ft (cos 6 ) 25.2111 ft 25.11 ft= → = = = =o

A2 in.

25 ft


1.0 - 1/20/2005 A-13

6. Answer: c. Either site A or site B Plans are being made to construct a short stack, 60 ft high, at site A or site B (Figure 7-19). Regulations require the stack to be a minimum of 300 ft away from point C across the river. Will either site meet the requirements and, if so, which one(s)? Select a, b, c, or d below.

a. Site A only b. Site B only c. Either site A or site B d. Neither site

Figure 7-19.

The Law of Sines may be used. Find angle C

Angle C = − − =180 53.87 55.82 70.31o o o o

Distance from B to C (side a)

csin C

asin A

a c sin Asin C

350 ft sin 53.87sin 70.31

300.24 ft= → = = =o

o

Distance from A to C (side b)

bsin B

csin C

c c sin Bsin C

350 ft sin 55.82sin 70.31

307.53 ft= → = = =o

o

Appendix A ______________________________________________________________________________________________________________________

A-14 1.0 - 1/20/2005

Lesson 8 1. For the equation 0 075 4 5. .x y= − , complete the following chart:

x 2 5 10 15 y 4.65 4.875 5.25 5.625

y 4.5 .075x x y 4.50.075

= + =−0

Let y = 4.65 x = 4.65 - 4.50.075

= 2

Let x = 5 ( ) y = 4.5 + 0.075 5 4.875=

Let y = 5.25 x = 5.25 - 4.50.075

10=

Let x = 15 ( ) y = 4.5 + 0.075 15 = 5 625.

2. Identify the intercepts in problem 1:

a. Answer: x intercept = (-60, 0)

y 0 x 0 4.50.075

= -60 = → =−

b. Answer: y intercept = (0, 4.5) ( )x = → =0 4 5 y = 4.5 + 0.075 0 .

3. A graph for a linear equation has the following coordinates: (0.15, 0.5) and (0.45, 3.5).

a. Answer: 10 Find the slope of the line. ∆∆

yx

3.5 .50.45 .15

3.00.30

10=−−

= =00

b. Answer: y 10x 1.0 or x y 110

= − =+

Find the equation.

( ) ( )y y m x x y .5 10 x .151 1− = − → − = −0 0

y= 10x - 1.5 + 0.5 → y = 10x - 1.0 or x = y +110


1.0 - 1/20/2005 A-15

4. Refer to Figure 8-9. Assume the line begins at (0.0) and crosses y = 0.5 exactly two-thirds of the distance between x = 0.4 and x = 0.5.

Figure 8-9.

a. Answer: 0.39 If [NO2]out = 0.35 ppm, what is the value of [NO2]conv in ppm?

( )x = + × = + =0 23 0 1 0 0 066 0.4 . .4 . .466

(Solve for the slope first.)

( ) ( )y y m x x y .0 1.1 x .0 y = 1.1x

1 1− = − → − = − →0 0

( )Let x 0.35y = 1.1 0.35 y 0.385 0.39

=

→ = =

Appendix A ______________________________________________________________________________________________________________________

A-16 1.0 - 1/20/2005

b. Answer: 1.1 What is the slope of the line? ∆∆

yx

0.5000.466

1.073 1.1= = =

5. A wet scrubber is being used to collect particulate matter from an industrial source. The state agency is interested in how the removal efficiency of cadmium compares to the total particulate removal. Data are available for particles ranging in size from 0-5 µm. Using the graph in Figure 8-10, plot the two curves using the data supplied in the following table. Add any necessary labels to the graph.

Removal Efficiency %

Particle Size (diameter in µm)

Cadmium

Total Particles

0.0 75 85

0.5 50 78

0.75 60 84

1.0 80 91

2.5 90 100

3.5 100 100

Figure 8-10a.


1.0 - 1/20/2005 A-17

Lesson 9 1. To meet government emission standards, automobile manufacturers are engaged in research

to improve gas mileage. In order to meet standards, 70% of a manufacturer’s five models, must conform to a 32 mpg average, ±1 SD. The average mileage obtained from six test cars from each of the five models yielded the following results (rounded to the nearest mile):

30 30 32 31 30 27

28 31 33 28 30 31

29 31 27 33 31 30

35 29 32 27 31 34

30 27 26 29 28 30 First, arrange the measurements in order (ascending). 26, 27, 27, 27, 27, 28, 28, 28, 29, 29, 29, 30, 30, 30, 30, 30, 30, 30, 31, 31, 31, 31, 31, 31, 32, 32, 33, 33, 34, 35

a. Answer: 30 Find the median.

There are 30 measurements, the median falls between the 15th and 16th measurements. Both measurements are 30, so the median is 30.

b. Answer: 30 Find the mode.

The most frequently occurring number is 30; therefore, the mode is 30 c. Answer: 30

Find the arithmetic mean.

x xn

26 27 27 27 27 28 ...34 3530

90030

30= =+ + + + + + +

= =Σ

d. Answer: 2.20 Find the arithmetic SD.

( )SD

x x

n 1 = 2.20

2

=−

−

Σ

e. Answer: 70% What percentage of the measurements fall within 1 SD?

21 measurements were between 27.8 and 32.2 → 2130 .70= 0

f. Answer: 97% What percentage of the measurements fall within 2 SDs?

29 measurements were between 25.6 and 34.4 → 2930 .9666 .97= =0 0

g. Answer: Yes Did the manufacturer meet the 32 mpg average within 1 SD?

Appendix A ______________________________________________________________________________________________________________________

A-18 1.0 - 1/20/2005

2. A study of ambient air concentrations of SO2 was conducted at a site. The following concentrations (in ppm) were recorded: 0.16, 0.12, 0.18, 0.13, 0.15, 0.14, 0.18.

a. Answer: 1.5 Find the arithmetic mean.

x xn

0.16 .12 .18 .13 .15 .14 .187

1.067

.15= =+ + + + + +

= =Σ 0 0 0 0 0 0 0

b. Answer: 0.023 Find the standard deviation.

( )SD

x x

n 1 = 0.023

2

=−

−

Σ

Lesson 10 1. An air pollution control system (Figure 10-2) has an inlet pollutant loading of 100 lb/hr. The

emissions are 1.6 lb/hr.

Figure 10-2.

a. Answer: 0.016 What is the penetration through this unit?

Penetration is: P pollutant outpollutant int =

P1.6 lb hr100 lb hr

0.016t = =


1.0 - 1/20/2005 A-19

b. Answer: 98.4% What is the collection efficiency of the unit?

Efficiency is: Eff pollutant in - pollutant outpollutant in

100=⎛⎝⎜

⎞⎠⎟ ×

Eff 100 lb hr 1.6 lb hr100 lb hr

100 98.4%=−⎛

⎝⎜

⎞⎠⎟ =

Since this problem was presented with 100 lb/hr input, conversions are very easy. In Figure 10-2, with 98.4% efficiency, 98.4 lb of pollutant are removed from each 100 pounds of air each hour. Or, conversely, 1.6 lb of pollutant are escaping each hour.

2. An air pollution control system (Figure 10-3) consists of three identical collection devices operating in series as follows:

Figure 10-3.

• Input to the first collection device is at 85 lb/hr.

• Each of the collection devices collects 70% of the entering particulate matter.

• The particulate matter weighs 15 lb/yd3.

• All three collection devices feed into one collection hopper capable of holding 7 tons of particulate matter.

Hint: Output from unit 1 is the input to unit 2, etc. a. Answer: 97.3%

What is the overall collection efficiency of the three collection devices?

Unit 1: 85 lb hr .70 59.5 lb hrOutput from unit 1 = 85 lb hr 59.5 lb hr 25.5lb hr

× =− =

0

Unit 2: 25.5 lb hr .70 17.85 lb hrOutput from unit 2 = 25.5 lb hr 17.85lb hr 7.65 lb hr

× =− =

0

Unit 3: 7.65 lb hr .70 5.355 lb hrOutput from unit 3 = 7.65 lb hr 5.355 lb hr 2.295 lb hr

× =− =

0

Total efficiency:

in outin

100 85 lb hr 2.295 lb hr85 lb hr

100 97.3%−⎛⎝⎜

⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟ =

Appendix A ______________________________________________________________________________________________________________________

A-20 1.0 - 1/20/2005

b. Answer: 0.027 What is the overall penetration through the three collection devices in series? Total penetration:

P outin

2.295 lb hr85 lb hr

.027t = = = 0

c. Answer: 169 hr How many hours does it take to fill the hopper? Particulate matter is collecting at the rate of: 59.5 lb hr 17.85 lb hr 5.355 lb hr 82.705 lb hr+ + =

7 tons 7 2000 lb 14,000 lb= × =

14,000 lb82.705 lb / hr

= 169.276 hr = 169 hr

Note: The fact that the particulate matter weighs 15 pounds per cubic yard may be an interesting fact, but it served no purpose in this problem.

Figure 10-3a.

3. Answer: 6.0 x 10-6

Write the expression “6 parts per million” in scientific notation. 6

1000000.000006 6.0 10 6= = × −0

4. An air quality specification allows a maximum of 3 parts per million (ppm) of pollutant A to be emitted during any 24-hour period. Readings from a monitoring device show that pollutant A is being emitted from the control device at the rate of 115 x 10-9 parts per hour. If the measured rate of emission remains constant, is the installation operating within allowable limits?


1.0 - 1/20/2005 A-21

a. Answer: 92% If yes, what percent of the allowable of A is being emitted?

Convert 115 10hr

9 parts× − to ppm

hr : 115 10hr

0.115 10hr

9 6× = ×− −

Emission for 24 hours = 24 hr 0.115 10hr 2.76 10 or 2.76 ppm

6 6× × = ×− −

2.76 ppm is less than 3.00 ppm, so the installation is within allowable limits. Calculate the percent of allowance: 2.76 ppm

3 ppm.92 92%= =0

b. Answer: Not applicable If no, what is the percentage of excess over the allowable?

5. Answer: 2.79 ppm or 2.79 x 106

In problem 4, the rate of emission was based on EPA standard conditions. What will the rate of emission be if the temperature in the control device rises 5oF ?

What is the new temperature?

EPA standard temperature = = → + =25 C 77 F 77 F 5 F 82 Fo o o o o

oo o

o oC 82 321.8

27.78 273 300.78 K=−

= + =F

Adjust to new temperature

2.76 ppm 300.78 K298 K

2.79 ppm× =o

o

6. A baghouse (Figure 10-4) has a compartment of 16 bags having diameters of 6 in. The bag height is 12 ft. The bags are placed on 8-in. centers. The outermost edges of the bags along the exterior walls are 2 in. from the wall. The gas flow rate into the compartment is 1200 acfm. The bottom of the bags are constructed of wooden plugs and are attached to and hang down from platform A . a. Answer: 382 ft/min

If the bags are all removed, what will the gas velocity be at platform A?

b. Answer: 245 ft/min What is the velocity of the gas stream near the bottom of the bags?

Area at platform A The gas passes through the 16 holes:

( )A 16 r 16 3.14 3 in. 452.16 in.area2 2 2= × = × × =π

Convert in.2 to ft2:

452.16 in.144

3.14 ft2

2=

Appendix A ______________________________________________________________________________________________________________________

A-22 1.0 - 1/20/2005

Velocity at A

V1200 ft

min3.14 ft

382.16 ftmin

382 ftminA

3

2= = =

Figure 10-4a. Using platform A to establish measurements, the platform is 34 in. on each side. There are 16 bags with 6-in. diameters.

Area at bottom B

( ) ( )( )

B 34 in. 34 in. 16 bags from part A

1156 in. 452 in. 703.84 in.area

2 2 2

= × −

= − =.16

Convert in.2 to ft2

703.84 in.144

4.89 ft2

2=

Velocity at B

V1200 ft

min4.89 ft

245.398 ftmin

245 ftminb

3

2= = =

7. Answer: Collection plate area is 70,000 ft2

A coal-fired power plant sends 240,000 acfm through its electrostatic precipitator. The particle migration velocity is known to be 0.35 ft/sec. Calculate the required collection plate area using the Deutsch-Anderson equation with a required efficiency of 99.78%..

The Deutsch-Anderson equation is:

( )ηω

= −−

1 eA

Q Where: η = efficiency / 100

ω = migration velocity (ft / sec) A = plate area (ft2 ) Q = gas flow (acfm)


1.0 - 1/20/2005 A-23

This problem may appear a little overwhelming. However, the secret to handling this type of problem is to simply break the processes down into basic operations. No new rules or methods are inherent in this problem, and the basics simplify to: a. Substituting the known values for symbols b. Using the rules of simplification to handle the exponent c. Recognizing and processing a natural log d. Rewriting the equation to solve for the unknown

Substituting the known values for symbols

0 9978 1

0 35240000 3

.

. secmin= −

−⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

e

A ft ft

Using the rules of simplification to handle the exponent

-0.35 ft ft 3

min sec -0.35 ft

ft 3min

-0.35 ft ft 3sec

minsec sec

A A A240000

604000 4000

× = =⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛⎝⎜

⎞⎠⎟

Now:

0.9978 1 e 0.9978 1 e0.35 ft

4000 ft30.35 ft

4000 ft3A A

= − → − = −− −⎛

⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

Recognizing and processing a natural log

Ln (0.0022) e

6.1193 0.35A ft4000 ft

6.1193 8.75 10 A ft

0.35 ft4000 ft3

A

3

5 2

=

− = −

− = − ×

− ⎛⎝⎜

⎞⎠⎟

−

Rewriting the equation to solve for the unknown

A 6.11938.8 10 ft

0.695375 10 ft 69,538 ft 70,000 ft5 25 2 2 2=

−− ×

= × = =−

Appendix A - Northern Arizona Universityitepsrv1.itep.nau.edu/itep_course_downloads/DAI resources/APTI... · 1.0 - 1/20/2005 A-1 Appendix A Practice Problem Answers and Explanations

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