Making the Grade ECI Exams: Math, Math, and More Math

Transcription

1 Making the Grade ECI Exams: Math, Math, and More Math Presented by: Kent O. McIntosh Senior Industrial-waste Inspector L.A. County Sanitation Districts CWEA P3S Training Riverside Tuesday, August 9, 2016

2 Using the Calculator Casio fx-115es PLUS SHIFT Cursor 10 x (-) S D DEL

3 Conversions

4 Conversions

5 Conversions

6 Casio fx-115es PLUS

7 Cursor

8

9

10 Calculations

11 Making the Grade: Calculating a GPA CLASS UNITS GRADE POINTS GRADE POINT Reading 4.0 A 4 16 Writing 3.0 B 3 9 Arithmetic 2.0 C 2 4 Quantum physics 3.0 F 0 0 TOTAL u i u i g i u g 29 i i C+ GPA 2.42 u 12 i

12 YOUR TURN Calculating a GPA UNITS GRADE POINTS GRADE POINT 3 F 0 2 D B TOTAL TOTAL

13 Calculating a GPA UNITS GRADE POINTS GRADE POINT 3 F D B TOTAL 10 TOTAL GPA u g i u i i u 1 g1 + u2g u + u u + u g 3

14 Pretreatment Facility Inspection: A Field Study Training Program Appendix II: Pretreatment Arithmetic Sample Problem 53

15 CWF (40 CFR 403.6)

16 CWF (40 CFR 403.6) F T average total daily flow F D average daily dilution flow

17 WASTESTREAM TYPE CWF (40 CFR 403.6) FLOW (GPD) DAILY MAX ZN LIMIT (MG/L) Regulated 50, Regulated 20, Dilution 30,000 0

18 CWF (40 CFR 403.6) C T , ,000 50, , ,000 30,000 50, ,000 50, , , , , mg / L

19 CWF Compare this answer, 157, , mg / to the GPA we calculated earlier, L GPA u g i u i i u g1 + u2g u + u u + u g 3

20 Mass Loading (sample problem 17 from Grade-3 Study Guide)

21 The Given Solution

22 Mass Loading As a GPA c 1 25 mg/l c 2 0 q 1 40,000 GPD 0.04 MGD q total 45 MGD c c q i q i i mg / But this is with no removal. If 45% is removed, then 55% (0.55) is left: 0.55 x mg/l L

23 YOUR TURN Mass Loading/Local Limit (sample problem 16 from Grade-3 Study Guide)

24 YOUR TURN Mass Loading/Local Limit (sample problem 16 from Grade-3 Study Guide) c mg/l q 1 24 MGD c 710 mg/l c 2 X q MGD c c 1 q 1 + c 2 q 2 q 1 + q 2

25 Mass Loading/Local Limit c mg/l q 1 24 MGD c 710 mg/l c 2 X q MGD c c 1 q 1 + c 2 q 2 q 1 + q X

26 Using the Calculator X SOLVE FOR X

27 Pretreatment Facility Inspection Appendix II: Pretreatment Arithmetic Sample Problem 14

28 Pretreatment Facility Inspection Appendix II: Pretreatment Arithmetic Sample Problem 14

29 THE GIVEN SOLUTION

30 Inconsistent units Maximum POTW BOD 150, mg / L

31 Mass Loading as a GPA Consistent units POTW Capacity 600 mg/l Industry Flow 48,000 GPD MGD Industry BOD 3500 mg/l POTW Flow 30 MGD POTW BOD 450 mg/l

32 Mass Loading as a GPA POTW Capacity 600 mg/l Industry Flow 48,000 GPD MGD Industry BOD 3500 mg/l POTW Flow 30 MGD POTW BOD 450 mg/l? c >

33 Mass Loading as a GPA POTW Capacity 600 mg/l Industry Flow 68,000 GPD MGD Industry BOD 3500 mg/l POTW Flow 30 MGD POTW BOD 450 mg/l c c 457 < 600 The industry will not cause problems.

34 Impact of Toxic Waste on Sewer System Waste-strength Monitoring Pretreatment Facility Inspection Sample Problem 11

35 THE GIVEN SOLUTION

36 THE GIVEN SOLUTION (CONTINUTED)

37 Impact of Toxic Waste on Sewer System Waste-strength Monitoring As a GPA c mg/l q 1 150,000 GPD c mg/l q 2 25,000 GPD Is the expected concentration (the GPA ) equal to 3.2 mg/l?

38 c mg/l Impact of Toxic Waste on Sewer System Waste-strength Monitoring As a GPA q 1 150,000 GPD c mg/l q 2 25,000 GPD Is the expected concentration (c, the GPA ) equal to 3.2 mg/l? TRY THIS

39 c mg/l Impact of Toxic Waste on Sewer System Waste-strength Monitoring As a GPA q 1 150,000 GPD c mg/l q 2 25,000 GPD Is the expected concentration (c, the GPA ) equal to 3.2 mg/l? c c 1 q q c q 2 2 q , , , , Since the actual concentration (3.2 mg/l) is greater than the expected, there does appear to be a leak.

40 Concentration and Volume Grade-1 Sample Question 19

41 Concentration and Volume Grade-1 Sample Question 19 The Given Solution: The final answer is wrong, and the equation is set up wrong. The correct equation and answer are: 25 mm x 0.01 N 100 mm N f N f N

42 Concentration and Volume TRY THIS What volume of a 3.2 N acid will neutralize 1.0 ml of a 7.9 N base?

43 Concentration and Volume What volume of a 3.2 N acid will neutralize 1.0 ml of a 7.9 N base? N 1 V 1 N 2 V x V x 1 V 1 7.9/ ml

44 ph (hydrogen-ion concentration) ph log[ H + ] log 1 [ H + ] poh log[ OH ] log 1 [ OH ] [ H + ] 10 ph [ OH ] 10 poh ph + poh 14 A change of 1 ph (or poh) unit is a 10-fold change in concentration; for example, a ph-4 solution is 10 times more acidic than a ph-5 solution.

45 ph

46 THE GIVEN SOLUTION

47 What was the answer to this question? What was your answer to this question? What volume of a 3.2 N acid will neutralize 1.0 ml of a 7.9 N base?

48 ph problems are often simply concentration-volume problems [ H + ] 10 ph [ OH ] 10 OH 10 (14 ph ) 10 ( ) q 1? q 2 1 gpm

49 ph problems are often simply concentration-volume problems [ H + ] 10 ph [ OH ] 10 OH 10 (14 ph ) 10 ( ) q 1? q 2 1 gpm Since normality is a measure of concentration and since flow volume/time, q v/t N 1 V 1 N 2 V 2 N 1 V 1 /t N 2 V 2 /t c 1 q 1 c 2 q 2

50 c c 1 2 ph problems are often simply concentration-volume problems [ H + ] [ OH ] q 1 X Since normality is a measure of concentration and since flow volume/time, q v/t N 1 V 1 N 2 V 2 N 1 V 1 /t N 2 V 2 /t q 2 1 gpm c 1 q 1 c 2 q X Solve for X

51 Compare the difficulty of the solutions X

52 Making the Grade with ph ± c excess ion c acid v acid v + acid c + base v ( v base base ) ± c x c a v v a a + c v b b v b If the term on the right side of the equation is negative, then the negative ion (OH-) is in excess, and the term on the left should also be negative.

53 Making the Grade with ph b a b b a a x v v v c v c c + ± What if there is no excess ion; that is, the final mixture is neutral (ph 7)? Then c x 0, and b a b b a a v v v c v c + 0 Which simplifies to b b a a v c v 0 c b b a a v c v c which is just our earlier concentration-volume equation

54 Making the Grade with ph ± c x c a v v a a + c v b b v b

55 The Given Solution

56 Making the Grade with ph OUR SOLUTION UNNECESSARY INFORMATION Because the desired ph is acidic (ph < 7), the excess ion concentration is positive. + c x c a v v a a + c v b b v b And if the ph of the caustic is 11.5, the poh

57 Making the Grade with ph OUR SOLUTION UNNECESSARY INFORMATION Because the desired ph is acidic (ph < 7), the excess ion concentration is positive. + c x 2 c v c v From the stoichiometric v a a a + v b b b equation, the acid has twice the concentration of hydrogen ions in a mole of the acid.

58 Making the Grade with ph OUR SOLUTION + c x UNNECESSARY INFORMATION 2 c v a a c v c x mol/l b b c a? X va + vb v a 1000 gal c b mol/l v b 45,000 gal TRY THIS

59 OUR SOLUTION Using the calculator X , ,000 SOLVE FOR X

60 OUR SOLUTION X ,000 45,000 X M c a

61 ph: Visualizing the Answer A caustic tank (ph 10.5) is being emptied at 45 gpm. It s combined with an acid-tank discharge (ph 0.5) so that the resulting combined discharge has a ph of 7.0. What is the acid-tank discharge rate?

62 THE GIVEN SOLUTION

63 ph: Visualizing the Answer How many ph units is the caustic (10.5) above neutral (ph 7)? How many units is the acid (ph 0.5) below neutral? What s the difference between these two?

64 ph: Visualizing the Answer How many ph units is the caustic (10.5) above neutral (ph 7)? 3.5 How many units is the acid (ph 0.5) below neutral? 6.5 What s the difference between these two? 3 Knowing that a difference of one ph unit is a 10-fold difference in concentration, how much stronger is the acid than the caustic? x 10 x So you would need only 1/1000 of the volume (or flow rate) of acid to neutralize the caustic. If the caustic is flowing at 45 gpm, what is the acid flow rate? 1/1000 x gpm

65 THE GIVEN SOLUTION

66 ph: Visualizing the Answer This technique can be used when the ph of the two solutions differ by a whole number and when complete neutralization (to ph 7) occurs. A modification of this technique can also be used even when complete neutralization does not occur.

67 ph: Visualizing the Answer

68 ph: Visualizing the Answer THE GIVEN SOLUTION

69 ph: Visualizing the Answer The acid is 5 ph units below neutral; the alkali is 4 units above. Because the acid is one unit farther from neutral than the alkali, it is 10 times more concentrated ( ) and so only 1/10 as much is needed to neutralize the alkali: 1/10 x gal. After neutralization, there will be gal of ph-2 acid in a volume of gal.

70 ph: Visualizing the Answer The acid is 5 ph units below neutral; the alkali is 4 units above. Because the acid is one unit farther from neutral than the alkali, it is 10 times more concentrated ( ) and so only 1/10 as much is needed to neutralize the alkali: 1/10 x gal. After neutralization, there will be gal of ph-2 acid in a volume of gal. The concentration is simply: c mol / 800 L And the ph is log( ) 2.5

71 MAKING THE GRADE This GPA (weighted averages) method has been used for Combined Wastestream Formula Plant loading Surveillance monitoring ph (neutralization) problems The solutions to some simple ph problems can be visualized without making any calculations An inexpensive, approved calculator can make the math even easier

72 MAKING THE GRADE By making the grade (calculating weighted averages), you will get a high grade (at least on the math problems) when you take your appropriate Grade-level certification test. GOOD LUCK!