Let’s finger it out together
John’s Math Wizard
1
Converting length (L) from inches to “fractions-of-a-foot”
Divide the number of inches by 12 = “fractions-of-a-foot”
!
Buy an “Engineer’s” ruler. It automatically converts for you. We will practice later!!
John’s Math Wizard
2
How do you convert temperatures from: • oF to oC That is degrees Fahrenheit to degrees Celsius • oC to oF That is degrees Celsius to degrees Fahrenheit
oC
= 5/9(oF – 32)
oF
John’s Math Wizard
= 9/5(oC) + 32
3
CONVERTING TEMPERATURES How to do this without a calculator (quickly). • oF to oC From F subtract 30 then divide in half = C • oC to oF Double C then add 30 = F Convert 50oF to equivalent oC? Subtract 30 from 50 = 20, then take ½ of 20 = 10 Answer is 50oF = 10oC Convert 10oC to equivalent oF? Double 10 = 20, then add 30 = 50 Answer is 10oC = 50oF
Why does this work? First, it is an approximation. Second, because wastewater temperatures on ABC Exams are between 38 and 80oF, this method will get you an answer FAST.
We will practice later!! John’s Math Wizard
4
Find the volume of a “shoe box” in gallons, knowing the dimensions for length (L), width (W), and height (H) all in terms of feet (ft). John’s Way
Real Way V,gal = 7.48xLxWxH
V,gal = 7½xLxWxH
W
7½
H
L
A “shoe box” could be a rectangular clarifier, a wet well, activated sludge basin, and has many other wastewater applications. We will practice later!! John’s Math Wizard
5
Find the volume of a “tuna can” in gallons, knowing the dimensions for diameter (D), and height (H) or length (L) all in terms of feet (ft). Real Way 2
V,gal = (7.48)(π/4)xD xL
D John’s Way V,gal = 6xDxDxL D
6
H
6
D
D
L
A “tuna can” could be a circular clarifier, a cylindrical wet well, a pipe (pipeline), and has many other wastewater applications. We will practice later!! John’s Math Wizard
6
Find the MASS of material contained in a basin (in TOTAL POUNDS or lbs) The “Davidson Pie Chart” has been used in water and wastewater math for many years. It is very useful in that you can visualize the parts required to find the MASS of material contained in a basin and express that mass in POUNDS (lbs). You need to know these facts: Concentration of the material in terms of mg/L. The fact that water weighs 8.34 pounds for each gallon in the basin. The Volume (V) of the basin expressed in terms of Million Gallons (MG).
This requires that you can find the volume (V) of a basin and express that volume in terms of Million Gallons (MG)
We will practice later!! John’s Math Wizard
7
Find the MASS of material flowing through a basin (in TOTAL POUNDS/DAY or ppd) You see here again that we will use the “Davidson Pie Chart”. It is very useful in that you can visualize the parts required to find the MASS of material flowing through a basin (aka MASS per UNIT TIME) and express that in terms of POUNDS (lbs) per DAY (d) or ppd. You need to know these facts: Concentration of the material entering the basin in terms of mg/L. The fact that water weighs 8.34 pounds for each gallon entering the basin. The flowrate (Q) of the water entering the basin expressed in terms of Million Gallons (MGD).
This requires that you can find the flowrate (Q) through a basin and express that flowrate in terms of Million Gallons per Day (MGD)
We will practice later!! John’s Math Wizard
8
“Practice isn’t the thing you do once you’re good. It’s the thing you do that makes you good.” Malcolm Gladwell Outliers: The Story of Success John’s Math Wizard
9
Convert inches to “fractions-of-a-foot”. Remember: Just divide the inches by 12
6” = ____ 2’ 10” = ____ ¾” = ____ 11’ 2¼” = ____ 134¼” = ____
6/12 = ½ ft or 0.5000 ft 2 + 10/12 = 2 ft + 10/12 ft or 2.8333 ft ¾/12 = 0.75/12 ft = 0.0625 ft 11 + 2¼/12 = 11 ft + 2.25/12 ft = 11.1875 ft 134¼/12 = 134.25/12 ft = 11.1875 ft
3” = ____
3” = 0.250 ft
John’s Math Wizard
10
oC
Convert
to
oF,
and Convert oF to oC.
• oC to oF Double C then add 30 = F 4oC
= ____ 38
oF
Actual Answers 39.2
8oC = ____ 46 oF
46.4
15oC = ____ 62 oF
59.0
20oC = ____ 72 oF
68.0
12oC = ____ 56 oF
53.6
• oF to oC From F subtract 30 then divide in half = C 36oF
= ____ 3
oC
Actual Answers 2.2
46oF = ____ 7 oC 60oF = ____ 15 oC
7.8 15.6
82oF = ____ 26 oC
25.6 John’s Math Wizard
11
Volume of a “Box” in gallons. • Volume, in gallons = 7.5 x L x W x H John’s “Shoe Box” Volume Calculation
Actual “Shoe Box” Volume Calculation
W
74 800
7.5 x 50 x 20 x 10 = 75 000 gal 7.5 x 80 x 30 x 15 = 270 000 gal
269280
7.5 x 100 x 50 x 18 = 675 000 gal
673 200
7.5 x 150 x 75 x 20 = 1 687 500 gal
1 683 000
H L BOX DIMENSIONS ARE GIVEN Length Width Height 50 20 10 80 30 15 100 150
John’s Math Wizard
50 75
18 20 12
Volume of a “Can” in gallons. • Volume, in gallons = 6 x D x D x H Actual “Tuna Can” Volume Calculation
John’s “Tuna Can” Volume Calculation V =6x Dx Dx H 6 x 50 x 50 x 10
=
150 000 gal
146 870
6 x 80 x 80 x 15
=
576 000 gal
563 980
6 x 100 x 100 x 18 = 1 057 000gal
1 057 463
6 x 150 x 150 x 20 = 2 700 000 gal
2 643 656
BOX DIMENSIONS ARE GIVEN
V,gal = (7.48)(π/4)xD2xH D D
6 Diameter 50 80 100 150
John’s Math Wizard
H
Height 10 15 18 20 13
Volume of a “Pipe” in gallons. • Volume, in gallons = 6 x D x D x H Actual “Tuna Pipe” Volume Calculation
John’s “Tuna Pipe” Volume Calculation V =6x Dx Dx H 6 x ½ x ½ x 1 000 = 1 500 gal
1 469
6x 1 x1
x
500 = 3 000 gal
2 937
6 x 1½ x 1½ x
150 = 2 025 gal
1 983
6 x 2 x 2 x 2 000 = 48 000 gal
46 998
V,gal = (7.48)(π/4)xD2xH
L
6
D
Diameter ½ 1 1½ 2
John’s Math Wizard
Length 1 000 500 150 2 000 14
Find the MASS of material contained in a basin (in TOTAL POUNDS or lbs) First, you must determine what kind of basin you have? Usually rectangular or cylindrical. Shoe Box
Tuna Can
X
X
Second, you find the Volume (V) of wastewater (in MG) found in the basin.
Third, establish the concentration (C) of material that you are looking to quantify (in mg/L) found entering or within the basin. Usually, given in the problem.
This requires that you can find the volume (V) of a basin and express that volume in terms of Million Gallons (MG)
Lastly, know that the unit weight of water is always 8.34 lbs/gallon; never changes. John’s Math Wizard
15
Find the MASS of material contained in a basin (in TOTAL POUNDS or lbs) You are the operator of an conventional activated sludge facility containing one rectangular aeration basin followed by a single circular clarifier. The average MLTSS in each basin was found to be 2 400 mg/L. You are to find the TOTAL POUNDS of MLTSS. • The one aeration basin measures 45 feet long by 30 feet wide with wastewater 12 feet deep. • The single circular clarifier measures 28 feet in diameter with a wastewater depth of 12 feet.
X
X
This requires that you can find the volume (V) of a basin and express that volume in terms of Million Gallons (MG)
Remember: We want Volumes in Million Gallons (MG), not gallons (gal). To do this we find the volume of each basin in gallons, then divide by 1 million (1 000 000) to obtain the desired MG. John’s Math Wizard
16
Find the MASS of material contained in a basin (in TOTAL POUNDS or lbs) Determine what kind of basin you have? Then calculate the basin’s volume. Examples below. Rectangular Shoe Box L = 45’ W = 30’ H = 12’
V, gal = (7½) x L x W x H
X
X
V, gal = (7 ½) x 45 x 30 x 12 V, gal = 121 500 V, MG = 0.122
Cylindrical
V, gal = (6) x D x D x H
Tuna Can D = 28’ H = 12’
V, gal = (6) x 28 x 28 x 12
This requires that you can find the volume (V) of a basin and express that volume in terms of Million Gallons (MG)
V, gal = 56 448 V, MG = 0.056 gal
MG, you divide by 1 000 000
John’s Math Wizard
17
Find the TOTAL MASS of material contained in both basins (in TOTAL POUNDS or lbs) Rectangular Aeration Basin (AB) C,MLTSS = 2 400 mg/L V, Basin = 0.122 MG
Cylindrical Secondary Clarifier (SC) C, MLTSS = 2 400 mg/L V, Clarifier = 0.056 MG
X
X
Aerator lbs = C, MLTSS, mg/L x 8.34 x V, AB, MG lbs = (2 400 mg/L)(8.34)(0.122) = 2 442 Clarifier lbs = C, MLTSS, mg/L x 8.34 x V, SC, MG
lbs = (2 400 mg/L)(8.34)(0.056) = 1 121 TOTAL POUNDS = 2 442 + 1 121 =
3 563
John’s Math Wizard
18
Find the MASS of material flowing through a basin (in TOTAL POUNDS/DAY or ppd) First, you must determine the flowrate (Q) entering the basin. Sometimes this is as simple as reading the treatment plant flowmeter. Bubbler Tube Assembly
2.73 Flowmeter, reading MGD
Open Channel Flume
John’s Math Wizard
This requires that you can find the flowrate (Q) through a basin and express that flowrate in terms of Million Gallons per Day (MGD)
19
Find the MASS of material (TSS) flowing through a basin (in TOTAL POUNDS/DAY or ppd) Second, establish the concentration (C) of material that you are looking to quantify (in mg/L) found entering or within the basin. Usually, given in the problem
Lastly, know that the unit weight of water is always 8.34 lbs/gallon; never changes.
Now for the Problem You are the operator of an conventional activated sludge facility containing one rectangular aeration basin followed by a single circular clarifier. The average daily flowrate into the treatment plant is 2.73 MGD. You have sampled and tested the wastewater for total suspended solids (TSS) and found the concentration to be 320 mg/L. You are to find the TOTAL POUNDS of TSS entering the treatment plant for the day (ppd of TSS).
John’s Math Wizard
20
Find the MASS of material (TSS) flowing through a basin (in TOTAL POUNDS/DAY or ppd) First, recall that the concentration of TSS in the treatment plant influent was found to be 320 mg/L. Next, recall that the unit weight of water is always 8.34 8.34 lbs/gal. Lastly, recall that the plant flowmeter read and totalized 2.73 MGD for the day.
John’s Math Wizard
21
Find the MASS of material (TSS) flowing through a basin (in TOTAL POUNDS/DAY or ppd)
LET’S DO THE MATH Recall the Pounds Formula (The Davidson Pie Chart)
320
2.73 8.34
Recall TSS Concentration = 320 mg/L Recall the unit weight of water = 8.34 lbs/gal
Recall the plant flowrate = 2.73 MGD
CALCULATE TSS, ppd = 320 x 8.34 x 2.73 = 7 John’s Math Wizard
286 22
John’s Math Wizard
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