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Math Final Exam Review

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Math Final Exam Review

Math for Water Technology

MTH 082

oC= 5 * (oF – 32)

9

oC= 5 * (188-32)/9

oC= 87

88 oF

oC= 5 * (oF – 32)

9

oC= 5 * (188-32)/9

oC= 87

- Given
- Formula:
- Solve:

- 31 OC
- 87 OC
- 122 OC
- 17 OC

DRAW:

- Given:
- Formula:
- Solve:

- 4295 gallons
- 5716 gallons
- 51670 gallons
- 7282 gallons

D= 16 in or 1.33 ft, L= 550 ft

V= 0.785(diameter2)(length)

V=(0.785)(1.33 ft)2(550 ft)

V= (0.785)(1.77ft2)(550 ft)

V= 764 ft3

V=(764ft3) (7.48 gal/1ft3)

V= 5716 gallons

D=16 in

l=550 ft

- 93%
- 2%
- 0.05%
- 0.75%

- Given
- Formula
- Solve:

10 ounces of polymer (1lb/16 oz) =0.625 lbs

% Strength = Chem lbs * 100

(# gal)(Water lb) + Chem, Lbs

% Strength = 0.625 lbs *100

(10 gal) (8.34 lb/gal) + 0.625 lbs

% Strength = 0.625 lbs * 100%

84.03 lbs Sol

% Strength = 0.75%

- 4 %
- 0.4 %
- 1 %
- 3 %

Solution #1= 20 lbs of 10% Solution #2 = 50 lbs of 1%

% Str of mix = Chem lbs in Sol #1 (% str/100) + Chem lbs in Sol#2 (% str/100) *100

Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix

% Str of mix = 25 lbs (10/100) + 50 lbs (1/100) *100

25 lbs Sol #1 + 50 lbs Sol #2 in Mix

% Str of mix = 3 lbs *100

75 lbs Sol

% Str of mix = 4 %

- 0.02 gpm
- 0.87 gpm
- 1.15 gpm
- 115 gpm

sec= 52 seconds = 52 sec/60sec/min=0.87 min

Flow, gpm= Volume (gal)

Time (min)

Flow, gpm = Vol (gal)

0.87 (min)

Flow, gpm = 1 (gal)

0.87 (min)

Flow, gpm= 1.15 gpm

Given

Formula

Solve:

1.36 gpm

2.46 gpm

3.3000 gpm

4. 155 gpm

#gal=?,Convert 3 hr=180 min+45 min=225 min

Rate=10,350 gallons

GPM(rate)= (gallons)/(minutes)

gpm= (10,350 g) = 46 gpm

(225 min)

- 3.4 min
- 6.34 min
- 12.68 min
- 111.8 min

D= ¾ inch (1ft/12 inch) =0.06 ft; L= 150 ft; Rate=0.5 gpm

A=0.785 (D)(D)(L)(7.48 gal/ft3)

Flow time, min= Pipe Volume (gal) (2)

Flow Rate, gpm

D= 0.75 in/12in/1ft=0.06 ft

Flow= 0.785 (0.06 ft)(0.06 ft)(150 ft)(7.48 gal/ft3)(2)

0.5 gpm

Flow = 12.68 min

Given

Formula

Solve:

- 3.9 gpd/ft3
- 3.9 gpm/ft2
- 0.25 gpm/ft2
- 0.25 gpd/ft3

Given

Formula

Solve:

L= 20 ft, W=25 ft; Rate 1940 gpm

A=L X W

Filtration Rate= (flow gpm)

Area (sq ft)

A= 20 ft X 25 ft = 500 ft2

Filtration Rate= (1940 gpm)

500 (sq ft)

Filtration Rate = 3.9 gpm/ft2

A water is tested and found to have a chlorine demand of 6 mg/L. The desired chlorine residual is 0.2 mg/L. How many lbs of chlorine will be required daily to chlorinate a flow of 8 mgd?

- Given
- Formula:
- Solve:

- 50 lb/d
- 400 lb/d
- 48 lb/d
- 414 lb/d

Demand= 6 mg/L, Residual = 0.2 mg/L, 8 mgd

Dose= Demand + Residual

Feed rate= (dosage)(flow rate)(conversion)

Dose = 6 mg/L + 0.2 mg/L

Dose= 6.2 mg/L

Feed rate= (dosage)(flow rate)(conversion)

FR= (6.2 mg/L)(8 mgd)(8.34 lbs/d)

FR= 414 lb/d

How many lbs of calcium hypochlorite (65% available chlorine) is required to disinfect a well if the casing is 18 inches in diameter and 220 ft long, with water level at 100 ft from the top of the well? The desired dose is 50 mg/L?

- 2 lbs
- 1 lbs
- .02 lbs
- 0.65 lbs

Given

Formula

Solve:

Cl= 65/100 D=18 in=1.5 ft Well 220-100 =120 ft

220 ft - 100 ft = 120 ft water in well

(0.785)(D2)(H) = ft3

(0.785)(1.5 ft)(1.5 ft) (120 ft)(7.48 gal/ft3)= 1585 gal

(50 mg/L)(.001585 MG)(8.34 lb/gal) = 1.01lbs

65/100

What is the motor horsepower (mph) for a pump with the following parameters? Motor Efficiency = 89%Total Head=144 ftPump Efficiency = 78% Flow=1.88 mgd

1.0.0068 mph

2.89 mph

3.68 mph

4. 78 mph

GPM= (1.88mgd)(1,000,000gal/1M)(1day/1,440 min) = 1,306 gpm

89%=0.89 and 78% =0.78

Motor horsepower: (flow,gpm)(total head, ft)

(3,960 )(Motor efficiency)(pump efficiency

Motor horsepower: (1,306 gpm)(144 ft)

(3,960)(0.89)(0.78)

Motor horsepower: 68 mph