Math Final Exam Review Math for Water Technology MTH 082

Given Formula: Solve: Given Formula: Solve: Convert 188 o F to o C? 88 o F o C= 5 * ( o F – 32) 9 o C= 5 * (188-32)/9 o C= o F o C= 5 * ( o F – 32) 9 o C= 5 * (188-32)/9 o C= 87 o C= 5 * ( o F – 32) 9 o C= 5 * (188-32)/9 o C= 87 o C= 5 * ( o F – 32) 9 o C= 5 * (188-32)/9 o C= O C 2.87 O C O C 4.17 O C 1.31 O C 2.87 O C O C 4.17 O C

DRAW: Given: Formula: Solve: DRAW: Given: Formula: Solve: A pipe is 16 inch in diameter and 550 ft long. How many gallons does the pipe contain? D= 16 in or 1.33 ft, L= 550 ft V= 0.785(diameter 2 )(length) V=(0.785)(1.33 ft) 2 (550 ft) V= (0.785)(1.77ft 2 )(550 ft) V= 764 ft 3 V=(764ft 3 ) (7.48 gal/1ft 3 ) V= 5716 gallons D= 16 in or 1.33 ft, L= 550 ft V= 0.785(diameter 2 )(length) V=(0.785)(1.33 ft) 2 (550 ft) V= (0.785)(1.77ft 2 )(550 ft) V= 764 ft 3 V=(764ft 3 ) (7.48 gal/1ft 3 ) V= 5716 gallons D=16 in l=550 ft gallons gallons gallons gallons gallons gallons gallons gallons

Given Formula Solve: Given Formula Solve: If a total of 10 ounces of dry polymer are added to 10 gallons of water, what is the percent strength (by weight) of the polymer solution? 10 ounces of polymer (1lb/16 oz) =0.625 lbs % Strength = Chem lbs * 100 (# gal)(Water lb) + Chem, Lbs % Strength = lbs *100 (10 gal) (8.34 lb/gal) lbs % Strength = lbs * 100% lbs Sol % Strength = 0.75% 10 ounces of polymer (1lb/16 oz) =0.625 lbs % Strength = Chem lbs * 100 (# gal)(Water lb) + Chem, Lbs % Strength = lbs *100 (10 gal) (8.34 lb/gal) lbs % Strength = lbs * 100% lbs Sol % Strength = 0.75% 1.93% 2.2% % % 1.93% 2.2% % %

If 25 lbs of a 10% strength solution are mixed with 50 lbs of a 1% solution, what is the percent strength of the new solution? 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) * lbs Sol # lbs Sol #2 in Mix % Str of mix = 3 lbs * lbs Sol % Str of mix = 4 % 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) * lbs Sol # lbs Sol #2 in Mix % Str of mix = 3 lbs * lbs Sol % Str of mix = 4 % 1.4 % % 3.1 % 4.3 % 1.4 % % 3.1 % 4.3 %

The flow from a faucet filled up a 1 gallon container in 52 seconds. What was the gpm flow rate from the faucet? 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 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: Given Formula Solve: gpm gpm gpm gpm gpm gpm gpm gpm

If a pump discharges 10,350 g, in 3 hr and 45 minutes, how many gallons per minute is the pump discharging? #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) #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) 1.36 gpm 2.46 gpm gpm gpm 1.36 gpm 2.46 gpm gpm gpm

How long (minutes) will it take to flush a 150 ft length of ¾ inch diameter service line if the flow through the line is 0.5 gpm? 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/ft 3 ) Flow time, min= Pipe Volume (gal) ( 2 ) Flow Rate, gpm D= 0.75 in/12in/1ft=0.06 ft Flow= (0.06 ft)(0.06 ft)(150 ft)(7.48 gal/ft 3 )( 2 ) 0.5 gpm Flow = 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/ft 3 ) Flow time, min= Pipe Volume (gal) ( 2 ) Flow Rate, gpm D= 0.75 in/12in/1ft=0.06 ft Flow= (0.06 ft)(0.06 ft)(150 ft)(7.48 gal/ft 3 )( 2 ) 0.5 gpm Flow = min Given Formula Solve: Given Formula Solve: min min min min min min min min

A filter 20 ft by 25 ft receives a flow of 1940 gpm. What is the filtration rate in gpm/ft 2 ? 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 ft 2 Filtration Rate= (1940 gpm) 500 (sq ft) Filtration Rate = 3.9 gpm/ft 2 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 ft 2 Filtration Rate= (1940 gpm) 500 (sq ft) Filtration Rate = 3.9 gpm/ft gpd/ft gpm/ft gpm/ft gpd/ft gpd/ft gpm/ft gpm/ft gpd/ft 3 Given Formula Solve: Given Formula Solve:

Given Formula: Solve: Given Formula: Solve: 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? Demand= 6 mg/L, Residual = 0.2 mg/L, 8 mgd Dose= Demand + Residual Feed rate= (dosage)(flow rate)(conversion) Dose = 6 mg/L 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 Demand= 6 mg/L, Residual = 0.2 mg/L, 8 mgd Dose= Demand + Residual Feed rate= (dosage)(flow rate)(conversion) Dose = 6 mg/L 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 1.50 lb/d lb/d 3.48 lb/d lb/d 1.50 lb/d lb/d 3.48 lb/d lb/d

Given Formula Solve: Given Formula Solve: 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? Cl= 65/100 D=18 in=1.5 ft Well =120 ft 220 ft ft = 120 ft water in well (0.785)(D 2 )(H) = ft 3 (0.785)(1.5 ft)(1.5 ft) (120 ft)(7.48 gal/ft 3 )= 1585 gal (50 mg/L)( MG)(8.34 lb/gal) = 1.01lbs 65/100 Cl= 65/100 D=18 in=1.5 ft Well =120 ft 220 ft ft = 120 ft water in well (0.785)(D 2 )(H) = ft 3 (0.785)(1.5 ft)(1.5 ft) (120 ft)(7.48 gal/ft 3 )= 1585 gal (50 mg/L)( MG)(8.34 lb/gal) = 1.01lbs 65/ lbs 2.1 lbs lbs lbs 1.2 lbs 2.1 lbs lbs lbs

What is the motor horsepower (mph) for a pump with the following parameters? Motor Efficiency = 89%Total Head=144 ft Pump Efficiency = 78% Flow=1.88 mgd 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 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 mph 2.89 mph 3.68 mph mph mph 2.89 mph 3.68 mph mph