A = L x w = 18 X 4 = 72 m2

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OIT PRACTICE MATH QUIZ 6 DETAILED CALCULATIONS ANSWERS

1. Calculate the surface area of a rectangular tank 18 m long and 4 m wide.

A = L x W

= 18 x 4

= 72 m²

2. Calculate the surface area of a circular sand filter that has a diameter of 15 m.

A = π R² or A = π D²

= 3.14 x 7.5^{2 4}

= 176.71 m ² = 3.14 x 15²

= 177 m²4

= 176.71 m²

= 177 m²
3. Calculate the volume of a raw water intake crib that is 8 m long, 3 m wide and 6 m

deep.

V = LxWxD

= 8 x 3 x 6

= 144 m³

4. What is the volume of a cylindrical storage tank that is 7 m in diameter and 15 m high?

V = π R² x H V = π D² x H

= 3.14 x 3.52 x 15 4

= 577.26m³ = 3.14 x 72 x 15

= 577 m³ 4

= 2309.07

4
= 577.26 m³

= 577 m³
5. What is the volume of water contained in 84 m of a 10 cm diameter pipe?
a) in m3 - .660 m³ V = π R² x H V = π O2 x H

= 3.14 x .052 x 84 OR 4

b) in L – 660 L = .660 m³ = 3.14 x .102 x 84

.660 m³ x 1000 = 660 L 4

= 2.639

4

= .660 m³ x 1000 =

660 L

if a pump delivers 1.44 m3in 20 min, what is the pumping rate

expressed in:
a) L/s? - 1.2 L/s a) 1.44 x 1000 = 1440 L divide 20 = 72 L/min

72 L = 1.2 L/s

b) m3/d? – 103.68 m³/d 60 sec.

b) 1.44 = .072 m³/min.

20

.072m³/min x 1440 = 103.68 m³

**NOTE PROBLEM 6 HAS SEVERAL SOLUTIONS**

How many m³of water will a 5L/s pump deliver in 5 min?

1000 L = 1 m³

5 L = .005 m³/s

.005 m³ x 60 sec. x 5 min.

1.5 m³/5 min

A 12 m³ storage tank supplies alum for coagulation at a rate of 330 mL/min. How often will the tank need refilling?

12 m³ x 1000 L = 12,000 L

m³
12,000 L = 3636.36 m

.330 mL/min

The prechlorination chamber at a water treatment plant has a volume of 225 m³. if the flow rate out of the tank is 11 L/s, what is the average detention time in hours.

225 m³ x 1000 = 225000 L

225000 L = 20454.545 s

11 L/s 340.91 min

5.68 hrs or 5.7 hrs

5 hr. 60 min. x .7 = 42 min = 5 hr 42 min

How many kg of chlorine are required each day to treat 18,000 m³ with chlorine at 5.0 mg/L?

5.0 mg/L = 5 kg/1000 m³

18000 m³ x 5 kg

1000 m³ 1
90 kg
5mg 18000 m³ 1000L 1kg

L day m³ 10000000 mg

11. A gas chlorinator treats 2,700 m³ with 2 kg of chlorine each day. Calculate the dosage rate. the residual is measured at 0.27 mg/L. What is the chlorine demand?

a) in mg/L - .74 mg/L 2700 x 1000 = 2700000 L

2 kg x 10000000 = 20,000000 mg

b) in kg/d

20 mg = .74 mg/L 2 kg = 1 kg

27 L 2700 m³ X
.74 mg/L – 0.27 mg/L = .74 mg/L 2 kg x = 2700 kg/m³

X = 2700 m³ kg

1m1.L = 1 kg/m³ 2 kg

.74 mg/L = .74 kg/m³ X = 1350 m³

12. A liquid solution with a total mass of 97 kg contains 84 kg of water and the remainder is alum

a) Calculate the percent that is water. – divide water by total multiply by 100
b) Calculate the percent that is alum. – 100 % - % of water = % of alum.
a) 84 x 100 = 8400 = 86.6 % water

97 1 97 100 % - 86.6 % = 13.4 % alum

A mixture of water and powdered carbon is to be 85% water. if the total volume of solution required is 3.6 m³, what is the mass of the powdered carbon?

3.6 m³ = 100%

3.6 m³ x 85% = 3.06 m³ water

3.6 m³ – 3.06 m³ = .54 m³

or

100% - 85% = 15%

3.6 m3 x .15 = .54 m3

14. A hypochlorite solution contains 12% available chlorine. if 3 kg of availible chlorine

are needed to disinfect a main:
a) how many kg of solution are required?
b) How many litres of solution are required
3 kg = 12%
a) 12% = 3 300 divide 12 = 25 kg required

b) 1 kg = 1 L – 25 kg = 25 L