Water Supply Part 5

The flowing back of used, contaminated, or polluted water from a plumbing fixture or vessel into a water supply pipe due to a pressure less than atmospheric in such pipe

• backflow

• back pressure

• back pressure back flow

• backflow preventer

• back siphonage

Answer explanation

back siphonage

backflow - the flow of water or other liquids, mixtures, or substances into the distributing pipes of a potable supply of water from any sources other than its intended source

backpressure backflow - backflow due to an increased pressure above the supply pressure, which may be due to pumps, boilers, gravity, or other sources of pressure

backpressure - pressure above the supply pressure

backflow preventer - a device or means to prevent backflow into the potable water system.

A liquid having a specific gravity of 2.0 is flowing in a 50 mm diameter pipe. The total head at a given point was found to be 17.5 J/N. The elevation of the pipe above the datum is 3meters and the pressure in the pipe is 65.6 kPa. Compute the velocity of flow

• 13.37 hp

• 14.79 m/s

• 11.16 m/s

• 17.37 m/s

• 11.16 m

Answer explanation

Total Energy = (v^2/2g)+(P/unit weight) + elevation z

E = 17.5 J/N = 17.5 m

(J/N = N-m/N)

17.5 m = (v^2/2g) +( 65.6 kPa/(9.81 kN/m3 x 2) + 3.0

v^2/2g = 11.156 m

v= 14.79 m/s

Determine then force acting on a 2m x 4m gate shown

• 733.16 kN

• 91.65 kN

• 733. 16 KPa

• 91.65 kPa

• 477.2 kN

Answer explanation

Pcg = 32 kPa +(0.8x9.81x1) +(9.81x1.5)+(1.26x9.81x3) =91.645 kPa

F =PxArea

F = 91.645 kN/m2 x (2x4) = 733.16 kN

Determine the magnitude of force on a inclined gate 1.5m x 0.5 m shown. The tank of water is completely closed and the pressure gage at the bottom reads 90,000 N/m2. Use 9.8kN/m3 in water.

• 64.03 kN

• 60.43 kPa

• 48.02 kN

• 48022.5 kN

• 64.03 kPa

Answer explanation

2.Pgage = 90000 Pa

Force will act at the center of gravity

Find the center of the plate

1.5/2 = 0.75

height = 0.75 sin60 = 0.65 m

height from the bottom = 2+0.65 m =2.65 m

P= unit weight x h

we need to find the Pressure at the center of the plate (Pc), thus, P2-P1 = unit weight x height

90000 - Pc = 9800 x 2.65

Pc = 64030 Pa

Solve for the Force,

F = PA

F= 64030 N/m2 x (1.5 x 0.5)

F = 48022.50 N= 48.02 kN

Flowing mostly north for 500 kilometers (300 miles), the Cagayan has wide channels and flood plains, with sizable sections deforested and covered by farmland. The main stem receives water from many streams and smaller rivers. In 2019, the flood-prone river valley was devastated by what was then called a “100-year flood.” What is the probability that the 100-year flood will

happen again within the next 15 years?

• 10%

• 11%

• 12%

• 13%

• 14%

Answer explanation

P(%) = 1-(1-(1/T))^n x 100%

P(%) = 1-(1-(1/100))^15 x 100%

P(%) = 14%

Determine the chlorinator setting (lbs/day) needed to treat a flow of 3 MGD with a chlorine dose of 4 mg/L

• 100 lbs/day

• 150 lbs/day

• 12 lbs/day

• 1.33 lbs/day

• 0.75 lbs/day

Answer explanation

(mg/L Cl2) (MGD flow) (8.34 lbs/gal) = lbs/day Cl

Thus,

(4 mg/L Cl2) (3 MGD flow) (8.34 lbs/gal) = 100 lbs/day Cl2

As per the Chapter VIII of PD 856, what is the minimum water level of swimming pool except for special purposes pool or for restricted areas in general swimming which are set aside primarily for the use of children

• 2 ft

• 3ft

• 4ft

• 5ft

• 1ft

Answer explanation

3 ft (91 cm)

An orifice has a coefficient of discharge of 0.62 and coefficient of contraction of 0.63. What is the coefficient of velocity of the orifice

• 0.984

• 0.391

• 1.016

• 1.25

• 0.01

Answer explanation

Coefficient of discharge (C) = Coefficient of contraction (Cc) and coefficient of velocity (Cv)

C = Cc x Cv

0.62 = 0.63 x Cv

Cv = 0.984

A cube of wood (sg=0.6) has 9-in. sides. Compute the magnitude of the force required to hold the wood completely submerged in water

• 26.33 lbs

• 15.80 lbs

• 23.63 lbs

• 10.53 lbs

• 15.03 lbs

Answer explanation

Force = Buoyant Force - Weight of wood

Weight of Wood = (62.4 x 0.6x (9/12)^3) = 15.80 lbs

Since completely submerged

Buoyant Force = (62.4x (9/12)^3) = 26.33 lbs

Force =26.33-15.80 = 10.53 lbs

In the syringe shown, the drug had a density of 900 kg/m3 and an absolute viscosity of 0.002 Pa-s. The flow through the needle is 0.4 mL/s. Neglect headloss in the larger cylinder. Determine the velocity at point B and the Reynold's number of the flow

• 8.15 m/s ; 916.88

• 8.15 m/s ; 961.88

• 8.15 m/s ; 968.88

• 8.51 m/s ; 916.88

• 8.51 m/s ; 961.88

Answer explanation

a. Q= AV

V =Q/A

V = 0.4 x 10^-6 / (pi/4 x 0.00025^2) = 8.15 m/s

b. Re = (velocity x diameter x density)/ absolute viscosity

Re = (8.15x 0.00025x 900)/(0.002) = 916.88