# Hydraulics Engineering Essay Example

1. Calculating coordinates for minimum and maximum pipeline system characteristics.

It is important to note the following equation:

1. Calculating the domestic Mdd.

Domestic mdd = Area of the graph/ 6

= {(1×200) + (1×800) + (2×1200) + (1×1000) + (1×400)}/6

l/h/d =

= 800 l/h/d

1. Maximum day pumping rate.

Mdd = 800 x 9000

Total capacity provided = 7.2 + 1.5 (industrial demand)

Applying energy equation between PS and D

For Hazen William constant c = 130, frictional head loss = 5m/km

20 + H = 168 + (2×5) + 0.3

H = 158.3m

1. Maximum day pumping rate.

Storage tank at D should be 20% the required total of 7.2 + 1.5 Ml

Maximum day pumping rate = 8.7/ (24x60x60) = 100.7 m/s

20% of 8.7 = 1.74 Ml

A fire storage allowance of 100.7l/s x 4 x 60 x60 = 1.45 Ml

A breakdown allowance of 50% = 50% of 8.7= 4.35 Ml

Allowed storage = 1.74+1.45+4.35 = 7.54 Ml

1. A has population of 4800 people;

Required water = 800×4800 l = 3.84Ml an equivalent of 44.4 l/s or 0.044m3/s

Required velocity is 0.4/s or thereabouts for diameter 375 mm

Proven by: 0.044 m3/s / 0.11m2 = 0.4m/s where 0.11 is the cross sectional area of the pipe. Hence the pipe is sufficient for delivery.

C has a population of 1200 people

Required water = 800×1200 l = 0.96Ml equivalent to 11.11 l/s or 0.011 m3/s

Required velocity for delivery is 0.3 for diameter 225mm

0.011 m3/s / 0.04m2 = 0.28

0.28 is close to 0/3 considering the pipe losses, hence it can deliver accordingly.

1. Considering that the piping within B is negligible in comparison to other areas, then, pressure shall be sufficient.

2. The current service is established as sufficient.

3. The maximum day pumping rate = 100.7 m/s

Considering the fraction pumped during time range 4am to 8pm;

= 4200/ 4800

= 0.875 of total water delivered.

If the pumping is done only from 4am to 8pm, then it means that no pumping shall be required for 8.00pm to 4.00am.

1. Future population distribution projections

Current total population = 9000 people.

Total after 20 years

=

= 13,374 people.

Area C = 1200 + 3000 = 4200 people

= 5486 peopleArea A =

=3686 peopleArea B =

 Area Code Population 123.7 l/s

Required Mdd = (13,374 x 800) + (Industrial demand x 1.5)

= 10.7 + 2.25 Ml

= 12.95 Ml

= 150 l/sRequired Mhd =

1. The acceptable domestic consumption water colour level is usually 20 – 25 HU. The fact that the river water reaches a maximum of 30-50 Hazen units is an automatic call for building a treatment plant before pumping. This especially applies when it is raining as the turbidity rises although not beyond the acceptable levels of 6.5 – 8.5 NTU. Another strong point as to why the people in these areas should be provided with a purification plant is due to the fact that deadly bacteria may infiltrate into the pipeline.

2. Water filtration methods: direct filtration and conventional clarification

• The direct filtration method applies hydraulic principles and mechanical coagulation in order to filter water rapidly to acceptable standards. This filtration method is suited for surface water with a turbidity of 10 – 20 NTU. On the other side, the conventional clarification or treatment plant utilizes a combination of methods to produce acceptable water standards including; coagulation, sedimentation, filtration and disinfection.

• The direct filtration method is advantageous in that it filters water to levels below 0.1NTU and is readily customizable. It is easy to operate with a low cost of maintenance and intervention. This system is also cheap when it comes to chemical usage as treatment comes later in the stages. This plant is however time consuming and therefore meant for low volume production.

• The conventional clarification or treatment plant is a high volume production plant that uses chemicals right from the initial processes. Water produced is highly purified with low levels of bacteriology therefore guaranteeing safety. This plant is however expensive to manage especially in terms of maintenance and chemical requirements.

• The dissolved air floatation method is utilized in the screening of floating objects and dispersed liquid air through use of saturated air-water mixture. This should be incorporated as the initial stage of clarification before the flocculation system especially in the conventional method in order to maximize the benefits attached to it. The main advantage is that this system is that it has a removal efficiency of above 95% hence reduces the cost of operation that applies in buying the coagulants and flocculants.

1. Design criteria

1. Modifications to be made to the pump in order to achieve as per the population after 20 years

1. In order to supply area A, B and C, Required Mhd =
= 150 l/s

Therefore one pump must pump at 75 l/s

At 75 l/s the one pump can effectively work on a head of 55m

Therefore this pumps require a booster pump that is capable of pumping from a head of 158.3m – (55×2) = 48.13 m between PS and D. This could also mean introduction of another pump worth the same power as the initial ones.

1. Maximum booster pump operational head = HD MAX – HC MAX

= 175 – 151 = 24m

Mdd = 800 l/h/d

Population = 4200 people

Amount of water required = 4200 x 800 = 3.36Ml

For 16 Hrs.

Flow rate required =
= 58.3 l/s

For 24 hours

Flow rate required =
= 38.9 l/s

1. _ Consider Option iv

2. Storage tank at D should be 20% the required total of 12.95 Ml

20% of 12.95 = 2.59Ml to allow for 24 hr. pumping,

And 24/16 x 2.59 = 3.9 Ml to allow for 16 hr. pumping

A fire storage allowance of 150l/s x 60 x60 = 0.54Ml

A breakdown allowance of 50% = 50% of 12.95 = 6.48 Ml

For 24 hr. Pumping the allowed storage = 2.59+0.54+6.48 = 9.61Ml

For 16 hr. Pumping the allowed storage = 3.9+0.54+6.48 = 10.92 Ml

Or a combination of the above at point B and D