Course - HND in Construction and the Built Environment

Unit 43 Hydraulics

Assignment - Hydraulics for Civil Engineering

LO1: Apply concepts of physics to develop solutions for hydrostatic and hydrodynamic problems

Question a) What affect will temperature have on the following properties of water? Density, viscosity and surface tension. Explain how can this affect the flow of water through a pipe and in an open channel such as a river.

Density of water gradually decreases with rise in temperature. This can be described with the support of the equation, p=m/V. Here, it can be seen that the volume of water (a liquid) increases gradually with the rise in temperature. On the other hand, the volume is inversely proportional to the density of water. Thus, the density increases with the rise in water volume.

Furthermore, the viscositydecreases with increase in pressure (Roy and Bhalla, 2017, p.104). This is because viscosity is the resistance exerted by water or any other liquid. This results to the fact that molecular gap increases with the increase in temperature. This leads to a fall of resistance. However, with the increase in the temperature, there is decrease of surface tension.

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b) Explain what the boundary layer is in fluid flow and show how the properties mentioned in part a) above influence this boundary layer and the distribution of velocity in a pipe or in a stream

The fluid's boundary layer is considered as the slim layer that is presentat its surface and where the flow of the fluid gets into contact with the fluid's surface. Here, under normal circumstances the flow would be considered laminar. The properties of the fluid like the surface tension, viscosity, and density vary as per the nature of the flow. There will be less density in the layer at the boundary having molecules that are less dense. The surface tension will be towards the surface along with a lower amount of viscosity. This is because it will be laminar boundary flow, which will be smooth, thereby having lesser resistance to it.

c) Explain the difference between laminar and turbulent flow and show how the Reynolds Number can be used to predict the type of flow.

In turbulent flow the adjacent layers to one another are mixed, whereas in laminar flow these layers adjacent to one another does not mix. There is straight fluid flow in laminar but zig zag in turbulent. Furthermore, the former is less than 2000 in terms of Reynolds number, whereas, a turbulent flow has a Reynolds number greater than 4000.

F) Currently the storm water/sewerage system is used. Briefly discuss

i) the disadvantages to the local population that arise from using the sewerage system to dispose of storm water

(i) When a sewerage system is used to dispose of storm water, the entire system is known as a combined sewer system. While using a combined sewer system numerous challenges and issues can arise (Movasat and Tomac, 2021, p.602). In the given system both the stormwater and sewage are moved to the sewage treatment plant in order to be finally disposed of. The most common sets of disadvantages to the local population from using the sewerage system to dispose of storm water are overflow of sewage water to the surroundings, rise of water borne diseases, the sewer structure itself may get damaged due to the heavy stormwater flow, an increased load on the water treatment plant, the stormwater cannot be reused if get mixed with sewage. Also, the system becomes highly uneconomical when the need for pumping arises.The flow and pressure exerted by the stormwater here is highly prone to damage the wall of the combined sewer system. A large part of this is due to the fact that the majority of a combined sewer system is enclosed from all sides, which makes it difficult for the stormwater to be contained in the enclosed space. In order to avoid this issue, open channel drains are used for disposing stormwater, which has been discussed elaborately in the next section.

ii) to discuss whether the storm water should be disposed of through pipes or an open channel .

(ii) The storm water should be disposed of through an open channel. This is the most commonly practiced way to dispose of rainwater in densely built-up or urban areas. The primary reason for this is closed pipes might not be able to withstand the force from the discharge of rainwater, which might damage these pipes in the long run, thereby requiring frequent maintenance (Zhanget al. 2019, p.501).Open channel systems can be described as vegetated swales that are developed in order to capture stormwater runoff along with treating and releasing the capture stormwater. Here, treatment via wet and dry cells occur via the installation of berms or check dams.Two primary types of an open channel system are as follows: dry swales and wet swales.

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LO2: Calculate forces related to fluids at rest and in motion

a) i) A reservoir is situated at a height of 50m above a town. The reservoir is connected to the town through a 1m diameter pipe. If the friction factor in the pipe is 0.008 and the distance the length of the pipe is 1000m and the normal delivery is 15m3s-1 calculate the pressure at the end of the pipe as it goes into the town. (Hint, think of the head loss )

Given Data,
The reservoir is situated at H = 50 m above the town
The diameter of the pipe = 1 m
Friction factor (f) = 0.008
Length of the pipe = 1000 m
Discharge (Q) = 15 m3 / sec
Velocity (v) = Q / h = 19.09859 m/sec
Head loss due to friction (hj) = Hj = flv^2 / 2gd = (0.008 * 1000 * 19.09809^2) / 2 * 9.81 * 1
Hj = 148.728343 m
Using Bernoulli's equation,
(P1/eg) + (V1^2/2j) + 2j = (P2/eg) + (V2^2/2j) + (22) + hjHere, (P1 = 0, V1 = 0)
0 + 0 + 50 = P2/eg + {(19.09859) ^2 / 2 * 9.81} + 0 + 148.72
(P2/eg) = 117.3193859 * 9810
Pressure @ end pipe, P2 = 1150.903175 KPa

ii) At that point in the pipe there is a valve. If the valve is turned off what is the pressure on the valve?

For the sake of doing the equation, the value that is closer has been considered as gradual or rapid as there is not mentioned in the question.
Here, rapid closure has been assumed.
Water hammer pressure (Ph) = PCV
Here, C = √ (K / P)
Where, K = bulk modulus of fluid medium
P = density of fluid
As the given liquid has been stated to be water,
C = √ {(2 * 10^6)/1000} = 1414.2 m/sec
Hence, Ph = 1000 * 1414.2 * 19.1 = 27009.31 KPa

b) Using the Manning Equation, explain how the hydraulic radius affects the rate at which water can flow through a rectangular channel by comparing the amount of water that would flow through a channel with a slope of 0.002, n value 0.008 if the channel were 1m wide and 4m deep to a channel 4m deep and 1 m wide.

As per Manning's equation,
v = (1/n) * R ^ (2/3) * S^ (1/2)
Here, R = hydraulic radius (A/P)
S = channel scope
A = welted perimeter
Thus, discharge (Q) = A * v
Q = (1/n) * R ^ (2/3) * S^ (1/2)
Q α R^ (2/3)
If,
R(up) = Q (up)
R (down) = Q (down)
Given,
S = 0.002
Manning's coefficient (n) = 0.008
Case 1: B = 1 m, D = 4 m
A = BD = 1 * 4 = 4 m2
P = B + 2D = 1 + 2 * 4 = 9 m
From equation 1,
Q = (A/n) * R ^ (2/3) * S^ (1/2) = (4/0.008) * (4/9) ^ (2/3) * (0.002) ^ (1/2)
Thus, Q = 13.02 m3/s

Case 2: B = 4 m, D = 1 m
A = BD = 4 m2
P = B + 2D = 4 + 2*1 = 6 m
From equation 1,
Q = (A/n) * R ^ (2/3) * S^ (1/2) = (4/0.008) * (4/6) ^ (2/3) * √ 0.002
Q = 17.06 m3/s
Thus, for case 2, P (down) = R (up) = Q (up)

c) A pipe is taking water to a remote property from a main pipe some 500m distant. The flow of water has to be no less than 0.01m3s-1 What is the smallest pipe diameter that flow would definitely be laminar?

The rate of flow of water = 0.01 m3/s
Let us assume dynamic viscosity of water = 0.001 Pa-s
In order for the pipe to have a laminar flow the maximum value of Reynolds number is 2000.
Once the value is crossed, the flow is going to the transition zone.
Re = PVD / μ
2000 = 1000 * {0.01/ (π/4 * D) * 0.001}
Hence, the diameter of the smallest pipe will be 6 mm (approx.)

d) i) What is the "energy principle" ?

The energy principle states that energy is neither created nor destroyed. However, energy may transform from one state to another.
∑k Ek = constant

ii) Use this to explain what happens to the depth of the water in a long horizontal channel of constant width lined with smooth concrete with an "n" value of 0.011 delivering water at 0.1m3s-1

Given,
n = 0.011
Q = 0.1 m3/s
Now,
Q = (1/h) * {A(R) ^ (2/3)} * {S ^ (1/2)}
Let us assume the shape of the channel to be rectangular.
Thus, A = By
R = (A/P) = (By/y) = y (As, B + 2y approx. to B, because it is along horizontal channel)
Therefore,
Q = (1/n) * By * y ^ (2/3) * S ^ (1/2)
Qn / {B(S) ^ (1/2)} = y ^ (5/3)
y = [{0.1 * 0.011} / {B * S ^ (1/2)}] ^ (3/5) = 0.0168 / {(B ^ 0.6) * (S ^ 0.3)} = ∞
(As, S = 0 for a horizontal channel)
According to energy principle,
E = y + (v^2 / 2g) = y + {Q / (A^2 * 2g)} = ∞ + Q^2 / (B^2 * y^2 * 2g) = ∞ + 0
E = ∞
As it is a horizontal channel its slope is 0, thereby, the depth of water (y) is unidentified here, and so is the energy as per the energy principle.

e) What are the advantages and disadvantages of bringing a water supply into a town using pipes rather than using open channels.

Water transportation by pipeline is known to reduce water loss from seepage, evaporation, and theft, as compared to the open channels of water transfer (Zhanget al. 2019, p.501). The society needs to have access to clean and safe drinking water, which can be greatly ensured if a water is transferred using a pipe.

Pipes at times can get clogged which might be difficult to open and frequent maintenance required can be challenging at times as the points of problems are occluded from the open sight. Also, in pipes when clogs and fungus left uncleared can lead to water borne diseases, such as diarrhea and typhoid.

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LO3: Develop practical solutions for the distribution of fluids within correctly sized pipes

b) Calculate the Reynolds number for the water in the above pipe line. Comment on whether the water is being delivered using the minimum of energy. (assume absolute viscosity of water is 0.008Pascal s-1 )

Re = QVD/ => V = Q //4* (d)^2

= {10^3*(5//4* (0.67) ^2) *0.67} / 0.008

Re = 1188325 * Q >4000 {Turbulent Flow}

LO4: Calculate the hydrostatic pressure exerted on substructures for a given context

a) A dam 80m long is holding water to a depth of 45m. Calculate what the total load on the dam is. Suggest how the water is kept at a safe level and what information is needed to design such a structure

Total load on the dam can be described as
Load = Area of pressure graph
= (1/2) * pressure at bottom * height * length
Pressure at bottom = γw* H = 9.81 * 45 = 441.45 KN / m2
Load = (1/2) * (441.45) * 45 * 80 = 794610 KN
Hence, the water is kept at a safe level.
The net forces should exist such that the dam must neither slide nor tip.
In order to design the structure, one needs the following:
• Uplift acting pressure
• Earthquake / wind forces
• Unit weight of materials of dam
• Materials in water, etc.

b) A two storey underground carpark has been designed to go down to a depth of 15m. The water table in that area is 3m below ground. The dimensions of the car park are 50m by 20m. Calculate the mean pressure on the walls and thus the total load that the water will exert on the structure from the sides.

Car parking dimensions = 20 * 50 m
Mean pressure on walls

Mean pressure on CB = GF = (0 + rw h) / 2 = (0 + 9.81 + 12) / 2 = 58.86 KPa
Mean pressure on BD = DF = 9.81 * 12 = 117.72 KPa
Force of water on the sides of the structure on CB = area of pressure diagram on GF = {(0 + 9.81 * 12) / 2} * 12 = 706.32 KN / unit length of wall
On BD = pressure * area = 117.72 * 50 = 5886 /m width

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c) Calculate the upward pressure and the total force on the floor of the car park exerted by the water.

Upward pressure = height of water raised in case piezometer is placed below the bottom of the parking
Thus, upward pressure = 9.81 * 12 = 117.72 KPa
Total force exerted by water = upward pressure * parking's contact area = 117.72 * 20 * 50 = 117720 KN

d) A caisson is needed to build piers for a new bridge across a wide river. Each caisson will have a footprint of 10 x 10m and will need to deal with a depth of water of 12m at maximum. Outline the main considerations that will allow piles to be sunk and the bridge pieces to be constructed safely.

A Cassion can be defined as a watertight retaining structure that is utilized for foundation activities for dams, bridge piers, and more (Esfeh and Kaynia, 2020, p.106). Its primary role is to keep a dry work environment, which is done by pumping out water.

How to install:-
To install a caisson in place, it is brought down through soft mud untill a suitable foundation material is encountered. While bedrock is preferred, a stable, hard mud is soometimes used when bedrock is too deep. The four main types of caisson are box caisson, open caisson, pneumatic caisson and monolithic caisson.

Main consideration that will allow piers to be sunk and bridge pieces to be constructed safely:-

When the soil contains large boulders, which obstruct penetration of piers. A massive substructure is required to extend to or below the area bed to provide resistance against destructive forces due to floating objects and source etc. Thefoundation is subjected to a large lateral load. Depth of water level in the river will be high at some places. There are river forces include in the load compositions. When the load is needed to carry at the end, caisson are preferred, mostly it will be required. The present ground levels of water is having aggressive in/ between, caissons are suitable. Finally, we can say that caisson construction needs to be accuracy for successful results.

Installation of caissons:
There are 4 types of caissons, namely open, box, monolithic, and pneumatic caisson. These are sunk to the needed level of depth by drudging or excavation. After the hole is created in the ground it is filled with concrete.
Main considerations:
• The suitability of caissonas per location and method of installation utilized along with its structural behavior.
• The water table level as the foundation can penetrate the level
• The nature of the soil
• The maximum scour depth
• The caisson's geometry based on mass
• An overall stability of the caisson
• Bedrock's bearing capacity
• Structure's durability

FAQ

• 1. What are Hydrostatics and Hydrodynamics?
• 2. Calculating Forces in Fluids at Rest (Hydrostatics):
• 3. Calculating Forces in Fluids in Motion (Hydrodynamics):
• 4. What Factors Affect Fluid Distribution in Pipes?
• 5. Practical Solutions for Efficient Fluid Distribution:
• 6. What is Hydrostatic Pressure on Substructures?
• 7. Steps to Calculate Hydrostatic Pressure on a Substructure

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