The density of sea water is 1029kN/m3, what would the pressure difference between the bottom of a freshwater lake of depth 30m and 30m below the surface of the sea?
Assumption – Considering fresh water density to be approx 1000kg/m3
Density of sea water = 1029 kN/m3
Density of fresh water = 1000 kg/m3
Depth of fresh water lake = 30 m
Depth of sea water lake = 30 m
Pressure at bottom of fresh water lake =
Pressure at the depth in sea =
Pressure difference
= = = 8534.7 Pa = 8.5347 kPa
Hence, the variation of the pressure of water at the bottom of fresh water lake which is situated 30m under the sea surface is 8.5347 kPa
Conclusion
The major factors that have effect on pressure in fluids are depth of the surface and density of the fluid. The fluid density and the value of pressure are directly proportional to each other.
A U-tube manometer is a device that would be helpful for calculating the difference of pressure among the two points. However, a strain gage pressure transducer is preferred for measuring the pressure in a water body at a depth of 30m.
Task 1.2. Hydrostatic forces.
Problem Description
Width of dam (w) = 80 m
Depth of water (h) = 30 m
Find force on acting on Dam (F)
Pressure at centroid of dam =
Surface area of dam (A) =
Force on the dam (F) =
Considering the spill height to be set at= h meter
Force acting upon dam (F)= 0.422GN
Therefore
The pressure situated at the plane surface’s submerged area and centroid is a measure of the force acting on the completely submerged plane surface.
LO2. Be able to solve engineering flow problems.
Task 2.1. Carry out calculations for Pipeline Flow problems.
Objective: To analyze and evaluate the comparison of the Froude and Reynolds number along with their significance
Reynolds Number- Reynolds Number can be defined as the ratio between the viscous force and the inertia force of the fluid flowing which becomes crucial for the fluid flow that in mainly manipulated by the viscous forces excluding the inertial force of the flowing fluid (for example, the flow of fluid within the pipe or completely submerged body.)
Reynolds number (Re) = (Inertial force / Viscous force)
Froude Number- Froude Number can be explained as the ratio between the inertial forces of the fluid to the gravitational force acting on the fluid flowing. The Froude Number gains significance when the flow of fluid is mainly dominated by the gravitational pull which nulls the impact of the inertial forces on the fluid such as flow over sluices, channels, spillways, and weirs (free surface flow).
Froude number = (Inertia force / Gravitational force)
Osborne Reynolds had done a tedious experiment in the year 1880 and he came to an inference that the process of transition of laminar flow to turbulent flow had been largely dependent apart from flow disturbance and surface roughness, on the value of Reynolds number. For most practical conditions, the transitional process of laminar flow to turbulent flow for varied types of flow are summarized as –
Internal flow – Laminar flow, Re ≤ 2300
Transition flow, 2300 ≤ Re ≤ 4000
Turbulent flow, Re ≥ 4000
Flow over smooth flat plate – Laminar Re ≤ 5 * 10^5
Turbulent Re ≥ 5 * 10^5
Flow over circular cylinder or sphere – Laminar, Re ≤ 2 * 10^5
Transition, 2 * 10^5 ≤ Re ≤ 2 * 10^6
Turbulent, Re ≥ 2 * 10^6
Open channel flow – Laminar, Re ≤ 500
Transition, 500 ≤ Re ≤ 2500
Turbulent, Re ≥ 2500
The three types of open channel flow as indicated by the value obtained from Froude number (Fr) are critical, subcritical or supercritical.
Fr < 1 Subcritical or tranquil flow
Fr = 1 Critical flow
Fr > 1 Supercritical or rapid flow
Mach number and the Froude number are similar for the open channel flow in compressible flow.
Task 2.2. Carry out calculations for Open Channel Flow problems.
Objective: To use Darcey Weisbach Equation for calculating the friction factor within the pipe flow
Given data-
Length of pipe, L = 10 km = 10000 m
Head loss, hf = 15 m
Discharge Q = 10 m3/s
Internal diameter D = 1400 mm = 1.4 m
Friction factor (f) = ?
Darcey Weisbach equation – hf = ………….eq. (2.2.1)
Vavg = ………….eq. (2.2.2)
Cross sectional area Ac = (π/4). D2
Using equation 2.2.1 the value of friction factor ‘f ’ is found to be
f = 0.000976
Objective: To calculate the head loss for fluid flow with the wearing of the pipe surface resulting in doubling the friction factor
Approach: The cross sectional area of task 2.2 A and a present fixed flow rate, the value of average flow velocity is same as calculated for task 2.2 A
Hence, utilizing the equation of 2.2.1, the inference that comes is that the head loss is in direct relation (proportional) to the friction factor i.e. hf α f
Hence, the head loss would also be doubled to 30 m when the friction factor is doubled.
Task 2.3. Analytical principles.
Objective: Using Bernoulli’s equation for analytical calculation of flow rate
Approach: Venturimeter (orifice meter) is a measuring instrument that can be mounted in pipe section for measuring the flow rate for the flowing fluid within a horizontal pipe. It would help in indirectly measuring the flow rate by calculation of the relative head difference among the two flow section.
Bernoulli equation has pointed out that the sum of flow energy, potential energy, and kinetic energy would be constant for incompressible, inviscid steady flow fluid
(P/ρ) + V2/2 + g.z = constant
The Venturimeter is considered for analysis of flow rate measurement,
The continuity equation in between section 1 and 2 for incompressible flow of fluid can be written as A1V1 = A2V2 ………eq. (2.3.1)
V1 = (A2V2 / A1)…….eq. (2.3.2)
V1 = V2 *β ………….eq. (2.3.3) where β = A2/ A1
Bernoulli’s equation for horizontal pipe between section 1 and 2
(P1/ ρg) + V12/2g = (P2/ ρg) + V22/2g
V22/2g = V12/2g + (P2/ ρg) – (P1/ ρg)
Denoting piezometric head ‘h’ = (P2/ ρg) – (P1/ ρg)
V22 = V12 + 2gh
Substituting for V1 from eq. 2.3.2 in above equation and solving for V2, we get
Hence, we would be able to measure the pipe’s flow rate with the help of measuring the pipe and throat area of venturimenter and piezometric head difference in compare to the manometric fluid.
Objective: Analysis of the impact of pipe diameter upon the flow velocity of fluids
Discussion: 1D steady flow mass balance equation for incompressible fluid
A1V1 = A2V2
Here the cross section area of circular cross section ‘A’ = (π/4).d2
The velocity V2 can be written as V2 = (d1/ d2)2.V1
Putting d1 = 100 mm and d1 = 200 mm
V2 = 0.25 V1
Conclusion:
The flow velocity would be reduced to 25% of its real value due to the increase in the diameter from 100 mm to 200 mm.
Objective: Calculation of force exerted by the fluid on the wall at the exit
Approach: Given
Velocity of flow of water in pipe V1 = 1m/s
Diameter of pipe ‘d’ = 100mm = 0.1 m
Taking density of sea water = 1029 Kg/m3
Area of cross section ‘A’ = (π/4).d2
Net Force (applied on wall by water) is ‘F’ = Rate of change of linear momentum in flow direction
Let’s assume for now that the fluid would turn for 90o after striking with respect to the wall
F = ρ.Q.(V1– V2)
= ρ.A.V1.(V1– V2)
V2 would be 0 as it would be perpendicular to V1.
F = 8.08 N
Increasing the diameter by a factor 2 means the new diameter is 200 mm.
Now for same mass flow rate ρQ, the flow velocity V1 will reduce to 0.25 m/s.
Therefore, the new force exerted on wall be 0.25 of previous force i.e. 2.02 N
Task 2.4. Closed Conduit Flow.
A spillway is needed for a dam. The spillway exit from the dam is 40m above the base of the dam and the bottom of the spillway is intended to be on the same height as the base of the dam but some 800m away. The channel is 4m wide. The maximum allowable depth is 3m and the maximum expected flow is 36m3s-1
Objective: Study of closed stream flow
Given:
Elevation difference between ends of spillways (z1– z2) = 40 m
Length of spillways L = 800m
Width of channel ‘b’= 4m
Max allowable depth dmax = 3 m
Max expected flow Q= 36 m3/s
Slope of channel ‘S’, tan (α) = Elevation difference between ends of spillways / Length of spillways
= 40/800
= 1/20
Max Cross sectional area of flow = Max allowable depth * width of channel
Minimum velocity of flow ‘V’= Expected flow / Max Cross sectional area of flow
= Q/(b.dmax)
= 3 m/s
From Manning’s equation
V = (1/n). Rh^(2/3). S^(1/2)
Rh = cross sectional area of flow(Ac) / wetted perimeter (p)
= b.dmax / (b + 2dmax)
Thus, the value of n is found to be, n = 0.0842
From Mannings equation
Q = (1/n). Rh^(2/3). S^(1/2). Ac
From Manning’s equation, we are required for calculating the depth of flow (d) for 36m3/s flow rate and 3 m channel width
We use Engineering Equation Solver for solving the mentioned equation for the value of ‘d’ to be,
d = 0.6906 m
flow area = b.d
for n = 0.011, the flow velocity, V = Q/(b.d)
V = 13.03 m/s
From solution of previous task 2.4 (D), the depth of flow for maximum flow rate corresponding to n=0.011 is 0.6906m i.e. d = 0.6906m
Task 2.5. Uniform Open Channel Flow.
Objective: Analysis of uniform open channel flow
Given:
Flow rate Q = 12m3/s
Maximum depth d = 2m
Slope of channel S =0.01
Value of manning coefficient n = 0.014
From Manning’s equation
Q = (1/n). Rh^(2/3). S^(1/2). Ac
Now solving above equation for width of flow ‘b’ using engineering equation solver
b = 1.333m
Given:
Slope S = 0.01
New manning coefficient n = 0.02
Max depth d= 2m
For the width of 1.333m the velocity of flow using manning’s equation
V = (1/n). Rh^(2/3). S^(1/2)
V = 3.1494 m/s
For uniform flow, the new discharge Q = V. Ac
Where Ac = b.d
= 1.333*2
Discharge Q = 8.396 m3/s
Discussion:
The flow rate decreases for the provided constant flow area due to the increased manning’s coefficient (higher than the friction)
Task 2.6. Flow Measurement in Open Channels.
Objective: Analysing difficulties in open channel flow measurement.
Discussion: The fluid is not bounded completely in the open channel flow that infers to that fact that it has free surface. It is easy for controlling the flow rate with the help of valves when the ducts and pipes flow confined within the wall boundary.
The flow rate is directed by the use of partial blocking of the flow area to overcome the fee surface problem of the fluid and it would measure the flow rate also. The obstruction placed in the path of flow and the fluid would be required for flowing above or below the obstruction. The measurement of the fluid height in a constant flow area would help in calculating the fluid’s flow rate. The weir is a term given to the phenomenon of fluid flowing over the obstruction whereas underflow gate (sluice gate or drum gate) is the term that points that the fluid flowing is under the obstruction.
LO3. Be able to match pumps to the demands of a specific system.
Task 3.1. Pump matching: energy and hydraulic gradients in pump-pipeline systems; pump performance and characteristic curves; pump selection to operate in a given system; pumps in series and parallel.
|
Discharge l/s |
0 |
50 |
100 |
15 |
200 |
250 |
300 |
350 |
400 |
450 |
500 |
|
Total head m |
55 |
54 |
52 |
50 |
46 |
41 |
36 |
32 |
29 |
25 |
20 |
|
Efficiency |
35 |
43 |
49 |
54 |
57 |
59 |
60 |
58 |
52 |
44 |
34 |
Fresh water is pumped to a service reservoir through a 0.3m diameter pipe 10km long. The static head is 20m. Determine the discharge through the pipe and the power requirement of the pump.
(This will need you to plot the pump – characteristic, system characteristic and efficiency curves against flow rate (Q).)
You must also find the operating point. Assume the friction factor f = 0.0015 and take minor losses at twenty times the velocity head.
Diameter of pipe (d) = 0.3 m
Length of pipe (L) = 10000 m
Static head = 20 m
Friction factor (f) = 0.0015
Volume flow rate = Q l/s
Therefore, the net head =
Plotting the net head loss against the discharge we get
Therefore, the operating point from the above graph
Q = 185 l/sec
H = 44.45 m
Pump power = 80.67 kW
Diameter of pipe (d) = 0.5 m
Length of pipe (L) = 10000 m
Static head = 20 m
Elevation head = 30 m
Friction factor (f) = 0.0015
Volume flow rate = Q l/s
Therefore the net head = Static head + Elevation head + total head loss
Plotting the system curve, pump curve and efficiency curve we get
The operating point of the pump is
Q = 100 l/sec
H = 50 m
Pump power = 49.05 kW
The installation of one more pump at the parallel location of net discharge would help in increasing the output discharge to,
= 200 l/sec
Conclusions
The intersection point of system and pump curve would act as the operating point for the pump
Task 3.2. Specific systems: hydrodynamic machines; classification of pumps and turbines (radial, axial, reaction)
Pump
Pump is a device with hydrodynamic machine and mechanical action for increasing the fluid energy. Its main function is adding the energy and increasing the pressure as a result for the incoming fluid.
Turbine
Turbine acts as a hydrodynamic machine that helps in extracting energy within the fluid for decreasing the overall pressure on the fluid.
Efficiency of the turbine is given as
Conclusions
It can be concluded from the above discussion that the efficiency of the pump is reciprocal of the turbine’s efficiency
Radial Turbine
The direction is radial for the fluid flow which is due to the fact that the flowing fluid enters the turbine in a radial direction. The radial turbine flow is classified into two type namely radial inward flow and outward flow inward. The Francis turbine is one types of radial turbine that can be used for medium specific speed and medium head.
Axial Turbine
An example of axial flow turbine is Kaplan turbine that allows the fluid flow in the turbine in parallel with the rotation axis of the rotor. The axial flow turbine is one type of radial turbine that can be used for high specific speed and low head.
Reaction Turbine
Reaction turbines are classified with the possession of both pressure and kinetic energy at the turbine’s inlet. The Reaction turbine is one type of radial turbine that can be used for high flow speed rate and low head.
LO4. Be able to undertake hydraulic experimental procedures
Task 4. You have participated in a number of different experiments in the hydraulics lab. Undertake brief reports as outlined below on four which much include:
In each experimental report provide the following:
Exhibition of the Principle of Bernoulli’s and analysis of its application for measurement of the pipe flow
Bernoulli’s Test apparatus, manometer, hydraulics bench, and hydrodynamic probe
The pressure head, dynamic head, and elevation head is related to the very famous fluid mechanics of Bernoulli’s equation.
The assumptions for using Bernoulli’s equations are
For a steady incompressible flow, Bernoulli’s equation shows that;
If the flow is irrotational the value of constant is same along all the streamline, therefore;
Where
For zero elevation head
Table 1. Pipe diameter at position where manometers were connected.
|
Tapping position |
Manometer Letter |
Pipe diameter at position (mm) |
|
A |
H1 |
0.025 |
|
B |
H2 |
0.0139 |
|
C |
H3 |
0.0118 |
|
D |
H3 |
0.0107 |
|
E |
H4 |
0.010 |
|
F |
H5 |
0.025 |
|
G |
H6 |
N/A |
Finding and discussions
The flow rate between the two points of the pipe was calculated by the help of the apparatus and the implication of the Bernoulli’s and continuity equations.
The assumptions we made are as follow
Using continuity equation between point 1 and 3 and assuming flow is incompressible for test 2
Putting the values of A1 and A2 we get
Using Bernoulli’s equation between point 1 and 3
The static head at point 1 and 3 were measured using manometer,
Therefore, the velocity at point 1 V1 = 0.3240 m/sec
Theoretical volume flow rate is given as
Error
Error due to the assumptions we made in calculating volume flow rate
Theoretical total head
At point 1 in test 2
Total head at point 1 = static head + velocity head = 0.225 + = 0.230 m
Hence, it can be said that the measured total head that we found out is more than our theoretical head calculated
Error due to assumptions in calculating total head
Conclusions
The demonstration of the Bernoulli’s equation and its application for measuring pipe’s flow rate in the experiment had shown the theoretical volume flow is calculated using frictionless flow equation in the pipe and the difference between the theoretical and actual rate. The comparison had shown that the assumptions were incorrect and the pipe flow had to face some frictional head loss.
The measurement of the flow rate for piping system such as oil pipelines and water supply is the primary application. The experimentation would help us in calculating the head loss within the pipe flow and the power required for pumping and overcoming the head loss.
The measurement of the flow rate and the comparison of the hypothetical flow rate in V-Notch and rectangular weirs is the primary objective of our experiment.
Table 1. Results for Rectangular notch weir
|
Height of water in Pool (mm) |
Volume (V) liters |
Time 1 ( sec) |
Time 2 ( sec) |
Time 3 ( sec) |
Mean Time ( sec) |
Flow rate Observed Q = (m3/sec) |
|
104.6 |
5 |
19.6 |
21.4 |
21.0 |
20.6 |
2.42 E-4 |
|
99.98 |
5 |
26.7 |
25.4 |
26.7 |
26.26 |
1.90 E-4 |
|
93.60 |
5 |
40.6 |
41.8 |
41.6 |
41.3 |
1.21 E-4 |
|
85.80 |
2 |
55.1 |
58.6 |
55.4 |
46.36 |
0.35 E-4 |
Table 2. Results for V notch weir
|
Height of water in Pool H (mm) |
Volume (V) liters |
Time 1 ( sec) |
Time 2 ( sec) |
Time 3 ( sec) |
Mean Time ( sec) |
Flow rate Qactual (m3/sec) |
|
154.5 |
5 |
23.37 |
20.12 |
18.27 |
20.58 |
0.243E-3 |
|
149.5 |
5 |
23.52 |
24.25 |
22.92 |
23.56 |
0.212E-3 |
|
139.2 |
5 |
41.20 |
46.02 |
44.97 |
44.06 |
0.113E-3 |
|
142.5 |
5 |
52.04 |
57.6 |
55.61 |
55.083 |
0.091E-3 |
Results and Calculations
Theoretical flow rate for rectangular weir
Where b = width of rectangular weir, in our experiment b = 33 mm
h = 0.063 m
H = Height of pool height
H-h = Height of water above weir
Therefore
For H = 0.1046 m, h = 0.063
Qth = 8.27E-4 m3/s
For V notch theoretical flow rate is given as
In our experiment angle of notch = 450
Where h = 0.104 m
Therefore for H = 0.1545 m
Similarly for other pool height we calculated and tabulated results as follows
Table 3. Comparison of actual flow rate and theoretical flow rate for rectangular weir
|
Pool height (H) m |
Theoretical flow rate (Qth) m3/sec |
Actual flow rate (Q) m3/sec |
Difference (%) |
|
|
|
0.1046 |
8.27E-04 |
2.42E-04 |
70.73% |
0.29 |
|
|
0.09998 |
6.93E-04 |
1.90E-04 |
72.58% |
0.27 |
|
|
0.0936 |
5.22E-04 |
1.21E-04 |
76.80% |
0.23 |
|
|
0.0858 |
3.35E-04 |
3.50E-05 |
89.57% |
0.10 |
Table 4. Comparison of actual flow rate and theoretical flow rate for V- notch weir
|
Pool height (H) |
Theoretical flow rate (Qth) m3/sec |
Actual flow rate (Q) m3/sec |
Difference (%) |
|
|
|
0.1545 |
5.33E-04 |
2.43E-04 |
54.45% |
0.46 |
|
|
0.1495 |
4.09E-04 |
2.12E-04 |
48.14% |
0.52 |
|
|
0.1392 |
2.12E-04 |
1.13E-04 |
46.61% |
0.53 |
|
|
0.1425 |
2.66E-04 |
9.10E-05 |
65.85% |
0.34 |
Conclusion
The comparison with the theoretical correlation with the flow rate in the open channel (that was measured by V-Notch and rectangle weir) is largely available in the literature for opening channel flow within the V-Notch and rectangle weir. In the experiment, the calculation of the value for the delivery coefficient of a flow range had formed the inference that the Cd value was almost a constant. The Cd value can be utilized for calculation of the open channel’s flow rate with the help of theoretical correlation.
The measurement of the flow rate for the open channel flow (for example water treatment plan, reservoir flow, and irrigation channel) is the primary application of the theory learnt from the experimentation.
Conclusion
I have learnt the concept for implementation of the hydraulics in the domain of engineering problems after the completion of the assignment. However, the most crucial point that I learnt from the study had been pointed out in the following paragraphs.
I became aware of the process of calculation of the force upon the dam along with the designing of the essential criteria for dam, and diverse sorts of pressure measuring devices. I also came to know about the process of using the devices for measuring.
I came to know about the open and closed channel flow in fluid flow for the second flow. The Reynolds and Froude number both are responsible for the governance of the open channel while Reynolds number is for the open channel fluid flow. The flow measurement and head loss is calculated for both open and closed channel governed by Reynolds and Froude number.
The comparison of the turbine and pump was understood by me after the completion of the study. The task had been very important and of great assistance for the process of selecting the pump. The theories included for the pump selection was characteristic curve and parallel and series combination concept for the pump.
I had been able for measuring the close channel and open channel for measuring the flow of fluid and its use for overcoming engineering problems related to the fluid mechanics.
The plotting of the system curve for the two pumps arranged in parallel arrangement was the most difficult part of my assignment. I had to learn and gain knowledge for a number of skills and knowledge.
References
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