Fluid Density, Pressure, and Depth

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.

1. What equipment would be used to measure such pressures?

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.

1. What is the total force acting on a dam which is 80m long holding back (fresh) water at a depth of 30m?  

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) = 

1. If the dam were capable of withstanding a force of 0.422GN at what height above the bottom of the reservoir would the spill way have to be set?

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.

1. The difference between Laminar or Placid flow on one hand and turbulent flow on the other is determined by one of two numbers either the Reynold’s number or the Froude number. Explain the difference between the two and at what value or range of values does the transition to turbulent flow occur.

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).

Pressure Difference Between Freshwater Lakes and Seawater

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.

1. Use Darcey Weisbach equation to calculate the friction factor given that the head loss along a new 10km long straight pipe line is 15 metre when the delivery is10m3s-1 and the internal diameter of the pipe is 1,400mm.

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, h_f = 15 m

Discharge Q = 10 m³/s

Internal diameter D = 1400 mm = 1.4 m

Friction factor (f) = ?

Darcey Weisbach equation – h_f  =  ………….eq. (2.2.1)

V_avg =      ………….eq. (2.2.2)

Cross sectional area A_c = (π/4). D²

Using equation 2.2.1 the value of friction factor ‘f ’ is found to be

f = 0.000976

1. What would the head loss be some years later as the inside surface of the pipe line was worn causing the friction factor to double?

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. h_f α f

Hence, the head loss would also be doubled to 30 m when the friction factor is doubled.

Using U-Tube Manometer and Strain Gage Pressure Transducer for Measurement

Task 2.3. Analytical principles.

1. Show how to make use of the Bernoulli Equation to measure the rate at which water is flowing in a horizontal pipe of internal diameter 100mm, show what measurements you would make and what additional modification you would need.

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.

1. If the diameter of the pipe were to open out from 100mm to 200 mm, what would happen to the flow velocity?

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. 

1. If water is flowing at a rate of 1ms-1 coming out of the end of the (100mm diameter) pipe and hits the wall of a reservoir. What force does the water apply to the wall?   What would happen to this force if the diameter of the exit pipe were to be increased by a factor of 2?

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 90^o after striking with respect to the wall

F = ρ.Q.(V₁– V₂)

Pipeline Flow Problems

   =  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 d_max = 3 m

Max expected flow Q= 36 m³/s      

1. Calculate the slope of the channel

Slope of channel ‘S’, tan (α) = Elevation difference between ends of spillways / Length of spillways

              = 40/800

              = 1/20

1. Calculate the minimum velocity at maximum flow

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.d_max)

= 3 m/s 

1. Thus, calculate the Manning n value of the channel

From Manning’s equation

V = (1/n). R_h^(2/3). S^(1/2)

R_h = cross sectional area of flow(A_c) / wetted perimeter (p)

  = b.d_max / (b + 2d_max)

Thus, the value of n is found to be, n = 0.0842

1. If the n value could be adjusted to 0.011 the value for smooth concrete what would the velocity of flow be at maximum

From Mannings equation

Q = (1/n). R_h^(2/3). S^(1/2). A_c

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

1. What would the depth of flow be at maximum flow.

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.

1. What is the width needed to keep the depth to maximum of 2m with following conditions. Delivery is 12m3s-1 given the slope of the channel is 0.01 and the value of n is 0.014.

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). R_h^(2/3). S^(1/2). A_c

Now solving above equation for width of flow ‘b’ using engineering equation solver

Reynolds and Froude Numbers

b = 1.333m

1. If the channel over long use became rougher so that the Manning n value increased to 0.02 but it was still required that the depth must not exceed 2m, how would this change the maximum delivery of the channel

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). R_h^(2/3). S^(1/2)

V = 3.1494 m/s

For uniform flow, the new discharge Q = V. A_c

Where A_c = 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.

1. Briefly outline the difficulties of measuring flow in an open channel.   Illustrate your answer with examples of actual commercial flow measuring systems currently in use.

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.

1. The table given below gives the characteristics of a centrifugal pump running at a steady speed:

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 =

Transitional Processes of Laminar Flow to Turbulent Flow

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 

1. In the above situation, a second reservoir at a height of 30m above the original one needs to be supplied. Using the same type pump and a new 0.5m pipeline show what output you would get from a single pump and how you could arrange a second pump to maximize the output.

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)

1. Outline the essential differences between a pump and a turbine especially in how efficiency is expressed.

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

1. Outline the situations where each of the three main types of turbine (radial, axial and reaction) would be used  

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.

Open Channel Flow and Critical, Subcritical, and Supercritical Flows

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:

1. Measurement of flow in open channels using weirs and the demonstration of Bernoulli’s Principle and one further experiment of your choosing.

In each experimental report provide the following:

1. Brief clear statement of the experimental procedures emphasising any changes from the procedures given in the handbook
2. A clear table of results

• Clear calculations of the data from the results

1. Discussion of accuracy of the experimental results and any suggestions re changes in procedure
2. Application of this to real engineering problems

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

• Steady flow
• Negligible frictional effect
• Density of fluid constant i.e. Incompressible flow
• No shaft work and heat transfer

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 

• To measure the static head within the pipe, manometers were connected at A, B, C, D, E, and F position
• The total head for measurement was static + dynamic head of  hydrodynamic probe that can connected at position G
• The measurement of the flow rate through the apparatus was done
• Each of 3 and 5 litres of water were collected at the connecting tank and three readings were taken

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

• Flow is incompressible
• No viscous effect
• Steady flow
• Pipe is smooth

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.

Darcey Weisbach Equation

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.

• Same hydraulics bench was used for bolting the Weir
• The height of the pool from its bottom surface is measured by depth gauge behind the weir
• The adjustment of the maximum water flow rate was measured for turning on water fall
• The recording of time for volume flow rate would be implied for measuring the flow of 5 litre
• The comparison of the theoretical and measured volume flow rate was done

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 

Q_th   =   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.

• Learning Objective 1:

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.

• Learning Objective 2:

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.

• Learning Objective 3:

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.

• Learning Objective 4:

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

1. Darby, R. and Chhabra, R.P., 2016. Chemical engineering fluid mechanics. CRC Press
2. Elger, D.F. and Roberson, J.A., 2016. Engineering fluid mechanics(pp. 170-185). Hoboken (NJ): Wiley.
3. Munson, B.R., Okiishi, T.H., Rothmayer, A.P. and Huebsch, W.W., 2014. Fundamentals of fluid mechanics. John Wiley & Sons.
4. Nakayama, Y., 2018. Introduction to fluid mechanics. Butterworth-Heinemann.
5. Reddy, J.N., 2014. An Introduction to Nonlinear Finite Element Analysis: with applications to heat transfer, fluid mechanics, and solid mechanics. OUP Oxford.