ME343 Lab Manual

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ME343

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ME343 Lab Manual

Offered in Fall Semester

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Faculty of MechanicalEngineering

Ghulam Ishaq Khan Instituteof Engineering Sciences and

Technology


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Permission in writing must be obtained from the Author before any part of this work may be

reproduced or transmitted in any form or by any means, electronic or mechanical, including

photocopying and recording, or by an information storage or retrieval system.


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Table ofContents

Instructions for the lab instructor....................................................... v

Instructions for the students.............................................................. vii

Grading Policy.......................................................... ix

Lab regulations............................................................ x

Dynamics Experiments

Experiment 1Determining center of pressure (Apparatus # FM11)........................ 1

Experiment 2

Investigation of stability (Apparatus # FM27) ............................12

Experiment 3

Analyze flow over weirs (Apparatus # FM05)....... .................... ......24

Experiment 4Determining friction factor for pipes (Apparatus # FM02)........... 33

Experiment 5Observing momentum transfer (Apparatus # FM07) ...........42

Experiment 6Determining discharge coefficient of an orifice (Apparatus # FM09) 50


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ME 343 Laboratory Manual

Experiment 7

Validity of Bernoullis equation (Apparatus # FM06).........................................................58

Experiment 8Investigating pressure drop through valves (Apparatus # FM02) .......................................65

Experiment 9Calculating Reynolds Number (Apparatus # FM12) ..........................................................73

Experiment 10

Calibrating a Bourdon Gauge (Apparatus # FM28) ...........................................................79

Experiment 11Determine Polytropic Index (Apparatus # HE4)................................................................85

Experiment 12Effect of Air/Fuel Ratio on Combustion (Apparatus # HE2)...............................................92

Appendix AGuidelines for project report writing...............................................................................100


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Instructions for theLaboratory Instructor

1. There will be six experiments going on simultaneously. So that only threestudents should perform one experiment each time. The normal strength

of class of students is 80 to 90. If we divide these students on 4 numbersof days from Monday to Thursday we are left with 20 students each day.For 20 students 6 experiments mean 3 students on each experiment.

2. There must be separate lecture of 2 hours to be taught to whole class inlecture hall to teach these 6 experiments. Only 2 classes will be neededin the whole semester because there are normally 13 to 14 experimentsin each lab.

3. During the 3 hours of lab the instructor must spend at least 30 minuteswith each group for demonstration of experiments to students.

4. Before each experiment there must be a small quiz to make students

start thinking about the theory involved in the experiment.

5. Before starting the experiment following steps should be followed:a) The students must study the experiment by themselves.b) The theory, objective and procedure to perform the experiment

must be understood fully by the students before starting theexperiment.

c) Instructor must be asked for any question, ambiguity or query ifany in theory, objective and procedure of the experiment.

d) Correct setting of the experiment is required like balancing theequipment, removing zero error etc.

e) Performing the experiment by keeping objectives of theexperiment in mind.

f) Analysis of data achieved by the experiment in lab beforeleaving the lab.

g) Again perform the experiment if data found is not satisfactory.

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6. There must be at least four quizzes, a mid and a final for the lab.

7. There must be a project of 10% of total weight age in the lab which is to design a newapparatus efficient and effective than existing for performing experiment on the basis oftheory of labs or to design a new experiment on existing apparatus available in the lab. Areport along with experiment apparatus must be submitted at the end of semester. Theguideline for the report writing is shown in appendix of the manual.

8. The lab groups (day of experiment and number of students) must be arranged, allocatedand controlled by the instructor.

9. Lab groups must be allocated according to registration numbers.

10. Lab reports must be submitted by each student independently. They must perform theexperiment and collect the data in a group collectively but the results must be interpretedand analyzed individually with their own reasoning.

11. At least, one software, related to the course content of lab must be taught to the students.The tutorial sessions on the software, practice sessions by student and their test ofcompetence in software must be ensured by the instructor. Lab software should be gradedand must be included in the final grading.


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Instructions for the

StudentsDear students;

The theory content and experimental procedures which are taught andwill be taught to you is solely for your learning. hands-on experience isthe best way to learn. This lab manual is designed to give hands-onlaboratory experience to better reinforce certain topics discussed inlecture as well as to present a number of other principles. Eachexperiment begins with a detailed discussion that provides all theinformation needed to understand that lab. The discussion section isfollowed by a detailed step-by-step procedure. Figures and graphs are

provided as and where required. Each experiment concludes with adetailed exercise to help the student interpret the results.

Your cooperation with lab engineer and other lab staff is highly required.

The general instruction is as follows for the students.

1. The Lab manual must be bought before the start of the semester.

2. Lab groups and days of lab will be provided by the instructor according toregistration numbers. The students must schedule their other labs andcommitments accordingly.

.3. Students are suggested to read the lab manual before coming to lab

because they will be a quiz in every lab before the start of experiment tomake start students thinking about the experiment.

4. Students must not leave the lab during the three hours before the priorpermission of lab instructor.

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5. Following steps must be followed to perform the lab.

a) The students must study the experiment by themselves.b) The theory, objective and procedure to perform the experiment must be

understood fully by the students before starting the experiment.c) Instructor must be asked for any question, ambiguity or query if any in

theory, objective and procedure of the experiment.d) Correct setting of the experiment is required like balancing the

equipment, removing zero error etc.e) Performing the experiment by keeping objectives of the experiment in

mind.f) Analysis of data achieved by the experiment in lab before leaving the lab.g) Again perform the experiment if data found is not satisfactory

6. The main objective of MEL lab is highest level of learning of each of the student which willbe achieved if experiment is fully performed by students by themselves.

7. There must be a project of 10% of total weight age in the lab which is to design a newapparatus efficient and effective than existing for performing experiment on the basis oftheory of labs or to design a new experiment on existing apparatus available in the lab. Areport along with experiment apparatus must be submitted at the end of semester. Theguideline for the report writing is shown in appendix of the manual.

8. Lab reports must be submitted by each student independently. They must perform theexperiment and collect the data in a group collectively but the results must be interpretedand analyzed individually with their own reasoning.

9. Individual lab manuals (completed) must be submitted the next day till 12 pm.

10. Retaking of labs, quizzes and software lab sessions must be administered according ofFME policy.


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Grading Policy

Lab attendance 10%

Lab performance and reports 10%

Lab project 10%

Software lab attendance and test 10%

Quizzes 10%

Mid exam 20%

Final exam 30%

The grading policy must be strictly followed by the instructor.

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Laboratory regulations

All students should bring their own lab manual available in servicescentre, pencil, ball point pen, graph paper etc.

Make up Lab: No Makeup lab. However, with the permission from theDean one can perform experiments. Such lab experiments will not begraded.

Late Comers: Students should come on time for the lab. Late comers will

be marked absent.

Lab Exam: Lab exam will be during last week of classes.

Schedule: Schedule will be provided at beginning of the course.

Duration: Duration of each practical experiment is 3hours and nostudents will be allowed to leave the Lab before time. The studentsshould keep themselves busy and get full understanding of the apparatusand the experiments. The student who leaves the Lab before the end oftime will be marked absent.

Cheating will be handled in accordance with FME policy, the details of

which are given o FME websitehttp://fme.giki.edu.pk .

ME343
http://fme.giki.edu.pk/http://fme.giki.edu.pk/http://fme.giki.edu.pk/http://fme.giki.edu.pk/


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Determining center ofpressure

(Apparatus # FM11)

Figure Error! No text of specified style in document.-1 Experimental setup

Objective:

The objective of this experiment is to determine the Center of Pressure

(COP) for both fully submerged and partially submerged plane surfaces

and also to compare the experimental and theoretical values of COP.

Center of Pressure:

The center of pressure is the point on a body where the sum of pressure

field acts, causing a resultant force but no moment about that point.

Mathematically it can be said that the net pressure force on the body acts

through this point. The net force applied at the center of pressure produces

the equivalent moment, equal in magnitude to the moment produced by

pressure field about any arbitrary point.

Total Resultant Force:

The magnitude of the total resultant force is equal to the pressure acting at

the centroid (geometric center) of the area multiplied by the total area. For

symmetric pressure fields this forces passes through the centroid of body

but for unsymmetric pressure fields force doesnt pass through centroid.

Experime

nt1


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Apparatus description:

A fabricated toroid is mounted on a balance arm, which pivots on knife edges. The line of contact of

the knifeedges coincides with the axis of the quadrant. Thus, of the hydrostatic forces acting on the

quadrant when immersed, only the force on the rectangular end face gives rise to a moment about

the knifeedge axis and an adjustable counterbalance to balance the moment produced. This

assembly is mounted on top of an acrylic stand, which may be leveled by leveling feet. Correct

alignment is indicated on a circular spirit level mounted on the base of the tank. Beam level

indication attached to the side of the tank shows when the balance arm is horizontal. Water isadmitted to the top of the tank by a flexible tube and may be drained through drain cock in the side

of the tank. The water level is indicated on a scale on the side of the quadrant.

Equipment Setup:

Figure Error! No text of specified style in document.-2 Schematic Diagram of a HydrostaticBench

Technical details:

Table Error! No text of specified style in document.-1 Important Parameters and their values

S/N Parameters Value

1 Tank Capacity 5.5 Liters

2 Distance b/w suspended mass & fulcrum 278 mm

3 Crosssectional area of quadrant (torrid) 7.5 x 10-3m24 Total depth of completely immersed quadrant 160 mm

5 Height of fulcrum above quadrant end face 100 mm

Analysis:

Because no shear stresses can exist in a static fluid, all hydrostatic forces on any element of a

submerged surface must act in a direction normal to the surface. The hydrostatic forces acting on

the two sides of the Toroid counterbalance themselves, and the forces exerted on the curved


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surfaces (the circular arc top and bottom faces) act through the pivot point of the moment arm of the

Toroid, hence contributing nothing to the net moment about the pivot point. The only hydrostatic

forces that act on the Toroid and have a net moment about the pivot point are those acting on the

plane end face of the Toroid. In fig. 2, forces acting on both surfaces parallel to the page will be

cancelled out, resulting in zero moment.

Nomenclature:

Table Error! No text of specified style in document.-2 Nomenclature

S/N Parameters Symbol Units

1 Resultant force applied F N

2 Density of water Kg/m

3 Distance between the pivot point and free surface q mm

4 Distance between the pivot point and edge of Toroid a mm

5 Width of Toroid b mm

6 Height of Toroid d mm

7 Height of water column y mm

8 Distance between the pivot point and applied weight L mm

9 Distance between the free surface and COP Z mm

10 Theoretical COP location XCT mm11 Actual COP location XCA mm

12 Center of Area mm13 Center of Pressure Hp mm

14 Average Pressure PaCalculating Center of Pressure:

In the following fig. 2, pressure distribution is shown at three surfaces of the fully submerged body in

water. Pressure builds up on the surfaces as water depth increases. Forces acting on curved

surfaces will pass through the pivot point will cause no moment. Only moment created will be from

the distributed force acting on the vertical plane surface. On the left side of the figure, side view of

the vertical plane surface is shown.

The point force equivalent to the distributed hydrostatic forces and the location of the force action

can be calculated from the following formulae:

Figure Error! No text of specified style in document.-3 Pressure

Distribution at different surfaces of Toroid

Balance arm

Knife edge

Center of

pressure

Center

of area

Water surface


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Eq. Error! No text of specified style in

document.-1

This is average value of force acting at the center of pressure under the effect of average

hydrostatics pressure.

For calculating actual location of center of pressure XCA, summation of moments about pivot point

(knife edge) is set equal to zero. Apart from average hydrostatic force acting on the plane surface,

hanging mass at the other end of the balance arm will also generate moment about pivot point.

Hence,

Mass m will produce counter clockwise moment whereas hydrostatic force F will produce clockwise

moment about the pivot point.

Eq. Error! No text of specified style indocument.-2

Where, F can be calculated from Eq. 1-1. XCA is distance between the center of pressure and pivot

point.

For Fully Submerged Plane Surface:

In this section, theoretical value of XCT, the distance between pivot point and center of pressure, will

be calculated.

Figure Error! No text of specified style in document.-4 completely immersed plane surface inwater

The force Fon any flat submerged surface is the pressure at the centre of area multiplied by the

area A of the submerged surface, which can be calculated using Eq. 1-1.

We know the magnitude of the distributed force F, which may be considered as a series of small

differential forces spread over the submerged surface. The sum of the moments of all thesesmall forces about any point must be equivalent to the moment about the same point of the

resultant force F (calculated using Eq. 1-1) acting through centre of pressure.


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Single strip with area and differential width is shown in the following figure.

In the above figure, is the distance from the center of differential strip to the free surface. ischanging over the area of plane surface. The differential force which will act on the center of the

strip will be:

Taking moments about an arbitrary point O at the free surface of water will leads to the following

relation:

Taking integral on both sides of the equation will yield:

Where second moment of area about an axis OO passing through point O is defined as:

Therefore, total moment becomes:


Moment produced by average hydrostatic force will be:


Where, F is calculated using Eq. 1-1 and Z is the distance from point O to center of pressure.

Hence comparing Eq. 1-3 and 1-4 will yield the following:

Therefore,

Figure Error! No text of specified style in

document.-5 Plane surface with strip of area dA

and pressure distribution is shown

O

COA

COP

dA = dx . b

Free Surface

x b

d


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Substitute F from Eq. 1-1 into Eq. 1-5.

Put in the above equation, reducing it to the following expression:


It can also be defined as follows:

From parallel axis theorem can be calculated as:


Where, is second moment of area about about axis gg passig through the geometric center ofthe area. For fully immersed surfaces is defined as:

Therefore substituting Eq. 1-7 into Eq. 1-6,


Therefore, distance between the center of pressure and the pivot point is defined as:


Where, q is the distance from free surface of water to the pivot point.

For partially submerged Plane Surface:

For this case the same equations as mentioned in previous section will apply except that the area ofthe plate varies as ( .Since for partial submerged surfaces is defined as:

is the measure of level of water and geometric center of the submerged surface is:


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After substituting and in Eq. 1-8, the equation for Z becomes:


Figure Error! No text of specified style in document.-6 partially submerged plane surface

Therefore, it can clearly be seen that the cop is always 2/3 down the section of the plate that is

submerged.

Location of the center of pressure can be found using the following formula:

Substitute from Eq. 1-10

Eq. Error! No text of specified style in document.-11Procedure:

1. With the acrylic tank on the bench, position the balance arm on the knifeedge pivot(fulcrum). Hang the balance pan from the end of the balance arm.

2. Connect a length of hose from the drain cock to the sump and a length from the bench feed

to the triangular aperture on the top of the acrylic tank.

3. Level the tank using the adjustable feet and spirit level. Move the counter balance weight

until the balance arm is horizontal.

4. Close the drain cock and admit water until the level reaches the bottom edge of the

quadrant. Place a mass on the balance pan, slowly adding water into the tank until the

balance arm is horizontal.


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5. Record the water level on the quadrant and the mass on the balance pan. Fine adjustment

of the water level can be achieved by overfilling and slowly draining, using the stopcock.

6. Repeat the above for each increment of mass until the water level reaches the top of the

quadrant end face. Then remove each increment of mass noting masses and water levels

until all the masses have been removed.

1 Report SheetName

Registration number

M T W R

Observations & Calculations

a = mm b = mm d = mm L = _____ mm

Table Error! No text of specified style in document.-3 changing mass with changing water levelheight

Level Obs. #Filling Tank Draining Tank Average

Mass m

(gm)

Height y

(mm)

Mass m

(gm)

Height y

(mm)

Mass m

(gm)

Height y

(mm)

Partially

Immersed 1

2

3

4

Fully

Immersed 5

6

7

8


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Table Error! No text of specified style in document.-4 Calculate Actual and Theoretical valuesfor XC

LevelObs.

#

Averagey

2

(mm2)

m/y2

(gm/mm2)

q(mm)

XCA(mm)

XCT(mm)

Mass m(gm)

Height y(mm)

Partially

Immersed 1

2

3

4

Fully

Immersed 5 --- --- ---

6 --- --- ---

7 --- --- ---

8 --- --- ---

Exercise:

1. For partial immersion , drive the following expression for moment produce byhydrostatic force:

* +_______________________________________________________________________________

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2. For partial immersion, what will be the pressure distribution at the partially immersed plane

surface? Draw figure and explain.

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3. Explain with the help of some figure, why centre of pressure is always below the centroid

geometric center of area?

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4. For partial immersion , plot against. The slope of this graph should be and the Intercept should be. Hint: Use equation described in exercise #1.

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5. Give reasons for any discrepancies between the actual and theoretical results.

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6. Plot XCAagainst XCTfor the partial and fully submerged cases on the same graph.


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2. . . . . . . . . .

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Investigation of stability


Objective:

The objective of the experiment is experimental investigation of stability,

and then how a theoretical calculation can be used to predict the results.

Center of Gravity (G):

Expe

riment2


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Center of mass of a system of particles is a specific point at which, the

system mass behaves as if it were concentrated. In the context of an

entirely uniform gravitational field, the center of mass is often called the

center of gravity the point where gravity can be said to act. The center of

mass of a body does not always coincide with its intuitive geometric center.

Figure Error! No text of specified style in document.-8 Forces acting on a floating body

Center of Buoyancy (B):

The buoyant force passes through the centroid of the displaced volume, so the point through which

the buoyant force acts is called center of buoyancy.

Introduction:

The question of stability of a body such as ship which floats on the surface of a liquid is one of

obvious importance. When designing a vessel such as a ship, it is clearly necessary to be able toestablish beforehand that it will float upright in stable equilibrium.

Fig. 2-2(a) shows such a floating body which is in equilibrium under the action of two equal and

opposite forces, namely its weight W acting vertically downwards through its centre of gravity G,

and the buoyancy force of equal magnitude W, acting vertically upwards at the centre of buoyancy

B. This centre of buoyancy is located at the centre of gravity of the fluid displaced by the vessel.

When in equilibrium, the points G and B lie in the same vertical line as in fig. 2-2(a).

Stable equilibrium:

The stability of a body can be determined by considering what happens when the body is displaced

from the equilibrium position. Consider a small angular displacement from the equilibrium positionas shown in fig. 2-2(b) and 2-2(c). As the vessel tilts, the centre of buoyancy moves sideways,

remaining always at the centre of gravity of the displaced liquid. Displaced liquid can adopt different

shapes depending on the position of the floating body. Hence, center of buoyancy changes

accordingly. If, as shown in fig. 2-2(b), the weight and the buoyancy forces together produce a

couple which acts to restore the vessel to its initial position, the equilibrium is stable.

Hence, for stable equilibrium one of the following conditions would hold:

1. If G lies below B, floating body at the surface of water will remain always be in stable

equilibrium.


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2. In floating bodies, the resulting couple (due to shifting of position of B) should restore the body

to its original position.

Unstable Condition:

If however, the couple acts to move the vessel even further from its initial position, as shown in fig.

2-2 (c), then the equilibrium is unstable.

The special case when the resulting couple is zero represents the condition of neutral stability asshown in fig. 2-2 (a).

Experimental determination of stability:

Fig. 2-3(a) shows a body of total weight , equal to weight of jockey plus weight of thepontoon, floating on even keel. In equilibrium state shown in fig. 2-3(b), the weight of thefloating body acts vertically downward and opposite to the buoyancy force equal to the displaced

volume of the fluid times specific weight of the fluid . The centre of gravity may be shiftedsideways by moving a jockey of weight across the width of the body. When the jockey is moveda distance as shown in fig. 2-4(a), the centre of gravity of the whole assembly moves to . Thedistance is denoted by is given from elementary statics as:

Eq. Error! No text of specified style in document.-12

Water Level

Jockey

Slot to move

jockey

G

B

W

Vg = W

(a) (b)

Figure Error! No text of specified style in document.-9 (a) Front view of floatingpontoon (b) Slot is omitted for simplicity

G G

B

B

M

M

BB

G G


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The shift of the centre of gravity causes the body to tilt to a new equilibrium position at a small angle

to the vertical as shown in fig. 2-4(a), with an associated movement of the centre of buoyancy

from to. The point must lie vertically below, since the body is in equilibrium in the tiltedposition. Let the vertical line of the up thrust through intersect the original line of up thrust atthe point, called the Metacentre. We may now regard the jockey movement as having caused thefloating body to swing about the point. Accordingly, the equilibrium is stable if the metacentre liesabove. Provided that is small, the distance GM is given by:

Where is very small:

Hence,


Where is in circular measure, Substituting for from Eq. 2-1 gives:

Eq. Error! No text of specified style indocument.-14The dimension is called the metacentric height. In the experiment described below, it ismeasured directly from the slope of a graph of against obtained by moving a jockey across apontoon.

Now the distance BG may be found from the following relationship:


This gives an independent check on the result obtained experimentally by traversing a jockeyweight across the floating body.

Analytical Determination of BM:

A quite separate theoretical calculation of the position of the metacentre can be made as described

in the following paragraphs.

The movement of the centre of buoyancy to produces a moment of the buoyancy force about theoriginal centre of buoyancy. To establish the magnitude of this moment, first consider the elementof moment exerted by a small element of change in displaced volume (V), as indicated on fig. 2-


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5(a) & (b). An element of width lying at distance x from B, has an additional depth due to thetilt of the body. Its length, as shown in the plan view on fig. 2- 5(b), is L. So the volume V of the

element is:

Whereas, the element of additional buoyancy force is:

Where, is the specific weight of water equal to g. The element of moment about B produced by

the element of forceis, calculated as:

The total moment about B is obtained by integration over the whole of the plan area of the body, in

the plane of the water surface:


Where, second moment of areais defined as:

Hence,

(a) (b)

x

.x

L

D

A = L.x

Axis of symmetry

Figure Error! No text of specified style in document.-11 (a) Differential

volume V = x.L .x(b) Bottom view of the pontoon


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is about the axis of symmetry of the water plane area of the body as shown in fig. 2-5(b). We canequate this moment, calculated in Eq. 2-6, with the moment produced by buoyancy force withmoment arm.Hence, moment M can also be defined as:

Eq. Error! No text of specified style in document.-18Equating Eq. 2-6 and Eq. 2-7:


document.-19

From triangle MBB in fig. 2-4(b), we see that:

Where, is very small:

Hence,


Substitute BB from Eq. 2-9 to Eq. 2-8:


For the particular case of a body with a rectangular platform of width D and length L as in this

experiment, the second moment of area is readily found as:

Now, BM calculated using Eq. 2-4 can be campared with the theoratical value of BM calculated from

Eq. 2-10.

Experimental Procedure:

The pontoon shown in fig. 2-1 and 2-6 has a rectangular platform, and is provided with a rigid sail. A

jockey weight may be traversed in preset steps and at various heights across the pontoon, along

slots in the sail. Angles of tilt are shown by the movement of a plumb line over an angular scale, as

indicated in fig. 2-6(a).


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1. The height of the centre of gravity of the whole floating assembly is first measured, for one

chosen height of the jockey weight. The pontoon is suspended from a hole at one side of the

sail, as indicated in fig 2-6(b), and the jockey weight is placed at such a position on the line of

symmetry as to cause the pontoon to hang with its base roughly vertical.

3. A plumb line is hung from the suspension point. The height of the centre of gravity G of the

whole suspended assembly then lies at the point where the plumb line intersects the line of

symmetry of the pontoon. This establishes the position of G for this particular jockey height.

2. The position of G for any other jockey height may then be calculated from elementary statics, as

will be seen later.

4. After measuring the external width and length of the pontoon, and noting the weights of the

various components, the pontoon is floated in water.5. With the jockey weight on the line of symmetry, small magnetic weights are used to trim the

assembly to even keel, indicated by a zero reading on the angular scale. The jockey is then

moved in steps across the width of the pontoon, the corresponding angle of tilt (over a range

which is typically 8) being recorded at each step. This procedure is then repeated with the

jockey traversed at a number of different heights.

Figure Error! No text of specified style in document.-12 (a)

Floating pontoon tilted by movement of jockey weight (b)Determination of position of center of gravity

Jockey

weight

Angular

Scale

Suspension

Plumb Line


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2 Report Sheet

Name

Registration number

M T W R

Results and Calculations:

Table Error! No text of specified style in document.-5 Important given parameters

S/N Parameter Symbol Value Units

1 Weight of pontoon WP 2.430 N

2 Weight of jockey Wj 0.391 N

3 Total Weight W=WP+Wj 2.821 N

4 Fluid volume displaced V=W/ 2.821x10-

m

5 Breadth of pontoon D 201.8 mm

6 Length of pontoon L 360.1 mm7 Area of pontoon in plane of water surface A=LD 7.267x10

- m

8 Second moment of area I=LD3/12 2.466x10

-4 m

4

9 Depth of immersion OC=V/A 38.8 mm

10 Height of center of Buoyancy B above O OB=OC/2=BC 19.4 mm

Height of Centre of Gravity:

Fig. 2-7 shows schematically the positions of the centre of buoyancy B, centre of gravity G, and

metacentre M. O is a reference point on the external surface of the pontoon, and C is the point

where the axis of symmetry intersects the plane of the water surface. The thickness of the material

from which the pontoon is made is assumed to be 2 mm. The height of G above the reference point

O is OG. The height of the jockey weight above O is yj.


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Figure Error! No text of specified style in document.-13 Pontoon sketch with important pointsmarked

When the pontoon is suspended as shown in fig. 2-7 and with the jockey weight placed in the

uppermost slot of the sail, the following measurements can be made:

1. The value of OG may now be determined for any other value of yj. If yjchanges by yj, then this

will produce a change OG in OG as:

2. The vertical separation of the slots in the sail (yj) is , so OG will change in steps as

calculated below:

Fill the Table below. The values of OG calculated in this way for the 5 different heights yj of the

jockey weight.

Table Error! No text of specified style in document.-6 Changing center of gravity with Jockeyheight

S/N yj(mm) OG (mm)

1

2

3

4

5

Find OG, using method described in fig. 2-6, for maximum jockey height. Then follow step-2, as

stated above, to fill tab. 2-2.

Experimental determination of metacentric height GM:


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Fill the following table, for different jockey heights yj, measuring tilt angles , when jockey weight is

displaced from centre by distance xj:

Table Error! No text of specified style in document.-7 Jockey displacement vs. tilt angle fordifferent jockey height

y1= ______mm y2= ______mm y3= ______mm y4= ______mm y5= ______mm

S/Nxj

(mm)

(Deg.)xj

(mm)

(Deg.)xj

(mm)

(Deg.)xj

(mm)

(Deg.)xj

(mm)

(Deg.)

1 -45 --- -45 --- -45 -45 -45

2 -30 --- -30 -30 -30 -30

3 -15 -15 -15 -15 -15

4 0 0 0 0 0

5 15 15 15 15 15

6 30 --- 30 30 30 30

7 45 --- 45 --- 45 45 45

Find the GM using eq. 2-3 from the slope of the graphs (five lines on same graph) between xjand .

Before calculating GM, convert units of slope from mm/deg to mm/rad.

Find BM from the relation below:

Table Error! No text of specified style in document.-8 BM for different jockey heights yj

S/N yj(mm) OG (mm) Slope (mm/deg) Slope (mm/rad) GM (mm) BM (mm)

1

2

3

4

5


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The result calculated in Tab. 2-4 can be compared with the value computed from theory. Eq. 2-10

can be used to fine theoretical value of BM for this particular apparatus.

Table Error! No text of specified style in document.-9 % difference between theoretical andexperimental values of BM

S/NBM (mm)

TheoreticalBM (mm)

Experimental%

Difference

1

2

3

4

5

Exercise:

1. Does the movement of the plumbbob over the angular scale affect the results in any way?Consider, for instance, a plumbbob of 0.005N weight, displaced sideways through a distance of90 mm. What effect does this have, as compared with that of a corresponding displacement of

the jockey weight?

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2. What accuracy do you consider you have achieved in obtaining the analytical value of BM? If,

for example, the possible uncertainty in measuring D and L is 2 mm, what is the corresponding

uncertainty in the calculated value of BM?

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3. How would the stability of the pontoon be affected if it were floated on a liquid with a greater

density than that of water?

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________

4. What suggestions do you have for improving the apparatus?

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3. . . . . . . . . . .

. . . . . . . . . . . . . . .

Analyze flow over

(Apparatus # FM05)

Figure Error! No text of specified style in document.-14 Hydraulic Bench, rectangular andtriangular weirs

Objective:

The objective of this experiment is to observe the characteristics of flow over a

Rectangular Notch and a Triangular or Vnotch, and to determine the Coefficient of

Discharge.

Apparatus:

The apparatus of this experiment includes a Hydraulic bench, Stilling baffle,

Vernier, V-notch plate, Rectangular notch plate, Stop watch.

Exp

eriment3


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Summary of Theory:

1. This lab is based upon the principle of conservation of mass.

2. Coefficient of discharge, Cd,is used in conjunction with an ideal velocity (or inviscid velocity) to

calculate the flow rate, through restricted passages. However, to determine Cdfor a device (i.e.

to calibrate the device), one needs to measure a key variable, H, for known flow rates, Q.

3. Generally, Cd accounts for the effect of contraction (which you will observe), velocity of

approach, viscosity, and surface tension. In other words, both the characteristics of the device

and the properties of the fluid influence the value of Cd.

Equipment Setup:

Rectangular Weir:

A weir is an obstruction in an open channel over which liquid flows. The discharge over the weir is a

function of the weir geometry and the head on the weir. The head on the weir is defined as the

vertical distance between the weir crest and the liquid surface taken far enough upstream of the

weir to avoid local free surface curvature as shown in fig. 3-2(a).

Figure Error! No text of specified style in document.-15Experimental Setup

Delivery

Nozzle

Stilling Baffle

Instrument

CarrierHook and

Point

Vernier

Scale

Locking and

Adjustment Nuts

Weir

Carrier


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The basic discharge equation for the weir is derived by integrating the following equation over the

total head on the weir:

Eq. Error! No text of specified style indocument.-22In Eq. 3-1, L is the length of the weir and V is the velocity at any given distance h below the free

surface. Neglecting streamline curvature and assuming negligible velocity of approach upstream of

the weir, we obtain an expression for V by writing Bernoullis equation between a point upstream of

the weir and the point in the plane of the weir as shown in fig. 3-4(a). Bernoullis equation between

point 1 and 2 will be as follow:

Where, P1and P2 are atmospheric pressures will be cancelled out from both sides, as referencepressure is atmospheric pressure. V1is assumed to be zero and height H 1 is equal to H and H2 is

equal to (H-h). Heights are measured from the reference elevation, crest of the weir. Velocity at

point 2 will be V2equal to V. The Bernoullis equation will reduce to:


document.-23

Velocity at the exit will depend upon variable height h.

(a) (b)

L

H

Figure Error! No text of specified style in document.-16 (a)Flow over rectangular weir (b) Front view


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From Eq. 3-1,


Substituting V from Eq. 3-2:

Integrate the above equation over the height of the weir to get total flow rate through the rectangular

weir.

Integrating will result in the following theoretical flow rate:


document.-25

Discharge coefficient Cdis defined as:



The actual value of flow rate can be calculated as:

Triangular Weir:

Figure Error! No text of specified style in document.-17 (a)

Velocity distribution over rectangular weir (b) Differential area

dA

(b)

L

Hh

Hh

dh

V=1

2

(a)

dA=Ldh


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The primary advantage of the triangular weir is that it has a higher degree of accuracy over much

wider range of flow than does the rectangular weir, because the average width of the flow section

increases.

Figure Error! No text of specified style in document.-18 (a) Triangular Weir (b) Change in weirlength with height

The discharge equation for the triangular weir is derived in the same manner as that of rectangularweir. The differential discharge from Eq. 3-3 is integrated over the total head on the weir. Thus we

have:

This integrates to:


document.-28

However, coefficient of discharge can be calculated from Eq. 3-5.

Hence we have:


document.-29

Actual flow rate will be:

Description:

The apparatus consists of five basic elements used in conjunction with the flow channel in the

molded bench top of the Hydraulics Bench as shown in fig. 3-1 and 3-2. A quick release connector

in the base of the channel is unscrewed and a delivery nozzle is screwed in its place. A stilling baffle

is slid into the slots in the walls of the channel. The inlet nozzle and stilling baffle in combination

promote smooth flow conditions in the channel. A vernier hook and point gauge is mounted on an

instrument carrier, which is located on side channels of the molded top. The carrier may be moved

along the channels to the required measurement position. The gauge is provided with a coarse

adjustment locking screw and a fine adjustment nut. Vernier is locked to the mast by screw and is

used in conjunction with the scale. The hook and point is clamped at the base of the vertical mast


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by a thumb screw. The rectangular notch weir or V notch weir to be tested is clamped to the weir

carrier in the channel by thumb nuts. The weir plates incorporate captive studs to aid assembly.

Technical Details:

Table Error! No text of specified style in document.-10 Technical details

S/N Parameter Value Units

1 Weir plate height 160 mm

2 Weir plate width 230 mm3 Weir plate thickness 4 mm

4 Rectangular notch height 82 mm

5 Rectangular notch width 30 mm

6 Angle of V-notch 90 degree

7 Hook-point gauge range 0-150 mm

Procedure:

1. Ensure that the hydraulics bench is located on a level floor, as the accuracy of the results will

be affected if the bench top is not level. Set up the equipment as shown in the diagram.

2. Set vernier height to a datum reading by placing the point on the crest of the weir / at the bottom

of the V notch on the weir. Take extreme care not to damage the weir plate with the pointgauge.

3. Position the gauge about half way between the notch plate and stilling baffle. Admit water to the

channel and adjust flow control valve to obtain heads, H, increasing in steps of 10mm.

4. For each flow rate, stabilize conditions, measure and record H. Take readings of volume and

time using the volumetric tank to determine the flow rate.

3 Report SheetName


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Registration number

M T W R

Observations & Calculations:

For Rectangular notch:

Height of notch = mm Breadth of notch = mm

Table Error! No text of specified style in document.-11 Readings for Rectangular weir

S/NHeadH(m)

TimeT(s)

VolumeV(m

3)

Flow RateQac(m

3/s)

H3/2

Flow RateQth(m

3/s)

Log Q Log H Cd

1

2

3

4

5

Average Value of Cdfor Rectangular weir= __________

For Triangular notch:

Angle of notch = mm

Table Error! No text of specified style in document.-12 Reading for Triangular weir

S/NHeadH(m)

TimeT(s)

VolumeV(m

3)

FlowRate

Qac(m3/s)

H5/2

Flow RateQth(m

3/s)

Log Q Log H Cd

1

23

4

5

Average Value of Cdfor Triangular weir= __________

Exercise:

1. Starting from Eq. 3-4 and 3-5 separately, derive equations of the form:

What will be the expressions for m and c?

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_______________________________________________________________________________

_______________________________________________________________________________

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_______________________

2. Plot Log Q vs. Log H for both Rectangular and Triangular weirs on the same graph below.

3. Compare y-intercept of the above graph with the expressions obtained in question #1 forconstant c. Find value of Cdfor both weirs.

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_______________________________________________________________________________

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4. For both weirs, plot Cdagainst H.

5. For both weirs, compare the average value of Cdobtained from the table and Cdcalculated in

question #3.

Table Error! No text of specified style in document.-13 Cdcomparison

S/N Weir CdAverage Cdfrom Graph % Difference

1 Rectangular

2 Triangular

6. Is Cdconstant for the weirs in this experiment? If yes, comment. If not, why?

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4. . . . . . . . . . .

. . . . . . . . . . . . . . .

Determining frictionfactor for pipes(Apparatus # FM02)

Experiment4


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Figure Error! No text of specified style in document.-19 Fluid friction apparatus

Objective:

The objective of this experiment is to determine the friction factorf as a function of

Reynolds Number for the smooth and rough pipes and compare them withempirical data contained in Moody chart.

Apparatus:

Fluid friction apparatus, Stop watch and vernier calipers

Head Loss for Fluid Flowing in the Pipe:

The overall head loss for the pipe system consists of the head loss due to viscous

effects in the straight pipes, termed the major loss and the head loss in the various

pipe components, termed the minor loss. Hence, overall head los is defined as:

Major Losses:

In this experiment our focus will be on Major Losses through the pipes. Minor Losses through other

components of piping e.g. valves elbows and fittings will come in greater detail in experiment #8 and

will not be discussed further in this experiment. Though there are many types of losses, yet the

major loss is due to shear stresses (w) between the fluid and pipe surface. The shear stress of the

pipe depends upon the roughness of the inside of the pipe. Shear stress for the turbulent flow is a

function of a fluid density (), whereas, for laminar shear stress is independent of density of the

fluid. In case of laminar flow, viscosity () is the only important fluid property. For laminar flow

pressure drop or head loss is independent of roughness of pipe but for turbulent flow there is a very

thin viscous sublayer formed in the fluid near the pipe wall. Thus for the turbulent flow the pressure

drop is expected to be function of the wall roughness. So turbulent flow properties depend on the

fluid density and the pipe roughness.

Critical Velocity

If the velocity of fluid inside the pipe is small, streamlines will be in straight parallel lines. As the

velocity of fluid inside the pipe gradually increase, streamlines will continue to be straight and

parallel with the pipe wall until velocity is reached when the streamlines will waver and suddenly

break into diffused patterns. The velocity at which this occurs is called "critical velocity". At velocities

higher than "critical", the streamlines are dispersed at random throughout the pipe.


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The regime of flow when velocity is lower than "critical" is called laminar flow (or viscous or

streamline flow). At laminar regime of flow the velocity is highest on the pipe axis, and on the wall

the velocity is equal to zero as shown in fig. 4-2.

Figure Error! No text of specified style in document.-20 Laminar and Turbulent Flow throughpipes and velocity profiles

When the velocity is greater than "critical", the regime of flow is turbulent. In turbulent regime of flow

there is irregular random motion of fluid particles in directions transverse to the direction on main

flow. Velocity change in turbulent flow is more uniform than in laminar. In the turbulent regime of

flow, there is always a thin layer of fluid at pipe wall which is moving in laminar flow. That layer is

known as the boundary layer or laminar sub-layer.

Professor Osborne Reynolds demonstrated that two types of flow may exist in a pipe:

1. Laminar flow at low velocities where (h u)2. Turbulent flow at higher velocities where (h un)Where h is the head loss due to friction and u is the fluid velocity. These two types of flow are

separated by a transition phase where no definite relationship between h and u exists.

Graphs of h versus u and log h versus log u show these zones. You will be asked to plot these

curves using your data and identify the low and high limits of critical velocity. Graphs of h vs u and

Log h vs Log u will look like fig. 4-3 (a) and (b), respectively.

Laminar Flow Turbulent Flow

Laminar Flow

Turbulent Flow

Smooth pipe

Rough pipe

V average


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Figure Error! No text of specified style in document.-21 (a) h vs. u (b) Log h vs. Log u

Friction Factor (f):

Friction factor (f) is a dimensionless quantity. For horizontal circular pipe, the friction factor f is

calculated by the following formula:


Where

= Length of pipe between tapings (m) = 1m for both pipesD= Internal diameter of the pipe (m)

u = Mean velocity of water through the pipe (m/s)

g = Acceleration due to gravity (m/s2) = 9.81 m/s

2

= Friction factorA large body of data exists on pressure drop in pipes of circular and noncircular crosssections.

This information is summarized empirically in the Moody friction factor chart shown in fig. 4-4. The

friction factoris plotted against Reynolds Number (Re) on LogLog scale. The plot clearly exhibitsthree flow regimes, laminar, transitional, and turbulent.

For laminar flow, friction factor, irrespective of the pipe roughness, is related to Re as:

Whereas, for turbulent Flow, friction factor depends upon relative roughness of the tube (/D) andvalue of Re, but for large value of Re, becomes independent of Re. In the transition zone, thebehavior is similar to that of turbulent regime.Moody chart:

Moody chart gives the friction factor in terms of Reynolds Number and relative roughness, because

it is possible to measure the effective relative roughness of typical pipes and thus to obtain the

friction factor by Moody chart.

(a) (b)


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Figure Error! No text of specified style in document.-22 Moody Diagram

Procedure to Conduct Experiment:

1. Calibrate the Rotameter by using a stop watch to record the time it takes to increase thevolume of the water in the tank from 0 to x. record the time/volume, compare collection flow rate

versus the Rota

meter flow rate reading. The system flow rate is changed by opening or closing

the V2 valve while valve V6 is completely open.

2. Measure the internal diameter of each test pipe sample using a set of calipers.

3. Close the inlet flow control valve V2 and open the outlet flow control V6.

4. Choose the appropriate pipe for the measurement of pressure drop. Open and close the

appropriate valves to obtain flow of water through the required test pipe.

5. For example: If you choose pipe 2, you should open V4 in pipe 2 and close V4 in Pipe 1, V4 in

pipe 3 and 7 in pipe 4.

6. Start the pump; use black button to turn it on and red to turn it off.

7. Gradually open the inlet flow control valve to allow water to flow along the test pipes and into

the volumetric tank.

8. Adjust V2 and V6 to obtain a suitable flow rate.

9. Take three measurements at three to four different flow rates for each pipe and the valves,meters, or bends.

10. To stop the operation, completely open valves V6 and V2, then switch off the pump.

Notes:

1. Make sure that there are no air bubbles in the pipe while the experiment is running.

2. The valves are 7, 10, and 11. The meters are the Pitot tube 16, the Venturi meter 17, and the

orifice meter 18. The bends are 9, 13, and 14.

3. For reading on mercury manometer, convert head of Mercury (Hg) to head of H2O. Specific

gravity of Hg is 13.6.


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Equipment Setup:

Figure Error! No text of specified style in document.-23 Fluid friction apparatus

4 Report Sheet


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Name

Registration number

M T W R

Observations and calculations:

Students are advised to take ten readings for each pipe. Fewer readings will result in insufficient

data to plot required graphs.

For smooth pipe:

Internal Diameter (D) = ______mm Pipe Area (A) = ______m2 Length of pipe () = _____m

Table Error! No text of specified style in document.-14 Readings for smooth pipe

S/NVolumeV(m

3)

TimeT(s)

Flow RateQ(m

3/s)

u=Q/A(m/s)

h(m of H2O) Log u Log h 1

2

3

4

5

67

8

9

10

For rough pipe:

Internal Diameter (D) = ______mm Pipe Area (A) = ______m2 Length of pipe () = _____m

Table Error! No text of specified style in document.-15 Readings for rough pipe

S/NVolumeV(m

3)

TimeT(s)

Flow RateQ(m

3/s)

u=Q/A(m/s)

h(m of H2O) Log u Log h 1

2

3


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4

5

6

7

8

9

10

It is assumed that the dynamic viscosity is 1.15 X 10 -3Ns/m2at 15C and the density is 999

kg/m3at 15C. However, you should measure the actual temperature of the water and adjust these

values accordingly.

Exercise:

1. Plot a graph of h versus u for the two pipes on the same graph paper. Identify the laminar,

transition and turbulent zones on the graph.

2. Plot a graph Log h versus Log u for the two pipes on the same graph paper. Determine the

slope of the straight line to find n for turbulent flow (see fig. 4-3).


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_______________________________________________________________________________

_______________________________________________________________________________

___________________________________________________________________

Table Error! No text of specified style in document.-16 Comparison of velocity exponent n forrough and smooth pipe

Flow Condition n for smooth pipe n for rough pipe Percentage Difference

Turbulent

3. From exercise #2, for the two pipes, determine the slope of the graph for both laminar andturbulent flows.

Table Error! No text of specified style in document.-17 Slope comparison

Slope of graph

S/NFlow Condition

Smooth Pipe(S1)

Rough Pipe(S2)

(S1>S2) or(S1


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5. Determine the pressure drop for the two pipes at the lowest and the highest flow rates. Express

your results in kPa (not in m of H2O).

Table Error! No text of specified style in document.-18 Pressure Drop comparison for both pipes

Pressure Drop (kPa)

FlowCondition

Flow Rate(m

3/s)

Smooth Pipe(P1)

Rough Pipe(P2)

PercentageDifference

Lowest

Highest

Comments:_______________________________________________________________________________

_______________________________________________________________________________

____________________________

6. On the Moody diagram (fig. 4-4), plot the points of friction factor against Reynolds number for

the two pipes. Compare the values of for both pipes (two values for each) in Laminar flow withthat of values on Moody diagram.

Table Error! No text of specified style in document.-19 Friction Factor comparison

S/N from Tab. 4-1 & 4-2 from fig. 4-4 PercentageDifference

1

23

4

Comments:

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

________________________

7. Using Moody Diagram (fig. 4-4), for the rough pipe in the turbulent flow regime, determine thevalue of relative roughness.Table Error! No text of specified style in document.-20 Relative roughness for rough pipe

S/N from Tab. 4-2 1

2

3

4


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5

. . . . . . . . . . . . . . . . . . . . . .

. . . .

Observing momentumTransfer (Apparatus #

FM07)

Figure Error! No text of specified style in document.-24 Impact of jet apparatus

Objective:

The objective of this experiment is to verify the theory of conservation of momentum in fluid

mechanics, and to measure the reaction force developed by a jet on different surface profiles.

Experiment

5


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Equipment Setup:

Figure Error! No text of specified style in document.-25 Experimental Setup of Jet apparatus

Summary of theory:

Starting from Newtons second Law of motion, it can be said that time rate of change of linear

momentum of a system is equal to the sum of all the external forces acting on system. Whereas,

time rate of change of linear momentum of the system is divided into two parts, one is the time rate

of change of linear momentum of the content of the control volume and second is the net rate of

flow of linear momentum in or out of the control surface. As the particles moves into and out of the

control volume (cv) through the control surface (cs), they carry linear momentum with them.

Hence, Newtons second Law of motion can be described as:

For steady state condition, the above equation in y-direction will reduce to:


The above equation is called general linear momentum equation. Linear momentum is a vector

quantity so it can have components in three orthogonal coordinate directions. The flow of the linear

momentum into the control volume involves a negative product. Momentum flow out of the CVinvolves a positive product. The correct algebraic sign (+ or ) to assign to momentum flow willdepend on the sense of the velocity. For steady system the time rate of change of linear momentum

of the control volume is zero.

Derivations:

It is assumed in the following derivations that the leaving streamlines are parallel to the edge

surface of the target. Targets used in this experiment are shown in fig. 5-3.

Weight pan

Target plate

Nozzle

Inlet pipe

Level gaugeSpring

Knurled screw


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Case 1 - Flat Target at 120oto the jet:

Using fig. 5-3 (a) and fig. 5-4, apply Eq. 5-1 in the direction of force vector R:

Where,

Substitute R in this equation:

Where, (cos30o) is equal to, and (V1A) is equal to the flow rate (Q) out of the nozzle.

Hence,


(a) (b) (c)

Figure Error! No text of specified style in document.-26 (a) Target

at 120o(b) Flat target (c) Hemispherical target


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Where, RYis the force exerted by the water jet to the 120oflat target and A is area of the nozzle.

Case 2 - Flat Target at right angles to the jet:

Using the same method as applied in Case 1, the following relation for resultant force applied on the

flat target is obtained:


Case 3 - Hemispherical Target:

In this case, the inlet velocity to the control volume is equal and opposite of the exit velocity. Using

the same method as applied in Case 1, the following relation for resultant force applied on the flat

target is obtained:


Technical details:



1 Nozzle diameter 8 mm

2 Distance b/w nozzle & target plate 20 mm

3 Diameter of target plate 36 mm

4 Target plates Hemispherical, 120o,90

o ---

Description:

Figure Error! No text of specified style in document.-27 (a) 120otarget (b) Inlet velocity components (c) Force components

RY

R

V2

V2

V1

RYR

30o

V130

o

V1cos30o

(a) (b) (c)


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This equipment allows the force developed by a jet of water impinging upon a stationary object to be

measured. The apparatus consists of a cylindrical clear acrylic fabrication which is positioned in the

bed of the bench top channel and the inlet pipe connected to the bench supply. The feet are

adjustable so that the apparatus can be leveled with the aid of the spirit level.

Water is fed through a nozzle and discharged vertically to strike a target carried on a stem which

extends through the cover. After striking the target plate, water leaves through the outlet holes in the

base. An air vent is provided so that the interior remains at atmospheric pressure.

Procedure:

1. Remove the top plate and transparent casing, measure the nozzle diameter. You may use

the diameter value provided by the manufacturer which is printed on the apparatus.

2. Place the flat target on the rod attached to the weight pan.

3. Reassemble the apparatus. Connect the inlet pipe to the bench, with the apparatus in the

open channel.

4. Level the base of the apparatus with the top plate loosely assembled.

5. Screw down the top plate to datum on the spirit level.

6. Adjust the level gauge to suit the datum on the weight pan.

7. A nominal mass (M) is placed on the weight pan, water is allowed to flow by operating the

control valve on the bench. The flow rate (Q) is then adjusted until the weight pan isadjacent to the level gauge. When testing for level, the weight pan should be oscillated to

minimize the effect of friction.

8. Take readings of volume (V) and time (T) to find the flow rate (Q). Note the mass on the

weight pan (M).

9. Repeat with additional masses on the weight pan. Repeat the above using the 120o

target

and the hemispherical target.


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5 Report SheetName

Registration number

M T W R

Observation and calculations:

Nozzle diameter = _______mm Area of nozzle: _________m2

Table Error! No text of specified style in document.-22 Reading for different targets andcomparison between forces

Target S/NVolumeV(m

3)

TimeT(s)

FlowRate

Q(m3/s)

Q2

Force onTargetRY(N)

MassM(g)

WeightW(N)

% ForceComparison

120o

Target

1

2

3

4

5

90oT

arget 1

2

3

4

5

Hemisphe

rical

1

2

3

4

5

Exercise:

1. Starting from Eq. 5-1, drive Eq. 5-3 and Eq. 5-4. Illustrate with the help of figures.


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2. For each target, plot W against Q2and compare with the theoretical slopes.

Table Error! No text of specified style in document.-23 Comparison

S/N Target Slope from formula Slope from graph%

Difference

1 120oFlat Plate

2 90oFlat Plate

3 Hemispherical


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3. If the slopes do not compare well, explain the reasons for the observed differences in the light

of your observation of the flow field and the assumptions made in deriving the expressions for

RY.

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4. For a given balancing mass M, compare the flow rates needed for each target achieve the

balance.

Table Error! No text of specified style in document.-24 Comparison between different targets

S/N Target Mass M Flow Rate Q Remarks1 120

oFlat Plate

2 90oFlat Plate

3 Hemispherical

Comments:

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6

. . . . . . . . . . .

. . . . . . . . . . . . . .

Determining dischargecoefficient of an orifice(Apparatus # FM09)

Figure Error! No text of specified style in document.-28 Orifice & jet apparatus

Objective:

The objective of this experiment is to determine the discharge coefficient

(Cd) for a small orifice for constant and varying heads.

Experimen

t6


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Equipment setup:


Orifice:

An opening, in a vessel, through which the liquid flows out, is called an orifice. The difference

between the weir and orifice is that for orifice liquid on the upstream side is always above the top

edge of the orifice. The usual purpose of the orifices is the measurement of flow. Due to friction the

real velocity will be somewhat lower than theoretical value.

Vena Contracta:

If the exit of the orifice is not smooth well contoured nozzle, but rather a flat plate, then the diameter

of the jet will be less than the diameter of the hole. This phenomenon, called a vena contracta

effect.

Jet:

A jet is a coherent stream of fluid that is projected into a surrounding medium (at atmospheric

pressure), usually from some kind of a nozzle or aperture. The exit pressure for an incompressible

fluid jet is equal to the surrounding pressure.

Coefficient of Contraction (CC):

The venacontracta is a function of geometry of the outlet resulting in contraction of the jet stream

as it moves out of the orifice. Contraction coefficient can always be measured experimentally if we

can measure the area of the jet stream after vena contracta. Following relation is used to calculate

CC:

Orifice plate with O ring

Scale Head tank

Adjustable

overflow

pipe

Locknut

Flexible

hose to

sump tank

Inlet pipe

Baffle

Thumb

nut

Screw

Needle

Paper clamp Backboard

Adjustable feet


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Coefficient of Discharge (Cd):

The discharge coefficient of orifice is used to account for nonideal effects. These nonideal effects

arise due to two reasons, first, the vena contracta and second, the swirling flow and turbulent

motion near the orifice plate. It can be determined experimentally. For a sharp edged opening it may

be as low as 0.6. For smooth orifices it may be between 0.95 and 1.

For constant head:

Orifices discharge coefficient, when tank head is maintained at constant, can be calculated from the

following relation:


Where, theoretical flow rate out of the orifice is calculated as:


Correlation for velocity through the orifice can be derived from Bernoulli s equation. Applying

Bernoullis equation between tank free surface (1) and orifice outlet (2) will result in the following

equation (see fig. 6-3):


Hence, after substituting Eq. 6-2 and Eq. 6-3 in Eq. 6-1, the discharge coefficient (Cd) for constant

head H will be as follows:

( ) Eq. Error! No text of specified style indocument.-37

Where, AOis the area of the orifice:

DO

1

2

H

Figure Error! No text of specified

style in document.-30 Water tankwith constant head H


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For varying head:

Figure 6-4 shows water tank with an orifice of diameter DO. Water drains out of the tank through the

orifice resulting in drop of water head (h). In small interval of time (dt), let the liquid level fall down by

an amount (dh). Therefore, volume of the liquid (dq) that has drained out of the tank in this time (dt)

is estimated as:


document.-38

Where, ATis area of the tank with diameter DT, calculated as:

Theoretical volume of liquid that has passed through the orifice in this small interval of time (dt) is

estimated as:

Actual value of the drained out liquid volume will be:


Equating Eq. 6-5 and Eq. 6-6 results:

After rearranging the above equation, it can be shown that the time required (dt) to drain the tank

from head h1to h2is given as:

Integrating above equation:

Figure Error! No text of

specified style in document.-31Water tank with variable head

DOh

dh

DT


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This on simplification gives discharge coefficient for variable head:

( ) Eq. Error! No text of specifiedstyle in document.-40

Technical details:



1 Orifice diameter 3 and 6 mm

2 Jet trajectory probes 8 Needles

3 Maximum constant head 410 mm

Procedure:

1. Measure the orifice diameter, removing the orifice plate if necessary; also measure the internal

dimensions of the header tank. Or use the values provided.

2. Connect the apparatus to the bench, leveling by adjusting the feet, ensuring the overflow piperuns into the sump tank.

3. Raise overflow pipe to a suitable level, release water into the head tank. Control the flow until

the water is just spilling into the overflow. Record the head h on the scale, measure the flow

rate using the volumetric tank, or by intercepting the jet with a measuring cylinder. Repeat for

different water levels.

4. For flow under varying head, the overflow pipe is raised to obtain maximum head, the header

tank filled to overflow level and the inlet water feed closed. Start a stopwatch when the level

reaches the first convenient scale mark (noted as h1). Take a reading of head h2at 20secondintervals or inter

Documents

ME343 Lab Manual