BY: Almoutaz Alsaeed

Cairo University

Faculty of Engineering

Civil Engineering Department

1. Weir Experiment (Rectangular and Triangular)

Objectives of the Experiment1. To demonstrate the flow over different weir types.2. To calculate the coefficient of discharge for different weir types.3. To study the variation and dependence of the relevant parameters.Theory

For the rectangular weir:Q=( 2

3 )CdB√2g H1.5

For the triangular weir:Q=( 8

15 )Cd tan ¿

where Cd = Coefficient of discharge

B = width of the rectangular weir (3 cm)

H = head above the weir crest or apex

θ = angle of the triangular weir

g = acceleration of gravity

Experimental Setup

Procedures and Readings 1. Make sure that the Hydraulic Bench is levelled.2. Set the Venire on the point gauge to a datum reading by placing the tip of the gauge on the crest or the apex of the weir. Take enough care not damage the weir plate and the point gauge.3. Put the point gauge half way between the stilling baffle plate and the weir plate.

4. Allow water to flow into the experimental setup and adjust the minimum flow rate by means of the control valve to have atmospheric pressure all around water flowing over the weir. Increase the flow rate incrementally such that the head above the weir crest increases around 1 cm for each flow rate increment

5. For each flow rate, wait until steady condition is attained then measure and record the head(H) above the weir.

6. For each flow rate, measure and record the initial and final volumes in the collecting tank and the time required to collect that volume. For each flow rate, take 3 different readings ofthe volumes and time and record the averages

For the rectangular weir:Q=Cd .

23. B .√2g H

32

Reading Crest level(C.L) (cm )

Water level (W.L)(cm)

Initial volume (I.V)(liter)

Final volume (F.V)(liter)

Time (T) (sec)

1 4 7.45 0 10 422 4 5.73 15 18 723 4 6.92 15 20 334 4 7.29 15 25 445 4 7.67 15 30 40

ReadingVolume=F.V.-I.V.(liter)

H=W.L.-C.L.(cm)

Time(sec)Q=volume/time(cm3/s)

Log QLog H

Cd

1103.4542238.09523

82.3767507

10.53781916.4080905

90.27960

911231.737241.666666

71.6197887

60.23804612.2754597

30.13779

988

352.9233151.515152

2.18045606

0.46538285

4.98969819

0.22851306

4103.2944227.272727

2.35654732

0.51719595.96751950.28660431

5153.67403752.57403127

0.56466606

7.03070857

0.40138526

slope=23∗Cd∗3∗√2∗981 Cd=

slope∗32

3∗√2∗981=0.46032

logQ=log( 23∗Cd∗3∗√2∗981)+ 3

2logH

Y=C+ 32X

0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.60

0.5

1

1.5

2

2.5

3

Y=0.950+2.735X

log H

logQ

the coefficient of X should be = 1.5 but it has another value which is (2.735) due to errors in readingsand from this we can calculate Cd❑

log ¿10log ¿¿

Cd=0.30873

2 3 4 5 6 7 80

50100150200250300350400

f(x) = 40.7818412385929 xR² = 0.949181087739477

slope=40.78

H^1.5 (cm)

Q (C

M3/

S)

1.5 2 2.5 3 3.5 4 4.5 50

50100150200250300350400450500

f(x) = 40.4459707872977 x^1.60789331485007R² = 0.996096937849252

k=40.44n=1.607

H

Q

K=23∗3∗Cd∗√2∗981 n=1.5

Cd is not constant

the normal range should be between (0.6-0.9) which is not valid bec the used hydraulics bench For the triangular weir:

Q=( 815 )Cd tan ¿

Reading Crest level (C.L) (mm)

Water level (W.L) (mm)

Initial volume (I.V)(liter)

Final volume(F.V) (liter0

Time (T) (sec)

1 40 76.7 15 30 402 40 74.5 0 10 423 40 72.9 15 25 404 40 69.2 15 20 335 40 57.3 15 18 74

Reading

Volume=F.V-I.V (liter)

H=W.L-C.L (cm)

TimeQ=volume/time

Log Q

Log HH 2.5

 Cd

(sec)(cm3/s)

1153.67403752.57403127

0.56466606

25.8027005

0.16405392

2103.4542238.09522.37675071

0.5378191

22.1079125

0.12156918

3103.29402502.397940.5171919.633130.14373

0015992772452.9233151.51522.18045

6060.46538

28514.56991

870.11738

685

Q=( 815 )Cd tan ¿

slope= 815

∗tan 45∗√2∗981∗Cd

Cd=

slope∗158

√2∗981=¿0.5985

the normal range should be between (0.6-0.9) which is not valid beck the used hydraulic bench

0 5 10 15 20 25 300

50

100

150

200

250

300

350

400

Slope=14.1399414

H^2.5

Q

logQ=2.5 logH+ log(Cd∗815

∗tan 45∗√2∗981)Y=2.5 X+C

100.923=Cd∗815

∗tan 45∗√2∗981

Cd=0.3545

1.5 2 2.5 3 3.5 40

50100150200250300350400

f(x) = 8.39108257077928 x^2.80571201638616R² = 0.980831306543745

n=2.805k=8.391

H

Q

Comment:Cd Is not constant

But the values of Cd should be close to each other

0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.60

0.5

1

1.5

2

2.5

3

f(x) = 2.80571206919487 x + 0.923817957425307R² = 0.98083129661639

logH

logQ

2. Impact of Jet

Objective of the Experiment

To demonstrate and investigate the validity of theoretical expressions for the calculation of the force exerted by a jet on objects of various shapes

Theory

From momentum principle

F y=ρQ (vout−v¿) , in our case θ=¿90 flat plate (90)

Then vout=0

Then F y=ρQ v¿ force of water on plate ,

v¿=vnozzle=QA, A=π∗(0.8)2

4=0.50265 cm2 , F y=ρ Q

2

A , M=ρ Q2

gA

Experimental Setup

1. The impact of jet apparatus is placed above the regular Hydraulic Bench as shown in the photographs.

2. A stopwatches.

Procedures and Readings1. Remove the stop plate and transparent casing to measure the nozzle diameter and place theflat plate (90º) on the rod attached to the weight pan. Then, reassemble the apparatus.2. Connect the inlet pipe of the apparatus to the outlet of the Hydraulic Bench.3. Level the base of the apparatus using the bubble balance.

4. Screw down the top plate to datum on the spirit level.5. Adjust the level gauge to suit datum on the weight pan.6. Add masses to the weight pan. Allow water to flow in the experiment and adjust the flowby the control valve of the Hydraulic Bench so that the pan will be re-adjacent to the levelgauge.7. Before taking readings the weight pan should be oscillated upwards and downwards androtated to minimize the effect of friction.8. Take the readings of the initial and final volumes and the time of accumulation.9. Record the masses on the weight pan.10. Repeat the experiment for different masses on the weight pan

Reading Mass on weight pan M (gm)

Initial volume (I.V.) (litre)

Final volume (F.V.) (litre)

Time(T) (Sec)

1 110 1 6 342 190 1 6 283 210 1 6 224 330 1 6 165 640 1 6 11

coefficient of impact

theoretical mass

Mass on weight pan M (gm)

Volume= F.V.-I.V (Liter)

Time (Sec)

Q=volume/time (cm3/s)

Q2

2.5081021 43.8578644 110 5 34 147.058824 21626.2976

2.938085 64.6679735 190 5 28 178.571429 31887.75512.0047459 104.751428 210 5 22 227.272727 51652.89261.6662823 198.045669 330 5 16 312.5 97656.251.5274255 419.005713 640 5 11 454.545455 206611.57

slope= ρgA

=0.00273738

0 50000 100000 150000 200000 2500000

200

400

600

800 slope=0.00273738

Q^2

M

Comment Coefficient of impact is greater than 1 (in our case)

And the real value of ρgA

=0.00203 so close to the value of the slope

3. Flow through Sharp Edged Orifice.-

Objective of the Experiment

1. To study the path of water jets issuing from orifices.2. To determine the coefficients of discharge, velocity and contraction from a sharp-edged circular orifice.3. To study the variation and dependence of the relevant parameters

TheoryThe coefficient of discharge Cd is the ratio of the actual discharge Q act to the theoretical discharge Qth.The theoretical discharge is given by the following relationship where A is the area of the orifice and His the total head on the orifice centerline and the actual discharge can be measured.

Qth=A√2gH Cd=Qact

Qth<1

The Path of the jet from the orifice is given by the following equation where x is the horizontal distance, y is the vertical distance and v is the flow velocity from the orifice

X=V act t. y=0.5g t 2

y=0.5g x2

vact2 from bernolli v th=√2 gH

y=0.5g x2

cv22 gH

CV=X

2√ y∗H

Experimental setup1. The orifice plate apparatus is placed above the regular Hydraulic Bench as shown in the photographs.2. A stopwatch is needed.3. The adjustable stainless steel overflow pipe near the top of the tank is used to adjust the level of water in the tank..

Procedures and Readings

1. Turn on the pump of the hydraulic bench and allow water into the constant head tank to build up above the orifice. Wait until steady condition is achieved

2. You can control the level of the water into the constant head tank by pulling up and down the adjustable stainless steel overflow pipe as shown in the photograph.

3. Measure the head (H) above the orifice using the graduated scale.

4. By setting the thin pins so that they just touch the issuing water jet, draw the path of the water jet on the given graph paper.

5. Measure and record the initial and final volumes and the time of accumulation for each reading of head (H).

6. Repeat the previous steps for at least four more different heads (H) by changing the position of the adjustable stainless steel overflow pipe.

Point 1 Point 2 Point 3 Point 4

Point 5

Point 6

H(cm)

Initial volum

Final volum

v Time

X (cm)

5 10 15 20 25 30 33.5

4 10 6 2.6 min

Y (cm)

0.2 0.7 1.6 3 4.7 7.2

X2 25 100 225 400 625 900

C v= (x2 / 4yH

0.96584

1.03252

1.02441

0.9975

0.9961

0.9658

).5

V th=√2gH=256.372cm / sec

Qth=AV th=72.48 cm3/sec , A=π 0.62

4=0.282743 cm2

Qact=6∗1000

156=38.46 cm3/sec

Cd=38.4672.48

=0.530

the normal range should be between (0.6-0.9)Slope=4HC v

2=128.4CV=0.9788814

Cc=CdCv

= 0.5300.9788814

=0.54143

0 1 2 3 4 5 6 7 80

100200300400500600700800900

1000

f(x) = 128.487561032318 xR² = 0.998757815074951

Y

X^2

Point 1 Point 2 Point 3 Point 4

Point 5

Point 6

H(cm)

Initial volume

Final volume

v Time (sec)

X (cm)

0 5 10 15 20 2536.

7 9 2 72.7

4

Y (cm)

0 0.5 1.1 2.7 4 6.15

X2 0 25 100 225 400 625

C v= (x2 / 4yH).5

0 1.172018 1.580349 1.51306 1.65748 1.67090

V th=√2gH=267.239 cmsec

Qth=AV th=75.56 , A=π 0.62

4=0.282743

Qact=2∗1000

72.7=27.5cm3/sec

Cd= 27.575.56

=0.36399

the normal range should be between (0.6-0.9)Cc=Cd

Cv= 0.36399

0.707854=0.5142

0 1 2 3 4 5 6 70

100200300400500600700

slope=103.0636 Cv=0.707854

Y

X^2

Point 1 Point 2 Point 3 Point 4

Point 5

Point 6

H(cm)

Initial volume

Final volume

v Time

X (cm)

0 5 10 15 20 2538.9

6 8 2 55.1

Y (cm)

0 0.5 1 2.6 2.9 5.8

X2 0 25 100 225 400 625

C v= (x2 / 4yH).5

0 1.13373147

1.603338 1.491522

1.883025

1.66437

V th=√2gH=276.264 cm /sec

Qth=AV th=78.111cm3/ sec, A=π 0.62

4=0.282743 cm2

Qact=2∗100055.11

=36.291 cm3/sec

Cd=36.29178.111

=0.4646

the normal range should be between (0.6-0.9)

Cc=CdCv

= 0.46460.720018

=0.6452

0 1 2 3 4 5 6 70

100200300400500600700

slope=112.0349 Cv=0.720018

Y

X^2

4- Buoyancy ExperimentObjectives of Experiment

1-To verify the water density 2-To calculate the wood density 3-To study the effect of C.G location on the stability condition

Theory

FB=γ f*∀disp

MG= Imin∀disp

−GB

Where

FB= Buoyant force γf = Specific weight of fluid

∀disp

MG = Fluid volume displaced by body / Distance

from metacenter to center of gravity of bodyIminGB

= Minimum moment of inertia of the fluid line area of the body / Distance from center of gravity to center

Procedures

1-Make sure that the nut is put at lowest position2-put the model on water surface3-wait until water disturbance disappear measure the depth of submergence by attached ruler on the model 4-Add weights to the screw and wait while then measure the depth of submergence5-Repeat step 4more than one time and measure the depth of submergence

Reading weight added ( gm) depth of submergence (mm)1 0 54.42 10 55.53 20 56.54 40 58.575 50 59.6

reading weight added (gm)

depth of submergence (mm)

volume initial (m^3)

volume final (m^3)

volume (m^3)

Total weight(N)

Ɣ fluid(N\m^3)

1 0 54.5 0.000214 0 0.000214 1.82466 6876.172 10 55.5 0.000214 0.000012 0.000226 1.92276 6945.133 20 56.5 0.000214 0.000024 0.000236 2.02086 7066.534 40 58.5 0.000214 0.000036 0.000254 2.6611 7338.1885 50 59.5 0.000214 0.00006 0.000274 2.31516 7160.58

0.00021 0.00022 0.00023 0.00024 0.00025 0.00026 0.00027 0.000280

0.5

1

1.5

2

2.5

3

f(x) = 8941.050109529 xR² = 0.991689790689127

∆ Volume (m^3)

Wei

ght (

gm)

Slope =γ water =8941.5 N /m3=894.15 kg /m3

Comment:

Gama water is not accurate because there is an error in reading or problem in

Must be 1000 kg /m3 or 9810N /m3

PART II

Procedures

1- Make the nut at certain position

2- Calculate the total volume of the wooden body

3-Put the model on water surface

4-Wait until water disturbance disappear and measure the depth of submergence

5-Calculate the submerged volume

6-Calculate the specific weight of wood by Archimedes law.

γwood∗∀total +W nut+W screw =γwater∗∀submerged

γwood (actual)=W /V=0.173/ (4∗10¿¿−4¿)¿¿ =432.5 kg /m3

γwood (th)=(894.15∗2.14∗10−4−0.063)/(4∗10−4¿)=320.870 kg/m3¿

Comment:

There is an error in (γ¿¿wood)¿ because there is an error in (γwater) from error in reading

PART III

Procedures

1-Make the nut at certain position and measure the height of the C.G. of nut w .r.t wooden body upper surface

2-Pat 50 gm weight specie on the nut measure the height of C.G. of weight specie w.r.t. wooden body upper surface

3-Put the model on water surface

4-Observe if the model is stable or un stable

5-Record the height of each trial and the status obtained stable, unstable

6-Calculate the required height of 50 gm weight specie to reach the neutral case and compare it with observed height

MG=IminV ¿

−BG

When it in natural case MG= 0 we add 50 gmImin=6.4*10−6 m4

V=2.68*10−4

BG=OG-OB

BG= 59.73-36.9=22.83 mmIminV

= 6.4∗10−6

2.68∗10−4 =0.02888 m = 23.88 mm

Comment:The body was stable just before turned over cannot calculated in lap cause error in reading.