## Purposes and Aims:

In this experiment the flow ( K ) and discharge ( Cd ) coefficients of a Venturi metre will be assessed by comparing the existent discharge measured in the experiment with the theoretical discharge calculated by deducing venturi discharge expression from Bernoulli and continuity equations.

Apparatus:

The setup used in this experiment are:

Stopwatch i?? It is used to cipher the clip required for a specific volume of H2O to be supplied, to mensurate the rate ( existent ) of flow ( Qa ) .

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Venturi setup i?? The Venturi setup is connected at six different points to the manometer

Thermometer i?? Thermometer is used to mensurate the temperature of the H2O, the Reynolds figure will be calculated by the ensuing kinematic viscousness.

Hydraulic bench i?? The hydraulic bench contains H2O pump and tracks the volume of H2O collected. The jet setup rests on it.

Venturi Meter

Method:

Topographic point the metre horizontally, in such a manner that at point 1 and 2 the omega values are the same.

Open the recess valve and leave at it maximal flow.

Let go of the air from the manometer and Venturi tubing.

Tightly near the air purging valve on the upper manifold.

Let the H2O to flux by exchanging on the bench supply valve.

Make sure to shut setups control valve.

Let some air to come in by easy opening the spindle one time more

With the halt lucifer step the discharge by entering the clip required for a specific volume of H2O to be collected.

To mensurate the discharge use the halt ticker to enter the clip required for a specific volume of H2O to be collected. ( Volume of the H2O is measured by a metre beside the supply valve. )

Repeat this 16 times and step the flow rate. Along with it, step h1, h2 and h3.

Where,

h1 = Height of H2O in manometer tubing A ( recess )

h2 = Height of H2O in manometer tubing D ( pharynx )

Results & A ; analysis

Theory

Harmonizing to the Bernoulli equation:

i??For a non-viscous, incompressible fluid in steady flow, the amount of force per unit area, possible and kinetic energies per unit volume is changeless at any pointi??

From this we can reason that the ground of fluctuation in force per unit area between two points is the convergent or divergent in the Venturi metre.

So, continuity equation can be used to mensurate the discharge between two points. At any clip per second, the discharge traveling through a specific cross sectional country at two points is equal, therefore:

Continuity Equation: V1A1 = V2A2 = V3A3 = Q ___________ ( 1 )

From the diagram of the Venturi metre it is clear that point 1 has a certain discharge due to contraction in the pipe, which minimizes the pharynx cross subdivision. From the diagram, for point 2 the discharge Q will alter because the speed did non increase which will do the country of this subdivision to alter. Now, harmonizing to the continuity equation, for Q to stay same as point 1 with the lessening in transverse sectional country, an addition in speed is required.

Bernoulli Equation: ( v^2 1 ) /2g + h1 + z1 = ( V^2 2 ) /2g + h2 + z2 = ( v^2 3 ) /2g + h3 + z3 ____ ( 2 )

Where:

z1 = possible energy, h1 = force per unit area energy and ( v^2 1 ) /2g = Kinetic energy

From the first equation the pharynx subdivision to be 2 and the upstream subdivision equal to 1:

V1 = A2V2/A1_____ ( 3 )

By replacing this equation in equation 2, presuming Z as changeless:

( A2V2/A1 ) ^2/2g+ h1= ? V2? ^2/2g+ h2 i??.. ( 3

( v^2 2 ) /2g- ( A2V2/A1 ) ^2/2g = h1-h2

( v^2 2 ) /2g X ( 1- ( A2/A1 ) ^2 ) = h1 i?? h2

V2 = ( 1- ( A2/A1 ) ) ^ ( -1/2 ) Ten V ( 2g ( h1-h2 ) ) ____ ( 4 )

We get a relative relationship between discharge ( Q ) andv ( h1-h2 ) as A1 and A2 are invariables, which means that the theoretical discharge does non include energy loss in the system e.g. clash.

The discharge of flow in the Venturi metre will be calculated as:

Qactual = Cd X ( ( A2 X v2g ) ) / ? V ( 1-A2/A1 ) ? ^2 X vh1-h2______ ( 5 )

In a pipe there are two types of flow brought about by changing speeds. These flows are called laminar and turbulent flow.

Flow Categorization:

Turbulent flow Re & gt ; 4000

Laminar flow Re & lt ; 2000

Using Reynolds equation,

Re = ( V x D ) /v_____ ( 6 )

Where

V = speed,

D = diameter,

V = kinematic viscousness.

Re is relative to the speed as V and D are changeless.

Coefficient of energy loss K must be found as the Bernoullii??s equation does non account for the energy loss in the existent life:

K = ( ( h1-h3 ) ) /v^ ( 2/2g ) _______ ( 7 )

Where,

h1-h3 is the force per unit area caput,

v^2 = kinetic energy

The coefficient of discharge is a comparing between world and theory, it includes the energy loss and the factors doing energy loss e.g. unsteady flow syrupy fluids and disruptive flow.

Cd = Qactual/Qth ____ ( 8 )

Where:

Qth = the theoretical discharge value of the Venturi metre,

Cd = Discharge coefficient

Consequences:

Run No h1 ( m ) h2 ( m ) h3 ( m ) h1-h2 ( m ) h1-h3 ( hL m ) t1 ( sec ) t2 ( sec ) T average Qa ( m^3/s ) Q theoretcial

1 0.226 0.005 0.191 0.219 0.035 16.87 17.12 16.995 0.000441306 0.000440753

2 0.225 0.022 0.193 0.2 0.032 18.13 17.94 18.035 0.000415858 0.0004212

3 0.224 0.043 0.195 0.179 0.029 19.19 19.16 19.175 0.000391134 0.000398474

4 0.221 0.06 0.195 0.159 0.026 20.34 20.06 20.2 0.000371287 0.000375554

5 0.22 0.081 0.197 0.137 0.023 21.91 21.75 21.83 0.000343564 0.000348605

6 0.219 0.102 0.198 0.115 0.021 24 23.81 23.905 0.000313742 0.000319391

7 0.215 0.119 0.199 0.094 0.016 26.38 26.31 26.345 0.000284684 0.00028876

8 0.225 0.14 0.21 0.083 0.015 27.72 27.69 27.705 0.000270709 0.000271339

9 0.206 0.158 0.198 0.046 0.008 38.59 38.59 38.59 0.000194351 0.000202

10 0.206 0.17 0.199 0.034 0.007 43.66 43.59 43.625 0.00017192 0.000173665

11 0.212 0.205 0.21 0.005 0.002 126.81 126.66 126.735 0.000059178 0.000066597

12 0.205 0.191 0.202 0.012 0.003 72.09 72.13 72.11 0.000104008 0.000103173

13 0.2075 0.2035 0.207 0.002 0.0005 182.94 182.91 182.925 0.000041 0.00004212

14 0.214 0.2025 0.211 0.0095 0.003 81.87 81.75 81.81 0.000091675 0.000091798

vh1-h2 1/t mean? h*t^2 V1 m/s Cd V1 ( speed m/s ) Reynolds K

0.467974358 0.058840836 10.10905088 0.87092895 1.001254763 0.87092895 19405.1679 0.905319

0.447213595 0.055447741 10.4083592 0.82070627 0.987317141 0.820706266 18286.1563 0.932124

0.423083916 0.052151239 10.66273813 0.7719133 0.981580525 0.771913299 17199.0002 0.954905

0.398748041 0.04950495 10.60904 0.73274443 0.988639344 0.732744431 16326.2787 0.950096

0.37013511 0.04580852 10.9606247 0.67803195 0.985538971 0.678031951 15107.2299 0.981582

0.339116499 0.041832253 12.00042953 0.61917747 0.982313592 0.619177473 13795.8933 1.074702

0.306594194 0.037957867 11.1049444 0.56183099 0.985883712 0.561830993 12518.1563 0.994506

0.288097206 0.036094568 11.51350538 0.53425149 0.997678517 0.534251489 11903.6574 1.031095

0.214476106 0.025913449 11.9135048 0.3835563 0.962130821 0.383556297 8546.01785 1.066917

0.184390889 0.022922636 13.32198438 0.33928797 0.989949231 0.339287966 7559.67516 1.193054

0.070710678 0.00789048 32.12352045 0.11678926 0.888590855 0.116789262 2602.18156 2.876889

0.109544512 0.013867702 15.5995563 0.20526193 1.008095562 0.205261926 4573.44098 1.397023

0.04472136 0.005466721 16.73077781 0.08091453 0.973409133 0.080914525 1802.85653 1.49836

0.097467943 0.012223445 20.0786283 0.1809229 0.998655444 0.180922904 4031.14323 1.79818

Calculation for the 4th Line on the spreadsheet:

Qa ( Discharge ) :

Volume Collected = 7.5 litres

T mean = 19.175 sec

Qa = ( volume collected ( m^3 ) ) / ( t mean ( s ) )

=0.0075/20.2= 0.000371287? m? ^3/s

V1 Calculation:

V1= Qa/A1

Qa = 0.000371287 m^3/s

A1 = ( p d^2 ) /4=

0.000371287 / ( p? ( 0.0254 ) ? ^2/4 ) =0.732744431 m/s

Reynolds figure Calculation:

Re= V1 x D/v

D/v = 22281

V = 0.732744431 m/s

0.732744431 x 22281=16326.278667111

Calculation of K:

K = ( ( h1-h3 ) ) /v^ ( 2/2g )

H1-h3 = 0.026 ( m )

0.026/ ( ? 0.732744431? ^2/ ( 2 x 9.81 ) ) =0.950096

Qtheoretical Calculation:

Qth= ( A2 v2g ) / ( v1- ? ( A2/A1 ) ? ^2 ) x vh1-h2

Qth =0.0000941832 x 0.398748041=0.000375554

Cd = Qa/Qth

0.000371287/0.000375554=0.988639344

Graphic analysis:

Plot Qactual against Qtheoretical and mensurate the swill of the best fit line:

From the Qactual-Qtheoretical secret plan, it is seeable that the best fit line is 0.988, which is rather reasonably accurate and close to, that enhance the efficiency of the operation.

From V ( h1-h2 ) -V-Qactual, it is seeable that the incline of the best line of the secret plan is 1087.3. Harmonizing to the theory:

V ( h1-h2 ) /Qactual=1087.3

By multiplying by 1061.7596 best estimation of Cd will be found.

1/ ( V ( h1-h2 ) /Qactual ) = 1/1087.3 x 1061.7596

Best estimation of Cd= 0.9886398004

Best discharge equation for the metre:

Qa = best estimation of Cd x ( A2 v2g ) / ( v1- ? ( A2/A1 ) ? ^2 ) x vh1-h2

i?? Qa =0.9886398004 x ( A2 v2g ) / ( v1- ? ( A2/A1 ) ? ^2 ) x vh1-h2

Simplest signifier of the discharge relationship for the metre

Qa = Best estimation of Cd x vh1-h2

Qa = 0.9886398004 ten vh1-h2

Discussion/conclusion:

By the consequences obtained in the research lab and the proceed informations, a fluctuation the values of Cd is seen throughout the process. From the secret plan of Qactual against Cd a alteration in Cd is seen due to the alteration of the value of K, and reasoning to the energy loss in the metre besides the discharge coefficient takes into history these alterations in energy loss.

The least value of K from the spread sheet which is run 1 of a 0.905319 value and a value of Cd of 1.001254763 was the 2nd highest recorded in the spreadsheet, besides run 11 shows the highs K value 2.876889 and Cd value of 0.888590855 was the lowest Cadmium value recorded.

Energy loss occurs in the metre because of the clash, turbulent and the viscousness of the flow.

Because of the Laminar flow and the turbulent flow, which causes energy loss the Venturi metre indicated changing values of Cd. From the best estimation of Cd 0.9886398004, we can presume that energy accumulated in the pipe are undistinguished and besides that the has incrementally small energy losingss

Evaluation:

The Venturi metre experiment has opened my head and it has helped me better understand the continuity equation and the Bernoulli equation.

The experiment has besides increased my cognition about the turbulent and laminar flow and their causes. I have besides learned that the value of Cd does non stay changeless.

I have learned how to outdo estimation the value of Cd, the value of K and the value of Re by utilizing either one of the two equations and using all the theory and computation required. It has increased my apprehension of the equations required to carry through the aims of this experiment.

Appendix

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