# Labaratory 0.5008 L 0.132 0.026 0.5309 ·

Labaratory

work

Venturi

meter

Ignotas

Virginis s16001586

Nick Burdon ENG458

·

Introduction

The

main objectives of this experiment is to find the coefficient of discharge from

experimental data using Venturi meter. Find relationship between flow area,

pressure and velocity.

With this laboratory work we get good understanding

of Bernoulli’s equation and its applications in flow analysis. In experiment,

we recorded fluid levels of 11 tubes.

· Materials

Venturi meter with fluid.

· Method

Check the schematic plate

on the back on the Venturi meter. It shows the cross- sectional areas and

distances to each point in the Venturi meter. Turn the pump on and adjust the

flow rate to a constant level using the valve on the tub. To adjust the

flowrate for the rest of the lab, use the valve located on the Venturi meter.

This will avoid introducing air into the system. ?Set the discharge as

high as possible (water in all piezometers must be readable on the each of

their scales) making sure there are no air bubbles in the piezometer tubes.

Record the water heights in all of the piezometers in Table 1. ?Find the flow rate using the Mag Meter. Record the results in

Table 2. ?Readjust both valves so that a difference in water heights in

cross-sections A and D is 3?4 of what it was in step 3. Record all levels and

find the discharge. ?Repeat step 5 two more times with the

difference in water levels between 1?2 and 1?4 of what it was in step 3. ?For a final check, shut off the flow into the Venturi meter and

make sure all of the water levels are the same. ?

· Data

Tubes

Height

(m)

Diameter

(m)

Area (m2)

A

0.154

0.026

0.5309

B

0.139

0.0232

0.4227

C

0.089

0.0184

0.2659

D

0.023

0.016

0.201

E

0.032

0.0168

0.2217

F

0.074

0.01847

0.268

G

0.99

0.02016

0.3188

H

0.112

0.02184

0.375

J

0.121

0.02353

0.435

K

0.128

0.02524

0.5008

L

0.132

0.026

0.5309

· Calculations

Formulas:

1)

2)

Va=

3)Cvb=

Mcalc=

Mreal= = 0.325

4) Qb=

B.

1) ( Pa – Pb ) = 1000 * 9.81 * ( ( 154 –

139 ) x10-3 ) =147.17 N/m2

2) Vb = = 0.897 m/s2

3) Cvb = = 1.167

4) Qb = 0.0004227*0.897 = 379.16×10-6

C.

1) ( Pa – Pc ) = 1000 * 9.81 * ( ( 154 –

89 ) x10-3 ) =637.65 N/m2

2) Vc = = 1.305 m/s2

3) Cvc = = 1.068

4) Qc = 0.0002659*1.305 = 347×10-6

D.

1) ( Pa – Pd ) = 1000 * 9.81 * ( ( 154 –

23 ) x10-3 ) =1285.11 N/m2

2) Vd = = 1.732 m/s2

3) Cvd = = 1.072

4) Qd = 0.0002011*1.732 = 347×10-6

E.

1) ( Pa – Pe ) = 1000 * 9.81 * ( ( 154 –

32 ) x10-3 ) =1196.82 N/m2

2) Ve = = 1.703 m/s2

3) Cve = = 1.162

4) Qe = 0.0002217*1.703 = 377.56×10-6

F.

1) ( Pa – Pf ) = 1000 * 9.81 * ( ( 154 –

74 ) x10-3 ) = 784.8 N/m2

2) Vf = = 1.451 m/s2

3) Cvf = = 1.917

4) Qf = 0.000268*1.451 = 388.87×10-6

G.

1) ( Pa – Pg ) = 1000 * 9.81 * ( ( 154 –

99 ) x10-3 ) = 539.55 N/m2

2) Vg = = 1.3 m/s2

3) Cvg = = 1.275

4) Qg = 0.0003188*1.3 = 414.12×10-6

H.

1) ( Pa – Ph ) = 1000 * 9.81 * ( ( 154 –

112 ) x10-3 ) = 417.02 N/m2

2) Vh = = 1.281 m/s2

3) Cvh = = 1.479

4) Qh = 0.000375*1.281 = 480.4×10-6

J.

1) ( Pa – Pj ) = 1000 * 9.81 * ( ( 154 – 121

) x10-3 ) = 323.73 N/m2

2) Vj = = 1.402 m/s2

3) Cvj = = 1.876

4) Qj = 0.000435*1.402 = 609.87×10-6

K.

1) ( Pa – Pk ) = 1000 * 9.81 * ( ( 154 – 128

) x10-3 ) = 255.1 N/m2

2) Vk = = 2.135 m/s2

3) Cvk = = 3.292

4) Qk = 0.0005008*2.135 = 1069.21×10-6

· Results and discussion

What

we get from results and data, graphs . whats good whats bad

· Conclusion