flow over a rectangular and vee notches

ATOOMUS احمد توفيق مؤمنالعتوم

1

OBJECT: To determine and comment on the coefficient of discharge for the Vee and rectangular notches provided.

APPARATUS:

1. Hydraulics Bench2. rectangular weirs 3. vee notch 4. Stop watch.

THEORY:

The theoretical discharge foe the rectangular weir is given by: Qtheo = (2/3) (2g)(1/2). B. H(3/2)

And for vee notch is given by: Qtheo = (8/15). (2g) (1/2) . tan(/2) . H(5/2)

Where B = breadth of rectangular notch H = hight of flow over notch = Angle of vee notch g = Gravitational acceleration

To the contraction of the flow area downstream of the notch, the actual discharge Q is considerably less and may be expressed as: Qact = Cd .(2/3). (2g)(1/2).B. H(3/2)

Where Cd : the coefficient of discharge for the rectangular notch.

And Qact = Cd (8/15). (2g) (1/2) . tan(/2) . H(5/2)

Where Cd : the coefficient of discharge for the vee notch.


2

PROCEDURE:

1. Slide the stilling baffle into the slots in the sides of the channel. Install the weir plate 4H 0n on the upstream side of the weir carrier and secure it using the thumb nuts.2. position the hook and point gauge, mounted on the instrument carrier, on the side channels adjacent to the weir plate.3.start he pump, and admit water to the channel by opening the flow control valve. Allow the level to rise until water discharge over the weir plate. Close the flow control valve and allow the water level to stabilize. Set the vernier height gauge to a datum reading using the top of hook.4. position the point gauge about half way between the notch plate and the stilling baffle. Admit water to the channel and adjust the flow control valve to obtain heads H increasing in steps of about 1cm.5. For each flow rate allow conditions to become steady, measure and record H and take readings of volume and time using the volumetric tank to determine the flow rate.6. For ach notch obtain five readings of H and Q.

RESULTS:

Breadth of rectangular notch = B = 3 cmApex angle of the vee notch = = 900

CALCULATION AND GRAPHS:

For a rectangular notch Q = K H(3/2)

K = Cd .(2/3). (2g)(1/2).B @ B = 0.03 mVol. m3

x10-

3

TimeSec.

Qact

m3/secx10

-4

Hmm

Qtheo

m3/sec x10

-

4

Cd K Log H Log Q

3 43.5 0.6896

13.6

1.45 0.47558 0.042 -1.8664

-4.16141

3 21.5 1.395 23.5

3.19 0.437 0.0387 -1.6299

-3.85543

10 22.5 4.44 42.5

7.76 0.572 0.05057 -1.3716

-3.3526

10 18.3 5.46 49.8

9.85 0.554 0.04908 -1.3027

-3.2627


3

15 16.41

9.14 68.7

15.95 0.5392 0.04776 -1.1630

-3.0390

For first reading:

Qact = 0.003/43.5 = 0.00006896 m3/sec Qtheo = (2/3)*(2*9.81) 0.5*0.03*0.01361.5 = 0.000145 m3/secCd = Q/ Qtheo = 0.47558 K = (2/3)x0.03x( 2*9.81) 0.5 x 0.47558 = 0.042 Should be n =~ (3/2 ) →→ if we take the log for the two sides of equation : log Q = log K + n log H , where n : the power of H = the slope.→→→→ ((-4.16141 +3.0390)/( -1.8664+1.1630))= 1.59 from table . log k = -1.2 from graph → k = 0.063 → Cd = 0.711 .


4

For Vee notch Q = K H(5/2)

K = Cd .(8/15). (2g)(1/2).tan (/2) Vol. m3

x10-

3

TimeSec.

Qact

m3/secx10

-4

Hmm

Qtheo

m3/sec x10

-

4

Cd K Log H Log Q

3 34 0.882 20.1

1.3531

0.6518 1.5392 -1.696 -4.0545

1 61.6 0.162 8 0.135 1.199 2.8335 -2.096- -4.790 ATOOMUS احمد توفيق مؤمن

العتوم 5

23 31.5 0.095

221 1.597 0.0596 0.1408 -1.677 -5.0213

5 63.52

0.787 19 1.1755

0.6695 1.582 -1.721 -4.10

5 37.48

1.33 24 2.108 0.639 1.5096 -1.62 -3.876

5 12.75

3.92 37.3

6.3477

0.618 1.4588 -1.428 -3.407

For a vee notch Q = K H(5/2)

K = Cd (8/15). (2g) (1/2) . tan(/2) For first reading: Qact = 0.003/34 = 0.0000882 m3/secQtheo = (8/15) (2x9.81)0.5 x tan(90/2) (0.00201)2.5 = 0.00013531 m3/sec CD = ( 0.0000882 / 0.000135310 ) = 0.6518 K = CD x (8/15) (2x 9.81) 0.5 x tan(90/2) = 1.5392 Should be n =~ (5/2 ) →→ if we take the log for the two sides of equation : log Q = log K + n log H , where n : the power of H = the slope.→→→→ ((-3.876 +3.407 )/(-1.62 +1.428))= 2.44 from table .

log k = - 0.9 from graph → k = 0.1259 → Cd = 0.0533 .


6

Conclusion : As we have seen the best lines drawn result werw quite near to the expected result for the ( n ) or the slope values but considering the K value the results were not expected due to the relatively large difference between the drawing interpolation and the calculated values .

Rectangular

0

2

4

6

8

10

020406080

H ( mm )

Q (

m3 /

s )

Note : Krect. < Kvee.


7

flow over a rectangular and vee notches

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