3 - web viewfigure 5.5z2z1xy( z1-y ) pqa1, v1, p1 and z1a2, v2, p2 and z2 = spec. wt of liquid in...

Figure 5.5

Z2Z1

X

y

( z1-y )

P Q

A1, v1, p1 and z1

A2, v2, p2 and z2

= spec. wt of liquid in pipeline

g

FLUID DYNAMICS J3008/5/1

BENDALIR DINAMIK (FLUID DYNAMICS)

5.4.4 Meter Venturi Condong (Inclined Venturi Meter)

From Bernoulli’s Equation,

z1+

v12

2 g+

p1

ω= z2+

v22

2 g+

p2

ω

v22−v

12=2 g {( p1−p2

ω )+(z1−z2)}——————(1)

For continuity of flow, Q1 = Q2

A1 v1=A2v2or

v2=

A1

A2v1=mv1


Nisbah luas muka keratan, m = area ratio =

A1

A2

Masukkan dalam (1) and dapatkan v1

m2 v12−v

12=2g {( p1−p2

ω )+(z1−z2 )}

v1=1

√ (m2−1 ) √ [2g {( p1−p2

ω )+ (z1−z2 )}]Kadaralir sebenar (Actual discharge) Qs = Cd×A1×v1

Qs=

Cd×A1

√ (m2−1 ) √ [2 g {( p1− p2

ω )+ (z1−z2 )}] ——— (2)

Where Cd = pekali kadaralir (coefficient of discharge.)

Untuk U-tube,

Bahagian kiri,pP=p1+w (z1− y )

Bahagian kanan,pQ=p2+ω ( z2− y−x )+wHg x

PP = PQ

p1+ω (z1− y )= p2+ω ( z2− y−x )+wHg x

p1+ω z1−ω z2= p2+ω z2−ω y−ω x+wHg x

p1−p2

ω+z1−z2=x ( ωHg

ω−1)

QS=Cd A1

√ (m2−1 ) √(2 gx )(ωHg

ω−1)


Example 5.5

Satu meter venturi condong mengukur aliran minyak berketumpatan bandingan 0.82 dan mempunyai diameter 125mm pada salur masuk dan 50mm pada leher. Bahagian leher berada 300mm lebih tinggi daripada bahagian salur masuk. Jika pekali kadaralir ialah 0.97 dan perbezaan tekanan 27.5 kN/m2. Tentukan kadaralir sebenar minyak?

Qactual=Cd×A1

√ (m2−1 ) √ [2 g {( p1−p2

ω )+(z1−z2)}]A1

=3 .142 (0 . 125 )2

4=0 .01226 m2

Cd = 0.97

p1−p2=27 .5×103 kN /m2 ω =0 . 82×9 . 81×103 N /m2

z1− z2=−0 . 3 m m =

d12

d22

=(12550 )

2=6 .25

Therefore,

QS=0 . 97×0 .01226

√( (6 .25 )2−1) √ [2×9 .81(27. 5×103

0 .82×9 .81×103 −0 .3)]= 0 .01535 m3 /s

1

2

z2z1


Example 5.6

Satu meter venturi dipasangkan pada sebatang paip berdiameter 7.5 cm di mana diameter pada bahagian leher 2.5 cm. Tentukan kadaralir yang melalui paip tersebut, ketika turus tekanan setinggi 41.2 cm air. Anggap Cd = 0.97.

(jwp : Qs = 1.362 x 10-3 m3/s)

Example 5.7

Satu meter venturi mengalirkan air berdiameter 1.2 m dan lehernya 0.6 m. Perbezaan tekanan di antara bahagian masukan dan leher diukur dengan jangka tekanan bezaan di mana ianya menunjukkan bacaan 51 mm. Cari kadaralir dan halaju air melalui leher.

(jwp : Qs = 0.292 m3/s , v = 1.03 m/s)

Example 5.8 (SOALAN FINAL JUL 04)

Satu meter venturi mempunyai diameter utama 65 mm dan diameter bahagian leher 26 mm. Apabila mengukur aliran sesuatu cecair yang berketumpatan 898 kg/m3, bacaan pada satu tolok tekanan pembeza raksa ialah 71 mm. Hitungkan aliran yang melalui meter dalam unit m3/jam. Ambil pekali kadaralir 0.97 dan graviti tentu raksa 13.6.

(jwp : Qs = 2.317 x 10-3 m3/s , = 8.34 m3/j) (15 m)

Example 5.9 (SOALAN FINAL JAN 04)

Garispusat salur masuk sebuah meter venturi mendatar ialah 0.2 m dan garispusat pada bahagian leher ialah 0.1 m. Ia digunakan untuk mengukur kadaralir minyak yang berketumpatan bandingan 0.8. Perbezaan manometer raksa/minyak yang digunakan adalah menunjukkan bacaan 0.2 m. Anggap pekali discas, Cd = 0.9. Tentukan :

i. Halaju aliran minyak.ii. Kadaralir teori minyak.iii. Kadaralir sebenar minyak.

(jwp : v = 2.046 m/s, QT = 0.064 m3/s, QS = 0.058 m3/s) (10 m)

X

Section 1 :A1, v1, p1

Section 2 :A2, v2, p2

Figure 5.7


5.4.5 Small Orifice

Section 1, given :

Section 2, given :

A1 = area of section 1

v1 = velocity of section 1

p1 = pressure of section 1

A2 = area of section 2

v2 = velocity of section 2

p2 = pressure of section 2


From Bernoulli’s Equation,

Total energy at section 1 = Total energy at section 2

p1

ω+

v12

2 g=

p2

ω+

v22

2 g ——————(1)

v22−v

12

2 g=

p1−p2

ω ——————(2)

z1 = z2 because the two parts are at the same level

We know that, Q=A×v


or A1v1 = A2v2

So,

v2 =

A1 v1

A2 ——————(3)

Putting (3) into (2),

v22−v

12

2 g=

p1−p2

ω ——————(2)

v2 =

A1 v1

A2 ——————(3)Then,

v12

2 g [ A12

A22

−1]= p1−p2

ω


So,

v1=√ 2 g( p1−p2

ω )( A

12

A22

−1)But,

H =

p1−p2

ω

And,

m =

A12

A22

So,

v1=√ 2 gH(m2−1 )

To determine the actual discharge, Qactual ;

Qactual=Cd×A1×v1So,

Qactual=Cd×A1√ 2gH

( m2−1 )

Where Cd = coefficient of discharge.Example 5.7

A meter orifice has a 100 mm diameter rectangular hole in the pipe. Diameter of the pipe is 250 mm. Coefficient of discharge, Cd = 0.65 and specific gravity of oil in the pipe is 0.9. The pressure difference that is measured by the manometer is 750 mm. Calculate the flow rate of the oil through the pipe.

Solution to Example 5.7

Given,

d1 = 100 mm = 0.10 md2 = 250 mm = 0.25


Cd = 0.65oil = 0.9p1 - p2 = 750 mm = 0.75 m

So,

A1=πd2

4

=

3 .124 (0 .25 )2

4=0 . 049m2

H=p1−p2

ωoil=x [ ωHg

ωoil−1]

=0.75[13 .6

0. 9−1]

=10 . 58 m

m=

d12

d22 =

(0 .25 )2

(0 .10 )2

=6.25

Therefore,

Qactual = Cd×A1 √ 2 gH

(m2−1 )

Qactual = 0 . 65×0 . 049√ 2×9. 81×10. 58

(6 .252)−1

=0.074 m3 /s

5.4.5.1 Types of orifice

1. Sharp-edged orifice, Cd = 0.62

2. Rounded orifice, Cd = 0.97


3. Borda Orifice (running free), Cd = 0.50

4. Borda Orifice (running full), Cd = 0.75

5.4.5.2 Coefficient of Velocity, Cv

hx

yAxB


Figure 5.8

From Figure 5.8 ,

x = horizontal falls = velocity ¿ time = v×t

y

= vertical falls =

12

gravity×time 2

=

12

g×t

h = head of liquid above the orificeCv

= Coefficient of Velocity = C v=

v√2 gH

t = time for particle to travel from vena contracta A to point B

Coefficient of Velocity, Cv =

Actual velocity at vena contactaTheoretical velocity

Cv=

v√2 gH

Example 5.8

A tank 1.8 m high, standing on the ground, is kept full of water. There is an orifice in its vertical site at depth, h m below the surface. Find the value of h in order the jet may strike the ground at a maximum distance from the tank.


x =v×tand

y =

12

g×t


Eliminating t these equation give,

x= 2v2 yg

y = 1.8 – h h = head of liquid above the orifice

Cv= v

√2 gHt = time for particle to travel from vena contracta A to point B

Putting y=1 . 8−h and v=C v √ (2 gh )

So,

x=√ 2 [Cv √(2 gh ) ]2× (1. 8−h )g

x=√ Cv2 4 gh (1. 8−h )

g

=2 C v [h (1.8−h ) ]

Thus x will be a maximum when h (1 .8−h ) is maximum or,[h (1 . 8−h ) ]

h=1. 8−2h=0

So,h=0 .9 m

Example 5.9

An orifice meter consists of a 100 mm diameter in a 250 mm diameter pipe (Figure 5.9), and has a coefficient discharge of 0.65. The pipe conveys oil of specific gravity 0.9. The pressure difference between the two sides of the orifice plate is measured by a mercury manometer, that leads to the gauge being filled with oil. If the difference in mercury levels in the gauge is 760 mm, calculate the flowrate of oil in the pipeline.

Pipe Area, A1

P1 P2 V1V2 X Orifice area A2 C C


Figure 5.9


Let v1 be the velocity and p1 the pressure immediately upstream of the orifice, and v2 and p2 are the corresponding values in the orifice. Then, ignoring losses, by Bernoulli’s theorem,

p1

ω+

v12

2 g=

p2

ω+

v22

2 g ——————(1)

v22−v

12

2 g=

p1−p2

ω ——————(2)

z1 = z2 because of the two parts are at the same level

We know that,Q=A×v


or A1v1 = A2v2

So,

v2 =

A1 v1

A2 ——————(3)

Putting (3) into (2),


v22−v

12

2 g=

p1−p2

ω ——————(2)

v2 =

A1 v1

A2 ——————(3)Then,

v12

2 g [ A12

A22

−1]= p1−p2

ω

So,

v1=√ 2 g( p1−p2

ω )( A

12

A22

−1)

This equation can therefore be written,

v1=A2

√( A12−a

22)2 g ( p1−p2

ω )——————(4)

So,

Actual disch arg e=coefficient of disch arg e×theoretical disch arge

Qactual=Cd×A 1×v 1 ——————(5)

Putting v1 into (5)

Qactual=Cd×A 1×¿ ¿A2

√( A12−a

22)2g ( p1− p2

ω )——————(6)

but,

m =

A12

A22

so putting m into (6),


Qactual=Cd×A1×Cd A1

√ (m2−1 ) √2 g( p1−p2

ω )Considering the U-tube gauge, where pressures are equal at level CC

p1+ω x= p2+ωq xp1−p 2

ω=x ( p 1−p2

ω )Putting x=760 mm=0 .76 m and,

ωg

ω=13 . 6

0 . 9=15. 1

p1−p2

ω=0 .76×14 .1=10 .72 m of oil

Cd=0 .65

A1=πd2

4=0.0497 m2

m= A 1A 2

=d

12

d22

=(0 .25 )2

(0 .10 )2=15 .1

m2=6 .17

Qactual=0 . 65×0.0497

6 .17 √(2×9 .81×10 .72 )

=0. 00524×14 . 5=0. 0762 m3/ s

5.4.6 Simple Pitot Tube


Figure 5.10 Pitot Tube

a

b

- The Pitot Tube is a device used to measure the local velocity along a streamline (Figure 5.10). The pitot tube has two tubes: one is a static tube (b), and another is an impact tube(a).

- The opening of the impact tube is perpendicular to the flow direction. The opening of the static tube is parallel to the direction of flow.

- The two legs are connected to the legs of a manometer or an equivalent device for measuring small pressure differences. The static tube measures the static pressure, since there is no velocity component perpendicular to its opening.

- The impact tube measures both the static pressure and impact pressure (due to kinetic energy).

- In terms of heads, the impact tube measures the static pressure head plus the velocity head.


Figure 5.11 Simple Pitot Tube A ctual Velocity, V

From Figure 5.11, if the velocity of the stream at A is v, a particle moving from A to the mouth of the tube B will be brought to rest so that v0 at B is zero. By Bernoulli’s Theorem : Total Energy at A = Total Energy at B or p1

ω+

v12

2 g=

p2

ω+

v22

2 g ——————(1)

Now d= p

ω and the increased pressure at B will cause the liquid in the vertical limb of the pitot tube to rise to a height, h above the free surface so

that h+d=

p0

ω .

Thus, the equation (1)

v 22 g

= p0−pω

=h or v=√2 gh

Although theoretically v=√ (2 gh ) , pitot tubes may require calibration.

The actual velocity is then given by v=C√ (2 gh ) where C is the coefficient of the instrument.

A B

h

H


Example 5.10

A Pitot Tube is used to measure air velocity in a pipe attached to a mercury manometer. Head difference of that manometer is 6 mm water. The weight density of air is 1.25 kg/m3. Calculate the air velocity if coefficient of the pitot tube, C = 0.94.


vair=C √2 gH

pwater= pairρ ghwater=ρ ghairhwater×ωwater = hair×ωair

hwater=0 . 006×

ωwater

ωair=0 .006×1000

1 .25

= 4 . 8 mSo,

v=0 . 94√2×9 . 81×4 . 8

=9. 12 m /s


ACTIVITY 5C

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

5.4 Fill in the blanks in the following statements.

1. The Orifice Meter consists of a flat orifice plate with a circular hole drilled in it. There is a _____________upstream from the orifice plate and another just downstream.

2. The reduction of pressure in the cross section of the flowing stream when passing through the orifice increases the __________________at the expense of the pressure head. The reduction in pressure between the taps is measured by a manometer.

3. The formula for Meter Orifice actual discharge, Qactual. =_______________

4. The Pitot Tube is a device used to measure the local velocity along a streamline. The pitot tube has two tubes which are the_______________and the ____________.

5. Although theoretically v=√ (2 gh ) , pitot tubes may require______________.

6. The actual velocity is given by __________ where C is the coefficient of the instrument.


FEEDBACK ON ACTIVITY 5C

5.4

1. The Orifice Meter consists of a flat orifice plate with a circular hole drilled in it. There is a pressure tap upstream from the orifice plate and another just downstream.

2. The reduction pressure in the cross section of the flowing stream when passing through the orifice increases the velocity head at the expense of the pressure head. The reduction in pressure between the taps is measured by a manometer.

3. The formula for Meter Orifice actual discharge, Qactual. =

Qactual=Cd×A1×v1 and Qactual = Cd×A1 √ 2 gH

(m2−1 )

4. The Pitot Tube is a device used to measure the local velocity along a streamline. The pitot tube has two tubes which are the static tube and the impact tube.

5. Although theoretically v=√ (2 gh ) , pitot tubes may require calibration.

6. The actual velocity is given by v=C√ (2 gh ) where C is the coefficient of the instrument.


SELF-ASSESSMENT

You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment. If you face any problems, discuss it with your lecturer. Good luck.

5.1 A venturi meter measures the flow of water in a 75 mm diameter pipe. The difference between the throat and the entrance of the meter is measured by the U-tube containing mercury which is being in contact with the water. What should be the diameter of the throat of the meter in order that the difference in the level of mercury is 250 mm when the quantity of water flowing in the pipe is 620 dm3/min? Assume coefficient of discharge is 0.97.

5.2 A pitot-static tube placed in the centre of a 200 pipe line conveying water has one orifice pointing upstream and the other perpendicular to it. If the pressure difference between the two orifices is 38 mm of water when the discharge through the pipe is 22 dm3/s, calculate the meter coefficient. Take the mean velocity in the pipe to be 0.83 of the central velocity.

5.3 A sharp-edged orifice, of 50 mm diameter, in the vertical side of a large tank, discharges under a head of 4.8 m. If Cc = 0.62 and Cv = 0.98, determine;

(a) the diameter of the jet, (b) the velocity of the jet at the vena contracta, (c) the discharge in dm3/s.


FEEDBACK ON SELF-ASSESSMENT

Answers :

5.1 40.7 mm

5.2 0.977

5.3 (a) 40.3 mm

(b) 9.5 m/s

(c) 12.15 dm3/s

3 - web viewfigure 5.5z2z1xy( z1-y ) pqa1, v1, p1 and z1a2, v2, p2 and z2 = spec. wt of liquid in...

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