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### Transcript of Bernoulli’s Equation - جامعة نزوى · PDF fileApplications of the Bernoulli...

• Chapter 4

• Bernoulli’s Equation

(Mechanical Energy Balance)

• Flow Measurements

1

• Mechanical Energy Balance From the energy equation, dividing by mass and assuming

incompressible flow, then rearranging

 

  

 

 

 

m

Q )uu(

m

W

)zz(g) 2

V

2

V ()

PP (

net 12

net shaft

12

2 1

2 212



Let’s define the

:friction heating per unit mass

m

Q u

 F

2

• Mechanical Energy Balance

 F/g (sometimes symbolized as hL) is the friction or head

loss

F m

W

)zz(g) 2

V

2

V ()

PP (

net shaft

12

2 1

2 212 

 

 

This is the Mechanical Energy Balance or Bernoulli’s

equation

By dividing above equation with g we obtain the “head form” of

the Mechanical Energy Balance:

g

F

mg

W

)zz() g2

V

g2

V ()

g

P

g

P (

net shaft

12

2 1

2 212 

 

 

3

• Simplifications Assumptions:

– Steady, incompressible, frictionless (inviscide) flow

– No heat transfer or change in internal energy, no shaft work

– Single input-output

Bernoulli’s equation may reduced to:

In terms of “heads” (units of length)

0)() 22

()( 12

2

1

2

212  zzg VVPP



0)zz() g2

V

g2

V ()

g

P

g

P ( 12

2

1

2

212  

 

4

• 5

• May be obtained directly from Euler’s equation:

dp/ρg + dz + dV2/2g = 0

• Integrating for incompressible flow:

p/ρg + z + V2/2g = constant

• Multiplying through by ρg:

i.e.

Bernoulli’s Equation

21 constant 2

p gz V   

2 2

1 1 1 2 2 2

1 1

2 2 p gz V p gz V       

• 6

Bernoulli’s Equation (Cont.)

Often interpreted as an energy conservation

equation, this may be represented as:

If incompressible and if elevation (z) terms are ignored:

Static pressure Elevation pressure Dynamic pressure

Total Pressure (p0)

21 constant 2

p gz V   

2 2

1 1 2 2

1 1

2 2 p V p V   

• 7

(Cont.) Bernoulli’s Equation

In this case the dimensions of all the terms are of length units (metres in SI).

2

constant 2

p V z

g g   

• 8

Interpretation of Bernoulli’s Equation

• 9

• 10

• • Applicable to large number of practical engineering problems though with several limitations.

– Simple form of equation only valid for incompressible fluids.

– No account taken of any mechanical devices in system which add/remove energy (e.g. pumps, turbines, etc.).

– No account taken of heat transfer into or out of fluid.

– No account taken of energy lost due to friction.

-Total energy of any stream line is constant.

-Link between the flow of the velocity and Pressure.

-If the Velocity increase, Pressure decrease.

Bernoulli’s Equation - Restrictions

11

• Bernoulli’s Equation 1Example

• Glycerine (SG = 1.26) flows in a processing plant pipe at a rate of 700 L/s. At a point where the pipe diameter is 60 cm, the pressure is 300 kN/m2.

• Find the pressure at a second point, 1 m lower than the first point, with a pipe diameter of 30 cm.

• Neglect any system head losses.

Z2=0 Z1=1

• Bernoulli’s Equation Solution • Using Q1 = A V1 (Continuity)

then 0.7 = ( x 0.62 / 4) V1,

V1 = 2.476 m/s

• Using Continuity: Q1=Q2 • then V2 = V1 (A1 / A2)

 V2 = 9.9 m/s

• Using Bernoulli:

p1 + ρgz1 + ½ρV1 2 = p2 + ρgz2 + ½ρV2

2

 p2/(ρg) = 300x10 3/(1260 x 9.81) +

(2.4762 – 9.92)/(2 x 9.81) + 1

 p2 = 254.5 kN/m 2 13

• 14

• 15

• 16

• 17

• 18

• 19

• 20

• Applications of the Bernoulli Theorem 1/5

•Flow and the streamline under consideration are shown in Fig. 5.4. Using the Bernoulli equation, we can form a relation between point (1) and point (2) as

p1 + ½V1 2 + gz1 = p2 + ½V2

2 + g z2

21

• Applications of the Bernoulli Theorem 2/5

At point (1), the pressure is atmospheric (p1 = po), or the gage

pressure is zero, and the fluid is almost at rest (V1 = 0).

At point (2), the exit pressure is also atmospheric (p2 = po), and

the fluid moves at a velocity V.

By using point (2) as the datum where z2 = 0 and the elevation of point (1) is h, the above relation can be reduced to

p0 + ½(0) 2 + gh = p0 + ½V 2 + g (0)

gh = ½V 2

Hence we can formulate the velocity V

to be

V2 =  2gh

22

• Applications of the Bernoulli Theorem 3/5 • Notice that we can also obtain the similar relation by using the

relation between point (3) and point (4).

• The pressure and the velocity V2 for point (4) is similar to point (2). However, the pressure for point (3) is the hydrostatic pressure, i.e.

• p3 = p0 + g (h - ) and the velocity V3 is also zero due to an assumption of a large tank. Hence, the relation becomes

[ p0 + g (h - )] + ½(0) 2 + g = p0 + ½V

2 + g (0)

and thus: V2 =  2gh gh = ½V 2

we can formulate the 5 & 1 , between Hence

velocity V to be

p1 + ½V1 2 + gz1 = p5 + ½V5

2 +g z5

p0 + ½(0) 2 + g(h+H) = p0 + ½V5

2

V5 =  2g ( h + H )

(h + H) is vertical distance from point 1 to 5

23

• Applications of the Bernoulli Theorem 4/5 For a nozzle located at the side wall of the tank, we can also form a

similar relation for the Bernoulli equation, i.e.

V1 = 2g(h – d/2), V2 = 2gh, V3 = 2g (h + d/2),

 For a nozzle having a small diameter (d < h), then we can conclude that :

V1  V2  V3 = V   2gh

24

• Applications of the Bernoulli Theorem 5/5

i.e., the velocity V is only dependent on the depth of the centre of the nozzle from the free surface h.

If the edge of the nozzle is sharp, as illustrated in Fig. 5.5b, flow contraction will be occurred to the flow. This phenomenon is known as vena contracta, which is a result of the inability for the fluid to turn at the sharp corner 90°. This effect causes losses to the flow.

25

• Example : Flow From a Tank－ Gravity • A stream of water of diameter d = 0.1m flows steadily from a

tank of Diameter D = 1.0m as shown below. Determine the

flowrate, Q, needed from the inflow pipe if the water depth

remains constant, h = 2.0m.

26

• Solution 1/3

2

2

221

2

11 zV

2

1 pzV

2

1 p gg

2

2

2

1 V

2

1 ghV

2

1 

For steady, inviscid, incompressible flow, the Bernoulli

equation applied between points (1) and (2) is

(1)

With the assumptions that p1 = p2 = 0, z1 = h, and z2 = 0, Eq.1 becomes

(2)

Although the water level remains constant (h = constant), there is an

average velocity,V1, across section (1) because of the flow from tank.

For the steady incompressible flow, conservation of mass requires

Q1 = Q2, where Q = AV. Thus, A1V1 =A2V2 , or

2

2

1

2 Vd 4

VD 4

 

 27

• Solution 2/3

2

2

1 V)

D

d (V 

s/m26.6 )m1/m1.0(1

)m0.2)(s/m81.9(2

)D/d(1

gh2 V

4

2

42 

 

 

Hence,

(3)

Equation 1 and 3 can be combined to give

Thus,

s/m0492.0)s/m26.6()m1.0( 4

VAVAQ 32 2211

 



28