Ch3 Bernoulli Equation

Elementary Fluid Dynamics – The Bernoulli Equation

Newton’s Second Law: F = ma • • • • • •

Consider inviscid, steady, two-dimensional flow in x-z plane Define streamlines Select coordinate systems based on streamlines Define acceleration Define forces Apply Newton’s second law of motion along and across streamline

Newton’s Second Law: F = ma • • • •

Prior to apply Newton’s second law of motion to fluid particle: Consider motion of an inviscid fluid Assume that fluid motion is governed by pressure and gravity forces Select an appropriate coordinate system. Consider two dimensional motion (x-z plane)

Coordinate System • • • •

Prior to apply Newton’s second law of motion to fluid particle: Consider motion of an inviscid fluid Assume that fluid motion is governed by pressure and gravity forces Select an appropriate coordinate system. Consider two dimensional motion (x-z plane)

Coordinate System •

Motion of a fluid particle is described by its velocity vector

Coordinate System • •

Motion of a fluid particle is described by its velocity vector As the particle moves, it follows a particular path, the shape of which is governed by velocity vector

Coordinate System • • •

Motion of a fluid particle is described by its velocity vector As the particle moves, it follows a particular path, the shape of which is governed by velocity vector If flow is steady, each successive particle that passes through given point (1) will follow the same path. For such cases the path is a fixed line in the x-z plane. The entire x-z plane is filled with such paths.

Coordinate System •

For steady flow each particle slides along its path and its velocity vector is everywhere tangent to the path

Streamlines • •

For steady flow each particle slides along its path and its velocity vector is everywhere tangent to the path The lines that are tangent to the velocity vectors throughout the flow field are called streamlines.

Streamlines • • •

For steady flow each particle slides along its path and its velocity vector is everywhere tangent to the path The lines that are tangent to the velocity vectors throughout the flow field are called streamlines. We will use coordinates based on streamlines

Particle Motion •

Particle motion is described in terms of its distance, s = s(t), along streamline, and local radius of curvature

Particle Motion • •

Particle motion is described in terms of its distance, s = s(t), along streamline, and local radius of curvature Distance s is related to particle’s speed V = ds/dt, and radius of curvature is related to the shape of streamline

Particle Acceleration •

Acceleration:

a  d V dt

Particle Acceleration a  d V dt

•

Acceleration:

•

Components of acceleration in s and n directions: V as  V , s

V2 an  

Forces •

To determine forces consider free-body diagram of small fluid particle

F = ma along a Streamline

Free-body diagram of a fluid particle

F = ma along a Streamline Equation of motion along streamline (details ) Change in fluid particle speed is accomplished by combination of pressure gradient and particle weight along streamline

Free-body diagram of a fluid particle

 sin  

p V  V   as s s

Example 3.1 Consider the inviscid, incompressible, steady flow along the horizontal streamline A–B in front of the sphere of radius a. From a more advanced theory of flow past a sphere, the fluid velocity along this streamline is 

a3  V  V0  1  3  x   Determine the pressure variation along the streamline from point A far in front of the sphere (xA = – ∞ and VA = V0) to point B on the sphere (xB = – a and VB = 0).

Example 3.1 Consider the inviscid, incompressible, steady flow along the horizontal streamline A–B in front of the sphere of radius a. From a more advanced theory of flow past a sphere, the fluid velocity along this streamline is 

a3  V  V0  1  3  x   Determine the pressure variation along the streamline from point A far in front of the sphere (xA = – ∞ and VA = V0) to point B on the sphere (xB = – a and VB = 0).

Solution Streamline is horizontal, then p V   V s s Acceleration

V V a3  a3 2 V V  3V0  1  3  4 s x x  x 

Example 3.1 Consider the inviscid, incompressible, steady flow along the horizontal streamline A–B in front of the sphere of radius a. From a more advanced theory of flow past a sphere, the fluid velocity along this streamline is 

a3  V  V0  1  3  x   Determine the pressure variation along the streamline from point A far in front of the sphere (xA = – ∞ and VA = V0) to point B on the sphere (xB = – a and VB = 0).

Solution Pressure gradient



3 2 3 3 p 3 a V0 1  a x  x x4



Pressure distribution



  a  3 a x6 p   V02     2   x 

V02 pB  2

  

Bernoulli Equation For incompressible fluid equation of motion along streamline reduces to Bernoulli equation (details)

p

1 V 2   z  constant along streamline 2

Restricted to: - inviscid flow - steady flow - incompressible flow - along streamline

Example 3.2 Consider the flow of air around a bicyclist moving through still air with velocity V0. Determine the difference in the pressure between points (1) and (2).

Example 3.2 Consider the flow of air around a bicyclist moving through still air with velocity V0. Determine the difference in the pressure between points (1) and (2). Solution Apply Bernoulli equation between (1) and (2) p1 

1 1 V12   z1  p2  V22   z2 2 2

Pressure difference p2  p1 

1 1 V12  V02 2 2

F = ma Normal to a Streamline Equation of motion along the normal direction (details)

Change in the direction of flow of a fluid particle is accomplished by combination of pressure gradient and particle weight normal to streamline

Free-body diagram of a fluid particle

dz p V 2    dn n 

Example 3.3 Shown in Fig. a, b are two flow fields with circular streamlines. The velocity distributions are V  r   C1r for case (a) C2 for case (b) r where C1 and C2 are constant. Determine the pressure distributions, p = p(r), for each, given that p = p0 at r = r0. V  r 

Example 3.3 Shown in Fig. a, b are two flow fields with circular streamlines. The velocity distributions are V  r   C1r for case (a) C2 for case (b) r where C1 and C2 are constant. Determine the pressure distributions, p = p(r), for each, given that p = p0 at r = r0. V  r 

Solution Assume steady, inviscid, and incompressible flow with streamlines in horizontal plane (dz/dn = 0) Since streamlines are circles, ∂/∂n = - ∂/∂r and = r Then equation of motion along the normal dz p V 2    dn n  becomes p V 2  r r

Example 3.3 Shown in Fig. a, b are two flow fields with circular streamlines. The velocity distributions are V  r   C1r for case (a) C2 for case (b) r where C1 and C2 are constant. Determine the pressure distributions, p = p(r), for each, given that p = p0 at r = r0. V  r 

Solution For case (a) p   C12 r r

1  C12 r 2  r02  p0 2





and

p

and

1 1 2 1 p   C2  2  2  p0 2  r0 r 

For case (b) p  C22  3 r r

Comments: (a) – free vortex, (b) forced vortex

F = ma Normal to a Streamline For steady, inviscid, incompressible flow (details) V2 p   dn   z  constant across streamline 

Restricted to: - inviscid flow - steady flow - incompressible flow - across streamline

Pressure variation across straight streamlines is hydrostatic

Physical Interpretation p

1 V 2   z  constant along streamline 2

Work done on a particle by all forces acting on the particle is equal to the change of the kinetic energy of the particle Each term of Bernoulli equation can be interpreted as: - head (elevation, pressure, velocity) - form of pressure (static, hydrostatic, dynamic)

Example 3.4

Pressure variation across straight streamlines is hydrostatic

Static, Stagnation, Dynamic, and Total Pressure Useful concept associated with the Bernoulli equation deals with the stagnation and dynamic pressures. As fluid is brought to rest its kinetic energy is converted to a pressure rise

Static, Stagnation, Dynamic, and Total Pressure Each term in Bernoulli equation can be interpreted as a form of pressure; static, p, hydrostatic, γ z, and dynamic, ρV 2/2 , p

1 V 2   z  constant along streamline 2

Point (2) is a stagnation point p2  p1 

1 V12 2

Pressure at the stagnation point is greater than static pressure by dynamic pressure

Static, Stagnation, Dynamic, and Total Pressure •

There is a stagnation point on any stationary body that is placed onto a flowing fluid

•

Some of the fluid flows “over” and some “under” the object. Dividing line is termed the stagnation streamline and terminates at the stagnation point on the body

•

Location of the stagnation point is function of body shape.

Static, Stagnation, Dynamic, and Total Pressure •

If elevation effect are neglected, stagnation pressure, p + ρV2/2, is the largest pressure obtainable along a given streamline. It represents the conversion of all of the kinetic energy into a pressure rise.

•

Sum of the static pressure, hydrostatic pressure, and dynamic pressure is termed the total pressure, pT

•

Bernoulli equation is a statement that total pressure remains constant along a streamline. p

1 V 2   z  pT  constant along streamline 2

Fluid Velocity Measurement Pitot-static tubes measure fluid velocity by converting velocity into pressure

V

2  p3  p4 



Pitot-static tube

Typical Pitot-Static Tube Designs

Measurement of Static Pressure

Incorrect and correct design of static pressure taps

Measurement of Static Pressure

Typical pressure distribution along a Pitot-static tube

Fluid Velocity Measurement Many velocity measurement devices use Pitot-static tube principle

If   0 and   29.5o p1  p3  p

V 2 p2  p  2 V

2  p2  p1 



Cross section of a directional-finding Pitot-static tube

Examples of Use of Bernoulli Equation

Free Jets Exit pressure for an incompressible fluid jet is equal to the surrounding pressure

Velocity:

V  2 gh

Vertical flow from a tank

Free Jets For horizontal nozzle velocity is not uniform If d « h centerline velocity can be used as an “average velocity”

Horizontal flow from a tank

Free Jets If exit is not smooth, diameter of the jet will be less than diameter of the hole. This phenomena, called a vena contracta effect, is a result of the inability of the fluid to turn the sharp 90º corner indicated by dotted lines in the figure

Vena contracta effect for a sharp-edged orifice

Free Jets Since streamlines in the exit plane are curved, the pressure across them is not constant. The highest pressure occurs along the centerline at (2), and lowest pressure, p1 = p3 = 0

Vena contracta effect for a sharp-edged orifice

Free Jets Assumption of uniform velocity with straight streamlines and constant pressure is not valid at the exit plane It is valid in the plane of vena contracta, sectшon a-a, provided dj « h

Vena contracta effect for a sharp-edged orifice

Free Jets Vena contracta effect is a function of the geometry of the ounlet. Contraction coefficient: Cc  Aj Ah

Typical flow patterns and contraction coefficients for various round exit configurations

Confined Flows In nozzles and pipes of variable diameter velocity changes from one section to another For such cases continuity equation must be used along with Bernoulli equation Continuity equation states that mass cannot be created or destroyed For incompressible fluid (details) AV 1 1  A2V2

or

Q1  Q2

Example 3.7 A stream of water of diameter d = 0.1 m flows steadily from a tank of diameter D = 1.0 m. Determine the flowrate, Q , needed from the inflow pipe if the water depth remains constant, h = 2.0 m

Example 3.7 A stream of water of diameter d = 0.1 m flows steadily from a tank of diameter D = 1.0 m. Determine the flowrate, Q , needed from the inflow pipe if the water depth remains constant, h = 2.0 m Solution Assume steady, inviscid, incompressible flow. Apply Bernoulli equation between points (1) and (2) 1 1 V12   z1  p2  V22   z2 2 2 With p1 = p2 = 0, z1 = h and z2 = 0 p1 

1 2 1 V1  gh  V22 2 2 2

 d V1    V2  D

From continuity equation Exit velocity and volume flowrate

V2 

2 gh 1  d D

4

 6.26 m/s

3 Q  AV 1 1  A2V2  0.0492 m /s

Example 3.7 A stream of water of diameter d = 0.1 m flows steadily from a tank of diameter D = 1.0 m. Determine the flowrate, Q , needed from the inflow pipe if the water depth remains constant, h = 2.0 m Solution If D » d, then we can assume V1 ≈ 0. Error associated with this assumption:

Q V2  Q0 V2 D 

4 2 gh  1   d D   1     4 2 gh 1  d D

Example 3.8

Example 3.8

Answers:

V3 

2 p1  69.0 m/s 

Q  0.00542 m3 / s

V2 =7.67 m/s p2  2963 N/m 2

Comments: V3 is determined strictly by the value of p1 In absence of viscous effect pressure throughout the hose is constant and equals to p2 Decrease in pressure from p1 to p3 accelerate the air and increase its kinetic energy Pressure change (density change) is within 3%. Hence, incompressibility assumption is reasonable

Example 3.9

 Q  1   A2 A1  h  A  2  2 g  1  SG  2

Answer:

2

Comments: For a given flowrate h does not depend on θ, but pressure difference, p1 – p2, as measured by pressure gage, does

Cavitation Cavitation occurs when the pressure is reduced to the vapor pressure Cavitation can cause damage to equipment

Pressure variation and cavitation in a variable area pipe

Cavitation

Tip cavitation from a propeller

Example 3.10

Answer:

H  28.2 ft

Comments: Results are independent of diameter and length of the hose (provided viscous effects are not important Proper design of hose is needed to ensure that it will not collapse due to the large pressure difference (vacuum) between the inside and outsides of the hose

Flowrate Measurement Various flow meters are governed by the Bernoulli and continuity equations We consider “ideal” flow meters – those devoid of viscous, compressibility, and other effects. The flowrate is a function of the pressure difference across the flow meter Q  A2

Typical devices for measuring flowrate

2  p1  p2  2   1   A2 A1    

Example 3.11

Answer:

1.16 kPa  p1  p2  116 kPa

Comments: These values represent “ideal” results, and these results are independent of flow meter geometry – an orifice, nozzle, or Venturi meter. Tenfold increase in flowrate requires one-hundredfold increase in pressure difference. This nonlinear relationship can cause difficulties when measuring flowrates over a wide range of values. An alternative is to use two flow meters in parallel

Flowrate Measurement. Sluice Gate The flowrate under a sluice gate depends on the water depths on either side of the gate

Q  z2b

2 g  z1  z2 

1   z2 z1 

2

In the limit of z1»z2

Q  z2b 2 gz1 A vena contracta occurs as water flows under a sluice gate

Sluice gate geometry

Flowrate Measurement. Sharp-crested Weir Flowrate over a weir is a function of the head on the weir

Q  C1 Hb 2 gH  C1b 2 g H 3 2

Rectangular, sharp-crested weir geometry

Energy Line and Hydraulic Grade Line Hydraulic grade line and energy line are graphical forms of the Bernoulli equation Energy line represents the total head available to the fluid Locus provided by a series of piezometric taps is termed the hydraulic grade line

Representation of the energy line and the hydraulic grade line

Energy Line and Hydraulic Grade Line If the flow is steady, incompressible, and inviscid, the energy line is horizontal and at the elevation of the liquid in the tank. Hydraulic grade line lies a distance of one velocity head below the energy line At the pipe outlet the pressure head is zero (gage) so the pipe elevation and hydraulic grade line coincide

Energy line and hydraulic grade line for flow from a tank

Energy Line and Hydraulic Grade Line The distance from pipe to hydraulic grade line indicates the pressure within the pipe For flow below the hydraulic grade line, the pressure is positive For flow above the hydraulic grade line, the pressure is negative

Use of the energy line and hydraulic grade line

Restriction on Use of the Bernoulli Equation Restrictions on use for the Bernoulli equation are imposed by the assumptions used in its derivation. To avoid incorrect use of Bernoulli equation one must take into account: - Compressibility effects; - Unsteady effects; - Rotational effects; - Viscosity effects; - Presence of mechanical devices (pumps, turbines)

“Change of scene, and absence of the necessity for thought, will restore the mental equilibrium” (Jerom K. Jerom, “Three Men In a Boat”)

END OF CHAPTER

Supplementary slides

F = ma along a Streamline Newton’s second law along streamline

  Fs   mas   mV

V V   V V s s

F = ma along a Streamline Gravity force

 Ws   W sin    V sin 

F = ma along a Streamline Pressure force

 Fps   p   ps   n y   p   ps   n y  

p V s

F = ma along a Streamline Net force

p    F   W   F    sin    s s ps   V s  

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Bernoulli Equation p V  V s s

Consider equation

 sin  

Along streamline

sin  

Also

2 V 1 d V V  s 2 ds

(a)

dz ds

 

Finally, along streamline value of n is constant (dn = 0) so that dp 

p p p ds  dn  ds s n s

Hence, along streamline ∂p/∂s = dp/ds . Then equation (a) becomes 1 dp   d V 2   dz  0 2

 

Integration at constant density gives Bernoulli equation

(along streamline) back

F = ma Normal to a Streamline Newton second law in normal direction

 mV 2  V V 2   Fn    

F = ma Normal to a Streamline Gravity force

 Wn   W cos    V cos 

F = ma Normal to a Streamline Pressure force

 Fpn   p   pn   s y   p   pn   s y  

p V n

F = ma Normal to a Streamline Net force

p    Fn   Wn   Fpn    cos    V n  

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Continuity Equation Consider a fluid flowing through a fixed volume. If the flow is steady, rate at which fluid flows into the volume must equal the rate at which it flows out of the volume (mass is conserved) Mass flow rate is given by Volume flow rate

m&  Q Q  VA

1 AV 1 1   2 A2V2

Conservation of mass requires If density remains constant

AV 1 1  A2V2

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Compressibility Effects Bernoulli equation can be modified for compressible flows. For compressible, inviscid, isothermal, steady flows: V12 RT  p1  V22  z1  ln   z2  2g g  p2  2 g Use of above equation is restricted by inviscid flow assumptions, since most isothermal flows are accompanied by viscous effects. For compressible, isentropic (no friction or heat transfer), steady flow of a perfect gas: 2 k  p1 V12  k  p2 V2  gz1    gz2      k  1  2 k  1  2   1   2



Compressibility Effects Bernoulli equation for compressible flow can be written for pressure ratio as 

k k 1

p2  p1   k 1    1 Ma12   p1 2  

  1  

Where Ma = V/c is the Mach number; c is local speed of sound A “rule of thumb” is that the flow of a perfect gas may be considered as incompressible provided the Mach number is less than about 0.3

Pressure ratio as a function of Mach number for incompressible and compressible (isentropic) flow

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Unsteady Effects Bernoulli equation can be modified for unsteady flows. For incompressible, inviscid, unsteady flows: s2 V V12 V22 p1    z1    ds  p2    z2 s1 t 2 2

Use of this equation requires knowledge of variation of ∂V/∂t along the streamline

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Example 3.12

Answers:

Comments: ?

Q  4.83 m 2 /s b

Example 3.13

Answers:

Q3 H 0 QH 0 Comments: ?

  C2 2 gh  C2 tan 2 ghH 5 2 2 2 52 C2 tan   2  2 g  3H 0 

Q  AV  H 2 tan 





C2 tan   2  2 g  H 0 

52

Example 3.14

Answers: Comments: ?

Example 3.16

Answers:

Comments: ?



2g l

Example 3.17

Answers: Comments: ?

V12 p2    h  518 kPa 2

Example 3.18