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Ch 5 Momentum and forces in fluid flow

Outline ¾ Momentum equation ¾ Applications of the momentum equation

Objectives

•

After completing this chapter, you should be able to • • •

Identify the various kinds of forces and moments acting on a control volume. Use control volume analysis to determine the forces associated with fluid flow. Use control volume analysis to determine the moments caused by fluid flow.

6.1 Development of the Momentum Principle

Start with a modified-form of Newton’s 2nd law:

( )

r r d mV ΣF = dt

r ∑F

= sum of forces on fluid system

r mV = linear momentum of system

r r Also, ΣFdt = d (mV ) s-------Impulse-momentum principle

6.1 Development of the Momentum Principle

Momentum entering = ρ A1u 1δ tu 1

Momentum leaving = ρ A 2 u 2 δ tu 2

Force = rate of change of momentum ( ρA2u 2δ tu 2 − ρA1u1δ tu1 ) F= δt Q= A1 u1 = A2 u2,

F = ρQ (u 2 − u1 )

Fig 6.1 Momentum in a flowing fluid

Momentum equation for twoand three-dimensional flow The force in the x-direction

Fig 6.2 Two dimensional flow in a streamtube

Momentum equation for twoand three-dimensional flow The force in the y-direction

Fig 6.2 Two dimensional flow in a streamtube

Momentum equation for twoand three-dimensional flow

Total force exerted on the fluid =

Rate of change of momentum through the control volume

r r r F = ρQ(u2 − u1 ) For steady flow with one inlet and one outlet, the momentum equation is

r r r F = ρQ( β 2u2 − β1u1 )

β—— momentum correction factor

Momentum correction factor True momentum per unit time = β × Mass per unit time × Mean velocity β = ∫ u 2 dA / V 2 A The momentum equation

r r r F = ρQ( β 2V2 − β1V1 )

the resultant force

F = Fx2 + Fy2

the angle which this force acts at is given by

What are the forces acting on the fluid in the control volume?

the total force, FT, is given by the sum of these forces

r r r r FT = FR + FB + FP

Step in Analysis with Momentum Equation

1. Draw a control volume: Based on the problem, selecting the stream between two gradually varied flow sections as the control volume; 2. Decide on co-ordinate axis system: Determining the directions of co-ordinate axis, magnitudes and directions of components of all forces and velocities on each axis. 3. Plotting diagram for computation : Analyzing the forces on control volume and plotting the directions of all forces on the control volume. 4. Writing momentum equation and solving it: Substituting components of all forces and velocities on axes into momentum equation and solving it. All the pressures are relative to the relative pressure.

Application of the Momentum Equation

1. Force due to the flow of fluid round a pipe bend. 2. Force on a nozzle at the outlet of a pipe. 3. Impact of a jet on a plane surface. 4. Force due to flow round a curved vane.

Force due to Jet Striking Surface

Force by Flow Round a Pipe-bend

Example 1 Find the horizontal thrust of the water on each meter of width of the sluice gate shown in the Fig., given y1=2.2 m, y2=0.4 m, and y3=0.5 m. Neglect friction. ( 6.4.2)

Solution

6.5.5

Solution The flow rate Momentum equation

6.8 A reducing right-angled bend lies in a horizontal plane. Water enters from the west with a velocity of 3 m/s and a pressure of 30 kPa, and it leaves toward the north. The diameter at the entrance is 500 mm and at the exit it is 400 mm. Neglecting any friction loss, find the magnitude and direction of the resultant force on the bend.

Solution

Energy

Momentum

Example Water flows through a reducing 180°bend. The bend is shown in plan. Determine the magnitude of the force exerted on the bend in the x-direction. Assume energy losses to be negligible.

Solution:

Example T=15 oC

• • •

Given: Figure Find: Horizontal force required to hold plate in position Solution:

Q=0.4 m3/s

F

B

pA=75 kPa

V A2 p B VB2 + zA + = + zB + γ γ 2g 2g

pA

VB2 = γ 2g

pA

r r V = Vi

− F = ρQ(V2 − V1 )

p VB = 2 A = 2 * 75000 / 999 = 12.3 m / s

ρ

F = ρQV = 999 * 0.4 *12.3 = 4.9 kN

question

If the value of force calculated from momentum equation is negative, what does that mean? Does the magnitude of the unknown force has anything or nothing to do with that of the control volume? How to select control volume in the application?

If the value of force calculated from momentum equation is negative, what does that mean? Does the magnitude of the unknown force has anything or nothing to do with that of the control volume? How to select control volume in the application? directions are inverse；independency（when there is no gravitation）；calculated crosssection and solid wall

Sluice Gate

Find: Force due to pressure on face of gate Solution: Assume: v1 and v2 are uniform (so pressure is hydrostatic)

Application of the Energy, Momentum, and Continuity Equations in Combination In general, when solving fluid mechanics problems, one should use all available equations in order to derive as much information as possible about the flow. For example, consistent with the approximation of the energy equation we can also apply the momentum and continuity equations

Forces on Transitions Example

Example: Energy Equation (energy loss) An irrigation pump lifts 50 L/s of water from a reservoir and discharges it into a farmer’s irrigation channel. The pump supplies a total head of 10 m. How much mechanical energy is lost? What is hL? cs2 4m 2.4 m 2m cs1 datum Why can’t I draw the cs at the end of the pipe? p in

g

+ z in + a

h p = z out + hL

in

V in2 p + hP = out + z out + a 2g g

hL = h p - z out

out

V o2u t + hT + h L 2g

Example: Energy Equation (pressure at pump outlet) The total pipe length is 50 m and is 20 cm in diameter. The pipe length to the pump is 12 m. What is the pressure in the pipe at the pump outlet? You may assume (for now) that the only losses are frictional losses in the pipeline. 50 L/s hP = 10 m 4m cs2 2.4 m 2m cs1 datum p in

g

+ z in + a

in

V in2 p + hP = out + z out + a 2g g

out

V o2u t + hT + h L 2g

We need _______ in the pipe, __, and ____ ____.

Example: Energy Equation (pressure at pump outlet) ¾How do we get the velocity in the pipe?

¾How do we get the frictional losses?

¾What about α?

Kinetic Energy Correction Term: α ¾α is a function of the velocity distribution in the pipe. ¾For a uniform velocity distribution ____ ¾For laminar flow ______ ¾For turbulent flow _____________ ¾Often neglected in calculations because it is so close to 1

Example: Energy Equation (pressure at pump outlet) V = 1.6 m/s α = 1.05 hL = 1.44 m 2.4 m 2m

hP =

p out

g

+ z out + a

out

V o2u t + hL 2g

50 L/s

hP = 10 m

4m

datum

2 ⎡ ⎤ (1.6 m/s) 3 p2 = (9810N/m )⎢(10m)− (2.4m)− (1.05) − (1.44m)⎥ 2 2(9.81m/s ) ⎣ ⎦

Example: Energy Equation (Hydraulic Grade Line - HGL) ¾We would like to know if there are any places in the pipeline where the pressure is too high (_________) or too low (water might boil - cavitation). ¾Plot the pressure as piezometric head (height water would rise to in a piezometer) ¾How?

Example: Energy Equation (Energy Grade Line - EGL) p = 59 kPa

Loss due to shear HP = 10 m

Entrance loss

Exit loss

2

p

V +a 2g g

50 L/s 4m

2.4 m 2m datum

What is the pressure at the pump intake? p in

g

+ z in + a

in

V in2 p out + hP = + z out + a 2g g

out

V o2u t + hT + h L 2g

EGL (or TEL) and HGL EGL =

p

γ

+ z +α

V

2

2g

HGL =

p

γ

+ z

What is the difference between EGL defined by Bernoulli and EGL defined here?

EGL (or TEL) and HGL ¾The energy grade line may never be horizontal or slope upward (in direction of flow) unless energy is added (______) ¾The decrease in total energy represents the head loss or energy dissipation per unit weight ¾EGL and HGL are ____________and lie at the free surface for water at rest (reservoir) ¾Whenever the HGL falls below the point in the system for which it is plotted the local

Example HGL and EGL velocity head α

V

2

2g pressure head

p

γ energy grade line hydraulic grade line

z elevation pump z=0 p in

g

+ z in + a

datum

2 in

in

V p + hP = out + z out + a 2g g

out

V o2u t + hT + h L 2g

See you next time.

View more...
Outline ¾ Momentum equation ¾ Applications of the momentum equation

Objectives

•

After completing this chapter, you should be able to • • •

Identify the various kinds of forces and moments acting on a control volume. Use control volume analysis to determine the forces associated with fluid flow. Use control volume analysis to determine the moments caused by fluid flow.

6.1 Development of the Momentum Principle

Start with a modified-form of Newton’s 2nd law:

( )

r r d mV ΣF = dt

r ∑F

= sum of forces on fluid system

r mV = linear momentum of system

r r Also, ΣFdt = d (mV ) s-------Impulse-momentum principle

6.1 Development of the Momentum Principle

Momentum entering = ρ A1u 1δ tu 1

Momentum leaving = ρ A 2 u 2 δ tu 2

Force = rate of change of momentum ( ρA2u 2δ tu 2 − ρA1u1δ tu1 ) F= δt Q= A1 u1 = A2 u2,

F = ρQ (u 2 − u1 )

Fig 6.1 Momentum in a flowing fluid

Momentum equation for twoand three-dimensional flow The force in the x-direction

Fig 6.2 Two dimensional flow in a streamtube

Momentum equation for twoand three-dimensional flow The force in the y-direction

Fig 6.2 Two dimensional flow in a streamtube

Momentum equation for twoand three-dimensional flow

Total force exerted on the fluid =

Rate of change of momentum through the control volume

r r r F = ρQ(u2 − u1 ) For steady flow with one inlet and one outlet, the momentum equation is

r r r F = ρQ( β 2u2 − β1u1 )

β—— momentum correction factor

Momentum correction factor True momentum per unit time = β × Mass per unit time × Mean velocity β = ∫ u 2 dA / V 2 A The momentum equation

r r r F = ρQ( β 2V2 − β1V1 )

the resultant force

F = Fx2 + Fy2

the angle which this force acts at is given by

What are the forces acting on the fluid in the control volume?

the total force, FT, is given by the sum of these forces

r r r r FT = FR + FB + FP

Step in Analysis with Momentum Equation

1. Draw a control volume: Based on the problem, selecting the stream between two gradually varied flow sections as the control volume; 2. Decide on co-ordinate axis system: Determining the directions of co-ordinate axis, magnitudes and directions of components of all forces and velocities on each axis. 3. Plotting diagram for computation : Analyzing the forces on control volume and plotting the directions of all forces on the control volume. 4. Writing momentum equation and solving it: Substituting components of all forces and velocities on axes into momentum equation and solving it. All the pressures are relative to the relative pressure.

Application of the Momentum Equation

1. Force due to the flow of fluid round a pipe bend. 2. Force on a nozzle at the outlet of a pipe. 3. Impact of a jet on a plane surface. 4. Force due to flow round a curved vane.

Force due to Jet Striking Surface

Force by Flow Round a Pipe-bend

Example 1 Find the horizontal thrust of the water on each meter of width of the sluice gate shown in the Fig., given y1=2.2 m, y2=0.4 m, and y3=0.5 m. Neglect friction. ( 6.4.2)

Solution

6.5.5

Solution The flow rate Momentum equation

6.8 A reducing right-angled bend lies in a horizontal plane. Water enters from the west with a velocity of 3 m/s and a pressure of 30 kPa, and it leaves toward the north. The diameter at the entrance is 500 mm and at the exit it is 400 mm. Neglecting any friction loss, find the magnitude and direction of the resultant force on the bend.

Solution

Energy

Momentum

Example Water flows through a reducing 180°bend. The bend is shown in plan. Determine the magnitude of the force exerted on the bend in the x-direction. Assume energy losses to be negligible.

Solution:

Example T=15 oC

• • •

Given: Figure Find: Horizontal force required to hold plate in position Solution:

Q=0.4 m3/s

F

B

pA=75 kPa

V A2 p B VB2 + zA + = + zB + γ γ 2g 2g

pA

VB2 = γ 2g

pA

r r V = Vi

− F = ρQ(V2 − V1 )

p VB = 2 A = 2 * 75000 / 999 = 12.3 m / s

ρ

F = ρQV = 999 * 0.4 *12.3 = 4.9 kN

question

If the value of force calculated from momentum equation is negative, what does that mean? Does the magnitude of the unknown force has anything or nothing to do with that of the control volume? How to select control volume in the application?

If the value of force calculated from momentum equation is negative, what does that mean? Does the magnitude of the unknown force has anything or nothing to do with that of the control volume? How to select control volume in the application? directions are inverse；independency（when there is no gravitation）；calculated crosssection and solid wall

Sluice Gate

Find: Force due to pressure on face of gate Solution: Assume: v1 and v2 are uniform (so pressure is hydrostatic)

Application of the Energy, Momentum, and Continuity Equations in Combination In general, when solving fluid mechanics problems, one should use all available equations in order to derive as much information as possible about the flow. For example, consistent with the approximation of the energy equation we can also apply the momentum and continuity equations

Forces on Transitions Example

Example: Energy Equation (energy loss) An irrigation pump lifts 50 L/s of water from a reservoir and discharges it into a farmer’s irrigation channel. The pump supplies a total head of 10 m. How much mechanical energy is lost? What is hL? cs2 4m 2.4 m 2m cs1 datum Why can’t I draw the cs at the end of the pipe? p in

g

+ z in + a

h p = z out + hL

in

V in2 p + hP = out + z out + a 2g g

hL = h p - z out

out

V o2u t + hT + h L 2g

Example: Energy Equation (pressure at pump outlet) The total pipe length is 50 m and is 20 cm in diameter. The pipe length to the pump is 12 m. What is the pressure in the pipe at the pump outlet? You may assume (for now) that the only losses are frictional losses in the pipeline. 50 L/s hP = 10 m 4m cs2 2.4 m 2m cs1 datum p in

g

+ z in + a

in

V in2 p + hP = out + z out + a 2g g

out

V o2u t + hT + h L 2g

We need _______ in the pipe, __, and ____ ____.

Example: Energy Equation (pressure at pump outlet) ¾How do we get the velocity in the pipe?

¾How do we get the frictional losses?

¾What about α?

Kinetic Energy Correction Term: α ¾α is a function of the velocity distribution in the pipe. ¾For a uniform velocity distribution ____ ¾For laminar flow ______ ¾For turbulent flow _____________ ¾Often neglected in calculations because it is so close to 1

Example: Energy Equation (pressure at pump outlet) V = 1.6 m/s α = 1.05 hL = 1.44 m 2.4 m 2m

hP =

p out

g

+ z out + a

out

V o2u t + hL 2g

50 L/s

hP = 10 m

4m

datum

2 ⎡ ⎤ (1.6 m/s) 3 p2 = (9810N/m )⎢(10m)− (2.4m)− (1.05) − (1.44m)⎥ 2 2(9.81m/s ) ⎣ ⎦

Example: Energy Equation (Hydraulic Grade Line - HGL) ¾We would like to know if there are any places in the pipeline where the pressure is too high (_________) or too low (water might boil - cavitation). ¾Plot the pressure as piezometric head (height water would rise to in a piezometer) ¾How?

Example: Energy Equation (Energy Grade Line - EGL) p = 59 kPa

Loss due to shear HP = 10 m

Entrance loss

Exit loss

2

p

V +a 2g g

50 L/s 4m

2.4 m 2m datum

What is the pressure at the pump intake? p in

g

+ z in + a

in

V in2 p out + hP = + z out + a 2g g

out

V o2u t + hT + h L 2g

EGL (or TEL) and HGL EGL =

p

γ

+ z +α

V

2

2g

HGL =

p

γ

+ z

What is the difference between EGL defined by Bernoulli and EGL defined here?

EGL (or TEL) and HGL ¾The energy grade line may never be horizontal or slope upward (in direction of flow) unless energy is added (______) ¾The decrease in total energy represents the head loss or energy dissipation per unit weight ¾EGL and HGL are ____________and lie at the free surface for water at rest (reservoir) ¾Whenever the HGL falls below the point in the system for which it is plotted the local

Example HGL and EGL velocity head α

V

2

2g pressure head

p

γ energy grade line hydraulic grade line

z elevation pump z=0 p in

g

+ z in + a

datum

2 in

in

V p + hP = out + z out + a 2g g

out

V o2u t + hT + h L 2g

See you next time.

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