Attachment(4).ppt

FLUID DYNAMICS- Fluids in Motion This section discusses the analysis of fluid in motion - fluid dynamics. The motion of fluids can be predicted in the same way as the motion of solids are predicted using the fundamental laws of physics together with the physical properties of the fluid Classification of fluid flow • There are many different types of fluid flow, although they may not be readily apparent • One of the first steps in investigating a problem involving moving fluids is to define the type of flow that you are dealing with • This gives an idea of which equations can be legitimately be applied to the problem. • This is to avoid applying wrong equations to wrong situations

FLUID DYNAMICS- Fluids in Motion Classification of fluid flow

uA

x t

FLUID DYNAMICS- Fluids in Motion Classification of fluid flow Laminar flow: if you were to put a tracer and watch it flow, you will see that it doesn’t disperse. It is usually associated with slow moving, viscous fluids. In laminar flow the motion of the fluid particles is very orderly with all particles moving in straight lines parallel to the pipe wall.Example is flow of water through an aquifer Turbulent flow: if you were to a tracer and watch it flow, you will see that it disperses and it is not uniform. It is usually associated with much faster and chaotic flow. Mixing is very significant in turbulent flow, in which fluid particles move haphazardly in all directions. It is therefore impossible to trace motion of individual particles in turbulent flow. A good example is water flowing in a mountain The Reynolds number is usually used to determine whether flow is laminar, transitional or turbulent.

FLUID DYNAMICS- Fluids in Motion Classification of fluid flow • After conducting experiments Reynolds argued that fluid flow is mainly influenced by viscosity (μ), density () and diameter of pipe (d). • These factors (μ,,d) were later combined with the velocity(V) of the fluid to form a dimensionless group called the Reynolds number, R.

For flow in circular pipes When R < 2000 – laminar flow R > 4000 – Turbulent flow A region of instability exists at the transition stage between these two flow regimes.(2000 ≤ Re ≤ 4000)

FLUID DYNAMICS- Fluids in Motion Worked Example • A pipe of diameter 0.1 m carries water at the rate of 0.025m3/s. Taking the density of water as 1000 kg/m3 and its dynamic viscosity as 1.005x10-3kg/ms, calculate the Reynolds number of the flow and determine whether it is laminar or turbulent Solution A = D2/4 = 3.14x(0.1)2/4 = 7.85x10-3m2 V= Q/A = 0.025/7.85x10-3m2 = 3.18 m/s Re = (1000x3.18x0.1)/1.005x10-5 = 316400 The flow is fully turbulent as 316400 is far much greater than 4000

FLUID DYNAMICS- Fluids in Motion Classification of fluid flow Uniform flow: · If the flow depth remains constant and velocity is the same magnitude and direction at every point in the fluid, then flow is said to be uniform. That is if flow attributes remain the same with respect to space Non Uniform flow: · If the flow depth and velocity change with space, then the flow is said to be non-uniform.

Steady flow: A steady flow is one in which the flow attributes (velocity, depth, pressure and cross-section area of flow) may differ from point to point but do not change with time. Thus flow attributes should be constant with respect to time. Unsteady flow: If at any point in the fluid, the conditions change with time, the flow is described as unsteady. (In practise there is always slight variations in velocity and pressure, but if the average values are constant, the flow is considered steady).

FLUID DYNAMICS- Fluids in Motion Classification of fluid flow .Combining the above we can classify any flow in to one of four types : A. Steady uniform flow. Conditions do not change with position in the stream or with time. An example is the flow of water in a pipe of constant diameter at constant velocity. B. Steady non-uniform flow. Conditions change from point to point in the stream but do not change with time. An example is flow in a tapering pipe with constant velocity at the inlet - velocity will change as you move along the length of the pipe toward the exit. C. Unsteady uniform flow. At a given instant in time the conditions at every point are the same, but will change with time. An example is a pipe of constant diameter connected to a pump pumping at a constant rate which is then switched off.

D. Unsteady non-uniform flow. Every condition of the flow may change from point to point and with time at every point. For example waves in a channel.

FLUID DYNAMICS- Fluids in Motion Fluid Flow Equations When analyzing fluid motion, we might take one of two approaches: (1) seeking to describe the detailed flow pattern at every point (x, y, z) in the filled volume of fluid or (2) working with finite region, making a balance of flow in versus flow out, and determining gross flow effects such as force, or torque on a body, or total energy exchange.

The second approach is the control volume concept and is the subject of this course.

FLUID DYNAMICS- Fluids in Motion Fluid Flow Equations We shall go through the three basic control volume relations in fluid mechanics:

 The principle of conservation of mass, from which the continuity equation is developed;  The principle of conservation of energy, from which the energy equation is derived;  The principle of conservation of linear momentum, from which equations evaluating dynamic forces exerted by flowing fluids may be established.

FLUID DYNAMICS- Fluids in Motion Fluid Flow Equations Control volume concept  A control volume is a finite region, chosen carefully by the analyst for a particular problem, with open boundaries through which mass, momentum and energy are allowed to cross . The analyst makes a budget, or balance, between incoming and outgoing fluid and the resultant changes within the control volume. Therefore, one can calculate the gross properties (net force, total power output, total heat transfer, etc) with this method.

FLUID DYNAMICS- Fluids in Motion Fluid Flow Equations Control volume concept  With this method however, we do not care about the details inside the control volume (in other words we can treat the control volume as a black box).

FLUID DYNAMICS- Fluids in Motion Continuity Equation o it is based on the conservation of mass principle

Q1 = A1V1 = A2V2 = Q2 Where A is the cross-sectional area of flow, V is the velocity of flow and Q is the flow rate

FLUID DYNAMICS- Fluids in Motion Continuity Equation Example Water flows through a branching pipeline shown below. diameter, D2, is 250 mm, V2 = 1.77 m/s and V3 = 1.43 m/s 2 1 3

(a) What diameter, D3, is required for Q3 = 2Q2? (b) What is the total discharge at section 1 or Q1?

If the

FLUID DYNAMICS- Fluids in Motion Continuity Equation Solution (a) Q3 =2Q2 so A3V3 2x(3.14x0.25x0.25/4)*1.77

=

2A2V2



(3.14xD3xD3/4)x1.43

Thus D3 = 0.393 m (b) Q2 =A2V2 = (3.14x0.25x0.25/4)x1.77 = 0.087 m3/s Q3 =2Q2 = 2x0.087 = 0.174 m3/s Total discharge = Q1 = Q2 + Q3 = 0.087 + 0.174 = 0.261 m3/s

=

FLUID DYNAMICS- Fluids in Motion Continuity Equation Example Two separate pipelines (1 and 2) join together to form a larger pipeline(3). It is known that D1 = 0.2 m, D2 = 1.0 m, Q2 = 0.23 m3/s and Q3 = 0.35 m3/s 1 3 2

(a) What is the value of Q1,V1 and V2 (b) If V3 must not exceed 3.00 m/s, what is the minimum diameter, D3, that can be used?

FLUID DYNAMICS- Fluids in Motion Continuity Equation Solution (a) Q1 = Q3 – Q2 = 0.35 - 0.23 = 0.12 m3/s V1 =Q1/A1 = (0.12)/(3.14xd2/4) = (4x0.12)/(3.14x(0.2)2) = 3.82 m/s V2 =Q2/A2 = (0.23)/(3.14xd2/4) = (4x0.23)/(3.14x(1.0)2) = 0.293 m/s (b) Q3 = A3V3 = 0.35  A3 =Q3/V3

i.e, 0.35/3 = (3.14D2/4)  D = (4x0.35)/(3x3.14) =0.385 m

FLUID DYNAMICS- Fluids in Motion Energy Equation

This means that the increase in velocity is countered by decrease in Pressure for example, such that the total energy of the system remains the same P1

Z1

P2 Z2

FLUID DYNAMICS- Fluids in Motion Energy Equation

Two points joined by streamline

The Bernoulli equation applies to conditions along a streamline. We can apply it between two points, 1 and 2, on the streamline as above

FLUID DYNAMICS- Fluids in Motion Energy Equation

Two points joined by streamline

Total energy per unit weigh t at 1  Total energy per unit weigh t at 2 Total head at 1  Total heat at 2 P1 V12 P2 V22   z1    z2 g 2 g g 2 g

FLUID DYNAMICS- Fluids in Motion Energy Equation Bernoulli’s equation has some restrictions in its applicability, they are: (assumptions)  Flow is steady;  Density is constant (which also means the fluid is incompressible)  Friction losses are negligible  The equation relates the states at two points along a single streamline, (not conditions on two different streamlines) All these conditions are impossible to satisfy at any instant in Reality! Fortunately for many real situations where the conditions are approximately satisfied, the equation gives very good results.

FLUID DYNAMICS- Fluids in Motion Energy Equation Bernoulli’s equation has some restrictions in its applicability, they are:  Flow is steady;  Density is constant (which also means the fluid is incompressible)  Friction losses are negligible  The equation relates the states at two points along a single streamline, (not conditions on two different streamlines) All these conditions are impossible to satisfy at any instant in reality. Fortunately for many real situations where the conditions are approximately satisfied, the equation gives very good results.

FLUID DYNAMICS- Fluids in Motion Energy Equation With real fluids and not ideal fluids the Bernoulli’s equation becomes:

2 1

2 2

P1 V P2 V   z1    z2  h f g 2 g g 2 g Where hf denotes the headloss/energy loss (friction etc)

FLUID DYNAMICS- Fluids in Motion Worked Example Water flows through a pipeline of constant diameter that is inclined upwards. On the centreline of the pipe, point 1 is 0.3 m below point 2. The pressure at point 1 is 9.3 x103 N/m2. What is the pressure at point 2 if there is no loss of energy

 0.3 m 

FLUID DYNAMICS- Fluids in Motion Worked Example The diameter of the pipe is constant so (V1)2/2g = (V2)2/2g Therefore the velocity heads are equal and therefore cancel out and the Bernoulli’s equation becomes Z1 +P1/g = Z2 + P2/g Taking the datum level through point 1, Z1 = 0 and Z2 = 0.3 m Therefore 0 + (9.3x103/(1000x9.81) = 0.3 + P2/g P2/g = 0.948 – 0.300 = 0.648 m of water P2 = 0.648x1000x9.81 = 6.36x103 N/m2

FLUID DYNAMICS- Fluids in Motion Worked Example Water flows through an expanding pipeline that is inclined upwards. On the centreline of the pipe, point 1 is 0.3 m below point 2. The velocities are V1 = 3.1 m/s and V2 = 1.7 m/s. The pressure at point 1 is 9.3 x103 N/m2. What is the pressure at point 2 if there is no loss of energy  0.3 m 

FLUID DYNAMICS- Fluids in Motion Worked Example Z1 +P1/g +(V1)2/2g = Z2 + P2/g + (V2)2/2g Taking the datum level through point 1, Z1 = 0 and Z2 = 0.3 m Therefore 0 + (9.3x103/(1000x9.81) + (3.12/19.62)= 0.3 + P2/g + (1.72/19.62) 0 +0.948 +0.490 = 0.3 + P2/g + 0.147 P2/g = 0.991 m of water

P2 = 0.991x1000x9.81 = 9.72x103 N/m2

FLUID DYNAMICS- Fluids in Motion Worked Example Water flows through a pipeline which reduces in cross section. The centreline of the pipe is horizontal. If V1 = 1.54 m/s and V2 = 2.65 m/s, P1 = 20.00 x103 N/m2 and P2 = 16.89 x103 N/m2 . What is the energy loss between sections 1 and 2? Give answer in m of water

FLUID DYNAMICS- Fluids in Motion Worked Example

Z1 +P1/g +(V1)2/2g = Z2 + P2/g + (V2)2/2g + loss The centreline of the pipe is horizontal so z1 = z2 and these terms cancel. Therefore (20.00x103/(1000x9.81) + (1.542/19.62)= (16.89x103/(1000x9.81) + (2.652/19.62) 0.121 + 2.039 = 0.358 + 1.722 + loss 2.160 = 2.080 + loss Energy loss = 0.080 m of water

FLUID DYNAMICS- Fluids in Motion Momentum Equation o The simplest definition of momentum is that momentum = mass x velocity

o Therefore a body has momentum by virtue of the fact that it is moving. o If the velocity is zero, then the momentum is zero

Review of Newton’s Laws of Motion 1. A body will remain in the same condition of rest or of motion with uniform velocity in a straight line until acted on by an external force 2. The rate of change of momentum of a body is proportional to the force acting upon it and it takes place in the line of action of the force. For a body of mass, M Force = rate of change of momentum = (MV2-MV1)/t = M(V2-V1)/t = Ma

FLUID DYNAMICS- Fluids in Motion Momentum Equation o Where V1 and V2 are the initial and final velocity of the body respectively, t is the time over which the change of velocity occurs, and a is the acceleration of the body. 3. To every action there is an equal and opposite reaction

FLUID DYNAMICS- Fluids in Motion Momentum Equation

FLUID DYNAMICS- Fluids in Motion Momentum Equation Therefore we can write the momentum equation suitable for hydraulic systems Since Q = A1V1 then F = Q(V2-V1) = A1V1(V2-V1)

Velocity as a Vector Quantity Velocity is a vector, so magnitude and direction must be considered when applying the momentum equation. This means that a force can arise in one or two ways

FLUID DYNAMICS- Fluids in Motion Momentum Equation Velocity as a Vector Quantity 1. Force due to change in velocity magnitude. An example could be a straight pipe along the x-axis that reduces in diameter so that V2x > V1x. A force would be exerted on the pipe taper section since there is a change of momentum and Fx = Q(V2x – V1x) 2. Force due to change in velocity direction Although the velocity in a pipeline of constant diameter does not change when the pipe bends through an angle  degrees the velocity component in say the x direction changes from V1 initially to V1cos(). So a force is still exerted when there is no change in velocity magnitude but there is change of direction

FLUID DYNAMICS- Fluids in Motion Momentum Equation Velocity as a Vector Quantity

2. Force due to change in velocity direction Although the velocity in a pipeline of constant diameter does not change when the pipe bends through an angle  degrees the velocity component in say the x direction changes from V1 initially to V1cos(). So a force is still exerted when there is no change in velocity but there is change of direction 3. Thus force can change due to change in both magnitude and direction

FLUID DYNAMICS- Fluids in Motion Momentum Equation Applying the control volume concept to momentum principle

Suppose we want to analyse the flow of water around the pipe bend as shown above with the objective of calculating the resultant force, FR

FLUID DYNAMICS- Fluids in Motion Momentum Equation

The control volume for the pipe bend has the same geometry as the real pipe. Then the control volume concept is that

FLUID DYNAMICS- Fluids in Motion Momentum Equation The control volume for the pipe bend has the same geometry as the real pipe. Then the control volume concept is that The algebraic sum of external forces acting on the fluid in the control volume in a given direction, F = The rate of change of momentum in the given direction as a result of fluid passing through the control volume, Q(V2 – V1)

FLUID DYNAMICS- Fluids in Motion Momentum Equation Thus we ignore the equal and opposite internal forces acting on the control volume. If we write the force in the x and y directions then they are as follows Fx = Q(V2x – V1x) Fy = Q(V2y – V1y)

FLUID DYNAMICS- Fluids in Motion Momentum Equation Applying the control volume concept to momentum principle Some steps to set up a problem prior to analysis 1. Draw the hydraulic system, then draw the imaginary control volume which represents the part of the system to be analysed. 2. Use arrows to show the direction of travel of the liquid entering and leaving the control volume. Label them V1 and V2 to represent the velocities 3. Label the axes, x and y for a two dimensional problem in the horizontal plane, x and z for a problem in the vertical plane . The axes are positive in the initial direction of the fluid as it enters the control volume. For example, in the diagram above , x is positive from left to right and y in the upward vertical direction

FLUID DYNAMICS- Fluids in Motion Momentum Equation Applying the control volume concept to momentum principle Some steps to set up a problem prior to analysis 4. Draw the external forces acting on the control volume. This includes the external pressure forces (PA) acting on the ends of the pipe, and the resultant force, FR.

5. Vector quantities such as velocity, pressure and the unknown resultant force must be resolved in the direction of the axes before the values are put into the momentum equation If you do not know the direction in which FRX and FRY act initially, guess. Having applied the momentum equation, your guess is correct if the answer obtained for FRY or FRX is positive and wrong if it is negative

FLUID DYNAMICS- Fluids in Motion Momentum Equation Applying the control volume concept to momentum principle Some steps to set up a problem prior to analysis 6. All forces acting in the same direction as positive axes are positive, those acting in the opposite direction are negative. Use these signs when evaluating FX, FY or FZ

FLUID DYNAMICS- Fluids in Motion Momentum Equation Applying the control volume concept to momentum principle

To calculate the resultant force, FR,the steps are

FLUID DYNAMICS- Fluids in Motion Momentum Equation 1. Draw the control volume and its external forces as outlined before 2. Work out the algebraic sum of external forces acting on the control volume in the x direction(FX). This includes the unknown FRX that we are calculating. 3. Equate FX from step 2 to the rate of change of momentum in the x direction calculated from the product of mass flow rate and the change of velocity, that is Q(V2X – V1X) 4. Solve for the unknown force, FRX 5. Repeat steps 2 to 4 but for the y direction to obtain FRY 6. Calculate FR = (FRX2 +FRY2)1/2 and its angle to the horizontal is given by  = tan-1 (FRY/FRX)

FLUID DYNAMICS- Fluids in Motion Applying the Momentum Equation Will use the momentum equation to evaluate forces acting on pipe bends and nozzles A. Pipe Bends Assuming there is no loss of momentum and that there is no turbulence at the bend then X direction: P1A1 –P2A2cos() – FRX = Q(V2cos() – V1) Y direction: FRY – P2A2Sin() = Q(V2sin()) This is when the flow is horizontal or xy plane

FLUID DYNAMICS- Fluids in Motion Momentum Equation Applying the control volume concept to momentum principle

This is when the flow is horizontal

FLUID DYNAMICS- Fluids in Motion Momentum Equation Applying the control volume concept to momentum principle

This is when flow is vertical

FLUID DYNAMICS- Fluids in Motion

Pipe bend

FLUID DYNAMICS- Fluids in Motion

Worked Example A pipeline of constant diameter of 0.30 m turns through an angle of 60°. The centreline of the pipe does not change elevation. The discharge through the pipeline is 0.1 m3/s of water, and the pressure at the bend is 30 m of water. Calculate the magnitude and direction of the resultant force on the pipe.

FLUID DYNAMICS- Fluids in Motion

Worked Example A pipeline of constant diameter of 0.30 m turns through an angle of 60°. The centreline of the pipe does not change elevation. The discharge through the pipeline is 0.1 m3/s of water, and the pressure at the bend is 30 m of water. Calculate the magnitude and direction of the resultant force on the pipe.

FLUID DYNAMICS- Fluids in Motion Worked Example STEP 1: Apply the continuity equation Q= A1V1 = A2V2 , where A1 =A2= (x0.32)/4 = 0.071 m2 0.1 = 0.071x V1  V1 = 1.41 m/s = V2 STEP 2: Apply the momentum equation in the x and y directions, remembering the sign convention P1 = P2 = gh = 1000x9.81x30 = 294.30 x 103 N/m2 For x direction:P1A1 –P2A2cos() – FRX = Q(V2cos() – V1) 294.30x103x0.071 – 294.30x103x0.071cos(60°) –FRX = 1000x0.1(1.41xcos(60°)-1.41) 20.90 x 103 – 10.45x103 –FRX = 1000x0.1x(-0.71) FRX =10.52 x103 N Note: the +ve answer indicates that FRX is acting in the direction assumed

FLUID DYNAMICS- Fluids in Motion Worked Example STEP 2: Y direction: FRY – P2A2Sin() = Q(V2sin()) : FRY –294.30x103x0.071Sin(60°) =1000x0.1x(1.41Sin(60°)) FRY – 18.10x103 = 1000x0.1x1.22 FRY = 18.22 x 103 N Note: the +ve answer indicates that FRY is acting in the direction assumed

FLUID DYNAMICS- Fluids in Motion Worked Example STEP 3: Calculate the magnitude and direction of the resultant force FR = (FRX2 +FRY2)1/2 = 103(10.522 + 18.222)1/2 = 21.0 x103N and its angle to the horizontal is given by  = tan-1 (FRY/FRX)= tan-1(18.22/10.52) = 60°. This is the magnitude and direction of the external force exerted by the pipe on the water.

The water exerts an equal force on the pipe in the opposite direction

FLUID DYNAMICS- Fluids in Motion Applying the Momentum Equation Will use the momentum equation to evaluate forces acting on pipe bends and nozzles B. Nozzles The momentum equation can also be used to investigate force exerted by a stream of moving liquid on a nozzle

FLUID DYNAMICS- Fluids in Motion Applying the Momentum Equation B. Nozzles

Notice how the one dimensional system greatly simplifies matters

FLUID DYNAMICS- Fluids in Motion Applying the Momentum Equation

The liquid tries to force the nozzle off the end of the pipe, so it must be held in place by for example , a number of bolts around the flange of the pipe. If the force, F, is calculated and is known what tensile force one bolt can withstand, then a suitable connection with an appropriate number of bolts can be designed. In this situation, there is change in velocity as a result of the reducing area of the nozzle, but no change in direction. So the momentum equation is applied along the direction of motion

FLUID DYNAMICS- Fluids in Motion Applying the Momentum Equation This means the subscripts are omitted. Using the same sign convention as before, forces are positive if acting from left to right, so P1A1 –P2A2 – FR = Q(V2 – V1) this can be simplified by assuming that because the nozzle is discharging to the atmosphere the pressure of jet P2, will equal the atmospheric pressure, i.e., P2 = 0. then FR = P1A1 – Q(V2 – V1)

FLUID DYNAMICS- Fluids in Motion

Worked Example

A pipeline reduces in diameter using a standard symmetrical taper as shown below. Given the following information, calculate the force exerted by the water on the taper section: Q = 0.42 m3/s, D1 = 0.60 m, D2 = 0.30 m, P1 = 25.30 m water, P2 = 9.13 m water,  = 1000 Kg/m3

FLUID DYNAMICS- Fluids in Motion Worked Example STEP 1: Apply the continuity equation Q= A1V1 = A2V2 , A1 =(x0.62)/4= 0.283 m2 and A2= (x0.32)/4 = 0.071 m2 0.42 =0.283V1 ,  V1 = 1.48 m/s 0.42 =0.071V2 ,  V2 = 5.92 m/s

STEP 2: Apply the momentum equation in the direction of motion P1 = P1 = gh = 1000x9.81x25.30 = 248.19 x 103 N/m2 P2 = P2 = gh = 1000x9.81x9.13 = 89.57 x 103 N/m2

P1A1 –P2A2 – FR = Q(V2– V1)

248.19x103x0.283 – 89.57x103x0.071–FR =1000x0.42(5.92-1.48) 70.24 x 103 – 6.36x103 –FR = 1.86x103 FR =62.0 x103 N The internal force exerted by the water on the taper is 62.0x103 N from left to right