Lecture 5 :1 : Fluid Motion dynamics I : Momentum Conservation Last Lecture we covered :

1/9

Using a stream tube as a basis, derived a force balance along the stream tube and showed that the acceleration of the fluid must be balanced by the pressure force and also any change in height in the presence of gravity. In the absence of a gravity force (ie in a horizontal stream tube) the change in velocity must be balanced by a change in pressure. We integrated this force balance to produce an energy balance equation, known as the Bernoulli equation. We showed that the total energy, the sum of kinetic, potential and pressure energy components is constant along a streamline. Because the Bernoulli equation uses a streamtube as a basis, it is only valid in steady flows. Also no viscous forces must be present and the relation presented here assumes an incompressible fluid. This lecture we are going to cover Newtons Second Law, Definition of Momentum, relation to acceleration and force Revisit the general Conservation law for mass (a scalar quantity) Derive the general Conservation law for momentum (a vector quantity)

1

5:1 : Mass, Velocity, Momentum, Force and Acceleration.

2/9

Scalar and Vector Quantities Mass (extensive) and its intensive quantity (density mass per unit volume) are scalars Momentum (extensive) and its intensive quantity (velocity momentum per unit mass) are vectors Force (extensive) and its intensive quantity (acceleration force per unit mass) are all vectors. So when we write a conservation equation for mass (or density) we write one equation. When we write a conservation equation for momentum (or velocity) we write three equations, one for each direction. Momentum in the x-direction is Rate of change of momentum is

So Newtons Second Law is a rate of momentum change equation.

mU x = M x , the momentum vector is d (mU x ) = Fx = max dt

r r mU = M

Before we get onto deriving a momentum conservation equation remind ourselves of mass conservation

2

5:1 : Deriving the general mass conservation equation (from lecture 4.2) Imagine we have some arbitrary 2D flowfield, defined by a set of streamlines. Now suppose we draw an imaginary area somewhere in this flowfield. Now suppose we want to ensure mass is conserved within this area (volume in 3D) We do this by defining the mass flow through a tiny elemental area

3/9

A .

r At this tiny elemental area A , the velocity vector, U passes through it at a different angle to the unit normal r of the area, n , a vector of unit magnitude.Conceptually, the mass flow through the area element is due only to the normal component of the velocity.

3

r r What we want to do is find out how much of U is going in the n direction. r U cos From simple trigonometry normal componentGeneral dot product definition :

5:1 : To find the normal component of the mass flow through a surface element

4/9

r r r r a b = a1b1 + a2 b2 + a3b3 = a b cos

r r r U n = U cos In this case :

r r & So the mass flow through the small element is m = U n A

(

)

All elements, and mass conservation over the entire surface (volume)

r r & m = U n A r r U n A = 0

(

)

For incompressible fluids the net mass flow must be zero, ie

(

)

4

5:1 : Mass Conservation : The velocity field carries mass information Density = mass per unit volume = specific mass The mass flow through the small element is

5/9

r r & m = U n AHere imagine the [steady] velocity field carrying density ( ) information into or out of each surface element of the entire volume. All elements, and mass conservation over the entire surface (volume)

(

)

r r & m = U n AMass conservation for the volume is obeyed if the net mass flow across the whole surface is zero.

(

)

5

5:1 : Force conservation : The velocity field carries momentum information

6/9

r U = momentum per unit volume = specific momentumMomentum flow in the x-direction through the element would be

r r & = U U n A M x x

(

)

[kg/m3.m2/s2.m2 = kg.m/s2 = N] Here, imagine the velocity field is carrying a small packet of x-momentum out of the volume All elements, and momentum conservation over the entire surface (volume)

r r & = U U n A M x xThis gives us the net momentum flow through the volume we now need to use Newtons Law and permit unsteady flow to define a general momentum conservation equation.

(

)

6

5:1 : Conservation of Momentum Equation Conservation of momentum is .

7/9

Rate of accumulation of momentum in the volume + net rate of momentum leaving the volume = force applied to the surface of the volume If the volume has a volume change of this is

V

, then the x-momentum in the volume is

U V , and the rate ofx

d dt

( U V )x

and the force applied we simply put

F

x

So the fundamental equation for the conservation of momentum in the x-direction is :

d dt

(

r r U xV + U x U n A = Fx d dt

)

(

)

The general form :

(

r r r r r UV + U U n A = F

)

(

)

which you will almost never use it this year.

In later years you will use it religiously, it is known as the Navier-Stokes Equations. It is the fluids equation

7

5:1 : Summary of this Lecture

8/9

Reminded ourselves that mass and density are scalar extensive and intensive quantities And that Force and acceleration are vector extensive and intensive quantities And that momentum and velocity are also vector extensive and intensive quantities Reviewed the derivation of the mass conservation equation Applied the same principle to the conservation of momentum across a small volume Invoked Newtons 2nd Law and unsteady conditions to define a general momentum conservation law.

8

5:1 : Suggested Reading/Reminders Crowe et al. Engineering Fluid Mechanics, 9th ed. SI version o Section 6.1-6.4,6.6 James A. Fay Introduction to Fluid Mechanics, 1994, MIT Press o Section 5.1-5.4 Fox, Pritchard and McDonald, Introduction to Fluid Mechanics, 7th ed, SI units o Section 4.4 Cengel and Cimbala, Fluid Mechanics : Fundamentals and applications, 2th ed, SI units o Section 6.1-6.4

9/9

No Reminders

9

Lecture 5:2 : Examples

1/2

Last Lecture we covered : Reminded ourselves that mass and density are scalar extensive and intensive quantities And that Force and acceleration are vector extensive and intensive quantities And that momentum and velocity are also vector extensive and intensive quantities Reviewed the derivation of the mass conservation equation Applied the same principle to the conservation of momentum across a small volume Invoked Newtons 2nd Law and unsteady conditions to define a general momentum conservation law.

This lecture we are going to cover Momentum Problem examples

10

5:2 : Suggested Reading/Reminders Crowe et al. Engineering Fluid Mechanics, 9th ed. SI version o Section X James A. Fay Introduction to Fluid Mechanics, 1994, MIT Press o Section X Fox, Pritchard and McDonald, Introduction to Fluid Mechanics, 7th ed, SI units o Section X

2/2

No Reminders

11

Lecture 5:3 : Fluid Motion dynamics I : Actuator Disk theory

1/12

Last Lecture we covered : Momentum Problem examples This lecture we are going to cover The careful combination of momentum and mechanical energy conservation Propeller performance, through use of : Actuator disk theory.

12

5:3 : A propeller and turbines..

2/12

Propellers : Move through the fluid

Turbines : Fluid moves through the turbine

U is the speed of the object speed of advance V is the speed of the oncoming fluid Pin is the power input to drive the propeller Pout is the power extracted from the fluid motion In both cases Power transfer occurs by net pressure forces acting on the blades of the propeller The power is either transferred into forward motion (propeller) or to electricity (turbine) To analyse the problem, in the propeller case, we should change the frame of reference

13

5:3 : Changing the frame of reference in the Propeller case

3/12

For the turbine, our coordinate system (ie where we sit) is on the unit itself, next to us, and stationary In the propeller case, it is moving with respect to us, so we change the frame of reference by moving with the propeller. This means, relative to us, the propeller is stationary and sees an incoming air flow of velocity V, speed V With this transformation in place, we can now treat both systems with the same bit of physics.

14

5:3 : Making the Propeller a black box the actuator disk Flow through the blades is extremely complicated, and we do not need to know all the complexity from the point of view of working out the power the propeller can impart to/extract from the fluid flow. So we treat it as a simple disk into which flow goes in at a certain pressure, and comes out a certain pressure. It is a thin disk of diameter D Consider a 1-D flow, steady. Just considering the disk at the moment, considering the fluid state at stations 1 and 2.

4/12

A = D 2 / 4

& Mass conservation : m = AU 1 = AU 2 = const . Therefore

& Momentum Conservation ( p1 p 2 ) A + FX = m(U 2 U 1 ) = 0 Pressure difference must exist it generates the Force (thrust) Bernoulli :

U1 = U 2

U s22

+ p + gz = Const

U 12: Here :

Violated because of lots of viscous losses due to turbulence. 15

2

+ p1 =

U 222

+ p2

is violated .

5:3 : Actuator Disk Theory Can apply Bernoulli from 0->1 Can apply Bernoulli from 2->3 Cannot between 1->2. Assume Stations 0 and 2 are far upstream and downstream. We can assume p0 = p3 = patm U 1 = U 2 known

5/12

U 02Bernoulli between 0->1, 2->3 :

2 2 2 2 U 02 U 32 U 22 U 12 + p2 + + p atm = + p atm + + p1 Adding these equations gives 2 2 2 2 1 D 2 p 2 p1 = U 32 U 02 , therefore the Force is F = ( p 2 p1 )A = U 32 U 02 Which gives 2 8

+ p atm =

U 12

+ p1 ,

U 22

+ p2 =

U 32

+ p atm

(

)

(

)

16

5:3 : Actuator Disk Theory Disk theory predicts that

6/12

F D2 F (U 32 U 02 ).

Now let us examine the velocity difference in more detail. Consider a x-momentum balance from 0->3 We know no force is applied 0>1 and 2->3, so the only force present 0->3 is the disk. x-momentum balance 0->3 is :

M x , out M x , in = Fx , m = const &F=

therefore

& F = m(U 3 U 0 )

& m= ChooseWhich gives

D 24

U 1

D 28

and from Bernoulli : .

U1 =

(U 3 + U 0 )2

(U 32 U 02 ) , so 2U 1 (U 3 U 0 ) = (U 32 U 02 )

Half the speed increase occurs upstream of the propeller

17

5:3 : Thrust generated, Power consumed and obtained, the theoretical efficiency. Force (thrust ) :

7/12

& F = m(U 3 U 0 )& m=

D 24

Substitute :

U 1

and

U 3 = 2U 1 U 0

gives :

Force (thrust)

F=

D 24

U1 (2(U1 U 0 )) = D 23 1

D 2

U U12 1 0 U 2 1

Power consumed

FU1 =

U0 U 1 U 2 1 FU 0

Useful power of the propeller

Thrust x Velocity of advance

Perfect (lossless) efficiency

Useful//Consumed Power

perfect =

FU 0 U 0 = FU 1 U 1

18

5:1 : Discussion on the theoretical characteristics of propellers

8/12

Thrust (force) :2

& F = m(U 3 U 0 ) =2

D 2

U U12 1 0 U 2 1

Proportional to D ,U 1 ,U 0 U 1 which means Large rotors and high wind speeds at the rotor required, as is the streamline curvature. Power consumed D ,U 1 ,U 0 U 1 - highly nonlinear with wind speed. Theoretical Efficiency directly proportional to2 3

U 0 U 1 , 100% efficiency is impossible. The more power you extract via reducing U 0 U 1 , the less efficiently you do so. Bigger slower propellers/turbines are more efficient.

19

5: 3 : Actual Efficiency of propellors In reality there are losses due to : Frictional effects on the propeller blade surface Rotational energy imparted to the fluid when it goes through the propeller Pressure variations on the surfaces of our black box The power losses are (roughly) proportional to the momentum entering the propeller,

9/12

& PLOSS mU12 actual

and therefore

& PLOSS = CmU12

Therefore the actual efficiency is

FU 0 U 0 1 U 0 U1 = = 1 U U + C / 2 FU 1 + PLOSS U 1 0 1

Typical aircraft have efficiencies of ~0.85.

20

5:3 : Discussion on the theoretical characteristics of wind turbines For wind turbines the analysis is the same with the exception of the efficiency definition. Here Perfect (lossless) efficiency definition : Generated Power/Wind Power

10/12

perfect

FU 1 1 U3 = = 1 3 1 2 A1U 0 2 U 0

U 3 1 + U 0

2

Known as the Betz Limit. Added to issues like Losses Intermittency Noise Means wind power is, from some standpoints, not a good choice for generating energy.

21

5:1 : Summary of this Lecture

11/12

Derived actuator disk theory, which makes use of careful application of the Bernoulli equation (before and after the disk where the assumption of no viscous losses are valid) and the Momentum equation (across the disk) Showed that to extract momentum from or deliver momentum to the fluid, the streamlines must expand or compress across the disk, and this leads to a loss in the theoretical efficiency of the device. Efficiency is improved in all cases by using larger blades Losses are due to frictional, rotational and pressure fluctuation effects The best propellers have efficiencies of 85% The theoretical efficiency of a wind turbine is low, and its maximum efficiency (the Betz Limit) is 0.59.

22

5:1 : Suggested Reading/Reminders Crowe et al. Engineering Fluid Mechanics, 9th ed. SI version o Nothing useful James A. Fay Introduction to Fluid Mechanics, 1994, MIT Press o Nothing useful Fox, Pritchard and McDonald, Introduction to Fluid Mechanics, 7th ed, SI units o Nothing useful Cengel and Cimbala, Fluid Mechanics : Fundamentals and applications, 2th ed, SI units o Page 822 onwards (wind turbines)

12/12

No Reminders

23