Basic Rotor Aerodynamics: 1D Momentum

8/13/2019 Basic Rotor Aerodynamics: 1D Momentum

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Basic Rotor Aerodynamics: 1D MomentumTheory

Mac Gaunaa, Ris-DTU

This short note is a short introduction to one of the simplest rotor aerodynamic models,the 1D momentum theory, or simple actuator theory, as it is also often called. For more

complete intoductions to 1D momentum theory, please refer to textbooks such as for

instance [1] or [2]. This note is intended for use in the Fluid Mechanics Course at

Faculty of Engineering at SDU (SydDansk Universitet). The reason for writing this note

is that the introduction to the theory as given in the textbook used in the course [3] is in

fact erroneous. Home tasks related to 1D momentum theory and a funny application of

the basic rotor theory: the wind turbine car, is given at the end of this note.

1 1D momentum theory for rotorsThe derivation of the basic 1D momentum equations employs the three most basic

fluid mechanic conservation laws in integral form for a control volume: conservationof mass, momentum and energy. Moreover, the Bernoulli equation, which can bederived from the former equations, is used. The present presentation adopts thenomenclature of [3] for the conservation laws. Conservation of mass means that the

amount of mass for an enclosed amount of particles cannot change in time.Expressed in control volume formulation for the incompressible case this read

0)( =CS

AdVrr

(1)

Note, that the vector for the area vector is perpendicular to the surface, positive outfrom the control volume. The additional restriction of steady flow does not furthersimplify the expression for mass conservation. Conservation of momentum in theintegral control volume formulation is basically Newtons second law for all fluid

particles in the CV. In a control volume without acceleration, for a steadyincompressible flow, the momentum equation read

=CS

AdVVF )(rrrr

(2)

Note that the formulation above is the vectorial version of the momentum equation.The left hand side force is the forces acting on the control volume, usually onlysurface forces (pressure forces and reaction forces through structures crossing thecontrol volume) and body forces (e.g. gravitation or magnetic forces). Moreover, if

the hydrostatic part of the pressure variation in the fluid is not needed (this gives nocontribution to the force in incompressible flow), the usual body force, gravity,

which gives rise to this, may be omitted. The integral form of the control volume

version of the energy equation in the steady case for a steady, incompressible flowread

( )

++=

CS

she arAdVgz

VpWP

rr&

2

2

(3)

If further an inviscid flow is assumed we arrive at


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( )

++=

CS

AdVgzVp

Prr

2

2

, (4)

where the shaft power, P, is positive out of the control volume. Employing this

equation for a streamtube where no shaft power is involved, we get the famousBernoulli equation:

constgzVp =++ 2

2

(5)

To sum up, the assumptions for this is: steady incompressible, inviscid flow along astreamline.

Now that the tools for deriving the 1D momentum results are ready, lets startderiving the classic 1D momentum theory. Consider the actuator disc for the powergenerating case sketched in Figure 1 below.

Figure 1: Representation of the rotor by an actuator disc, and the three control volumes

used for derivation of 1D momentum results. This illustrates the power generation case.

The rotor area is A and the cross-sectional area of the low velocity region in the far

wake is AW.

Even though the derivations are shown for the power generation case, the results areapplicable in the propulsive case too. This is commented after the derivation. Theinfinitely thin rotor is inside CV2, the green control volume, of which the extension

in the axial direction is infinitesimally small. The sides of CV1 follow thestreamlines, go through the tip of the rotor, and extend far up and downstream. The

cylindrical CV3 extends far up and downstream, as well as far out in the radialdirection. The magnitude of the inflow is V, at the rotor disc the flow velocity isV(1-a), and far downstream of the rotor, at 4, the flow velocity is V(1-b). The rotor

area is A, and the density of the fluid is . CV1 and CV3 extend far enough up anddownstream such that the streamlines are straight at the inlet and outlet sides.

Therefore there is no pressure variation along the front side or back sides of CV3,where the pressure is equal to the ambient pressure, patm. On top of this we assume

steady, incompressible, inviscid flow, and do not consider hydrostatic pressurevariations in the flow. Due to continuity, the axia l velocity through the rotor is

1 2 3 4

CV 1

CV 3

CV 2

V

V(1-a)

T

V(1-b)

V


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continuous, but the pressure has a discontinuity in the axial direction at the rotordisc. The slow velocity in the rotor wake results in a mass flow rate through the sides

of the cylindrical CV3. This can be evaluated from the mass conservation equationon CV3 (AWis the area of the rotor wake at position 4):

WsideCV VbAm =3& (6)

Application of the axial component of the momentum equation for CV3 yields

322 ))1(1( sideCVW mVbAVT &= (7)

Since everything later on will be related to the rotor area,A, the relation between thisand the rotor wake area, AW, will come in handy. This is obtained from massconservation in the part of CV1 from the rotor (position 3) to the exit (position 4):

b

aAA

W =

1

1 (8)

In order to get a relation between the axial induction at the rotor, a, and in the farwake, b, we derive a second expression for the thrust force, T. Now consider the

axial component of the momentum equation for the control volume enclosing onlythe infinitesimally thin rotor, CV2. Note, that since the length of CV2 in the axial

direction is infinitesimal, the mass flow through the upper and lower sides must bealso infinitesimal because the radial velocity component can not be singular.Therefore, the fluxes in and out of CV2 are identical, resulting in vanishing fluxintegrals in the momentum equation, leading to a second expression for the thrust on

the rotor

ApT = (9)

In this expression, p is the pressure difference over the rotor. Application of theBernoulli equation from position 1 to 2 in combination with the analog expressionbetween position 3 to 4, noting that the pressures at position 1 and 4 are identical

(patm), yields for the pressure jump over the rotor disc :

))1(1(2

1 22bVp = (10)

A combination of Equations (6) to (10) results in the important relation between band a:

ab 2= (11)

That is, the induction at the rotor is half of the induction in the far wake, a resultwhich is independent of the sign of a, meaning that this result is applicable also inthe case of a propeller. This result can now be used with for instance Equations (6) to

(8) (or (9) and (10)) to arrive at the final expression for the thrust force T

)1(42

1 2aaAVT = (12)

This corresponds to the non-dimensional thrust coefficient

)1(4

2

1 2aa

AV

TCT ==

(13)

An interesting feature, which is used as an argument in the Blade Element

Momentum theory, appears when evaluating the axial component of the pressure


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forces acting on the sides of the stream-tube going through the tip of the rotor. Thisis done using the axial momentum equation on CV1, where in this case the total

surface forces include both the tower reaction force T as well as the resultantpressure force in the axial direction from the sides of CV1, where the pressure is notconstant. In this case, the relation between the inlet area and the rotor area is

obtained from mass conservation between position 1 and 2 in CV1:

)1( aAAin = (14)

Using this in the axial momentum equation for CV1, and combination withEquations (11) and (12) results in the simple result:

0, =streamtubeaxialF (15)

This means that there is no axial component of the pressure forces acting on the sides

of the stream-tube going through the tip of the rotor. This was assumed in thederivation of the theory in [3], but cannot be assumed without justifying it somehow ,

since in fact the pressure varies along the sides of CV 1 in figure 1. This is the proofthat is needed in order to be able to derive the theory as done in [3].

The derivation of the expression for the power is posed as a problem for the reader,

the result is

2

3

)1(4

2

1aa

AV

PCP ==

(16)

As noted in the beginning of this section, the derivations in the present formulationwas performed for the power generation case. The results are, however, applicable to

the propulsive case too. In that case the axial induction factors obtain negativevalues.

The thrust and power as obtained from the 1D momentum theory is the absoluteupper limit of what can be achieved for propulsion and generation, respectively,since all losses due to viscous effects (profile drag, separation, etc), wake rotation

(due to less-than infinite rotation velocity of the rotor) and a finite number of blades(so-called tip-loss) are disregarded. The interested readers are referred to standardtextbooks (etc [1] or [2]) for discussion of these effects.

It should be pointed out that since the equations on which the 1D momentum theoryis based on only requires the coordinate system to be non-accelerating, the results areapplicable to any control volume moving with constant velocity, so the 1D

momentum results as derived above is applicable to vehicles using rotors for energyconversion, as long as one remembers, that the velocity used in the non-

dimensionalisation of the thrust and power is the undisturbed velocity as seen from acoordinate system following the rotor.

The classical result of Betz for the optimal loading of the wind turbine formaximizing the power is obtained by finding the maximum of Equation (16), whichis seen to occur for a=1/3. This correspond to CP=16/27 and CT=8/9. This, however,

is only the optimum when optimizing for the power output of a wind turbine at rest.If the goal is not just a simple maximization of power, as for instance in the case ofvehicle applications, the optimization should take that into account. Therefore, theoptimum axial induction factor in the vehicle application cases is generally not as in

Betzs optimum.


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2 Home tasksRequirements: Task A + minimum 2 of the other ones. You pick which ever two (ormoreJ) you find the most interesting among Tasks B to F.

Task A:Show that the power output of an idealized rotor is indeed as in Equation

(16). You will have to use the energy equation (4) and some convenient choice ofcontrol volume.

Task B to F are all on the wind car.

The question that needs an answer is: How fast will a wind car be able to go?

Task B:How fast can an idealwind car go?

Ideal here means 1D momentum results for rotor aerodynamics, no rollingresistance, car drag or transmission loss.

Comment on the result.

Task C:Now include rolling resistance.

How fast will it go now? Comment on the result.

Task D:As B, but include drag on the car.

How fast now? And at what axial induction factor? Comment on the result.

Task E:As B, but include mechanical loss in the transmission.

How fast now? And at what axial induction factor? Comment on the result.

Task F:Now include all losses mentioned above.How fast now & at what axial induction factor? Comment on the result.


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References[1]: Hansen, M.O.L., Aerodynamics of Wind Turbines

[2]: Burton, Sharpe & Bossanyui, Wind Energy Handbook.

[3]: Fox, R.W., McDonald, A.T. & Pritchard, P.J., Introduction to FluidMechanics, Wiley, Sixth Edition, 2006 JustAsk! Edition.

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Basic Rotor Aerodynamics: 1D Momentum