Basic Rotor Aerodynamics: 1D Momentum Theory Mac Gaunaa, Risø-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.
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1D momentum theory for rotors
The derivation of the basic 1D momentum equations employs the three most basic fluid mechanic conservation laws in integral form for a control volume: conservation of mass, momentum and energy. Moreover, the Bernoulli equation, which can be derived from the former equations, is used. The present presentation adopts the nomenclature 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
r r ( V ∫ ⋅ dA) = 0
(1)
CS
Note, that the vector for the area vector is perpendicular to the surface, positive out from the control volume. The additional restriction of steady flow does not further simplify the expression for mass conservation. Conservation of momentum in the integral control volume formulation is basically Newton’s second law for all fluid particles in the CV. In a control volume without acceleration, for a steady incompressible flow, the momentum equation read
r r r r F = ρ ∫ V (V ⋅ dA)
(2)
CS
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 only surface forces (pressure forces and reaction forces through structures crossing the control 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 no contribution 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 flow read
(
)
p V2 r r − P − W& shear = ρ ∫ + + gz V ⋅ dA 2 CS ρ
(3)
If further an inviscid flow is assumed we arrive at
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(
)
p V2 r r P = − ρ ∫ + + gz V ⋅ dA , 2 CS ρ
(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 famous Bernoulli equation:
p V2 + + gz = const ρ 2
(5)
To sum up, the assumptions for this is: steady incompressible, inviscid flow along a streamline. Now that the tools for deriving the 1D momentum results are ready, let’s start deriving the classic 1D momentum theory. Consider the actuator disc for the power generating case sketched in Figure 1 below. V
CV 3
V
CV 2 CV 1 1
2
3
4 V(1-b)
V(1-a)
T 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 are applicable in the propulsive case too. This is commented after the derivation. The infinitely 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 the streamlines, 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 radial direction. The magnitude of the inflow is V, at the rotor disc the flow velocity is V(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 ρ. CV1 and CV3 extend far enough up and downstream 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 pressure variations in the flow. Due to continuity, the axia l velocity through the rotor is
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continuous, but the pressure has a discontinuity in the axial direction at the rotor disc. 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 equation on CV3 (AW is the area of the rotor wake at position 4):
m& sideCV3 = ρVbAW
(6)
Application of the axial component of the momentum equation for CV3 yields
& sideCV 3 T = ρ V 2 AW (1 − (1 − b ) 2 ) − V m
(7)
Since everything later on will be related to the rotor area, A, the relation between this and the rotor wake area, AW, will come in handy. This is obtained from mass conservation in the part of CV1 from the rotor (position 3) to the exit (position 4):
AW = A
1−a 1− b
(8)
In order to get a relation between the axial induction at the rotor, a, and in the far wake, 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 only the 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 be also infinitesimal because the radial velocity component can not be singular. Therefore, the fluxes in and out of CV2 are identical, resulting in vanishing flux integrals in the momentum equation, leading to a second expression for the thrust on the rotor
T = ∆p A
(9)
In this expression, ∆p is the pressure difference over the rotor. Application of the Bernoulli equation from position 1 to 2 in combination with the analog expression between position 3 to 4, noting that the pressures at position 1 and 4 are identical (patm), yields for the pressure jump over the rotor dis c:
∆p =
1 ρ V 2 (1 − (1 − b) 2 ) 2
(10)
A combination of Equations (6) to (10) results in the important relation between b and a:
b = 2a
(11)
That is, the induction at the rotor is half of the induction in the far wake, a result which is independent of the sign of a, meaning that this result is applicable also in the 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
T=
1 ρV 2 A4a (1 − a) 2
(12)
This corresponds to the non-dimensional thrust coefficient
CT =
T 1 ρV 2 A 2
= 4 a (1 − a)
(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. This is 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 resultant pressure force in the axial direction from the sides of CV1, where the pressure is not constant. 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:
Ain = A(1 − a)
(14)
Using this in the axial momentum equation for CV1, and combination with Equations (11) and (12) results in the simple result:
Faxial , streamtube = 0
(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 the derivation 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 proof that 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
CP =
P 1 ρV 3 A 2
= 4 a (1 − a ) 2
(16)
As noted in the beginning of this section, the derivations in the present formulation was performed for the power generation case. The results are, however, applicable to the propulsive case too. In that case the axial induction factors obtain negative values. The thrust and power as obtained from the 1D momentum theory is the absolute upper 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 standard textbooks (etc [1] or [2]) for discussion of these effects. It should be pointed out that since the equations on which the 1D momentum theory is based on only requires the coordinate system to be non-accelerating, the results are applicable to any control volume moving with constant velocity, so the 1D momentum results as derived above is applicable to vehicles using rotors for energy conversion, as long as one remembers, that the velocity used in the nondimensionalisation of the thrust and power is the undisturbed velocity as seen from a coordinate system following the rotor. The classical result of Betz for the optimal loading of the wind turbine for maximizing the power is obtained by finding the maximum of Equation (16), which is seen to occur for a=1/3. This correspond to C P=16/27 and C T=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 of vehicle applications, the optimization should take that into account. Therefore, the optimum axial induction factor in the vehicle application cases is generally not as in Betz’s optimum.
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2 Home tasks Requirements: Task A + minimum 2 of the other ones. You pick which ever two (or more J) 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 of control 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 ideal wind car go? Ideal here means 1D momentum results for rotor aerodynamics, no rolling resistance, 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 Fluid Mechanics”, Wiley, Sixth Edition, 2006 JustAsk! Edition.
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