Simple Model of an Ideal Wind Turbine and the Betz Limit

Transcription

1 Simple Model of an Ideal Wind Turbine and the Betz Limit Niall McMahon January 5, 203 A straightforward one-dimensional model of an ideal wind turbine; the model is used to determine a value for the maximum power obtainable by a rotor from an airflow, i.e. as determined by Betz. See the accompanying notes, diagrams and references, for context. You can find these in the notes from the first aerodynamics lecture at: One-Dimensional Model of an Ideal Wind Turbine The mass flow of a fluid, ṁ, which flows along a stream-tube (which is itself a control volume) of cross-sectional area A d at the rotor plane can be written as: ṁ = ρa d U d () where ρ is the fluid density. This is shown diagrammatically on the website. This is another way of writing that the amount of material passing through the area A d per unit time, e.g. per second if the units are all SI, is given by the volume of the material passing through A d per second multiplied by the density of the material. With a little reflection, it s evident that the volume of material passing through A d per second is A d U or, in another way, the volume of material passing through A d, in a time t, is A d U t. The mass of material passing through an area A d in a time t is: m = ρa d U d t (2) A quantity called the axial induction factor is defined so that: U d = U ( a) (3)

2 That is, the air velocity at the rotor disc, U d, is some factor ( a) times the free-stream air velocity U. Another way of thinking about this is to say that the flow at the rotor disc comprises the free-stream flow, which has a velocity of U, and a superimposed, induced flow in the opposite direction, which has a magnitude au, or U d = U au. Assuming that no material crosses the streamwise limits of the control volume, the mass flow, ṁ, is constant everywhere in the flow. This is a commonsense observation and simple means that what goes into the system in unit time must come out of the system in unit time. Otherwise material, i.e. air, is being added or taken away somewhere. A consequence of this is that the mass flow, defined in terms of the cross-sectional area at the rotor plane in Eqn. (), applies everywhere in the flow. Momentum (M), i.e. mass velocity, is constant in an undisturbed, ideal flow. For the flow under consideration, we can say that the momentum of the fluid entering the system far upstream in a time t is: M = (ρa d U d t) U (4) Since ṁ is constant and U is the velocity far upstream. Similarly, the momentum of the fluid leaving the system in the wake is: M w = (ρa d U d t) U w (5) Since ṁ is constant and U w is the velocity far downstream. For the system under consideration, momentum is extracted by the rotor, i.e. M M w or, better, M = M w + M, where M is the momentum extracted by the rotor. We can write: M = M M w Substituting in Eqn. (4) and Eqn. (5), M = ρa d U d (U U w ) t (6) For the control volume defined, and assuming that no external force acts on its boundaries, the axial momentum equation, which is simply an expression of Newton s second law, and which governs fluid motion in a control volume, can be written in a simple way as: T = ma = (mu) t = M dt 2

3 where T is the streamwise force, or thrust, exerted by the rotor on the flow; it is associated with the pressure drop, i.e. the extraction of momentum ( M), across the rotor. m is the mass of fluid involved and U is the velocity of the fluid. This expression is equivalent to F = ma, which you ll know. From our expression for M, we can write: Or: T = ρa du d (U U w ) t t T = ρa d U d (U U w ) (7) Now the T, the thrust exerted by the rotor on the flow (and vice versa), is equivalent to the pressure drop across the rotor (P + d P d ), by the area of the rotor disc (A d ). P + d and P d are the air pressures just upstream of, and just downstream of, the rotor disc, respectively. We can write: T = (P + d P d )A d = ρa d U d (U U w ) Using Eqn. (3) in place of U d : T = (P + d P d )A d = ρa d U d (U U w ) P + d P d = ρu ( a)(u U w ) (8) Bernoulli s equation is assumed valid upstream of the rotor and downstream of the rotor, but not at the rotor. This is because of the sudden pressure drop across the rotor. We can use the equation from far upstream to the rotor and then from just beyond the downstream face of the rotor to the far wake. The Bernoulli Equation can be written: 2 ρu 2 + p + ρgh = C Where ρ and U are the density and velocity of the flow, p is the pressure, g is acceleration due to gravity, h is the height of the flow above ground level and C is a constant. It means that the sum of the three terms must remain constant at each point in a flow where the equation is valid. Applying this equation to the upstream portion of the flow: 3

4 2 ρu 2 + p + ρgh = 2 ρu 2 d + p+ d + ρgh d where quantities and subscripts maintain their meaning. Re-writing and neglecting the effect of height since the flow is at the same level at both points, i.e. h = h d : p + d = 2 ρu 2 2 ρu 2 d + p (9) Similarly, for the downstream part of the flow: Re-writing: 2 ρu 2 w + p + ρgh w = 2 ρu 2 d + p d + ρgh d p d = 2 ρu 2 w 2 ρu 2 d + p (0) Combining these two expressions for p + d and p d, p + d p d = 2 ρ(u 2 U 2 w) () Inserting this expression into Eqn. (8), 2 ρ(u 2 U 2 w) = (U U w )ρu ( a) (U 2 U 2 w) = 2(U U w )U ( a) (U U w )(U + U w ) = 2(U U w )U ( a) (U + U w ) = 2U ( a) U w = U ( 2a) (2) 4

5 This gives an expression for U w, the air velocity in the wake, in terms of U and a. The most obvious observation is that when a 2, U w 0. Since this implies a zero or negative wake velocity, which is not possible, we can say that a should never be greater than or equal to a half. 2 Deriving an Expression for U d in Terms of U and U w From Eqn. (8), (P + d P d )A d = ρa d U d (U U w ) Substituting P + d P d the other, from Eqn. () and swapping terms from one side to ρa d U d (U U w ) = 2 A dρ(u 2 U 2 w) U d (U U w ) = 2 (U 2 U 2 w) U d (U U w ) = 2 (U U w )(U + U w ) U d = 2 (U + U w ) (3) Which is an interesting result. 3 Betz Limit The power coefficient C p can be defined as: C p = P a P t (4) Where P a is the actual power captured by the rotor and P t is the theoretical power available in the idealised air flow. The theoretically available power from the air across a sectional area A is, P t = 2 ρau 3 (5) 5

6 This can be derived by considering that kinetic energy is defined as E k = 2 mu. 2 E Power is the rate of energy delivery, i.e. ; we can write that t the total, theoretical maximum, power content of a flow of cross section A d, moving with a velocity U, is P t = 2 2ṁU = 2 (ρa du )U 2 = 2 ρa du, 3 as defined. We can write, C p = P a 2 ρa du 3 As described earlier, the thrust force acting on the rotor (and acting on the fluid, slowing the flow) can be written as: T = (p + d p+ d )A d Using the expression for p + d p+ d from Eqn. (8) and for U w in Eqn. (2), and re-arranging, T = 2ρA d U 2 a( a) Energy can be written as the product of a force F acting on a body and the distance d through which the body moves under forcing, i.e. E = F d. Power E can be thought of as a force by the velocity of the body, i.e. t = P = F U. Applying this to the system under consideration, the power capture at the rotor, P a = T U d, where T is the thrust and U d is the fluid velocity at the rotor disc, i.e. P a = 2ρ A d U 3 a( a) 2 Substituting this into the expression for the power coefficient, C p = 2ρA du 3 a( a) 2 2 ρau 3 Simplifying and re-arranging yields an expression for C p in terms of a, the axial induction factor, that is, C p = 4a( a) 2 (6) 6

7 Using this result, how do we go about determining the maximum C p? The maximum (or minimum) value of C p occurs when dc p da = 0, C p = 4a( a) 2 C p = 4a( 2a + a 2 ) C p = 4a 8a 2 + 4a 3 dc p da = 4 6a + 2a2 dc p da = 4(3a2 4a + ) dc p da = 4( 3a + )( a + ) If we set dc p da = 0, then we can write, 0 = 4( 3a + )( a + ) This means that there are two possible solutions, 0 = 3a + a = 3 (7) 0 = a + a = (8) 7

8 Only a = 3 is sensible as a = exceeds a =, discussed earlier. Additionally, a = results in C p = 0, which cannot be the maximum power efficiency 2 of the rotor if wind turbines work! If a = 3 is substituted into C p = 4a( a) 2, C p = 6 27 = This is the Betz Limit. This concludes this set of notes. References online. Niall McMahon, September 20. 8