Wind energy management with battery storage

Abstract

This article considers short-term energy management of a wind power plant with battery storage, in order to smooth the variations of power output to the external grid. An optimal online heuristic is developed for a control environment, in which decision making is independent of wind historical data and forecasts.

Appendix

At various stages of the discussion about 𝕋, we made use of inequalities (3), (6) and (8), whose proof was relegated to Lemmas 10 to 12 below.

Lemma 10

• For an instance I in M̄^t(C), we have W _max>ρ(C)b _i , where W _i =W _max for all i with i=1,2,…,t.

Proof.

• This lemma is for instances in the set M̄^t(C), and thus S ₀=C and W _i =W _max for 1⩽i⩽t. Furthermore, we have b _i =min{(nW _max+C)/n,(iW _max+C)/i}=W _max+C/n for 1⩽i⩽t. Therefore, to obtain the required result, we need only to show that ρ(C)⁻¹>1+C/(nW _max)=1+f ^C/n, where f ^C=C/W _max. The expression of ρ(C)⁻¹ can be written in terms of f ^C as follows.

  

The above equality can be also written as:

By assumption, we have , which implies that f ^C⩽2n/3. Now we consider the sum of the last two terms in the above expression.

because and which follows that

Hence we obtain ρ(C)⁻¹>1+f ^C/n+1/2n>1+f ^C/n. Now the proof is complete. □

Lemma 11

• Given that , where f ⁰=⌈S ₀/W _max⌉, we have ρ(S ₀)⩾p(S ₀′) if S ₀⩾S ₀′⩾0.

Proof

• We let function h(S ₀)=[ρ(S ₀)]⁻¹ and 0<x ₁⩽x ₂, where x ₁ and x ₂ are constants. Also, let f ₁=⌈x ₁/W _max⌉ and f ₂=⌈x ₂/W _max⌉. Obviously, f ₁⩽f ₂, since x ₁⩽x ₂. Then we have ρ(x ₁)⩽ρ(x ₂), if we can show that h(x ₁)⩾h(x ₂), which is derived from

  

Lemma 12

• For the instance I ^t(C), we see that ρ(C)∑ _i=t+1 ⁿ b _i ^t<C, where b _i ^t=min{[(n+t−i)W _max+C]/n,(C+tW _max)/i} for t+1⩽i⩽n, with t=1,2,…,n−1.

Proof

• This lemma is valid for the instance I ^t(C), which is defined as: S ₀=C, W _i =W _max for 1⩽i⩽t and W _i =0 for t+1⩽i⩽n. Thus, by the definition of b, we have b _i ^t=min{[(n+t−i)W _max+C]/n,(C+tW _max)/i}.

The main proof consists of three steps.

Step 1: :

First, we prove that for n⩾60, where f ^C=⌈C/W _max⌉. We recall the hypotheses we made for the model of minimizing the minimal output, namely, ⌈C/W _max⌉=C/W _max,n⩾60 and C⩾2n/5. Thus, f ^C=C/W _max⩾⌈2nW _max/5⌉. We partition the set of n, excluding n=60, as follows:

where n _x ^l=55+5l+x for x=1,2,…,5 and l=1,2,….

Now we show that for x=1,2,3,4 and .

We see that and , since ⌈2n ₁ ^l/5⌉=⌈2n ₂ ^l/5⌉ and ⌈2n ₃ ^l/5⌉=⌈2n ₄ ^l/5⌉=2n ₅ ^l/5. Thus, to prove that for 1⩽x⩽4, we need only show that , which can be obtained by:

Then we consider the difference between and .

Consequently, we obtain for n⩾61. Also, because , we conclude that for n⩾60.

Step 2: :

Next we show, by induction, that for 1⩽t⩽n. When t=1, we have

as .

We assume that for some t with 1⩽t⩽n−1. Then,

The inductive proof is now complete.

Step 3: :

Now we recall that f ^t is the smallest integer for which A _i ^t⩾B _i ^t for I ^t(C). We indicated before that f ^t>t (in Argument (ii) of Lemma 7), and thus, for t+1⩽i⩽n,

So f ^t=t+C/W _max.

Finally, we consider the difference between C/ρ(C) and ∑ _i=t+1 ⁿ b _i ^t.

Consequently, we obtain the required result: ρ(C)∑ _i=t+1 ⁿ b _i ^t<C. □