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At Traverse we perform micro-siting for wind farms by using IRR, NPV or LCOE as the target optimization variables. If you look at our landing page, it says we help project developers increase greenfield IRR by 0.5%. To fulfill this promise, we have to perform three key tasks.

- A. energy and wake modelling which covers revenue
- B. roads and foundation CAPEX optimization which covers costs and
- C. electrical array cable optimization which affects both revenue and cost.

These 3 topics are intertwined as every time you move a turbine it affects the optimality of all other turbines in a recursive cascade.

Our previous article was about B, road network optimization. This article is a showcase on how we perform C, electrical array cable optimization, where we optimize the electrical collection system for the lowest amount of CAPEX resulting in the lowest amount of operational losses.

Power from wind turbines are daisy-chained via many medium-voltage (MV around 33kV) electrical cables which converge at an internal substation. This substation collects all these cables, steps up the voltage to high-voltage (HV greater than 66kV) and proceeds to inject into the regional power grid via overhead cables.

Electrical losses occur in the MV collection system and directly impacts revenue. Although this is a generalization, the MV cable configuration contributes the largest losses that is optimizable on a technical basis. In onshore wind farms, these MV electrical cables are typically laid underground in trenches under the roads that connect the turbines. In offshore wind farms, cables lay either on the seafloor or are buoyant in the water. This article tries to optimize the following equation:

The building blocks to this optimization are the following key ideas:

- [hard constraint] Cables overheat when they start to carry too many turbines behind their back - the largest cables can only carry the load of 4 to 5 large modern turbines (for onshore cases).
- [hard constraint] Cables placed side by side lose even more carrying capacity as they heat each other up!
- [soft losses] Thicker cables are more expensive upfront but lose less power during operations.
- [soft losses] Turbines are at varying distances from each other and losses are directly proportional to distance.
- [soft losses] Where is the best place to put the wind farm substation such that it results in the highest IRR / lowest LCOE?
- [soft losses] Junction boxes reduce "loop backs" and therefore reduce operational losses, but they cost additional CAPEX.

*Note we also optimize transformer sizing and HV system design in the full electrical design process which we will write about in another article*.

The discussion in 1.1 assumes some hand of God has placed all the turbines, substations and junction boxes for us already. But in reality they can be permutated too. For the sake of this argument, assume turbine positions have been affixed. That leaves us with:

- Substation location: the optimal substation location will result in the highest benefits as eventually all cables and the power they carry within must arrive at the substation. If there is more than one substation then it gets worse.
- Junction boxes: junction boxes reduce cable distance by allowing multiple turbines to simply arrive at the junction box vs. cables looping back from one turbine to another (see section 1.3). Worse than the substation problem, you can arbitrarily place many junction boxes all over the place. What is the least number of junction boxes that will reduce total electrical losses? Where should they be placed in the electrical network?

Assume our simple example wind farm array roads are fixed as below with a trifurcated fork.

If we let the cables simply follow the roads, then we'll have to handle the fork by installing an additional junction box, which will contribute a certain cost:

Alternatively, we may first connect T1 to T2, and then double back along the road to connect to T3:

Or, have T1 and T2 both connect to T3 directly:

There are many other configurations. Which is the best?

Secondly, every cable also has an ampacity which is the maximum allowable current it can carry before it starts to overheat. In the simplified example before, we assume:

- We have many turbines along a long mountain ridge connected by a single road with no forks.
- A single cable size (we will discuss cable sizing later).

We may be able to connect at most 4 of them before the cable starts to overheat. We'd then have to connect the subsequent turbines with a separate circuit with cable running parallel to the first:

If we were solely optimizing for cable length, then this would be the optimal configuration, requiring (assuming equidistance for simplicity) 10 units of cable:

But another factor to consider is electrical losses, which is \(I^{2}R\). Assuming again for simplicity that each turbine produces the same 1 unit of current, in this case, our electrical losses would be 92R:

We may instead "share the burden equally", which gives us a lower electrical loss of 72R, but a higher cable length of 13 units of cable:

We see that a configuration that has a higher cable CAPEX due to longer distances may have a lower electrical loss (OPEX) and vice versa. To be able to compare and sum up these costs properly, we'll have to get precise details such as:

- Cable sizing and pricing: a thicker cable is more expensive, can carry more current and has less losses.
- Energy production profile which results in different currents at different times of the year.
- Electricity pricing that changes according to market conditions.

Shown below is a example of the specific calculations that would have resulted if using the correct resistivity multipliers of thicker cables.

Finally, there is the the issue of cable derating. Simply put, multiple cables laid side-by-side will affect one another, resulting in the cables having lowered ampacities due to the heat dissipation of the cables. The image below shows how to space cable groups according to IEC60502-2:

In the above example, although the limit of one cable is 4 turbines, two cables side-by-side may have their ampacities derated by up to 83% of their original value, and thus are only able to handle 83% * 4 = 3.3 turbines each.

So, upon introducing the second purple cable, we have to backtrack to the first red cable and reduce its load to 3 turbines. But wait! We have 7 turbines to handle, and our red and purple cables can only handle 3+3=6 cables, which means that we'll need a third cable (blue color below).

But wait! Now in the last section just before entering the substation, we now have 3 cables side-by-side where their ampacities in that section is derated to 73% of their original value. So the red and purple cables can only handle 73% * 4 = 2.9 turbines each. Assuming you take high safety factors, this may mean each cable can only support 2 cables. So our red, purple and blue can support 2+2+2=6 turbines. This means we need a fourth cable 😭 .

If we then extend this logic to a 500MW or 1,000MW project with 100 turbines with any number of junction boxes, the total number of combinations and permutations spiral out of control. How would we go about finding the optimum configuration of cable connections to minimize losses and CAPEX?

We turn to mixed-integer linear programming (MILP). Linear programming is used to solve many problems such as:

- Economics: optimizing the allocation of scarce resources to production of various goods and services for maximum economic output.
- Transportation: routing these goods across a road network for maximum throughput with minimum cost.
- Scheduling: managing supply and demand schedules, and minimizing inventory costs.

MILP is linear programming (LP) *but with integers*: i.e. we can choose to connect two turbines with either one cable, or no cable, but not *half* a cable and linear programming is basically solving an optimization problem where the cost and constraints can all be expressed as linear inequalities.

Suppose that a farmer has a piece of farm land, say \(L~km^{2}\), to be planted with either wheat or barley or some combination of the two. The farmer has a limited amount of fertilizer, \(F~kg\), and pesticide \(P~kg\). Every square kilometer of wheat requires \(F_{1}~kg\) of fertilizer and \(P_{1}~kg\) of pesticide, while every square kilometer of barley requires \(F_{2}~kg\) of fertilizer and \(P_{2}~kg\) of pesticide. Let \(S_{1}\) be the selling price of wheat per square kilometer, and \(S_{2}\) be the selling price of barley. If we denote the area of land planted with wheat and barley by \(x_{1}\) and \(x_{2}\) respectively, then profit can be maximized by choosing optimal values for \(x_{1}\) and \(x_{2}\).Simple Linear Programming Problem

This problem can be expressed with the following linear programming problem in the standard form:

In our case, we use variables \(x_{mnc}\) which represent whether ** turbine m** is connected to

The cost coefficient would then be the length of the road-distance between ** turbine m and n** multiplied by the "cost per length", which as previously introduced, is the sum of the:

- Capex cost: the $/m pricing of
*cable c* - Opex cost: \(\int I^{2}R\cdot\text{tariff(t)}~dt\), adjusted for net present value for
*t*over the lifetime of the wind farm

Astute readers may notice that *I²* is not linear (and not a constant coefficient either), since the *I-value* passing through each cable depends on how the cables are connected. We deploy a series of tricks to discretize I and transform \(x_{mnc}\) to \(x_{mnci}\). This increases the search space dramatically and further mechanisms have to be deployed to ensure convergence.

After laying out the cost function, we then lay out the constraints accordingly, e.g.:

- Current is conserved and doesn't just appear out of or disappear into thin air.
- Cable ampacity is not exceeded.
- There is a feasible number of cables entering and exiting turbines.
- Turbines don't try to connect to themselves in a loop.
- The
*I²*binning constraints, as described above. - Derating constraints: similar to
*I²*binning, but modify the*i*in \(x_{mnci}\) to the*derated I-value*, and adjust the constraints accordingly for the new promised derated*I-value*(this part gets extremely hairy).

When all is said and done, we end up with ~100,000 variables and constraints, +/- a few *orders of magnitude* depending on things like the number of turbines and cable types.

Now that we have converted our real world electrical cable optimization problem into an MILP problem, the next step is actually solving it.

The traditional way of solving linear programming problems is with the simplex method, which has exponential time-complexity in the worst-case, but is remarkably efficient in practice. With N variables, we are dealing with an N-dimensional search space. Each constraint bisects this N-dimensional space into two half-spaces, one feasible and the other infeasible. The intersection of all the constraints' feasible half-spaces is always convex:

- With 2 variables, each constraint would be a line bisecting ℝ². The feasible solution space is a convex 2D polygon.
- With 3 variables, each constraint would be a plane bisecting ℝ³, and the feasible solution space a convex 3D polyhedron.
- What the simplex method does is to first locate a vertex in this feasible polyhedron, and then repeatedly move to neighboring vertices until the cost function can no longer be improved. This is guaranteed to result in a globally optimal solution to the linear programming problem.

At this point, recall that we actually have not just an LP but an MILP problem, i.e., some of our variables are binary values (cable is connected vs. not connected). However, the optimal solution produced by the simplex method may not be an integer solution. What we then do is apply branch-and-bound, we check the "nearby" integer solutions in a smart manner.

Can we within the framework of MILP also further solve the advanced questions posed in section 1.2? The answer is yes but with many subtle strategies and tricks. A small summary of this is:

- [Jon welp] A turbine is just a junction box where up to 3 cables come in and 1 comes out ???
- [Jon welp] A substation is just a turbine where 50 cables can come in ???
- [Jon welp] For the substation problem, it is mathematically provable that the optimal substation is very close to a turbine. This reduces the total number of possible substation locations to the wind turbine tower fall radius. In an example 50 turbine configuration, we can parallel process 50 different substations using a 50 core machine.
- [Jon welp] For the junction box problem, it is simply to add more constraints to the MILP equations and allow cables to connect

We finally have our optimal solution: take those *x_mnc* variables which are 1, and for each, connect the corresponding turbine *m*'s to turbine *n*'s using cable *c*'s. Here is a visualization of how the electrical cable setup would look like in a real project:

The different colors represent the selection of different cable types: thinner cables gradually make way for thicker cables as more and more turbines are connected. Ampacity constraints force us to keep starting afresh with separate cables, so that as we move closer to the center of the wind farm, we see more and more cables side-by-side.

It looks like we're nearing the end of the electrical cable optimization, but often times this is just the beginning of the next design iteration: tariff values change as power purchase agreements are negotiated, and wind turbine position change and roads are rerouted due to engineering constraints or environmental concerns. The fact that this entire process of electrical optimization is automated and integrated into a comprehensive wind farm development pipeline gives Traverse the capability to simply tweak some parameters on demand and produce an updated electrical collection design with little effort.

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