Create Two New Evaluator Nodes
Amount of Electricity per SF with Surface PV Panels
As the first evaluation metric, I calculated the amount of electricity per SF that would be generated if all the building's surface are covered with photovoltaic (PV) panels. I calculate this value with the following logic. I first computed the insolation on the building surface using the SolarAnalysis.Analyze node. I assumed the insolation for one year (1/1-12/31, 12:00AM -11:59PM, as shown on the right). Then I multiplied the values by a typical PV efficiency of 20% [1] to calculate the amount of electricity generated during the year, and divided it by the gross surface area.
Unit Cost of the Facade Panels
As a second metric, I calculated the unit cost of the building's facade panels. The amount was set at $80/SF [2] and multiplied by the gross surface area to calculate the total cost of the facade. I divided this by the number of panels to calculate the unit cost of the panels.
The table below shows the results of the calculations. The amount of PV electricity depends on the rotation degrees in the middle, and is not proportional to the height or surface area of the building. I think this is because the amount of insolation that hits the building surface changes with the twist of the building. Meanwhile, the panel cost increases as the surface area increases. This is not surprising because I have set the number of panels (u and v values) constant, and as the building surface increases, the panel size also increases. I assume that if the logic is set up so that the size of the panels is constant, the number of panels will change depending on the variable and I can see variations in the cost. I also think it will be useful to design the building if the energy used for air conditioning the building according to the isolation and the amount of the isolation for each direction are calculated.
Develop a Single-Objective Optimization Scheme
When looking at the Stage 1 metrics, surface area, floor area, and volume are proportional to height. However, the rotation in the middle is not proportional, with the highest value at 30 degrees. Also, the surface area has the lowest value when the rotation is 0 degrees, but the amount of PV energy has the highest value at that time. In other words, the PV efficiency is higher at 0 degrees, but the total floor area is smaller, which is a trade-off.
From Stage 1, the area exceeds 2.5M SF regardless of the angle when the building height is 750 feet. So, in this case, I fixed the height at 750 feet. Testing with the middle rotation variable from 0 to 90 degrees in 10 degree increments, the total floor area was the largest at 40 degrees and the amount of PV energy is the largest at 0 degrees. To come up with the best alternative, I computed values for floor area and PV energy for each degree of rotation, weighting each value as 100 when it was the largest and 0 when it was the smallest. Then, I took the best alternative to be the one with the largest sum of those weighting values.
As a result of calculation, the best alternative is when the rotation degree is 0, the second is 10 degrees, and the third is 30 degrees. The PV energy at 0 degrees has the largest value, but the floor area has a smaller value than the other angles. There is also a disadvantage that when the rotation is 0 degrees, the design is inferior to when the twist is applied.
References