Storage costs¶
Costs by state of charge¶
General description¶
Example that shows the parameter storage_costs of GenericStorage. A battery is used to make profit from fluctuating electricity prices. For a battery without storage costs, it is beneficial to be empty the end of the time horizon of the optimisation. For a battery that assumes the average revenue, energy is kept at the end.
Code¶
Download source code: storage_costs.py
Click to display code
import numpy as np
import pandas as pd
from matplotlib import pyplot as plt
from oemof import solph
def storage_costs_example():
# create an energy system
idx = pd.date_range("1/1/2023", periods=13, freq="H")
es = solph.EnergySystem(timeindex=idx, infer_last_interval=False)
# power bus
bel = solph.Bus(label="bel")
es.add(bel)
es.add(
solph.components.Source(
label="source_el",
outputs={bel: solph.Flow()},
)
)
es.add(
solph.components.Sink(
label="sink_el",
inputs={bel: solph.Flow()},
)
)
electricity_price = np.array(
[
0.38,
0.31,
0.32,
0.33,
0.37,
0.32,
0.33,
0.34,
0.39,
0.38,
0.37,
0.35,
0.35,
]
)
# Electric Storage 1
# Costs are designed in a way that storing energy is benificial until the
# last four time steps but emptying it is not a good option.
battery1 = solph.components.GenericStorage(
label="battery 1",
nominal_storage_capacity=10,
inputs={
bel: solph.Flow(
nominal_value=1,
variable_costs=electricity_price,
)
},
outputs={
bel: solph.Flow(
nominal_value=1,
variable_costs=-electricity_price,
)
},
initial_storage_level=0.5,
balanced=False,
)
es.add(battery1)
# storages that balance our fluctuating costs
# Electric Storage 2
battery2 = solph.components.GenericStorage(
label="battery 2",
nominal_storage_capacity=10,
inputs={
bel: solph.Flow(
nominal_value=1,
variable_costs=electricity_price,
)
},
outputs={
bel: solph.Flow(
nominal_value=1,
variable_costs=-electricity_price,
)
},
storage_costs=12 * [0] + [-np.mean(electricity_price)],
initial_storage_level=0.5,
balanced=False,
)
es.add(battery2)
# create an optimization problem and solve it
model = solph.Model(es)
# solve model
model.solve(solver="cbc")
# create result object
results = solph.processing.results(model)
plt.plot(
results[(battery1, None)]["sequences"],
label="content w/o storage costs",
)
plt.plot(
results[(battery2, None)]["sequences"],
label="content w/ storage revenue",
)
plt.legend()
plt.grid()
plt.show()
if __name__ == "__main__":
storage_costs_example()
Installation requirements¶
This example requires oemof.solph (v0.5.x) and matplotlib, install by:
pip install oemof.solph[examples] matplotlib
