The discussion started in issue #822 but seems rather misplaced there so I am creating a new issue to have a clearer overview on this. Looking deeper into hydrogen as an energy carrier, the efficiency of transformers between two energy vectors seems rather tricky.
In this applied example, we are looking at the efficiency of an electrolyzer that transforms electricity into H2. The electrical efficiency of an electrolyzer (E_eff) is can be calculated with the Lower Heating Value (LHV) or the Higher Heating Value (HHV) see this source, p.5, or this source. For this example we will use the LHV of H2 for the E_eff of the electrolyzer.
Considering an electrolyzer with a consumption of 55.5 kWh / kgH2 and the LHV of H2 of 33.3 kWh / kgH2 we would have an E_eff of 0.6. With this efficiency as input in energyConversion.csv, the optimization would perform:
*Electrolyzer_input E_eff = Electrolyzer_output**
This results in an output per kWh which is higher than the inverse of the LHV of H2 even with an efficiency lower than 1 which seems to be a problem.
To solve this, it could be possible to assume a conversion efficiency (C_eff) which would be:
The discussion started in issue #822 but seems rather misplaced there so I am creating a new issue to have a clearer overview on this. Looking deeper into hydrogen as an energy carrier, the efficiency of transformers between two energy vectors seems rather tricky.
In this applied example, we are looking at the efficiency of an electrolyzer that transforms electricity into H2. The electrical efficiency of an electrolyzer (E_eff) is can be calculated with the Lower Heating Value (LHV) or the Higher Heating Value (HHV) see this source, p.5, or this source. For this example we will use the LHV of H2 for the E_eff of the electrolyzer.
E_eff = (LHV) / (electrolyzer_consumption)
Assuming: [Electrolyzer_input] = kWh [Electrolyzer_output] = kgH2 [LHV] = kWh / kgH2 [Weighting factor WH2] = kWh / kgH2 [E_eff] = factor
Considering an electrolyzer with a consumption of 55.5 kWh / kgH2 and the LHV of H2 of 33.3 kWh / kgH2 we would have an E_eff of 0.6. With this efficiency as input in energyConversion.csv, the optimization would perform:
*Electrolyzer_input E_eff = Electrolyzer_output**
This results in an output per kWh which is higher than the inverse of the LHV of H2 even with an efficiency lower than 1 which seems to be a problem.
To solve this, it could be possible to assume a conversion efficiency (C_eff) which would be:
C_eff = (E_eff) / (LHV)
Which would result in an output per kWh that is:
*Electrolyzer_input ( (E_eff) / (LHV) ) = Electrolyzer_output**
The same assumption is made in this source, p.19