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I am on the process of making assumption how much capacitance/joule/watt is required on average supercapacitor as household' energy storage system. I plan the design for energy hub to be charged by electrical grid on night, and by the sunlight on daylight. I've found that for normal household energy use, the power supplied by meter (9.2 kVA on average) should suffice. In theory, this allows you to simultaneously supply devices with a maximum power of 9.2 kW or 9200 watts. Does it mean that ideal household supercapacitor for renewable energy resource (sunlight) on energy hub is 50% or...? Any help would be appreciated. Thank you.

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    \$\begingroup\$ I think you need to figure out a few specifications. (1) Understand that the capacitors (or batteries) can only store DC. You can't store AC. (2) Understand that batteries give a fairly constant voltage until they approach full discharge and then the voltage drops. Capacitor voltage drops exponentially from the very start. That's why we generally use batteries. (3) "Does it mean that ideal household supercapacitor for renewable energy resource (sunlight) on energy hub is 50% or...?" I think you need an edit to clarify what you mean by this. \$\endgroup\$ – Transistor Aug 13 '18 at 13:05
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    \$\begingroup\$ 9.2 kVA on average but then maximum power of 9.2 kW or 9200 watts. That's unclear, do you want your system to be able to deliver 9.2 kW on average (so peaks can be much higher) or is the peak 9.2 kW? Also relevant is for how much time that power is needed, 1 second, 5 minutes or the whole day? That makes quite a difference. \$\endgroup\$ – Bimpelrekkie Aug 13 '18 at 13:07
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    \$\begingroup\$ In addition to what Transistor comments: supercaps can generally store less energy per volume when compared to SLA or LiIon batteries. Also supercaps are more expensive per stored Joule of energy. Also supercaps can only handle low voltages (like 2.7 V per cap) so need you extra circuits to protect the caps from overvoltage and complex DC-AC converters to handle the wide voltage range. What is your motivation to choose supercaps over more conventional solutions using SLA or LiIon batteries? \$\endgroup\$ – Bimpelrekkie Aug 13 '18 at 13:12
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    \$\begingroup\$ Maxwell makes some nice cells for storing bulk energy. However, above what others have commented, one of these DuraBlue cells reads "2.85v" and "3.84w/h". How many would you need? 9200W / 3.84w/h = 2552 cells to supply 9200W for 1 hour. And this is assuming zero loss. Supercaps just are not feasible yet. \$\endgroup\$ – rdtsc Aug 13 '18 at 13:19
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    \$\begingroup\$ supercapacitor has high prospective Sure but one can do an easy calculation how much it would cost in $$ and volume to store a certain amount of energy. Then supercaps have the extra handicap of being more difficult to handle due to their voltage changing directly proportional to their charge level. So more complex charge/discharge circuits are needed. That increases cost. Supercaps would need to get much better at basic energy storing than chemical batteries to overcome their handicap. At this moment, supercaps simply aren't good enough. \$\endgroup\$ – Bimpelrekkie Aug 13 '18 at 13:36
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Here's a quick calculation, using an example 500 F, 16.5 volt supercapacitor (https://www.mouser.com/datasheet/2/257/Maxwell_16VModule_DS_1009363-10-1179674.pdf) Assume your converter electronics can work between 5 and 15 volts with a 90% efficiency. At 5 volts, this capacitor would have energy ( 1/2 * C * V^2) of 12,500 joules. At 15 Volts it would have an energy 112,500 joules. Subtract and you would have 100,000 joules to work with if the capacitor was charged to 15V and discharged to 5V.

Since a joule is a watt-second, you have 100,000 watt-seconds, and at 9200 watts you have a theoretical 10.9 seconds (100,000 / 9200) of capacity in each capacitor of this size. Throw in your 90% efficiency and you are just under 10 seconds of capacity for each 500 F, 16.5 volt capacitor. So calculate the number of seconds you want to run the household, and you will see how many capacitors you need.

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