Hi,
I am modeling thermal storage (DAYNITE) with self-discharge of 20% (S_LOSS=0.2) and an efficiency of 70% (S_EFF=0.7) when I get the results, only the efficiency seems to be taken into consideration (Ex. in year 2035 FOut(0.0126)/FIn(0.0180)=0.7).
PART II of the manual says "Timeslice storage process (EQ_STGTSS): applied to the average storage level (VAR_ACT) between two consecutive timeslices" but it does not seem to be affecting my results in this way.
In the Demos and Advance Demos there is not an example with it.
I would like to know how the S_LOSS is calculated and how can I enter 20% of self-discharge using S_LOSS attribute.
Thanks.

From the documentation:
STG_LOSS: Coefficient that represents the annual storage losses of a storage process p in region r, as a fraction of the (average) amount stored, corresponding to a storage time of one year. If the value specified is negative, the corresponding annual losses are interpreted as an annual equilibrium loss (under exponential decay).
Thus, the storage loss is proportional to the time the amount of energy is stored in the storage. For a DAYNITE storage, the storage time is typically only a few hours, and therefore the annual loss of 0.2 would be very small indeed: with a 10 hours average storage time the loss would be only 0.02%, so hardly visible at all. If you want 20% loss per day, you could try multiplying 0.2 × 365. However, if the losses are very high, they might easily even exceed the storage level, and therefore one might recommend using the equilibrium loss approach instead when the losses are very high. So, for 20% hourly loss I would recommend the negative value: −0.2 × 8760.

Returning back to your explicit request:

(28-08-2019, 03:11 AM)Dianaprz7 Wrote: [ -> ]I would like to know how the S_LOSS is calculated and how can I enter 20% of self-discharge using S_LOSS attribute.

See Part II, pages 254–255. The standard way of calculating the loss in a timeslice storage is given on page 254 (STG_LOSS>0). If defining it on the basis of equilibrium loss (STG_LOSS<0), the loss is calculated differently (exponentially), but taking into account the storage time in the same way as shown on page 254:
(1-EXP(STG_LOSS×G_YRFR(r,s)/RS_STGPRD(r,s)))
As shown above, for each timeslice, the storage time is G_YRFR(r,s)/RS_STGPRD(r,s) (in years), where RS_STGPRD is the number of cycles under the parent timeslice.
STG_LOSS>0 defines the loss in proportion to the initial storage level during one year’s storage time.
STG_LOSS<0 defines an equilibrium loss, i.e. how much the annual losses would be if the storage level is kept constant.
It is not obvious to say how you could enter a 20% of self-discharge using S_LOSS attribute, because you are not telling in which time frame your 20% self-discharge would occur. For a 20% self-discharge in

one year you can enter STG_LOSS=0.2. For a 20% self-discharge in

one hour you could enter STG_LOSS=‒0.2×8760 (equilibrium loss).

Dear Antti,
I tried the "−0.2 × 8760" and now I can see an impact (energy loss proportional to the time stored). Since my TS represent two hours, the impact is too big but now I can work with it.
Thanks!.