Returning back to your explicit request:

(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).

(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).