TES energy for parabolic trough with molten salt

Dear Paul,

Thanks for your reply regarding the issue of the linear fresnel model. I would like to submit a new question on TES energy for a parabolic trough with molten salt.

The problem is related to the TES thermal capacity. This is the energy that must be stored in TES to operate the plant for the "Full load hours of TES" without sun. This means that the dump of energy should occur only when the maximum TES capacity is reached.

I create a new CSP Parabolic trough (physical), no financial model, using Hitec Solar Salt in the range of temperature of 290 and 540°C as heat transfer fluid in the solar field. The design gross output is 55 MWe (estimated gross to net conversion factor of 0.9), with a rated cycle conversion efficiency of 0.41. The TES has 6 hr of full load, with a capacity of 805 MWht. The tank height is 10 m with 1 m of minimum fluid level, and a tank diameter of 23 m.

After calculation I made the energy balance in each part of the plant (Solar field, TES and Power block) in order to evaluate the related efficiencies and also to use the thermal energy for the "Parasitic thermal field freeze protection" instead of electricity as done in SAM and then calculate the effective electricity production.

I found a strange behaviour of the energy stored in TES. I reported in attach the annual graph of the energy stored and of the energy dumped. As we can see the maximum value of the thermal energy stored is about 670 MWht, instead of 805 MWht, -16.8%.

I think that this could depend on the calculation of the storage volume. The formula 4.2 of pag. 81 of SAM manual, calculates the volume associated only with the TES thermal capacity and it isn’t the total TES volume included the "dead volume". In the file "sam_mw_trough_Type251.f90", the maximum HTF volume is calculated as :

vol_max = vol*(1.-h_min/h) ![m3] Maximum HTF volume among all tanks

and then is reduced by the ratio h_min/h, in this case of 10%.

I reduced the minimum height of .1 m, the results are in figure 2. In this case the maximum TES energy increases above 800 MWht.
I attach also the comparison for the 25 of May of the TES energy and of the storage volume considering only the volume above the minimum level.

Furthermore I think that in the formula 4.2, in evaluating the volume should be used the density of the hot tank instead of the average density. During the discharge phase there is a transfer of mass from the hot tank via the SG to the cold tank, so the specific volume of the storage should be :

V = 1000/(cp,av*(Tsf,out – Tsf,in))*3600/rho,ht = 5.44 m3/MWht

SAM calculates a value of 5.21 m3/MWht.

I would like to know what do you think about these consideration.

Best regards

Domenico

Dear Paul

I have a question with regards to the TES system for the physical parabolic trough model.

When I compare the output statistics "TES charge thermal power", "TES thermal losses" and "TES discharge thermal power" (the cumulative values, which would be the total energy), I observe that the cumulative charge thermal power - losses >> discharge thermal power. 

This tells me that there is some energy, in addition to losses, that is not being discharged, but instead is withheld? To my knowledge, an energy balance says that the energy discharged should be equal to the energy accumulated during charge less the energy lost. According to the balance, more energy should then be discharged.

As an example, for 5 hours of storage I get:
- TES charge thermal power (cumulative) = 354275 MWh
- TES thermal losses (cumulative) = 8412 MWh
- TES discharge thermal power (cumulative) = 344021 MWh

Then, 354275 - 8412 = 345863 - 344021 = 1842 MWh withheld?

Is there perhaps some physical constraint which prevents all the charged energy, less losses, to be discharged?

I look forward to hearing from you.

Best,
Louw