TES Storage Volume

How does SAM calculate the storage volume in the TES page? Which parameter is using to calculate it?

Thank you in advance

For the CSP models that include thermal storage, SAM calculates the storage volume shown on the Storage page as follows:
Storage Volume =

Equiv Full Load Hours of TES * 3.6e6

/ Storage Fluid Density

/ Storage Fluid Specify Heat

/ (Solar Field Outlet Temp - Solar Field Inlet Temp)
[Update May 8, 2015: Please see the correct equation in my reply below]
I have added this and other equations for the "calculated values" that appear in the user interface to SAM's Help system, which will be part of an update to SAM 2012.5.11.
Thanks,

Paul.

Hello Paul,

Two comments:

1.- Your formula seems not to include the design turbine thermal input capacity. Therefore the units do not match.

2.- Correcting your formula and including in the numerator the design turbine gross output divided by the power cycle efficiency, still gives me a volume smaller than the Storage Volume provided by SAM.

For my understanding your formula only considers the TES Volume from which energy is withdrawn. However, the total TES Volume is this previous volume plus the Minimum fluid volume (2 tanks) plus whatever is left inside the circuit.
Can this explain the differences between SAM your formula?

Thanks in advance.

The correct equation for the physical trough storage volume in SAM 2014.11.24 and later is:

storage volume (m3) =
thermal capacity (MWh) * 3,600,000 (kJ/MW)
/ fluid density (kg/m3)
/ specific heat (kJ/kg-K)
/ heat exchanger derate
/ ( field outlet temperature (C) - field inlet temperature (C) )

For the default case:

1,764.71 MWh * 3,600,000 kJ/MWh / 1,872.49 kg/m3 / 1.50182 kJ/kg-K / 0.877551 / ( 391 C - 293 C ) = 26,268.79 m3

SAM displays all of those values on the Thermal Storage page, except for the field inlet and outlet temperatures, which are on the Solar Field page. Note that these are design values used for system sizing, not operating values reported in the simulation results.

Best regards,
Paul.

Thanks, but this equation for time of full load storage =1 hr and Cp of fluid should be in J/kg.k TO unify the units.
isnot right?

for default, Themral capacity= Wdesign gross*tfull load/efficiency
wdesign gross=111 Mwe, tfull load=6 hrs, eff.=0.3774
then Thermal capacity=1764.71 Mwth
storage volume= 1764.71*10E06*6*3600(Wh.s/h and W=J/s, so it is in Joule) divided by (1872.49 kg/m3*1.5018*1000j/kg.c*heat derate (0.877551 if indirect storage or 1 if direct storage*(391-293)c

so storage volume= 6.353*10^12 J/(2.4184*10^=26269.32 m3