Question 11.7: An SHS storage system is required for a high-temperature CSP......

An SHS storage system is required for a high-temperature CSP plant that is operating as follows:

• CSP plant electric power output PelP_{el} is 30 MW
• Overall (electrical) efficiency of the CSP power plant ηel\eta_{el} is 20%
• Energy storage duration τ\tau is 6 h a day
• Storage medium is high-temperature concrete
• Storage medium temperature varies in a storage cycle from tmint_{min} of 520°C to tmaxt_{max} of 760°C
• Properties of storage medium: specific heat c = 0.9 kJ/kg K, density ρ\rho = 2400 kg/m³

Calculate (a) the required thermal capacity of the storage, (b) the storage energy density, and (c) the required mass and volume of storage material.

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1. Required thermal capacity of the SHS storage

Qs=τPel/ηel=6h×30  MW/0.2=900  MWh=900  MWh×3600s/h=3240  GJQ_{s}=\tau\,P_{\mathrm{el}}/\eta_{\mathrm{el}}=6\,\mathrm{h}\times30\,\,\mathrm{MW/0.2}=900\,\,\mathrm{MW}\mathrm{h}=900\,\,\mathrm{MW}\mathrm{h}\times3600\,\mathrm{s}/\mathrm{ h}=3240\,\,\mathrm{GJ}

2. Energy density of the storage

qs=c(tmaxtmin)=0.9×(760520)=225 kJ/kgq_{s}=c\left(t_{\mathrm{max}}-t_{\mathrm{min}}\right)=0.9\times(760-520)=225\ \mathrm{kJ/kg}

3. Required mass of the storage material

ms=Qs/qs=3240×106 kJ/225 kJ/kg=14,400,000 kg=14,400 tm_{s}=Q_{s}/q_{s}=3240\times10^{6}\mathrm{~kJ} /225\mathrm{~kJ}/\mathrm{kg=14,400,000~kg=14,400 ~t }

4. Required volume of the storage material

Vs=ms/ρ=14,400  t/2.4  t/m3=6000  m3V_{s}=m_{s}/\rho=14,400\;\mathrm{t}/2.4\;\mathrm{t}/\mathrm{m}^{3}=6000\;\mathrm{m}^{3}

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