Question 11.8: A latent heat storage (LHS) system is required for an integr......
A latent heat storage (LHS) system is required for an integrated solar combined-cycle (ISCC) power plant operating under the following conditions:
• ISCC electric power output P_{el} is 39 MW
• Solar fraction in the total primary energy input f_{sol} is 0.24
• Overall (electrical) efficiency of the ISCC power plant η_{ISCC} is 52%
• Solar energy storage duration \tau is 6 h a day
• LHS storage material is an inorganic salt that undergoes melting during storage charging and solidification during discharging
• Maximum and minimum temperatures in the storage cycle are t_{max} = 1050°C and t_{min} = 770°C
• PCM fusion (melting) temperature and enthalpy are t_{f} = 800°C and h_{f} = 520 kJ/kg
• Storage efficiency is \eta_{s} = 0.9
• PCM specific heat c is 0.95 kJ/kg K and density ρ is 2160 kg/m³ in both solid and liquid states
Calculate (a) the required thermal capacity of the storage, (b) the storage energy density, (c) the required mass and volume of the PCM material, and (d) the diameter d and height H of the storage tank if H = 0.5 d.
1. Total primary energy (fuel + solar) input rate
E_{\mathrm{in}}=P_{\mathrm{el}}/\eta_{\mathrm{ISCC}}=39/0.52=75~M W2. Solar energy contribution to the total energy input
E_{\mathrm{sol}}=f_{\mathrm{sol}}\ E_{\mathrm{in}}=0.24\times75=18\ \mathrm{MW}3. Required thermal capacity of the storage (only for solar portion)
Q_{s}=\tau\,E_{s ol}/\eta_{s}=6\,\mathrm{h}\times18\,\,\mathrm{MW/0.9}=120\,\,\mathrm{MW}\mathrm{h}=120\,\,\mathrm{MW}\mathrm{h}\times3600\,\,\mathrm{s/h}= 432,000 MJ
3. Energy density of the TES system
q_{s}=c\left(t_{\mathrm{max}}-t_{\mathrm{min}}\right)+h_{f}=0.95\times(1050-770)+520=786\ {\mathrm{kJ}} /{\mathrm{kg}}4. Required mass of the PCM material
m_{s}=Q_{s}/q_{s}=432,000~MJ/0.786~MJ/\mathrm{kg}=549,618~\mathrm{kg}5. Required volume of the PCM material
V_{s}=m_{s}/\rho=549,618\ {\mathrm{kg/2160\,kg/m^{3}}}=254.5\ {\mathrm{m^{3}}}6. Diameter and height of a cylindrical storage tank (H = 1.7 d): V_{s} = π/4 d² H = 1.7 π/4 d³
d=(4\ { V}_{s}/(1.7\pi))^{1/3}=[4\times254.5/(1.7\ \pi)]^{1/3}\approx5.76\ \mathrm{m}H = 1.7 d = 1.7 × 5.76 m = 9.8 m