Question 11.2: A CAES storage plant comprising an air compressor, undergrou......
A CAES storage plant comprising an air compressor, underground compressed air storage, gas turbine with a combustor, and an HRSG is operating as follows:
• Electric power output of peaking gas turbine–generator train is P_{el} = 100 MWe
• Storage cavern charging period (compression) duration is \tau_{c} = 5 h
• Power generation period duration is \tau_{g} = 8 h
• Compressor intake is at t_{1} = 15°C and p_{1} = 100 kPa and discharge is at p_{2} = 6.4 MPa
• Supplementary firing raises the compressed-air temperature to a turbine inlet temperature t_{3} of 1050°C
• Fuel: natural gas with an LHV of 49 MJ/kg
• Compressor and turbine isentropic efficiencies \eta_{ic} and \eta_{it} are 0.85 and 0.8, respectively
Assuming a constant specific heat for the air of 1.05 kJ/(kg K) in the entire temperature range, calculate (i) the required energy storage capacity, (ii) the mass of air required for the storage, (iii) the total fuel requirements for the generation period per hour, (iv) the volume of the air storage cavern, (v) the air flow rate, and (vi) the fuel flow rate.
1. The required energy storage capacity is equal to the total energy input to the gas turbine train during the power generation period. Thus, taking into account the energy losses during generation
Q_{s}=P\mathbf{\mathrm{\bf~t}}_{g}/\mathbf{h}_{it}\,=\,100\times8/0.8\,=\,1000\,\,\mathrm{MW}\,\mathrm{h}2. With T_{1} = 15°C + 273 = 288 K, air temperature after isentropic (reversible) and actual (irreversible) zero compression with a pressure ratio \beta = p_{2}/p_{1} = 6.4 MPa/0.1 MPa = 64, respectively,
T_{2s}\,=\,T_{1}\mathbf{b}^{(k-1)/k}\,=\,288\times64^{(1.4-1)/1.4}\,=\,945\ \mathrm{K}T_{2}\,=\,T_{1}+(T_{2s}-T_{1})/\mathbf{h}_{ic}\,=\,288+(945-288)/0.85=\,1061\,\mathrm{K}
3. With the gas turbine inlet temperature T_{3} = 1050°C + 273 = 1323 K, the fuel–air ratio is given by
{f}\,=\,c_{p}(T_{3}-T_{2})/\mathrm{LHV}\mathrm{ }=\,1.05\times(1323-1061)/49,000=\,0.0056\mathrm{1\,kg/kg}4. Mass of air required
m_{a}\,=\,Q_{s}/c_{p}\Delta T\,=\,1\times1 0^{6}\,\mathrm{kW}\,\mathrm{h}\times3600\,\,\mathrm{s/h}\,/[1.05\,\mathrm{kJ} /\mathrm{kg}\,\,\mathrm{k}\times(1323-288)\,\mathrm{K}]= 3,312,629 kg
5. Total fuel requirements
m_{f}\;=\;f\;m_{a}\;=\;0.00561\times3,312,629=18,584\;{\mathrm{kg}}6. Fuel mass flow rate per hour of generation
m_{f} = 18,584 kg/8 h = 2323 kg/h
7. Air density at p_{2} = 6.4 MPa and T_{2} = 1061 K
{\bf r}_{2}\,=\,p_{2}/\mathrm{RT}_{2}\,=\,6.4\times10^{6}\mathrm{~Pa~}/\left[287 \,\mathrm{J}/(\mathrm{kg~}\mathrm{K})\times1061\mathrm{~K}\right]\,=\,21.02\mathrm{~kg/m^{3}}\,8. Cavern volume required
V_{c}\,=\,m_{a}/{\bf r}_{2}\,=\,3,312,629\;\mathrm{kg/21.}02\;\mathrm{kg/m^{3}}\,=\,157,594\;\mathrm{m^{3}}9. Average air flow to the cavern during 5 h of compression
V_{a,a v}=157,594\mathrm{~m^{3}/5~h=31},518.8\mathrm{~m^{3}/h}=8.755\mathrm{~m^{3}/s}It is more reasonable to employ a two-stage intercooled compression and two-stage reheat expansion.
Example 11.3 presents the performance calculation of a CAES system for a utility-scale power plant.