Question 11.3: A peaking power plant uses a CAES system that comprises a tw......
A peaking power plant uses a CAES system that comprises a two-stage air compressor with intercooling, an underground air storage, a two-stage gas turbine with natural gas-fired combustor and reheater, and an electric generator. Its operating conditions are as follows:
• Plant electric power output P_{el} = 320 MW
• Compressor intake air pressure p_{1} = 100 kPa
• Compressor intake air temperature T_{1} = 300 K, air IC to T_{3} = 300 K
• Compressor stage and turbine stage pressure ratio is the same: \beta = 8
• HP compressor discharge pressure: p_{3} = 6.4 MPa
• Isentropic efficiency of compressor and turbine: \eta_{ic} = 0.85 and \eta_{it} = 0.9
• Thermal efficiency of gas turbine train \eta_{th} = 0.4
• Gas turbine inlet temperature T_{5} = 1450 K and reheat temperature T_{7} = 1450 K
• Duration of storage charging and discharging (power generation) periods \tau_{c} = \tau_{gen} = 6 h
• Fuel in combustor and reheater: natural gas, LHV = 50 MJ/kg
• Isobaric specific heat of the air and gas c_{p} = c_{pg} = 1.1 kJ/(kg K)
Calculate (i) the air and gas temperatures, (ii) the air and fuel requirements, and (iii) the CAES volume, thermal capacity, and efficiency.
1. Compressor discharge air temperature (LP and HP stages), reversible and irreversible
T_{2s}=T_{1}\;{\beta}^{(k -1)/k}=300\times{8^{(1.4-1)/1.4}}=543.4\;K\;\mathrm{and}\;T_{4s}=T_{2s}T_{2}=T_{1}+(T_{2s}-T_{1})/\eta_{ic}=300+(543.4-300)/0.85=586.4\,\mathrm{K\,and}\ T_{4}=T_{2}
2. Gas turbine exhaust temperatures, reversible and irreversible, with \beta_{t} = 8
T_{6s}=T_{5}/\beta_{t}^{(k-1)/k}=1450/8^{(1.4-1)/1.4}=800.5\mathrm{~K~and~}T_{8s}=T_{6s}T_{6}=T_{5}-(T_{5}-T_{6s})\times\eta_{\mathrm{it}}=1450-(1450-800.5)\times0.9=865.4\mathrm{~K~and~}T_{8}=T_{6}
3. Compression work in a two-stage compressor with IC, per kg air
w_{c}=2~c_{\mathrm{p}}(T_{2}-T_{1})=2\times1.1\times(586.4-300)=630~{\mathrm{kJ}}/{\mathrm{kg}}4. Rate of heat removal from the air in intercooler, per kg air
q_{\mathrm{ic}}=c_{p}(T_{2}-T_{3})=1.1\times(586.4-300)=315\ {\mathrm{kJ}}/{\mathrm{kg}}5. Heat added by fuel combustion in combustor and reheater, per kg air
q_{f}=q_{f,\mathrm{cc}}+q_{f,\mathrm{rh}}=c_{p}[(T_{5}-T_{4})+(T_{7}-T_{6})]= 1.1 × [(1450 – 586.4) + (1450 – 865.4)] = 1593 kJ per kg air
6. Mass of fuel required for combustor and reheater, per kg air
f = q_{f}/HV = 1593 kJ/kg/50,000 kJ/kg = 0.0319 kg fuel per kg air
7. Work of HP and LP turbine stages, per kg air (fuel mass ignored)
w_{t}=c_{p}(T_{5}-T_{6})+c_{p}(T_{7}-T_{8})=2~c_{p}(T_{5}-T_{6})=2\times1.1\times(1450-865.4)= 1286 kJ/kg
8. Mass flow rate of the air (with fuel mass ignored)
m_{a}=P_{\mathrm{el}}/w_{t}=160,000\;\mathrm{kW/}1286\;\mathrm{kJ}/\mathrm{kg=}124.4\;\mathrm{kg/s}9. Fuel required for combustor and reheater per unit time
m_{f}={{f}}\,m_{a}=0.0319\times124.4=3.964{\mathrm{~kg/s}}10. Air density at p_{2} = \beta, p_{1} = 800 kPa, and T_{2} = 586.4 K, gas constant of air R = 0.287 kJ/kg K
\rho_{2}=p_{2}/\mathrm{R}T_{2}=800\mathrm{~kPa}/(0.287\mathrm{~kJ}/\mathrm{kg~K}\times586.4\mathrm{~K})=4.754\mathrm{~kg/m}^{3}11. Volume of underground air storage
V_{s}=m_{a}\ \tau/\rho_{2}=124.4\,{\mathrm{kg/s}}\times6\times3600\,{\mathrm{s}}/4.754\,{\mathrm{kg/m}}^{3}=565,288\,{\mathrm{m}}^{3}12. CAES thermal capacity (with specific heat of the air c_{v} = 0.718 kJ/kg K)
Q_{s}=\rho_{2}\,c_{v}\,V_{s}\,(T_{2}-T_{1})=4.754\times0.718\times565,288\times(586.4-300)= 552,619,750 kJ ≈ 552.62 GJ
13. Efficiency of the CAES storage with both charging and generation period duration \tau of 6 h
\eta_{s}=P_{\mathrm{el}}\ \tau/(m_{\mathrm{a}}\mathrm{w_{c}}\tau+m_{f}\ \mathrm{HV}\ \tau)=P_{\mathrm{el}}/(m_{\mathrm{a}}w_{c}+m_{f}\ HV)= 80,000/(62.2 × 630 + 1.982 × 50,000) = 0.58
14. Fuel consumption during power generation period, \tau = 6 h
M_{f}=m_{f}\ \tau=3.964\ {\mathrm{kg/s}}\times6\ {\mathrm{h}}\times3600\ {\mathrm{s/h}}=85,618\ {\mathrm{kg}}