Question 11.AE.6: Design a compressed-air energy storage (CAES) plant for Pout......
Design a compressed-air energy storage (CAES) plant for P_{out\_generator}= P_{rated}=100 MW, V_{L-L}=13,800 V, and cosφ=1 which can deliver rated power for two hours. The overall block diagram of such a plant [20,21] is shown in Figure E11.6.1. It consists of a compressor, cooler, booster compressor, booster cooler, booster-compressor three-phase, 2-pole induction motor with a 1:2 mechanical gear to increase the compressor speed, underground air-storage reservoir, combustor fired either by natural gas or oil or coal, a modified (without compressor) gas turbine, and a three-phase, 2-pole synchronous generator/ motor.
Design data:
Air inlet temperature of compressor: 50 ° F≡283.16 K at ambient pressure 1 atm≡14.696 psi≡101.325 kPa(scal).
Output pressure of compressor: 11 atm≡161 psi≡1114.5 kPa.
Output temperature of booster cooler: 120 ° F≡322.05 K at 1000 psi≡6895 kPa.
Output temperature of combustor: 1500 ° F≡1089 K at 650 psi≡4482 kPa.
Output temperature of gas turbine: 700 ° F≡644 K at ambient pressure 1 atm≡14.696 psi≡101.325 kPa.
Generation operation: 2 h at 100 MW.
Re-charging (loading) operation of underground air storage reservoir: 8 h at 20 MW.
Capacity of air storage reservoir: 3.5 .10^6 ft³ ≡97.6 10³ m³ ≡(46 m x 46 m x 46 m).
Start-up time: 6 minutes.
Turbine and motor/generator speed: n_{s\ mot/gen}=3600 rpm
Compressor speed: n_{s\ comp}=7000 rpm.
Cost per 1 kW installed power capacity: $2000.
a) Calculate the Carnot efficiency of the gas turbine η_{carnot\ gas\ turbine}≡η_{compressor}\ \eta_{\text {turbine }}=\frac{T_1-T_2}{T_1}
b) Note that the compressor of a gas turbine has an efficiency of η_{compressor}=0.50. In this case the compressor is not needed because pressurized air is available from the underground reservoir. The absence of a compressor increases the overall efficiency of the CAES, if “free” WP is used for charging the reservoir. Provided the synchronous machine (motor/generator set) has an efficiency of η_{generator}=0.9 calculate
b1) the efficiency of the turbine without the compressor η_{compressor},
b2) P_{out\ turbine},
b3) P_{out\ combustor},
b4) For a heat rate of the combustor of 5500 BTU/kWh, determine the input power of the combustor required (during 1 hour) P_{in\ combustor}.
b5) The two air compressors (charging the reservoir during 8 hours) including the two coolers require an input power of 20 MW (whereby wind energy is not free) calculate the overall efficiency of the CAES (during 1 hour of operation)
\eta_{\text {overall }}^{\text {wind energy is not free }}=\frac{P_{out\_generator }}{P_{\text {in combustor }}+P_2 \text { compresoss }+2 \_ \text {coolers }}
b6) Compute the overall efficiency if wind energy is free, that is, \eta_{\text {overall }}^{\text {wind energy is free }}=\frac{P_{out\_ generator }}{P_{\text {in combustor }}}.
c) What is the overall construction cost of this CAES plant, if the construction price is $2000 per installed power capacity of 1 kW?
d) The CAES plant delivers the energy E=200 MWh per day (during 2 hours) for which customers pay $0.25/kWh due to peak-power generation. What is the payback period of this CAES plant, if the cost for pumping (loading, recharging) is neglected (free wind power), the cost of 1 cubic foot of natural gas is $0.005, and the interest rate is 3%?
e) Repeat the analysis for a heat rate of 4000 BTU/kWh at a gas turbine output temperature of 600 K.
Hint: For the calculation of the Carnot efficiency of the gas turbine and of the compressors you may use the software available on the Internet address:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html#c3
a) The Carnot efficiency of the gas turbine is
\eta_{\text {carnot gas turbine }}=\eta_{\text {compressor }} \cdot \eta_{\text {turbine }}=\frac{T_1-T_2}{T_1}=\frac{1089 \mathrm{~K}-644 \mathrm{~K}}{1089 \mathrm{~K}}=0.41b1) The efficiency of the gas turbine without the compressor η_{compressor} is \eta_{\text {turbine }}=\frac{\eta_{\text {camot gas turhine }}}{\eta_{\text {compressor }}}=\frac{0.41}{0.5}=0.82
b2) P_{\text {out turbine }}=\frac{P_{\text {generator }}}{\eta_{\text {generator }}}=\frac{100 \mathrm{MW}}{0.9}=111 \mathrm{MW}
b3) P_{\text {out turbine }}=\frac{P_{\text {out turbine }}}{\eta_{\text {turbine }}}=\frac{111 \mathrm{MW}}{0.82}=135.37 \mathrm{MW}
b4) For a heat rate of the combustor of 5500 BTU/kWh, determine the input power of the combustor required (during 1 hour) P_{in\ combustor}. Figure E11.6.2 shows the output powers and the associated efficiencies of generator, gas turbine without compressor, and combustor.
The heat rate =5500 BTU/kWh indicates the thermal input energy to the combustor required to generate 1 kWh of electrical energy, that is, 5500 BTU=1.612 kWh of thermal energy is required to generate 1 kWh of electrical output energy.
The input energy delivered within 1 hour to the combustor is identical to the input power of the combustor P_{in\ combustor} = P_{out\_generator }\left(\frac{1.612 \mathrm{kWh}}{1 \mathrm{kWh}}\right)=100 \mathrm{MW} \frac{1.612 \mathrm{kWh}}{1 \mathrm{kWh}}=161.2 \mathrm{MW}.
The efficiency of the combustor is \eta_{\text {combustor }}=\frac{135.37}{161.2 .}=0.84
b5) Two air compressors (charging the reservoir during 8 hours) including the two coolers require an input power of 20 MW (whereby wind energy is not free) calculate the overall efficiency of the CAES (during 1 hour of operation). The energy required during an 8 hour period for the 2 compressors and 2 coolers is
E_2 \text { compressons }+2 \text { coolers }=20 \mathrm{MW} \cdot 8 \mathrm{~h}=160 \mathrm{MWh}. If we want to deliver 100 MW during 1 h then the power required is P_{2\ compressors+2\ coolers}=80 MW. The overall efficiency is
\eta_{\text {overall }}^{\text {wind engy is not free }}=\frac{P_{out\_generator }}{P_{\text {ln combustor }}+P_{2 \_compressors +2 \_coolers }}=\frac{100 \mathrm{MW}}{(161.2+80) \mathrm{MW}}=0.415
b6) If wind energy is free, then \eta_{\text {overall }}^{\text {wind energy is free }}=\frac{P_{out\_generator }}{P_{\text {in combustor }}}=\frac{100 \mathrm{MW}}{161.2 \mathrm{MW}}=0.62
c) The construction cost of this CAES plant is $200 M.
d) Earnings per day by discharging the CAES
Earning=200 000 kWh\cdot$0.25=$50 k/day.
Natural gas cost: 5500 BTU =1.6117\cdot10^3 Wh or 1 Wh=3.413 BTU resulting in 1 MWh=3.413.10^6 BTU.
The fuel energy delivered to the combustor is E_{natural\ gas}=161.2 MWh.2.3.413.10^6 BTU/MWh=1100.10^6 BTU. 1(ft)³ of natural gas contains 1021 BTU, and the volume of natural gas required is \operatorname{Vol}_{\text {natural gas }}=\frac{1100 \cdot 10^6}{1021}(\mathrm{ft})^3=1.08 \cdot 10^6(\mathrm{ft})^3 within a 2 h period or per day.The fuel costs are then Cost_{fuel}=1.08.10^6 $0.005=$5400 for 2 h or per day.
Payback period:
$(1.03)^y 100000kW.$2000/kW+$5400.365.y=$50000.365.y or (1.03)y 200¼ 16.28.y resulting in approximately y=30 years.
e) Repetition of the analysis for a heat rate of 4000 BTU/kWh at a gas turbine output temperature of 600 K.
The Carnot efficiency of the gas turbine is determined from Figure E11.6.3.
The Carnot efficiency of the gas turbine is
\eta_{\text {carnot gas turbine }}=\eta_{\text {compressor }} \cdot \eta_{\text {turbine }}=\frac{T_1-T_2}{T_1}=\frac{1089 \mathrm{~K}-600 \mathrm{~K}}{1089 \mathrm{~K}}=0.449, where the turbine power is 488974.29 J/3600 s=135.83 MW.
The efficiency of the gas turbine without the compressor η_{compressor} is
For a heat rate of the combustor of 4000 BTU/kWh at a gas turbine output temperature of 600 K, determine the input power of the combustor required (during 1 hour) P_{in\ combustor}.
The heat rate =4000 BTU/kWh indicates the thermal input energy to the combustor required to generate 1 kWh of electrical energy, that is, 4000 BTU=1.1723 kWh of thermal energy is required to generate 1 kWh of electrical energy.
The input energy delivered within one hour to the combustor is identical to the input power of the combustor P_{in\ combustor}=P_{out\_generator} \left(\frac{1.1723 k W h}{1 k W h}\right) = =100 M W\left(\frac{1.1723 k W h}{1 k W h}\right)=117.23 \mathrm{MW}.
The efficiency of the combustor is \eta_{\text {combustor }}=\frac{123.73}{117.23}>1.0 which is not possible.
This design is not feasible.
