Analysis of pumped heat energy storage (PHES) process using explicit exponential matrix solutions (EEMS)
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1 Analysis of pumped heat energy storage (PHES) process using explicit exponential matrix solutions (EEMS) Fan Ni and Hugo S. Caram* Department of Chemical Engineering, Lehigh University, 18015
2
3 Contents Background Introduction Mathematical Model Results and Discussion Conclusions
4 Background Electric Energy Storage Incentives: 1. provide a way of grid management to compensate the peak demand; 2. promote the utilization of intermittent renewable energy like wind, solar and wave energy ;
5 Pumped heat energy storage (PHES) Basic difference compared with traditional thermal energy storage: W T 0 T H W mc T T Q mc dt p( H 0) H p Traditional thermal energy storage T H W T 0 Carnot efficiency for fixed cold sink temperature W ' Q H T T T 0 W PHES T H T 0 If the process is reversible : W =W T H T 0 W
6 Comparison with other energy storage methods Advantages: No special geological formation requirement Less environmental concerns
7 Loading step: Electric Energy T 2nom = 500 o C Heater 433 ~ 500 o C 870 ~ 980 o C Electric energy is converted into sensible heat stored in the solid material Turbomachines Temperature is regulated at heater and cooler to maintain a constant operating condition and eliminate turbomachine irreversiblities -72 ~ -5 o C Low pressure tank (LP tank) T 0nom = 20 o C Cooler 20 ~ 136 o C High pressure tank (HP tank) T. Desrues, et al., Applied Thermal Engineering (2010). A. White, et al., Applied Thermal Engineering (2012).
8 Delivery step: Electric Energy T 2nom = 500 o C Heater 433 ~ 500 o C 870 ~ 980 o C Electric energy is retrieved from sensible heat stored in the solid material Turbomachines -72 ~ -5 o C Low pressure tank (LP tank) T 0nom = 20 o C Cooler 20 ~ 136 o C High pressure tank (HP tank) T. Desrues, et al., Applied Thermal Engineering (2010).
9 Irreversibilities of Turbomachines Compressors: PH w Tout / Tin ( ) P Turbines: L 1 PH Tin / Tout ( ) P Polytropic efficiency: ζ=0.9 PH PH ( ) delivery ( ) P P L L 1 2 Loading L P L T in P H T in = C / C p P H T out v P L T out Different pressure ratios for Loading and Delivery Specific heat ratio: PH ( ) P L 1 T. Desrues, et al., Applied Thermal Engineering (2010).
10 Thermodynamic limit Work done during loading: 1 p 2nom 0nom W c [ T ( 1) T (1 )] Work received during delivery: 1 p 2nom 0nom W '' c [ T ( ' 1) T ( ' )] Efficiency: W '' T Eff W 1 2nom T0nom 1 T2nom ( 1) T0nom (1 ) When T 0nom = 293K T 2nom = 773K ψ = 1.55 ψ = 1.72 ζ=0.9 Eff = 0.83 ( ' 1) ( ' ) T. Desrues, et al., Applied Thermal Engineering (2010).
11 Assumptions: Ideal gas law Adiabatic system No axial dispersion term Model details No accumulation of heat in the gas phase of each compartment Constant pressure and mass flow rate The properties are independent on temperature or pressure The pressure change between steps is neglected for the current analysis
12 Discretized heat transfer model Loading wt 2nom T 0nom /α n=1 n=n-1 n=n Delivery T 2nom n=n+1 n=2n-1 n=2n n=2n n=2n-1 n=n n=n-1 n=2 n=1 T 0nom U C ( ) hal( T ) g g pg n 1 n n n Dimensionless transform n Tn n 1 dtn scps ha( n Tn ) dt θ n is the temperature of gas, T n is the temperature of solid dt d n T n n / N 1 1 / N Dimensionless length: ha Uc g g pg L Dimensionless time: ha t c s ps
13 Loading step: DAEs: T T T T d n1 n Tn Tn d T0 T n T n1 T2 0 n1 T 2n T n 2n T 1 1 0nom T 2 2 T0 nom Explicit exponential matrix solution: A F nom nom 2nom / / / w w d T( ) ( ) d AT A 1 1 F ( ) e ( 0 ) T T A F A F
14 Delivery step: DAEs: T T T T d n1 n Tn Tn d T T n T n1 T2 0 n1 T 2n T n 2n T 1 1 0nom T 2 T0 nom Explicit exponential matrix solution: d T( ) ( ) ' d AT F A 0nom 2nom F ' nom T T A F A F A 1 1 ( ) e ( 0 ') '
15 Cyclic steady state solution T ssc : The solid temperature distribution after the delivery step T ssh : The solid temperature distribution after the loading step A 1 1 Tssh e ( T ssc A F) A F A 1 1 Tssc M e ( MT ssh A F') A F' A A 1 A A ( ) ' ' Tssc e e e e I M M M M A F A F A F A F A A ssh e e A A T I M M e Me A F ' A F ' A F A F where M 1 0
16 ,,,,, Global efficiency The energy stored in the process during the loading step is: E Q W Q W Q Q tot hot1 net1 cool1 net 2 hot 2 cool 2 Neglecting the energy change of the gas we have: Change of internal energy E E E c LS( T T ) tot HP LP s ps HP LP Q hot1 W net1 Q hot2 W net2 Defined by Desrues et al.*: Eff Q Q W Q 1 = E Q W Q cool1 cool 2 net 2 hot 2 tot cool1 net1 hot1 Modified definition: Wnet 2 Qcool 1 Qcool 2 Eff 1 W Q Q E Q Q T tot = net1 hot1 hot 2 tot cool1 hot 2 U ggcpg Tcool 1t U ggcpg Tcool 2t 1 c L( T T ) U c T t U c T t s ps HP LP g g pg cool1 g g pg hot 2 ( T T ) cool1 cool 2 1 ( THP TLP ) Tcool 1 Thot 2 ( T T ) HP LP Q cool1 Q cool2 Loading Delivery where Q cool1, Q cool2, Q hot1, Q hot2, W net1, W net2 and T tot are defined as Tcool1, Tcool 2, Thot1, Thot 2, ( TComp TTub ),( TTub TComp ), THP T ( ) is used as the total temperature difference to quantify the storage capacity *T. Desrues, et al., "A thermal energy storage process for large scale electric applications," Applied Thermal Engineering, vol. 30, pp , LP
17 Results and Discussion Cyclic steady state temperature distribution of the solid in the LP tank and the HP tank Gas flow direction Gas flow direction Temperature/K =0 =/4 =/2 = =3/4 =0 =/4 =/2 =3/4 = Temperature/K = =5/4 =3/2 =7/4 =2 =2 =7/4 =3/2 =5/4 = from the bottom of the LP tank to the bottom of the HP tank from the bottom of the LP tank to the bottom of the HP tank Figure 1 Temperature profile during the loading and the delivery step after the cyclic steady state is reached when PR=3, PR =3.88, T 0nom =298.15K, T 2nom = K, π=100 and Λ=200
18 Comparison with literature reported Data The solid temperature distribution at cyclic steady state After loading step After delivery step After loading step * After delivery step * Λ=100 π=40 Temperature/K from the bottom of the LP tank to the bottom of the HP tank Eff=0.678 Eff*=0.667 *T. Desrues, et al., "A thermal energy storage process for large scale electric applications," Applied Thermal Engineering, vol. 30, pp , 2010.
19 Process gas selection Comparison of Air and Argon as the process gas γ air =1.40 γ argon =1.67 Table 1 Comparison of Argon and air when PR=3, PR =3.88, T 0nom =298.15K, T 2nom = K, π=100 and Λ=200 * 10 3 K T tot * Eff Q cool1 * Q cool2 * Q hot1 * Q hot2 * W net1 * W net2 * Argon Air Bernoulli s principle 2 Gas density : Pv 1.2 kg m -3 for Air 1.7 kg m -3 for Argon (20 o C 1 atm) For the same ΔP the peripheral velocity of the rotors is lower
20 Effects of changing the loading step pressure ratio: PR 0.84 x Global efficiency Total temperature difference/ o C The maximum solid temperature / K PR PR Figure 2 Effects of PR on the global efficiency, the total temperature difference and the maximum solid temperature when T 0nom =298.15K, T 2nom = K, π=150 and Λ=300
21 Effects of changing the dimensionless numbers: π and Λ on the capacity and efficiency The dimensionless step duration: ha t c s ps step The dimensionless tank length: ha Uc g g pg L 16 x Total temperature difference T tot / K =50 =100 =200 =300 =400 Global efficiency =50 =100 =200 =300 = Figure 3 Effects of π and Λ on the total temperature difference and global efficiency when T 0nom =298.15K, T 2nom = K, PR 1 =3 and PR 2 =3.88
22 Ideal thermal wave Neglecting the thermal gradient when the tank is fully used: t step /t 0 =50 =100 =200 =300 =400 Ideal thermal wave Figure 4 The relationship between the average temperature difference and π/λ when T0nom=298.15K, T2nom= K, PR1=3 and PR2=3.88 t step U t c Lc 0 g pg g ps s t Lc 0 ps s Uc g pg g Average temperature difference: Ttot THP TLP= 386 K t U c Lc 0 step g pg g / t / ps s
23 Effects of the heat transfer coefficient and solid surface area 0.74 x Global efficiency Total temperature difference/ o C / o or / o Figure 5 Effects of ha on the efficiency and the total temperature difference when T 0nom =298.15K, T 2nom = K, PR 1 =3, PR 2 =3.88, π o =150 and Λ o =200
24 Case study Data* for long heat regenerators packed with uniformly sized spherical basaltic beach stones For the solid: d p =0.08m ε=0.4 a=3(1- ε )/R=45 m -1 k s =0.5 W m -1 K -1 ρ s =912 kg m -3 C ps =1000 J kg -1 K -1 U g =0.4 m s -1 at 20 o C For Argon at 20 o C and 1 atm: µ= kg m -1 s -1 k g =0.017 W m -1 K -1 ρ g =1.7 kg m -3 C pg =521 J kg -1 K -1 For the process parameters: PR=3, PR =3.88, T 0nom =298.15K, T 2nom = K Using the correlation from Levenspiel: W m -2 K -1 Suppose we choose π=30 The simulation results show that the global efficiency is 0.64 and the average temperature difference is T T 159.7K HP LP The stored energy density: Etot EHP ELP c ( T T ) / 2=22.2 kwh / t 2V 2V energy ps HP LP Lead acid battery : 30 kwh/ t c ( T T ) / kwh / t *O. Levenspiel, "Design of long heat regenerators by use of the dispersion model," Chemical Engineering Science, vol. 38, pp , max ps HP LP max
25 Conclusions 1. Cyclic steady state solutions of the PHES process are obtained using the explicit exponential matrix form A A 1 A A ( ) ' ' Tssc e e e e I M M M M A F A F A F A F A A 1 A A ' ' T I M M M A F A F A F A F ssh e e e e 2. A thorough dimensionless analysis procedure is established Dimensionless length: hal Uc g g pg 3. Increasing PR and ha will increase the capacity and efficiency, gas with higher heat capacity ratio and density is prefered. Dimensionless time: ha t c s ps
26 Air Reversible Energy Storage:Loading 23 bar, 50 C C 22.1 bar, 44 C 1.13 bar, -115 C 23.1 bar, 537 C 1.1 bar, 30 C
27 Air Reversible Energy Storage:Delivering Small expander, lower power of compression 23.1 bar, 532 C
28 Questions?
A thermal energy storage process for large scale electric applications
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