1. Exergy analysis of CuCl cycle
J.M. Borgard, D. Doizi, P. Carles*

2. CuCl hybrid cycle 5-step version
All reactions demonstrated in laboratory

3. CuCl Cycle efficiency evaluation is difficult
Large uncertainties on many relevant thermodynamical data : Cu2OCl2 enthalpy data CuCl2 and CuCl mixing in HCl/H2O mixture
HCl/H2O azeotrope : - non linear heat exchanges => flowsheet errors are common => choice to use exergy analysis to identify critical points

4. Thermal efficiency  Exergetic efficiency
Equivalence :
DSi: Entropy generation
T: temperature of available heat, Ta : temperature of rejected heat
Q: heat input of the process, W: work input of the process
ηel:efficiency of work generation

5. Some minimal exergy losses Exmin
Heat exchangers
Q: heat,T1 => T2
gas separation
N: total mole number, R: gas constant, x: mole fraction, T : temperature
(theoretical value, usually 5-20 times higher)
Electricity use
W: work, ηv: isentropic efficiency, hel: electrical efficiency
Tmax, Tmin ? Ex_min of chemical reaction ?
Ex = 2 F(E-Ereversible) Ta/T
Exmin ~ kJ (120 = alkalin electrolysis)
Electrolytic step
Irreversible
chemical reaction
Exmin ~ Ta DGR/T

6. Exergetic calculation assumptions :
Reference temperature : Ta =298 K => DG(Ta) = -229 kJ/mol => DH(Ta) = -286 kJ/mol
Average temperature of secondary loop for nuclear heat: Tav = 773 K
Heat to Electrical conversion : efficiency : 0.42
Therefore, we have in our case the relation: hT ≈ 0.68 hex
T = 955 K not 955OC ?

7. Exergy minimal losses evaluation Application to CuCl cycle
For detailed calculations please refer to proceeding

8. (1) Cu(s) + 2 HCl(g) = H2(g) + CuCl(l)
Exergy loss type
Estimation
Heat exchange
6-10 kJ
Chemical reaction
Negligible (DG ≈ 0)
H2/HCl Separation
Min= 13 kJ
More likely > 100 kJ
- Heat release (100 kJ including CuCl solidification) can be recovered for step 4
- Reaction is always incomplete and H2 and HCl can not be easily separated

9. (2): 4 CuCl = 2 CuCl2(aq) + 2 Cu(s)
Exergy loss type
Estimation of exergy loss
Heat release from electrolyser
19 kJ (may be used for cogeneration)
Electrolyser voltage kJ
CuCl dissolution
5-10 kJ
Remaining CuCl
can be done in next step
Exergetic losses are important because of the electrolysis step, which requires heat to be first transformed into electricity
highly dependent on effective voltage

10. (3): 2 CuCl2(aq) = 2 CuCl2(s) Exergy loss type
Estimation of exergy loss
Heat exchange with nuclear heat
9 kJ
Internal heat exchanges
Min = 25 kJ
Heat release
Min = 18 kJ
Too much H2O with CuCl2 => very large heat exchanges for evaporation/condensation of H2O
Global heat demand is important because of the heat of mixing of CuCl2 with HCl/H2O

11. (4): 2 CuCl2(s) + H2O(g) = Cu2OCl2(s) + 2 HCl(g)
Exergy loss type
Estimation of exergy loss
Heat exchange with nuclear heat
3 kJ
Separation of H2O/HCl
Min = 33 kJ
More likely > 150 kJ
Incomplete reaction, even with 15 H2O for 2 HCl (DG >>0) separation of HCl/H2O mixture very energy intensive because of azeotrope

12. (5): Cu2OCl2(s) = 2 CuCl(s) + 0.5 O2 (g)
Exergy loss type
Estimation
Heat exchange
3 kJ
Chemical reaction
5 kJ
Separation issues
0 kJ
Small exergetic loss due to nuclear coupling and irreversibility of reaction

13. Minimal Exergy losses (summary)
stage
Main issues
Exergy minimal loss
Step 1
HCl/H2 separation
20 kJ/mol
( kJ/mol more likely)
Step 2
Electrolysis overvoltage
101 kJ/mol ( kJ/mol more likely)
Step 3
Too low CuCl2 concentration
52 kJ/mol ( kJ more likely)
Step 4
H2O/HCl separation
33 kJ/mol
( kJ expected)
Step 5
none
8 kJ/mol
For comparison : industrial alkaline electrolysis : 285 kJ on same basis

14. Possible Improvements :
« mix » steps to avoid exergy losses

15. Combining steps 1 and 4 No Match !!
Step 4 : 2 CuCl2 + H2O = Cu2OCl2 + 2 HCl
Step 1 :
2 Cu + 2 HCl = 2 CuCl + H2
2 Cu + 2 CuCl2 + H2O = 2 CuCl + Cu2OCl2 + H2
Gain : HCl separation issues easier
Trouble :
Reaction 1: high Pressure reaction 4 : low Pressure
No Match !!

16. Dokiya cycle 2 CuCl(aq) + 2 HCl(g) = 2 CuCl2(aq) + H2(g)
Combining steps 1 and 2
Step 1 :
2 Cu + 2 HCl = 2 CuCl + H2
Step 2 : 4 CuCl = 2 CuCl2 + 2 Cu
Dokiya cycle 2 CuCl(aq) + 2 HCl(g) = 2 CuCl2(aq) + H2(g)
Gain : lower H2O with CuCl2 => less exergy loss in step 3 no separation issues in step 1
Inconvenients :
Higher voltage Cu formation in electrolysis step
Use of high HCl concentration to avoid Cu formation

17. Step 5 : Cu2OCl2 => 2 CuCl + 0.5 O2
Combining steps 4 and 5
Step 4 :
2 CuCl2 + H2O = Cu2OCl2 + 2 HCl
Step 5 : Cu2OCl2 => 2 CuCl O2
CuCl2 direct decomposition 2 CuCl2 + H2O(g) = 2 CuCl + 2 HCl(g) +0.5 O2(g)
Gain : no parasit reaction to handle
Less separation issue
with H2O/HCl
Inconvenients :
separation of remaining Cl2
from O2/HCl
Do it in liquid phase at high pressure ?

18. Towards a liquid-gas cycle?
CuCl2 direct decomposition in liquid phase
2 CuCl2(aq) + H2O(l) = 2 CuCl(aq) + 2 HCl(aq) +0.5 O2(g)
T=350°C
Could perhaps work around 350°C and high pressure
Then combined with
CuCl decomposition 2 CuCl(aq) + 2 HCl (aq) = 2 CuCl2(aq) + H2(g)
T=90°C, electrolysis
Liquid-gas cycle with temperature below 400°C
Only two reactions
HCl/H2O concentrations need to be adjusted between the two steps !

19. Conclusion Cycle evolution unavoidable due to separation issues
Many variations possible, need to be checked precisely
Proposition of a two reaction liquid-gas variation, with temperature compatible with SFR heat