Ba#eries: Part 1 1 Battery chemistries: Reactions
Discharge
Charge
+ e-
Oxidation
Anode
e-
Reduction
Cathode
Negative electrode
-­‐ Positive electrode
Negative electrode
Load -­‐ Positive electrode
Power supply + e-
Reduction
Cathode
e-
Oxidation
Anode
2 Ba#ery figures of merit •  Voltage Ecell (V); Current I (A); Power P (W) = IV
•  Capacity (Ah): a measure of the amount of electric charge o  1 Ah = the charge transferred by a steady current of 1 Amp (1 C/s) flowing for one hour = 3600 Coulombs o  A ba#ery with a capacity of xAh can deliver xA for 1 hr. •  Energy density o  measured in Wh = 1 Js−1 × (3600s) = 3.6 kJ o  1 Wh = the amount of energy converted if work is done at an average rate of 1 W for one hour o  A ba#ery with energy density xWh can deliver xW of power for 1 hr o  By mass (Wh kg−1) o  By volume (Wh dm−3) – 1 dm−3 = 1 litre = 0.001 m−3 •  TheoreVcal and achievable capacity/energy 3 Ba#ery requirements: Driving force for choice of chemistry (redox couples) and materials •  Economic and pracVcal –  Low cost, easy to process materials –  Easy to manufacture electrode, cell and pack assemblies –  Widely available, safe and non-­‐toxic materials •  Must compete with alternaVve power units •  High cell voltage –  Large difference between formal potenVals of +ve and –ve electrode reacVons P=I×V
–  Low overpotenVals at both electrodes –  High soluVon/electrolyte conducVvity 4 Ba#ery requirements •  High energy efficiency –  Low overpotenVals and low ohmic drops –  High charge and voltage efficiency •  High power density –  Important in some applicaVons (e.g. portable electronics/EV/HEV) –  Reduce bulk (reduces energy and interior space losses) 5 Electrode reacVons on discharge •  ReacVon at posiVve electrode: ν a A + ne ↔ ν cC
−
•  ReacVon at negaVve electrode: ν b B ↔ ν d D + ne
−
•  Overall cell reacVon: ΔG = −nFE
X
r
ν a A + ν b B ↔ ν cC + ν d D
0
Recall EquaVon (15b) in thermodynamics slides ⌫ r Sr ⌦
X
⌫ p Sp
p
0
Ecell = E+ve − E−ve
6 Nernst equilibrium Eeq,+ ve
Eeq,−ve
•
•
•
•
νc
⎛
⎞
RT
α
0
C
= E+ ve −
ln ⎜ ν a ⎟
nF ⎝ α A ⎠
νb
⎛
RT
αB ⎞
0
= E−ve −
ln ⎜ ν d ⎟
nF ⎝ α D ⎠
Non-­‐standard condiVons Relates cell potenVal to acVviVes of reactants Applies to zero current (open circuit) only Usually replace acVvity with concentraVon (acVvity for water, solid = 1) 7 Nernst equation: cell
EOCV = Eeq,+ ve − Eeq,−ve
E
EOCV
0
cell
=E
0
+ ve
−E
0
−ve
νc
νb
⎡
⎛
⎞
⎛
RT
αC
RT
αB ⎞ ⎤
0
0
= E+ ve −
ln ⎜ ν a ⎟ − ⎢ E−ve −
ln ⎜ ν d ⎟ ⎥
nF ⎝ α A ⎠ ⎣
nF ⎝ α D ⎠ ⎦
νc νd
⎛
⎞
RT
α
0
C αD
= Ecell −
ln ⎜ ν a ν b ⎟
nF ⎝ α A α B ⎠
8 Charge and discharge voltages
Charge Discharge -­‐ve electrode +ve electrode -­‐ve electrode +ve electrode cathode anode anode cathode I > 0 (i > 0)
I<0
I<0
I > 0 Anodic current η > 0 η < 0 η < 0 η > 0 nFη
−(1−α )nFη
⎛ αRT
⎞
RT
i = i0 (CO CO,0 )
(CR CR,0 ) ⎜⎝ e − e
⎟⎠
During charge the cell voltage increases above the OCV because more than the theoreVcal amount of work (reversible) is required to charge the ba#ery ν O (1−α )
ν Rα
9 Charge and discharge voltages
bB ↔ dD + ne
Charge
−
aA + ne ↔ cC
Discharge
Eeq,−ve
During charge: −
Potential, E / V
Charge
Eeq,+ ve
Ecell = EOCV − η+ ve + η−ve + ∑ k ΔVohm,k
During discharge: Ecell = EOCV − η+ ve + η−ve − ∑ k ΔVohm,k
10 where 'IV = DIV /d and 'V = DV /d. ı is the average inter-fibre distance in the porous carbon electrode and, as before, Di is the
diffusion coefficient for species i in solution. The quantities 'i ,
Charge and discharge cycles
Vanadium flow ba5ery Fig. 2. A comparison of simulated and measured charge–discharge curves for
0 = 1080 mol m−3 and c 0 = 1440 mol m−3 . The charge times are 2017 s and 2714 s,
cIII
III
respectively, and in each case 2 min at zero current were simulated before discharge.
See Tables 2-5 for the other parameter values.
11 C-­‐Rate •  Discharge current is olen expressed as a C-­‐rate to normalise against capacity •  A C-­‐rate is a measure of the rate at which a ba1ery is discharged rela6ve to its maximum capacity – it is a way of implicitly specifying the discharge current •  A 1C rate means that the chosen discharge current will discharge the enVre ba#ery in 1 hour; a 2C rate in 30 mins, a 3C rate in 20 mins, etc. •  A C/x rate or (1/x)C rate will discharge in x hours •  For a ba#ery with a capacity of 100 Ah, a 1C rate equates to a discharge current of 100 A •  A 5C rate (12 mins) for a 100 Ah ba#ery is 500 A •  A C/2 (or 0.5C) rate for a 100 Ah ba#ery is 50 A 12 TheoreVcal capacity and energy •  TheoreVcal capacity and energy are based only on ac6ve species taking part in electrochemical reac6on o  Use stoichiometric reacVons o  Wa# hour (Wh) = voltage (V) x ampere hour (Ah) o  TheoreVcal capacity (coulombic) = Ah g-­‐1 o  TheoreVcal energy = Wh g-­‐1 •  TheoreVcal weight (gravimetric) capacity = nF/M (Ah g-­‐1) o  M = molecular weight (g mol-­‐1) •  TheoreVcal volumetric capacity (Ah cm-­‐3) 13 aday’s law states that if x moles of reagent Ox are r
TheoreVcal capacity [54]
unt of charge (Q) spent is given by
:
•  Faraday’s law states that if x moles of oxidant/reductant are reduced, the total amount of charge (Q) spent is •  n is the number of electrons needed per mole of oxidant •  E.g., during discharge in a Li-­‐ion ba#ery, 1 mole of lithium is intercalated 1 mole o(96485
f FePO4 (iron phosphate) is the
Faraday into constant
C/mol),
n isin the
the cathode, resulVng in the flow of 1 mole of electrons (n = 1per
) through ircuit. QFaraday’s then needed
molethe ofexternal Ox, cand
is the law said
charg
gives us the total charge passed: Q = (1 × F × 1) C per (C). mol of FePO 4
ple calculation of the theoretical capacity of amorphou
14 TheoreVcal capacity •  Calculate theoreVcal capacity per g of the Li host material FePO4 since the the Li ions reside in this material •  The theoreVcal capacity (in C per g of FePO4) is obtained by dividing the charge by the molecular mass (MFePO4 = 151 g mol−1) of FePO4 •  ConverVng the theoreVcal capacity from C g−1 to A h g−1 (the more common measure of capacity) involves a simple conversion of charge from C to Ah, by dividing by 3600. Thus for FePO4 we get the theoreVcal capacity: Q/MFePO4 = nFx/MFePO4/3600 = 1×96485×1/151/3600 = 0.178 A h g−1 15 Example: lead acid 10.3.1 Theory of Operation
The chemical reactions that occur in a lead-acid battery are represented by the following equations:
EsVmate theoreVcal capacity and energy density for the lead-­‐acid ba#ery discharge
$
!
---------------------- PbsO 4 $ 2H 2 O
Positive electrode: PbO 2 $ H 2 SO 4 $ 2H $ 2e
(1)
charge
discharge
$
!
Negative electrode: Pb $ H 2 SO 4 ---------------------- PbSO 4 $ 2H $ 2e
(2)
charge
discharge
Overall cell reaction: PbO 2 $ pb $ 2H 2 SO 4 ---------------------- 2PbSO 4 $ 2H 2 O
(3)
charge
•  NegaVve acVve material = Pb As the cell is charged, the sulfuric acid (H2SO4) concentration increases and becomes highest when the
•  PosiVve acVve material = PbO2 is discharged, the acid concentration decreases and becomes
cell is fully charged. Likewise, when the cell
most dilute
•  Voltage = 2when
.1 Vthe
cell is fully discharged. The acid concentration generally is expressed in terms of
specific gravity, which is weight of the electrolyte compared to the weight of an equal volume of pure water.
•  Pb electrode (negaVve) © 2001 by CRC Press LLC
–  MPb = 207 g mol-­‐1; Density of Pb = 11.3 g cm-­‐3; n = 2 •  TheoreVcal capacity of the Pb electrode (per mole of Pb): nF/MPb/3600= (2 x96485)/207/3600 = 0.26 Ah g-­‐1 Volumetric capacity = 0.26 x 11.3 = 2.9 Ah cm-­‐3 (0.34 cm3 Ah-­‐1) 16 Example: lead acid
•  PbO2 –  MPbO2 = 239 g mol-­‐1; Density = 9.4 g cm-­‐3; n = 2 per mol PbO2 also •  TheoreVcal capacity: nF/MPbO2/3600= (2 x 96485)/3600/239 = 0.22 Ah g-­‐1 Volumetric: 0.22 x 9.4 = 2.1 Ah cm-­‐3 (0.48 cm3 Ah-­‐1) •  TheoreVcal capacity determined by the limiVng electrode. The lowest capacity (0.22 Ah g-­‐1) is taken as the ba#ery capacity •  Energy density = 2.1 x 0.12 = 0.252 Wh g-­‐1 Note: Practical specific energy (measured) = 0.040 Wh g-1
17 Practical batteries
•  Not standard condiVons E0cell ≠ operaVonal voltage •  Use addiVonal materials (not just acVve materials) •  Not 100 % acVve material uVlisaVon •  Internal resistance •  Once a current flows, there are losses (same as those for a fuel cell): o  AcVvaVon losses o  Ohmic losses o  Mass transport losses •  Analysis very similar except Vme dependent 18 Practical batteries
•  The pracVcal (specific) capacity measured when pass a constant current Vll the ba#ery reaches its cutoff voltage iAt cutoff
−1
practical capacity =
[Ah g ]
3600 × W
•  where i is the current density in Am−2, A the acVve area in m2, tcutoff is the Vme to reach the cut off voltage (Ecutoff) in seconds and W is the weight of the acVve material in kg. 19 Efficiencies
•
•
•
•
Charge, voltage and energy efficiencies of the VRFB are obtained from the following formulae (for constant current during charge and discharge): td
Charge efficiency = td
discharge capacity ∫tc
i (t −t )
tc
= tc
=
= d d c
tc
charge capacity
ic tc
∫ Ic dt A ∫ ic dt
I d dt
A ∫ id dt
0
0
The charge is to the state of charge (at tc) as pertains to the start of discharge (at td) t
1
Voltage efficiency = average Ecell during discharge
average Ecell during charge
(td −tc ) ∫tc
1
t c ∫0
tc
d
Ecell dt
Ecell dt
td
•
•
td
∫tc (Aid )Ecell dt id ∫tc Ecell dt
discharged energy
= tc
= × tc
Energy efficiency = energy
to charge
to same
SOC
)Ecell dt ic ∫ Ecell dt
∫0 (Ai
!###c"###
$
0
power
energy efficiency = charge efficiency × voltage efficiency 20 Challenges: Cell performance •  Increase energy density –  High energy density acVve materials –  Reduce packaging –  AcVve material uVlisaVon –  Novel (lightweight) current collectors •  Increase cycle life (lead acid + others) •  New ba#ery types 21