All posts

Deriving the Bernardi Heat Generation Equation from First Principles

Why Joule heating alone can be off by 20–40% — and how the First Law plus the Nernst equation give the complete picture in two measurable terms.

6 min read

The Bernardi equation (1985) gives the volumetric heat generation rate of a lithium-ion cell in measurable quantities alone:

q=IVb(EocVT)qirr    IVbTEocTqrevq = \underbrace{\frac{I}{V_b}(E_{oc} - V_T)}_{q_{irr}} \;-\; \underbrace{\frac{I}{V_b}\,T\frac{\partial E_{oc}}{\partial T}}_{q_{rev}}
Symbol Meaning Units
II Current (positive on discharge) A
VbV_b Cell volume
EocE_{oc} Open-circuit voltage at current SOC and TT V
VTV_T Terminal voltage under load V
TT Cell temperature K
Eoc/T\partial E_{oc}/\partial T Entropic coefficient at fixed SOC V K⁻¹

Energy balance

Define the system as the entire cell. Two quantities cross the boundary: electrical work VTIV_T I leaves through the load; heat Q˙in\dot{Q}_{in} exchanges with the environment. All matter stays inside. The First Law in rate form:

dUsysdt=Q˙inVTI(1)\frac{dU_{sys}}{dt} = \dot{Q}_{in} - V_T I \tag{1}

The internal energy splits as Usys=Uchem+UthermU_{sys} = U_{chem} + U_{therm}. Evaluating dUchem/dtdU_{chem}/dt requires the reaction rate — moles of lithium transferred per second — which is set by the current.

Faraday's Law. Define:

  • NA=6.022×1023N_A = 6.022\times10^{23} mol⁻¹ — Avogadro's number
  • e=1.602×1019e = 1.602\times10^{-19} C — charge per electron → Faraday constant F=NAe=96485F = N_A e = 96\,485 C mol⁻¹
  • n=1n = 1 — electrons per Li atom (LiLi++e\text{Li} \to \text{Li}^+ + e^-, one electron per formula unit for all Li-ion)
  • NLiN_{Li} — moles of Li transferred (variable)

Total charge: Q=nFNLiQ = nFN_{Li}. Differentiating (dQ/dt=IdQ/dt = I):

N˙Li=InF(2)\dot{N}_{Li} = \frac{I}{nF} \tag{2}

Each mole transferred changes chemical energy by ΔHrxn\Delta H_{rxn} (using ΔUrxnΔHrxn\Delta U_{rxn} \approx \Delta H_{rxn} since PΔV0P\Delta V \approx 0 for solid/liquid electrodes). Defining H˙rxn(I/nF)ΔHrxn>0\dot{H}_{rxn} \equiv -(I/nF)\,\Delta H_{rxn} > 0 (positive because discharge is exothermic, ΔHrxn<0\Delta H_{rxn} < 0):

dUchemdt=H˙rxn(3)\frac{dU_{chem}}{dt} = -\dot{H}_{rxn} \tag{3}

At steady state (dT/dt=0dT/dt = 0), dUtherm/dt=0dU_{therm}/dt = 0, so dUsys/dt=H˙rxndU_{sys}/dt = -\dot{H}_{rxn}. Substitute into Eq. 1 and define Q˙gen=Q˙in\dot{Q}_{gen} = -\dot{Q}_{in}:

Q˙gen=H˙rxnVTI(4)\boxed{\dot{Q}_{gen} = \dot{H}_{rxn} - V_T I} \tag{4}

Heat generated = enthalpy released by reaction − electrical work extracted. H˙rxn\dot{H}_{rxn} is still unknown. The next section expresses it in measurable electrical terms.


H˙rxn\dot{H}_{rxn} in measurable form

Nernst. Define the Gibbs free energy GHTSG \equiv H - TS — the portion of a system's energy available to do useful work at constant temperature and pressure. For the discharge reaction, ΔGrxn=GproductsGreactants<0\Delta G_{rxn} = G_{\text{products}} - G_{\text{reactants}} < 0 (spontaneous). The Clausius inequality shows the maximum non-PV work the reaction can deliver is Wmax=ΔGrxnW'_{max} = -\Delta G_{rxn}. Per mole, nn electrons flow through EocE_{oc}, so Wmax=nFEocW'_{max} = nFE_{oc}. Equating:

Eoc=ΔGrxnnF(5)E_{oc} = \frac{-\Delta G_{rxn}}{nF} \tag{5}

OCV is Gibbs energy per coulomb — voltage is thermodynamics in electrical units.

Entropic coefficient. The Gibbs relation (G/T)P=S(\partial G/\partial T)_P = -S applied to the reaction gives (ΔGrxn/T)P=ΔSrxn(\partial\Delta G_{rxn}/\partial T)_P = -\Delta S_{rxn}. Substituting Eq. 5:

EocT=ΔSrxnnF(6)\frac{\partial E_{oc}}{\partial T} = \frac{\Delta S_{rxn}}{nF} \tag{6}

The OCV slope with temperature at fixed SOC directly measures ΔSrxn\Delta S_{rxn} — no calorimetry needed.

Enthalpy. Substitute Eq. 5 and Eq. 6 into ΔGrxn=ΔHrxnTΔSrxn\Delta G_{rxn} = \Delta H_{rxn} - T\Delta S_{rxn}, then multiply by N˙Li=I/nF-\dot{N}_{Li} = -I/nF (Eq. 2); nFnF cancels:

H˙rxn=I ⁣[EocTEocT](7)\boxed{\dot{H}_{rxn} = I\!\left[E_{oc} - T\frac{\partial E_{oc}}{\partial T}\right]} \tag{7}

All quantities on the right are directly measurable.


The Bernardi equation

Write Q˙gen=qVb\dot{Q}_{gen} = q\,V_b. Substitute Eq. 7 into Eq. 4 and divide by VbV_b:

q=IVb(EocVT)IVbTEocT(9)\boxed{q = \frac{I}{V_b}(E_{oc} - V_T) - \frac{I}{V_b}\,T\frac{\partial E_{oc}}{\partial T}} \tag{9}

qirr=IVb(EocVT)q_{irr} = \frac{I}{V_b}(E_{oc}-V_T) — the overpotential gap (EocVT)(E_{oc}-V_T) is all the voltage lost inside: ohmic resistance, charge-transfer activation, concentration polarisation. Since EocVTIRintE_{oc}-V_T \approx IR_{int}, this scales as I2I^2. Always 0\geq 0.

qrev=IVbTEocTq_{rev} = -\frac{I}{V_b}T\frac{\partial E_{oc}}{\partial T} — not a dissipation loss. It is the thermodynamically required heat exchange TΔSrxnT\,\Delta S_{rxn} per mole (the Clausius reversible heat). A cell with zero resistance still has it. It scales as I1I^1, so at low C-rates it dominates the quadratic qirrq_{irr}. Its sign depends on SOC — the same cell can cool the environment at one SOC and heat it at another. Viswanathan et al. (2010) measured qrevq_{rev} at 20–40% of total heat at C/5 and below.

Why Eoc/T\partial E_{oc}/\partial T changes sign with SOC

Entropic coefficient vs SOC for NMC811/graphite (LG M50). Top: electrode half-cell contributions. Bottom: full-cell dEoc/dT. Data: O'Regan et al. (2022).

Full cell: Eoc/T=dU+/dTdU/dT\partial E_{oc}/\partial T = dU_+/dT - dU_-/dT. The graphite anode dominates because it intercalates lithium in ordered staging phases — in Stage nn, Li occupies every nn-th graphene layer (Stage 4: dilute, x<0.12x < 0.12; Stage 3; Stage 2: x0.25x \approx 0.250.50.5; Stage 1: dense, x>0.5x > 0.5). Each stage has a distinct configurational entropy.

  • 0–47% SOC — graphite is in Stage 4/3 (dilute, disordered). Removing Li collapses disorder sharply → dU/dT0dU_-/dT \gg 0Eoc/T<0\partial E_{oc}/\partial T < 0qrev>0q_{rev} > 0 (extra heat).
  • 47–100% SOC — graphite enters Stage 2/1 (ordered, single-phase). Anode entropy contribution shrinks; NMC cathode (solid-solution, no staging) slightly dominates → Eoc/T>0\partial E_{oc}/\partial T > 0qrev<0q_{rev} < 0 (cell absorbs heat).

Numerical example

18650 NMC/graphite, 1C, 50% SOC, 298 K (I=3I=3 A, Vb=16.5×106V_b=16.5\times10^{-6} m³, Eoc=3.85E_{oc}=3.85 V, VT=3.72V_T=3.72 V, Eoc/T=1.0×104\partial E_{oc}/\partial T = -1.0\times10^{-4} V K⁻¹):

qirr=316.5×106×0.13=23636  W m3q_{irr} = \frac{3}{16.5\times10^{-6}}\times 0.13 = 23\,636\;\text{W m}^{-3} qrev=316.5×106×298×(1.0×104)=+5418  W m3q_{rev} = -\frac{3}{16.5\times10^{-6}}\times 298\times(-1.0\times10^{-4}) = +5\,418\;\text{W m}^{-3} q=29054  W m3q = 29\,054\;\text{W m}^{-3}

The Joule-only model (I2Rint/VbI^2R_{int}/V_b) gives 23 636 W m⁻³ — a 23% underestimate.


References

  1. D. M. Bernardi, E. Pawlikowski, J. Newman, "A General Energy Balance for Battery Systems," J. Electrochem. Soc., 132(1), 5–12, 1985. DOI: 10.1149/1.2113792.
  2. V. V. Viswanathan et al., "Effect of entropy change of lithium intercalation in cathodes and anodes on Li-ion battery thermal management," J. Power Sources, 195(11), 3720–3729, 2010. DOI: 10.1016/j.jpowsour.2009.12.034.
  3. K. O'Regan, F. Brosa Planella, W. D. Widanage, E. Kendrick, "Thermal-electrochemical parameters of a high energy lithium-ion cylindrical battery," Electrochimica Acta, 425, 140700, 2022. DOI: 10.1016/j.electacta.2022.140700.
  4. A. Jokar et al., "Evaluation of accuracy for Bernardi equation under pulse-discharge protocols," Appl. Therm. Eng., 201, 117794, 2022. DOI: 10.1016/j.applthermaleng.2021.117794.
  5. G. L. Plett, Battery Management Systems, Vol. 1, Artech House, 2015.
  6. J. Newman, K. E. Thomas-Alyea, Electrochemical Systems, 3rd ed., Wiley, 2004.

Comments