Energy Storage Systems: Batteries, Thermal Stores, Pumped Hydro, Compressed Air and Round-Trip Efficiency

This chapter answers Section XE8.2 of XE8 Energy Science, the sub-heading “Energy storage systems”: “Batteries (capacity, C-rate, state of charge, state of health, depth of discharge, energy and power densities); conservation of energy and mass for thermal energy storage, pumped hydro storage, and compressed air storage; charging, discharging, and roundtrip efficiency.” It belongs to the Engineering Sciences (XE) paper, as Section XE8.2 of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam. The chapter teaches battery ratings and ageing measures, the energy balances of sensible and latent thermal stores, pumped hydro and compressed air, and how charging, discharging and round-trip efficiencies multiply. The numericals are pack energy and C-rate current, usable energy from a state-of-charge window, thermal store sizing, pumped-hydro energy and compressed-air work from stated data.

1. Battery ratings: capacity, C-rate, SoC, DoD, SoH, densities

A cell stores charge. Its capacity Q is the charge it can deliver between full and empty under stated conditions, quoted in ampere-hours (Ah); 1 Ah = 3600 C. Its energy is the capacity times the mean discharge voltage, E = V × Q in Wh (a pack of 48 V nominal and 200 Ah holds 48 × 200 = 9600 Wh = 9.6 kWh). Capacity is not a constant of the cell: it falls at high current, at low temperature and with age.

  • C-rate expresses current as a multiple of the rated capacity: I = (C-rate) × Q_rated, with Q in Ah and I in A. At 1C a 200 Ah cell draws 200 A and empties in 1 h; at 0.5C it draws 100 A and lasts 2 h; at 2C it draws 400 A and lasts 0.5 h. Ideal run time is 1/(C-rate) hours; real time is shorter at high C-rate.
  • State of charge SoC = (charge now)/(present full-charge capacity), as a percentage; depth of discharge DoD = 1 − SoC, the fraction removed from the full state. Both are relative to the present capacity under the convention used here, and in practice SoC is estimated, by current integration (coulomb counting) corrected against open-circuit voltage, not measured directly.
  • State of health SoH, on the capacity definition, = Q_present/Q_rated (end of life is commonly taken near 80 %); a resistance-based SoH, from the rise of internal resistance, is also used. Deep cycling, high temperature and high C-rate all lower SoH faster.
  • Specific energy (Wh/kg) and energy density (Wh/L) say how much energy a unit mass or volume holds; specific power (W/kg) and power density (W/L) say how fast it can be delivered. The two are different axes: an energy-type cell stores much and delivers slowly; a power-type cell (or a capacitor) the reverse.
⚠️ Which capacity is the denominator?
C-rate is always referred to the rated capacity, but SoC is referred to the present capacity. An aged cell with SoH = 0.8 at 1C still draws the rated 1C current, and so empties in about 0.8 h, not 1 h.
Worked example: 48 V nominal, 200 Ah rated, 60 kg pack
QuantityWorkingResult
Energy48 × 2009.6 kWh
Specific energy9600/60160 Wh/kg
0.5C current, power, time0.5 × 200 = 100 A; 48 × 100; 1/0.5100 A, 4.8 kW, 2 h
At SoH 0.8 and SoC 50 %0.8 × 200 = 160 Ah; × 0.5 × 483.84 kWh stored

2. Charging, discharging and round-trip efficiency

Every store loses energy on the way in, while it waits, and on the way out. The round-trip efficiency is the energy delivered on discharge divided by the energy drawn from the source to charge it, over a complete cycle that returns the store to its starting state: η_RT = E_out/E_in. When the stages are in series, the efficiencies multiply: η_RT = η_charge × η_storage × η_discharge. Two stages of 90 % give 0.90 × 0.90 = 81 %, not 90 %.

  • Battery: the coulombic efficiency is Ah out/Ah in (close to 1 for a good cell); the voltage efficiency is the mean discharge voltage over the mean charge voltage (below 1 because charging needs V above the open-circuit value and discharge delivers V below it, the gap being I × R_internal); the energy efficiency is their product. Losses rise as I²R, so efficiency falls with C-rate.
  • Self-discharge and standby losses act with time, so a store that sits longer returns less. A thermal tank loses heat in proportion to its surface and temperature excess; a battery self-discharges slowly.
  • Energy accounting: if a plant charges using power P_c for time t_c and discharges P_d for t_d, then E_in = P_c t_c, E_out = P_d t_d, and the energy lost is E_in − E_out. Never form a round-trip figure by adding percentages, and never quote it without saying which boundary (cell, pack with converter, or whole plant with auxiliaries) it covers.
⚠️ A common slip
Round-trip efficiency of 79.2 % does not mean 20.8 % of the energy is lost in each direction. With 88 % up and 90 % down the loss is 12 % on charge, 10 % of what remains on discharge, and the product is 0.88 × 0.90 = 0.792.

3. Thermal energy storage: energy and mass balance

A thermal store holds energy as heat in a mass m of storage medium. Sensible storage raises the temperature of a liquid or solid with no phase change: Q = m c ΔT (c in kJ/kg·K, ΔT in K). Latent storage uses a phase change at nearly constant temperature, as in a phase-change material (PCM): Q = m L_f. A store that heats a solid PCM, melts it and superheats the liquid has Q = m [c_s (T_m − T₁) + L_f + c_l (T₂ − T_m)]. A latent store gives a larger energy per unit mass over a narrow temperature window; a sensible one is simple and cheap.

Conservation of energy for a tank of fixed mass m, charged by a heat input Q̇_in and losing heat to ambient through an overall conductance UA, is m c dT/dt = Q̇_in − UA (T − T_amb). With no input, the excess temperature decays exponentially with time constant τ = m c/(UA). Conservation of mass matters for open stores: charging a water tank by running hot water in and drawing cold out at the same flow ṁ keeps the mass constant, and the energy carried in is ṁ c (T_hot − T_cold). A tall tank keeps hot water above cold by density difference (stratification), so the delivery temperature stays high until the hot layer is used up; mixing destroys this and lowers the useful energy.

🎯 Why a PCM store needs only a small temperature swing
Most of a PCM store’s energy is the latent term m L_f at one temperature. A PCM with L_f = 200 kJ/kg and c = 2 kJ/kg·K stores 200 kJ/kg in melting alone; to store the same by sensible heating you would need ΔT = 100 K. The same property is a limit: the delivery temperature is pinned at the melting point.
Worked example: 50 m³ of water, ρ = 1000 kg/m³, c = 4.18 kJ/kg·K, heated from 40 °C to 90 °C
StepWorkingResult
Mass50 × 100050 000 kg
Heat stored50 000 × 4.18 × 5010 450 MJ
In kWh10 450/3.6≈ 2903 kWh
Boiler input at 85 % heating efficiency2903/0.85≈ 3415 kWh

4. Pumped hydro storage

Surplus electricity pumps water from a lower to an upper reservoir; at demand peaks the water falls back through a turbine. The energy is gravitational potential energy. For a volume V raised through a mean head H, the gross energy is E = ρ g H V, with ρ = 1000 kg/m³ and g = 9.81 m/s². Raising 1.0 × 10⁶ m³ through 200 m stores 1000 × 9.81 × 200 × 10⁶ = 1.962 × 10¹² J = 545 MWh (divide joules by 3.6 × 10⁹ for MWh).

Mass conservation fixes the volume moved: the volume leaving the upper reservoir equals that entering the lower (minus evaporation and seepage). Energy conservation across a pump of efficiency η_p gives the electrical input E_in = ρ g H V/η_p, and across a turbine-generator of efficiency η_t the output E_out = η_t ρ g H V. Hence η_RT = η_p η_t when friction in the tunnel is lumped into the machine efficiencies. If it is treated separately, the pump must overcome H + h_f and the turbine receives H − h_f, where h_f is the head lost to pipe friction, so the loss is paid twice.

Worked: with η_p = 0.88 and η_t = 0.90, the 545 MWh store above takes 545/0.88 = 619 MWh to fill and returns 0.90 × 545 = 490.5 MWh, so η_RT = 490.5/619.3 = 0.792. Power is a separate rating: P = ρ g H Q η, and the time the store can run is energy divided by power. Energy depends on V and H; power on the flow rate and the machines. A site can be sized for hours of storage by choosing V.

5. Compressed air energy storage

Compressed air energy storage (CAES) uses surplus power to compress air into a cavern or tank and later expands it through a turbine. Treat air as an ideal gas. The minimum (isothermal, reversible) work to compress a gas from p₁ to p₂ is W = p₁V₁ ln(p₂/p₁) = mRT ln(p₂/p₁), where p₁V₁ equals the final pV of the stored gas. The reversible adiabatic work per kg is w = [γ/(γ − 1)] R T₁ [(p₂/p₁)^((γ−1)/γ) − 1], always larger, because the gas heats up and resists compression.

Worked, air with R = 287 J/kg·K, γ = 1.4, from 100 kPa and 300 K to 1 MPa (r = 10): isothermal w = 287 × 300 × ln 10 = 198.3 kJ/kg; adiabatic w = 3.5 × 287 × 300 × (10^0.2857 − 1) = 280.5 kJ/kg, with exit temperature 300 × 10^0.2857 = 579 K. In diabatic CAES this compression heat is thrown away and the air must be reheated (usually by burning fuel) before expansion, so the electrical round-trip figure is low and fuel is also used. Adiabatic CAES stores the compression heat in a thermal store (sensible heat in a solid or liquid) and returns it to the air before expansion, raising the round trip; isothermal schemes try to compress and expand slowly with heat exchange. Intercooled multistage compression moves the work towards the isothermal value.

⚠️ Stored air energy is small per volume
A cavern of V = 1000 m³ at 7 MPa that was filled from 0.1 MPa isothermally needed p V ln(p/p₀) = 7 × 10⁹ × ln 70 = 29.7 GJ = 8.26 MWh. Much of the pressure energy cannot be recovered, since a fixed-volume store loses pressure as it empties and the turbine must be fed above a minimum pressure.

6. Choosing a store: scale, duration and traps

Qualitative comparison
StoreEnergy is stored asEnergy set byMain losses
BatteryElectrochemical energyAh × V, SoHI²R, side reactions
Thermal tank / PCMSensible or latent heatm c ΔT or m L_fHeat loss to ambient, mixing
Pumped hydroGravitational potential energyρ g H VPump, turbine, friction
Compressed airPressure (and heat)p V ln(p/p₀) at bestCompression heat, fuel, throttling

Power versus energy is the first choice. Batteries and flywheels suit fast response and minutes to hours; pumped hydro, CAES and thermal tanks suit large energy with a long duration. Duration = energy capacity/power rating: a 100 MW unit with 400 MWh of usable energy runs 4 h at full power. Usable energy is less than rated, because the usable SoC window is limited, SoH has fallen, and the round trip loses part of every charge.

Worked: a pack rated 400 V and 50 Ah, at SoH 0.9, operated between 20 % and 90 % SoC, gives 400 × 50 × 0.9 × (0.90 − 0.20) = 12 600 Wh = 12.6 kWh usable. The window width (0.70) multiplies the present capacity, which is itself SoH times rated.

Key takeaways

  • Battery energy is V × Ah; current is C-rate × rated Ah, ideal run time is 1/C-rate hours; SoC and DoD refer to present capacity, SoH is present over rated capacity.
  • Specific energy (Wh/kg) and specific power (W/kg) are independent axes; losses in a battery grow as I²R, so efficiency falls at high C-rate.
  • Round-trip efficiency is E_out/E_in over a full cycle and is the product of the series efficiencies; always state the boundary it covers.
  • Thermal stores: Q = m c ΔT (sensible) and m L_f (latent, constant temperature); a tank cools as exp(−t/τ) with τ = m c/UA, and stratification preserves delivery temperature.
  • Pumped hydro stores ρ g H V, with η_RT = η_p η_t; compressed air needs at least p₁V₁ ln(p₂/p₁) isothermally and more adiabatically, so compression heat is the key loss that adiabatic CAES recovers.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A cell has a rated capacity of 40 Ah. The current corresponding to a discharge rate of 0.25C is

    1. 10 A, which would ideally empty the cell in 4 h
    2. 0.25 A, which would ideally empty the cell in 160 h
    3. 160 A, which would ideally empty the cell in 15 min
    4. 40 A, which would ideally empty the cell in 1 h
    Show answer

    Answer: A — 10 A, which would ideally empty the cell in 4 h

    I = C-rate × Q_rated = 0.25 × 40 = 10 A, and the ideal time is 1/0.25 = 4 h. A current of 40 A is the 1C value, the tempting wrong answer when the rate is ignored.

  2. State of health of a cell, on the capacity-based definition used in this chapter, is

    1. the charge now divided by the present full-charge capacity
    2. the present full-charge capacity divided by the rated capacity
    3. one minus the fraction of charge removed from the full state
    4. the rated current divided by the present discharge current
    Show answer

    Answer: B — the present full-charge capacity divided by the rated capacity

    SoH = Q_present/Q_rated measures ageing. The charge-now-over-present-capacity ratio is SoC, and one minus the fraction removed is also SoC (one minus DoD); neither says anything about how much capacity has been lost.

  3. A storage plant draws 100 MWh from the grid to charge. Its charging stage is 92 % efficient, its standby losses take 4 % of the stored energy, and its discharge stage is 90 % efficient. The energy delivered is

    1. 78.0 MWh, from subtracting the 8, 4 and 10 percentage-point losses from 100
    2. 79.5 MWh, from the product 0.92 × 0.96 × 0.90
    3. 86.4 MWh, from applying only the standby and discharge stages, 0.96 × 0.90
    4. 92.7 MWh, from taking the simple average of the three efficiencies, 92, 96 and 90 %
    Show answer

    Answer: B — 79.5 MWh, from the product 0.92 × 0.96 × 0.90

    The stages are in series, so the efficiencies multiply: 0.92 × 0.96 × 0.90 = 0.795, so 79.5 MWh. Subtracting the percentage losses from 100 (giving 78) treats every loss as acting on the original 100 MWh, whereas each acts only on what is left.

  4. A well-insulated hot-water tank of mass m, specific heat c and loss conductance UA is left standing with no heat input. If its mass doubles at the same UA and same specific heat (a larger tank of the same surface conductance), the time for the excess temperature over ambient to fall to half

    1. doubles, because τ = m c/(UA) is proportional to m
    2. halves, because a bigger tank loses heat faster
    3. is unchanged, because the loss depends only on UA
    4. becomes four times larger, because mass enters squared
    Show answer

    Answer: A — doubles, because τ = m c/(UA) is proportional to m

    From m c dT/dt = −UA (T − T_amb), the excess decays as exp(−t/τ) with τ = m c/(UA). Doubling m at fixed UA doubles τ, and so the half-time (τ ln 2). The same loss rate acts on twice the stored heat, so it falls twice as slowly in relative terms.

  5. In a pumped hydro scheme the static head between reservoirs is 100 m. The pump and the turbine are ideal (efficiency 1), but pipe friction costs a head of 5 m in each direction, at the same flow. The round-trip efficiency is

    1. 95.0 %, because only 5 m is lost
    2. 90.5 %, from (100 − 5)/(100 + 5)
    3. 100 %, because the same water returns
    4. 81.9 %, from squaring the ratio 95/105
    Show answer

    Answer: B — 90.5 %, from (100 − 5)/(100 + 5)

    Pumping needs a head of 100 + 5 = 105 m and generation receives 100 − 5 = 95 m for the same volume, so η_RT = 95/105 = 0.905. The loss is incurred twice, which is why 95 % is wrong; squaring the ratio would count the friction a second time on top.

  6. For the same ideal gas, the same initial state and the same pressure ratio, the reversible work of compression is

    1. smaller for adiabatic than for isothermal compression, since the rising temperature lets the gas do part of the work itself
    2. equal for adiabatic and isothermal compression, since both processes are reversible and end at the same pressure
    3. larger for adiabatic than for isothermal compression, since the gas heats and resists compression
    4. zero for isothermal compression, since the internal energy of an ideal gas does not change
    Show answer

    Answer: C — larger for adiabatic than for isothermal compression, since the gas heats and resists compression

    The adiabatic p-V curve is steeper than the isothermal one, so more area (work) is needed to reach the same final pressure. Zero internal-energy change in the isothermal case means the work equals the heat rejected, not that the work vanishes.

  7. A latent-heat store using a PCM is compared with a water tank (sensible heat) of the same mass, charged over a temperature window of 10 K around the PCM melting point. Which statement is correct?

    1. The PCM store always holds less energy, since the latent heat of any material is small compared with c ΔT
    2. The PCM store holds more when m L_f exceeds the sensible heat that water takes up over that window, and it delivers heat near one temperature
    3. Both hold equal energy over that window, since the temperature window is the same and the masses of the two media are equal
    4. The PCM store delivers its heat at a temperature that falls steadily throughout the discharge, like a sensible store
    Show answer

    Answer: B — The PCM store holds more when m L_f exceeds the sensible heat that water takes up over that window, and it delivers heat near one temperature

    The comparison is between m L_f (plus a little sensible heat) and m c ΔT. Water with c = 4.18 kJ/kg·K over 10 K takes about 42 kJ/kg, so a PCM with L_f of 150 to 250 kJ/kg holds far more; and during melting the temperature is nearly constant, not steadily falling.

  8. In one full cycle a cell’s coulombic efficiency is 0.99, the mean charging voltage is 3.90 V and the mean discharge voltage is 3.60 V. Its energy efficiency is nearest to

    1. 99.0 %, the coulombic efficiency taken alone
    2. 92.3 %, the voltage ratio 3.60/3.90 taken alone
    3. 107.3 %, from the charge voltage over the discharge voltage
    4. 91.4 %, the product 0.99 × 3.60/3.90
    Show answer

    Answer: D — 91.4 %, the product 0.99 × 3.60/3.90

    Energy out/in = (Ah out × V_d)/(Ah in × V_c) = 0.99 × 3.60/3.90 = 0.914. The coulombic and voltage factors multiply; either alone overstates the efficiency, and a value above 100 % would violate energy conservation.

  9. A battery management system estimates SoC only by integrating the measured current over time (coulomb counting) with a fixed value of full capacity. Over many cycles the estimate becomes unreliable mainly because

    1. current-sensor offset errors accumulate in the integral and the true full capacity falls with ageing
    2. charge is not conserved inside a cell, so the integral of current over time has no physical meaning
    3. SoC depends only on the terminal voltage and never on the charge that has been passed
    4. coulomb counting works only for discharge and fails on charge
    Show answer

    Answer: A — current-sensor offset errors accumulate in the integral and the true full capacity falls with ageing

    Integration accumulates any small bias, and a fixed denominator ignores the fall of capacity with SoH, so the estimate drifts unless corrected (for example against open-circuit voltage). The claim that terminal voltage alone fixes SoC is false: it also depends on current and temperature.

  10. Which of the following statements about a 30 Ah rated cell are correct?

    1. A 2C discharge corresponds to 60 A
    2. An ideal 2C discharge empties the cell in 30 minutes
    3. Usable capacity at a very high C-rate is often less than at a low C-rate
    4. The current at a given C-rate does not depend on the rated capacity
    Show answer

    Answer: A — A 2C discharge corresponds to 60 A; B — An ideal 2C discharge empties the cell in 30 minutes; C — Usable capacity at a very high C-rate is often less than at a low C-rate

    I = C × Q_rated gives 60 A, the ideal time is 1/2 h, and internal resistance and diffusion limits cut usable capacity at high rates. The current clearly does depend on Q_rated, since I = C-rate × Q_rated.

  11. Which of the following statements about pumped hydro and compressed air storage are correct?

    1. For a fixed volume of water, the stored energy ρ g H V is proportional to the head
    2. In diabatic compressed air storage the heat of compression is rejected and heat must be supplied to the air before it is expanded
    3. The reversible adiabatic compression work for a given pressure ratio is less than the isothermal work
    4. With friction lumped into the machine efficiencies, the pumped hydro round-trip efficiency is the product of the pump and turbine-generator efficiencies
    Show answer

    Answer: A — For a fixed volume of water, the stored energy ρ g H V is proportional to the head; B — In diabatic compressed air storage the heat of compression is rejected and heat must be supplied to the air before it is expanded; D — With friction lumped into the machine efficiencies, the pumped hydro round-trip efficiency is the product of the pump and turbine-generator efficiencies

    E = ρ g H V is linear in H, diabatic CAES wastes compression heat and must reheat, and η_RT = η_p η_t. The adiabatic work exceeds the isothermal work, so the statement that the adiabatic work is smaller is reversed.

  12. Which of the following statements about sensible hot-water thermal storage are correct?

    1. Stratification, hot water above cold, keeps the delivery temperature high until the hot layer is used up
    2. Reducing the loss conductance UA lengthens the time constant m c/UA
    3. Stored heat for a given temperature rise is proportional to the mass of water
    4. Mixing the hot and cold layers raises the useful energy that can be delivered at the original hot temperature
    Show answer

    Answer: A — Stratification, hot water above cold, keeps the delivery temperature high until the hot layer is used up; B — Reducing the loss conductance UA lengthens the time constant m c/UA; C — Stored heat for a given temperature rise is proportional to the mass of water

    Q = m c ΔT is linear in m, τ = m c/UA grows when UA falls, and stratification preserves the delivery temperature. Mixing spreads the heat over the whole tank at a lower mean temperature, which reduces the energy available at the original hot temperature.

  13. A battery pack has a nominal voltage of 350 V and a rated capacity of 80 Ah. Its state of health (capacity basis) is 0.85, and it is operated only between 10 % and 95 % SoC, both measured relative to its present capacity, at a constant mean voltage equal to the nominal value. The usable energy per cycle is ____ kWh (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 20.23

    Present capacity = 0.85 × 80 = 68 Ah. Window = 0.95 − 0.10 = 0.85, so usable charge = 0.85 × 68 = 57.8 Ah. Energy = 350 × 57.8 = 20 230 Wh = 20.23 kWh. Taking the rated capacity instead gives 23.8 kWh, which ignores ageing.

  14. A phase-change material has a latent heat of fusion of 200 kJ/kg and a specific heat of 2.0 kJ/kg·K in both solid and liquid states, with a melting point of 58 °C. A store is charged from 50 °C (solid) to 70 °C (liquid). The mass of PCM needed to store 1.0 GJ is ____ kg (nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 4167

    Per kg: 2.0 × (58 − 50) = 16 kJ sensible solid, + 200 kJ latent, + 2.0 × (70 − 58) = 24 kJ sensible liquid = 240 kJ/kg. Mass = 1.0 × 10⁶ kJ/240 = 4166.7 kg, nearest integer 4167. Latent heat alone would give 5000 kg.

  15. An upper reservoir holds 2.0 × 10⁶ m³ of water whose mean height above the turbine is 300 m. Take ρ = 1000 kg/m³ and g = 9.81 m/s², treat the head as constant, and let the combined turbine and generator efficiency be 0.88. The electrical energy produced when all of the water is discharged is ____ MWh (nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 1439

    Gross energy = 1000 × 9.81 × 300 × 2.0 × 10⁶ = 5.886 × 10¹² J = 1635 MWh (÷ 3.6 × 10⁹). Electrical output = 0.88 × 1635 = 1438.8 MWh, so 1439.

  16. Air, treated as an ideal gas, is compressed isothermally and reversibly from 0.1 MPa into a rigid cavern of volume 1000 m³ until the pressure is 7 MPa. The minimum compression work is ____ MWh (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 8.26

    Final pV = 7 × 10⁶ × 1000 = 7 × 10⁹ J equals p₁V₁ for the isothermal process. W = p₁V₁ ln(p₂/p₁) = 7 × 10⁹ × ln 70 = 7 × 10⁹ × 4.2485 = 2.974 × 10¹⁰ J = 8.26 MWh (÷ 3.6 × 10⁹).

  17. A pumped hydro plant must deliver 500 MWh to the grid in a day. Its overall pumping efficiency (grid to water) is 0.88 and its overall generating efficiency (water to grid) is 0.90, with no other losses. The energy the plant must draw from the grid to refill the upper reservoir is ____ MWh (nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 631

    η_RT = 0.88 × 0.90 = 0.792. E_in = 500/0.792 = 631.3 MWh, so 631. Dividing 500 by 0.90 alone gives 556, which omits the pumping stage.