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Energy Storage Technologies Questions and Answers

41.

Frequency regulation asks of a store a response measured in

Answer: A

Correct answer: seconds

Frequency regulation corrects the second-by-second imbalance between generation and demand, so the store must respond within a second or two and sustain the output for seconds to a few minutes. Speed matters far more than capacity, which is why flywheels, supercapacitors and lithium ion batteries are used for it and pumped hydro, with its minutes-long start, historically was not.

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Answer: B

Correct answer: energise a network with no external supply

Black start is the ability to begin generating without any incoming supply, and so to bring the rest of the system back after a total shutdown. Most large thermal plant cannot do it, since it needs auxiliary power to start. A store with a grid-forming inverter can establish a voltage and frequency reference from nothing and then energise the network step by step, which is why black start capability is increasingly required of new storage.

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Answer: C

Correct answer: meeting the peak that would otherwise need a larger circuit

Distribution equipment is sized for the maximum demand it must carry, and that maximum may occur for only a few hours a year. A store placed downstream can supply the excess during those hours, so the circuit never sees the peak and the reinforcement is postponed - sometimes for many years. The annual energy through the circuit is barely changed; it is the peak that is the constraint.

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44.

A reservoir holding 10\( ^6 \) m\( ^3 \) of water at a head of 200 m stores approximately

Answer: D

Correct answer: 545 MWh

The stored potential energy is

\[ E = \rho g V h = 1000 \times 9.81 \times 10^6 \times 200 = 1.962 \times 10^{12} \ \mathrm{J} \]\[ \frac{1.962 \times 10^{12}}{3.6 \times 10^9} \approx 545 \ \mathrm{MWh} \]

A million cubic metres at 200 metres is a large but not exceptional scheme, and 545 MWh would supply a small city for an hour or so.

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45.

If that scheme has a round trip efficiency of 75 per cent, the energy recovered is about

Answer: A

Correct answer: 409 MWh

The round trip efficiency applies to what comes back out:

\[ 545 \times 0.75 \approx 409 \ \mathrm{MWh} \]

The 136 MWh lost appears as heat in the pumps, turbines, pipework and electrical machines. Note that the efficiency is a round trip figure, covering both the pumping and the generating passes; each individual conversion is nearer 87 per cent.

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46.

A 48 V battery bank rated 200 Ah stores

Answer: B

Correct answer: 9.6 kWh

Energy is the product of voltage and ampere hours:

\[ E = 48 \times 200 = 9600 \ \mathrm{Wh} = 9.6 \ \mathrm{kWh} \]

The ampere hour rating alone says nothing about energy until the voltage is known, which is why bank capacity is quoted both ways - and why comparing a 12 V and a 48 V bank on ampere hours alone is meaningless.

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47.

If that bank may be discharged only to 50 per cent, its usable capacity is

Answer: C

Correct answer: 4.8 kWh

Half the nameplate capacity may be used:

\[ 9.6 \times 0.5 = 4.8 \ \mathrm{kWh} \]

The other half is left in the battery deliberately, to obtain the cycle life that a shallow discharge gives. This is why a system specified to deliver a given usable energy must be bought at twice that nameplate, and why usable rather than rated capacity is the figure that matters commercially.

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48.

A 100 Ah cell discharged at 0.5C passes a current of

Answer: D

Correct answer: 50 A

The C-rate multiplies the capacity in ampere hours:

\[ I = 0.5 \times 100 = 50 \ \mathrm{A} \]

and at that rate the cell would be discharged in two hours. Expressing current this way makes the limits transferable: 0.5C is a moderate rate for almost any cell, whatever its size, whereas 50 A means nothing without knowing the capacity.

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49.

A store absorbing 100 kWh and returning 80 kWh has a round trip efficiency of

Answer: A

Correct answer: 80 per cent

Round trip efficiency is what comes out over what went in:

\[ \eta = \frac{80}{100} = 0.80 = 80\% \]

The 20 kWh lost appears as heat in the store itself and in the power conversion equipment at each end. The figure must always be less than one, so the 125 per cent answer can be dismissed without calculating - it inverts the ratio.

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50.

A flywheel of moment of inertia 10 kg\( \cdot \)m\( ^2 \) spinning at 1000 rad/s stores

Answer: B

Correct answer: 1.39 kWh

The stored kinetic energy is

\[ E = \tfrac{1}{2}I\omega^2 = 0.5 \times 10 \times 1000^2 = 5 \times 10^6 \ \mathrm{J} \]\[ \frac{5 \times 10^6}{3.6 \times 10^6} \approx 1.39 \ \mathrm{kWh} \]

Modest as energy, but that 5 MJ can be delivered in a second or two if required, which is the flywheel’s point - it is a power device, not an energy one.

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