= Solution
A pumped-storage plant moves water uphill using electrical <power>, then returns it through a turbine and generator when electricity is needed. The upper and lower reservoirs set the usable <volume> and head; pumps, penstocks, valves and reversible pump-turbines set the <power> and flow. The plant stores <energy> rather than generating net <energy> over a cycle.
\Image[/past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2018-pumped-storage-energy-flow.png]
{title=Gravitational storage separates reservoir energy capacity from pump and turbine power}
{description=An original functional diagram shows an upper and a lower reservoir connected through a penstock and reversible pump-turbine. Charging uses grid electricity to lift water; discharge returns gravitational potential energy through the generator.}
For constant head, the gravitational <energy> is $E_g=\rho Vgh$. With the stated generation efficiency,
$$
\boxed{E_{\rm usable}=0.89(1000)(1.4\times10^7)(9.81)(400)
\simeq4.89\times10^{13}\,\mathrm J=13.6\,\mathrm{GWh}.}
$$
At $3$ GW, $P=\eta\rho gQh$ gives
$$
\boxed{Q\simeq8.59\times10^2\,\mathrm{m^3\,s^{-1}},\qquad
t_{\rm full}=E_{\rm usable}/P\simeq4.53\,\mathrm h.}
$$
Head variation, dead <volume> and hydraulic losses beyond the stipulated efficiency have been neglected.
To deliver $9$ kWh with the same head and efficiency needs $V\simeq9.3\,\mathrm{m^3}$ of water, roughly a $2.1$ m cube, together with a $400$ m elevation difference and a lower reservoir. At a $10$ m head it would need about $370\,\mathrm{m^3}$. This contrasts with the small fuel <volume>: the useful <energy> <density> of lifted water depends on its external head, not its chemical composition.
<Pumped storage> offers long service life, fast dispatch, substantial <power>, and separate choices of storage <volume> and turbine rating. Its drawbacks include geographic constraints, construction cost and <time>, ecological impacts, evaporation and water requirements, plus round-trip losses. The quoted generation efficiency alone is not the <round-trip efficiency> because pumping also consumes more than the stored <potential energy>.
The rock cylinder has <mass> $\rho_r\pi r^2(2r)$ and rises through $r$. Equating its useful <energy> to the reservoir result cancels the common efficiency and <gravity>:
$$
2\pi\rho_r r^4=\rho_wVh,\qquad
\boxed{r=\left[\frac{1000(1.4\times10^7)(400)}{2\pi(3000)}\right]^{1/4}
\simeq131\,\mathrm m.}
$$
Its height is about $262$ m and its lift about $131$ m, ignoring buoyancy as instructed. Stored <energy> scales as $r^4$, excavated <volume> roughly as $r^3$, and lined area as $r^2$. If excavation dominates cost, cost per unit stored <energy> decreases approximately as $r^{-1}$. Seals, increasing <pressure>, geology, leakage, supporting structure and <power> equipment can change that scaling; a <volume> law is not a complete cost model.
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