= Solution
{c}
Under spherical symmetry, the <continuity equation> and <Darcy's law> give
$$
\frac1{r^2}\frac{d}{dr}(r^2q_r)=0,\qquad
q_r=-\frac{\Pi_1}{\eta}\frac{dp}{dr}.
$$
Hence $q_r=C/r^2$, and integrating the <pressure gradient> from infinity gives
$$
p(r)-p_0=\frac{\eta C}{\Pi_1r}.
$$
The <boundary condition> at $r=a$ fixes $C=\Pi_1a\Delta p/\eta$. Therefore <spherical Darcy injection> has
$$
\boxed{p(r)=p_0+\frac{a\Delta p}{r},\qquad
q_r(r)=\frac{\Pi_1a\Delta p}{\eta r^2},\qquad
Q=4\pi r^2q_r=\frac{4\pi\Pi_1a\Delta p}{\eta}.}
$$
Here $q_r$ is the superficial <Darcy velocity>, not the pore <velocity> $q_r/\varphi$.
For the comparison, assume geometrically similar pore networks, equal <porosity>, the same source radius and far boundary geometry, and equal pump efficiency. <Dimensional analysis> then gives $\Pi_1=C_{\mathrm{geom}}d^2$ with the same dimensionless factor. Consequently
$$
\frac{\Pi_{1A}}{\Pi_{1B}}=4,\qquad
\frac{(\Pi_1/\eta)_A}{(\Pi_1/\eta)_B}=2.
$$
At a fixed desired <volumetric flow rate>, the required excess <pressure> and pumping <power> are
$$
\Delta p=\frac{\eta Q}{4\pi\Pi_1a},\qquad
\mathcal P=Q\Delta p=\frac{\eta Q^2}{4\pi\Pi_1a}.
$$
\b[Aquifer A requires half the pumping power] at equal throughput, or half the <energy> to inject the same volume over the same time. This compares the excess <pressure-volume work> dissipated by the <Darcy flow>; any different baseline compression or other plant costs need separate accounting.
The imposed comparison matters. At fixed $\Delta p$, A takes twice the <volumetric flow rate> and twice the instantaneous <power>, but the same excess <energy> per injected volume. The pore spacing alone does not determine <permeability of a porous medium> if network geometry or <porosity> changes; without the geometric assumptions the preference cannot be inferred from $d$ alone.
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