= Solution
{c}
Take $z$ vertically upward and the porous skeleton stationary. <Darcy's law> and <incompressibility> are
$$
\boxed{\mathbf u=-\frac{k}{\mu}(\nabla p+\rho g\mathbf e_z),\qquad
\nabla\cdot\mathbf u=0.}
$$
Here $\mathbf u$ is the <Darcy velocity>, a <volume flux> per unit total cross-sectional area, including solid and void. If $\mathbf v$ is the liquid <pore velocity> and all the pore space is saturated and connected, then
$$
\boxed{\mathbf u=\phi\mathbf v.}
$$
The <porosity> $\phi$ therefore enters storage and the conversion to <pore velocity>; it does not multiply the <Darcy velocity> again when calculating discharge.
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