= Solution
{c}
For the <unidirectional flow> $\mathbf u=(0,0,w)$, <incompressibility> gives $\partial_z w=0$. Hence $w=w(x,y,t)$, and the <convective acceleration> $w\partial_z w$ vanishes. The transverse components of the <Navier-Stokes equation> give $p_x=p_y=0$, while the axial component gives
$$
\rho\partial_t w=G+\mu(w_{xx}+w_{yy}).
$$
\b[A steady-flow assumption is implicit in the question.] Neither <unidirectional flow> nor a time-independent <pressure gradient> alone eliminates $\partial_t w$: a decaying transient can be superposed on the final pipe profile. The native PDF does not explicitly say <steady flow> in this part. For the intended <steady flow>, $w_t=0$, and therefore
$$
\boxed{w_{xx}+w_{yy}=-\frac G\mu.}
$$
Taking the pipe walls to be stationary, the <no-slip boundary condition> is
$$
\boxed{w(\pm a,y)=0\quad(-b\leq y\leq b),\qquad
w(x,\pm b)=0\quad(-a\leq x\leq a).}
$$
The <no-penetration boundary condition> is already satisfied by the assumed <velocity field>. This is the <Poisson equation> defining <rectangular duct flow>.
Back to article page
