= Solution
For the intended ideal <Venturi effect>, take <steady flow>, <incompressible flow>, negligible viscous losses, and the same height at the two stations shown. The <Bernoulli equation> along a <streamline> gives
$$
p_1+\frac12\rho v_1^2=p_2+\frac12\rho v_2^2.
$$
Using water <mass density> $\rho\simeq1000\ \mathrm{kg\,m^{-3}}$,
$$
\boxed{p_2-p_1=\frac12(1000)(2^2-6^2)=-1.60\times10^4\ \mathrm{Pa}.}
$$
Thus the outlet <pressure> is lower by $16\ \mathrm{kPa}$. The increased <kinetic energy> per unit volume comes from the drop in <pressure>.
The source's assumption of <laminar flow> alone does not imply negligible <viscosity> or losses. The numerical result uses the usual lossless <Bernoulli equation> approximation, with the quoted speeds interpreted as representative streamline speeds. A real tube with significant viscous losses requires additional data; in a uniform-velocity energy balance its pressure drop would include an extra positive loss term.
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