Question 1.3: Lagrangian vs. Eulerian Flow Description of River Flow In th......
Lagrangian vs. Eulerian Flow Description of River Flow
In the Eulerian fixed coordinate frame, river flow is approximated as steady 1-D, i.e.,
\mathrm{v}\bigl({ \mathrm{x}}\bigr)=\mathrm{v}_{0}+\Delta \mathrm{v}\Bigl(1-e^{-\mathrm{ax}}\Bigr)
which implies that at x=0, say, the water surface moves at \mathrm{v}_0 and then accelerates downstream to \mathrm{v}(x→∞)= \mathrm{v}_0 +Δ\mathrm{v} . Derive an expression for \mathrm{v} = \mathrm{v}( \mathrm{v}_0 , t) in the Lagrangian frame.
Recall: \vec{\mathrm{v}} =d\vec{r} /dt and in our 1-D case
{\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}\mathbf{t}}}=\mathbf{v}(\mathbf{x})=\mathbf{v}_{0}+\Delta\mathbf{v}{\Big(}1-\mathbf{e}^{-\mathbf{a}\mathbf{x}}{\Big)}
Separation of variables and integration yield:
\int_{0}^{x}\frac{\mathrm{d}\mathbf{x}}{(\mathbf{v}_{0}+\Delta\mathbf{v})-\Delta\mathbf{v}\mathbf{e}^{-\mathbf{ax}}}=\int_0^tdt
so that
\mathbf{x}+{\frac{1}{\mathbf{a}}}\ln\left[1+{\frac{\Delta\mathbf{v}}{\mathbf{v}_{\mathrm{0}}}}\left(1-\mathbf{e}^{-\mathbf{a}\mathbf{x}}\right)\right]=\left(\mathbf{v}_{0}+\Delta\mathbf{v}\right)\mathbf{t}
Now, replacing the two x-terms with expressions from the v(x)-equation, i.e., \mathbf{x}=-{\frac{1}{\mathbf{a}}}\ln\left\lgroup1-{\frac{\mathbf{v}-\mathbf{v}_{0}}{\Delta\mathbf{v}}}\right\rgroup and e^{-ax}=1-{\frac{\mathbf{v}-\mathbf{v}_{0}}{\Delta\mathbf{v}}}, we can express the Lagrangian velocity as:
\mathrm{{v(t)}}={\frac{{\mathrm{v}_{0}}({\mathrm{v}_{0}+\Delta{\mathrm{v}}})}{{\mathrm{v}_{0}}\ +\Delta{\mathrm{v}}\exp[{\mathrm{-a}}{}({\mathrm{v}}_{0}\ +\Delta{\mathrm{v}}){\mathrm{t}}]}}
Graphs:
Comments:
Although the graphs look quite similar because of the rather simple v(x)-function considered, subtle differences are transparent when comparing the velocity gradients (i.e., dv/dx and dv/dt) rather than just the magnitudes v(x) and v(t). Clearly, the mathematical river flow description is much more intuitive in the Eulerian frame-ofreference.
