The Conjugate flow equations

Consider flow in a 2-layer system, where in one region we have uniform flow where the lower layer has velocity, c₁, depth h₁ and density $\rho_1$ and the upper layer has c₂, h₂ and $\rho_2$. For our purposes here we have that c₁=c₂=c i.e. both outskirts layer flow velocities are equal to (minus) the wave velocity. Suppose over some ``surge-region'', this uniform flow transformed into a distinct uniform flow where the corresponding velocities are now $c_1^\prime$ and $c_2^\prime$ and the depths $h_1^\prime$ and $h_2^\prime$. For our purposes here Benjamin (1966) defined such regions of uniform 2-layer flows as being ``conjugate'' to each other if

1.

the fluid flux in every layer is equal in both regions i.e.

$$c_1^\prime h_1^\prime = c_1 h_1 = c h_1 \mbox{\ \ \ \ and\ \ \ } c_2^\prime h_2^\prime = c_2 h_2 = c h_2$$ (1)

2.

the momentum flux in both regions are equal i.e.

 

$$\int_0^{h_2} dy\;(p_t+\rho_2 gy+\rho_2 c^2) + \int_0^{h_1} dy (p_t+\rho_2 gh_2+\rho_1 gy+\rho_1 c^2)$$

$$= \int_0^{h_2^\prime} dy\;(p_t^\prime+\rho_2 gy+\rho_2 c_2^{\prim... ...h_1^\prime} dy (p_t^\prime+\rho_2 gh_2^\prime+\rho_1 gy+\rho_1 c_1^{\prime\,2})$$

$${\it i.e.\ } p_t(h_1+h_2)+\frac{1}{2}\rho_2 gh_2^2+ \rho_2 gh_1h_2 +\frac{1}{2}\rho_1 gh_1^2 +c^2(\rho_1 h_1+\rho_2 h_2)$$

$$= p_t^\prime(h_1^\prime+h_2^\prime)+\frac{1}{2}\rho_2 gh_2^{\prim... ...h_2^\prime +\frac{1}{2}\rho_1 gh_1^{\prime\,2}+\rho_1 c_1^{\prime\,2}h_1^\prime$$ (2)

where p_t and $p_t^\prime$ are the pressures at the top surfaces of the upper layers in both uniform flow regions respectively.

3.

the energy flux in both regions are equal or, equivalently, we can obtain a relation by summing all the pressure changes round a large loop bounded by the upper and lower streamlines and with vertical ``arms'' in the regions of uniform flow. This latter form is more convenient when we consider ``free top-surface'' boundary conditions. In the rigid lid flat-topped case we have, from Bernoulli's equation

$$p_t - p_t^\prime = \frac{1}{2}\rho_2 (c_2^{\prime\,2}-c^2) \mbox{\ \ \ \ and\ \ \ } h_1 + h_2 = h_1^\prime + h_2^\prime$$ (3)

or, alternatively, for the free surface case,

$$p_t - p_t^\prime =0 \mbox{\ \ \ \ and\ \ \ } \rho_2 g(h_1+h_... ...^\prime-h_2^\prime) = \frac{1}{2}\rho_2 (c_2^{\prime\,2}-c^2)$$ (4)

Applying Bernoulli's equation along the bottom streamline gives the difference between the bottom pressures, p_b and $p_b^\prime$ viz.

$$p_b - p_b^\prime = \frac{1}{2}(\rho_1 c_1^{\prime\,2} - \rho_1 c^2)$$

In either case, as the sum of the pressure changes round the closed path must add to zero,

$$\{p_t-p_t^\prime\} + \{\rho_2 gh_2+\rho_1 gh_1\} + \{p_b^\prime-p_b\} - \{\rho_2 gh_2^\prime+\rho_1 gh_1^\prime\} = 0$$ (5)

With h₁, h₂ given, the above effectively constitute seven equations in seven unknowns, $h_1^\prime$, $h_2^\prime$, c, $c_1^\prime$, $c_2^\prime$, $p_t-p_t^\prime$ and $p_b - p_b^\prime$. However they are easily reducible to virtually two equations in two unknowns, that can be taken as c²/gh₂ and $h_1^\prime/h_2$, that can then be easily solved by the Newton method.