Question 6 of 6: Question 6 (Part B, Question B3): Two Immiscible Liquids in a Pressure-Driven Two-Dimensional Channel

Question 6 (Part B, Question B3): Two Immiscible Liquids in a Pressure-Driven Two-Dimensional Channel (30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The boundary conditions, the two-layer velocity profile, the shear stresses at both walls and the interface, the interface speed, and the location and value of the maximum velocity.

Approach. With the flow steady, fully developed and two-dimensional, the Navier–Stokes equations reduce in each layer to a balance between the constant pressure gradient and the viscous term; integrate twice per layer and fix the four constants with no-slip at the two walls plus continuity of velocity and of shear stress at the interface.

1. Part (a) — reduce the governing equations and state the boundary conditions. The channel is long, horizontal and two-dimensional, and the flow is steady and fully developed, so $v = w = 0$, $u = u(y)$ only, and the $x$-momentum equation in each layer collapses to $$\mu_i \frac{\mathrm{d}^{2}u_i}{\mathrm{d}y^{2}} = \frac{\mathrm{d}P}{\mathrm{d}x} = \text{constant}$$ Four conditions are needed. Because the liquids wet the non-porous walls, no-slip applies at both: $u_1(h) = 0$ at the top wall and $u_2(-h) = 0$ at the bottom. At the interface the two fluids neither separate nor slip past one another, so the velocity is continuous, $u_1(0) = u_2(0) = u_I$; and a massless interface can support no net force, so the shear stress is continuous, $$\mu_1\left.\frac{\mathrm{d}u_1}{\mathrm{d}y}\right|_{0} = \mu_2\left.\frac{\mathrm{d}u_2}{\mathrm{d}y}\right|_{0}$$ Note that it is the stress, not the velocity gradient, that matches: the gradients differ by the viscosity ratio of ten, which is what makes the profile kinked.
2. Part (b) — integrate twice in each layer. Writing $G = -\mathrm{d}P/\mathrm{d}x = 1$ Pa/m for the favourable gradient, integration gives $$u_i(y) = -\frac{G}{2\mu_i}y^{2} + A_i y + u_I, \qquad i = 1 \text{ (top)},\ 2 \text{ (bottom)}$$ where the common constant term already imposes velocity continuity at $y = 0$. Stress continuity then requires $\mu_1 A_1 = \mu_2 A_2$, so $A_2 = (\mu_1/\mu_2)A_1$.
3. Part (b) — apply the wall conditions and solve. Substituting $u_1(h) = 0$ and $u_2(-h) = 0$ and subtracting eliminates $u_I$, leaving one equation for $A_1$: $$A_1 = \frac{G h}{2}\,\frac{\dfrac{1}{\mu_1}-\dfrac{1}{\mu_2}}{1+\dfrac{\mu_1}{\mu_2}} = \frac{(1)(0.01)}{2}\,\frac{1000-100}{1.1} = 4.0909\ \text{s}^{-1}$$ whence $A_2 = 0.1 A_1 = 0.40909$ per second. Back-substitution into either wall condition gives the interface speed, and the profile is complete: $$\boxed{\;u_1(y) = -500\,y^{2} + 4.0909\,y + 0.009091 \quad (0 \le y \le h)\;}$$ $$\boxed{\;u_2(y) = -50\,y^{2} + 0.40909\,y + 0.009091 \quad (-h \le y \le 0)\;}$$ both in metres per second with $y$ in metres. Each parabola vanishes at its own wall, as required: $u_1(0.01) = 0$ and $u_2(-0.01) = 0$.
4. Part (c) — shear stress at the interface. The Newtonian stress is $\tau = \mu\,\mathrm{d}u/\mathrm{d}y$, and at $y = 0$ the gradient in the top layer is $A_1$: $$\tau_I = \mu_1 A_1 = (0.001)(4.0909) = \boxed{4.091\ \text{mPa}}$$ The bottom layer returns the identical value, $\mu_2 A_2 = (0.01)(0.40909)$, which is the stress-continuity condition satisfying itself and a useful check on the algebra.
5. Part (c) — shear stress at the two walls. Differentiating each profile at its wall, $$\tau_{top} = \mu_1\left[-\frac{G}{\mu_1}h + A_1\right] = (0.001)(-10 + 4.091) = -5.909\ \text{mPa}$$ $$\tau_{bot} = \mu_2\left[\frac{G}{\mu_2}h + A_2\right] = (0.01)(1 + 0.4091) = 14.091\ \text{mPa}$$ The signs simply record that the two walls drag on the fluid from opposite sides; both retard the flow. Their magnitudes are the physically useful numbers, $5.91$ and $14.09$ mPa. A global check closes the question: the pressure force per unit wall area on a slice of channel is $G(2h) = 0.02$ Pa, and the two wall stresses sum to $0.020$ Pa, exactly balancing it. The more viscous bottom fluid carries the larger wall stress even though it moves more slowly.
6. Part (d) — speed at the interface. The constant term found in step 3 is precisely the interface speed: $$u_I = \frac{G h^{2}}{2\mu_1} - A_1 h = 0.05 - 0.04091 = \boxed{9.091\ \text{mm/s}}$$ Computing it instead from the lower layer, $u_I = Gh^{2}/2\mu_2 + A_2 h = 0.005 + 0.004091$, returns the same value, confirming the solution.
7. Part (e) — where the maximum sits. The maximum is where the shear stress vanishes, so it lies in whichever layer contains the root of $\mathrm{d}u/\mathrm{d}y = 0$. In the top layer, $$\frac{\mathrm{d}u_1}{\mathrm{d}y} = -\frac{G}{\mu_1}y + A_1 = 0 \;\Longrightarrow\; y_{max} = \frac{\mu_1 A_1}{G} = \boxed{4.091\ \text{mm above the interface}}$$ which does fall inside $0 \le y \le h$, whereas the bottom layer's root lies outside its own domain, so $u_2$ increases monotonically from the lower wall to the interface. The peak value is $$u_{max} = u_1(y_{max}) = 17.459\ \text{mm/s}$$ almost twice the interface speed. As the figure shows, the profile is a pair of parabolas joined with a kink at the interface, strongly biased toward the thin, low-viscosity upper fluid; the maximum sits above the centreline for exactly that reason. Integrating each layer gives volume flows of 1.288e-04 and 5.379e-05 cubic metres per second per metre of width, a total of 1.826e-04, so roughly seventy per cent of the throughput is carried by the upper layer.

Final Results.