 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/6c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let the upper plate in flow A move at speed

U

, with the lower plate fixed. The Couette flow profile is

u_{A} (y) = \frac{U}{h} y,

so its mid-plane speed is

U /2

and its total viscous dissipation per unit plate area is

\dot{Q}_{A} = \int_{0}^{h} μ (\frac{U}{h})^{2} d y = \frac{μ U ^{2}}{h} .

Write

G = - d p / d x > 0

for the constant pressure-gradient magnitude in flow B. The plane Poiseuille flow profile is

u_{B} (y) = \frac{G}{2 μ} y (h - y) .

Equality of the mid-plane speeds gives

\frac{G h ^{2}}{8 μ} = \frac{U}{2}, G = \frac{4 μU}{h ^{2}} .

Since

u_{B}^{'} (y) = G (h - 2 y) / (2 μ)

\dot{Q}_{B} = \frac{G ^{2}}{4 μ} \int_{0}^{h} (h - 2 y)^{2} d y = \frac{G ^{2} h ^{3}}{12 μ} = \frac{4 μ U ^{2}}{3 h} .

Hence

\frac{Q ˙ _{A}}{Q ˙ _{B}} = \frac{3}{4} .

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