 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2019/paper-2/section-a-iii/19/iii/solution

 Top articles  Latest articles New article in topic

Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 2 / Section A(iii) / 19 / iii / Solution by

Codex  0 2026-10-20

For the hydraulic ladder network, using the frontier star just defined, start at a single groove with

(x_{1}, y_{1}, z_{1}) = (0, 0, R_{g})

. Put

B = y_{n} + R_{c}

,

C = z_{n} + R_{c}

and

S = B + C + R_{g}

. The exact update is

x_{n + 1} = x_{n} + \frac{BC}{S}, y_{n + 1} = \frac{B R _{g}}{S}, z_{n + 1} = \frac{C R _{g}}{S} .

After all grooves have been included, the left frontier is a blind end, so no external flow passes through

y_{N}

. The inlet-to-outlet resistance is $R_{N} = x_{N} + z_{N}$ . Iterate 49 times to obtain

R_{50}

.

An equivalent closed form, for

R_{c}, R_{g} > 0

, is

R_{N} = \frac{( N - 1 ) R _{c}}{2} + \frac{R _{c}}{α + sinh κ tanh [( N - 1 ) κ /2 ]}, α = \frac{R _{c}}{R _{g}}, cosh κ = 1 + α .

Thus substitute

N = 50

; no single number follows without

R_{c} / R_{g}

. Checks are

R_{1} = R_{g}

,

R_{2} = (R_{g} + R_{c}) /2

, and

R_{N} \to R_{g} / N

R_{c} \to 0

. The formula counts only the

N - 1

inter-groove segments in each rail; extra inlet/outlet lead resistances, if present, must be added separately. All elements obey linear laminar-flow pressure-drop laws.

 Read the full article

 New to topics? Read the docs here!