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25-Nav-A2 Hydrodynamics of Ships (I)_ Resistance and Propulsion · December 2019

Question 3 of 9: Minimum Model Size for a Resistance Test

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2019 — 16-Nav-A2 Hydrodynamics of Ships I: Resistance and Propulsion. Three-hour, closed-book exam; a data sheet, a propeller (Wageningen B4-55) chart and a Burrill cavitation chart are supplied. Format: Questions 1–7 are compulsory (attempt all seven), then one of Questions 8 or 9. All nine are solved below for completeness. Units follow the paper (mixed SI, with the historic Imperial-unit legend that duplicates on the supplied Burrill sheet noted where relevant).

Reference texts: Larsson & Raven, Ship Resistance and Flow (SNAME) — model-scale resistance testing, Froude/Reynolds scaling and the ITTC 1978 performance-prediction method; Lewis (ed.), Principles of Naval Architecture, Vol. II — Resistance, Propulsion and Vibration (SNAME) — propeller geometry, open-water B-series design and cavitation; Carlton, Marine Propellers and Propulsion (Butterworth-Heinemann) — Wageningen B-series charts and the Burrill back-cavitation criterion.

Question 3: Minimum Model Size for a Resistance Test (10 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. Ship length $L_S=160$ m, design speed $V_S=20.0$ kn, wetted surface $S_S=6650\ \text{m}^2$. At the lowest model test speed, $F_n=0.10$ and $R_n\ge1\times10^6$ is required. Fresh water at 15 °C: $\nu=1.139\times10^{-6}\ \text{m}^2/\text{s}$.

QuantityValue
Ship length $L_S$160 m
Design speed $V_S$20.0 kn = 10.288 m/s
Wetted surface $S_S$6650 m²
Lowest test $F_n$ / min. $R_n$ there0.10 / $1\times10^6$
Kinematic viscosity $\nu$ (fresh water, 15°C)$1.139\times10^{-6}\ \text{m}^2/\text{s}$

Find. (i) the minimum model length $L_M$; (ii) the model speed and Reynolds number that Froude-correspond to 20.0 kn full scale, for that model.

Approach. At the lowest test point the model speed is fixed by $F_n=0.10$ on the (unknown) model length; requiring $R_n=V_ML_M/\nu\ge1\times10^6$ there gives one equation in $L_M$ alone. Having fixed $L_M$, the scale ratio $\lambda=L_S/L_M$ then gives the Froude-corresponding model speed and Reynolds number at the ship's 20.0 kn design point.

  1. Express $R_n$ at the lowest test point in terms of $L_M$ alone. At $F_n=0.10$, the model speed is $V_M=F_n\sqrt{gL_M}=0.10\sqrt{gL_M}$. Substituting into $R_n=V_ML_M/\nu$: $$R_n=\frac{0.10\sqrt{gL_M}\,L_M}{\nu}=\frac{0.10\sqrt{g}}{\nu}\,L_M^{3/2}.$$
  2. Solve for the minimum $L_M$. Setting $R_n=1\times10^6$ and rearranging, $$L_M^{3/2}=\frac{R_n\,\nu}{0.10\sqrt{g}}=\frac{(1\times10^6)(1.139\times10^{-6})}{0.10\sqrt{9.806}}=3.638,$$ $$L_M=(3.638)^{2/3}=\boxed{2.37\ \text{m (minimum model length)}}.$$ That is part (i).
  3. Scale ratio. $\lambda=L_S/L_M=160/2.365=67.65.$
  4. Model speed Froude-corresponding to 20.0 kn full scale. Because $F_n$ is matched, $V_M=V_S/\sqrt{\lambda}$: $$V_M=\frac{10.288}{\sqrt{67.65}}=\boxed{1.25\ \text{m/s}\ (2.43\ \text{kn})}.$$
  5. Reynolds number at that speed. $$R_n=\frac{V_ML_M}{\nu}=\frac{(1.251)(2.365)}{1.139\times10^{-6}}=\boxed{2.60\times10^{6}}.$$ This exceeds the $1\times10^6$ minimum, as it must — the design-speed point is not the constraining (lowest-speed) condition. That is part (ii).
QuantityResult
(i) Minimum model length $L_M$2.37 m
Scale ratio $\lambda$67.6
(ii) Model speed at 20.0 kn full-scale correspondence1.25 m/s (2.43 kn)
(ii) Reynolds number there$2.60\times10^{6}$