25-Nav-A2 Hydrodynamics of Ships (I)_ Resistance and Propulsion · Undated paper
Question 9 of 10: Planning a Model Open-Water Propeller Test
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2019 — 16-Nav-A2 Hydrodynamics of Ships I: Resistance and Propulsion. Three-hour, closed-book exam; a data sheet, a Wageningen B4-55 propeller chart and a Burrill cavitation chart are supplied. Format: Questions 1–8 are compulsory (attempt all eight), then one of Questions 9 or 10. All ten are solved below for completeness.
Reference texts: Larsson & Raven, Ship Resistance and Flow (SNAME) — model-scale resistance testing, Froude/Reynolds scaling, boundary-layer estimates and the ITTC 1978 performance-prediction method; Lewis (ed.), Principles of Naval Architecture, Vol. II — Resistance, Propulsion and Vibration (SNAME) — propeller geometry, wave-pattern interference, open-water B-series design and wake-induced blade loading; Carlton, Marine Propellers and Propulsion (Butterworth-Heinemann) — Wageningen B-series charts, the Burrill back-cavitation criterion and open-water model testing.
Question 9: Planning a Model Open-Water Propeller Test (20 marks)
Given. Model diameter $D_M=0.240$ m; full-scale diameter $D_S=4.2$ m; full-scale chord at $0.7R$, $c_{0.7R,S}=1.1$ m; ship shaft speed $n_S=165$ rpm; ship design-speed advance velocity $V_{A,S}=7.0$ m/s; as-fitted propeller's bollard coefficients $K_{T,0}\approx0.30$, $K_{Q,0}=0.04$; towing tank max carriage speed 5 m/s; test boss max shaft speed 20 rps; dynamometer limits: thrust $\le300$ N, torque $\le15$ N·m. Fresh water, 15 °C ($\nu=1.139\times10^{-6}\ \text{m}^2/\text{s}$, $\rho=999\ \text{kg/m}^3$).
Quantity
Value
Scale ratio $\lambda=D_S/D_M$
17.5
Ship shaft speed $n_S$
2.750 rps
Ship $V_{A,S}$
7.0 m/s
Bollard $K_{T,0}$, $K_{Q,0}$
0.30, 0.04
Dynamometer limits
300 N, 15 N·m
Find. (i) the similitude laws that must be obeyed; (ii) the model shaft speed $n_M$ and carriage speed $V_M$ to use.
Approach. An open-water propeller test (fully submerged, no free surface at the disk) is governed by geometric similarity and equal advance coefficient $J$, together with a Reynolds number floor at the blade sections rather than exact Reynolds equality (which Question 2's argument shows is unachievable at model scale). The dynamometer and shaft-speed limits are checked at bollard pull ($J=0$), where $K_T$ and $K_Q$ — and hence the loads — are highest, to find the fastest safe shaft speed; the corresponding carriage speed then follows from matching $J$ to the ship's design condition.
(i) Laws of similitude for an open-water propeller test.Geometric similarity — the model is an exact scaled replica ($\lambda=D_S/D_M=17.5$). Kinematic similarity via equal advance coefficient, $J_M=J_S=V_A/(nD)$ — this fixes the same ratio of axial to rotational velocity, and hence the same blade angle of attack, at model and full scale; it is what actually needs to be matched for $K_T$, $K_Q$, $\eta_o$ to transfer between scales. Froude number is not a controlling requirement here — the propeller disk is deeply submerged with no free surface, so there is no wave-making similarity to satisfy (unlike a resistance test). Reynolds number cannot be matched exactly at this scale (as in Question 2, equal $R_n$ would need a model fluid with viscosity $\approx\!\lambda^{1.5}\approx\!73\times$ lower than water); instead the model test is run at a shaft speed high enough to keep the local blade Reynolds number above a minimum threshold ($\approx\!2\text{–}3\times10^5$ at the 0.7R section), avoiding laminar-separation scale effects on the measured $K_T$, $K_Q$.
(ii) Reference values. Model chord at $0.7R$: $c_M=c_{0.7R,S}/\lambda=1.1/17.5=0.0629$ m. Ship advance coefficient at the design condition (this is the $J$ the model test must reproduce): $$J=\frac{V_{A,S}}{n_SD_S}=\frac{7.0}{(2.750)(4.2)}=\boxed{0.606}.$$
Maximum shaft speed from the dynamometer's thrust limit (worst case at bollard, $J=0$). $$T_{bollard}=K_{T,0}\,\rho\,n^2D_M^4\le300\ \text{N}\ \Rightarrow\ n\le\sqrt{\frac{300}{(0.30)(999)(0.240)^4}}=17.4\ \text{rps}.$$
Check against the torque limit and test-boss limit. $$Q_{bollard}=K_{Q,0}\,\rho\,n^2D_M^5\le15\ \text{N}\cdot\text{m}\ \Rightarrow\ n\le\sqrt{\frac{15}{(0.04)(999)(0.240)^5}}=21.7\ \text{rps},$$ which is less restrictive than the thrust limit, as is the test boss's 20 rps ceiling. The thrust dynamometer governs: $n_M\approx17.4$ rps.
Carriage speed from matching $J$ at the design condition. $$V_M=J\,n_M\,D_M=(0.606)(17.4)(0.240)=\boxed{2.53\ \text{m/s}},$$ well within the tank's 5 m/s carriage-speed limit.
Confirm the Reynolds-number floor is met. $$V_{R,M}=\sqrt{V_M^2+(0.7\pi n_MD_M)^2}=\sqrt{(2.53)^2+(9.17)^2}=9.51\ \text{m/s}.$$ $$R_{n,M}=\frac{c_MV_{R,M}}{\nu}=\frac{(0.0629)(9.51)}{1.139\times10^{-6}}=5.25\times10^{5}\ \gg\ 2\times10^{5}.$$ The chosen speed comfortably clears the minimum-Reynolds-number floor, so no further reduction in $n_M$ is needed.
Quantity
Result
Target advance coefficient $J$
0.606
Model shaft speed $n_M$ (governed by dynamometer thrust limit)