25-Nav-A2 Hydrodynamics of Ships (I)_ Resistance and Propulsion · Undated paper
Question 1 of 10: Flat-Plate Towing Force and Boundary-Layer Thickness
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 1: Flat-Plate Towing Force and Boundary-Layer Thickness (10 marks)
Given. Plate length $L=5.0$ m, width $B=0.5$ m, deeply (doubly) submerged so both faces are wetted; tow speed $V=5.0$ m/s; freshwater at 15 °C ($\nu=1.139\times10^{-6}\ \text{m}^2/\text{s}$, $\rho=999\ \text{kg/m}^3$).
Quantity
Value
Length $L$
5.0 m
Width $B$
0.5 m
Speed $V$
5.0 m/s
Wetted area $S=2LB$
5.0 m²
Fluid
freshwater, 15 °C
Find. (a) towing (drag) force at $V=5.0$ m/s; (b) boundary-layer thickness $\delta$ at $x=2.0$ m; (c) qualitative effect of an ideal (inviscid) fluid on $\delta$.
Approach. Check the Reynolds number to decide whether the plate is effectively all-turbulent, then apply the ITTC 1957 model-ship correlation line (the standard turbulent flat-plate friction estimate on the data sheet) to the doubly-wetted area to get the towing force; use the companion turbulent boundary-layer growth formula for part (b); part (c) follows directly from the physical meaning of a boundary layer.
Check the flow regime. $$R_n=\frac{VL}{\nu}=\frac{(5.0)(5.0)}{1.139\times10^{-6}}=2.19\times10^{7}.$$ Laminar-to-turbulent transition ($R_{n,x}\approx5\times10^{5}$) occurs at $x_{crit}=\dfrac{5\times10^{5}\,\nu}{V}=0.114$ m — only 2.3% of the plate length — so the plate is treated as fully turbulent over essentially its whole length.
Select the friction line and justify it. Because the flow is fully turbulent at this Reynolds number, the Blasius laminar solution ($C_F=1.327\,R_n^{-0.5}$, valid only below $R_n\approx5\times10^5$) does not apply. The data sheet's ITTC 1957 model-ship correlation line, $$C_F=\frac{0.075}{(\log_{10}R_n-2)^2},$$ is the standard turbulent flat-plate friction estimate used throughout ship hydrodynamics for exactly this kind of towing-tank prediction, so it is the appropriate choice here (the Prandtl–Schlichting line $C_F=0.072\,R_n^{-0.2}$ gives an almost identical estimate, $C_F=0.00245$ vs. $0.00263$, confirming the result is not sensitive to which turbulent line is used).
Compute $C_F$ and the towing force. $$C_F=\frac{0.075}{(\log_{10}(2.19\times10^{7})-2)^2}=\frac{0.075}{(5.341)^2}=0.00263.$$ With both faces wetted, $S=2LB=2(5.0)(0.5)=5.0\ \text{m}^2$: $$R=C_F\cdot\tfrac12\rho SV^2=(0.00263)(0.5)(999)(5.0)(5.0)^2=\boxed{164\ \text{N}}.$$
Boundary-layer thickness at $x=2.0$ m (turbulent formula from the data sheet). $$R_{n,x}=\frac{Vx}{\nu}=\frac{(5.0)(2.0)}{1.139\times10^{-6}}=8.78\times10^{6},\qquad \delta=0.371\,x\,(Vx/\nu)^{-1/5}.$$ $$\delta=0.371(2.0)(8.78\times10^{6})^{-1/5}=\boxed{0.0303\ \text{m}\approx30\ \text{mm}}.$$
Ideal fluid. A boundary layer exists only because viscosity enforces the no-slip condition at the wall. In an ideal (inviscid) fluid there is no shear stress and no no-slip requirement — potential flow slips freely along the surface — so no boundary layer forms at all: $\delta\to0$ everywhere on the plate, and (consistent with d'Alembert's paradox) the plate would experience zero drag.
Quantity
Result
(a) Towing force at 5.0 m/s (ITTC-57 turbulent line)
