Question 2 of 6: Question 2 (Part A, Question A2): Skin-Friction Drag and Power for a Super-Tanker

Question 2 (Part A, Question A2): Skin-Friction Drag and Power for a Super-Tanker (20 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.

Find. The total skin-friction force on the wetted hull and the propulsive power needed to overcome it.

Approach. Treat the wetted bottom and the two sides as flat plates of length L in a stream of speed U, integrate the aid-sheet local friction coefficient along the plate in two pieces — laminar up to the transition point, then the 1/7-law turbulent form — to obtain a mean coefficient, and convert that to force and power.

1. Wetted area and free-stream speed. The flat bottom contributes $B L$ and each of the two vertical sides contributes $T L$, so $$A_w = BL + 2TL = (70)(360) + 2(25)(360) = 25{,}200 + 18{,}000 = 43,200\ \text{m}^{2}$$ and the ship speed converts to $U = 24\ \text{km/h} \div 3.6 = 6.6667$ m/s.
2. Reynolds numbers and the transition point. Over the full hull length $$Re_L = \frac{U L}{\nu} = \frac{(6.6667)(360)}{1.37\times 10^{-6}} = 1.752 \times 10^{9}$$ and transition occurs where $Re_x = 3\times10^{5}$, that is at $$x_{tr} = \frac{Re_{x,tr}\,\nu}{U} = \frac{(3\times10^{5})(1.37\times10^{-6})}{6.6667} = 0.06165\ \text{m}$$ Transition is complete within the first 62 mm of a 360 m hull, so the layer is turbulent over essentially the whole ship; the laminar piece is retained anyway because the question asks for a naturally developing layer.
3. Mean friction coefficient, laminar portion. The mean coefficient is the length average of the local one, $C_F = (1/L)\int_0^L C_{fx}\,\mathrm{d}x$. Over the laminar run the aid sheet gives $C_{fx} = 0.67\,Re_x^{-1/2}$, and since $Re_x^{-1/2} = (\nu / U x)^{1/2}$ the integral is elementary: $$\int_0^{x_{tr}} 0.67\left(\frac{\nu}{Ux}\right)^{1/2}\mathrm{d}x = 1.34\,\frac{x_{tr}}{\sqrt{Re_{x,tr}}} = 1.34\,\frac{0.06165}{547.7} = 1.5083 \times 10^{-4}\ \text{m}$$
4. Mean friction coefficient, turbulent portion. With $C_{fx} = 0.0266\,Re_x^{-1/7}$ the same integration from the transition point to the stern gives $$\int_{x_{tr}}^{L} 0.0266\left(\frac{\nu}{Ux}\right)^{1/7}\mathrm{d}x = \frac{7}{6}(0.0266)\left[L\,Re_L^{-1/7} - x_{tr}\,Re_{x,tr}^{-1/7}\right]$$ Substituting $Re_L^{-1/7} = 0.04781$ and $Re_{x,tr}^{-1/7} = 0.16503$ yields $0.5338$ m. Dividing the sum of the two pieces by the hull length, $$C_F = \frac{0.00015 + 0.5338}{360} = \boxed{1.4832 \times 10^{-3}}$$ The laminar strip supplies only about 0.03 per cent of the total, so an all-turbulent estimate $C_F = \tfrac{7}{6}(0.0266)Re_L^{-1/7} = 0.00148$ differs by less than a tenth of a per cent — a reassuring cross-check.
5. Skin-friction force. The dynamic pressure is $q = \tfrac12\rho U^{2} = \tfrac12(1020)(6.6667)^{2} = 22,667$ Pa, and the drag follows directly from the definition of the mean coefficient: $$F_D = C_F\,\tfrac12 \rho U^{2} A_w = (0.001483)(22,667)(43,200) = \boxed{1,452\ \text{kN}}$$ that is about 1.45 MN, equivalent to the weight of roughly 148 tonnes.
6. Propulsive power. The power delivered against a steady drag force is the product of force and speed: $$P = F_D\,U = (1,452,336\ \text{N})(6.6667\ \text{m/s}) = \boxed{9.68\ \text{MW}}$$ This is the effective power needed for skin friction alone; a real tanker also pays for wave-making and form drag and for propulsive inefficiency, so the installed shaft power would be appreciably larger.

Final Results.