Question 10.1: A super tanker is 360 m long and has a beam width of 70 m an......
A super tanker is 360 m long and has a beam width of 70 m and a draft
of 25 m. Estimate the force and power required to overcome skin-friction drag at a cruising speed of 13 knots in seawater at 10°C.
[1 knot = 1852 m per hour; at 10°C viscosity v = 1.4 × 10^{-6} m²/s]
Step-by-Step
L=360~{\mathrm{m}},~U={\frac{1852\times13}{3600}}=6.69~{\mathrm{m}}/s,~~v=1.4\times10^{-6}~{\mathrm{m}}^{2}/s \\ R_{L}=\frac{6.69\times360}{1.4\times10^{-6}}=1.72\times10^{9} \\ C_{D f}=\frac{0.455}{(1\mathrm{og}\,R_{L})^{2.58}}-\frac{1610}{R_{L}}\qquad{(\text{Valid for }5\times10^{5}\lt R_{L}\lt 10^{9})} \\ = 0.00147 – 0.0000016 = 0.00147 \\ {\frac{1}{2}}\rho U^{2}={\frac{1}{2}}\times1020\times6.69^{2}=22825.6\ N/{\mathrm{m}}^{2}
Total area to be considered = bottom + sides
= (360 × 70) + (360 × 25) × 2
= 25,200 + 18,000 = 43,200 m²
Therefore, F=C_{D f}\times{\frac{1}{2}}\times\rho\times U^{2}\times A=0.00147\times43,200\times22,825.6
=1.45 × 10^6 N = 1.45 MN
Power = F · U = 1.45 × 10^6 × 6.69 = 9.7 × 10^6 Nm/s = 9.7 MW
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