Question 4 of 9: ITTC 1978 Extrapolation to Full Scale

Question 4: ITTC 1978 Extrapolation to Full Scale (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. $L_M=6$ m, $L_S=160$ m, $V_S=20.0$ kn; measured model resistance $R_{TM}=77.00$ N at the Froude-corresponding model speed; form factor $k=0.230$; correlation allowance $C_A=0.0004$; air resistance $C_{AA}=0$. Tow-tank water: fresh water at 15 °C ($\nu=1.139\times10^{-6}\ \text{m}^2/\text{s}$, $\rho=999\ \text{kg/m}^3$). Full-scale prediction water: salt water at 15 °C ($\nu=1.188\times10^{-6}\ \text{m}^2/\text{s}$, $\rho=1025\ \text{kg/m}^3$).

Find. The full-scale ship total resistance $R_{TS}$ and effective power $P_E$ at 20.0 kn.

Approach. Compute the model's wetted surface and Froude-corresponding speed from the geometric scale ratio, form the model's total and frictional coefficients to isolate the residuary coefficient $C_R$ (assumed to scale unchanged with Froude number), then rebuild the ship's total coefficient from the ship-scale friction line plus $C_R$, $C_A$ and $C_{AA}$, per $C_{TS}=(1+k)C_{FS}+C_{TM}-(1+k)C_{FM}+C_A+C_{AA}$.

1. Scale ratio, model speed and model wetted surface. $$\lambda=\frac{L_S}{L_M}=\frac{160}{6}=26.67,\qquad V_M=\frac{V_S}{\sqrt\lambda}=\frac{10.288}{\sqrt{26.67}}=1.992\ \text{m/s},$$ $$S_M=\frac{S_S}{\lambda^2}=\frac{6650}{26.67^2}=9.352\ \text{m}^2.$$
2. Model Reynolds number and ITTC-57 frictional coefficient. $$R_{n,M}=\frac{V_ML_M}{\nu_{fw}}=\frac{(1.992)(6)}{1.139\times10^{-6}}=1.049\times10^{7},\qquad C_{FM}=\frac{0.075}{(\log_{10}R_{n,M}-2)^2}=0.002975.$$
3. Model total coefficient and residuary coefficient. $$C_{TM}=\frac{R_{TM}}{\tfrac12\rho_{fw}S_MV_M^2}=\frac{77.00}{\tfrac12(999)(9.352)(1.992)^2}=0.004153,$$ $$C_R=C_{TM}-(1+k)C_{FM}=0.004153-(1.230)(0.002975)=\boxed{0.000494}.$$
4. Ship Reynolds number and frictional coefficient (salt water, 15 °C). $$R_{n,S}=\frac{V_SL_S}{\nu_{sw}}=\frac{(10.288)(160)}{1.188\times10^{-6}}=1.386\times10^{9},\qquad C_{FS}=\frac{0.075}{(\log_{10}R_{n,S}-2)^2}=0.001471.$$
5. Ship total coefficient (ITTC 1978 form-factor extrapolation). $$C_{TS}=(1+k)C_{FS}+C_R+C_A=(1.230)(0.001471)+0.000494+0.0004=\boxed{0.002703}.$$
6. Ship resistance and effective power. $$R_{TS}=C_{TS}\cdot\tfrac12\rho_{sw}S_SV_S^2=(0.002703)\left[\tfrac12(1025)(6650)(10.288)^2\right]=\boxed{975\ \text{kN}},$$ $$P_E=R_{TS}\,V_S=(975{,}000)(10.288)=\boxed{10{.}0\ \text{MW}}.$$