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Chapter 7: Problem 44

A 250 -m-long ship has a wetted area of \(8000 \mathrm{m}^{2} .\) A \(\frac{1}{100}\) -scale model is tested in a towing tank with the prototype fluid, and the results are: $$\begin{array}{|l|l|l|l|} \text { Model velocity }(\mathrm{m} / \mathrm{s}) & 0.57 & 1.02 & 1.40 \\ \hline \text { Model drag }(\mathrm{N}) & 0.50 & 1.02 & 1.65 \end{array}$$ Calculate the prototype drag at \(7.5 \mathrm{m} / \mathrm{s}\) and \(12.0 \mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
The prototype drag at 7.5 m/s is approximately 1.42 MN (Mega Newton), and at 12.0 m/s is approximately 3.62 MN.

Step by step solution

01

Determine the length ratio and velocity ratios for each case

The ship and its model are related by a scale of \(1/100\), so the length ratio \(L_{r} = L_{ship} / L_{model} = 100\). The velocity ratios are \(V_{r1} = V_{ship1} / V_{model1} = 7.5 / 0.57\) and \(V_{r2} = V_{ship2} / V_{model2} = 12.0 / 1.02\).
02

Determine the wetted area ratio

The wetted areas are proportional to the square of the length ratio, so \(A_{r} = (L_{r})^{2} = 100^{2}\).
03

Calculate the force ratios using the Froude scaling

The force ratios are \(F_{r1} = (V_{r1})^{2} \times A_{r}\) and \(F_{r2} = (V_{r2})^{2} \times A_{r}\).
04

Calculate the prototype drag forces

Use the force ratio to scale up the drag forces from the model to the ship. The drag forces for the ship are \(F_{ship1} = F_{r1} \times F_{model1}\) and \(F_{ship2} = F_{r2} \times F_{model2}\).

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Most popular questions from this chapter

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