A ship has a length of 180 m and travels in the sea where ρs=1030 kg / m^3. A model of the shi

step 1 So in this problem, I'm going to write my information down. We're asked to find the drag in a sh

A ship has a length of 180 m and travels in the sea where...

A ship has a length of $180 \mathrm{~m}$ and travels in the sea where $\rho_{s}=1030 \mathrm{~kg} / \mathrm{m}^{3}$. A model of the ship is built to a $1 / 60$ scale, and it displaces $0.06 \mathrm{~m}^{3}$ of water such that its hull has a wetted surface area of $3.6 \mathrm{~m}^{2}$. When tested in a towing tank at a speed of $0.5 \mathrm{~m} / \mathrm{s}$, the total drag on the model was $2.25 \mathrm{~N}$. Determine the drag on the ship and its corresponding speed. What power is needed to overcome this drag? The drag due to viscous (frictional) forces can be determined using $\left(F_{D}\right)_{f}=\left(\frac{1}{2} \rho V^{2} A\right) C_{D},$ where $C_{D}$ is the drag coefficient determined from $C_{D}=1.328 / \sqrt{\operatorname{Re}}$ for $\operatorname{Re}<10^{6}$ and $C_{D}=0.455 /\left(\log _{10} \mathrm{Re}\right)^{2.58}$ for $10^{6}<\mathrm{Re}<10^{9}$. Take $\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}$ and $\nu=1.00\left(10^{-6}\right) \mathrm{m}^{2} / \mathrm{s}$.

Key Concepts

Drag Coefficient and Its Empirical Formulas

The drag coefficient is a dimensionless number that encapsulates the influence of shape, surface roughness, and flow regime on the drag force experienced by an object. Empirical formulas for the drag coefficient, which vary with Reynolds number, provide a means to estimate viscous (frictional) drag in different flow regimes. These formulas adjust the drag coefficient based on whether the flow is within a low or high Reynolds number range.

Drag Force

Drag force is the resistance a body experiences as it moves through a fluid. It arises due to both form and viscous (frictional) effects. In the context of fluid dynamics, drag is typically modeled as proportional to the fluid’s density, the square of the velocity, and a characteristic area, with a drag coefficient accounting for shape and flow conditions.

Power to Overcome Drag

This concept addresses the calculation of the mechanical power required to maintain a body’s motion against drag forces. It is obtained by multiplying the drag force by the velocity of the object. This calculation is essential for determining the energy necessary for propulsion in marine and other fluid-dynamic applications.

Model Scaling and Similarity

This concept involves creating a smaller, geometrically similar model of a full?scale ship so that the behavior in a controlled setting can predict the real system’s performance. It includes understanding how key parameters like length, area, and volume scale, and ensuring that both geometric similarity and dynamic similarity (through appropriate non-dimensional numbers) are maintained between the model and the full size ship.

Reynolds Number

Reynolds number is a dimensionless parameter that characterizes the relative importance of inertial forces to viscous forces in a fluid flow. It is crucial in determining the flow regime—whether laminar or turbulent—and affects the scaling of drag forces. In similarity studies, matching or properly accounting for Reynolds number differences between model and full scale is essential for accurate predictions of viscous phenomena.

Transcript

00:01 So in this problem, i'm going to write my information down.

00:03 We're asked to find the drag in a shift, the corresponding speed, and the power needed to overcome the drag.

00:08 We are given the drag force, and we're told right here, and here's how to calculate my two cds.

00:16 We're given these two values.

00:17 So i'm going to write down the rest of my values here.

00:20 My lp is given as 180 meters.

00:25 Vm was 0 .5 meters per second, and my fd is 2 .25.

00:32 My little vm is 1 times 10 to the minus 6 meters squared per second.

00:45 My lm is 1 over 60 lp.

00:52 My density m is 1 ,000 kilograms per meter cubed.

01:00 And my density of my ship is 130 kilograms per meter cubed.

01:11 My a is 3 .6 meters squared.

01:20 And then my other values are this, this, and this.

01:30 Okay.

01:37 So we know that f -r -m equals frp.

01:58 Okay.

01:59 And my gs will cross -out, solving for vp.

02:13 So this will be equal to, this will equal three, 0 .872 meters per second.

02:33 Then my drag force will be, and that will be equal to, so this will be equal to 0 .5 times 3, divided by 1 times 10 to the minus 6, equals 1 .5 times 10 to the 6th.

03:18 And this is because my lm will be 1 over 60 lp, be 160.

03:26 Whoops, one divided by 160 times 180 will equal three.

03:39 So let's get back over here.

03:47 And then when we evaluate our reynolds number, and this will equal, that'll equal...

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