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Chapter 16: Problem 9

On the basis of fluid power scaling as the cube of velocity, show the supporting math for the claim in the text that a water current at \(1 \mathrm{~m} / \mathrm{s}\) delivers the equivalent power (per rotor area) as a wind speed of about \(9 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
Answer: Approximately 9.34 m/s.

Step by step solution

01

Recall the formula for Fluid Power

The formula for fluid power, given the density of the fluid \(\rho\), the area of the rotor \(A\), and the velocity of the fluid \(v\), is given by: \(P = \frac{1}{2}\rho Av^3\)
02

Determine the equivalent power for water and wind

In order to determine the equivalent power for water and wind, we will use the fluid power formula for both fluids and then set them equal to each other, like this: \(\frac{1}{2}\rho_{water} A_{water} (1\,\mathrm{m/s})^3 = \frac{1}{2}\rho_{wind} A_{wind} (v_{wind}\,\mathrm{m/s})^3\) To find the equivalent wind speed, we must find the ratio between the two densities and rotor areas. Notice that the fraction \(\frac{1}{2}\) can be canceled.
03

Plug in the density values for water and air and the rotor areas

The density of water is approximately \(\rho_{water} \approx 1,000\,\mathrm{kg/m^3}\) and the density of air is approximately \(\rho_{wind} \approx 1.225\,\mathrm{kg/m^3}\). Also since we need to compare equivalent power per rotor area we can assume the rotor areas are the same for both fluids \(A_{water}= A_{wind} = A\). We can plug these values into our equation from Step 2: \(1,000 A (1\,\mathrm{m/s})^3 = 1.225 A (v_{wind}\,\mathrm{m/s})^3\)
04

Solve for v_wind

In order to solve for \(v_{wind}\), we need to rearrange the equation from Step 3: \(v_{wind}^3 = \frac{1,000 (1\,\mathrm{m/s})^3}{1.225}\) \(v_{wind}^3 \approx 816.33\) \(v_{wind} \approx 9.34\,\mathrm{m/s}\) The equivalent wind speed is approximately \(9.34\,\mathrm{m/s}\), which is close to the claim in the text that the equivalent power is delivered by a wind speed of about 9 m/s.

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