Question

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

Answer: Approximately 9.34 m/s.

Step-by-step solution

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: \(P = \frac{1}{2}\rho Av^3\)

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.

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: \(1,000 A (1\,\mathrm{m/s})^3 = 1.225 A (v_{wind}\,\mathrm{m/s})^3\)

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.

Key concepts

Velocity Scaling

When discussing fluid power, velocity scaling stands out as a core component. Fluid power is closely related to the velocity of the fluid, whether it's water or air. The relationship between fluid power and velocity isn't linear; rather, it's linked to the cube of the velocity of the fluid. This means that even small changes in velocity can lead to significant differences in fluid power. In mathematical terms, if you double the velocity, the power increases by a factor of eight (\(2^3 = 8\)).
In practical terms, this cubic relationship is crucial when comparing different mediums like water and wind. A small stream of water can generate more power than a similar stream of air if the velocities are the same, due to water's higher density and the cubic velocity effect.
For any fluid power system, optimizing the velocity of the fluid can result in more efficient power generation. When designing systems that generate power from fluid movement, understanding and leveraging velocity scaling can lead to more effective designs.

• Fluid power depends on the cube of velocity.
• Small changes in velocity result in large power changes.
• This principle helps compare different fluids, like water and air.

Density of Fluids

Density plays an integral role in determining the power potential of a fluid. It defines how much mass can be found in a specific volume of a fluid. The greater the density, the more energy can be harnessed from a given volume moving at a specific velocity. This concept is vital when comparing the power potentials of water and air in generating energy.
In the step-by-step solution, we note that water has a density of approximately 1,000 kg/m³, significantly higher than air's density of around 1.225 kg/m³. This difference explains why a slower-moving stream of water can deliver the same power as a much faster-moving stream of air. The higher density of water compensates for its slower velocity when comparing energy output.
Understanding fluid density is essential for engineers and physicists when designing energy conversion systems. By choosing the right fluid and accounting for its density, systems can be optimized for maximum energy extraction.

• Density determines power potential of a fluid.
• Water is much denser than air.
• Higher density allows lower velocity to produce equivalent power.

Rotor Area Comparison

The rotor area is a crucial factor in the fluid power equation. It represents the surface area through which the fluid interacts with an energy-generating device, like a turbine or rotor. Intuitively, a larger rotor area can capture more fluid, thus potentially generating more power, provided that other factors like fluid velocity and density are constant.
In energy systems, maintaining a consistent rotor area across different mediums enables us to compare their power-generating capabilities directly. This was done in the demonstration where the rotor area for water and air was assumed to be equal. It allows for a fair comparison of how other factors, such as velocity and density, influence power generation.
When engineers design turbines or other renewable energy systems, an important consideration is how much fluid can be harnessed by the rotor area. A well-designed rotor can maximize energy extraction from the fluid passing through it. Understanding how to optimize rotor area can therefore mean the difference between a successful and inefficient energy generation system.

• Rotor area determines the amount of fluid captured.
• Equally compared rotor areas help evaluate power from different fluids.
• Designing for the right rotor area is key to optimizing energy capture.