/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 20.56. The maximum power that ca... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 56

20.56. The maximum power that can be extracted by a wind turbine from an air stream is approximately $$ P=k d^{2} v^{3} $$ where \(d\) is the blade diameter \(v\) is the wind speed, and the constant \(k=0.5 \mathrm{W} \cdot \mathrm{s}^{3} / \mathrm{m}^{5} .\) (a) Explain the dependence of \(P\) on \(d\) and on \(v\) by considering a cylinder of air that passes over the turbine blades in time \(t(\text { Fig. } 20.31)\) . This cylinder has diameter \(d .\) length \(L=v t\) and density \(\rho .\) (b) The Mod-SB wind turbine at Kahaku on the Hawaiian island of Oahu has a blade diameter of 97 \(\mathrm{m}\) (slightly longer than a football field sits atop a \(58-\mathrm{m}\) tower. It can produce 3.2 \(\mathrm{MW}\) of electric power. Assuming 25\(\%\) efficiency, what wind speed is required to produce this amount of power? Give your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\) . (c) Commercial wind turbines are commonly located in or downwind of mountain passes. Why?

Short Answer

Expert verified
Dependence: P varies with \(d^2\) and \(v^3\). Required wind speed: around 13.7 m/s (or 49.3 km/h). Mountain passes accelerate winds, enhancing energy capture.

Step by step solution

01

Understanding the Dependence of P on d and v

The power extracted by the wind turbine is given by the expression \( P = k d^2 v^3 \). Here, \( d \) represents the blade diameter and \( v \) represents the wind speed. The cylinder of air passing over the turbine has diameter \( d \), length \( L = vt \), and density \( \rho \). The mass hitting the turbine in time \( t \) is the cylinder's volume times its density: \( m = \rho \times (\pi (\frac{d}{2})^2 \times vt) \). The kinetic energy of this air mass is \( \frac{1}{2}mv^2 \). The power, being energy per unit time, is therefore \( P = \frac{1}{2} \rho \pi (\frac{d}{2})^2 v^3 \), which explains the \( v^3 \) and \( d^2 \) dependence in the given formula.
02

Calculate Required Wind Speed (v)

Given the power produced \( P = 3.2 \text{ MW} = 3.2 \times 10^6 \text{ W} \) and efficiency \( \eta = 0.25 \), the actual power from the wind is \( \frac{P}{\eta} = \frac{3.2 \times 10^6}{0.25} = 1.28 \times 10^7 \text{ W} \). Using the formula \( P = k d^2 v^3 \), solve for \( v \):\[1.28 \times 10^7 = 0.5 \times 97^2 \times v^3\]\[ v^3 = \frac{1.28 \times 10^7}{0.5 \times 97^2} \]Calculate \( v^3 \), then take the cube root to find \( v \).
03

Convert v from m/s to km/h

After calculating \( v \) in m/s from Step 2, convert it to km/h using the conversion factor. Since 1 m/s equals 3.6 km/h, the conversion is \( v_{\text{km/h}} = v_{\text{m/s}} \times 3.6 \). Compute the wind speed in both units for the final answer.
04

Explain Use of Wind Turbines in Mountain Passes

Mountain passes are ideal for wind turbines because they naturally channel and accelerate wind flows due to the funnel-like geography. The wind speed increases as it is funneled through the narrower pass, providing more kinetic energy that can be harnessed by turbines, making these locations more effective for power generation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Blade Diameter
The blade diameter of a wind turbine plays a crucial role in determining the amount of energy that can be extracted from the wind. The diameter essentially defines the size of the "wind-capturing" area. To visualize, imagine a large circle in the air where the turbine blades sweep through.

In the formula for wind power, given as \( P = k d^2 v^3 \), the \( d^2 \) term indicates that the power is proportional to the square of the blade diameter. This means that even a small increase in the diameter can lead to a significant increase in power capture. Hence, larger blades are often sought after to maximize energy output.

  • Blade diameter defines the volume of air the turbine interacts with.
  • Larger diameters capture more wind, leading to greater potential energy.
Increasing the blade diameter can substantially boost the turbine's total energy production, making it a key factor in optimizing wind power generation.
Wind Speed
Wind speed is another critical factor in determining how much power a wind turbine can generate. In the power formula \( P = k d^2 v^3 \), wind speed \( v \) appears raised to the power of three. This cubic relationship indicates that power increases dramatically with wind speed.

To put it simply, if the wind speed doubles, the amount of power generated by the wind turbine increases by a factor of eight. This shows the importance of selecting locations with consistently high wind speeds for installing wind turbines.

  • Small changes in wind speed have large impacts on power generation.
  • Locations with stable and strong winds are ideal for turbine placement.
Because of its significant impact, accurately measuring and analyzing wind speed is vital in the planning stages for wind energy projects.
Energy Efficiency
Energy efficiency measures how much of the wind's kinetic energy is converted into electrical power by the turbine. It is expressed as a percentage and is influenced by several factors, including technology and design.

For example, if a wind turbine is 25% efficient, only a quarter of the wind's energy is converted into usable power. The rest is lost due to various inefficiencies. In our exercise, given the efficiency and turbine's power production, one can calculate the real required wind speed.

  • Higher efficiency means more energy is converted and less is wasted.
  • Design improvements can boost efficiency, making turbines more effective.
Efficiency is a critical consideration in the development and deployment of sustainable energy systems.
Mountain Pass Wind Turbines
Mountain passes provide an advantageous setting for wind turbines due to the natural funneling effect they create. As wind is forced through these constricted spaces, it speeds up, enhancing the potential kinetic energy for the turbines to capture.

These geographical features are naturally predisposed to higher wind speeds, making them more effective for wind power generation as compared to open fields. Increased wind speeds mean more electrical energy can be harvested, which is why so many wind farms are strategically located near or in mountain passes.

  • Mountain passes act as natural wind accelerators due to their terrain.
  • This results in optimal conditions for generating significant wind power.
The informed placement of turbines in such locations helps maximize energy output and ensure the efficient use of resources.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

20.42. Heat Pump. A heat pump is a heat engine run in reverse. In winter it pumps heat from the cold air outside into the warmer air inside the building, maintaining the building at a comfortable temperature. In summer it pumps heat from the cooler air inside the building to the warmer air outside, acting as an air conditioner. (a) If the outside temperature in winter is \(-5.0^{\circ} \mathrm{C}\) and the inside temperature is \(17.0^{\circ} \mathrm{C}\) , how many joules of heat will the heat pump deliver to the inside for each joule of electrical energy used to run the unit, assuming an ideal Carnot cycle? ( b) Suppose you have the option of using electrical resistance heating rather than a heat pump. How much electrical energy would you need in order to deliver the same amount of heat to the inside of the house as in part (a)? Consider a Carnot heat pump delivering heat to the inside of a house to maintain it at \(68^{\circ} \mathrm{F}\) . Show that the beat pump delivers less heat for each joule of electrical energy used to operate the unit as the outside temperature decreases. Notice that this behavior is opposite to the dependence of the efficiency of a Carnot heat engine on the difference in the reservoir temperatures. Explain why this is so.

20.13. A Camot engine whose high-temperature reservoir is at 620 \(\mathrm{K}\) takes in 550 \(\mathrm{J}\) of heat at this temperature in each cycle and gives up 335 \(\mathrm{J}\) to the low-temperature reservoir. (a) How much mechanical work does the engine perform during each cycle? (b) What is the temperature of the low-temperature reservoir? (c) What is the thermal efficiency of the cycle?

20.36 . A lonely party balloon with a volume of 2.40 \(\mathrm{L}\) and containing 0.100 \(\mathrm{mol}\) of air is left behind to drift in the temporarily uninhabited and depressurized Intermational Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is 425 \(\mathrm{m}^{3}\) . Calculate the entropy change of the air during the expansion.

20.50. A stirling-cycle Engine. the Otto cycle, except that the compression and expansion of the gas are done at constant temperature, not adiabatically as in the Otto cycle. The Stirling cycle is used in external combustion engines (in fact, burning fuel is not necessary; any way of producing a temperature difference will do -solar, geothermal, ocean temperature gradient, etc. \(.\) which means that the gas inside the cylinder is not used in the combustion process. Heat is supplied by burning fuel steadily outside the cylinder, instead of explosively inside the cylinder as in the Otto cycle. For this reason Stirling-cycle engines are quieter than Otto-cycle engines, since there are no intake and exhaust valves (a major source of engine noise). While small Stirling engines are used for a variety of purposes, Stiring engines for automobiles have not been successful because they are larger, heavier, and more expensive than conventional automobile engines. In the cycle, the working fluid goes through the following sequence of steps (Fig. 20.30\()\) : (i) Compressed isothermally at temperature \(T_{1}\) from the initial state \(a\) to state \(b\) , with a compression ratio \(r .\) (ii) Heated at constant volume to state \(c\) at temperature \(T_{2}\) . (iii) Expanded isothermally at \(T_{2}\) to state \(d\) . (iv) Cooled at constant volume back to the initial state \(a\) . Assume that the working fluid is \(n\) moles of an ideal gas (for which \(C_{V}\) is independent of temperature). (a) Calculate \(Q, W,\) and \(\Delta U\) for each of the processes \(a \rightarrow b, b \rightarrow c, c \rightarrow d,\) and \(d \rightarrow a\) . (b) In the Stirling cycle, the heat transfers in the processes \(b \rightarrow c\) and \(d \rightarrow a\) do not involve external heat sources but rather use regeneration: The same substance that transfers heat to the gas inside the cylinder in the process \(b \rightarrow c\) also absorbs heat back from the gas in the process \(d \rightarrow a\) . Hence the heat transfers \(Q_{b \rightarrow c}\) and \(Q_{d \rightarrow a}\) do not play a role in determining the efficiency of the engine. Explain this last statement by comparing the expressions for \(Q_{b \rightarrow c}\) and \(Q_{d \rightarrow a}\) calculated in part (a). (c) Calculate the efficiency of a Stirling-cycle engine in terms of the temperatures \(T_{1}\) and \(T_{2}\) . How does this compare to the efficiency of a Carnot-cycle engine operating between these same two temperatures? (Historically, the Stirling cycle was devised before the Carnot cycle.) Does this result violate the second law of thermodynamics? Explain. Unfortunately, actual Stirling-cycle engines cannot achieve this efficiency due to problems with the heat-transfer processes and pressure losses in the engine.

A gasoline engine has a power output of 180 \(\mathrm{kW}\) (about 241 \(\mathrm{hp}\) ). Its thermal efficiency is 28.0\(\%\) . (a) How much heat must be supplied to the engine per second? (b) How much heat is discarded by the engine per second?
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Particle Model of Matter

Read Explanation

Astrophysics

Read Explanation

Kinematics Physics

Read Explanation

Work Energy and Power

Read Explanation

Fluids

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.