/*! 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 10 You recently purchased a large p... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 10

You recently purchased a large plot of land in the Amazon jungle at an extremely low cost. You are quite pleased with yourself until you arrive there and find that the nearest source of electricity is 1500 miles away, a fact that your brother-in-law, the real estate agent, somehow forgot to mention. since the local hardware store does not carry 1500 -mile-long extension cords, you decide to build a small hydroelectric generator under a 75-m high waterfall located nearby. The flow rate of the waterfall is 10 \(^{5} \mathrm{m}^{3} / \mathrm{h}\), and you anticipate needing \(750 \mathrm{kW} \cdot \mathrm{h} / \mathrm{wk}\) to run your lights, air conditioner, and television. Calculate the maximum power theoretically available from the waterfall and see if it is sufficient to meet your needs.

Short Answer

Expert verified
The maximum power theoretically available from the waterfall is about 20.5 MW, which is well above the needed power (~1.38 W). Therefore, the energy obtainable from the waterfall should be more than enough to meet the needs.

Step by step solution

01

Convert energy requirements to the correct unit

First, convert the energy needs from kWh per week to watts. There are 1,000 watts in a kilowatt, and 1 week has 604800 seconds (7 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute). So, \[750 \, \mathrm{kW \cdot h/wk} = 750,000 \, \mathrm{W \cdot h/wk} \approx 1.38 \, \mathrm{W} \]
02

Calculate potential energy of the waterfall

Second, calculate the power available from the waterfall. Power is the rate at which energy is transferred and can be calculated from \[P = \frac{E}{t}\] where \(P\) is power, \(E\) is energy and \(t\) is time. We can calculate energy using the formula for gravitational potential energy: \[E = m \cdot g \cdot h\] where \(m\) is mass, \(g\) is the gravitational constant (approximated as 9.8 m/s虏 on the surface of the Earth), and \(h\) is height. But, we don't have mass. However, we know that \[ m = V \cdot \rho \] where \(V\) is volume and \(\rho\) is the water density (about 1000 kg/m鲁 at room temperature). However, volume flow rate is given, not the volume. We know that \[Q = \frac{V}{t}\] where \(Q\) is the volume flow rate. So, \[V = Q \cdot t\]. Let's substitute \(V\) in the previous equation, we get: \[m = Q \cdot t \cdot \rho\]. Going back to the power equation, we get: \[P = Q \cdot t \cdot \rho \cdot g \cdot h \, / \, t\]. As \(t\) cancels out, we're left with \[P = Q \cdot \rho \cdot g \cdot h\]. Now we can calculate the power. Convert the volumetric flow rate from m鲁/hr to m鲁/sec. There are 3600 seconds in an hour, so \[10^5 \, \mathrm{m}^3/\mathrm{h} = 27.78 \, \mathrm{m}^3/\mathrm{s}\]. Using these numbers, we obtain \[P = 27.78 \, \mathrm{m}^3/\mathrm{s} \cdot 1000 \, \mathrm{kg/m}^3 \cdot 9.8 \,\mathrm{m/s}^2 \cdot 75 \, \mathrm{m} \approx 2.05 \times 10^7 \, \mathrm{W}\] or about 20.5 MW.
03

Comparing the results

Finally, we compare the power needed (1.38 W) to the power available (20.5 MW). The available power from the waterfall is well above the necessary power requirement, so it should be sufficient to power all the appliances.

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Key Concepts

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

Potential Energy
When water in a waterfall is at a height, it has stored energy due to its position, known as potential energy. This energy depends on three factors:
  • Mass of the water ( \(m\))
  • Gravitational acceleration ( \(g\), which is approximately 9.8 m/s虏)
  • Height ( \(h\))
This can be mathematically expressed as \(E = m \cdot g \cdot h\).
In simple terms, the higher the water is, and the more mass it has, the greater its potential energy. When the water falls, this potential energy is converted into other forms of energy, like kinetic energy, which can then be utilized by hydroelectric generators to produce electricity.
Gravitational Force
Gravitational force is a natural phenomenon by which all things with mass or energy are brought toward one another, including the water in waterfalls. It essentially pulls the water down between the crest and base of the waterfall, converting gravitational potential energy into kinetic energy.

On Earth, this force is approximated by the constant 9.8 m/s虏. This constant allows us to predict how fast the object, or in this case, water, will accelerate towards the Earth when it falls. Moreover, it is the driving force behind the potential energy in the water, facilitating its transformation into electrical energy as it descends.
Volume Flow Rate
The volume flow rate refers to the amount of water flowing per unit of time, which is pivotal in determining the energy potential of a waterfall. It is measured in units such as cubic meters per hour ( \(\mathrm{m}^3/\mathrm{h}\)) or cubic meters per second (\( \mathrm{m}^3/\mathrm{s}\)).

Understanding the flow rate is crucial because it describes how much water will fall in a certain period.
  • A higher volume flow rate means more water, hence more mass, contributing to greater potential energy.
  • By converting the given flow rate to the desired unit (e.g., from hours to seconds), we ensure accurate calculations of the available power. In the original problem, a conversion was needed from 10^{5} \(\mathrm{m}^3/\mathrm{h}\) to approximately 27.78 \(\mathrm{m}^3/\mathrm{s}\).
Energy Conversion
Energy conversion refers to the process of changing one form of energy into another. In the context of hydroelectric power, potential energy from the water above is converted into kinetic energy as it falls.

This kinetic energy is then turned into mechanical energy and eventually into electrical energy using turbines and generators.
  • The machine performs mechanical work using the water's kinetic energy.
  • The generator then converts this mechanical energy into electricity.
Energy conversion processes are crucial for harnessing natural resources like waterfalls for practical uses such as generating electricity. The efficiency of this conversion determines how much of the original potential energy is translated into usable electrical power.
Power Comparison
Power comparison involves analyzing whether the energy available from a source meets the required demand. In our scenario, we wish to determine if the hydroelectric generator powered by the waterfall can meet a specified energy requirement.

By calculating the available power using the waterfall's volume flow rate and height, we found a power output of approximately 20.5 MW.
  • Ensure that the available power is compared against the power necessity. In this problem, only 1.38 watts was needed, showing a large surplus of energy available.
  • This comparison validates whether the resource can adequately supply the intended usage.
Such feasibility studies are essential before implementing energy solutions in real-world settings.

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Most popular questions from this chapter

Steam produced in a boiler is frequently "wet"-that is, it is a mist composed of saturated water vapor and entrained liquid droplets. The quality of a wet steam is defined as the fraction of the mixture by mass that is vapor. A wet steam at a pressure of 5.0 bar with a quality of 0.85 is isothermally "dried" by evaporating the entrained liquid. The flow rate of the dried steam is \(52.5 \mathrm{m}^{3} / \mathrm{h}\). (a) Use the steam tables to determine the temperature at which this operation occurs, the specific enthalpies of the wet and dry steams, and the total mass flow rate of the process stream. (b) Calculate the heat input (kW) required for the evaporation process. (c) Suppose leaks developed in the feed pipe to the dryer and in the dryer exit pipe. Speculate on what you would see at each location.

Saturated steam at \(100^{\circ} \mathrm{C}\) is heated to \(350^{\circ} \mathrm{C}\). Use the steam tables to determine (a) the required heat input (J/s) if a continuous stream flowing at \(100 \mathrm{kg} / \mathrm{s}\) undergoes the process at constant pressure and (b) the required heat input (J) if \(100 \mathrm{kg}\) undergoes the process in a constant-volume container. What is the physical significance of the difference between the numerical values of these two quantities?

Superheated steam at 40 bar absolute and \(500^{\circ} \mathrm{C}\) flows at a rate of \(250 \mathrm{kg} / \mathrm{min}\) to an adiabatic turbine, where it expands to 5 bar. The turbine develops \(1500 \mathrm{kW}\). From the turbine the steam flows to a heater, where it is reheated isobarically to its initial temperature. Neglect kinetic energy changes. (a) Write an energy balance on the turbine and use it to determine the outlet stream temperature. (b) Write an energy balance on the heater and use it to determine the required input (kW) to the steam. (c) Verify that an overall energy balance on the two-unit process is satisfied. (d) Suppose the turbine inlet and outlet pipes both have diameters of 0.5 meter. Show that it is reasonable to neglect the change in kinetic energy for this unit.

Write and simplify the closed-system energy balance (Equation \(7.3-4\) ) for each of the following processes, and state whether nonzero heat and work terms are positive or negative. Begin by defining the system. The solution of Part (a) is given as an illustration. (a) The contents of a closed flask are heated from \(25^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). (b) A tray filled with water at \(20^{\circ} \mathrm{C}\) is put into a freezer. The water tums into ice at \(-5^{\circ} \mathrm{C}\). (Note: When a substance expandsit does work on its surroundings and when it contracts the surroundings do work on it.) (c) A chemical reaction takes place in a closed adiabatic (perfectly insulated) rigid container. (d) Repeat Part (c), only suppose that the reactor is isothermal rather than adiabatic and that when the reaction was carried out adiabatically, the temperature in the reactor increased.

Agricultural irrigation uses a significant amount of water, and in some regions it has overwhelmed other water needs. Suppose water is drawn from a reservoir and delivered into an irrigation ditch. For most of the length of the ditch, the delivery is through a \(10-\mathrm{cm}\) ID pipe, and in the last few meters the pipe diameter is \(7 \mathrm{cm} .\) The exit from the pipe is \(300 \mathrm{m}\) lower than the pipe inlet. (a) Assume that the pipe is smooth (i.e., ignore friction) and that the delivery rate is 4000 \(\mathrm{kg} / \mathrm{h}\). Estimate the required pressure difference between pipe inlet and outlet. How far below the surface of the reservoir is the pipe inlet? (b) How would your answer be different if the pipe were not smooth? Explain. Exploratory Exercise- Research and Discover (c) What are possible environmental impacts of diverting significant quantities of river water for use in irrigation? Cite at least two sources for your response.
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