/*! 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

Eight fluid ounces \((1 \mathrm{qt}=32 \mathrm{oz})\) of a beverage in a glass at \(18.0^{\circ} \mathrm{C}\) is to be cooled by adding ice and stirring. The properties of the beverage may be taken to be those of liquid water. The enthalpy of the ice relative to liquid water at the triple point is \(-348 \mathrm{kJ} / \mathrm{kg} .\) Estimate the mass of ice (g) that must melt bring the liquid temperature to \(4^{\circ} \mathrm{C},\) neglecting energy losses to the surroundings.

A perfectly insulated cylinder fitted with a leakproof frictionless piston with a mass of \(30.0 \mathrm{kg}\) and a face area of \(400.0 \mathrm{cm}^{2}\) contains \(7.0 \mathrm{kg}\) of liquid water and a 3.0 -kg bar of aluminum. The aluminum bar has an electrical coil imbedded in it, so that known amounts of heat can be transferred to it. Aluminum has a specific gravity of 2.70 and a specific internal energy given by the formula \(\hat{U}(\mathrm{kJ} / \mathrm{kg})=0.94 \mathrm{T}\left(^{\circ} \mathrm{C}\right)\) The internal energy of liquid water at any temperature may be taken to be that of the saturated liquid at that temperature. Negligible heat is transferred to the cylinder wall. Atmospheric pressure is 1.00 atm. The cylinder and its contents are initially at \(20^{\circ} \mathrm{C}\). Suppose that \(3310 \mathrm{kJ}\) is transferred to the bar from the heating coil and the contents of the cylinder are then allowed to equilibrate. (a) Calculate the pressure of the cylinder contents throughout the process. Then determine whether the amount of heat transferred to the system is sufficient to vaporize any of the water. (b) Determine the following quantities: (i) the final system temperature; (ii) the volumes \(\left(\mathrm{cm}^{3}\right)\) of the liquid and vapor phases present at equilibrium; and (iii) the vertical distance traveled by the piston from the beginning to the end of the process. (Suggestion: Write an energy balance on the complete process, taking the cylinder contents to be the system. Note that the system is closed and that work is done by the system when it moves the piston through a vertical displacement. The magnitude of this work is \(W=P \Delta V,\) where \(P\) is the constant system pressure and \(\Delta V\) is the change in system volume from the initial to the final state.) (c) Calculate an upper limit on the temperature attainable by the aluminum bar during the process, and state the condition that would have to apply for the bar to come close to this temperature.

A steam trap is a device to purge steam condensate from a system without venting uncondensed steam. In one of the crudest trap types, the condensate collects and raises a float attached to a drain plug. When the float reaches a certain level, it "pulls the plug," opening the drain valve and allowing the liquid to discharge. The float then drops down to its original position and the valve closes, preventing uncondensed steam from escaping. (a) Suppose saturated steam at 25 bar is used to heat \(100 \mathrm{kg} / \mathrm{min}\) of an oil from \(135^{\circ} \mathrm{C}\) to \(185^{\circ} \mathrm{C}\). Heat must be transferred to the oil at a rate of \(1.00 \times 10^{4} \mathrm{kJ} / \mathrm{min}\) to accomplish this task. The steam condenses on the exterior of a bundle of tubes through which the oil is flowing. Condensate collects in the bottom of the exchanger and exits through a steam trap set to discharge when 1200 g of liquid is collected. How often does the trap discharge? (b) Especially when periodic maintenance checks are not performed, steam traps often fail to close completely and so leak steam continuously. Suppose a process plant contains 1000 leaking traps (not an unrealistic supposition for some plants) operating at the condition of Part (a), and that on the average 10\% additional steam must be fed to the condensers to compensate for the uncondensed steam venting through the leaks. Further suppose that the cost of generating the additional steam is \$7.50 per million Btu, where the denominator refers to the enthalpy of the leaking steam relative to liquid water at \(20^{\circ} \mathrm{C}\). Estimate the yearly cost of the leaks based on \(24 \mathrm{h} /\) day, 360 day/yr operation.

A turbine discharges \(200 \mathrm{kg} / \mathrm{h}\) of saturated steam at \(10.0 \mathrm{bar}\) absolute. It is desired to generate steam at \(250^{\circ} \mathrm{C}\) and 10.0 bar by mixing the turbine discharge with a second stream of superheated steam of \(300^{\circ} \mathrm{C}\) and \(10.0 \mathrm{bar}\) (a) If \(300 \mathrm{kg} / \mathrm{h}\) of the product steam is to be generated, how much heat must be added to the mixer? (b) If instead the mixing is carried out adiabatically, at what rate is the product steam generated?

A rigid 6.00-liter vessel contains 4.00 L of liquid water in equilibrium with 2.00 L of water vapor at \(25^{\circ} \mathrm{C} .\) Heat is transferred to the water by means of an immersed electrical coil. The volume of the coil is negligible. Use the steam tables to calculate the final temperature and pressure (bar) of the system and the mass of water vaporized (g) if 3915 kJ is added to the water and no heat is transferred from the water to its surroundings. (Note: A trial-and-error calculation is required.)
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