John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 84
Текст из файла (страница 84)
By way of illustration, suppose that we sought to photovoltaically convert 615 W/m2of solar energy into electric power with a 15% efficiency (which is notpessimistic) during 8 hr of each day. This would correspond to a dailyaverage of 31 W/m2 , and we would need almost 26 square kilometers (10square miles) of collector area to match the steady output of an 800 MWpower plant.581582Radiative heat transfer§10.6Other forms of solar energy conversion require similarly large areas.Hydroelectric power — the result of evaporation under the sun’s warminginfluence — requires a large reservoir, and watershed, behind the dam.The burning of organic matter, as wood or grain-based ethanol, requiresa large cornfield or forest to be fed by the sun, and so forth.
Any energysupply that is served by the sun must draw from a large area of theearth’s surface. Thus, they introduce their own kinds of environmentalcomplications.A second problem stems from the intermittent nature of solar devices.To provide steady power—day and night, rain or shine—requires thermalstorage systems, which add both complication and cost.These problems are minimal when one uses solar energy merely toheat air or water to moderate temperatures (50 to 90◦ C).
In this case theefficiency will improve from just a few percent to as high as 70%. Suchheating can be used for industrial processes (crop drying, for example),or it can be used on a small scale for domestic heating of air or water.Figure 10.26 shows a typical configuration of a domestic solar collector of the flat-plate type. Solar radiation passes through one or more glassplates and impinges on a plate that absorbs the solar wavelengths. Theabsorber plate would be a selective solar absorber, perhaps blackenedcopper or nickel. The glass plates might be treated with anti-reflectivecoatings, raising their solar transmissivity to 98% or more. Once the energy is absorbed, it is reemitted as long-wavelength infrared radiation.Glass is almost opaque in this range, and energy is retained in the collector by a greenhouse effect.
Multiple layers of glass serve to reduce bothreradiative and convective losses from the absorber plate.Water flowing through tubes, which may be brazed to the absorberplate, carries the energy away for use. The flow rate is adjusted to givean appropriate temperature rise.If the working fluid is to be brought to a fairly high temperature, thedirect radiation from the sun must be focused from a large area down toa very small region, using reflecting mirrors.
Collectors equipped with asmall parabolic reflector, focused on a water or air pipe, can raise the fluidto between 100 and 200◦ C. In any scheme intended to produce electricalpower with a conventional thermal cycle, energy must be focused in anarea ratio on the order of 1000 : 1 to achieve a practical cycle efficiency.A question of over-riding concern as we enter the 21st century is“How much of the renewable energy that reaches Earth, can we hopeto utilize?” Of the 1.74×1014 kW arriving from the sun, 33% is simply§10.6Solar energyFigure 10.26 A typical flat-plate solar collector.reflected back into outer space.
If we were able to collect and use theremainder, 1.16×1014 kW, before it too was reradiated to space, each ofthe 6 billion or so people on the planet would have 19 MW at his or herdisposal. Of course, the vast majority of that power must be used tosustain natural processes in the world around us.In the USA, total energy consumption in 2002 averaged roughly 3.2 ×109 kW, and, dividing this value into a population of 280 million peoplegives a per capita consumption of roughly 11 kW. Worldwide, energywas consumed at a rate just over 1010 kW.
That means that world energyconsumption was just under 0.01% of the renewable energy passing intoand out of Earth’s ecosystem. Since many countries that once used verylittle energy are moving toward a life-style which requires much greaterenergy consumption, this percentage is rising at an estimated rate of2%/y.We must also bear in mind two aspects of this 0.01% figure. First, itis low enough that we might aim, ultimately, to take all of our energyfrom renewable sources, and thus avoid consuming irreplaceable terrestrial resources.
Second, although 0.01% is a small fraction, the absolute583Chapter 10: Radiative heat transfer584amount of power it represents is enormous. It is, therefore, unclear justhow much renewable energy we can claim before we create new ecologicalproblems.There is little doubt that our short-term needs can be met by fossilfuel reserves. However, continued use of those fuels will clearly amplifythe now-well-documented global warming trend. Our long-term hope foran adequate energy supply may be at least partially met with solar powerand other renewable sources. Nuclear fission remains a promising optionif one or more means for nuclear waste disposal is deemed acceptable.Nuclear fusion—the process by which we might manage to create minisuns upon the earth—may also be a hope for the future. Under almostany scenario, however, we will surely be forced to limit the continuinggrowth of energy consumption.Problems10.1What will ελ of the sun appear to be to an observer on theearth’s surface at λ = 0.2 µm and 0.65 µm? How do theseemittances compare with the real emittances of the sun? [At0.65 µm, ελ 0.77.]10.2Plot eλb against λ for T = 300 K and 10, 000 K with the helpof eqn.
(1.30). About what fraction of energy from each blackbody is visible?10.3A 0.6 mm diameter wire is drawn out through a mandril at950◦ C. Its emittance is 0.85. It then passes through a longcylindrical shield of commercial aluminum sheet, 7 cm in diameter. The shield is horizontal in still air at 25◦ C. What is thetemperature of the shield? Is it reasonable to neglect naturalconvection inside and radiation outside? [Tshield = 153◦ C.]10.4A 1 ft2 shallow pan with adiabatic sides is filled to the brim withwater at 32◦ F. It radiates to a night sky whose temperature is−18◦ F, while a 50◦ F breeze blows over it at 1.5 ft/s.
Will thewater freeze or warm up?10.5A thermometer is held vertically in a room with air at 10◦ C andwalls at 27◦ C. What temperature will the thermometer read ifeverything can be considered black? State your assumptions.10.6Rework Problem 10.5, taking the room to be wall-papered andconsidering the thermometer to be nonblack.Problems58510.7Two thin aluminum plates, the first polished and the secondpainted black, are placed horizontally outdoors, where they arecooled by air at 10◦ C. The heat transfer coefficient is 5 W/m2 Kon both the top and the bottom.
The top is irradiated with750 W/m2 and it radiates to the sky at 250 K. The earth belowthe plates is black at 10◦ C. Find the equilibrium temperatureof each plate.10.8A sample holder of 99% pure aluminum, 1 cm in diameter and16 cm in length, protrudes from a small housing on an orbital space vehicle.
The holder “sees” almost nothing but outerspace at an effective temperature of 30 K. The base of the holders is 0◦ C and you must find the temperature of the sample atits tip. It will help if you note that aluminum is used, so thatthe temperature of the tip stays quite close to that of the root.[Tend = −0.7◦ C.]10.9There is a radiant heater in the bottom of the box shown inFig. 10.27. What percentage of the heat goes out the top? Whatfraction impinges on each of the four sides? (Remember thatthe percentages must add up to 100.)Figure 10.27 Configuration forProb. 10.9.10.10With reference to Fig. 10.12, find F1–(2+4) and F(2+4)–1 .10.11Find F2–4 for the surfaces shown in Fig.
10.28. [0.315.]10.12What is F1–2 for the squares shown in Fig. 10.29?10.13A particular internal combustion engine has an exhaust manifold at 600◦ C running parallel to a water cooling line at 20◦ C.If both the manifold and the cooling line are 4 cm in diameter, their centers are 7 cm apart, and both are approximatelyblack, how much heat will be transferred to the cooling line byradiation? [383 W/m.]Chapter 10: Radiative heat transfer586Figure 10.28 Configuration forProb. 10.11.Figure 10.29 Configuration forProb. 10.12.10.14Prove that F1–2 for any pair of two-dimensional plane surfaces,as shown in Fig.
10.30, is equal to [(a + b) − (c + d)]/2L1 .This is called the string rule because we can imagine that thenumerator equals the difference between the lengths of a setof crossed strings (a and b) and a set of uncrossed strings (cand d).Figure 10.30 Configuration forProb.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
