John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook, страница 7

Furthermore, the energy emitted from ablack body reaches a theoretical maximum, which is given by the StefanBoltzmann law. We look at this next.The Stefan-Boltzmann law. The flux of energy radiating from a bodyis commonly designated e(T ) W/m2 . The symbol eλ (λ, T ) designates thedistribution function of radiative flux in λ, or the monochromatic emissivepower:de(λ, T )or e(λ, T ) =eλ (λ, T ) =dλThuse(T ) ≡ E(∞, T ) =λ0eλ (λ, T ) dλ(1.27)∞0eλ (λ, T ) dλThe dependence of e(T ) on T for a black body was established experimentally by Stefan in 1879 and explained by Boltzmann on the basis ofthermodynamics arguments in 1884.

The Stefan-Boltzmann law iseb (T ) = σ T 4(1.28)where the Stefan-Boltzmann constant, σ , is 5.670400 × 10−8 W/m2 ·K4or 1.714 × 10−9 Btu/hr·ft2 ·◦ R4 , and T is the absolute temperature.eλ vs. λ. Nature requires that, at a given temperature, a body will emita unique distribution of energy in wavelength. Thus, when you heat apoker in the fire, it first glows a dull red—emitting most of its energyat long wavelengths and just a little bit in the visible regime. When it is§1.3Modes of heat transfer31Figure 1.15 Monochromatic emissivepower of a black body at severaltemperatures—predicted and observed.white-hot, the energy distribution has been both greatly increased andshifted toward the shorter-wavelength visible range.

At each temperature, a black body yields the highest value of eλ that a body can attain.The very accurate measurements of the black-body energy spectrumby Lummer and Pringsheim (1899) are shown in Fig. 1.15. The locus ofmaxima of the curves is also plotted. It obeys a relation called Wien’slaw:(λT )eλ=max = 2898 µm·K(1.29)About three-fourths of the radiant energy of a black body lies to the rightof this line in Fig.

1.15. Notice that, while the locus of maxima leanstoward the visible range at higher temperatures, only a small fraction ofthe radiation is visible even at the highest temperature.Predicting how the monochromatic emissive power of a black bodydepends on λ was an increasingly serious problem at the close of thenineteenth century. The prediction was a keystone of the most profoundscientific revolution the world has seen. In 1901, Max Planck made theIntroduction32§1.3prediction, and his work included the initial formulation of quantum mechanics.

He found thateλb =2π hco25λ [exp(hco /kB T λ) − 1](1.30)where co is the speed of light, 2.99792458 × 108 m/s; h is Planck’s constant, 6.62606876×10−34 J·s; and kB is Boltzmann’s constant, 1.3806503×10−23 J/K.Radiant heat exchange. Suppose that a heated object (1 in Fig. 1.16a)radiates only to some other object (2) and that both objects are thermallyblack.

All heat leaving object 1 arrives at object 2, and all heat arrivingat object 1 comes from object 2. Thus, the net heat transferred fromobject 1 to object 2, Qnet , is the difference between Q1 to 2 = A1 eb (T1 )and Q2 to 1 = A1 eb (T2 )(1.31)Qnet = A1 eb (T1 ) − A1 eb (T2 ) = A1 σ T14 − T24If the first object “sees” other objects in addition to object 2, as indicatedin Fig. 1.16b, then a view factor (sometimes called a configuration factoror a shape factor ), F1–2 , must be included in eqn. (1.31):Qnet = A1 F1–2 σ T14 − T24(1.32)We may regard F1–2 as the fraction of energy leaving object 1 that isintercepted by object 2.Example 1.5A black thermocouple measures the temperature in a chamber withblack walls. If the air around the thermocouple is at 20◦ C, the wallsare at 100◦ C, and the heat transfer coefficient between the thermocouple and the air is 75 W/m2 K, what temperature will the thermocoupleread?Solution. The heat convected away from the thermocouple by theair must exactly balance that radiated to it by the hot walls if the system is in steady state.

Furthermore, F1–2 = 1 since the thermocouple(1) radiates all its energy to the walls (2):44hAtc (Ttc − Tair ) = −Qnet = −Atc σ Ttc− TwallModes of heat transfer§1.3Figure 1.16another.33The net radiant heat transfer from one object toor, with Ttc in ◦ C,75(Ttc − 20) W/m2 =5.6704 × 10−8 (100 + 273)4 − (Ttc + 273)4 W/m2since T for radiation must be in kelvin. Trial-and-error solution ofthis equation yields Ttc = 28.4◦ C.We have seen that non-black bodies absorb less radiation than blackbodies, which are perfect absorbers.

Likewise, non-black bodies emit lessradiation than black bodies, which also happen to be perfect emitters. Wecan characterize the emissive power of a non-black body using a propertycalled emittance, ε:enon-black = εeb = εσ T 4(1.33)where 0 < ε ≤ 1. When radiation is exchanged between two bodies thatare not black, we haveQnet = A1 F1–2 σ T14 − T24(1.34)where the transfer factor, F1–2 , depends on the emittances of both bodiesas well as the geometrical “view”.Introduction34§1.3The expression for F1–2 is particularly simple in the important specialcase of a small object, 1, in a much larger isothermal environment, 2:F1–2 = ε1forA1 A2(1.35)Example 1.6Suppose that the thermocouple in Example 1.5 was not black andhad an emissivity of ε = 0.4.

Further suppose that the walls werenot black and had a much larger surface area than the thermocouple.What temperature would the thermocouple read?Solution. Qnet is now given by eqn. (1.34) and F1–2 can be foundwith eqn. (1.35):44hAtc (Ttc − Tair ) = −Atc εtc σ Ttc− Twallor75(Ttc − 20) W/m2 =(0.4)(5.6704 × 10−8 ) (100 + 273)4 − (Ttc + 273)4 W/m2Trial-and-error yields Ttc = 23.5◦ C.Radiation shielding.

The preceding examples point out an importantpractical problem than can be solved with radiation shielding. The ideais as follows: If we want to measure the true air temperature, we canplace a thin foil casing, or shield, around the thermocouple. The casingis shaped to obstruct the thermocouple’s “view” of the room but to permitthe free flow of the air around the thermocouple. Then the shield, likethe thermocouple in the two examples, will be cooler than the walls, andthe thermocouple it surrounds will be influenced by this much coolerradiator. If the shield is highly reflecting on the outside, it will assume atemperature still closer to that of the air and the error will be still less.Multiple layers of shielding can further reduce the error.Radiation shielding can take many forms and serve many purposes.It is an important element in superinsulations.

A glass firescreen in afireplace serves as a radiation shield because it is largely opaque to radiation. It absorbs heat radiated by the fire and reradiates that energy(ineffectively) at a temperature much lower than that of the fire.A look ahead§1.4Experiment 1.4Find a small open flame that produces a fair amount of soot. A candle,kerosene lamp, or a cutting torch with a fuel-rich mixture should workwell. A clean blue flame will not work well because such gases do notradiate much heat. First, place your finger in a position about 1 to 2 cmto one side of the flame, where it becomes uncomfortably hot. Now takea piece of fine mesh screen and dip it in some soapy water, which will fillup the holes.

Put it between your finger and the flame. You will see thatyour finger is protected from the heating until the water evaporates.Water is relatively transparent to light. What does this experimentshow you about the transmittance of water to infrared wavelengths?1.4A look aheadWhat we have done up to this point has been no more than to reveal thetip of the iceberg. The basic mechanisms of heat transfer have been explained and some quantitative relations have been presented.

However,this information will barely get you started when you are faced with a realheat transfer problem. Three tasks, in particular, must be completed tosolve actual problems:• The heat diffusion equation must be solved subject to appropriateboundary conditions if the problem involves heat conduction of anycomplexity.• The convective heat transfer coefficient, h, must be determined ifconvection is important in a problem.• The factor F1–2 or F1–2 must be determined to calculate radiativeheat transfer.Any of these determinations can involve a great deal of complication,and most of the chapters that lie ahead are devoted to these three basicproblems.Before becoming engrossed in these three questions, we shall firstlook at the archetypical applied problem of heat transfer–namely, thedesign of a heat exchanger.

Chapter 2 sets up the elementary analyticalapparatus that is needed for this, and Chapter 3 shows how to do such35Introduction36§1.5design if h is already known. This will make it easier to see the importance of undertaking the three basic problems in subsequent parts of thebook.1.5ProblemsWe have noted that this book is set down almost exclusively in S.I. units.The student who has problems with dimensional conversion will findAppendix B helpful.

The only use of English units appears in some of theproblems at the end of each chapter. A few such problems are includedto provide experience in converting back into English units, since suchunits will undoubtedly persist in the U.S.A. for many more years.Another matter often leads to some discussion between students andteachers in heat transfer courses. That is the question of whether a problem is “theoretical” or “practical”.