John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 13
Текст из файла (страница 13)
Thus, the equivalent thermal resistance isRtequiv =1= 386.8 K/W(1.33 × 10−4 )(13 + 6.44)SinceQ=Tresistor − TairRtequivWe findTresistor = Tair + Q · Rtequiv = 35 + (0.1)(386.8) = 73.68 ◦ CThermal resistance and the electrical analogy§2.377TresistorQconvQrad1R t conv=–hAQconvTresistorTairR t rad =1hradAFigure 2.17 An electrical resistor cooledby convection and radiation.QradWe guessed a resistor temperature of 50◦ C in finding hrad . Recomputing with this higher temperature, we have Tm = 327 K andhrad = 7.17 W/m2 K. If we repeat the rest of the calculation, we get anew value Tresistor = 72.3◦ C. Further iteration is not needed.Since the use of hrad is an approximation, we should check itsapplicability:14∆TTm21=472.3 − 35.03272= 0.00325 1In this case, the approximation is a very good one.Example 2.9Suppose that power to the resistor in Example 2.8 is turned off.
Howlong does it take to cool? The resistor has k 10 W/m·K, ρ 2000 kg/m3 , and cp 700 J/kg·K.Solution. The lumped capacity model, eqn. (1.22), may be applicable. To find out, we check the resistor’s Biot number, noting thatthe parallel convection and radiation processes have an effective heat78Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.4transfer coefficient heff = h + hrad = 18.44 W/m2 K. Then,Bi =(18.44)(0.0036/2)heff ro== 0.0033 1k10so eqn.
(1.22) can be used to describe the cooling process. The timeconstant isT =ρcp V(2000)(700)π (0.010)(0.0036)2 /4== 58.1 sheff A(18.44)(1.33 × 10−4 )From eqn. (1.22) with T0 = 72.3◦ CTresistor = 35.0 + (72.3 − 35.0)e−t/58.1 ◦ CNinety-five percent of the total temperature drop has occured whent = 3T = 174 s.2.4Overall heat transfer coefficient, UDefinitionWe often want to transfer heat through composite resistances, such asthe series of resistances shown in Fig. 2.18. It is very convenient to havea number, U , that works like this4 :Q = U A ∆T(2.32)This number, called the overall heat transfer coefficient, is defined largelyby the system, and in many cases it proves to be insensitive to the operating conditions of the system.In Example 2.6, for instance, two resistances are in series.
We can usethe value Q given by eqn. (2.25) to get1Q (W)"=U=!ro ln(ro /ri )12π ro l (m2 ) ∆T (K)+kh(W/m2 K)(2.33)We have based U on the outside area, Ao = 2π ro l, in this case. We mightinstead have based it on inside area, Ai = 2π ri l, and obtainedU=1ri ln(ro /ri )+khrori(2.34)4This U must not be confused with internal energy. The two terms should alwaysbe distinct in context.Overall heat transfer coefficient, U§2.479Figure 2.18 A thermal circuit with manyresistances in series.
#The equivalentresistance is Rtequiv = i Ri .It is therefore important to remember which area an overall heat transfer coefficient is based on. It is particularly important that A and U beconsistent when we write Q = U A ∆T .In general, for any composite resistance, the overall heat transfer coefficient may be obtained from the equivalent resistance. The equivalentresistance is calculated taking account of series and parallel resistors,as in Examples 2.4 and 2.8.
Then, because Q = ∆T /Rtequiv = U A ∆T , itfollows that U A = 1/Rtequiv .Example 2.10Estimate the overall heat transfer coefficient for the tea kettle shownin Fig. 2.19. Note that the flame convects heat to the thin aluminum.The heat is then conducted through the aluminum and finally convected by boiling into the water.Solution. We need not worry about deciding which area to base Aon because the area normal to the heat flux vector does not change.We simply write the heat flowTflame − Tboiling water∆T=Q= #11LRt++hA kAl A hb Aand apply the definition of UU=Q1=L11A∆T++h kAlhbLet us see what typical numbers would look like in this example: hmight be around 200 W/m2 K; L kAl might be 0.001 m/(160 W/m·K)or 1/160,000 W/m2 K; and hb is quite large— perhaps about 5000W/m2 K. Thus:U1= 192.1 W/m2 K111++200 160, 000 500080Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.4Figure 2.19 Heat transfer through the bottom of a tea kettle.It is clear that the first resistance is dominant, as is shown in Fig.
2.19.Notice that in such casesU A → 1/Rtdominant(2.35)where A is any area (inside or outside) in the thermal circuit.Experiment 2.1Boil water in a paper cup over an open flame and explain why you cando so. [Recall eqn. (2.35) and see Problem 2.12.]Example 2.11A wall consists of alternating layers of pine and sawdust, as shownin Fig. 2.20). The sheathes on the outside have negligible resistanceand h is known on the sides. Compute Q and U for the wall.Solution.
So long as the wood and the sawdust do not differ dramatically from one another in thermal conductivity, we can approximatethe wall as a parallel resistance circuit, as shown in the figure.5 The5For this approximation to be exact, the resistances must be equal. If they differradically, the problem must be treated as two-dimensional.Overall heat transfer coefficient, U§2.4Figure 2.20 Heat transfer through a composite wall.equivalent thermal resistance of the circuit is1Rtequiv = Rtconv + 1+Rtpine + Rtconv1RtsawdustThusQ=∆T=Rtequiv1hAT∞1 − T∞r+1kp ApL+ks As+1hALandU=Q=A∆T2h1+1kp ApL A+ks AsL AThe approach illustrated in this example is very widely used in calculating U values for the walls and roofs houses and buildings. The thermalresistances of each structural element — insulation, studs, siding, doors,windows, etc.
— are combined to calculate U or Rtequiv , which is then usedtogether with weather data to estimate heating and cooling loads [2.5].8182Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.4Table 2.2 Typical ranges or magnitudes of UHeat Exchange ConfigurationWalls and roofs dwellings with a 24 km/houtdoor wind:• Insulated roofs• Finished masonry walls• Frame walls• Uninsulated roofsSingle-pane windowsAir to heavy tars and oilsAir to low-viscosity liquidsAir to various gasesSteam or water to oilLiquids in coils immersed in liquidsFeedwater heatersAir condensersSteam-jacketed, agitated vesselsShell-and-tube ammonia condensersSteam condensers with 25◦ C waterCondensing steam to high-pressureboiling water†U (W/m2 K)0.3−20.5−60.3−51.2−4∼ 6†As low as 45As high as 60060−55060−340110−2, 000110−8, 500350−780500−1, 900800−1, 4001, 500−5, 0001, 500−10, 000Main heat loss is by infiltration.Typical values of UIn a fairly general use of the word, a heat exchanger is anything thatlies between two fluid masses at different temperatures.
In this sense aheat exchanger might be designed either to impede or to enhance heatexchange. Consider some typical values of U shown in Table 2.2, whichwere assembled from a variety of technical sources. If the exchangeris intended to improve heat exchange, U will generally be much greaterthan 40 W/m2 K. If it is intended to impede heat flow, it will be less than10 W/m2 K—anywhere down to almost perfect insulation. You shouldhave some numerical concept of relative values of U , so we recommendthat you scrutinize the numbers in Table 2.2. Some things worth bearingin mind are:• The fluids with low thermal conductivities, such as tars, oils, or anyof the gases, usually yield low values of h.
When such fluid flowson one side of an exchanger, U will generally be pulled down.Overall heat transfer coefficient, U§2.4• Condensing and boiling are very effective heat transfer processes.They greatly improve U but they cannot override one very smallvalue of h on the other side of the exchange. (Recall Example 2.10.)In fact:• For a high U , all resistances in the exchanger must be low.• The highly conducting liquids, such as water and liquid metals, givehigh values of h and U .Fouling resistanceFigure 2.21 shows one of the simplest forms of a heat exchanger—a pipe.The inside is new and clean on the left, but on the right it has built up alayer of scale. In conventional freshwater preheaters, for example, thisscale is typically MgSO4 (magnesium sulfate) or CaSO4 (calcium sulfate)which precipitates onto the pipe wall after a time. To account for the resistance offered by these buildups, we must include an additional, highlyempirical resistance when we calculate U .
Thus, for the pipe shown inFig. 2.21,U older pipebased on Ai=11hi+ri ln(ro /rp )kinsul+ri ln(rp /ri )kpipe+Figure 2.21 The fouling of a pipe.riro ho+ Rf8384Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.4Table 2.3 Some typical fouling resistances for a unit area.Fouling ResistanceRf (m2 K/W)Fluid and SituationDistilled waterSeawaterTreated boiler feedwaterClean river or lake waterAbout the worst waters used in heatexchangersNo.
6 fuel oilTransformer or lubricating oilMost industrial liquidsMost refinery liquidsSteam, non-oil-bearingSteam, oil-bearing (e.g., turbineexhaust)Most stable gasesFlue gasesRefrigerant vapors (oil-bearing)0.00010.0001 − 0.00040.0001 − 0.00020.0002 − 0.0006< 0.00200.00010.00020.00020.0002 − 0.00090.00010.00030.0002 − 0.00040.0010 − 0.00200.0040where Rf is a fouling resistance for a unit area of pipe (in m2 K/W). AndclearlyRf ≡11−UoldUnew(2.36)Some typical values of Rf are given in Table 2.3. These values havebeen adapted from [2.6] and [2.7].
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
