John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 14
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Notice that fouling has the effect ofadding a resistance in series on the order of 10−4 m2 K/W. It is rather likeanother heat transfer coefficient, hf , on the order of 10,000 W/m2 K inseries with the other resistances in the exchanger.The tabulated values of Rf are given to only one significant figure because they are very approximate. Clearly, exact values would have to bereferred to specific heat exchanger configurations, to particular fluids, tofluid velocities, to operating temperatures, and to age [2.8, 2.9].
The resistance generally drops with increased velocity and increases with temperature and age. The values given in the table are based on reasonableOverall heat transfer coefficient, U§2.4maintenance and the use of conventional shell-and-tube heat exchangers.With misuse, a given heat exchanger can yield much higher values of Rf .Notice too, that if U 1, 000 W/m2 K, fouling will be unimportantbecause it will introduce a negligibly small resistance in series. Thus,in a water-to-water heat exchanger, for which U is on the order of 2000W/m2 K, fouling might be important; but in a finned-tube heat exchangerwith hot gas in the tubes and cold gas passing across the fins on them, Umight be around 200 W/m2 K, and fouling will be usually be insignificant.Example 2.12You have unpainted aluminum siding on your house and the engineerhas based a heat loss calculation on U = 5 W/m2 K. You discover thatair pollution levels are such that Rf is 0.0005 m2 K/W on the siding.Should the engineer redesign the siding?Solution.
From eqn. (2.36) we get1Ucorrected=1Uuncorrected+ Rf = 0.2000 + 0.0005 m2 K/WTherefore, fouling is entirely irrelevant to domestic heat loads.Example 2.13Since the engineer did not fail you in the preceding calculation, youentrust him with the installation of a heat exchanger at your plant.He installs a water-cooled steam condenser with U = 4000 W/m2 K.You discover that he used water-side fouling resistance for distilledwater but that the water flowing in the tubes is not clear at all. Howdid he do this time?Solution.
Equation (2.36) and Table 2.3 give1Ucorrected=1+ (0.0006 to 0.0020)4000= 0.00085 to 0.00225 m2 K/WThus, U is reduced from 4,000 to between 444 and 1,176 W/m2 K.Fouling is crucial in this case, and the engineer was in serious error.8586Chapter 2: Heat conduction, thermal resistance, and the overall heat transfer coefficient2.5SummaryFour things have been done in this chapter:• The heat diffusion equation has been established. A method hasbeen established for solving it in simple problems, and some important results have been presented.
(We say much more aboutsolving the heat diffusion equation in Part II of this book.)• We have explored the electric analogy to steady heat flow, payingspecial attention to the concept of thermal resistance. We exploitedthe analogy to solve heat transfer problems in the same way wesolve electrical circuit problems.• The overall heat transfer coefficient has been defined, and we haveseen how to build it up out of component resistances.• Some practical problems encountered in the evaluation of overallheat transfer coefficients have been discussed.Three very important things have not been considered in Chapter 2:• In all evaluations of U that involve values of h, we have taken thesevalues as given information. In any real situation, we must determine correct values of h for the specific situation. Part III deals withsuch determinations.• When fluids flow through heat exchangers, they give up or gainenergy.
Thus, the driving temperature difference varies throughthe exchanger. (Problem 2.14 asks you to consider this difficultyin its simplest form.) Accordingly, the design of an exchanger iscomplicated. We deal with this problem in Chapter 3.• The heat transfer coefficients themselves vary with position insidemany types of heat exchangers, causing U to be position-dependent.Problems2.1Prove that if k varies linearly with T in a slab, and if heat transfer is one-dimensional and steady, then q may be evaluatedprecisely using k evaluated at the mean temperature in theslab.Problems872.2Invent a numerical method for calculating the steady heat fluxthrough a plane wall when k(T ) is an arbitrary function. Usethe method to predict q in an iron slab 1 cm thick if the temperature varies from −100◦ C on the left to 400◦ C on the right.How far would you have erred if you had taken kaverage =(kleft + kright )/2?2.3The steady heat flux at one side of a slab is a known value qo .The thermal conductivity varies with temperature in the slab,and the variation can be expressed with a power series ask=i=n$Ai T ii=0(a) Start with eqn.
(2.10) and derive an equation that relatesT to position in the slab, x. (b) Calculate the heat flux at anyposition in the wall from this expression using Fourier’s law.Is the resulting q a function of x?2.4Combine Fick’s law with the principle of conservation of mass(of the dilute species) in such a way as to eliminate j1 , andobtain a second-order differential equation in m1 . Discuss theimportance and the use of the result.2.5Solve for the temperature distribution in a thick-walled pipeif the bulk interior temperature and the exterior air temperature, T∞i , and T∞o , are known. The interior and the exteriorheat transfer coefficients are hi and ho , respectively. Followthe method in Example 2.6 and put your result in the dimensionless form:T − T ∞i= fn (Bii , Bio , r /ri , ro /ri )T∞i − T∞o2.6Put the boundary conditions from Problem 2.5 into dimensionless form so that the Biot numbers appear in them.
Let the Biotnumbers approach infinity. This should get you back to theboundary conditions for Example 2.5. Therefore, the solutionthat you obtain in Problem 2.5 should reduce to the solution ofExample 2.5 when the Biot numbers approach infinity. Showthat this is the case.88Chapter 2: Heat conduction, thermal resistance, and the overall heat transfer coefficientFigure 2.22 Configuration forProblem 2.8.2.7Write an accurate explanation of the idea of critical radius ofinsulation that your kid brother or sister, who is still in gradeschool, could understand. (If you do not have an available kid,borrow one to see if your explanation really works.)2.8The slab shown in Fig. 2.22 is embedded on five sides in insulating materials.
The sixth side is exposed to an ambient temperature through a heat transfer coefficient. Heat is generatedin the slab at the rate of 1.0 kW/m3 The thermal conductivityof the slab is 0.2 W/m·K. (a) Solve for the temperature distribution in the slab, noting any assumptions you must make. Becareful to clearly identify the boundary conditions. (b) Evaluate T at the front and back faces of the slab. (c) Show that yoursolution gives the expected heat fluxes at the back and frontfaces.2.9Consider the composite wall shown in Fig. 2.23. The concreteand brick sections are of equal thickness. Determine T1 , T2 ,q, and the percentage of q that flows through the brick. Todo this, approximate the heat flow as one-dimensional.
Drawthe thermal circuit for the wall and identify all four resistancesbefore you begin.2.10Compute Q and U for Example 2.11 if the wall is 0.3 m thick.Five (each) pine and sawdust layers are 5 and 8 cm thick, re-Problems89spectively; and the heat transfer coefficients are 10 on the leftand 18 on the right. T∞1 = 30◦ C and T∞r = 10◦ C.2.11Compute U for the slab in Example 1.2.2.12Consider the tea kettle in Example 2.10.
Suppose that the kettle holds 1 kg of water (about 1 liter) and that the flame impinges on 0.02 m2 of the bottom. (a) Find out how fast the water temperature is increasing when it reaches its boiling point,and calculate the temperature of the bottom of the kettle immediately below the water if the gases from the flame are at500◦ C when they touch the bottom of the kettle.
Assume thatthe heat capacitance of the aluminum kettle is negligible. (b)There is an old parlor trick in which one puts a paper cup ofwater over an open flame and boils the water without burningthe paper (see Experiment 2.1). Explain this using an electricalanalogy. [(a): dT /dt = 0.37◦ C/s.]2.13Copper plates 2 mm and 3 mm in thickness are processedrather lightly together. Non-oil-bearing steam condenses under pressure at Tsat = 200◦ C on one side (h = 12, 000 W/m2 K)and methanol boils under pressure at 130◦ Con the other (h =9000 W/m2 K).
Estimate U and q initially and after extendedservice. List the relevant thermal resistances in order of decreasing importance and suggest whether or not any of themcan be ignored.2.140.5 kg/s of air at 20◦ C moves along a channel that is 1 m fromwall to wall. One wall of the channel is a heat exchange surfaceFigure 2.23 Configuration forProblem 2.9.90Chapter 2: Heat conduction, thermal resistance, and the overall heat transfer coefficient(U = 300 W/m2 K) with steam condensing at 120◦ C on its back.Determine (a) q at the entrance; (b) the rate of increase of temperature of the fluid with x at the entrance; (c) the temperatureand heat flux 2 m downstream. [(c): T2m = 89.7◦ C.]2.15An isothermal sphere 3 cm in diameter is kept at 80◦ C in alarge clay region. The temperature of the clay far from thesphere is kept at 10◦ C.
How much heat must be supplied tothe sphere to maintain its temperature if kclay = 1.28 W/m·K?(Hint: You must solve the boundary value problem not in thesphere but in the clay surrounding it.) [Q = 16.9 W.]2.16Is it possible to increase the heat transfer from a convectivelycooled isothermal sphere by adding insulation? Explain fully.2.17A wall consists of layers of metals and plastic with heat transfer coefficients on either side. U is 255 W/m2 K and the overalltemperature difference is 200◦ C. One layer in the wall is stainless steel (k = 18 W/m·K) 3 mm thick. What is ∆T across thestainless steel?2.18A 1% carbon-steel sphere 20 cm in diameter is kept at 250◦ C onthe outside.
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