Hutton - Fundamentals of Finite Element Analysis (523155), страница 54
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The actual heat transfer is less than qmax , so weHutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 2004Problemsdefine efficiency as=7.10q actq maxUse this definition to determine the efficiency of the pin of Problem 7.5.Figure P7.10 represents one tube of an automotive engine’s radiator. Theengine coolant is circulated through the tube at a constant rate determined bythe water pump. Cooling is primarily via convection from flowing air aroundthe tube as a result of vehicle motion. Coolant enters the tube at a temperatureof 195◦F and the flow rate is 0.3 gallons per minute (specific weight is68.5 lb/ft3). The physical data are as follows: L = 18 in., d = 0.3125 in.,k x = 225 Btu/hr-ft-◦F, h = 37 Btu/ft2-hr-◦F.
Determine the stiffness matrixand load vector for an element of arbitrary length to use in modeling thisproblem, assuming steady-state conditions. Is the assumption of steady-stateconditions reasonable?mⴢTair 20 CLt 0.03 in.dFigure P7.107.117.12Use the results of Problem 7.10 to model the tube with four equal lengthelements and determine the nodal temperatures and the total heat flow fromsurface convection.Consider the tapered heat transfer pin shown in Figure P7.12a. The base of thepin is held at constant temperature Tb , while the lateral surfaces and tip aresurrounded by a fluid media held at constant temperature Ta . Conductance k xand the convection coefficient h are known constants.
Figure P7.12b shows thepin modeled as four tapered finite elements. Figure P7.12c shows the pinmodeled as four constant cross-section elements with the area of each elementequal to the average area of the actual pin. What are the pros and cons of the twomodeling approaches? Keep in mind that use of four elements is only a starting287Hutton: Fundamentals ofFinite Element Analysis2887. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferpoint, many more elements are required to obtain a convergent solution.
Howdoes the previous statement affect your answer?h, TaTbh, Ta(a)1234545(b)123(c)Figure P7.127.13A cylindrical pin is constructed of a material for which the thermal conductivitydecreases with temperature according tok x = k 0 − cTand c is a positive constant. For this situation, show that the governing equationfor steady-state, one-dimensional conduction with convection isk07.14dd2 T−dx 2dxcTdTdx+Q=hP(T − Ta )AThe governing equation for the situation described in Problem 7.13 is nonlinear.If the Galerkin finite element method is applied, an integral of the formx 2NiddxcTdTdxdxx17.15appears in the conductance matrix formulation. Integrate this term by parts anddiscuss the results in terms of boundary flux conditions and the conductancematrix.A vertical wall of “sandwich” construction shown in Figure P7.15a is held atconstant temperature T1 = 68 ◦ F on one surface and T2 = 28 ◦ F on the othersurface.
Using only three elements (one for each material), as in Figure P7.15b,determine the nodal temperatures and heat flux through the wall per unit area.The dimensions of the wall in the y and z directions are very large in comparisonto wall thickness.Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 2004ProblemsT1abc5兾8 in.2 in.5兾8 in.T2x(a)1123432(b)Figure P7.157.16The wall of Problem 7.15 carries a centrally located electrical cable, which isto be treated as a line heat source of strength Q ∗ = constant , as shown inFigure P7.16.
With this change, can the problem still be treated as onedimensional? If your answer is yes, solve the problem using three elementsas in Problem 7.15. If your answer is no, explain.Q*Figure P7.167.17A common situation in polymer processing is depicted in Figure P7.17, whichshows a “jacketed” pipe. The inner pipe is stainless steel having thermalconductivity k; the outer pipe is carbon steel and assumed to be perfectlyinsulated. The annular region between the pipes contains a heat transfer mediumat constant temperature T1 . The inner pipe contains polymer material flowingCarbon steelT1tStainless steelFlowRxFigure P7.17289Hutton: Fundamentals ofFinite Element Analysis2907. Applications in HeatTransferCHAPTER 77.18Text© The McGraw−HillCompanies, 2004Applications in Heat Transferat constant mass flow rate ṁ .
The convection coefficient between the stainlesssteel pipe wall and the polymer is h. Polymer specific heat c is also taken to beconstant. Is this a one-dimensional problem? How would you solve this problemusing the finite element method?An office heater (often incorrectly called a radiator, since the heat transfer modeis convection) is composed of a central pipe containing heated water at constanttemperature, as depicted in Figure P7.18. Several two-dimensional heat transferfins are attached to the pipe as shown. The fins are equally spaced along thelength of the pipe. Each fin has thickness of 0.125 in. and overall dimensions4 in. × 4 in.
Convection from the edges of the fins can be neglected. Consider thepipe as a point source Q ∗ = 600 Btu/hr-ft2 and determine the net heat transfer tothe ambient air at 20◦C, if the convection coefficient is h = 300 Btu/(hr-ft2-◦F).Use four finite elements with linear interpolation functions (consider symmetryconditions here).Figure P7.187.197.20One who seriously considers the symmetry conditions of Problem 7.18 wouldrealize that quarter symmetry exists and four elements represent 16 elements inthe full problem domain. What are the boundary conditions for the symmetricmodel?The rectangular fins shown in Figure P7.20 are mounted on a centrally locatedpipe carrying hot water. Temperature at the contact surface between fin andpipe is a constant T1 .
For each case depicted, determine the applicable symmetryconditions and the boundary conditions applicable to a finite element model.T1TaTa(a)Figure P7.20TaTaTaT2T2TaT1(b)(c)(d)Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 2004Problems7.217.22Note that the cross-hatched edges are perfectly insulated and it is assumed thatthe convection coefficients on all surfaces is the same constant.Solve Example 7.5 using the two-element model in Figure 7.11b. How do theresults compare to those of the four-element model?The rectangular element shown in Figure P7.22 contains a line source of constant(e)strength Q ∗ located at the element centroid.
Determine { f Q }.4(2, 4)3(4, 4)Q*1(2, 2)y2(4, 2)xFigure P7.227.237.24(e)Determine the forcing function components of { f Q } for the axisymmetricelement of Example 7.10 for the case of uniform internal heat generation Q.A steel pipe (outside diameter of 60 mm and wall thickness of 5 mm) containsflowing water at constant temperature 80 ◦ C , as shown in Figure P7.24. Theconvection coefficient between the water and pipe is 2000 W/m2-◦C.
The pipeis surrounded by air at 20◦C, and the convection at the outer pipe surface is20 W/m2-◦C. The thermal conductivity of the pipe material is 60 W/m-◦C.Determine the stiffness matrices and nodal forcing functions for the twoaxisymmetric elements shown.Flow122.5 mm2.5 mmzrFigure P7.24291Hutton: Fundamentals ofFinite Element Analysis2927.
Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transfer7.25 Assemble the global equations for the two elements of Problem 7.24. Use theglobal node numbers shown in Figure P7.25. Compute the nodal temperatures,and find the net heat flow (per unit surface area) into the surrounding air.45116223Figure P7.257.26 Solve the problem of Example 7.11 using a time step t = 0.01 sec. How do theresults compare to those of the example solution?Hutton: Fundamentals ofFinite Element Analysis8.
Applications in FluidMechanicsText© The McGraw−HillCompanies, 2004C H A P T E R8Applications in FluidMechanics8.1 INTRODUCTIONThe general topic of fluid mechanics encompasses a wide range of problems ofinterest in engineering applications. The most basic definition of a fluid is to statethat a fluid is a material that conforms to the shape of its container. Thus, bothliquids and gases are fluids. Alternately, it can be stated that a material which,in itself, cannot support shear stresses is a fluid. The reader familiar with the distortion energy theory of solids will recall that geometric distortion is the result ofshear stress while normal stress results in volumetric change.
Thus, a fluid readily distorts, since the resistance to shear is very low, and such distortion resultsin flow.The physical behavior of fluids and gases is very different. The differences inbehavior lead to various subfields in fluid mechanics.
In general, liquids exhibitconstant density and the study of fluid mechanics of liquids is generally referredto as incompressible flow. On the other hand, gases are highly compressible(recall Boyle’s law from elementary physics [1]) and temperature dependent.Therefore, fluid mechanics problems involving gases are classified as cases ofcompressible flow.In addition to considerations of compressibility, the relative degree to whicha fluid can withstand some amount of shear leads to another classification of fluidmechanics problems. (Regardless of the definition, all fluids can support someshear.) The resistance of a fluid to shear is embodied in the material propertyknown as viscosity.
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