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Heat and Mass Transfer Solutions Manual Second Edition Download Full Solution Manual for Heat and Mass Transfer 2nd Edition by Kurt Rolle https://getbooksolutions.com/download/solution-manual-for-heat-and-mass-transfer2nd-edition-by-kurt-rolleThis solutions manual sets down the answers and solutions for the Discussion Questions, Class Quiz Questions, and Practice Problems. There will likely be variations of answers to the discussion questions as well as the class quiz questions. For the practice problems there will likely be some divergence of solutions, depending on the interpretation of the processes, material behaviors, and rigor in the mathematics. It is the author’s responsibility to provide accurate and clear answers. If you find errors please let the author know of them at rolle@uwplatt.edu.Chapter 2 Discussion Questions Section 2-1 1. Describe the physical significance of thermal conductivity. Thermal conductivity is a parameter or coefficient used to quantitatively describe the amount of conduction heat transfer occurring across a unit area of a bounding surface, driven by a temperature gradient. 2. Why is thermal conductivity affected by temperature? Conduction heat transfer seems to be the mechanism of energy transfer between adjacent molecules or atoms and the effectiveness of these transfers is strongly dependent on the temperatures. Thus, to quantify conduction heat transfer with thermal conductivity means that thermal conductivity is strongly affected by temperature. 3. Why is thermal conductivity not affected to a significant extent by material density? Thermal conductivity seems to not be strongly dependent on the material density since thermal conductivity is an index of heat or energy transfer between adjacent molecules and while the distance separating these molecules is dependent on density, it is not strongly so. Section 2-2 4. Why is heat of vaporization, heat of fusion, and heat of sublimation accounted as energy generation in the usual derivation of energy balance equations? Heats of vaporization, fusion, and sublimation are energy measures accounting for phase changes and not directly to temperature or pressure changes. It is17 Š 2016 Cengage LearningÂŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.convenient, therefore, to account these phase change energies as lumped terms, or energy generation. Section 2-3 5. Why are heat transfers and electrical conduction similar? Heat transfer and electrical conduction both are viewed as exchanges of energy between adjacent moles or atoms, so that they are similar. 6. Describe the difference among thermal resistance, thermal conductivity, thermal resistivity and R-Values. Thermal Resistance is the distance over which conduction heat transfer occurs times the inverse of the area across which conduction occurs and the thermal conductivity, and thermal resistivity is the distance over which conduction occurs times the inverse of the thermal conductivity. The R-Value is the same as thermal resistivity, with the stipulation that in countries using the English unit system, 1 R-Value is 1 hr∙ft2 ∙0F per Btu. Section 2-4 7. Why do solutions for temperature distributions in heat conduction problems need to converge? Converge is a mathematical term used to describe the situation where an answer approaches a unique, particular value. 8. Why is the conduction in a fin not able to be determined for the case where the base temperature is constant, as in Figure 2-9? The fin is an extension of a surface and at the edges where the fin surface coincides with the base, it is possible that two different temperatures can be ascribed at the intersection, which means there is no way to determine precisely what that temperature is. Conduction heat transfer can then not be completely determined at the base. 9. What is meant by an isotherm? An isotherm is a line or surface of constant or the same temperature. 10. What is meant by a heat flow line? A heat flow line is a path of conduction heat transfer. Conduction cannot cross a heat flow line. Section 2-5 11. What is a shape factor? The shape factor is an approximate, or exact, incorporating the area, heat flow paths, isotherms, and any geometric shapes that can be used to quantify conduction heat flow between two isothermal surfaces through a heat conducting media. The product of the shape factor, thermal conductivity, and temperature difference of the two surfaces predicts the heat flow.18 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.12. Why should isotherms and heat flow lines be orthogonal or perpendicular to each other? Heat flow occurs because of a temperature difference and isotherms have no temperature difference. Thus heat cannot flow along isotherms, but must be perpendicular or orthogonal to isotherms. Section 2-6 13. Can you identify a physical situation when the partial derivatives from the left and right are not the same? Often at a boundary between two different conduction materials the left and the right gradients could be different. Another situation could be if radiation or convection heat transfer occurs at a boundary and then again the left and right gradients or derivatives could be different. Section 2-7 14. Can you explain when fins may not be advantageous in increasing the heat transfer at a surface? Fins may not be a good solution to situations where a highly corrosive, extremely turbulent, or fluid having many suspended particles is in contact with the surface. 15. Why should thermal contact resistance be of concern to engineers? Thermal contact resistance inhibits good heat transfer, can mean a significant change in temperature at a surface of conduction heat transfer, and can provide a surface for potential corrosion.Class Quiz Questions 1. What is the purpose of the negative sign in Fourier’s law of conduction heat transfer? The negative sign provides for assigning a positive heat transfer for negative temperature gradients. 2. If a particular 8 inch thick material has a thermal conductivity of 10 Btu/ hr∙ft∙0F, what is its R-value? The R-value is the thickness times the inverse thermal conductivity; R − Value = thicks / κ = 8in / (12in/ ft)(10Bu/ hr ⋅ ft ⋅0 F) = 0.0833hr ⋅ ft ⋅ 0 F / Btu 3. What is the thermal resistance of a 10 m2 insulation board, 30 cm thick, and having thermal conductivity of 0.03 W/m∙K? The thermal resistance isx / A ⋅ κ = 0.3m  / 10 m 2 0.03W / m ⋅ K  = 1.0 K / W4. What is the difference between heat conduction in series and in parallel between two materials?19 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.The thermal resistance, or thermal resistivity are additive for series. In parallel the thermal resistance needs to be determined with the relationshipReq=R 1 R2  / R1 + R2 5. Write the conduction equation for radial heat flow of heat through a tube that has inside diameter of Di and outside diameter of D0. iiTQ=2πκ L ln D0 Di  6. Write the Laplace equation for two-dimensional conduction heat transfer through a homogeneous, isotropic material that has constant thermal conductivity.∂ 2T (x, y) + ∂2T(x, y) = 0 ∂x 2 ∂y2 7. Estimate the heat transfer from an object at 1000F to a surface at 400F through a heat conducting media having thermal conductivity of 5 Btu/hr∙ft∙0F if the shape factor is 1.0 ft.Qi = S κ T = 1 .0 ft   5 Btu / hr ⋅ ft ⋅0 F  60 0 F  = 300 Btu / hr 8. Sketch five isotherms and appropriate heat flow lines for heat transfer per unit depth through a 5 cm x 5 cm square where the heat flow is from a high temperature corner and another isothermal as the side of the square.9. If the thermal contact resistance between a clutch surface and a driving surface is 0.0023 m2 -0C/W, estimate the temperature drop across the contacting surfaces, per unit area when 200 W/m2 of heat is desired to be dissipated. The temperature drop is iT = QRTCR =  200W / m 2  0.0023m 2 ⋅0 C / W  = 0.460 C 10. Would you expect the wire temperature to be greater or less for a number 18 copper wire as compared to a number 14 copper wire, both conducting the same electrical current?20 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.A number 18 copper wire has a smaller diameter and a greater electrical resistance per unit length. Therefore the number 18 wire would be expected to have a higher temperature than the number 14 wire.Practice Problems Section 2-1 1. Compare the value for thermal conductivity of Helium at 200C using Equation 2-3 and the value from Appendix Table B-4.Solution Using Equation 2-3 for heliumκ = 0.8762 x10 −4 T = 0.0015W / c m ⋅ K = 0.15W / m ⋅ Kκ = 0.152W / m ⋅ KFrom Appendix Table B-42. Predict the thermal conductivity for neon gas at 2000F. Use a value of 3.9 Ǻ for the collision diameter for neon. Solution Assuming neon behaves as an ideal gas, with MW of 20, converting 2000F to 367K, and using Equation 2-1κ = 8.328 x10 −4= 8.328 x10 − 4 367K  20  3.9 MW ⋅ Γ T= 18.05 x10 −4 W / cm ⋅ K3. Show that thermal conductivity is proportional to temperature to the 1/6-th power for a liquid according to Bridgeman’s equation (2-6).21 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Solution From Bridgeman’s equation κ = 3.865 x10 that-1/2~ρ−23Vsxm2 Also, Vs (sonic velocity) ~ - ⁄κ ∼ ρ−2/3+1/2 = ρ−1/6 ∼T1/6the mean separation distance between molecules= ⁄ 2/3/~ρso4. Predict a value for thermal conductivity of liquid ethyl alcohol at 300 K. Use the equation suggested by Bridgman’s equation (2-6).Solution Bridgeman’s equation (2-6) uses the sonic velocity in the liquid,⁄ , which for ethylalcohol at 300 K is nearly 1.14 x 105 cm/s from Table 2-2. The equation also uses the mean distance between molecules, assuming a uniform cubic arrangement of the molecules, which is⁄ , mm being the mass of one molecule in grams, the molecular mass divided by Avogadro’s number. Using data from a chemistry handbook the value of xm is nearly 0.459 x 10-7 cm. Using Equation 2-6,κ = 3.865 x10 − 23 Vs xm2  = 20.9 x10 −4 W / c m ⋅ K = 0.209W / m ⋅ K5. Plot the value for thermal conductivity of copper as a function of temperature as given by Equation 2-10. Plot the values over a range of temperatures from -400F to 1600F.Solution Using Equation 2-10 and coefficients from Appendix Table B-8Eκ = κ TO + α  T − T0  = 227Btu 0Btu− 0.0061hr ⋅ ft ⋅ R02 T − 4920 Rhr ⋅ ft ⋅ RThis can be plotted on a spreadsheet or other modes.6. Estimate the thermal conductivity of platinum at -1000C if its electrical conductivity is 6 x 107 mhos/m, based on the Wiedemann-Franz law. Note: 1 mho = 1 amp/volt = 1 coulomb/volt-s, 1 W = 1 J/s = 1 volt-coulomb/s.22 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Solution Using the Wiedemann-Franz law, Equation 2-9 gives κ = Lz ⋅ T = 2.43 x10 − 8 V 2 K 2 6 x10 7 amp V ⋅ m  173 K  = 252.2W / m ⋅ K7. Calculate the thermal conductivity of carbon bisulfide using Equation 2-6 and compare this result to the listed value in Table 2-2.Solution Equation 2-6 uses the sonic velocity in the material. This is =⁄ = 1.1810/ , where Eb is the bulk modulus. The mean distance between adjacent molecules, assuming a uniform cubic arrangement, is also used. This is =/ where mm is the mass of one molecule; MW/Avogadro’s number. − 23Vs0This gives = 0.46610. thenκ = 3.865 x10xm2 = 0.0021W / cm ⋅ CSection 2-2 8. Estimate the temperature distribution in a stainless steel rod, 1 inch in diameter that is 1 yard long with 3 inches of one end submerged in water at 400F and the other end held by a person. Assume the person’s skin temperature is 820F, the temperature in the rod is uniform at any point in the rod, and steady state conditions are present.23 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Solution Assuming the heat flow to be axial and not radial and also 400F for the first 3 inches of the rod, the temperature distribution between x = 3 inches and out to x = 36 inches we can use Fourier’s law of conduction and then for 3in ≤ x ≤ 36 inches, identifying theT(x)  1.2727 x  36.1818slope and x-intercept . The sketched graph is here included. One could now predict the heat flow axially through the rod, using Fourier’s law and using a thermal conductivity for stainless steel.9. Derive the general energy equation for conduction heat transfer through a homogeneous, isotropic media in cylindrical coordinates, Equation 2-19.Solution Referring to the cylindrical element sketch, you can apply an energy balance, Energy in – Energy Out = Energy Accumulated in the Element. Then, accounting the energies in and out as conduction heat transfer we can write24 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.i−κ r zqθrr iq θ ir ran in energy1 T −κ rz −κr θ θ T r  θ rθqTrzzz2i −κ  rqqT rzθ 1 Ti −κ rθ θr r +Δrr θ θ +Δθθ +Δθr −κ r rρ r 2θ⋅θr 2z +Δzz ⋅ Δr  cpan out energyziz zan in energyzr r +Δrqan in energyTan out energy Txz +Δzan out energytThe rate of energy accumulated in the element. If you put the three energy in terms and the three out terms on the left side of the energy balance and the accumulated energy on the right, divide all terms by - + -⁄2 !" ∙ !$ ∙ !, and take the limits as Δr →0, Δz → 0, and Δθ→ 0 gives, using calculus, Equation 2-19 T 1 κ r T  1  κ T κ  ρc T p θ z z t r r r r 2 θ25 Š 2016 Cengage LearningÂŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.10. Derive the general energy equation for conduction heat transfer through a homogeneous, isotropic media in spherical coordinates, Equation 2-20.Solution Referring to the sketch of an element for conduction heat transfer in spherical coordinates, you can balance the energy in – the energy out equal to the energy accumulated in the element. Using Fourier’s law of conduction iθ r sinθ φ −κ rQrTr i−κQ θθ iQ1r⋅ r r sinθθrφφ Tr sinθ ian in termθ θ 1 T−κ rφan in termrran in termφφ −κ  r  r  θ  r  r  sinθ φQ r rr +Δri −κQ θ θθ +ΔθT1Trr +Δran out termr sinθ φ r θ rθ θ +Δθan out term26 Š 2016 Cengage LearningÂŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Qφ +Δφρ Vc p∂Tri= −κ r θ φ +Δφ∂Tr sinθ∂φφ +Δφan out termφ⋅ r= ρ  r sinθ ⋅ rθ ∂T ∂t∂tWhich is the accumulated energy. Inserting the three in terms as positive on the left side of the energy balance, inserting the three out terms as negative on the left side of the balance, inserting the accumulated term on the right side, and dividing all terms by the quantity -%&"!' ∙ !- ∙ !" gives the followingκ  r + r  sin θ θ 2φ∂T− κ  r 2∂r r +Δrrr sinθ r θ ∂Tκ r sin θ φκrθrθ∂T∂T ∂θ θφ∂T −κ θ r r sin θ r sinθ ∂φ φ = ρc ∂T φ +Δφ 2 p r sinθ r θ φ ∂t∂φ∂T∂r r+φ− κ r sinθ φ θ∂θ θ +Δθ r 2 sinθ r θsinθ θ φ+27 Š 2016 Cengage LearningÂŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Taking the limits as Δr →0, Δθ →0, Δφ → 0 and reducing1∂ 2∂rrκr2∂T+∂r r1 2∂ κ sinθsin θ ∂θ∂T+∂1 2∂θ r sin2θ ∂φκ∂T= ρcp∂φ∂T∂twhichis Equation 2-20, conservation of energy for conduction heat transfer in spherical coordinates.11. Determine a relationship for the volume element in spherical coordinates. Solution Referring to the sketch for an element in spherical coordinates, and guided by the concept of a volume element gives,V = r sinθ φ  ⋅ r  ⋅  r θ 28 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Section 2-3 12. An ice-storage facility uses sawdust as an insulator. If the outside walls are 2 feet thick sawdust and the sideboard thermal conductivity is neglected, determine the R-Value of the walls. Then, if the inside temperature is 250F and the outside is 850F, estimate the heat gain of the storage facility per square foot of outside wall. Solution Assuming steady state conditions and that the thermal conductivity is the value listed in Appendix Table B-2E,x2 ft R − Value = κ = 0.034 Btu / hr ⋅ ft ⋅0 F = 58.8 hr ⋅ ft 2 ⋅0 F / Btu = 58.8R − Value Tiq=κ A85 0 F − 250 FT= x= R − Value= 1.02 Btu / hr ⋅ ft22058.8 hr ⋅ ft⋅F / Btu13. The combustion chamber of an internal combustion engine is at 8000C when fuel is burnt in the combustion chamber. If the engine is made of cast iron with an average thickness of 6.4 cm between the combustion chamber and the outside surface, estimatethe heat transfer per unit area if the outside surface temperature is 500C and the outside air temperature is 300C. Solution Assuming steady state one-dimensional conduction and using a thermal conductivity that is assumed constant and has a value from Table B-2, iT qA κx39W 800 0 − 500 K = 450 kW / m m⋅ K 0.065m214. Triple pane window glass has been used in some building construction. Triple pane glass is a set of three glass panels, each separated by a sealed air gap. Estimate the R-Value for triple pane windows and compare this to the R-Value for single pane glass. Solution Assume the air in the gaps do not move so that they are essentially conducting media. Then the R-Value is29Š 2016 Cengage LearningÂŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.xR − Value  3κx2air  3κglass20.0020.0060.4658 m 2 K / W 2.647 hr ft 2 0 F / Btu ⋅ ⋅ ⋅ 0.0261.4The R-Value for a single pane window isxR − Value ==kglass0.002m 1.4W / m ⋅ K2 = 0.1429 m2⋅ K / W = 0.008 hr ⋅ ft0⋅ F / Btu = 0.008R − ValueThe ratio of the R-Value for the triple pane to the R-Value for a single pane is roughly 324.15. For the outside wall shown in Figure 2-50, determine the R-Value, the heat transfer through the wall per unit area and the temperature distribution through the wall if the outside surface temperature is 360C and the inside surface temperature is 150C.Solution The R-Value is the sum of the three materials; pine, plywood, and limestone, with thermal conductivity R − Value = RV =x κx + pineκ0.04x + plywoodk= limestone0.02 +0.150.06 +0.122= 0.462 m2.15values obtained from Appendix Table B-2.The conversion to English units is 0.176mi2K/W = 1 R-Value so that ( − *+,- = 2.62 . The heat transfer per unit area isT = 36 − 15 = 45.45W / m2 0.462 The temperature distribution is determined by V noting that the heat flow is the same through each material. For the pine,q= AR30⋅ K/WŠ 2016 Cengage LearningÂŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.T − 150 Ciq A , pine = 45.45W / m2 =RT −15 =11V .pine0.04 / 0.15so that T1 at the surface between the pine and the plywood, is 27.10C. Similarly, to determine the temperature between the plywood and the limestone, again noting that the heat flow is the same as before iq= 45.45 = T2 − T1ARV , plywood= T2 − 27.100.02 / 0.12so that T2 is 34.6 C. This is sketched in thefigure.16. Determine the heat transfer per foot of length through a copper tube having an outside diameter of 2 inches and an inside diameter of 1.5 inches. The pipe contains 1800F ammonia and is surrounded by 800F air. Solution Assuming steady state and only conduction heat transfer, for a tube cylindrical coordinates is the appropriate means of analysis. Then iq= 2πκ T = l ln  D0 Di 0 2π 231.16 Btu / hr ⋅ ft ⋅ R 180ln  2in 1.5in− 80 R= 504,870 Btu / hr ⋅ ft 017. A steam line i
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