1、材料科学基础课后习题答案13Solutions for Chapter 131. FIND: Cite 5 applications where thermal properties are key. SOLUTION: Here are but a few of the many applications where thermal properties are critical: gaskets in space ships particularly near liquid oxygen; oven-to-freezer dishes; insulation for homes and c
2、oats; solar storage medium; bi-metallic strip; and a catalytic converter support.2. FIND: Calculate the energy of a photon. GIVEN: Its wavelength is 1 micrometer. DATA: Planks constant, h = 6.63 x 10-33 J-s. The speed of light, c = 3 x 108 m/s. SOLUTION: The equation that is used to calculate the en
3、ergy of a wave is:E = hc / = (6.63 x 10-33 J-s)(3 x 108 m/s) / 10-6 m = 2.0 x 10-18 J.3. Find: Verify that xv3xth. Data: Equation 13.2-3 v/vo = vT Equation 13.2-1 th = th/T Solution: vT = v/vo = (Lo+L)3-Lo3)/Lo3 = (3Lo2L+3LoL2+L3)/Lo3 We neglect the terms of order L2 and L3 because they are very sma
4、ll compared to the term 3Lo2L, so vT (3Lo2L)/Lo3 = 3(L/Lo). Recognizing that th = L/Lo = thT, we can write vT 3 (thT) or v 3th.4. Find: Relationships between bond-energy curves and thermal properties. Solution: A. Since material II has the more asymmetric bond-energy well it will have the larger coe
5、fficient of thermal expansion and, therefore, the larger thermal strain for a given T. B. Since material I has the deeper bond-energy curve it will have a higher melting temperature.5. FIND: Do ceramics have low thermal expansion coefficients because they have open structures? DATA: A look at the li
6、st of thermal expansion coefficients of metals in Table 13.2-1 shows a range of 5 to 83 x 10-6/C. Ceramics range from 0.5 to 14 x 10-6/C. In general, the polymers are higher than the metals or ceramics. SOLUTION: The huge range of thermal expansion coefficient in metals, many being less than that of
7、 MgO, and the fact that polymers are perhaps the most open structures suggest the argument is tentative.6. FIND: Why do polymer fibers generally have negative axial thermal expansion coefficients? SOLUTION: Polymer molecules want to coil on themselves. In textile fibers, and even more so in high per
8、formance fibers, the molecules are stretched out along the fiber axis. When the polymer is heated the molecules gain some mobility, they seek to lower their energy state by shrinking to move towards coiling.7. Find: Maximum T for v 3th to give (TS). The reason for this trend is, of course, the shape
9、s of the respective bond-energy curves. Deep curves, such as those for ceramics, tend to be symmetric while shallow curves, such as those for thermoplastic polymers, are very asymmetric.13. Find: Which has a higher melting temperature: Fe or W? Data: From Table 13.2-1: (Fe) = 12x10-6 oC-1 and (W) =
10、5x10-6 oC-1 Solution: The lower value for W suggests that this metal has a comparatively symmetric bond-energy well. Since such wells are usually deep (indicating high Tm) and display high curvature (indicating high E) we suspect that W has a higher Tm and a higher E than Fe. Comment: Tm(Fe) = 1535o
11、C while Tm(W) = 3410oC E(Fe) 170 Gpa while E(W) = 408 Gpa 14. FIND: Calculate the sag in the wire. GIVEN: The sag at -28C, the minimum temperature, is 20 feet. The maximum allowable sag is significantly less than 75 feet. The maximum temperature is 35C. the pole span is 1000 ft.ASSUMPTIONS: Creep is
12、 not accounted for. APPROXIMATIONS: The shape is catenary. I dont know the equation of a catenary. I will therefore approximate (crudely) the shape as a straight line from the pole to the midpoint. At the lowest temperature, it is 500 feet across and 20 feet down to the lowest point. Perhaps you did
13、 the problem more accurately. DATA: The thermal expansion coefficient of Cu is 17 x 10-6/C, according to Table 13.2-1. The thermal expansion coefficient of Kevlar is -4 x 10-6/C (S.B. Warner, Fiber Science, Prentice-Hall, 1995, p.241). SKETCH: SOLUTION: Lets just probe the problem without trying to
14、get accurate values. We begin by using the Pythagorean theorem to determine the length of the wire that enables a sag of 20 ft as per the sketch:5002 + 202 = length2 length, l = 500.4 ft. Only a 0.4 ft. = 5 inch growth causes the deflection shown in the figure! (The catenary is more forgiving than i
15、s the triangle shape.) Next we use the equation: = l / l = th T.Here, T = 35C - (-28C) = 63C and l = 500.4 ft, the half-length.Hence, l = 500.4 ft x th x 63C.Substitution for Cu, Kevlar, and glass gives:l (Cu) = 500.4 ft 17 x 10-6/C x 63C = 0.54 ft.l (Kevlar) = 500.4 ft -4 x 10-6/C x 63C = -0.12 ft.
16、l (glass) = 500.4 ft 0.55 x 10-6/C x 63C = 0.005 ft.Using the Pythagorean theorem again:500.942 - 5002 = sag2 sag = 31 ft with copperThe sag with Kevlar alone would be reduced, but the glass wants to increase the sag slightly. The net change in length with the increased temperature, which is difficu
17、lt to compute, would be small.15. Find: An example of a material that contracts when heated. Solution: As discussed in Section 6.6.5, elastomers contract when heated.16. Find: Influence of amorphous/crystalline structure on the coefficient of thermal expansion. Solution: Since amorphous structure ar
18、e more open than their crystalline counterparts they are able to accommodate some of the temperature induced thermal vibrations without expansion. Thus, for TTg, however, (amorphous) is approximately equal to that of the corresponding liquid and is therefore about three times larger than (crystallin
19、e).17. Find: What T is required to increase the room temperature volume of Si by 1%? What about a decrease of 1%? Data: From Table 13.2-1: (Si) = 2.6x10-6 oC-1 From Eqn. 13.2-2: V/Vo = vT From Eqn. 13.2-3: v 3 Solution: v(Si) 3( (Si) = 3(2.6x10-6 oC-1) = 7.8x10-6 oC-1 For a 1% expansion: V/Vo = 0.01
20、 T = 0.01/v = 0.01/7.8x10-6 oC-1, therefore T = 1282oC T = 1282oC+25oC = 1307 oC. For a 1% contraction: T =- 0.01/v = 0.01/7.8x10-6 oC-1= -1282oC. Comment: The 1% volume contraction is not possible since the required temperature is below absolute zero.18. Find: Should you use hot or cold water to re
21、move a metal lid from a glass jar? Solution: Since metals generally have larger expansion coefficients than (oxide) glasses, the dimensions of the lid will change more than the dimensions of the jar for any temperature change. If cold water is used the lid will contract more than the jar and the fit will be tighter. If, however, hot water is used then the lid expands more than the jar and it will be easier to remove.19. Find: Influence of material type on accuracy of a measuring tape. Solution:
