292 8 Thermal Properties M M Stress Martensite Austenite Temperature Figure 8.10.Effect of stress on the start and finish temperatures of the change from austenite to martensite (Ms and M) and the start and finish temperatures of the change from martensite to austenite(As and Ar) A shape memory coupling for connecting two tubes is placed over the joint area in an expanded state,so that it fits over the tubing (Fig.8.12a and b).Upon subsequent heating,it shrinks back to its original diameter,thereby squeezing the tubing and providing a tight fit (Fig.8.12c). Examples of shape memory alloys (abbreviated to SMAs)are copper-zinc- aluminum,copper-aluminum-nickel and nickel-titanium alloys.NiTi with 50 at. Ti is known as nitinol,which stands for nickel titanium Naval Ordnance Labora- tory (the place where it was discovered).NiTi alloys are generally more expensive and exhibit better mechanical properties than copper-based SMAs.Disadvantages Strain Strain Strain Heat Heat Figure 8.11.Manifestations of the shape memory effect,depending on the temperature.a Temperatures below As,as shown in Fig.8.7,b temperatures above Ar,as shown in Fig.3.33a,ctemperatures between As and Af,showing behavior intermediate between those shown in (a)and(b) doriginal dexpanded b dtubing<dexpanded Tubing Heat OD Figure8.12.Theuseoftheshapememoryeffect(Fig.8.7)toachieveacoupling thatfitstightly aroundtubing.aDeformation of the martensite to cause expansion of the coupling,b fitting the expanded coupling around the tubing,c heating to change the martensite back to austenite,thereby restoring the coupling its original,undeformed shape with a smaller diameter
292 8 Thermal Properties Stress Temperature Austenite Martensite Mf Ms As Af Figure 8.10. Effect of stress on the start and finish temperatures of the change from austenite to martensite (Ms and Mf) and the start and finish temperatures of the change from martensite to austenite (As and Af) A shape memory coupling for connecting two tubes is placed over the joint area in an expanded state, so that it fits over the tubing (Fig. 8.12a and b). Upon subsequent heating, it shrinks back to its original diameter, thereby squeezing the tubing and providing a tight fit (Fig. 8.12c). Examples of shape memory alloys (abbreviated to SMAs) are copper-zincaluminum, copper-aluminum-nickel and nickel-titanium alloys. NiTi with 50at.% Ti is known as nitinol, which stands for nickel titanium Naval Ordnance Laboratory (the place where it was discovered). NiTi alloys are generally more expensive and exhibit better mechanical properties than copper-based SMAs. Disadvantages Figure 8.11. Manifestations of the shape memory effect, depending on the temperature. a Temperatures below As, as shown in Fig. 8.7, b temperatures above Af, as shown in Fig. 3.33a, c temperatures between As and Af, showing behavior intermediate between those shown in (a) and (b) Figure8.12. Theuseoftheshapememoryeffect(Fig.8.7)toachieveacouplingthatfitstightlyaroundtubing.aDeformation of the martensite to cause expansion of the coupling, b fitting the expanded coupling around the tubing, c heating to change the martensite back to austenite, thereby restoring the coupling its original, undeformed shape with a smaller diameter
8.4 Thermal Conductivity 293 of shape memory alloys include high cost,poor machinability and poor fatigue resistance compared to steel. 8.3.3 Calorimetry Investigations of phase transitions are commonly conducted by calorimetry,which allows the transition temperature to be measured during heating and cooling,and the latent heat of the transformation during heating and cooling.A first-order transition gives a peak in the heat flow vs.temperature plot such that the area under the peak is the latent heat of the transformation.In contrast,a second-order transition does not give a peak,just a change in slope in this plot,due to the absence of latent heat. One widely used form of calorimetry is differential scanning calorimetry(DSC), which involves scanning the temperature at a controlled heating/cooling rate, while the temperature of the specimen is kept the same as a reference(typically an empty pan).At a first-order phase transition,the absorption or release of latent heat from the specimen means that the calorimeter must add heat to or remove heat from the specimen in order to maintain the specimen and the reference at the same temperature as the temperature is scanned.The amount of heat applied or removed is the latent heat,which is therefore measured. An alternative form of DSC involves measuring the temperature difference be- tween the specimen and the reference when the two are subjected to the same constant heating/cooling rate.The higher the heat capacity of the specimen com- pared to that of the reference,the greater the amount of heat that needs to be applied to the specimen in order to maintain the specimen and reference at the same constant heating/cooling rate.The application of heat gives rise to a small temperature difference between the specimen and the reference. 8.4 Thermal Conductivity The thermal conductivity(k)is a material property that describes the ability of the material to conduct heat.It is defined as △Q/△t=kA(△T/x), (8.22) where Q is the heat (with units of J),t is the time,AQ/At is the heat flow (i.e.,the amount of heat flowing per unit time),A is the cross-sectional area of the material perpendicular to the direction of heat flow,and AT is the temperature difference between the two ends of the material of length x in the direction of heat flow,as illustrated in Fig.8.13,where one-dimensional heat flow is assumed(i.e.,there is no loss of heat to the surrounding).The temperature gradient is AT/x.Equation 8.22 means that the thermal conductivity is the heat flow per unit cross-sectional area per unit temperature gradient;i.e.,the heat flux per unit temperature gradient.The heat flow per unit cross-sectional area is also called the heat flux.Since W=J/s, the units of k are W/(mK),which are the same as W/(mC)
8.4 Thermal Conductivity 293 of shape memory alloys include high cost, poor machinability and poor fatigue resistance compared to steel. 8.3.3 Calorimetry Investigations of phase transitions are commonly conducted by calorimetry, which allows the transition temperature to be measured during heating and cooling, and the latent heat of the transformation during heating and cooling. A first-order transition gives a peak in the heat flow vs. temperature plot such that the area under the peak is the latent heat of the transformation. In contrast, a second-order transition does not give a peak, just a change in slope in this plot, due to the absence of latent heat. One widely used form of calorimetry is differential scanning calorimetry (DSC), which involves scanning the temperature at a controlled heating/cooling rate, while the temperature of the specimen is kept the same as a reference (typically an empty pan). At a first-order phase transition, the absorption or release of latent heat from the specimen means that the calorimeter must add heat to or remove heat from the specimen in order to maintain the specimen and the reference at the same temperature as the temperature is scanned. The amount of heat applied or removed is the latent heat, which is therefore measured. An alternative form of DSC involves measuring the temperature difference between the specimen and the reference when the two are subjected to the same constant heating/cooling rate. The higher the heat capacity of the specimen compared to that of the reference, the greater the amount of heat that needs to be applied to the specimen in order to maintain the specimen and reference at the same constant heating/cooling rate. The application of heat gives rise to a small temperature difference between the specimen and the reference. 8.4 Thermal Conductivity The thermal conductivity (k) is a material property that describes the ability of the material to conduct heat. It is defined as ΔQ/Δt = kA(ΔT/x) , (8.22) where Q is the heat (with units of J), t is the time, ΔQ/Δt is the heat flow (i.e., the amount of heat flowing per unit time), A is the cross-sectional area of the material perpendicular to the direction of heat flow, and ΔT is the temperature difference between the two ends of the material of length x in the direction of heat flow, as illustrated in Fig. 8.13, where one-dimensional heat flow is assumed (i.e., there is no loss of heat to the surrounding). The temperature gradient is ΔT/x. Equation 8.22 means that the thermal conductivity is the heat flow per unit cross-sectional area per unit temperature gradient; i.e., the heat flux per unit temperature gradient. The heat flow per unit cross-sectional area is also called the heat flux. Since W = J/s, the units of k are W/(mK), which are the same as W/(m°C)
294 8 Thermal Properties Area A Tnot Q→ Tcold △T=Thot-Tcold Figure 8.13.Heat flowing from the hot end to the cold end of a material of cross-sectional area A and lengthx Q Tcold Figure 8.14.A thermal conductor in the form of a composite with components labeled 1(dotted regions)and 2(white regions)in the form of strips that are parallel and oriented in the direction of heat flow The thermal conductivity of a composite(ke)can be calculated from the thermal conductivities and volume fractions of each of the components(labeled 1 and 2)in the composite.Consider the case where the components are positioned in parallel, as shown in Fig.8.14.The heat flow in component 1 (all the strips of component 1 together)plus that in component 2(all the strips of component 2 together)equals the heat flow in the overall composite.Hence,based on Eq.8.22, k1A1(△T/x)+k2A2(△T/x)=kA(△T/x), (8.23) where k and k2 are the thermal conductivities of components 1 and 2 respectively, A and A2 are the cross-sectional areas(all the strips ofeach component considered together)of components 1 and 2,respectively,and A is the cross-sectional area of the overall composite.Obviously, A=A1+A2· (8.24) Eq.8.23 can be rewritten as kc =(K1A1/A)+(K2A2/A)=vik1 +v2k2, (8.25) where vi=A1/A and is the volume fraction of component 1,and v2 =A2/Aand is the volume fraction of component 2.Equation 8.25 is the rule of mixtures for the parallel configuration
294 8 Thermal Properties x Thot Q Tcold Area A T = Thot Tcold Figure 8.13. Heat flowing from the hot end to the cold end of a material of cross-sectional area A and length x Figure 8.14. A thermal conductor in the form of a composite with components labeled 1 (dotted regions) and 2 (white regions) in the form of strips that are parallel and oriented in the direction of heat flow The thermal conductivity of a composite (kc) can be calculated from the thermal conductivities and volume fractions of each of the components (labeled 1 and 2) in the composite. Consider the case where the components are positioned in parallel, as shown in Fig. 8.14. The heat flow in component 1 (all the strips of component 1 together) plus that in component 2 (all the strips of component 2 together) equals the heat flow in the overall composite. Hence, based on Eq. 8.22, k1A1(ΔT/x) + k2A2(ΔT/x) = kcA(ΔT/x) , (8.23) where k1 and k2 are the thermal conductivities of components 1 and 2 respectively, A1 andA2 are the cross-sectional areas (all the strips of each component considered together) of components 1 and 2, respectively, and A is the cross-sectional area of the overall composite. Obviously, A = A1 + A2 . (8.24) Eq. 8.23 can be rewritten as kc = (k1A1/A)+(k2A2/A) = v1k1 + v2k2 , (8.25) where v1 = A1/A and is the volume fraction of component 1, and v2 = A2/Aand is the volume fraction of component 2. Equation 8.25 is the rule of mixtures for the parallel configuration
8.4 Thermal Conductivity 295 Q hot cold Figure 8.15.A thermal conductor in the fomm of a composite with components labeled 1(dotted regions)and 2(white regions)in the series configuration;i.e.,the components are in the form ofstrips that are parallel and oriented perpendicular to the direction of heat flow Next,consider the case where the components are in series,as shown in Fig.8.15. The temperature difference between the two ends of the composite is given by △T=△T1+△T2, (8.26) where ATi is the contribution to the temperature difference from component 1(all the strips of component 1 together)and AT2 is the contribution to the temperature difference from component 2(all the strips of component 2 together).Based on Eq.8.22,Eq.8.26 becomes (△Q/△t)[x1/k1A)]+(△Q/△t)[x2/(k2A)]=(△Q/△t)[x/(kA)】, (8.27) assuming that both components have the same cross-sectional area A.Simplifica- tion of Eq.8.27 gives x1/k1+x/k2 =x/kc (8.28) Dividing by x gives y1/k1+V2/k2=1/ke, (8.29) where v=x1/x is the volume fraction of component 1,and v2=x2/x is the volume fraction of component 2.Equation 8.29 is the rule of mixtures for the series configuration. The thermal resistivity is defined as the reciprocal of the thermal conductiv- ity.Thus,the units of thermal resistivity are m K/W.From Eq.8.22,the thermal resistivity 1/k is given by 1/k=[A(△T/x)]V(△Q/△t). (8.30) In other words,the thermal resistivity is the temperature gradient divided by the heat flux. The thermal resistance is defined as the temperature gradient divided by the heat flow(not flux).Hence,it is given by the thermal resistivity divided by the cross-sectional area A: Thermal resistance=l/(kA)=(△T/x)/(△Q/△t). (8.31)
8.4 Thermal Conductivity 295 Figure 8.15. A thermal conductor in the form of a composite with components labeled 1 (dotted regions) and 2 (white regions)intheseriesconfiguration;i.e.,thecomponentsareintheformofstripsthatareparallelandorientedperpendicular to the direction of heat flow Next, consider the case where the components are in series, as shown in Fig. 8.15. The temperature difference between the two ends of the composite is given by ΔT = ΔT1 + ΔT2 , (8.26) where ΔT1 is the contribution to the temperature difference from component 1 (all the strips of component 1 together) and ΔT2 is the contribution to the temperature difference from component 2 (all the strips of component 2 together). Based on Eq. 8.22, Eq. 8.26 becomes (ΔQ/Δt)[x1/(k1A)] + (ΔQ/Δt)[x2/(k2A)] = (ΔQ/Δt)[x/(kcA)] , (8.27) assuming that both components have the same cross-sectional area A. Simplification of Eq. 8.27 gives x1/k1 + x2/k2 = x/kc . (8.28) Dividing by x gives v1/k1 + v2/k2 = 1/kc , (8.29) where v1 = x1/x is the volume fraction of component 1, and v2 = x2/x is the volume fraction of component 2. Equation 8.29 is the rule of mixtures for the series configuration. The thermal resistivity is defined as the reciprocal of the thermal conductivity. Thus, the units of thermal resistivity are mK/W. From Eq. 8.22, the thermal resistivity 1/k is given by 1/k = [A(ΔT/x)]/(ΔQ/Δt) . (8.30) In other words, the thermal resistivity is the temperature gradient divided by the heat flux. The thermal resistance is defined as the temperature gradient divided by the heat flow (not flux). Hence, it is given by the thermal resistivity divided by the cross-sectional area A: Thermal resistance = 1/(kA) = (ΔT/x)/(ΔQ/Δt) . (8.31)