Materials Science Assignment #7 Solutions Chapter 9 Principles of Solidification The Science and Engineering of Materials

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Clyde Shepherd
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1 Materials Science Assignment #7 Solutions Chapter 9 Principles o Solidiication The Science and Engineering o Materials, by Donald Askeland, Pradeep Phule and Wendylin Wright 1) Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate: a. the critical radius o the nucleus required; and b. the number o nickel atoms in the nucleus. Assume that the lattice parameter o the solid FCC nickel is 0.56 nm. The critical radius is ound rom equation 9-2 (5 th edition) * 2σ sl T r = m H T We can ind the values to plug in rom the table 7 * 2* 255*10 J / cm*( ) K 8 r = = 6.65*10 cm 2756 J / cm *480K We ind the corresponding number o Ni atoms by irst inding the volume o the nucleus

2 V nucleus = 4/* π * r 8 ( ) 4* π * 6.65*10 21 = = 1.218*10 cm and inding the volume o a unit and dividing which gives us the number o unit s ( ) V = a0 =.56*10 cm = *10 cm 8 2 N _ s = = 27.s V Notice that the lattice parameter used is not the same as that reported in the textbook appendix. This is because the textbook values are at room temperature but nucleation occurs at the melting point in this case 145 C. Now to ind the number o atoms we need to realize that the FCC structure contains 4 atoms per unit N _ atoms = 4 atoms / *27.s = 109atoms 2) Suppose that solid nickel was able to nucleate homogeneously with an undercooling o only 22 0 C. How many atoms would have group together spontaneously or this to occur: Assume that the lattice parameter o the solid FCC nickel is 0.56 nm. This problem is similar to problem one. The only dierence is the amount o undercooling r σ T = and H T * 2 sl m r 2* 255*10 J / cm*( ) K 2756 J / cm *22K 7 * 8 = = *10 cm 8 ( ) 4* π * *10 = 4 / * π * r = = 1.282*10 cm 17 ( ) V = a =.56*10 cm = *10 cm N _ s = = 2.84*10 V N _ atoms = 4 atoms / * 2.84*10 s = 1.16*10 atoms

3 ) Calculate the raction o solidiication that occurs dendritically when iron nucleates a. at 10 0 C undercooling b. at C undercooling c. homogeneously The speciic heat o iron is 5.78 J/cm 0 C. The dendritic raction can be estimated with c T = H or 10 degrees o undercooling 0 0 = 5.78 J / cm C *10 C J / cm = or.% Similarly or 100 degrees o undercooling 0 0 = 5.78 J / cm C *100 C J / cm = or % The amount o undercooling or homogeneous nucleation is 420 C (Table 9-1) J / cm C *420 C = J / cm =, which is clearly ridiculous. It says that 140% o the iron is dendritic. We should interpret this result as all (100%) o the iron is dendritic 4) What are the two steps encountered in the solidiication o molten metals? As a unction o time, can they overlap with one another? The two steps are nucleation and growth and yes they can overlap each other. 5) During solidiication, speciic heat o the material and the latent heat o usion need to be removed. Deine each o these terms. The speciic heat is the heat necessary to change the temperature o the material. The latent heat is the heat necessary or a phase change. In particular, the latent heat o usion is the heat that needs to be removed to go rom the liquid phase to the solid phase. 6) Describe under what conditions we expect molten metals to undergo dendritic solidiication. Molten metals undergo dendritic solidiication when the melt is undercooled. This happens when the material is not adequately inoculated. 7) Describe under what conditions we expect molten metals to undergo planar ront solidiication. Planar ront solidiication occurs when there are plenty o places or nucleation to occur in other words when the melt is well inoculated. In this case there is no undercooling.

8) Analysis o a nickel casting suggests that 28% o the solidiication process occurred in a dendritic manner. Calculate the temperature at which nucleation occurred. The speciic heat o nickel is 4.

1 J / cm C Since the melting point o Nickel is 145 C, the temperature at which nucleation occurred is T = T T T nucleation melting undercooling nucleation 0 = 145 188 = 1265 C 9) Consider this

4 8) Analysis o a nickel casting suggests that 28% o the solidiication process occurred in a dendritic manner. Calculate the temperature at which nucleation occurred. The speciic heat o nickel is 4.1 J/cm 0 C. Recalling that the dendritic raction can be approximated as c T = we can calculate the amount o undercooling H * H 0.28* 2756 J / cm 0 T = = = 188 C 0 c 4.1 J / cm C Since the melting point o Nickel is 145 C, the temperature at which nucleation occurred is T = T T T nucleation melting undercooling nucleation 0 = = 1265 C 9) Consider this photograph o an aluminum alloy a) Estimate the secondary arm spacing and b) the local solidiication time or that area o the casting. This is tricky, since it will depend on what size the image is when you print it out. Find a place where there are several arms in a row, measure the length o a row o 5 or 6 arms, then divide by the number o arms. 16 mm / 6 arms = 2.67 mm 9 mm / 5 arms = 1.80 mm 1 mm / 7 arms = 1.85 mm 18 mm / 9 rms = 2.00 mm average = 2.08 mm = cm

Dividing by the magniication o 50: SDAS = 0.208 cm / 50 = 4.

5 Dividing by the magniication o 50: SDAS = cm / 50 = cm (b) From Figure 9 6, we ind that local solidiication time (LST) = 90 s 10) What is meant by the terms total and local solidiication time? Total solidiication time is the amount o time it takes or the melt to solidiy starting rom the time it is poured into the mold Local solidiication time is the time required or complete solidiication, starting rom when the irst crystal o solid orms 11) Consider the cooling curve below. Determine: a) the pouring temperature approximately 475 C b) the solidiication temperature approximately 0 C c) the superheat The amount o superheat is the dierence between the pouring temperature and the solidiication temperature =145 degrees o superheat d) the cooling rate, just beore solidiication begins The cooling rate is the slope o the line (475-60)/170 min = 0.7 degrees/min e) the total solidiication time approximately 470 minutes ) the local solidiication time ( )=00 minutes g) the probable identity o the metal Since the melting temperature is approximately 0 we can go to Appendix A and ind a metal with a similar melting point. Cadmium melts at 21 C and Lead melts at 27 C either would be a good guess. Better melting point inormation would allow us to select between the two. h) Is the solidiication process homogeneous or non-homogeneous? Heterogeneous, (non-homogeneous) because there is no undercooling

6 12) Consider the cooling curve below. Determine: i) the pouring temperature 900 C j) the solidiication temperature 420 C k) the superheat ( )= 480 C l) the cooling rate, just beore solidiication begins ( )/1.6 min = 12 degrees C/min m) the total solidiication time 9.7 min n) the local solidiication time ( )=8.1 min o) the probable identity o the metal Since the melting point is approximately 420 C, a good guess would be Zn p) Is the solidiication process homogeneous or non-homogeneous? Homogeneous, because there is undercooling 1) What is a riser? Why should it reeze ater the casting? A riser is a reservoir o molten metal that connects to the casting. I it reezes ater the casting it can provide liquid metal to compensate or shrinkage. 14) How can gas porosity in molten alloys be removed or minimized? Gas porosity is caused by bubbles o gas trapped within a casting during solidiication, caused by the lower solubility o the gas in the solid, compared with that in the liquid. There are a number o ways to minimize the problem. Keeping the liquid temperature low will prevent the unwanted gas rom dissolving in the irst place. This can also be accomplished by working in a vaccuum chamber. You could add materials that will react with the gas and orm a solid. or you can remove the gas by bubbling a third gas through the melt in which the unwanted gas is more soluable. 15) Deine the terms brazing and soldering Both processes involve joining two pieces o metal, using a iller which is a dierent material. The iller acts like a glue. Brazing occurs at temperatures above 450 C and soldering occurs at temperatures below 450 C. In neither case does the base metal melt. 16) What is the dierence between usion welding and brazing and soldering? In usion welding, unlike brazing and soldering, there is no iller. Some o the base metal is melted to act as the glue to join the pieces

7 17) What is a heat aected zone? The heat aected zone is the portion o the base metal in a weld that was heated enough to cause a change in properties, such as stress relie, recrystallization and grain growth. 18) Explain why, while using low intensity heat sources, the strength o the material in weld regions can be reduced. Creating a weld with a low intensity heat source requires more time. and thereore the surrounding material heats up, which causes a large heat aected zone. One result is that the strength o the material is decreased due to recrystallization an grain growth. Welding with a high intensity source can generally be accomplished more quickly, and limits the heat aected area. 19) Why do laser and electron-beam welding processes lead to stronger welds? See above

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