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Chapter 9: Principles of Solidification

9-5 What is the difference between homogenous nucleation and heterogeneous nucleation?

Solution:

Homogenous nucleation occurs when nuclei of the resulting phase have the same probability of forming at every location. Heterogeneous nucleation of the resulting phase occurs at certain preferred locations in the original phase. In practice, homogenous nucleation does not occur. Heterogeneous nucleation commonly occurs at the surfaces for molds of castings, surfaces of inclusions and/or induced by inoculation.

9-6 From the thermodynamic point of view, what two things must occur for solidification from liquid to solid to proceed?

Solution:

Thermodynamically speaking, the first thing that must occur is the formation of a critical-sized nucleus. At this critical size, there is an activation energy that must be overcome. At the melting or freezing point, this activation energy will approach infinity. Thus the second thing that must occur is an undercooling in the liquid that is to solidify. A greater undercooling will lower the activation energy resulting in a greater tendency to solidify.

9-9 Derive the formula for the change in total energy change (i.e. the derivative of ΔG with respect to r) as a function of the radius r of the solid nucleus. Begin with Equation 9-1. Numerically, how is the critical point at which solidification proceeds defined?

Solution:

This is a calculus problem. Starting with the equation:

The critical point for solidification is when the radius reaches the critical radius. This is also when ΔG reaches an inflection point, so:

9-10 Using Equation 9-2, prove that interfacial energy (σsl) has units of J/m2 in SI.

Solution: Equation 9-2 is

∆ = 4 3 ∆ +4 Differentiating: ∆ = 4 3 ∆ +4 ∆ = 4 3 ∆ 3 +4 2 ∆ =4 ∆ +8

∆ =0;

9-11

Solving for the interfacial energy:

In SI, the units are as follows:

Inserting these into the previous equation:

Calculate the total interfacial surface energy for 1016 spheres of copper, each with a radius of r*.

Solution:

Example 9-1 found the critical radius of copper to be 12.51 × 10-8 centimeters. Calculating the total surface area of a single sphere: =4 =4 12.51×10 cm =1.9110 cm

The total surface area for all spheres:

10 1.9 10 cm =2 10 cm

Using the interfacial energy from Table 9-1:

9-12 Of the ferrous elements, which has the lowest undercooling required for homogeneous nucleation? Does this have any practical significance?

Solution:

Between iron and nickel, the lowest ΔT in Table 9-1 belongs to iron at 420 °C. This does not have a great practical significance in itself because homogeneous nucleation is not seen in industry or commerce. Heterogeneous nucleation is seen in production.

*With apologies, as written, this problem statement does not make sense. Perhaps it should have been of the “ferromagnetic elements.”

9-13 If the total change in free energy of a molten metal being cooled is 5.34 × 10-17 J upon formation of the first stable solid and the free energy per unit volume is -17.7 J/cm3 , approximate the radius of the spherical solid. The surface free energy of the solid-liquid interface is 100 × 10-7 J/cm2 .

= ∗∆ ∆ 2

Name SI units Critical radius m Freezing temperature K Heat of fusion J/m

Temperature difference K

= m 1 J m K 1 1 K = J m

2 10 cm 1 17710 1 J cm =35 mJ

Solution:

This is a straightforward use of Equation 9-1, however making it explicit in r is difficult algebraically. Instead a numerical solution using a spreadsheet’s goal seek function or calculator’s SOLVE application is advised. The result will be 1.16 × 10-6 cm.

9-14

Using the densities in Appendix A, convert the heats of fusion in Table 9-1 from units J/cm3 to kJ/kg.

Solution:

While doing the conversions, note that [kJ/kg] are the same as [J/g]

9-15

Assume that instead of a spherical nucleus, we have a nucleus in the form of a cube of length x. Calculate the critical dimension x* of the cube necessary for nucleation. Write an equation similar to Equation 9-1 for a cubical nucleus, and derive an expression for x* similar to Equation 9-2. Solution:

Note that ∆G v is negative such that x * is positive.

A similar treatment for Equation 9-1 for a spherical nucleus produces

Element Heat of fusion (kJ/kg) Ga 83 Bi 55 Pb 21 Ag 92 Cu 183 Ni 310 Fe 195

∆G = x 3∆G v + 6x 2 σsl ∂G ∂x = 3x 2∆G v +12xσsl = 0 ∂G ∂x x=x * = 3 x * ()2 ∆G v +12x * σsl = 0; x > 0 x * =− 12σ 3∆G v =− 4σsl ∆G v

∂G ∂r r=r * = 4π r * ()2 ∆G v + 8πr * σsl = 0; r > 0 r * =− 2σsl ∆G v

Comparison to Equation 9-2 shows that

9-16 Why is undercooling required for solidification? Derive an equation showing the total free energy change as a function of undercooling when the nucleating solid has the critical nucleus radius r*.

Solution: Undercooling is required for solidification because the energy to create a new solid-liquid interface needs to be overcome for a solid phase to begin forming. The undercooling below the melting temperature is the driving force that overcomes this barrier. From Equation 9–1, ∆G = (4/3) πr 3 ∆Gv + 4πr 2 σsl, where ∆G is the total change in free energy upon solidification, r is the radius of spherical solid, ∆Gv is the free energy change per unit volume for solidification, and σsl is the surface energy per unit area of the solid–liquid interface. From Equation 9–2, the critical radius for solidification r* is given by r* = 2σslTm/(∆Hf∆T), where Tm is the melting temperature, ∆Hf is the latent heat of fusion, and ∆T is the undercooling. At r = r*,

2 33333222 (4/3)[2/()]4[2/()] (32/3)(/)(1/)16(/)(1/)

9-17 Why is it that nuclei seen experimentally are often sphere-like but faceted? Why are they sphere-like and not like cubes or other shapes?

Solution:

Nuclei are sphere-like because spheres have the largest volume to surface area ratio. Nuclei are faceted to favor crystallographic planes with low surface energies.

9-18 Explain the meaning of each term in Equation 9-2.

Solution:

∆H f is the latent heat of fusion per unit volume, T m is the equilibrium solidification temperature in kelvin, and ∆T is the undercooling when the liquid temperature is T. The latent heat of fusion represents the heat given up during the liquid-to-solid transformation. σsl Is the surface energy per area of the solid–liquid interface.

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2016 Cengage Learning. May not be scanned,

∆G v =− ∆H f ∆T T m Thus, x * =− 4σsl ∆G v = 4σsl T m ∆H f ∆T

slmfvslmfsl slmvfslmf GTHTGTHT GTGHTTHT πσπσσ πσπσ ∆=∆∆∆+∆∆ ∆=∆∆∆+∆∆

9-19 Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate (a) the critical radius of the nucleus required and (b) the number of nickel atoms in the nucleus. Assume that the lattice parameter of the solid FCC nickel is 0.356 nm.

Solution: From Table 9–1, ∆Tmax = 480°C

(2)(25510 J/cm)(1453273) *6.6510cm (2756J/cm)(480)

ao = 3.56 Å V = 45.118 × 10–24 cm 3 Vnucleus = (4π/3)(6.65 × 10–8 cm)3 = 1232 × 10–24 cm 3 number of unit cells = 1232/45.118 = 27.3 atoms per nucleus = (4 atoms/cell)(27.3 cells) = 109 atoms

9-20 Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate (a) the critical radius of the nucleus required and (b) the number of iron atoms in the nucleus. Assume that the lattice parameter of the solid BCC iron is 2.92 Å.

Solution: 72 8 3 (2)(20410 J/cm)(1538273) *10.12810cm (1737J/cm)(420)

V = (4π/3)(10.128)3 = 4352 Å3 = 4352 × 10–24 cm 3

Vuc = (2.92 Å)3 = 24.897 Å3 = 24.897 × 10–24 cm 3 number of unit cells = 4352/24.897 = 175 atoms per nucleus = (175 cells)(2 atoms/cell) = 350 atoms

9-21 Suppose that solid nickel was able to nucleate homogeneously with an undercooling of only 22°C. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid FCC nickel is 0.356 nm.

Solution: 72 8 3 (2)(25510J/cm)(1453273) *145.1810cm (2756J/cm)(22)

Vuc = 45.118 × 10–24 cm 3 (see Problem 9–11)

Vnucleus = (4π/3)(145.18 × 10–8 cm)3 = 1.282 × 10–17 cm 3 number of unit cells = 1.282 × 10–17 / (45.118 × 10–24) = 2.84 × 105

atoms per nucleus = (4 atoms/cells)(2.84 × 105 cells) = 1.136 × 106

9-22 Suppose that solid iron was able to nucleate homogeneously with an undercooling of only 15°C. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid BCC iron is 2.92 Å.

Solution: 72 8 3 (2)(20410 J/cm)(1538273) *283.610cm (1737J/cm)(15) r

Vuc = 24.897 × 10–24 cm 3 (see Problem 9–12)

Vnucleus = (4π/3)(283.6 × 10–8 cm)3 = 95,544,850 × 10–24 cm 3

72 8 3

r ×+ ==×

×+ ==×

r ×+ ==×

==×

×+

number of unit cells = 95,544,850/24.897 = 3.838 × 106 atoms per nucleus = (2 atoms/cells)(3.838 × 106 cells) = 7.676 × 106

9-23 Explain the term inoculation.

Solution:

Inoculation is the addition of particles to the liquid to induce the nucleation of equiaxed grains in a casting. The common inoculants are refractory elements with melting points much higher than the liquid melt. The inoculants provide surfaces on which heterogeneous nucleation of solid crystals may occur.

9-35 What is a dendrite and why do dendrites form during solidification?

Solution:

A dendrite is a “tree-like” formation of solid crystals from liquid. The primary dendrite (main trunk) is due to a preferred growth of crystals in specific crystallographic directions which occurs opposite the direction of heat extraction. From the main trunk, branches (secondary dendrites) or dendritic arms form and grow also along preferred directions into the liquid. For cubic crystals, the easiest growth path are along the <100> directions.

9-36 Use the data in Table 9–1 and the specific heat data given below to calculate the undercooling required to keep the dendritic fraction at 0.5 for each metal.

Specific heat [J/(cm3–K)]

Solution: The dendritic fraction f is given by Equation 9–3:

f = c∆T/∆Hf, where c is the specific heat, T is the temperature, and ∆Hf is the latent heat of fusion.

For f = 0.5, the undercooling required is

∆T = 0.5 ∆Hf/c.

The results for the various metals are shown in the table.

Metal Heat of Fusion ∆Hf (J/cm3) Specific heat c [J/(cm3–K)] ∆T for f = 0.5 (K)

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Bi 1.27 Pb 1.47 Cu 3.48 Ni 4.75

Metal

Bi 543 1.27 213.8 Pb 237 1.47 80.6 Cu 1628 3.48 233.9 Ni 2756 4.75 290.1

9-37 Calculate the fraction of solidification that occurs dendritically when silver nucleates (a) at 10°C undercooling; (b) at 100°C undercooling; and (c) homogeneously. The specific heat of silver is 3.25 J/(cm3·°C).

Solution:

9-38 Calculate the fraction of solidification that occurs dendritically when iron nucleates (a) at 10°C undercooling; (b) at 100°C undercooling; and (c) homogeneously. The specific heat of iron is 5.78 J/(cm3 · °C). Solution:

Therefore, essentially all of the solidification occurs dendritically.

9-39 Analysis of a nickel casting suggests that 28% of the solidification process occurred in a dendritic manner. Calculate the temperature at which nucleation occurred. The specific heat of nickel is 4.1 J/(cm3 · °C).

9-40 It is desired to increase the dendritic fraction from 0.012 to 0.025. Assuming that the specific heat of the metal being solidified is constant, determine the increase in the amount of undercooling required.

Solution:

Equation 9-3 is linear, so we can simply divide the two fractions to find the ratios of ΔT

cT f Hf ∆⋅°° === ∆ (b) 3 3 [3.25

0.337

f cT H ∆⋅°° == ∆ (c)

f cT H ∆⋅°° == ∆

(a) 3 [3.25J/cmC](10C) 0.0337 965J/cm

J/(cmC)](100C)

965 J/cm

[3.25 J/(cmC)](250C) 0.842 965 J/cm

(a) 3 3 [5.78

0.0333 1737 J/cm f cT f H ∆⋅°° === ∆ (b) 3 3 [5.78

0.333 1737 J/cm f cT H ∆⋅°° == ∆ (c) 3 3 [5.78 J/(cmC)](420C) 1.40 1737 J/cm f cT H ∆⋅°° == ∆

J/(cmC)](10C)

J/(cmC)](100C)

Solution: 3 3 [4.1 J/(cmC)]() 0.28 2756 J/cm f cTT f H ∆⋅°∆ === ∆ 188Cor14531881265C n TT ∆=°=−=°

= ∆ ∆ = ∆ ∆ ∆ ∆

So, the undercooling is to be increased by a factor of 2.1. Note that this means the temperature difference, not the actual temperature the metal is being undercooled to.

9-42 Find the mold constant B and exponent n in Chvorinov’s rule using the following data and a log–log plot.

Solution: Chvorinov’s rule gives the solidification time ts as B n s V t A

where B is the mold constant, V is the volume of the casting, A is the surface area of the casting in contact with the mold, and n is a constant.

Taking the natural logarithm of both sides, ln (ts) = ln (B) + n ln (V/A)

The values of n and B can be found by plotting ln (ts) versus ln (V / A).

The volume and surface area of each shape is given in the table.

Plate Length = 30, Width = 20, Height = 1



 

Shape Dimensions (cm) Volume (cm3) Area (cm2) Volume / Area (cm) Cylinder Radius

9425 2513 3.75 Sphere Radius

3054 1018 3

= 10, Length = 30

= 9

Cube Length =

216 216 1

600 1300 0.46

From the graph, the slope n = 2.21 and ln (B) = 5.3407 such that B = 209 s/cm2 .

9-43 A 2-in. cube solidifies in 4.6 min. Assume that n = 2. Calculate (a) the mold constant in Chvorinov’s rule; and (b) the solidification time for a 0.5 in. × 0.5 in. × 6 in. bar cast under the same conditions.

Solution: (a) We can find the volume and surface area of the cube: V = (2)3 = 8 in.3 A = 6(2)2 = 24 in.2 t = 4.6 = B(8/24)2 B = 4.6/(0.333)2 = 41.48 min/in.2 (b) For the bar, assuming that B = 41.48 min/in.2 : V = (0.5)(0.5)(6) = 1.5 in.2 A = 2(0.5)(0.5) + 4(0.5)(6) = 12.5 in.2 t = (41.48)(1.5/12.5)2 = 0.60 min

9-44 A 5-cm diameter sphere solidifies in 1050 s. Calculate the solidification time for a 0.3 cm × 10 cm × 20 cm plate cast under the same conditions. Assume that n = 2.

2 (1512)(0.31020) 1512[60/418]31.15s [2(0.3)(10)2(0.3)(20)2(10)(20)]

9-45 Find the constants B and n in Chvorinov’s rule by plotting the following data on a log-log plot:

Solution: 2 3 22 2 1050(4/3)(2.5)

t π π  ====   2 2

t ×× ===

sBB[2.5/3]orB15124(2.5)s/cm

9-46 Find the constants B and n in Chvorinov’s rule by plotting the following data on a log-log plot:

A (in.2

V/A (in.) 48 212 0.226 60 112 0.536 15.6 37.5 0.416 36 98 0.367

find

B = 48 min/in.

n = 1.72

)

From the graph, we

that

2 and

Solution: V (cm

A (cm2

V/A (cm) 6 26 0.23 32 64 0.5 64 96 0.67 240 236 1.02 From

B = 305 s/cm2

n = 1.58

)

the graph, we find that

and

9-47 A 3-in.-diameter casting was produced. The times required for the solid-liquid interface to reach different distances beneath the casting surface were measured and are shown in the following table:

Determine (a) the time at which solidification begins at the surface and (b) the time at which the entire casting is expected to be solid. (c) Suppose the center of the casting actually solidified in 720 s. Explain why this time might differ from the time calculated in part (b).

Solution: We could plot d versus t , as shown, finding surfacet from where the plot intersects the x-axis and centert where the plot intersects d = 1.5 in. Or we could take two of the data points and solve for c and k k dtc

0.1k32.6c 0.5k130.6c 0.4k[32.6130.6] 5.718 k

Distance from surface d (in.) Time t (s) t 0.1 32.6 5.71 0.3 73.5 8.57 0.5 130.6 11.43 0.75 225.0 15.00 1.0 334.9 18.22

=−

=− =− −=− =− k0.070 c0.07032.60.10.30 = =−=

00.0700.30dt==−

tsurface = (0.3/0.07)2 = 18.4 s

1.50.0700.3 t =−

tcenter = (1.8/0.07)2 = 661 s

The mold gets hot during the solidification process, and consequently heat is extracted from the casting more slowly. This in turn changes the constants in the equation and increases the time required for complete solidification.

9-48 What solidification time is required for a casting of 9.5 cm3 and a mold contact area of 10.0 cm2? The mold constant is 838 s/cm2.01

Solution:

This is an easy application of Chvorinov’s rule. The only curveball is that the exponential constant must be read from the units of the mold constant.

9-49 An aluminum alloy plate with dimensions 20 cm × 10 cm × 2 cm needs to be cast with a secondary dendrite arm spacing of 10–2 cm (refer to Figure 9–6). What mold constant B is required (assume n = 2)?

= =838 s cm . 9.5 10.0 cm cm . =756 s

Solution: From Figure 9–6, for a secondary dendrite arm spacing of 0.01 cm, the solidification time is 500 s. Chvorinov’s rule for solidification time (Equation 9–4) is ts = B(V/A)n, where B is the mold constant, V is the volume of the casting, A is the surface area of the casting in contact with the mold, and n is a constant. In this case, the volume of the casting is V = 20 × 10 × 2 = 400 cm3 , and the surface area of the casting is

A = 2 × 20 × 10 + 2 × 2 × 10 + 2 × 2 × 20 = 520 cm2

Substituting into the equation above and taking n = 2,

ts = B(V/A)n

500 = B(400/520)2 B = 845 s/cm2

9-50 Figure 9–5(b) shows a micrograph of an aluminum alloy. Estimate (a) the secondary dendrite arm spacing and (b) the local solidification time for that area of the casting.

Solution: The distance between adjacent dendrite arms can be measured. Although most people doing these measurements will arrive at slightly different numbers, the author’s calculations obtained from four different primary arms are

16 mm / 6 arms = 2.67 mm

9 mm / 5 arms = 1.80 mm

13 mm / 7 arms = 1.85 mm

18 mm / 9 rms = 2.00 mm

average = 2.08 mm = 0.208 cm

Dividing by the magnification of ×50:

SDAS = 0.208 cm / 50 = 4.16 × 10–3 cm

From Figure 9–6, we find that the local solidification time (LST) = 90 s.

9-51 Find the constants k and m relating the secondary dendrite arm spacing to the local solidification time by plotting the following data on a log-log plot:

Solution: The secondary dendrite arm spacing (SDAS) is given by m SDASk s t = , where k and m are constants and ts is the solidification time. Taking the natural logarithm of both sides, ln (SDAS) = ln (k) + m ln (ts).

The values of k and m can be found by plotting ln (SDAS) versus ln (ts).

From the graph, the slope m = 0.35 and ln (k) = –5.8013 such that k = 0.003 cm / s.

9-52 Figure 9–25 shows dendrites in a titanium powder particle that has been rapidly solidified. Assuming that the size of the titanium dendrites is related to solidification time by the same relationship as in aluminum, estimate the solidification time of the powder particle.

Solution: The secondary dendrite arm spacing can be estimated from the photomicrograph at several locations. The author’s calculations, derived from measurements at three locations, are

11 mm / 8 arms = 1.375 mm

13 mm / 8 arms = 1.625 mm

average = 1.540 mm

Dividing by the magnification of 2200:

SDAS = (1.540 mm)(0.1 cm/mm) / 2200 = 7 × 10–5 cm

The relationship between SDAS and solidification time for aluminum is

SDAS = 8 × 10–4 t0.42 = 7 × 10–5 t = (0.0875)1/0.42 = 0.003 s

9-53 The secondary dendrite arm spacing in an electron-beam weld of copper is 9.5 × 10–4 cm. Estimate the solidification time of the weld.

Solution: From Figure 9–6, we can determine the equation relating SDAS and solidification time for copper: m = 20/50 = 0.4 k = 4 × 10–3 cm Then for the copper weld: 9.5 × 10–4 = 4 × 10–3(LST)0.4

(Note: LST is local solidification time)

0.2375 = (LST)0.4 or –1.4376 = 0.4 ln LST ln LST = –3.5940 or LST = 0.03 s

9-54 A zinc alloy has a SDAS of 0.01 cm and m = 0.40. What is the solidification time?

Solution:

The m value is a distraction here. By referring to Figure 9-6, we directly read the solidification time as being about 120 seconds.

9-55 In Figure 9-7, what is the equation (slope intercept form) of the tensile strength line, and what SDAS would appear to give a tensile strength of zero if the line is extrapolated? Does extrapolating the trend like this make sense?

Solution:

Since the graph is purely linear, we can easily take two points and find the equation of the line. Points that land on a gridline can be more accurately read, such as (0.005 cm, 44 ksi) and (0.010 cm, 39.5 ksi) that both fall on vertical gridlines.

Taking the difference to find the slope:

= 39.5 ksi−44 ksi 0.010 cm−0.005 cm =−900 ksi cm

Applying the slope to find the x-intercept:

= 39.5 ksi− 0.010 cm−0 =−900 ksi cm 39.5 ksi− 0.010 cm =−900 ksi cm

= ∆ ∆

3950 ksi

cm 0.010 cm =−900 ksi cm

4850 ksi

cm = 0.010 cm =48.5 ksi

Together: = −900ksi cm +48.5 ksi

Where x is in cm and y is in ksi.

To find out what SDAS (x) results in a strength of zero, we set y to zero and solve:

0 ksi= −900ksi cm +48.5 ksi

48.5 ksi= 900ksi cm

=SDAS=0.0539 cm

The tensile strength for the aluminum alloy will never be zero regardless of how large the SDAS spacing is. Thus we see that extrapolation can be dangerous since trends change. The curve will plateau at a value for which the SDAS becomes sufficiently large that the presence of dendrites no longer causes strength to be increased relative to a casting that solidified without dendrites.

9-60 In Figure 9-8, what do the slopes of the lines A-B and E-onward represent? Do these quantities represent physical properties?

Solution:

No. The slopes of these lines represent how fast the temperature is falling in the solid and liquid phases respectively. They can be changed at will by increasing or decreasing the rate of heat removal from the samples.

9-61 A cooling curve is shown in Figure 9–26. Determine (a) the pouring temperature; (b) the solidification temperature; (c) the superheat; (d) the cooling rate, just before solidification begins; (e) the total solidification time; (f) the local solidification time; and (g) the probable identity of the metal. (h) If the cooling curve was obtained at the center of the casting sketched in the figure, determine the mold constant, assuming that n = 2.

Solution:

(a) Tpour = 475°C

(b) Tsol = 320°C

(e) ts = 470 s

(f) LST = 470 – 160 = 310 s (c) ∆Ts = 475 – 320 = 155°C (g) Cadmium (Cd) (d) 475320 /1.0C/s 1600 Tt ∆∆==°

(h) ts = 470 = B[38.4/121.6]2 B = 4713 s/cm2

9-62 A cooling curve is shown in Figure 9–27. Determine (a) the pouring temperature; (b) the solidification temperature; (c) the superheat; (d) the cooling rate, just before solidification begins; (e) the total solidification time; (f) the local solidification time; (g) the undercooling; and (h) the probable identity of the metal. (i) If the cooling curve was obtained at the center of the casting sketched in the figure, determine the mold constant, assuming n = 2.

Solution: (a) Tpour = 900°C

(b) Tsol = 420°C

(e) ts = 9.7 min

(f) LST = 9.7 – 2.5 = 7.2 min

(d) 900400 /250C/min 20 Tt ∆∆==°

(h) Zn

(i) 229.7B[8/24]orB87.3 min/in. s t ===

9-63 Figure 9–28 shows the cooling curves obtained from several locations within a cylindrical aluminum casting. Determine the local solidification times and the SDAS at each location, then plot the tensile strength versus distance from the casting surface. Would you recommend that the casting be designed so that a large or small amount of material must be machined from the surface during finishing? Explain.

Solution: The local solidification times can be found from the cooling curves and can be used to find the expected SDAS values from Figure 9–6. The SDAS values can then be used to find the tensile strength, using Figure

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posted to a publicly accessible website, in whole

in part.

not be

9–7.

Surface: LST = 10 s ⇒ SDAS = 1.5 × 10–3 cm ⇒ TS = 47 ksi

Mid-radius: LST = 100 s ⇒ SDAS = 5 × 10–3 cm ⇒

= 44 ksi

Center: LST = 500 s ⇒ SDAS = 10 × 10–3 cm ⇒ TS = 39.5 ksi If high strength is desirable, you prefer to machine as little material off the surface of the casting as possible; the surface material has the finest structure and highest strength; any excessive machining simply removes the “best” material.

9-70 Calculate the volume, diameter, and height of the cylindrical riser required to prevent shrinkage in a 1 in. × 6 in. × 6 in. casting if the H/D of the riser is 1.0.

Solution: The riser should take longer to solidify than the casting: t

where tr is the solidification time for the riser, tc is the solidification time for the casting, and ( )r V A and ( )c V A are the volume (V) to surface area (A) ratios of the riser (r) and casting (c), respectively. The mold constant B is the same for the riser and casting. ( )c V A is given by

where D is the diameter of the riser and H is its height. Note that the riser area in contact with the casting is not included in either the riser or casting surface area; no heat is lost across this interface. Since H / D is 1.0,

rc BB VV AA  >   ,

22 c 16636 21621626696 44 V A DDππ ××  ==   ××+××+××−− , and r V A    is given by 2 2 r 4 4 DH V A DDH π π π  =   + ,

r > tc,

and solving for D, D > 1.93 in.

For this minimum diameter, the minimum height is 1.93 in., and the minimum volume is

(1.93 in.)(1.93 in.)5.7

9-71 Calculate the volume, diameter, and height of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H/D of the riser is 1.5.

Solution: The riser should take longer to solidify than the casting: t

or rc

Thus,

36 596 4 D D π   >   

Thus, 2

V π =×=

4in.

r >

BB VV AA  >   ,

22 c 41020800 2410242021020640 44 V A DDππ ××  ==   ××+××+××−−

r V A    is given by 2 2 4 4 DH V A DDH π π π  =   + ,

r 3 14 VD A  =   .

tc,

and

Therefore, solving for D, D > 6.11 in.

For this minimum diameter, the minimum height is 9.17 in., and the minimum volume is

9-72 Figure 9–29 shows a cylindrical riser attached to a casting. Compare the solidification times for each casting section and the riser and determine whether the riser will be effective.

(8)(6)(3) (/)0.889 (3)(6)2(3)(8)2(6)(8) (6)(6)(6) (/)1.13 (6)(3)5(6)(6)(/4)(3) (/4)(3)(7) (/)0.68 (3)(7)(/4)(3)

Note that the riser area in contact with the casting is not included in either the riser or casting surface area; no heat is lost across this interface. In a like manner, the area of contact between the thick and thin portions of the casting are not included in the calculation of the casting area.

The riser will not be effective; the thick section of the casting has the largest V/A ratio and therefore requires the longest solidification time. Consequently the riser will be completely solid before the thick section is solidified; no liquid metal will be available to compensate for the solidification shrinkage.

23 r (6.11 in.)(9.17in.)269 4in. V π =×=

thin thick2 2 riser2

VA VA VA π π ππ == ++ == +− == +

Solution:

9-73 Figure 9–30 shows a cylindrical riser attached to a casting. Compare the solidification times for each casting section and the riser and determine whether the riser will be effective.

Solution:

2 riser2 (4)(4)(4) (/)0.73 5(4)(4)1(2)(4) (2)(2)(4) (/)0.50 3(2)(4)2(2)(2) (/4)(4)(8) (/)0.8 (4)(8)2(/4)4

The area between the thick and thin sections of the casting are not included in calculating casting area; no heat is lost across this interface.

The riser will not be effective; the thin section has the smallest V/A ratio and therefore freezes first. Even though the riser has the longest solidification time, the thin section isolates the thick section from the riser, preventing liquid metal from feeding from the riser to the thick section. Shrinkage will occur in the thick section.

9-74 A hollow cylindrical mold for casting aluminum ingots has a 300 mm inside diameter and is 2 m high. If the mold is filled with liquid aluminum at 935 K, what is the largest spherical cavity that may form in the ingot?

Solution:

Since aluminum has 7% volume shrinkage, if the top of the ingot freezes first before the entire ingot solidifies, there will be a cavity within the ingot. The largest cavity forms when the entire shrinkage volume is converted to one cavity, which we shall assume to be spherical.

Volume of initial liquid = D2H = (0.3)2(2) = 0.1414 m3

Shrinkage volume = 0.1414 (0.07) = 0.009896 m3

Assuming a spherical cavity: 0.009896 m3 = πr 3 ; we find that r = 0.1332 m or 13.32 cm, so the diameter of the cavity would be 26.64 cm.

If we assume that the solidification process proceeds from the bottom up with no cavities formed within the ingot, the top of the ingot will be lower than the inside height of the ingot mold. Assuming the outside diameter of the ingot to be the same as the

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thick thin

VA VA VA π ππ == + == + == +

inside diameter of the ingot mold, the height of the ingot produced would be 93% (100 – 7% from shrinkage) of the mold height.

Ingot height = 0.93 (2 m mold height) = 1.86 m Diameter of the ingot = 0.3 m

9-75 A 4-in.-diameter sphere of liquid copper is allowed to solidify, producing a spherical shrinkage cavity at the center of the casting. Compare the volume and diameter of the shrinkage cavity in the copper casting to that obtained when a 4-in. sphere of liquid iron is allowed to solidify.

Solution: Cu: 5.1% Fe: 3.4% rsphere = 4/2 = 2 in.

Cu: Vshrinkage = (4π/3)(2)3 (0.051) = 1.709 in.3

(4π/3)r 3 = 1.709 in.3 or r = 0.742 in. dcavity = 1.48 in.

Fe: Vshrinkage = (4π/3)(2)3 (0.034) = 1.139 in.3

(4π/3)r 3 = 1.139 in.3 or r = 0.648 in. dcavity = 1.30 in.

9-76 A 4-in. cube of a liquid metal is allowed to solidify. A spherical shrinkage cavity with a diameter of 1.49 in. is observed in the solid casting. Determine the percent volume change that must have occurred during solidification.

Solution: Vliquid = (4 in.)3 = 64 in.3

Vshrinkage = (4π/3)(1.49/2)3 = 1.732 in.3

Vsolid = 64 – 1.732 = 62.268 in.3 %6462.268 volume change1002.7% 64 =×=

9-77 A 2 cm × 4 cm × 6 cm magnesium casting is produced. After cooling to room temperature, the casting is found to weigh 80 g. Determine (a) the volume of the shrinkage cavity at the center of the casting and (b) the percent shrinkage that must have occurred during solidification.

Solution: The density of magnesium is 1.738 g/cm3 (a) Vinitial = (2)(4)(6) = 48 cm3 Vfinal = 80 g/(1.738 g/cm3) = 46.03 cm3 (b)

9-78 A 2 in. × 8 in. × 10 in. iron casting is produced and, after cooling to room temperature, is found to weigh 43.9 lb. Determine (a) the percent shrinkage that must have occurred during solidification and (b) the number of shrinkage pores in the casting if all of the shrinkage occurs as pores with a diameter of 0.05 in.

Solution: The density of the iron is 7.87 g/cm3 3 actual3 (43.9 lb)(454 g/lb)2532.5 cm 7.87 g/cm

Vintended = (2)(8)(10) = 160 in.3 × (2.54 cm/in)3 = 2621.9 cm3 2621.92532.5

Vpores = 2621.9 – 2532.5 = 89.4 cm3

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shrinkage100%4.1% 48

%4846.03

=×=

shrinkage100%3.4% 2621.9 =×=

rpores = (0.05 in./2)(2.54 cm/in.) = 0.0635 cm 3 3 89.4 cm # pores83,354(4/3)(0.0635pores cm)

9-79 If you cool an open vessel of liquid gallium until a thin solid “cake” forms, will the cake form on the top or the bottom? Assume the air above the liquid gallium is stagnant and does not cool the surface at all.

Solution:

The cake will float on top. Table 9-2 shows that gallium expands when it solidifies. This means that a given volume of solid gallium is lighter than the same volume of liquid gallium. In other words, it is less dense.

9-80 Give examples of materials that expand upon solidification.

Solution:

From Table 9-2, gallium, gray cast iron and water.

9-81 What is Sievert’s Law? How can gas porosity in molten alloys be removed or minimized?

Solution:

Sievert’s Law expresses the solubility of gases in liquid metals as a function of the partial pressure of the gas over the liquid. In vacuum degassing of the melt, the partial pressure of the dissolved gas over the melt is reduced and therefore the concentration of the gas in the liquid is reduced.

9-83 Liquid magnesium is poured into a 2 cm × 2 cm × 24 cm mold and, as a result of directional solidification, all of the solidification shrinkage occurs along the 24-cm length of the casting. Determine the length of the casting immediately after solidification is completed.

Solution: Vinitial = (2)(2)(24) = 96 cm3 3 % contraction4or0.04963.84 cm =×=

Vfinal = 96 – 3.84 = 92.16 cm3 = (2)(2)(L) Length (L) = 23.04 cm

9-84 A liquid cast iron has a density of 7.65 g/cm3. Immediately after solidification, the density of the solid cast iron is found to be 7.71 g/cm3. Determine the percent volume change that occurs during solidification. Does the cast iron expand or contract during solidification?

Solution: 1/7.71 1/7.65 1/7.65 ×100% = 0.1297 cm3 0.1307 cm3 0.1307 cm 3 ×100% =−0.78%

The casting contracts.

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π ==

9-85

Molten copper at atmospheric pressure contains 0.01 wt% oxygen. The molten copper is placed in a chamber that is pumped down to 1 Pa to remove gas from the melt prior to pouring into the mold. Calculate the oxygen content of the copper melt after it is subjected to this degassing treatment.

Solution: According to Sievert’s law, the amount of gas that can be dissolved in a molten metal is given by gas

Percent of gasK p = ,

where K is a constant for a particular metal-gas system at constant temperature and pgas is the partial pressure of the gas in contact with the metal. Taking the ratio of the oxygen concentrations,

%OK

%OK pp pp == .

1 Pa1 Pa1 Pa

Solving for the oxygen content at 1 Pa, 1 Pa

%O%O p p =

1 Paatm atm

Recalling that 1 atm = 1.013 × 105 Pa,

Pa5 1 %O0.013.1410 wt%oxygen 1.01310 ==× × .

9-86 From Figure 9–14, find the solubility of hydrogen in liquid aluminum just before solidification begins when the partial pressure of hydrogen is 1 atm. Determine the solubility of hydrogen (in cm3/100 g Al) at the same temperature if the partial pressure were reduced to 0.01 atm.

atmatmatm

Solution: 0.46 cm3 H2/100 g aluminum 1 0.46/ 0.01 x = 3 0.460.010.046 cm/100 g Al x ==

9-87 The solubility of hydrogen in liquid aluminum at 715°C is found to be 1 cm3/(100 g Al). If all of this hydrogen precipitated as gas bubbles during solidification and remained trapped in the casting, calculate the volume percent gas in the solid aluminum.

Solution: (1 cm3 H2/100 g Al)(2.699 g/cm3) = 0.02699 cm3 H2/cm3 Al = 2.699%

9-88 Review Example 9-7. The usual method of measuring a vacuum in the United States vacuum pump industry is in inches of mercury. For example, atmospheric pressure is 29.92 inches of mercury. Covert the pressure found in the example to in. Hg.

Solution:

The pressure required is 10-6 atm. An atmosphere conversion factor is given, so: = 10 atm 1 29.92 1 in.Hg atm =0.00002992 in.Hg

Note that the method usually used in industry is more confusing: A perfect vacuum is 29.92 in. Hg and atmospheric pressure is zero. This is the absolute value of the reading of a vacuum gage.

9-98 Why has continuous casting of steels and other alloys assumed increased importance?

Solution:

The advantages of the continuous casting process are 1) more uniform composition and less segregation across the section of the ingot, 2) produces higher yield (less cropping) in primary shapes, 3) the shapes may be immediately deformed after casting with minimal reheating of the slabs, blooms or billets to form other desired shapes and 4), saves energy cost and increases productivity.

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