Guide to Exercises Hints A in Chapters 3 6

Transcription

1 Guide to Exercises Hints A in Chaters 3 6 HINT A1 Exercise 3.1 What comonent casting methods, which give high accuracy and good surface finish, can be considered? If you have no suggestion, look f recision casting methods in Chater 1. Hint A14 HINT A5 Exercise 6.4 A solidified ingot has three different crystal zones, surface crystal zone, columnar crystal zone and central crystal zone. Divide the curve into a number of relevant time intervals and relate them one by one to the macrostructure of the alloy. Hint A138 HINT A HINT A6 Exercise 4.1 In this case the solidification is one-dimensional and the contact between the Cu late and the solidified metal is good. Draw a figure of the temerature rofile and suggest a strategy to calculate the solidification time. Hint A75 Exercise 6.1 Draw a sketch of art of the wire. HINT A7 Hint A HINT A3 Exercise 5.1a What basic equation do you use to calculate the excess radiation losses Q cool rad from the ingot? Hint A173 Exercise 6.7 The condition f grey solidification is that the solidification rate v growth dy L = < 4: 1 4 m/s. Make a sketch of the temerature distribution at the surface of and inside the rod and set u an analytical exression of the solidification rate dy L = as a function of the osition y L of the solidification front. Hint A61 HINT A4 Exercise 6.1 What haens in the melt when steel owder of room temerature is oured into the melt? Hint A17 Hint A8 (A7) Q sol rad Aes BðT L 4 T 4 Þ t ð3þ The liquidus temerature is 145 þ K. Introduction of known values gives Q sol rad :4 1 : 5: ð Þ :3 1 7 J Materials Processing during Casting Guide to Exercises and U. Åkerlind # 6 John Wiley & Sons, Ltd. H. Fredriksson The answer is given in Hint A6. Hint A6

2 Guide to Exercises HINT A9 Exercise 6. You want the dendrite arm distance l den as a function of the thickness of the columnar crystal zone of an ingot. You know the growth rate v growth as a function of the time and the relation between v and l den. How do you find a relation between thickness y L of the columnar crystal zone and time? Hint A51 HINT A1 Exercise 3. Set u an exression f the casting time as a function of the dimensions of the srue in case of uhill casting. Hint A81 HINT A11 Exercise 6.3 The heat transfer coefficient is mentioned in the text. It is therefe reasonable to set u a heat balance equation, which might be useful. Hint A176 HINT A1 The solidification time can be calculated from Chvinov s rule [Equation (A4.74a) on age 8 in Chater 4]. In this case we obtain t total C V metal C Ay L A A þ A Find an exression f the constant C. HINT A14 Hint A93 (A1) The best choice may be either the Shaw rocess (Chater 1, age 7) the investment casting method (wax melting method) (Chater 1, ages 7 8). Comare them. Hint A45 HINT A15 Exercise 5.6 The heat transfer coefficient h w deends on the water flow and the temerature of the cooling water. Only emirical relations are available. Use one of them. Hint A114 HINT A16 Exercise 5.3 Why is the solidification front non-lanar? HINT A13 Hint A178 (A35) The flow is assumed to be laminar. In this case, Bernoulli s equation is valid [Equation (3.) on age 9 in Chater 3]. Suggest suitable oints as oints 1 and. Set u Bernoulli s equation, introduce convenient variables and aly it in the resent system. Hint A88 (A137) Figure 4. on age 78 in Chater 4 shows the temerature distribution in the mould and the melt during casting in a sand mould. HINT A17 (A44) The outlet seed v 3 can be calculated if you use the continuity rincile: A v A 3 v 3 where A and A 3 are known. Inserting the v function, you obtain v 3 A v A ffiffiffiffiffiffiffi :1 ffiffiffiffiffiffiffi gh gh A 3 A 3 :14 :14 ð3þ

3 Materials Processing during Casting 3 The casting rate is v 3 :1 ffiffi h, where v3 and h are measured in SI units. HINT A18 (A133) The heat lost to the surroundings er unit time is, with nmal designations dq R dy r L c L dt dt < HINT A Exercise 3.4 Draw a figure of the tundish with molten steel. What equation is valid f the steel flow? Think of a strategy to solve the roblem. Hint A95 HINT A1 Exercise 5.1a Describe the rocess and make reasonable assumtions. Hint A19 To continue, you have to set u a heat balance. Hint A54 HINT A19 V = πr dy dy R HINT A (A6) Consider a wire element with radius r and length z. It solidifies inwards from radius r þ dr to radius r during the time. The solidification heat at the solidification front is removed by a heat flow in the oosite direction. z r r y r (A) (A154) The driving fce f heat transt during and after solidification is rotional to the temerature difference ðt L T Þ. H T L T Solidification time Quantity (kj/kg) (K) (h) Aluminium Steel Set u a heat balance. HINT A3 r + dr Hint A179 Exercise 3.9a Set u a heat balance f a volume element of the siral. Hint A195 (a) The solidification time of the aluminium ingot is about 1.5 h. (b) The solidification time of the steel ingot is about 1 h. In site of its much higher mass, the steel ingot solidifies me raidly than the aluminium ingot, because Fe has a lower heat of fusion and a higher driving fce f heat transt than Al. HINT A4 (A35) Equation () gives A outlet d outlet 4 A strandv cast v outlet ð5þ

4 4 Guide to Exercises rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 d outlet A strandv cast 4 1:5::9 :33m 3:13 v outlet The casting rate is.9 m/s.54 m/min. The outlet diameter of the tundish has to be 33 mm. HINT A5 (A16) The solidification time is obtained if you combine Equations (1) and (): t 1 4a metal y r metal c metal y l 4 k metal l where y thickness of solidified shell t 7: :1 63 s 4 3 :9 The solidification time of a turbine blade is about 1.5 min (l :9). HINT A6 (A165) (A8) (A1) (A38) (a) The fraction radiated excess heat is 6%. (b) The total radiated heat during the solidification is 1:5 1 MJ. (c) The thickness of the maximum layer is calculated to be 15 cm, which is an unrealistic value. (A e is often fmed below the solidified layer. It insulates the melt from the solidified shell.) (d) Very little radiation heat is emitted rovided that the uer surface is roerly insulated. The main advantage is that the uer surface solidifies later than the centre of the ingot. HINT A7 (A8) You insert these values into Equation (1), which gives t : R þ R 1: t 1: R :88 þ R 3:36 and lot are given in Hint A79. HINT A8 ð11þ Hint A79 (A191) Comosition of the molten alloy (age 49 in Chater 3). Small additions of imurities increase the viscosity strongly and reduce the fluidity. Pure metals and eutectic alloys have low viscosities and high fluidity lengths. The larger the liquidus solidus interval is, the higher will be the viscosity and the lower the maximum fluidity length. The best fluidity is obtained if the metal freezes on the mould wall and the solidification front becomes lanar. An alloying element leads to a solidification interval of the alloy, which revents a lanar front. Instead, so-called dendrites (netwk of crystal arms) are fmed. The viscosity increases and the fluidity decreases. The answer is given in Hint A164. Hint A164 HINT A9 (A7) Accding to the text, you have y L þ Y L :6 m ð5þ Combine equations (4) and (5), introduce known values and material constants and solve Y L. Accurate calculations are necessary! Hint A156 HINT A3 Exercise 4. How do you classify the tye of casting conditions? What model can be used to the solidification rocess? Hint A81

5 Materials Processing during Casting 5 HINT A31 Exercise 6.8a What is the influence of convection on the nucleation of equiaxed crystals in a solidifying ingot? Hint A3 HINT A3 (A199) Introduce Equation (4) into Equation (3) and relace v ingot by dh=: A srue ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi gðh hþ 6A ingot dh Integrate Equation (5) and solve the cooling time t 3. Hint A18 HINT A35 Exercise 3.3 Make some reasonable assumtions about the steel flow. Is it turbulent laminar? What equation is alicable? Hint A16 HINT A36 (A116) The heat flux from the melt to the solid hase can be written as How do you solve this differential equation? Hint A75 dq h ðt melt T L Þ HINT A33 (A4) You choose some values of l and use Table 4.4 on age 67 in Chater 4. l erf(l).35 þ erf(l) ffiffiffi le l ffiffiffi le l ½:35 þ erfðlþš The heat flux through the solid shell is dq k T L T i metal s y ðþ The heat flux from the outer surface to the surroundings is dq h 1ðT i metal T Þ ð3þ You want y as a function of the excess temerature, T T melt T L. How do you roceed? Hint A184 Interolation gives l :79. How do you roceed? HINT A34 Hint A1 HINT A37 (A3) Accding to geometry and Figure 4 in Hint A3, the radius r is two-thirds of the height in the equilateral triangle T 1 T T 3 : (A119) The heat flow from the suerheated melt to the surroundings can be written in rvc dt Cooling heat er unit time sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k mould r mould c mould A ðt i T Þ t Heat flow across the interface mould/ melt Comare Equation ð4:7þ on age 79 in Chater 4 ð5þ r R ffiffi ffiffiffi 3 3 R ffiffi 6 3 R r ffiffiffi 6 Now you are close to the final answer! ð5þ Hint A183

6 6 Guide to Exercises HINT A38 Exercise 5.1d As you will see later (Chater 1), it is imossible to accet that the uer surface of the melt solidifies befe the centre of the ingot. The thickness you got in Exercise 5.1c is also unreasonably large and unrealistic. If the uer surface were to solidify, then a e would be fmed below the solid layer owing to shrinkage. This e would sto the further growth of the solid layer. The answer is given in Hint A6. Hint A6 HINT A39 and different equations are valid. What equation is valid f the sand mould? Hint A16 HINT A41 (A344) sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffi 1 l den 1 t 1:5 1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 l den 1 y L 1:5 1 3: 1 4: ffiffiffiffiffi y L (A178) Heat flux across the solidification front: k s T L T i metal y L Heat flux through the solidified shell k L T melt T L d Heat flux through the boundary layer þ rð HÞ dy L Solidification heat er unit time l den 4:7 1 4 ffiffiffiffiffi y L T melt T L T T i metal HINT A4 T y L y Exercise 6.8c Comare the values of v front and v crystal. You can find them in Hints A39 and A171. Hint A67 Discuss the convection flow within the boundary layer and its influence on the shae of the solidification front. Hint A46 HINT A4 Exercise 4.6 The solidification conditions are different in the two cases HINT A43 (A58) sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r L Lðsin 5 Þc L dt k mould r mould c mould ðt E T Þ ð8þ t cool Calculate the cooling rate as a function of L. Hint A91

7 Materials Processing during Casting 7 HINT A44 (A95) 1 þ rgh 1 þ rv 1 þ rgh þ rv figure and make reasonable assumtions in der to set u an exression f the heat flow to the surroundings. Hint A47 HINT A47 Point 1 is chosen at the uer surface of the steel melt and oint at the outlet from the tundish. This level is chosen as zero level. As A 1 is very large, v 1. Inserting 1 atm, h and h 1 h into Equation (1), you obtain atm þ rgh þ atm þ þ rv ) v ffiffiffiffiffiffiffi gh ðþ How do you roceed when the v function is known? Hint A17 HINT A45 (A14) (A75) Inserting h h fill gives t fill 6A ingot ffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi ð H H h fill Þ g A srue Inserting the values given in the figure in the text in SI units, you obtain 6 :5 ffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi t fill ð:15þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð 1:7 1:7 1:5Þ51 9:81 The filling time is 5 s. HINT A48 Shaw rocess Model built of wood gysum Mould of refracty material. Silicon acid as binding agent. Mould dried and heated in an oven (1 C). Parted mould Good accuracy Large comonents. Small series single comonents ossible from economic oint of view Investment casting method (wax melting) Model of wax The mould built by diing the model in a mixture of ceramic material with silicon acid as binding agent. The mould is dried and the wax is melted away. The mould is burnt Very good accuracy Small comounds,.1.5 kg Large series necessary f economic reasons Exercise 4.1b How do you estimate the value of the heat transfer coefficient? Hint A155 HINT A49 Exercise 4.9a Can you find a relation between the surface temerature and the thickness of the solidified shell? Hint A157 HINT A5 What would you choose on the basis of this knowledge? Hint A7 Exercise 3.6 Set u a mass balance f sulfur. Hint A11 HINT A46 Exercise 6.9a How do you get an equation f calculation of the cooling rate in the centre of the wedge-shaed samle? Start with a HINT A51 (A9) Integrate the relation v growth ðtþ. Then you get the thickness y L of the columnar zone as a function of t. Hint A8

8 8 Guide to Exercises HINT A5 Exercise 6.5a The coarseness of the structure is determined by the growth rate. Describe this fact mathematically. Hint A185 HINT A56 (A3) Other facts are temerature, structure of the solidified melt, thermal conductivity, heat of fusion and magnitude of the flow. How do they affect the fluidity? Hint A164 HINT A53 Exercise 3.5 The casting rate v cast is the velocity of the strand when it leaves the chill-mould. How do you calculate v cast? Hint A6 HINT A54 (A7) v max D t D 4h av ðt L T Þ arð HÞ v max Dh avðt L T Þ arð HÞ ð7þ The average heat transfer coefficient has been calculated as 95 W/m K in Hint A31. The temerature of the cooling water is assumed to be 1 C. D : m and a m. These values and material constants, given in the text, are inserted into Equation (7): v max : 95ð183 1Þ : :1 m=s HINT A57 (A9) The exression of T melt in Equation (5) is introduced into Equation (4): Ah con dr v front dr r 4r ð HÞm Introduction of known values gives dr 1 N 1: : 1 4 dr ð8þ 1 7: 1 3 4ð1 1 6 Þ :1 N Solve dr=. HINT A58 Hint A171 (A173) It is reasonable to use an average value of T because the temeratures are high. T 15 þ K If you insert the given values into Equation (1) you obtain the radiated excess heat: Q cool rad Aes BðT 4 T 4 Þt :4 1: : 5: ð Þ1 6 : J The answer is given in Hint A31. Hint A31 How do you calculate the total excess heat? Hint A96 HINT A55 Exercise 5.a Draw a sketch of the temerature rofile of a cross-section of the inlet. Hint A116 HINT A59 Exercise 6.6 Consider the temerature rofile of the casting and illustrate it grahically. Set u the heat balance of the casting when convection is taken into consideration. Hint A59

9 Materials Processing during Casting 9 HINT A6 Exercise 5.7 Discuss the heat transt in the stri, illustrate its temerature rofile during cooling and solidification rocesses and discuss reasonable aroximations. Hint A15 HINT A61 (A149) Region VI. When all the melt has solidified the cooling rate becomes constant. dt= <. The solid hase cools at a constant cooling rate. The cooling rocess is controlled by the heat flow to the surroundings. It is described by the heat caacity c s of Cu. dq Vrcs dt HINT A6 Exercise 4.5 What equation can be used to calculate the solidification time in case (a)? Hint A159 HINT A63 (A1) Assume that the melt has the volume V. Set u a heat balance. Hint A11 HINT A64 (A56) rlrð HÞ dr k LðT melt T i Þ ln r R ð4þ The temerature T i is not a constant, but is a function of the osition of the solidification front. To find an exression of T i, you need another equation. Which one? Hint A163 HINT A65 (A339) Inserting given data and material constants, you obtain t t rð HÞ s B eðt 4 L T4 Þ R 7: : ð Þ R : 13 R Provided that the wire is so thin that the temerature gradient can be neglected, the solidification time of the wire is rotional to its radius: t rð HÞ s B eðtl 4 T4 Þ R The given data reresents the function t : 1 3 R which is illustrated in the figure. t (s) HINT A ðsi unitsþ R (µm) (A11) T i metal can be solved from Equation (5): T i metal T L T 1 þ h þ T k y L ð6þ which is identical with Equation (4.45) on age 74 in Chater 4. It can be seen from Equation (6) that if y L then T i metal T L. Hence T i metal can be read at the T-axis where y L. What is y L? How do you obtain cresonding T i metal and y L values? Hint A31

10 1 Guide to Exercises HINT A67 Exercise 6.8b The natural thing to do is to set u a heat balance. Assume that the equiaxed crystals can be regarded as sheres. Hint A16 How do you roceed? HINT A7 (A45) Hint A14 HINT A68 Exercise 3.1a Consider the fces that act on the curved metal surface in the cner. Hint A115 Two methods fulfil the technical demands. In this situation, the economic asects become imtant. The text says nothing about the number of comonents. In the case of small series, choose the Shaw rocess. In the case of large series, choose investment casting. HINT A69 (A76) Remember that T i T E ffiffiffiffiffiffiffiffi t cool r L Lðsin 5 Þc L ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ðt E T Þ k mould r mould c mould Introduction of numerical values gives ffiffiffiffiffiffiffiffi 7: 1 3 :87 4 ffiffiffi 1 t cool ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L ð1153 Þ :63 1: : :43 L How do you roceed to find the cooling rate as a function of L? Hint A58 HINT A7 HINT A73 Exercise 5.1b If you can calculate the heat radiated from the uer steel surface during the solidification time, you have solved the roblem. F this urose you must calculate the solidification time. How? Hint A141 HINT A74 Exercise 5.1b Maybe is not reasonable to use an average of the h values? Me careful calculations will answer this question. In what way do you have to revise your calculations above? Hint A13 (A161) The distance d is given by d ut 8 :1 :8 m The answer is given in Hint A189. HINT A71 (A16) v crystal dr Hint A189 HINT A75 (A3) Searate the variables and integrate the differential equation: 6A ingot A srue which gives dh ffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffi ) g H h t 6A ingot A srue ð t 6A ingot A srue ffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi ð Þ½ H h g ffiffiffiffiffi g Š h ð h dh ffiffiffiffiffiffiffiffiffiffiffiffi H h

11 Materials Processing during Casting 11 t 6A ingot ffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi ð H H hþ g A srue How do you obtain the filling time? Hint A47 (a) t rð HÞ R ðt melt T Þ 4k þ R h HINT A76 Exercise 4.8b How do you obtain the solidification rate? Hint A14 HINT A77 (A78) At the distance that cresonds to the maximum of the curve, the film of evaated oil molten casting owder is gone and there is good, direct contact between the steel and the chill-mould. (b) dy L T melt T rð HÞ 1 R y L k þ 1 h HINT A78 (A31) Because n growth dr=! Derivatization of Equation (6) with resect to time gives dr 1 6 Ct 5=6 N 1=3 ð7þ Use the relation (1) which you set u in Hint A185. How do you roceed? Hint A51 HINT A8 (A16) (A341) On inoculation, the length of the columnar zone decreases. This reduces the anisotroy of the mechanical roerties, which is an advantage during rolling and fging. This is esecially the case in the roduction of aluminium sheets. With the aid of basic equations the relation HINT A79 (A7) (A188) can be derived. dr 1 6 C t 5=6 N 1=3

12 1 Guide to Exercises (a) From this equation, in combination with the equation m growth l constant, it can be shown that when N is increased, owing to inoculation, the lamella distance l also increases, i.e. the structure becomes coarser, which was to be roved. (b) From this equation, in combination with the equation n growth mðt E TÞ n, it can be shown that the undercooling decreases when N is increased. This reduces the risk of white solidification and is the main reason f inoculation of a cast iron melt. Another reason, valid f all metals, is better mechanical roerties of the cast metal when the number of crystals in the melt is increased. Inoculation imroves the quality of the cast metal. the heat caacity c L of Cu. dq VrcL dt The melt cools and region I lasts until the excess temerature of the melt is gone. Equiaxed crystals are fmed at the outer surface of the ingot, close to the mould, as soon as T T (critical temerature f nucleation) and the surface crystal zone is fmed during the casting oeration. The transition from the equiaxed crystals in the surface zone to dendrites in the columnar zone cannot be observed in this cooling curve as it shows the temerature of the centre of the ingot. Comare Figure 6.38 at age 163 in Chater 6. The columnar crystals start to grow inwards. Characterize region II. Hint A99 HINT A81 (A1) The desired exression is Equation (3.13) on age 34 in Chater 3: t fill A ffiffiffiffiffiffiffiffiffi casting ffiffiffiffiffi ð h total ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h total h casting Þ g A srue Which quantities in Equation (1) are known can be calculated? Hint A166 HINT A8 (A138) A uer h total A lower A runner A casting h casting Region I. The sloe of the curve is determined by the heat transt to the surroundings, which is controlled by the natural convection in the melt. The cooling rocess is described by HINT A83 (A15) Introduce known values and the measured valued of y L and t used above into Equation (1) and roceed as above. Transfmation of Equation (1) gives th II ðt L T Þr metal ð HÞy L þ h IIr metal ð HÞ k y L h II tðt L T Þ r metal ð HÞ y L k ð15 5Þ 7: ð Þ W=m K y L ð4þ If you use another air of values to check the value, f examle y L 16 mm and t 1 min 6 s, and introduce these values into Equation (1), you obtain h II y L t sol ðt L T Þ r metal ð HÞ y L k : ð15 5Þ 7: ð:16þ 3 This value does not agree with that obtained with the aid of the knee values. The reason is that the mould is heated by the metal at high values of t. In this case, Equation (1) is not valid as T i is no longer constant and equal to T L.

13 Materials Processing during Casting 13 (b) The heat transfer coefficient h is of magnitude 1:5 1 W=m K in region I and at the beginning of region II. HINT A84 (A1) Accding to Equation (), dr=, i.e. the nuclei cannot grow. Nuclei can robably be fmed ahead of the solidification front but they cannot grow because the melt is not undercooled. Neither the free crystals n the solidification front can grow. HINT A85 Exercise 3.9b To get a numerical value of L f you need a value of the velocity of the melt. How do you get v? Hint A148 HINT A86 (A155) Region I. Equation () can be written as h I r metalð HÞ y L ð3þ T L T t The cresonding values ffiffi of y L and t at the knee of the curve are 14 mm and t 1:5 min ½, resectively, the latter cresonding to t 1:5 min :3 min 14 s. Calculate h I and try to find h II in a similar way. Hint A15 HINT A87 Exercise 5.1c You want to calculate the maximum thickness of the solidified uer layer. Set u a heat balance. Hint A48 HINT A88 (A16) The natural choice of oints 1 and can be seen in the figure. Bernoulli s equation can be written as 1 þ rgh 1 þ rv 1 þ rgh þ rv constant If you choose the inlet level as the zero level you obtain atm þ rg þ rv srue atm þ rgðh hþþ rv ingot What additional equations can you find? HINT A89 (A71) The two exressions must be equal. Hint A13 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r L Lðsin 5 Þc L dt k mould r mould c mould ðt i T Þ ð5þ t Equation (5) is a differential equation. Solve it and calculate the time required to remove the excess temerature of the melt by choice of convenient limits of the integrals. Hint A76 HINT A9 (A34) H = 1.7 m d uer 4 d lower 4 A srue = m 1 Zero level A ingot =.5 m Srue Ingot h h fill = 1.5 m A uer ) d uer rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A uer 1: :4 1 m A lower ) d lower rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A lower 1: :19 1 m

14 14 Guide to Exercises The calculated difference between the uer and lower diameters is small and of the der of.5 mm. This value is robably lower than the uncertainty limits. The difference can be neglected. A straight srue can be used. HINT A93 (A176) dy L = reresents the solidification rate growth rate v growth. By means of the relation v growth l den 1: 1 1 m 3 =s in the text you obtain HINT A91 dy L v growth 1 1 l den ðþ (A187) which gives 1:5 1 ffiffi l den 1 1 t sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffi 1 l den 1 t 1:5 1 ð4þ What use can you make of the other relation in the text? Hint A98 HINT A94 Exercise 5.1d Discuss the otimal rocess arameters. Hint A189 How do you roceed? Hint A344 HINT A95 HINT A9 (A15) The heat transfer coefficient h between the rolls and the metal is high but the thermal conductivity k of Al (and most metals) is high and the distance between the solidification front and the melt is small. We have reasons to assume that Nu hs k hy L k 1 (A) The outlet velocity v deends on the height h of the uer steel surface above the bottom of the tundish. h 1 A 1 v 1 Tundish A A v A 3 v 3 T T L To find this relation, you may use Bernoulli s equation [Equation (3.) on age 9 in Chater 3]. Hint A44 T Mould Solid Liquid HINT A96 y L In this case the temerature rofile is the one seen in the figure. Set u a balance of the heat flux (energy er unit time and unit cross-sectional area). Hint A174 y (A13) Accding to Equation (4.48) on age 74 in Chater 4, you obtain t rð HÞ y L 1 þ h T L T w h k y L ð3þ How do you get the desired distance? Hint A7

15 HINT A97 (A84) Materials Processing during Casting 15 ffiffiffi and calculate the cresonding values of le l ð:41 þ erfðlþþ. The final choice of l can be made by interolation. The discontinuity of the curve after about 45 s is due to the shrinkage of the cooling shell. An air ga is fmed which reduces the heat transfer instantly. Start f examle with l :5. HINT A1 Hint A38 Exercise 4.1a There are two alternatives treated in Chater 4 when heat transt across a mould/metal interface is concerned. Which ones? Hint A153 HINT A11 The solidification rate is the derivative of the curve. It increases at the end of the solidification rocess owing to the decreasing area of the solidification front. The heat flow is reduced inside the ingot while the cooling at the solid/mould interface is unchanged. Exercise 4.7 If you rotate the wedge 9 and relace it with a late with two vertical surfaces at a distance y L from each other, it will be easier to associate the resent roblem with the they of solidification in Section in Chater 4. Draw a figure and try to find an exression f the solidification time. Hint A169 HINT A98 (A61) Nu 1 means that ðh=kþy L is small comared with 1 and can be neglected in Equation (1). Hence you simly obtain HINT A1 (A95) The cooling curve consists of three intervals, two cooling eriods and one solidification eriod. Consider each time interval searately. Calculate h. h rð HÞ T L T v growth ðþ Hint A134 T excess T L Cooling of the melt Solidification rocess at constant temerature. Phase transfmation Cooling of the solid down to room temerature HINT A99 T (A18) 1. Choose an arbitrary value of l, use Table 4.4 on age 67 in Chater 4 to find erf(l) and use Figure 4.15 on age ffiffiffi 7 in Chater 4 to read an aroximate value of le l.. Fm the roduct (4). If the roduct is 4.33 you have found the right value of l. If not, try other l values to come close to ffiffiffi When you are close, use Table 4.4 and calculate le l. List some values of l T t 1 t t 3 t How do you calculate the cooling time of the melt? Hint A63

16 16 Guide to Exercises HINT A13 (A88) Consider the ingot. The inlet velocity is v srue. The total ingot area cresonds to six moulds. A ingot A srue HINT A16 (A38) Hence you obtain V metal A r L rl rl L þ r 116 7: ffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffi :63 1: :5 1 3 t total þ 1 :63t total r The velocity of the ingot area is v ingot. It can also be written v ingot dh ðþ The rincile of continuity is valid f an incomressible liquid [Equation (3.1) on age 9 in Chater 3]. A srue v srue A total v ingot 6A ingot dh ð3þ How many variables and equations do you have? Is the number of equations sufficient to find the filling time? If the answers are satisfacty, derive the time required to fill the ingots to the level h. Hint A14 HINT A14 Exercise 5.5a Accding to the advice in the text, it might be a good idea to calculate the average heat transfer coefficient h av. Hint A198 :15 :6 ð :6Þþð:15Þ 116 7: ffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffi :63 1: :5 1 3 t total þ 1 :63t total :15 6: 1 : ð1: ffiffiffiffiffiffiffiffi þ :1t total Þ t total :1t total þ 1: ffiffiffiffiffiffiffiffi 6: 1 t total : t total þ 535 ffiffiffiffiffiffiffiffi t total 86 1 which gives ffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi t 67:5 67:5 þ þ t total 49 s 4 min The solidification time of the cylinder is about 4 min. HINT A15 Exercise 5.4a It might be helful to consider Figure 5.18 on age 16 in Chater 5. The chill-mould is strongly water cooled. The figure in the text shows that the heat flux increases with the distance from the to of the mould in region 1. Why? Hint A194 HINT A17 (A8) Now you know y L as a function of time. If you can find l den as a function of time, you will be close to the solution of the roblem. How? Hint A187

17 Materials Processing during Casting 17 HINT A18 (A11) sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð TL k mould r mould c mould rvc dt A ðt i T Þ T start ð t1 ffiffit osition of the solidification front y L time t: where ffiffiffiffiffiffiffiffiffiffiffiffiffiffi y L ðtþ l 4a metal t a metal k metal r metal c metal as a function of ðþ ffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rvc ðt start T L Þ t 1 AðT i T Þ k mould r mould c mould [Equation (4.11) on age 6 in Chater 4]. l is a constant. How can you determine its value? Hint A313 t 1 rvc ðt start T L Þ AðT i T Þ k mould r mould c mould ðþ HINT A11 (A5) Insert material constants and other given values and calculate t 1. Hint A33 Electrode HINT A19 (A1) The wire solidifies raidly in air. After comlete solidification, it starts to cool to the temerature of the air. The wire is so thin that you may neglect the radial temerature gradient in it. The consequence is that each wire element has a unifm temerature, i.e. the surface and the centre have the same temerature. Discuss ossible ways of heat removal. Hint A77 HINT A11 Exercise 5.8 Consider the heat flow in the stri. HINT A111 Hint A19 (A81) Equation (4.6) on age 7 in Chater 4 describes the V melt V Solid metal The sulfur balance in the molten metal when a dro with volume dv is added to the melt and a volume of equal size solidifies: c dv c melt dv Contribution of S Loss of S to the melt due to in the melt addition of the due to solidification of volume dv from the electrode a volume dv Slag bath Molten metal Water-cooled Cu mould Water-cooled base late V melt dc melt Change of S content in the melt V and c melt are variables while V melt is ket constant. Equation (1) is searable. Solve it! Hint A1

18 18 Guide to Exercises HINT A113 r dv ð HÞ k mould r mould c mould A ðt i T Þ t Heat flow Heat flow across the generated mould=metal interface: by solidification Comare Equation ð4:7þ of the melt on age 79 in Chater 4 Integrate Equation (3) and solve the solidification time t. Hint A4 HINT A114 ð3þ The ressure i inside the shere is the ressure of the metal melt. It is the sum of the atmosheric ressure and the static ressure rgh of the melt, where h is the height u to the free metal surface. It is reasonable to use an average value of h. The ressure outside the shere is the external atmosheric ressure. Pressures are erendicular to the surface, indeendent of its direction. In addition, there are surface tension fces acting in the tangent lane along the boundary line, i.e. along the circle with radius r. Find the exression f the resulting surface tension fces in terms of r and y by use of a fce balance. Hint A144 HINT A116 (A55) (A15) Accding to the emirical relation (5.3) on age 11 in Chater 5, the heat transfer coefficient will be h 1:57 w:55 ð1 :75T w Þ a where a a machine arameter 4, w water flux and T w water temerature ( C). To calculate the h values f the different zones you must know the water flux w f the different zones. Hint A193 T T melt T L T i metal S L T y y L HINT A115 (A68) F symmetry reasons, the fces that act on the curved surface have a resultant fce in the direction of the symmetry line OA, which is marked in the figure. z R O q R x r A y In the figure shown, T temerature of the surroundings, T i metal temerature of the outer metal surface, T L liquidus temerature of the melt and T melt temerature of the melt, including the excess temerature. At steady state, the heat flux through the system must be the same at all interfaces. No freezing occurs at the inlet. The shell thickness is constant. Use conventional designations and set u the heat transt equations. Assume that the interface areas are equal. Hint A36 HINT A117 (A94) The solidification is comlete when the solidification fronts of two oosite sides meet. Which air, 41 3, will meet first?

19 Materials Processing during Casting 19 4 Solidification time 79.3 min 1.3 min 78 min Set u a heat balance f the final cooling rocess. Hint A HINT A1 As h 4 < h, you can conclude from Equation (1) that y 4 < y. Hence side 4 will grow me slowly than the other three equivalent sides. The conclusion is that sides 1 3 meet befe 4, which is illustrated in the figure. When the solidification fronts of the sides 1 and 3 meet, the casting becomes comletely solid. F comarison with the result in Exercise 5.1a, you have to calculate the solidification time and solidification rate f sides 1 3. Set u an analogous heat balance f sides 1 and 3. Hint A39 Exercise 4.11b List cresonding values of time and thickness y L of the solidified shell. Plot y L as a function of time. Hint A84 HINT A11 (A64) The heat flows are equal, which gives ka dt dy haðt i metal T Þ HINT A118 (A3) which gives h k dt T i metal T dy ð3þ The solidification time is about 13 s during sand mould casting and about 5 s during metal mould casting. The roduction caacity will be me than doubled in the latter case. HINT A119 (A4) ffiffiffi ffiffiffi ð t t 1Þ :7 13 : :5 ð66 Þ :63 1: : s which gives ffiffiffi ffiffiffi ffiffiffiffiffiffiffiffiffiffi t t :8 which gives ffiffiffi ffiffiffiffiffi t 79 þ 6:8 68:97 t 4757 s 79:3 min The temerature gradient is aroximately constant in the shell and can be relaced by dt dy T L T i metal y L This exression is introduced into Equation (3): h k y L T L T i metal T i metal T ð4þ ð5þ If you use the figure to find cresonding T i metal and y L values, you can calculate h f each curve. How do you read T i metal from the curve? Hint A66 HINT A1 (A33) You will use Equation (1), hence you need a value of the thermal diffusion coefficient a metal [Equation (4.1) on age 6 in Chater 4]: a metal k metal 3 r metal c metal : m =s

20 Guide to Exercises Then you solve the solidification time by squaring Equation (1): R ffiffiffiffiffiffiffiffiffiffiffiffiffiffi y L ðtþ l 4a metal t ) t 1 y L ð5þ 4a metal l What value of y L has to be inserted into Equation (5) to find the solidification time? Hint A44 y L R HINT A13 (A74) Equation () will not be introduced. You have to calculate searate solidification times f the different side airs of the casting. HINT A15 Exercise 5.11 Check Nusselt s number! Hint A34 a y 4 z 4 y 3 y 1 a z 3 z 1 a z y a HINT A16 (A38) l erf(l).41 þ erf(l) ffiffiffi le l ffiffiffi le l ½:41 þ erfðlþš The thicknesses of the solidified shells at time t are shown in the figure. Owing to a different heat transfer coefficient h 4, the thickness of shell 4 is less than that of the other shells. Assume that Interolation gives l :9. How do you calculate the value of the solidification time when l is known? Hint A5 y y 1 y 3 y and y 6 y 4 You also have z 1 z 3 a ðyþy 4 Þ z z 4 a y ð8þ ð9þ Set u the material balances f sides and 4. Hint A94 HINT A17 (A4) Assume that you add the m kg of steel owder er kg of melt. It will be heated by the melt, which cools. Aly the energy law! Hint A66 HINT A14 (A76) Relace R by R y L in Equation (1) in Hint A8 and derive it with resect to t. Hint A97 HINT A18 (A75) The material constants of Fe and Cu are introduced into Equation (3) in der to solve l. The temerature values shown in the figure are reasonable assumtions.

21 Materials Processing during Casting 1 Cu Solid Liquid Steel belt 188 K Roller Casting Roller The highest ossible value of the temerature at the external side of the Cu late is 1 C. With the given values inserted into Equation (3), you obtain 83ð188 4Þ 71 3 ffiffiffi le l 373 K rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! þerf ðlþ :33 ffiffiffi le l ð:41 þ erf ðlþþ ð4þ You have available Table 4.4 with the err function erf(z) on age 67 and Figures 4.14 and 4.15 on ages 71 7 in Chater 4. Use them to find a numerical solution of l by trial and err. Hint A99 HINT A19 Exercise 5.b Use Equation (8) rather Equation (7) in Hint A37 to calculate the excess temerature with the aid of the given values. Hint A HINT A13 Exercise 5.1b Consider the solidification rocess and set u a heat balance. Hint A88 HINT A131 ~ 4 K Exercise 5.1a The square casting solidifies laterally from all four sides. Check Nusselt s number to find the tentative temerature rofile. Hint A197 HINT A13 (A6) The maximum casting rate is v max l t D t ð17þ where D is the diameter of the wheel and l is half the erihery. You introduce the exression f t into Equation (17): v max D hðt L T Þ arð HÞ þ h ð18þ 4 h Insert values of material data, calculate v max and comare the result with that in 5.1a. Hint A31 HINT A133 Exercise 3.8 Consider a small volume element dy and discuss the energy change during the time. Hint A18 HINT A134 Roller (A98) The maximum heat transfer coefficient is obtained if you introduce the maximum solidification rate into Equation (): h max rð HÞ v max growth ð3þ T L T If you insert the given values and the material constants you obtain h max 7: :1 4 6:71 W=m K 115

22 Guide to Exercises The heat transfer coefficient must not exceed.6 kw/m K (.67 kw/m K). HINT A135 Exercise 4.8a Consider the heat flow through a cylinder element with radius r and height L and set u a heat balance. Hint A315 The temerature time curve f the centre of the ingot deends on cooling rate (rate of heat losses to the surroundings) nucleation rate growth rate heat of solidification and heat caacitivities. The cooling rates and the heat of solidification are constant. Characterize region I. Hint A8 HINT A136 (A1) Cooling caacity of the mould: A large cooling caacity results in a small maximum fluidity length as heat is transted away quickly and the metal solidifies quickly. The maximum fluidity length decreases with increasing cooling caacity. What about the influence of surface tension? Hint A191 HINT A137 Exercise 4.3a Owing to the dimensions of the casting it can be regarded as a late with a thickness of 1 mm, which solidifies onedimensionally. The Al ingot is cast in a sand mould. Draw a figure which shows the temerature distribution in the mould and metal. What equation is valid f casting in a sand mould? Hint A13 HINT A138 (A5) HINT A139 (A311) The casting rate equals the ratio of the cooling length to the solidification time: v cast l t total Exression (4) is introduced into Equation (5): v cast l lhðt L T Þ t total rdð HÞ ð5þ ð6þ Find the distance d between the rollers as a function of the casting rate and material constants. Hint A5 HINT A14 (A71) You derive Equation () and introduce the derivative into Equation (1): Ah con T melt Nr 4r dr ð HÞ ð3þ which gives dr Ah cont melt r 4r ð HÞ 1 N ð4þ All quantities excet A, T and N are known. How do you calculate the area A? Hint A38

23 Materials Processing during Casting 3 HINT A141 (A73) A smart way is to use the so-called rule of thumb. You can find it on age 97 in Chater 5. Hint A7 The external ressure is i þ rgh Combine Equations (1) and () and solve r. ðþ Hint A45 HINT A14 HINT A145 (A13) Variables are v srue, v ingot, h and t. The number of indeendent equations is three, which is sufficient to eliminate v srue and v ingot and obtain the time t as a function of the height h. Derive the time t as a function of the height h. Hint A199 HINT A143 (A36) From the figure in the text, it can be seen that the cooling length L is.5 m. With the aid of Equation (3) you can find the maximum casting rate: v max L t LhðT L T Þ 1 rð HÞ d ð4þ Introduce material constants and other values, given in the text, and calculate the relation between v max and d. Hint A34 HINT A144 (A115) The surface tension fces have tangential directions along the circle with radius r. Their resultant in the direction OA is rscosy (comare the figures in Examle 3.6 on age 5 in Chater 3). Three fces act on the curved metal surface. r Resulting fce from the external ressure directed towards the centre þ rscosy Resulting fce of the surface tension fces directed towards the centre i r Resulting fce from the internal ressure; directed outwards from the centre (A156) rð HÞy L 1 þ h T L T w h k y L 1 rð HÞ s B eðt L4 T 4 Þ ð:6 y LÞ 1 y L þ 1 44 y L 1 5: ð Þ ð:6 y LÞ y L 1 þ 1 44 y L : ð Þ ð:6 y LÞ y L þ :77y L : : ð:6 y LÞ y L þ :77y L 15:5ð:6 y L Þ y L þ :44y L : :6y L y L þ 7:6y L :39733 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y L 3:531 3:531 þ : :531 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1: þ :3973 You want the ositive root y L 3:531 þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1: :531 þ 3:5366 :56m Check: The sum of Y L (Hint 156) and Y L is 6. mm, in agreement with the text in Exercise 5.7. The solidification fronts meet at.5 mm from the uer surface of the stri.

24 4 Guide to Exercises HINT A146 (A333) (A68) (a) Figure 4.17 is valid f all cases. Figure 4.7 is valid f small values of Nusselt s number hy L =k. (b) Diagrams f steel and coer f the two h values are shown below. F steel you have T L 153 C k Fe 3 W=mK h 1 W=m K h 1 3 W=m K y L T ife y L T ife HINT A147 Exercise 5.5b What condition determines the casting rate? HINT A148 (A85) You use the equation v ffiffiffiffiffiffiffi gh and read the height of the melt from the figure: Hint A8 ð4þ h ð45 þ 5 þ 5 þ 8Þ mm 18 mm :18 m v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9:81 :18 1:58 m=s Find the required material constants of Al and use Equations () and (3). Hint A3 HINT A149 (A9) F coer you have T L 183 C k Cu 398 W=mK h 1 W=m K h 1 3 W=m K y L T icu y L T icu Region V. Region V is characterized by dt= <. The total heat of solidification is smaller than the heat losses to the surroundings. The heat of solidification decreases owing to decreasing area A of the solidification front, when the crystals iminge against each other. Region V lasts until all melt has solidified, f s 1. Characterize region VI. Hint A61

25 Materials Processing during Casting 5 HINT A15 Exercise 5.9 Check Nusselt s number to estimate the tye of temerature distribution in the solid metal. Hint A9 HINT A151 (A93) If you insert the given material constants and temerature values, you obtain C r metal ð HÞ 1 4 T L T k mould r mould c mould C : :63 1: :5 1 3 :1 16 s=m Calculate the solidification time! HINT A15 Hint A (A6) Radiation and heat transfer to the late occur simultaneously and cool the melt to the melting-oint temerature. Equation (4.48) on age 74 in Chater 4 gives the solidification time: t r metalð HÞ y L 1 þ h T L T h k y L Case : hy L =k 1 If h and/ y L are/is small and k is large, then the term hy L =k 1 and can be neglected in Equation (1). In this case the solidification time can be written as t r metalð HÞ y L T L T h ðþ Aly this infmation to exlain the aearance of the curve. Hint A57 HINT A154 (A86) Comletely analogous calculations as in 4.3a with other material data give t total C y L 1:5 1 6 :5 1:4 h List the values of solidification time, heat of fusion f aluminium and steel and a measure of the driving fce f heat transt. Hint A19 The stri is thin and it is reasonable to assume that its temerature is constant as long as the excess temerature ersists. The temerature rofile is sketched in the figure. How do you roceed? Hint A36 HINT A153 (A1) Case 1: T excess T L T w T Po contact between mould and metal HINT A155 (A48) The straight way is to use reasonable values of material data, temeratures and measured values of y L and t as a basis f calculation of the heat transfer coefficient f the two alternatives. Hint A86 HINT A156 (A9) 1 :6 Y L t 1 þ h T L T w h k ð:6 Y LÞ 1 s B eðt L4 T 4 Þ Y L ð6þ

26 6 Guide to Exercises Solve Equation (6) after introduction of known values and the material constants f aluminium. Assume that e 1. 1 :6 Y L þ 1 ð:6 Y LÞ 1 5: ð Þ Y L ð:6 Y L Þ 1 þ 1 ð:6 Y LÞ : ð Þ Y L :6ð1 þ :73 :6Þ :6Y L Y L ð1 þ :73 :6ÞþY L : Y L Y L :6Y L Y L 1: :5Y L þ :6 1:1364 Y L 16:7Y L þ :6818 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Y L þ8:35 ð8:35þ :68 As Y L < :6, you want the smallest root: Y L 8:35 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 64:5615 :68 8:35 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 64: :35 8:346 :4 m The accuracy of the known quantities involved is not articularly satisfacty. As a check you may eliminate Y L between Equations (5) and (6) and solve y L. If you do not want to ractice solving another equation of second degree, you can look at the next hint. Hint A145 HINT A157 (A49) Equation (4.45) on age 74 in Chater 4 gives T i metal T L T 1 þ h þ T k y L Use this relation and the figures mentioned in the text to discuss the validities of the two figures. Hint A333 HINT A158 (A3) 1. Owing to the motion in the melt, dendrite arms are tn away, which gives crystal multilication due to inhomogeneous nucleation.. The hot melt causes artial remelting of already solidified dendrite arms. The new articles in the melt give crystal multilication in the same way as in oint The convection gives increased heat transt from the melt to the solidification front and a faster decrease of the suerheat of the melt. The last effect results in better ossibilities f the nuclei to survive and grow when the temerature of the melt has decreased to the nucleation temerature T. HINT A159 (A6) The dimensions of the cylinder are given. It is easy to calculate its volume and the area. Hence you can use Chvinov s rule. Are there any comlications? Hint A38 HINT A16 (A67) Two exressions of the heat flow can be obtained: dq Ah cont N dðrvþ ð HÞ where V 4r3 ðþ 3 How do you define the growth rate of the equiaxed crystals? Hint A71 HINT A161 (A88) t 7: : ð Þ :1 s Now you know the total solidification time. How do you obtain the distance? Hint A7

27 Materials Processing during Casting 7 HINT A16 (A4) Sand mould: If you cast a metal in a dry sand mould, Equation (4.7) on age 79 in Chater 4 gives the relation between shell thickness and solidification time: y L ðtþ T i T qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffi ffiffiffi k mould r r metal ð HÞ mould c mould t (b) Pure metals and eutectic alloys ) high L f. Increased temerature ) low viscosity ) high L f. Fmation of dendrites revents the flow strongly ) low L f. Low thermal conductivity and high heat of fusion increase L f. Laminar flow gives a larger fluidity length than turbulent flow. t r metal ð HÞy L 1 4 T i T k mould r mould c mould ðþ HINT A165 (A96) The maj art of the excess heat is removed by convection in the steel melt to the surrounding cast iron mould. The fraction f removed with the aid of radiation is Use the material data and calculate the solidification time. Hint A73 f Qcool rad Q cool :68 total 4 :64 The answer is given in Hint A6. Hint A6 HINT A163 (A64) The heat transfer between the solid metal and the mould can be described by the relation dq h RLðT i T Þ ð5þ Combine Equations () and (5) to eliminate dq=. Hint A316 HINT A166 (A81) t fill can be calculated with the aid of the emirical equation in the text, where A casting is the cross-sectional area of the cylinder r ðr 5cmÞ, h casting 3 cm and h total 8 cm. A srue A runner A lower is unknown and can be derived from Equation (1) as all other quantities are known. Calculate t fill and then A srue. Hint A85 HINT A164 (A8) (A56) (a) Increased cooling caacity of the mould ) low L f Increased surface tension ) low L f. Increased viscosity ) low L f. Wide solidification interval of the alloy ) low L f. HINT A167 (A343) You derive Equation (6) and change to SI units, which gives v front dy :5 1 ffiffiffiffiffiffi 6t ð7þ To find v front you must find the value of t. How do you find t? Hint A41