Heating the Metal <ul><li>Heating furnaces are used to heat the metal to molten temperature sufficient for casting  </li><...
Equation 10.1 <ul><li>H =  ρ V{C s (T m  – T o ) + H f   + C l (T p  –T m )} </li></ul><ul><li>H = Total heat required </l...
Problem 10.2 <ul><li>A sufficient amount of pure copper is to be heated for casting a large plate in an open mold. The pla...
First Step <ul><li>Separate the facts from the fiction </li></ul>©2007 John Wiley & Sons, Inc.  M P Groover,  Fundamentals...
Problem 10.2 Solution <ul><li>Dimensions are 20 in x 10 in x 3in </li></ul><ul><li>T p  = 2150 F </li></ul><ul><li>Volume ...
Problem 10.2 Solution <ul><li>Volume  V  = (20 x 10 x 3)(1 + 10%) = 600(1.1) = 660.0 in 3 </li></ul><ul><li>Assuming  T o ...
Pouring the Molten Metal <ul><li>For this step to be successful, metal must flow into all regions of the mold, most import...
Sand Casting Mold <ul><li>Figure 10.2 (b) Sand casting mold. </li></ul>©2007 John Wiley & Sons, Inc.  M P Groover,  Fundam...
Problem 10.5 <ul><li>The flow rate of liquid metal into the downsprue of a mold = 1 liter/sec. The cross-sectional area at...
Equations <ul><li>Equation 10.4 </li></ul><ul><li>v = (2gh) 1/2 </li></ul><ul><li>v = velocity of liquid metal at the base...
Equations  <ul><li>Equation 10.5 </li></ul><ul><li>Q = v 1 A 1  = v 2 A 2 </li></ul><ul><li>Q = Volumetric flow rate cm 3 ...
Solution 10.5 <ul><li>Flow rate  Q  = 1.0 l/s = 1,000,000 mm 3 /s </li></ul><ul><li>Velocity  v  = (2 x 9815 x 175) 0.5  =...
Problem 10.7 <ul><li>Molten metal can be poured into the pouring cup of a sand mold at a steady rate of 1000 cm 3 /s. The ...
Solution Problem 10.7 <ul><li>Velocity at base  v  = (2 gh ) 0.5  = (2 x 981 x 25) 0.5  = 221.5 cm/s  </li></ul><ul><li>As...
Solidification Time <ul><li>Solidification takes time </li></ul><ul><li>Total solidification time  T TS  = time required f...
Problem 10.17 <ul><li>A disk-shaped part is to be cast out of aluminum. The diameter of the disk = 500 mm and its thicknes...
Solution Problem 10.17 <ul><li>Volume  V  = π D 2 t /4 = π(500) 2 (20)/4 = 3,926,991 mm 3 </li></ul><ul><li>Area  A  = 2π ...
Mold Constant in Chvorinov's Rule <ul><li>Mold constant  C m  depends on:  </li></ul><ul><ul><li>Mold material  </li></ul>...
What Chvorinov's Rule Tells Us <ul><li>A casting with a higher volume‑to‑surface area ratio cools and solidifies more slow...
Problem 10.21 <ul><li>The total solidification times of three casting shapes are to be compared: (1) a sphere, (2) a cylin...
Solution Problem 10.21 <ul><li>For ease of computation, make the substitution 10 cm = 1 decimeter (1 dm). Thus 1000 cm 3  ...
Solution 10.21 cont <ul><li>(1) Sphere volume  V  = π D 3 /6 = 1.0 dm 3 .  D 3  = 6/π = 1.910 dm 3 .  D  = (1.910) 0.333  ...
Solution 10.21 cont <ul><li>(2) Cylinder volume  V  = π D 2 H /4 = π D 3 /4 = 1.0 dm 3 .  D 3  = 4/π = 1.273 dm 3   </li><...
Solution 10.21 cont <ul><li>(3) Cube:  V  =  L 3  =1.0 dm 3 .  L  = 1.0 dm  </li></ul><ul><li>Cube area = 6 L 2  = 6(1) 2 ...
Solution 10.21 cont part b <ul><li>Chvorinov’s Rule  Sphere  =  0.0428 C m </li></ul><ul><li>Chvorinov’s Rule  Cylinder  =...
Solution 10.21 cont part c <ul><li>(c) Given that  C m   = 3.5 min/cm 2  = 350 min/dm 3   </li></ul><ul><li>Sphere:  T TS ...
Shrinkage in Solidification and Cooling <ul><li>Figure 10.8  Shrinkage of a cylindrical casting during solidification and ...
Shrinkage in Solidification and Cooling <ul><li>Figure 10.8  (2) reduction in height and formation of shrinkage cavity cau...
Solidification Shrinkage <ul><li>Occurs in nearly all metals because the solid phase has a higher density than the liquid ...
Shrinkage Allowance <ul><li>Patternmakers account for solidification shrinkage and thermal contraction by making mold cavi...
Problem 10.10 <ul><li>The cavity of a casting mold has dimensions:  L  = 250 mm,  W  = 125 mm and  H  = 20 mm. Determine t...
Solution Problem 10.10 <ul><li>For aluminum, solidification shrinkage = 6.6%, solid contraction during cooling = 5.6%.  </...
Upcoming SlideShare
Loading in …5
×

Ch10 math

8,275 views

Published on

Published in: Business, Technology

Ch10 math

  1. 1. Heating the Metal <ul><li>Heating furnaces are used to heat the metal to molten temperature sufficient for casting </li></ul><ul><li>The heat required is the sum of: </li></ul><ul><ul><li>Heat to raise temperature to melting point </li></ul></ul><ul><ul><li>Heat of fusion to convert from solid to liquid </li></ul></ul><ul><ul><li>Heat to raise molten metal to desired temperature for pouring </li></ul></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  2. 2. Equation 10.1 <ul><li>H = ρ V{C s (T m – T o ) + H f + C l (T p –T m )} </li></ul><ul><li>H = Total heat required </li></ul><ul><li>ρ = density, g/cm 3 or lbs/in 3 </li></ul><ul><li>V = volume of metal being heated, cm 3 or in 3 </li></ul><ul><li>C s = weight specific heat of solid metal J/g-C or Btu/lb-F </li></ul><ul><li>T m = melting temperature, C or F </li></ul><ul><li>T o = starting temperature, C or F </li></ul><ul><li>T p = pouring temperature, C or F </li></ul><ul><li>H f = heat of fusion J/g or Btu/lb </li></ul><ul><li>C l = weight specific heat of liquid metal J/g-C or Btu/lb-F </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  3. 3. Problem 10.2 <ul><li>A sufficient amount of pure copper is to be heated for casting a large plate in an open mold. The plate has dimensions: length = 20 in, width = 10 in, and thickness = 3 in. Compute the amount of heat that must be added to the metal to heat it to a temperature of 2150°F for pouring. Assume that the amount of metal heated will be 10% more than what is needed to fill the mold cavity. Properties of the metal are: density = 0.324 lbm/in 3 , melting point = 1981°F, specific heat of the metal = 0.093 Btu/lbm-F in the solid state and 0.090 Btu/lbm-F in the liquid state; and heat of fusion = 80 Btu/lbm. </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  4. 4. First Step <ul><li>Separate the facts from the fiction </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  5. 5. Problem 10.2 Solution <ul><li>Dimensions are 20 in x 10 in x 3in </li></ul><ul><li>T p = 2150 F </li></ul><ul><li>Volume + 10% </li></ul><ul><li>ρ = 0.324 lbs/in 3 </li></ul><ul><li>T m = 1981 F </li></ul><ul><li>C s = 0.093 Btu/lbm-F </li></ul><ul><li>C l = 0.090 Btu/lbm-F </li></ul><ul><li>H f = 80 Btu/lbm </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  6. 6. Problem 10.2 Solution <ul><li>Volume V = (20 x 10 x 3)(1 + 10%) = 600(1.1) = 660.0 in 3 </li></ul><ul><li>Assuming T o = 75 °F </li></ul><ul><li>H = 0.324 x 660{0.093(1981 - 75) + 80 + 0.090(2150 - 1981)} </li></ul><ul><li>H = 213.84{177.26 + 80 + 15.21} </li></ul><ul><li>H = 58,265 Btu </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  7. 7. Pouring the Molten Metal <ul><li>For this step to be successful, metal must flow into all regions of the mold, most importantly the main cavity, before solidifying </li></ul><ul><li>Factors that determine success </li></ul><ul><ul><li>Pouring temperature </li></ul></ul><ul><ul><li>Pouring rate </li></ul></ul><ul><ul><li>Turbulence </li></ul></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  8. 8. Sand Casting Mold <ul><li>Figure 10.2 (b) Sand casting mold. </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  9. 9. Problem 10.5 <ul><li>The flow rate of liquid metal into the downsprue of a mold = 1 liter/sec. The cross-sectional area at the top of the sprue = 800 mm 2 and its length = 175 mm. What area should be used at the base of the sprue to avoid aspiration of the molten metal? </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  10. 10. Equations <ul><li>Equation 10.4 </li></ul><ul><li>v = (2gh) 1/2 </li></ul><ul><li>v = velocity of liquid metal at the base of the sprue cm/s or in/sec </li></ul><ul><li>g = gravity 981 cm/sec/sec or 386 in/sec/sec </li></ul><ul><li>h = height of the downsprue </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  11. 11. Equations <ul><li>Equation 10.5 </li></ul><ul><li>Q = v 1 A 1 = v 2 A 2 </li></ul><ul><li>Q = Volumetric flow rate cm 3 /sec or in 3 /sec </li></ul><ul><li>v = velocity </li></ul><ul><li>A = cross-sectional area cm 2 or in 2 </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  12. 12. Solution 10.5 <ul><li>Flow rate Q = 1.0 l/s = 1,000,000 mm 3 /s </li></ul><ul><li>Velocity v = (2 x 9815 x 175) 0.5 = 1854 mm/s </li></ul><ul><li>Area at base A = 1,000,000/1854 = 540 mm 2 </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  13. 13. Problem 10.7 <ul><li>Molten metal can be poured into the pouring cup of a sand mold at a steady rate of 1000 cm 3 /s. The molten metal overflows the pouring cup and flows into the downsprue. The cross-section of the sprue is round, with a diameter at the top = 3.4 cm. If the sprue is 25 cm long, determine the proper diameter at its base so as to maintain the same volume flow rate. </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  14. 14. Solution Problem 10.7 <ul><li>Velocity at base v = (2 gh ) 0.5 = (2 x 981 x 25) 0.5 = 221.5 cm/s </li></ul><ul><li>Assuming volumetric continuity, area at base A = (1000 cm³/s)/(221.5 cm/s) = 4.51 cm 2 </li></ul><ul><li>Area of sprue A = π D 2 /4; rearranging, D 2 = 4 A /π = 4(4.51)/π = 5.74 cm 2 </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  15. 15. Solidification Time <ul><li>Solidification takes time </li></ul><ul><li>Total solidification time T TS = time required for casting to solidify after pouring </li></ul><ul><li>T TS depends on size and shape of casting by relationship known as Chvorinov's Rule </li></ul><ul><li>where TST = total solidification time; V = volume of the casting; A = surface area of casting; n = exponent with typical value = 2; and C m is mold constant. </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  16. 16. Problem 10.17 <ul><li>A disk-shaped part is to be cast out of aluminum. The diameter of the disk = 500 mm and its thickness = 20 mm. If the mold constant = 2.0 sec/mm 2 in Chvorinov's Rule, how long will it take the casting to solidify? </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  17. 17. Solution Problem 10.17 <ul><li>Volume V = π D 2 t /4 = π(500) 2 (20)/4 = 3,926,991 mm 3 </li></ul><ul><li>Area A = 2π D 2 /4 + π Dt = π(500) 2 /2 + π(500)(20) = 424,115 mm 2 </li></ul><ul><li>Chvorinov’s Rule: T TS = C m ( V / A ) 2 = 2.0(3,926,991/424,115) 2 = 171.5 s = 2.86 min </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  18. 18. Mold Constant in Chvorinov's Rule <ul><li>Mold constant C m depends on: </li></ul><ul><ul><li>Mold material </li></ul></ul><ul><ul><li>Thermal properties of casting metal </li></ul></ul><ul><ul><li>Pouring temperature relative to melting point </li></ul></ul><ul><li>Value of C m for a given casting operation can be based on experimental data from previous operations carried out using same mold material, metal, and pouring temperature, even though the shape of the part may be quite different </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  19. 19. What Chvorinov's Rule Tells Us <ul><li>A casting with a higher volume‑to‑surface area ratio cools and solidifies more slowly than one with a lower ratio </li></ul><ul><ul><li>To feed molten metal to main cavity, TST for riser must greater than TST for main casting </li></ul></ul><ul><li>Since mold constants of riser and casting will be equal, design the riser to have a larger volume‑to‑area ratio so that the main casting solidifies first </li></ul><ul><ul><li>This minimizes the effects of shrinkage </li></ul></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  20. 20. Problem 10.21 <ul><li>The total solidification times of three casting shapes are to be compared: (1) a sphere, (2) a cylinder, in which the length-to-diameter ratio = 1.0, and (3) a cube. For all three geometries, the volume = 1000 cm 3 . The same casting alloy is used in the three cases. (a) Determine the relative solidification times for each geometry. (b) Based on the results of part (a), which geometric element would make the best riser? (c) If the mold constant = 3.5 min/cm 2 in Chvorinov's Rule, compute the total solidification time for each casting. </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  21. 21. Solution Problem 10.21 <ul><li>For ease of computation, make the substitution 10 cm = 1 decimeter (1 dm). Thus 1000 cm 3 = 1.0 dm 3 . </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  22. 22. Solution 10.21 cont <ul><li>(1) Sphere volume V = π D 3 /6 = 1.0 dm 3 . D 3 = 6/π = 1.910 dm 3 . D = (1.910) 0.333 = 1.241 dm </li></ul><ul><li>Sphere area A = π D 2 = π(1.241) 2 = 4.836 dm 2 </li></ul><ul><li>V/A = 1.0/4.836 = 0.2067 dm </li></ul><ul><li>Chvorinov’s Rule T TS = (0.2067) 2 C m = 0.0428 C m </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  23. 23. Solution 10.21 cont <ul><li>(2) Cylinder volume V = π D 2 H /4 = π D 3 /4 = 1.0 dm 3 . D 3 = 4/π = 1.273 dm 3 </li></ul><ul><li>Therefore, D = H = (1.273) 0.333 = 1.084 dm </li></ul><ul><li>Cylinder area A = 2π D 2 /4 + π DL = 2π(1.084) 2 /4 + π(1.084)(1.084) = 5.536 dm 2 </li></ul><ul><li>V/A = 1.0/5.536 = 0.1806 dm </li></ul><ul><li>Chvorinov’s Rule T TS = (0.1806) 2 C m = 0.0326 C m </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  24. 24. Solution 10.21 cont <ul><li>(3) Cube: V = L 3 =1.0 dm 3 . L = 1.0 dm </li></ul><ul><li>Cube area = 6 L 2 = 6(1) 2 = 6.0 dm 2 </li></ul><ul><li>V/A = 1.0/6.0 = 0.1667 dm </li></ul><ul><li>Chvorinov’s Rule T TS = (0.1667) 2 C m = 0.02778 C m </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  25. 25. Solution 10.21 cont part b <ul><li>Chvorinov’s Rule Sphere = 0.0428 C m </li></ul><ul><li>Chvorinov’s Rule Cylinder = 0.0326 C m </li></ul><ul><li>Chvorinov’s Rule Cube = 0.02778 C m </li></ul><ul><li>(b) Sphere would be the best riser, since V/A ratio is greatest. </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  26. 26. Solution 10.21 cont part c <ul><li>(c) Given that C m = 3.5 min/cm 2 = 350 min/dm 3 </li></ul><ul><li>Sphere: T TS = 0.0428(350) = 14.98 min </li></ul><ul><li>Cylinder: T TS = 0.0326(350) = 11.41 min </li></ul><ul><li>Cube: T TS = 0.02778(350) = 9.72 min </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  27. 27. Shrinkage in Solidification and Cooling <ul><li>Figure 10.8 Shrinkage of a cylindrical casting during solidification and cooling: (0) starting level of molten metal immediately after pouring; (1) reduction in level caused by liquid contraction during cooling (dimensional reductions are exaggerated for clarity). </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  28. 28. Shrinkage in Solidification and Cooling <ul><li>Figure 10.8 (2) reduction in height and formation of shrinkage cavity caused by solidification shrinkage; (3) further reduction in height and diameter due to thermal contraction during cooling of solid metal (dimensional reductions are exaggerated for clarity). </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  29. 29. Solidification Shrinkage <ul><li>Occurs in nearly all metals because the solid phase has a higher density than the liquid phase </li></ul><ul><li>Thus, solidification causes a reduction in volume per unit weight of metal </li></ul><ul><li>Exception: cast iron with high C content </li></ul><ul><ul><li>Graphitization during final stages of freezing causes expansion that counteracts volumetric decrease associated with phase change </li></ul></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  30. 30. Shrinkage Allowance <ul><li>Patternmakers account for solidification shrinkage and thermal contraction by making mold cavity oversized </li></ul><ul><li>Amount by which mold is made larger relative to final casting size is called pattern shrinkage allowance </li></ul><ul><li>Casting dimensions are expressed linearly, so allowances are applied accordingly </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  31. 31. Problem 10.10 <ul><li>The cavity of a casting mold has dimensions: L = 250 mm, W = 125 mm and H = 20 mm. Determine the dimensions of the final casting after cooling to room temperature if the cast metal is aluminum. Assume that the mold is full at the start of solidification and that shrinkage occurs uniformly in all directions. Use the shrinkage values given in Table 10.1. </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e
  32. 32. Solution Problem 10.10 <ul><li>For aluminum, solidification shrinkage = 6.6%, solid contraction during cooling = 5.6%. </li></ul><ul><li>Total volumetric contraction = (1-0.066)(1-0.056) = 0.8817 </li></ul><ul><li>Linear contraction = (0.8817) 0.333 = 0.9589 </li></ul><ul><li>Final casting dimensions: L = 250(0.9589) = 239.725 cm </li></ul><ul><li>W = 125(0.9589) = 119.863 cm </li></ul><ul><li>H = 20(0.9589) = 19.178 cm </li></ul>©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e

×