CH 15
CH 15
CH 15
15 POWDER METALLURGY
Review Questions
15.1 Name some of the reasons for the commercial importance of powder metallurgy technology.
Answer. PM is important because (1) parts can be made to net or near net shape, (2) parts can be
made with a controlled level of porosity, (3) certain metals difficult to process by other methods can
be processed by PM, and (4) PM allows the formulation of unusual alloys not easily obtained by
traditional alloying methods.
15.2 What are some of the disadvantages of PM methods?
Answer. Disadvantages include (1) high tooling costs, (2) metal powders are expensive, (3)
difficulties in storing and handling metallic powders, (4) certain limitations on part geometry
imposed by the uniaxial press methods, and (5) variations in density in a PM component can be
troublesome.
15.3 In the screening of powders for sizing, what is meant by the term mesh count?
Answer. The mesh count of the screen is the number of openings per linear inch.
15.4 What is the difference between open pores and closed pores in a metallic powders?
Answer. Open pores are air spaces between particles, while closed pores are voids internal to a
particle.
15.5 What is meant by the term aspect ratio for a metallic particle?
Answer. The aspect ratio of a particle is the ratio of the maximum dimension to the minimum
dimension of the given particle.
15.6 How would one measure the angle of repose for a given amount of metallic powder?
Answer. One measure would be to let the powders flow through a small funnel and measure the
angle taken by the resulting pile of powders relative to the horizontal.
15.7 Define bulk density and true density for metallic powders.
Answer. Bulk density refers to the weight per volume of the powders in the loose state, while true
density is the weight per volume of the true volume of metal in the powders (the volume that would
result if the powders were melted).
15.8 What are the principal methods used to produce metallic powders?
Answer. The powder production methods are (1) atomization - the conversion of molten metal into
droplets which solidify into powders; (2) chemical reduction - reducing metallic oxides by use of
reducing agents which combine with the oxygen to free the metals in the form of powders; and (3)
electrolysis - use of an electrolytic cell to deposit particles of the metal onto the cathode in the cell.
15.9 What are the three basic steps in the conventional powder metallurgy shaping process?
Answer. The steps are (1) blending and/or mixing, (2) pressing, and (3) sintering.
15.10 What is the technical difference between blending and mixing in powder metallurgy?
Answer. Blending means combining particles of the same chemistry but different sizes, while
mixing refers to the combining of metal powders of different chemistries.
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Solutions for Fundamentals of Modern Manufacturing, 5e (published by Wiley) MPGroover 2012
15.11 What are some of the ingredients usually added to the metallic powders during blending and/or
mixing?
Answer. The additives include (1) lubricants, (2) binders, and (3) deflocculants.
15.12 What is meant by the term green compact?
Answer. The green compact is the pressed but not yet sintered PM part.
15.13 Describe what happens to the individual particles during compaction.
Answer. Starting with the initial powder arrangement, the particles are first repacked into a more
efficient arrangement, followed by deformation of the particles as pressure is increased.
15.14 What are the three steps in the sintering cycle in PM?
Answer. The three steps in the cycle are (1) preheat, in which lubricants and binders are burned off,
(2) sintering, and (3) cool down.
15.15 What are some of the reasons why a controlled atmosphere furnace is desirable in sintering?
Answer. Some of the purposes of a controlled atmosphere furnace are (1) to protect against
oxidation, (2) to provide a reducing atmosphere to remove existing oxides, (3) to provide a
carburizing atmosphere, and (4) to remove lubricants and binders from pressing.
15.16 What is the difference between impregnation and infiltration in powder metallurgy?
Answer. Impregnation is when oil or other fluid is permeated into the pores of a sintered PM part.
Infiltration is when a molten metal (other than the PM metal) is permeated into the pores of a
sintered part.
15.17 What is the difference between powder injection molding and metal injection molding?
Answer. Metal injection molding is a subset of powder injection molding, in which the powders are
metallic. The more general term includes powders of ceramic.
15.18 How is isostatic pressing distinguished from conventional pressing and sintering in PM?
Answer. Isostatic pressing applies hydrostatic pressure to all sides of the mold, whereas
conventional pressing is uniaxial.
15.19 Describe liquid phase sintering.
Answer. Liquid phase sintering occurs when two metals of different melting temperatures are
sintered at a temperature between their melting points. Accordingly, one metal melts, thoroughly
wetting the solid particles and creating a strong bonding between the metals upon solidification.
15.20 What are the two basic classes of metal powders as far as chemistry is concerned?
Answer. The two classes are (1) elemental powders - powders of pure metal such as iron or copper,
and (2) pre-alloyed powders - powders of alloys such as stainless steel or brass.
15.21 Why is PM technology so well suited to the production of gears and bearings?
Answer. The reasons are (1) the geometries of these parts lend themselves to conventional PM
pressing, which consists of pressing in one direction, and (2) the porosity allows impregnation of the
PM parts with lubricants.
Problems
Answers to problems labeled (A) are listed in an Appendix at the back of the book.
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(c) Cylinder with L/D = 1.0. For this cylinder shape, L = D. Thus, A = 2D2/4 + DL = 0.5L2 + L2
= 1.5L2, and V = (D2/4)L = 0.25L3.
Find diameter D of a sphere of equivalent volume.
V = D3/6 = 0.25L3
D3 = 6(0.25L3)/ = 1.5L3
D = (1.5 L3)0.333 = 1.1447 L
Ks = AD/V = (1.5L2)(1.1447 L)/0.25L3 = 6.868
(d) Cylinder with L/D = 2.0. For this cylinder shape, 0.5L = D. Thus, A = 2D2/4 + DL =
0.5(0.5L)2 + (0.5L)L = 0.125L2 + 0.5L2 = 0.625L2, and V = (D2/4)L = 0.25(0.5L)2 L =
0.0625L3
Find diameter D of a sphere of equivalent volume.
V = D3/6 = 0.0625L3
D3 = 6(0.0625L3)/ = 0.375L3
D = (0.375 L3)0.333 = 0.721 L
Ks = A D/V = (0.625L2)(0.721 L)/0.0625L3 = 7.211
15.6 (A) Determine the shape factors for particles that are disk-shaped flakes with thickness-to-diameter
ratios of (a) 1:10 and (b) 1:20.
Solution: (a) Disk with t/D = L/D = 1/10 = 0.10. For this shape, 10L = D. Thus, A = 2D2/4 + DL
= 0.5(10L)2 + (10L)L = 50L2 + 10L2 = 60L2, and V = (D2/4)L = 0.25(10L)2 L = 25L3
Find diameter D of a sphere of equivalent volume.
V = D3/6 = 25L3
D3 = 6(25L3)/ = 150L3
D = (150 L3)1/3 = 5.313 L
Ks = A D/V = (60L2)(5.313 L)/25L3 = 12.75
(b) Disk with L/D = 0.05. For this shape, 20L = D. Thus, A = 2D2/4 + DL = 0.5(20L)2 + (20L)L
= 200L2 + 20L2 = 220L2, and V = (D2/4)L = 0.25(20L)2 L = 314.16L3
Find diameter D of a sphere of equivalent volume.
V = D3/6 = 314.16L3
D3 = 6(314.16L3)/ = 1884.96 L3
D = (1884.96 L3)1/3 = 12.353 L
Ks = A D/V = (220L2)(12.353 L)/314.16L3 = 8.65
15.7 (USCS units) A pile of iron powder weighs 2 lb. The particles are spherical in shape and all have the
same diameter of 0.002 in. (a) Determine the total surface area of all the particles in the pile. (b) If
the packing factor = 0.6, determine the volume taken by the pile. Note: the density of iron = 0.284
lb/in3.
Solution: (a) For a spherical particle of D = 0.002 in, V = D3/6 = (0.002)3/6
= 0.00000000418 = 4.18 x 10-9 in3/particle
Weight per particle W = V = 0.284(4.18 x 10-9 in3) = 1.19 x 10-9 lb/particle
Number of particles in 2 lb = 2.0/(1.19 x 10-9) = 1.681 x 109
A = D2 = (0.002)2 = 0.00001256 in2 = 12.56 x 10-6 in2
Total surface area = (1.681 x 109)(12.56 x 10-6) = 21.116 x 103 in2
(b) With a packing factor of 0.6, the total volume taken up by the pile = (2.0/0.284)/0.6 = 11.74 in3
15.8 (USCS units) Solve Problem 15.7, except that the diameter of the particles is 0.004 in. Assume the
same packing factor.
Solution: (a) For a spherical particle of D = 0.004 in, V = D3/6 = (0.004)3/6
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