Degarmo's Materials And Processes In Manufacturing
13th Edition
ISBN: 9781119492825
Author: Black, J. Temple, Kohser, Ronald A., Author.
Publisher: Wiley,
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Chapter 38, Problem 73RQ
What is driving the conversion to lead-free solders?
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Chapter 38 Solutions
Degarmo's Materials And Processes In Manufacturing
Ch. 38 - What are some joining conditions where welding...Ch. 38 - What are some of the lower-temperature methods of...Ch. 38 - In what ways is a thermit weld similar to the...Ch. 38 - What is the source of the welding heat in thermit...Ch. 38 - For what types of applications might thermit...Ch. 38 - What is the source of the welding heat in...Ch. 38 - What are some of the various functions of the slag...Ch. 38 - Electroslag welding would be most attractive for...Ch. 38 - What is the source of heat in electron-beam...Ch. 38 - Why is a high vacuum required in the electron-beam...
Ch. 38 - What types of production limitations are imposed...Ch. 38 - What are the major assets and negative features of...Ch. 38 - What are some of the attractive features of...Ch. 38 - What is unique about the fusion zone geometry of...Ch. 38 - Describe the weld pool geometry and size of the...Ch. 38 - What is an autogenous weld?Ch. 38 - What might be necessary to permit the laser...Ch. 38 - What are some of the Ways in which laser-beam...Ch. 38 - What are the three common types of industrial...Ch. 38 - Which type of laser light can be transmitted...Ch. 38 - What are some of the attractive features of a...Ch. 38 - Why is laser-beam welding an attractive process...Ch. 38 - What are the attractive properties of hybrid...Ch. 38 - Prob. 24RQCh. 38 - What is the function of the assist gas in...Ch. 38 - What is the difference between exothermic cutting...Ch. 38 - Which type of laser is preferred for cutting...Ch. 38 - Prob. 28RQCh. 38 - Prob. 29RQCh. 38 - What features have made lasers a common means of...Ch. 38 - What are some of the attractive features of laser...Ch. 38 - What are some common objectives of surfacing...Ch. 38 - What types of materials are applied by surfacing...Ch. 38 - Prob. 34RQCh. 38 - What is the benefit of high-velocity oxyfuel...Ch. 38 - What are some of the arc or plasma techniques that...Ch. 38 - How is thermal spraying similar to surfacing? How...Ch. 38 - Prob. 38RQCh. 38 - Prob. 39RQCh. 38 - Provide a reasonable definition of brazing?Ch. 38 - What are some key differences between brazing and...Ch. 38 - What kinds of materials or combinations can be...Ch. 38 - What advantages can be gained by the lower...Ch. 38 - Why do brazed joints have an enhanced...Ch. 38 - What is the most important factor influencing the...Ch. 38 - How does capillary action relate to joint...Ch. 38 - Why is it necessary to adjust the initial...Ch. 38 - What is wettability? Fluidity? How do each relate...Ch. 38 - What are the two most common types of brazed...Ch. 38 - How do the butt-lap and scarf joint configurations...Ch. 38 - What are some important considerations when...Ch. 38 - What are some of the most commonly used brazing...Ch. 38 - Why are eutectic alloys attractive as brazing...Ch. 38 - What special measures should be taken when brazing...Ch. 38 - What are the three primary functions of a brazing...Ch. 38 - Why is it important to preclean brazing surfaces...Ch. 38 - How might braze metal be preloaded into joints?Ch. 38 - What is the purpose of brazing jigs and fixtures?Ch. 38 - 59- What are some factors to consider when...Ch. 38 - What are the advantages and disadvantages of torch...Ch. 38 - What is the primary attraction of furnace-brazing...Ch. 38 - Why might reducing atmospheres or a vacuum be...Ch. 38 - What are some of the attractive features of...Ch. 38 - Why is dip brazing usually restricted to use with...Ch. 38 - What are some of the attractive features of...Ch. 38 - Why is flux removal a necessary part of many...Ch. 38 - What benefits can be achieved through fluxless...Ch. 38 - How does braze welding differ from traditional...Ch. 38 - What is the primary difference between brazing and...Ch. 38 - What are the six steps of a soldering operation?Ch. 38 - Why is soldering unattractive if a high-strength...Ch. 38 - For many years, the most common solders have been...Ch. 38 - What is driving the conversion to lead-free...Ch. 38 - What are some of the difficulties encountered when...Ch. 38 - What are the two basic families of soldering flux?Ch. 38 - What are some of the more common heat sources for...Ch. 38 - Why is wave soldering attractive for making the...Ch. 38 - Describe the vapor-phase soldering process.Ch. 38 - A common problem with brazed or soldered joints is...Ch. 38 - When molten metal deposition is applied to a...Ch. 38 - Prob. 3P
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- Hello I was going over the solution for this probem and I'm a bit confused on the last part. Can you please explain to me 1^4 was used for the Co of the tubular cross section? Thank you!arrow_forwardBlood (HD = 0.45 in large diameter tubes) is forced through hollow fiber tubes that are 20 µm in diameter.Equating the volumetric flowrate expressions from (1) assuming marginal zone theory and (2) using an apparentviscosity for the blood, estimate the marginal zone thickness at this diameter. The viscosity of plasma is 1.2 cParrow_forwardQ2: Find the shear load on bolt A for the connection shown in Figure 2. Dimensions are in mm Fig. 2 24 0-0 0-0 A 180kN (10 Markarrow_forward
- determine the direction and magnitude of angular velocity ω3 of link CD in the four-bar linkage using the relative velocity graphical methodarrow_forwardFour-bar linkage mechanism, AB=40mm, BC=60mm, CD=70mm, AD=80mm, =60°, w1=10rad/s. Determine the direction and magnitude of w3 using relative motion graphical method. A B 2 3 77777 477777arrow_forwardFour-bar linkage mechanism, AB=40mm, BC=60mm, CD=70mm, AD=80mm, =60°, w1=10rad/s. Determine the direction and magnitude of w3 using relative motion graphical method. A B 2 3 77777 477777arrow_forward
- The evaporator of a vapor compression refrigeration cycle utilizing R-123 as the refrigerant isbeing used to chill water. The evaporator is a shell and tube heat exchanger with the water flowingthrough the tubes. The water enters the heat exchanger at a temperature of 54°F. The approachtemperature difference of the evaporator is 3°R. The evaporating pressure of the refrigeration cycleis 4.8 psia and the condensing pressure is 75 psia. The refrigerant is flowing through the cycle witha flow rate of 18,000 lbm/hr. The R-123 leaves the evaporator as a saturated vapor and leaves thecondenser as a saturated liquid. Determine the following:a. The outlet temperature of the chilled waterb. The volumetric flow rate of the chilled water (gpm)c. The UA product of the evaporator (Btu/h-°F)d. The heat transfer rate between the refrigerant and the water (tons)arrow_forward(Read image) (Answer given)arrow_forwardProblem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and (y2), respectively. Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s]. Givens: y1 = 4.112 m y2 = 0.387 m b = 0.942 m Answers: ( 1 ) 1880.186 lit/s ( 2 ) 4042.945 lit/s ( 3 ) 2553.11 lit/s ( 4 ) 3130.448 lit/sarrow_forward
- Problem (14): A pump is being used to lift water from an underground tank through a pipe of diameter (d) at discharge (Q). The total head loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h where (V) is the flow velocity in the pipe. The elevation difference between the pump and tank surface is (h). Given the values of h [cm], d [cm], and K [-], calculate the maximum discharge Q [Lit/s] beyond which cavitation would take place at the pump entrance. Assume Turbulent flow conditions. Givens: h = 120.31 cm d = 14.455 cm K = 8.976 Q Answers: (1) 94.917 lit/s (2) 49.048 lit/s ( 3 ) 80.722 lit/s 68.588 lit/s 4arrow_forwardProblem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…arrow_forwardThe flow rate is 12.275 Liters/s and the diameter is 6.266 cm.arrow_forward
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