
DeGarmo's Materials and Processes in Manufacturing
12th Edition
ISBN: 9781118987674
Author: J. T. Black, Ronald A. Kohser
Publisher: WILEY
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Textbook Question
Chapter 35, Problem 16RQ
What are some of the factors that influence the cost of making a weldment?
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Problem (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/s
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
4
Chapter 35 Solutions
DeGarmo's Materials and Processes in Manufacturing
Ch. 35 - What types of design features favor manufacture as...Ch. 35 - What types of manufacturing processes fall under...Ch. 35 - Define welding.Ch. 35 - What conditions are required to produce an ideal...Ch. 35 - What are some of the ways in which welding...Ch. 35 - What are some possible problems associated with...Ch. 35 - What are the three primary aspects required to...Ch. 35 - How are welding processes identified by the...Ch. 35 - What is thermal cutting?Ch. 35 - What are some of the common types of weld defects?
Ch. 35 - What are the four basic types of fusion welds?Ch. 35 - What are some of the common edge configurations...Ch. 35 - What is the role of an insert in welding?Ch. 35 - What types of weld joints commonly employ fillet...Ch. 35 - What are the five basic joint tvpes for fusion...Ch. 35 - What are some of the factors that influence the...Ch. 35 - Why is it important to consider welded products as...Ch. 35 - How does the fracture resistance and temperature...Ch. 35 - How might excessive rigidity actually be a...Ch. 35 - What is autogenous welding?Ch. 35 - In what way is the weld-pool segment of a fusion...Ch. 35 - Why is it possible for the fusion zone to have a...Ch. 35 - Why is it not uncommon for the selected filler...Ch. 35 - What are some of the defects or problems that can...Ch. 35 - Why can the material properties vary widely within...Ch. 35 - What are some of the structure and property...Ch. 35 - Why do most welding failures occur in the...Ch. 35 - Discuss the various regions within the...Ch. 35 - What are some of the characteristics and...Ch. 35 - What process features can increase the size of the...Ch. 35 - What are some of the difficulties or limitations...Ch. 35 - What is the purpose of pre- and postheating in...Ch. 35 - What heat-related metallurgical effects can...Ch. 35 - What causes weld-induced residual stresses?Ch. 35 - What is the cause of reaction-type residual...Ch. 35 - How are reaction stresses affected by the distance...Ch. 35 - What are some of the techniques that can reduce...Ch. 35 - How can the surfaces of weldments be put into...Ch. 35 - In what ways might welding create geometric...Ch. 35 - 40. Why might a welded structure warp if the...Ch. 35 - Why might a stress relief heat treatment be...Ch. 35 - What are some of the techniques that can be...Ch. 35 - Why are the terms weldability and joinability...Ch. 35 - Prob. 1PCh. 35 - Two pieces of AISI 1025 steel are being...Ch. 35 - Figure 35.A schematically depicts the design of a...Ch. 35 - Investigate and summarize some of the important...Ch. 35 - What do you foresee as the major difficulties when...
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- Problem (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_forwardAn experimental setup is being built to study the flow in a large water main (i.e., a large pipe). The water main is expected to convey a discharge (Qp). The experimental tube will be built at a length scale of 1/20 of the actual water main. After building the experimental setup, the pressure drop per unit length in the model tube (APm/Lm) is measured. Problem (20): Given the value of APm/Lm [kPa/m], and assuming pressure coefficient similitude, calculate the drop in the pressure per unit length of the water main (APP/Lp) in [Pa/m]. Givens: AP M/L m = 590.637 kPa/m meen Answers: ( 1 ) 59.369 Pa/m ( 2 ) 73.83 Pa/m (3) 95.443 Pa/m ( 4 ) 44.444 Pa/m *******arrow_forward
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