Foundations of Materials Science and Engineering
Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Chapter 11.13, Problem 14KCP
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Explain the arrangement of bonding in the cristobalite silica structure.

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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
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…

Chapter 11 Solutions

Foundations of Materials Science and Engineering

Ch. 11.13 - Prob. 11KCPCh. 11.13 - Prob. 12KCPCh. 11.13 - Prob. 13KCPCh. 11.13 - Prob. 14KCPCh. 11.13 - Describe the feldspar network structure.Ch. 11.13 - Prob. 16KCPCh. 11.13 - Prob. 17KCPCh. 11.13 - Describe two methods for preparing ceramic raw...Ch. 11.13 - Prob. 19KCPCh. 11.13 - Prob. 20KCPCh. 11.13 - Prob. 21KCPCh. 11.13 - Prob. 22KCPCh. 11.13 - Prob. 23KCPCh. 11.13 - Prob. 24KCPCh. 11.13 - Prob. 25KCPCh. 11.13 - Prob. 26KCPCh. 11.13 - What are the purposes of drying ceramic products...Ch. 11.13 - Prob. 28KCPCh. 11.13 - What is the vitrification process? In what type of...Ch. 11.13 - Prob. 30KCPCh. 11.13 - Prob. 31KCPCh. 11.13 - Prob. 32KCPCh. 11.13 - Prob. 33KCPCh. 11.13 - Prob. 34KCPCh. 11.13 - Prob. 35KCPCh. 11.13 - Prob. 36KCPCh. 11.13 - Prob. 37KCPCh. 11.13 - Prob. 38KCPCh. 11.13 - Why do most ceramic materials have low thermal...Ch. 11.13 - Prob. 40KCPCh. 11.13 - Prob. 41KCPCh. 11.13 - Prob. 42KCPCh. 11.13 - Prob. 43KCPCh. 11.13 - Prob. 44KCPCh. 11.13 - Prob. 45KCPCh. 11.13 - Prob. 46KCPCh. 11.13 - How is a glass distinguished from other ceramic...Ch. 11.13 - Prob. 48KCPCh. 11.13 - Prob. 49KCPCh. 11.13 - Prob. 50KCPCh. 11.13 - Prob. 51KCPCh. 11.13 - Prob. 52KCPCh. 11.13 - Prob. 53KCPCh. 11.13 - Prob. 54KCPCh. 11.13 - Prob. 55KCPCh. 11.13 - Prob. 56KCPCh. 11.13 - Prob. 57KCPCh. 11.13 - Prob. 58KCPCh. 11.13 - Prob. 59KCPCh. 11.13 - Prob. 60KCPCh. 11.13 - Prob. 61KCPCh. 11.13 - Prob. 62KCPCh. 11.13 - Prob. 63AAPCh. 11.13 - Prob. 64AAPCh. 11.13 - Prob. 65AAPCh. 11.13 - Prob. 66AAPCh. 11.13 - Prob. 67AAPCh. 11.13 - Prob. 70AAPCh. 11.13 - Calculate the ionic packing factor for (a) MnO and...Ch. 11.13 - Prob. 72AAPCh. 11.13 - Prob. 73AAPCh. 11.13 - Prob. 74AAPCh. 11.13 - Prob. 75AAPCh. 11.13 - Prob. 77AAPCh. 11.13 - Prob. 78AAPCh. 11.13 - Prob. 79AAPCh. 11.13 - Prob. 80AAPCh. 11.13 - Prob. 81AAPCh. 11.13 - Why are triaxial porcelains not satisfactory for...Ch. 11.13 - Prob. 83AAPCh. 11.13 - Prob. 84AAPCh. 11.13 - Prob. 85AAPCh. 11.13 - What causes the lack of plasticity in crystalline...Ch. 11.13 - Prob. 87AAPCh. 11.13 - Prob. 88AAPCh. 11.13 - Prob. 89AAPCh. 11.13 - A reaction-bonded silicon nitride ceramic has a...Ch. 11.13 - Prob. 91AAPCh. 11.13 - Prob. 92AAPCh. 11.13 - Prob. 93AAPCh. 11.13 - Prob. 94AAPCh. 11.13 - How does the silica network of a simple silica...Ch. 11.13 - Prob. 96AAPCh. 11.13 - Prob. 97AAPCh. 11.13 - Prob. 98AAPCh. 11.13 - Prob. 99AAPCh. 11.13 - Prob. 100AAPCh. 11.13 - Prob. 101AAPCh. 11.13 - Prob. 102AAPCh. 11.13 - Prob. 103AAPCh. 11.13 - Prob. 104AAPCh. 11.13 - Prob. 105AAPCh. 11.13 - Prob. 106AAPCh. 11.13 - Prob. 107AAPCh. 11.13 - Prob. 108SEPCh. 11.13 - Prob. 109SEPCh. 11.13 - Prob. 110SEPCh. 11.13 - Prob. 111SEPCh. 11.13 - Prob. 112SEPCh. 11.13 - Alumina (A12O3) and chromium oxide (Cr2O3) are...Ch. 11.13 - (a) How are the ceramic tiles used in the thermal...Ch. 11.13 - The nose cap and the wing leading edges of the...Ch. 11.13 - Prob. 116SEPCh. 11.13 - Prob. 117SEPCh. 11.13 - Prob. 118SEPCh. 11.13 - Prob. 119SEPCh. 11.13 - Prob. 120SEPCh. 11.13 - Prob. 121SEPCh. 11.13 - Prob. 122SEPCh. 11.13 - Prob. 123SEPCh. 11.13 - Prob. 124SEPCh. 11.13 - Prob. 125SEPCh. 11.13 - Prob. 126SEPCh. 11.13 - Prob. 127SEPCh. 11.13 - Prob. 128SEPCh. 11.13 - Prob. 129SEP
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