Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Chapter 4.8, Problem 59SEP
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What is the significance or impact of Schottky and/or Frenkel imperfection of the properties and behavior of ionic materials.
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A concrete wall retains water as shown. Assume that the wall is fixed at the base. Given: H = 3 m, t = 0.5m, Concrete unit weight = 23 kN/m3Unit weight of water = 9.81 kN/m3(Hint: The pressure of water is linearly increasing from the surface to the bottom with intensity 9.81d.)1. Find the maximum compressive stress (MPa) at the base of the wall if the water reaches the top.2. If the maximum compressive stress at the base of the wall is not to exceed 0.40 MPa, what is the maximum allowable depth(m) of the water?3. If the tensile stress at the base is zero, what is the maximum allowable depth (m) of the water?ANSWERS: (1) 1.13 MPa, (2) 2.0 m, (3) 1.20 m
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A short plate is attached to the center of the shaft as shown. The bottom of the shaft is fixed to the ground.Given: a = 75 mm, h = 125 mm, D = 38 mmP1 = 24 kN, P2 = 28 kN1. Calculate the maximum torsional stress in the shaft, in MPa.2. Calculate the maximum flexural stress in the shaft, in MPa.3. Calculate the maximum horizontal shear stress in the shaft, in MPa.ANSWERS: (1) 167.07 MPa; (2) 679.77 MPa; (3) 28.22 MPa
A counter flow double pipe heat exchanger is being used to cool hot oil from 320°F to 285°F
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Chapter 4 Solutions
Foundations of Materials Science and Engineering
Ch. 4.8 - Prob. 1KCPCh. 4.8 - Define the homogeneous nucleation process for the...Ch. 4.8 - In the solidification of a pure metal, what are...Ch. 4.8 - In the solidification of a metal, what is the...Ch. 4.8 - During solidification, how does the degree of...Ch. 4.8 - Distinguish between homogeneous and heterogeneous...Ch. 4.8 - Describe the grain structure of a metal ingot that...Ch. 4.8 - Distinguish between equiaxed and columnar grains...Ch. 4.8 - How can the grain size of a cast ingot be refined?...Ch. 4.8 - Prob. 10KCP
Ch. 4.8 - Prob. 11KCPCh. 4.8 - Prob. 12KCPCh. 4.8 - Distinguish between a substitutional solid...Ch. 4.8 - What are the conditions that are favorable for...Ch. 4.8 - Prob. 15KCPCh. 4.8 - Prob. 16KCPCh. 4.8 - Prob. 17KCPCh. 4.8 - Prob. 18KCPCh. 4.8 - Describe the structure of a grain boundary. Why...Ch. 4.8 - Describe and illustrate the following planar...Ch. 4.8 - Prob. 21KCPCh. 4.8 - Describe the optical metallography technique. What...Ch. 4.8 - Prob. 23KCPCh. 4.8 - Prob. 24KCPCh. 4.8 - Prob. 25KCPCh. 4.8 - Prob. 26KCPCh. 4.8 - Prob. 27KCPCh. 4.8 - Prob. 28KCPCh. 4.8 - Prob. 29KCPCh. 4.8 - Prob. 30KCPCh. 4.8 - Prob. 31KCPCh. 4.8 - Calculate the size (radius) of the critically...Ch. 4.8 - Prob. 33AAPCh. 4.8 - Prob. 34AAPCh. 4.8 - Calculate the number of atoms in a critically...Ch. 4.8 - Prob. 36AAPCh. 4.8 - Prob. 37AAPCh. 4.8 - Prob. 38AAPCh. 4.8 - Prob. 39AAPCh. 4.8 - Prob. 40AAPCh. 4.8 - Prob. 41AAPCh. 4.8 - Prob. 42AAPCh. 4.8 - Determine, by counting, the ASTM grain-size number...Ch. 4.8 - Prob. 44AAPCh. 4.8 - For the grain structure in Problem 4.43, estimate...Ch. 4.8 - Prob. 46AAPCh. 4.8 - Prob. 47SEPCh. 4.8 - Prob. 48SEPCh. 4.8 - Prob. 49SEPCh. 4.8 - Prob. 50SEPCh. 4.8 - In Chapter 3 (Example Problem 3.11), we calculated...Ch. 4.8 - Prob. 52SEPCh. 4.8 - Prob. 53SEPCh. 4.8 - Prob. 54SEPCh. 4.8 - Prob. 55SEPCh. 4.8 - Prob. 56SEPCh. 4.8 - Prob. 57SEPCh. 4.8 - Prob. 58SEPCh. 4.8 - Prob. 59SEPCh. 4.8 - Prob. 60SEPCh. 4.8 - Prob. 61SEPCh. 4.8 - Prob. 62SEPCh. 4.8 - Prob. 63SEPCh. 4.8 - Prob. 64SEPCh. 4.8 - Prob. 65SEPCh. 4.8 - Prob. 66SEP
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- ! Required information A one-shell-pass and eight-tube-passes heat exchanger is used to heat glycerin (cp=0.60 Btu/lbm.°F) from 80°F to 140°F by hot water (Cp = 1.0 Btu/lbm-°F) that enters the thin-walled 0.5-in-diameter tubes at 175°F and leaves at 120°F. The total length of the tubes in the heat exchanger is 400 ft. The convection heat transfer coefficient is 4 Btu/h-ft²°F on the glycerin (shell) side and 70 Btu/h-ft²°F on the water (tube) side. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Determine the rate of heat transfer in the heat exchanger before any fouling occurs. Correction factor F 1.0 10 0.9 0.8 R=4.0 3.0 2.0.15 1.0 0.8.0.6 0.4 0.2 0.7 0.6 R= T1-T2 12-11 0.5 12-11 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 (a) One-shell pass and 2, 4, 6, etc. (any multiple of 2), tube passes P= T₁-11 The rate of heat transfer in the heat exchanger is Btu/h.arrow_forward! Required information Air at 25°C (cp=1006 J/kg.K) is to be heated to 58°C by hot oil at 80°C (cp = 2150 J/kg.K) in a cross-flow heat exchanger with air mixed and oil unmixed. The product of heat transfer surface area and the overall heat transfer coefficient is 750 W/K and the mass flow rate of air is twice that of oil. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air Oil 80°C Determine the effectiveness of the heat exchanger.arrow_forwardIn an industrial facility, a counter-flow double-pipe heat exchanger uses superheated steam at a temperature of 155°C to heat feed water at 30°C. The superheated steam experiences a temperature drop of 70°C as it exits the heat exchanger. The water to be heated flows through the heat exchanger tube of negligible thickness at a constant rate of 3.47 kg/s. The convective heat transfer coefficient on the superheated steam and water side is 850 W/m²K and 1250 W/m²K, respectively. To account for the fouling due to chemical impurities that might be present in the feed water, assume a fouling factor of 0.00015 m²-K/W for the water side. The specific heat of water is determined at an average temperature of (30 +70)°C/2 = 50°C and is taken to be J/kg.K. Cp= 4181 Water Steam What would be the required heat exchanger area in case of parallel-flow arrangement? The required heat exchanger area in case of parallel-flow arrangement is 1m².arrow_forward
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- Problem 18-26 Added Extension Springs Spring Material ASTM A227 Modulus of Elasticity of the Material in Shear: G 1.150E+07 psi Average Service Max Operating Load: F₁ = 100 lb Max Length between attachment points: L₁ = 60.00 in 20.00 lb 26.00 1.400 Min Operating Load: F₁ = Min Length between attachment points: L₁ = Maximum Outside Diameter = in in Results Note: you select a wire diameter from the "US steel wire gage" column in table 18-2 Formula k = AF/AL k = (F0-F1)/(Lo - L₁) Spring Rate: k = lb/in Assumed Trial Outside Diameter: OD = Assumed Trial Mean: D ma Assumed Design Stress in Spring: Tda in 1.070 in 102,000 psi Assumed Wahl Factor: K = 1.2 Calculated Wire Diameter: Dwa Actual Wire Diameter: Dw Actual outer diameter: OD = Actual inner diameter: ID= Spring Index: C = See Figure 18-8 Dw= [8KF Dm πTd 1/3 in 5' 5' 5' 5' This corresponds to US Steel 9 wire gage ID = Dm - Dw C = Dm/Dw 4C - 1 0.615 K = + 4C - с Wahl Factor: K = 8KFDm 8KFC T = TD πD Stress in Spring at F = Fo: To psi…arrow_forwardCHAIN DRIVE DESIGN Initial Input Data: Application: Garage Door Opener Drive type: AC Motor Driven machine Chain and Sprocket to pull the door up Degrees per Radian: 57.2958 degrees Sprocket Diameter: D = 1.690 in Number of strands: Chain number: 1 40 Service factor: 1.3 Table 7-10 No. of teeth Computed Data: Actual Motor Power Input: 0.000 hp Sprocket Speed (for sprocket attached to gear shaft) Design power: 0.00 rpm 0 hp 11 12 0.06 0.15 0.29 0.56 0.99 1.09 1.61 2.64 TABLE 7-7 Horsepower Ratings-Single Strand Roller Chain No. 40 0.500 inch pitch 10 25 50 100 180 200 300 500 700 900 1000 1: 0.06 0.14 0.27 0.52 0.91 1.00 1.48 2.42 3.34 4.25 4.70 ! 3.64 4.64 5.13 13 0.07 0.16 0.31 0.61 1.07 1.19 1.75 2.86 3.95 5.02 5.56 Design Decisions-Chain Type and Teeth Numbers: 14 Chain number: Use Table 7-7 Chain pitch: p = in 15 Number of Teeth: N = Per Table 7-7 16 0.08 0.20 0.39 0.75 1.32 1.46 2.15 3.52 0.07 0.17 0.34 0.66 1.15 1.28 1.88 3.08 0.08 0.19 0.36 0.70 1.24 1.37 2.02 3.30 4.55 5.80…arrow_forwardInput Data: Torque needed to overcome rolling friction in rollers, slides and other moving parts, except for Motor and Worm Gear the worm gear T₁ = Length of travel of door: Time for door to open or close: LD = 50 lb-in. 90 in t= 12.5 seconds Pitch diameter for chain sprocket: DPC 1.690 in Weight of Door: P = No. of worm threads: Nw = Worm Pitch diameter: Dw Diametral pitch: Pd Normal pressure angle: Degrees per Radian: Number of gear teeth: Calculated Data: Linear velocity of door and chain (in/sec): Linear velocity of door and chain (ft/min): Output Speed of Gear and Sprocket: Upward Force due to Weight of Door: Фо = = NG= 240 lb 2 1.250 in 12 14.5 degrees 57.2958 degrees 28 Vα= in/sec VC= ft/min NG = rpm FD lb Net Upward Force on Door: Fou lb Torque on gear ignoring rolling friction: TG = lb-in. Formula = FDU FD-2 x Fo (note: Fo is the Max Operating load of the extension springs). This is also the initial tension in the chain. TG = FDU X DPC/2 This is the also the torque on the…arrow_forward
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