Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 16.13, Problem 42KCP
To determine
For a hard magnetic material, state its maximum energy product and how it is calculated. Write the SI unit and cgs unit for the energy product.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
CORRECT AND DETAILED HANDWRITTEN SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED.
The cantilevered spandrel beam shown whose depth tapers from d1 to d2, has a constant width of 120mm. It carries a triangularly distributed end reaction.Given: d1 = 600 mm, d2 = 120 mm, L = 1 m, w = 100 kN/m1. Calculate the maximum flexural stress at the support, in kN-m.2. Determine the distance (m), from the free end, of the section with maximum flexural stress.3. Determine the maximum flexural stress in the beam, in MPa.ANSWERS: (1) 4.630 MPa; (2) 905.8688 m; (3) 4.65 MPa
CORRECT AND DETAILED HANDWRITTEN SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED.
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
CORRECT AND DETAILED HANDWRITTEN SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED.
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
Chapter 16 Solutions
Foundations of Materials Science and Engineering
Ch. 16.13 - Prob. 1KCPCh. 16.13 - Prob. 2KCPCh. 16.13 - Prob. 3KCPCh. 16.13 - Prob. 4KCPCh. 16.13 - What is the relationship between B and H?Ch. 16.13 - Prob. 6KCPCh. 16.13 - Prob. 7KCPCh. 16.13 - Prob. 8KCPCh. 16.13 - Prob. 9KCPCh. 16.13 - Prob. 10KCP
Ch. 16.13 - Prob. 11KCPCh. 16.13 - Prob. 12KCPCh. 16.13 - Prob. 13KCPCh. 16.13 - Prob. 14KCPCh. 16.13 - Prob. 15KCPCh. 16.13 - Prob. 16KCPCh. 16.13 - Prob. 17KCPCh. 16.13 - Prob. 18KCPCh. 16.13 - Prob. 19KCPCh. 16.13 - Prob. 20KCPCh. 16.13 - Prob. 21KCPCh. 16.13 - Prob. 22KCPCh. 16.13 - Prob. 23KCPCh. 16.13 - Prob. 24KCPCh. 16.13 - Prob. 25KCPCh. 16.13 - Prob. 26KCPCh. 16.13 - Prob. 27KCPCh. 16.13 - Prob. 28KCPCh. 16.13 - Prob. 29KCPCh. 16.13 - Prob. 30KCPCh. 16.13 - Prob. 31KCPCh. 16.13 - Prob. 32KCPCh. 16.13 - Prob. 33KCPCh. 16.13 - Prob. 34KCPCh. 16.13 - Prob. 35KCPCh. 16.13 - What are eddy currents? How are they created in a...Ch. 16.13 - Prob. 37KCPCh. 16.13 - Prob. 38KCPCh. 16.13 - Prob. 39KCPCh. 16.13 - Prob. 40KCPCh. 16.13 - What compositions of NiFe alloys are especially...Ch. 16.13 - Prob. 42KCPCh. 16.13 - Prob. 43KCPCh. 16.13 - Prob. 44KCPCh. 16.13 - Prob. 45KCPCh. 16.13 - Prob. 46KCPCh. 16.13 - Prob. 47KCPCh. 16.13 - Prob. 48KCPCh. 16.13 - Prob. 49KCPCh. 16.13 - Prob. 50KCPCh. 16.13 - Prob. 51KCPCh. 16.13 - Prob. 52KCPCh. 16.13 - Prob. 53KCPCh. 16.13 - Prob. 54KCPCh. 16.13 - Prob. 55KCPCh. 16.13 - Prob. 56KCPCh. 16.13 - Prob. 57KCPCh. 16.13 - Prob. 58KCPCh. 16.13 - Prob. 59KCPCh. 16.13 - Prob. 60KCPCh. 16.13 - Prob. 61KCPCh. 16.13 - Prob. 62AAPCh. 16.13 - Prob. 63AAPCh. 16.13 - Prob. 64AAPCh. 16.13 - Prob. 65AAPCh. 16.13 - Prob. 66AAPCh. 16.13 - Gadolinium at very low temperatures has an average...Ch. 16.13 - Prob. 68AAPCh. 16.13 - Prob. 69AAPCh. 16.13 - Prob. 70AAPCh. 16.13 - Prob. 71AAPCh. 16.13 - Prob. 72AAPCh. 16.13 - Prob. 73AAPCh. 16.13 - Prob. 74AAPCh. 16.13 - Prob. 75AAPCh. 16.13 - Draw a hysteresis B-H loop for a ferromagnetic...Ch. 16.13 - Describe what happens to the magnetic induction...Ch. 16.13 - What happens to the magnetic domains of a...Ch. 16.13 - What are desirable magnetic properties for a soft...Ch. 16.13 - What are hysteresis energy losses? What factors...Ch. 16.13 - How does the AC frequency affect the hysteresis...Ch. 16.13 - How can eddy currents be reduced in metallic...Ch. 16.13 - Why does the addition of 3% to 4% silicon to iron...Ch. 16.13 - What disadvantages are there to the addition of...Ch. 16.13 - Why does a laminated structure increase the...Ch. 16.13 - Prob. 86AAPCh. 16.13 - Prob. 87AAPCh. 16.13 - Prob. 88AAPCh. 16.13 - Prob. 89AAPCh. 16.13 - Prob. 90AAPCh. 16.13 - Prob. 91AAPCh. 16.13 - Prob. 92AAPCh. 16.13 - Prob. 93AAPCh. 16.13 - Prob. 94AAPCh. 16.13 - Prob. 95AAPCh. 16.13 - Prob. 96AAPCh. 16.13 - Prob. 97AAPCh. 16.13 - Prob. 98AAPCh. 16.13 - Prob. 99AAPCh. 16.13 - Prob. 100AAPCh. 16.13 - Prob. 101AAPCh. 16.13 - Prob. 102AAPCh. 16.13 - Prob. 103SEPCh. 16.13 - Prob. 104SEPCh. 16.13 - Prob. 105SEPCh. 16.13 - Prob. 106SEP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- A counter flow double pipe heat exchanger is being used to cool hot oil from 320°F to 285°F using cold water. The water, which flows through the inner tube, enters the heat exchanger at 70°F and leaves at 175°F. The inner tube is ¾-std type L copper. The overall heat transfer coefficient based on the outside diameter of the inner tube is 140 Btu/hr-ft2-°F. Design conditions call for a total heat transfer duty (heat transfer rate between the two fluids) of 20,000 Btu/hr. Determine the required length of this heat exchanger (ft).arrow_forward! 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_forward
- In 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_forwardA single-pass crossflow heat exchanger is used to cool jacket water (cp = 1.0 Btu/lbm.°F) of a diesel engine from 190°F to 140°F, using air (Cp = 0.245 Btu/lbm.°F) at inlet temperature of 90°F. Both air flow and water flow are unmixed. If the water and air mass flow rates are 85500 lbm/h and 400,000 lbm/h, respectively, determine the log mean temperature difference for this heat exchanger. Assume the correction factor F to be 0.92. Air flow (unmixed) Water flow (unmixed) The log mean temperature difference of the heat exchanger is °F.arrow_forwardusing the theorem of three moments, find all the reactions and supports, I need concise calculations only. the answers are at the bottom, I need concise steps and minimal explanationsarrow_forward
- In 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 Cp J/kg-K. Water Steam Determine the heat exchanger area required to maintain the exit temperature of the water to a minimum of 70°C. The heat exchanger area required isarrow_forwardStress, ksi 160 72 150- 140 80 70 ༄ ྃ ༈ ཎྜ རྦ ༅ ཎྜ ྣཧྨ ➢ 130 120 110 100 90 2.0 2.8 3.6 4.4 5 Wire diameter, mm 6.0 6.8 2 7.6 8.4 Compression and extension springs. ASTM A227 Class II Light service Average service 0.020 0.060 0.100 0.140 0.180 0.220 0.260 0.300 0.340 0.380 0.420 0.460 0.500 Wire diameter, in Torsional stress due to initial tension, ksi 10 ४ 20 Preferred range 100 Stress, MPa 9.2 10.0 10.8 11.6 12.4 1100 1035 965 895 825 760 Severe service 690 620 550 50 150 3456789 10 11 12 13 14 15 16 Spring index, C = DJD FIGURE 18-21 Recommended torsional shear stress in an extension spring due to initial tension (Data from Associated Spring, Barnes Group, Inc.) 50 200 485 Stress, MPaarrow_forwardBolted Joint Design Bolted Frames Total Force due to door weight: P = 240 lb Number of Bolts: N = Distance to Bolt C/L: a = 4 N/A Bolt Material - Allowable shear stress of bolt material: T₂ = x Distance from Bolt centroid to bolt: x = y Distance from Bolt centroid to bolt: y = Degrees per Radian- Results y-Load on each bolt: F, = Moment resisted by bolt pattern: M = Radial distance from Bolt centroid to bolt: r = Sum squares of all radial distances: Σr² Force on each bolt to resist moment: F, - Angle for force composition: e= X-Force on each bolt to resist moment: F- y-Force on each bolt to resist moment: Fly Total y-Force on each bolt: Fy = Resultant force on bolt 1: R₁ = Required shear stress area for a bolt: A₂ = ASTM Grade A307 Steel 10,000 0 psi from Table 20-1 3.0 57.296 in degrees lb per bolt lb-in Formula FS-P/N M-Px XB r = (x² + y²)0.5 in² Σ 4r² Mr F₁ = Στ lb degrees lb lb lb Minimum Bolt Diameter: Din = Rounded up Bolt Diameter: D = 55 P. 1.5 in 2 in (3x) 1 in This bracket…arrow_forward
- University of Babylon Collage of Engineering/ Al-Musayab Department of Automobiles Final Examination/ Stage: 3rd Notes: Answer 4 questions only 2023-2202 Subject: Theory of vehicles Date: 2023\06\10-Saturday Time: Three Hours Course 2nd Attempt 1st Q1: A Hooke's coupling connects two shafts whose axes are inclined at 30°. The of the driven shaft? Find the maximum value of retardation or acceleration and driving shaft rotates uniformly at 600 rpm. What are the extreme angular velocities state the angle where both will occur. (12.5 Marks) Q2: Four masses, A, B, C, and D), revolve at equal radii and are equally spaced along a shaft. The mass B is 7 kg, and the radius of C and D make angles of 90° and 240°, respectively, with the radius of B. Find the magnitude of the masses A, C, and D and the angular position of A so that the system may be completely balanced. (12.5 Marks) Q3: A cam has straight worked faces that are tangential to a base circle of diameter 90 mm. The follower is a roller…arrow_forwardProblem 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_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Understanding Motor ControlsMechanical EngineeringISBN:9781337798686Author:Stephen L. HermanPublisher:Delmar Cengage Learning
Understanding Motor Controls
Mechanical Engineering
ISBN:9781337798686
Author:Stephen L. Herman
Publisher:Delmar Cengage Learning
Material Properties 101; Author: Real Engineering;https://www.youtube.com/watch?v=BHZALtqAjeM;License: Standard YouTube License, CC-BY