
Describe the classic model for electrical conduction in metals.

The description of the classic model for electrical conduction in metals.
Explanation of Solution
In metals, the arrangement of the atoms in the crystal structure is kept together with the metallic bonds that make it possible the movement of valance electrons. In the classic model of the electrical conductivity of a metal, the valance electrons present at the outermost layer are assumed to freely displace between the cores of positive ions in the lattice. Having some kinetic energy at the room temperature, the cores of positive ions vibrate in their lattice positions; however, with the rise in temperature the intensity of the vibrations rise and interchange of energies between electrons and the cores occurs. The electrons vibrate without any electric potential and randomly moves producing no current flow. But when a potential difference is applied, the electrons begin to travel uniformly by producing electric currents.
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Chapter 14 Solutions
Foundations of Materials Science and Engineering
- 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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- Understanding Motor ControlsMechanical EngineeringISBN:9781337798686Author:Stephen L. HermanPublisher:Delmar Cengage LearningAutomotive Technology: A Systems Approach (MindTa...Mechanical EngineeringISBN:9781133612315Author:Jack Erjavec, Rob ThompsonPublisher:Cengage Learning

