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Fayetteville State University | EML 3050
15 sample documents related to EML 3050
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Numerical Examples Example: Find the system of finite difference equations for the following problem. Steady state heat conduction through a plane wall with no internal generation. L=1 2 3 4 5 x=0.25 m=1 2T Heat Equation: 2T = 2 = 0 x Finite diffe
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Numerical Methods for Unsteady Heat Transfer Unsteady heat transfer equation, no generation, constant k, twodimensional in Cartesian coordinate: 1 T 2T 2T = 2 + 2 t x y We have learned how to discretize the Laplacian operator into system of finit
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EML 3050 Analytical Tools in Mechanical Engineering Course Syllabus: This class introduces mathematical and numerical tools relevant to practical applications in mechanical engineering. Emphasis is placed on learning how to model real physical syst
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Outcome-Specific Syllabus for Analytical Tools in Mechanical Engineering COURSE #: EML 3050 TERMS OFFERED: Fall and Spring COURSE TITLE: Analytical Tools in Mechanical Engineering PREREQUISIES: MAP 3305, Engr. Math I; EML 3002C, ME Tools; EML 3004C,
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Fourier Series Example Determine the Euler coefficients of the Fourier series of the function f(x)=x for - x : 1 1 a0 = f(x)dx= xdx=0 2 - 2 - 1 1 an = f(x)cos(nx)dx= xcos(nx) dx - - 1 x = [ 2 cos( nx ) + sin( nx )] = 0 - n n 1 1 bn = f(
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Example Evaluate the following integral: 0 ( / 2) cos( x ) if x < / 2 cos( w / 2) cos( xw) dw = 2 0 if x > / 2 1- w A( w) cos( wx )dw, it can be clearly seen that if From Fourier cosine integral: f ( x) = 0 cos( w / 2) A( w) = , then 2 1- w f (
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Partial Differential Equation (PDE) An ordinary differential equation is a differential equation that has only one independent variable. For example, the angular position of a swinging pendulum as a function of time: =(t). However, most physical syst
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Series Solution Since the equation is linear and homogeneous, the solution is a superposition of all possible solutions: n x u( x, t ) = un ( x, t ) = ( Bn cos( n t ) + B sin( n t ) ) sin L n =1 n =1 * In order to determine the coefficents Bn
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Heat Diffusion Equation All go to zero dE dE g - E - dW Energy balance equation: = + qin - q out + E1 2 dt dt dt Apply this equation to a solid undergoing conduction heat transfer: E=mcpT=(V)cpT=(dxdydz)cpT dx dy y qx qx+dx T q x = - KA x x x
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Heat Equation Solution 2 cu u 2 One-dimensional unsteady heat equation: = c t x2 Need one initial condition and two boundary conditions Example: A bar of length L with both ends insulated, that is, no heat transfer in or out of the end: cu u ( x = 0,
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2-D Steady Heat Equation Heat Equation 2 2 2 T k T T T k 2 2 = ( 2 + 2 + 2)= T = c 2T 0t cp x y z cp 2 02T T 2 Steady state, 2-D: T = 0, 2 + 2 = 0 0x y Bounday Conditions: T ( x = 0, y ) = 0, T ( x = a, y ) = 0 T ( x, y = 0) = 0, T ( x, y = b) = f
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2-D Wave Equation 2 c 2u u 2 The one-dimensional wave equation: 2 = c can easily be 2 t x 2 u extended to 2 or 3-dimensional that: 2 = c 2 2 u t 2 where x 2 + 2 y 2 + 2 z 2 is the 3-D Laplacian operator. 2 2 2u u u 2 T 2 In 2-D, = c
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LEARNING-THROUGH-TEACHING PROJECT Project Description: The ABET EC2000 accreditation criteria specify a new set of requirements for engineering graduates. Among other traditional attributes (see the EC 2000 ME graduate outcomes list in the last page)
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Analytical Tools in ME Finite Element Method (FEM) Example: one-dimensional spring system Consider a two elements, three nodes system as shown above. The generalized nodal equations can be written using the following graph: Where subscript ip denot
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Finite Element Method (FEM) Different from the finite difference method (FDM) described earlier, the FEM introduces approximated solutions of the variables at every nodal points, not their derivatives as has been done in the FDM. The region of intere
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