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Foundations of Fluid Mechanics I

by: Chester Goldner III

Foundations of Fluid Mechanics I M E 521

Marketplace > Pennsylvania State University > Mechanical Engineering > M E 521 > Foundations of Fluid Mechanics I
Chester Goldner III
Penn State
GPA 3.79


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This 0 page Class Notes was uploaded by Chester Goldner III on Sunday November 1, 2015. The Class Notes belongs to M E 521 at Pennsylvania State University taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/233073/m-e-521-pennsylvania-state-university in Mechanical Engineering at Pennsylvania State University.

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Date Created: 11/01/15
Review of Complex Variables Author John M Cimbala Penn State University Latest revision 23 October 2007 Basic De nitions in the Complex Plane Define zE xijl where x and y are real numbers and z is a complex number made up of a real part x and an imaginary part iy The imaginary unity number is Complex number z is often represented graphically on the complex xy plane as shown in the sketch Complex number z can also be written in terms of r and 6 where r is the magnitude of z often thought of as a radius and 6 is the angle between the x axis and the ray z as shown on the sketch Mathematically 9 E irctan lx Some Rules and Review a Complex Conjugate b Magnitude of a Complex Variable The magnitude also called the modulus of a com lex variable z x iy is obtained by taking the square root of the Note that this can be expanded to give 7 product of z and its complex conjugate ie c Miscellaneous Equations 51mg s a 70 e9cos lsm e equot elf 391 7 191 d 7 52 7 2i 0121quot an 22 29 39gtZ1 Zz irze VET 39cos 8 is ir 9 39 O V Separating a Complex Function into Real and Imaginary Parts A trick to separate a complex function into its real and imaginary parts is to multiply and divide by the complex conjugate of the denominator of the function This always works because it guarantees that the denominator will become real Example Calculate the real and imaginary parts of F z cz where c is a real constant x 7139 39 SolutionF c c y cx 7 my xiy xiy xiiy x2 y2 x2 y2 So erealpartis Zcx 2 x y Derivatives of Complex Functions x2y239 and lhe imaginary part is a V Differentiation of complex functions is relatively simple because it follows the same basic rules as does differentiation of real functions product rule exponent rules etc Note F z is an analytic function in any region where a nite unique derivative dFdz can be de ned everywhere within the region lfFz is an analytic function then dFdz is independent of the orientation of dz in the xy plane Examples Fz z2 g 2z exponent rule z F z lnz E 1 natural log rule dz z Fz w E lnz 7i product rule combined with above rules z dz z z 22 Review Solutions of Ordinary Differential Equations Author John M Cimbala Penn State University atest revision 06 September 2007 First Order Senarable Ordinary Differential Emm nm Example dy 7 C I Separate variables Cdt yyt y dl 7 I Solve yz 2 Example dy I 7 C y ya 1y 7 C I Separate variables ydy 7 7d I Solve yz K2Ch1tC139 First Order Linear Non l Ordinary Differential Emm nm E l y 3323 e Px y Qx I Multiply each term by an integrating factor Solve This leads to an exact differential on the LHS which can be solved easily Higher 0rd 3r Linear l rdinary Differential En uations with Constant Coef cients Example d3 d2 d y yx 7 Lick Jr y 6y 0 I Rewrite in terms of a linear operator Dquot D34D2D6y0 I Factor the LHS Here D l D 2 D 3y 0 I Find the roots of the LHS Here the roots are l 2 and 3 Solve yi CleFX39 TGZ eZquotC C e3t Example d y 2 I Rewrite in terms of the linear operator Here D2 a y 0 yyx 2 7 y0 I Factor the LHS Here Da Day0 I Find the roots of the LHS Here the roots are a and a Solve y 166 295 or y C 9fdshegt G4 si th Example dzy 2 I Rewrite in terms of the linear operator Here D2 a y 0 y yx dXz a y 0 I Factor the LHS Here D ia D iay 0 Find the roots of the LHS Here the roots are ia and ia Solve 3 Cl m r CzefiwC or vy39C3cosaxfC4 sina5cj Higher 0rd er Nonlinear Non l Ordinary Differential Emmtinm Example y yx d3 d2 d 2 dPxdx fay Qx Define N unknowns where N is the highest order derivative of the 39 uation Here N 3 and define ori inal e dy J i dx acl dx dx Define derivative array in standard RungeKutta form in terms of the N unknowns F1 F 2 Solve simultaneously for F1 F2 F Nusing the RungeKutta numerical technique and the boundary conditions d3 d2 Qx FN D3 939 2 2TF 2 i 3 D1 2 El Qxiech a 717 Nearly Incompressible Laminar Flow Equations and Solution Technique Author John M Cimbala Penn State University Latest revision 31 October 2007 Assumptions and Approximations o The uid is Newtonian with constant properties 01 1 k a K C P o The ow is laminar rather than transitional or turbulent o The uid is nearly incompressible 7 either an incompressible liquid or an ideal gas at very low Mach numbers Differential Equations of Motion for Nearly Incompressible Flow general review 0 Conservation of mass o Momentum equation 7 figag azu pDi pa 39Oxj 05c pg axpr an an p p Dt39 fixI 0x1 0 Conservation of energy first law For incompressible liquid DT OZT DT 82T For ideal gas at very low Ma OCp k K V Di xxf xx Dt MaxI Dwk auk 0 wk 0 VortiCity equation a V Differential Equations of Motion for Nearly Incompressible Flow with Buoyancy o Boussinesq approximation See Kundu Section 418 Assume that chanes in densit 0 are negligible everywhere except in the gravity term buoyancy where we let 0 po 17 0i T 7T0 where a is the thermal 39 1 and n is a reference dens1ty corresponding to p reference temperature T 0 T is assumed to vary only slightly from T 0 so that density is nearly constant but does lead to buoyancy in the ow The density is assumed to equal a in all other terms in the equation expansion coef cient for an ideal gas F o The continuity and vorticity equati ns are the same as above since density does not appear in these eq lations Du 0a 0a 0p 0 Th t t39 b u 7 17a TT r o e momen um equa 1011 ecomes p0 Dr p0 at J ax 0x1 pi 0g u axjax D a I an 71 a 32 or i ifu l17aT7glv u where Dr 01 0x p0 6xI OxJOx D39T air 0 The energy equation then becomes For incompreSSible liqu1d 30639 k b F Di mag For ideal gas at very low Mach number amp viscous dissipation is negligible Solution Technique for Nearly Incompressible Laminar Flow without Buoyancy 1 Write the continuity and momentum equations Note The energy equation is uncoupled if there is no buoyancy If buoyancy is important the energy equation must be solved simultaneously with mass and momentum 2 Simplify the equations as much as possible cross off zero terms etc 7 always justi your simpli cations 3 Solve for u and p This step will generate some constants from the integration 4 Apply BCs on u and p to solve for the unknown constants Now a and p are known everywhere 5 Write the energy equation 6 Simplify as much as possible 7 Solve for T This step will generate some constants from the integration 8 Apply BCs on T to solve for the unknown constants Now T is known everywhere and we are finished


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