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Chapter 7 First-order Differential Equations

Chapter 7 Application of First-order Differential Equations in Engineering Analysis( Chapter 7 First order DEs) Tai-Ran Hsu*Based on the book of Applied Engineering Analysis , by Tai-Ran Hsu, published byJohn Wiley & Sons, 2018 Applied Engineering Analysis- slides for class teaching* Chapter Learning Objectives Learn to solve typical first order ordinary Differential Equations of both homogeneous and non homogeneous types with or without specified conditions. Learn the definitions of essential physical quantities in fluid mechanics analyses. Learn the Bernoulli s equation relating the driving pressure and the velocities of fluids in motion.

For space and variable t for time. First order differential equations are the equations that involve highest order derivatives of order one. They are often called “ the 1st order differential equations Examples of first order differential equations: Function σ(x)= the stress in a uni-axial stretched metal rod with tapered cross section (Fig. a),

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Transcription of Chapter 7 First-order Differential Equations

1 Chapter 7 Application of First-order Differential Equations in Engineering Analysis( Chapter 7 First order DEs) Tai-Ran Hsu*Based on the book of Applied Engineering Analysis , by Tai-Ran Hsu, published byJohn Wiley & Sons, 2018 Applied Engineering Analysis- slides for class teaching* Chapter Learning Objectives Learn to solve typical first order ordinary Differential Equations of both homogeneous and non homogeneous types with or without specified conditions. Learn the definitions of essential physical quantities in fluid mechanics analyses. Learn the Bernoulli s equation relating the driving pressure and the velocities of fluids in motion.

2 Learn to use the Bernoulli s equation to derive Differential Equations describing the flow of non compressible fluids in large tanks and funnels of given geometry. Learn to find time required to drain liquids from containers of given geometry and dimensions. Learn the Fourier law of heat conduction in solids and Newton s cooling law for convective heat transfer in fluids. Learn how to derive Differential Equations to predict required times to heat or cool small solids by surrounding fluids. Learn to derive Differential Equations describing the motion of rigid bodies under the influence of on Differential EquationsTypes of Differential Equations :We have learned in Chapter 2 that Differential Equations are the Equations that involve derivatives.

3 They are used extensively in mathematical modeling of engineering and physical are generally two types of Differential Equations used in engineering analysis. These are:1. Ordinary Differential Equations (ODE): Equations with functions that involve only one variable and with different orders of ordinary derivatives , and2. Partial Differential Equations (PDE): Equations with functions that involve more than one variable and with different orders of partial Differential Equations are derived?They are derived from the three fundamental laws of physics of which most engineering analyses involve.

4 These laws are: (1) The law of conservation of mass,(2) The law of conservation of energy, and (3) The law of conservation of momentum. on Differential Equations Cont dDifferential Equations for mechanical engineering:For mechanical engineering analyses, frequently used laws of physics include the following: The Newton s laws for statics, dynamics and kinematics of solids. The Fourier s law for heat conduction in solids. The Newton cooling law for convective heat transfer in fluids. The Bernoulli s principle for fluids in motion. Fick s law for diffusion of substances with different densities Hooke s law for deformable of Solution Methods forFirst Order Differential EquationsIn real-world, there are many physical quantities that can be represented by functionsinvolving only one of the four independent , (x, y, z, t), in which variables (x,y,z)For space and variable t for order Differential Equations are the Equations that involve highest order derivatives of order one.

5 They are often called the1storder Differential equationsExamples of first order Differential Equations :Function (x)= the stress in a uni-axial stretched metal rod with tapered cross section (Fig. a), or Function v(x)=the velocity of fluid flowing in a straight channel with varying cross-section (Fig. b):xFigure axv(x) (x)Figure bMathematical modeling using Differential Equations involving these functions are classified asFirst Order Differential Methods for Separable First Order ODEs)()()(xgdxxduuh Typical form of the first order Differential Equations :( )in which h(u) and g(x) are given re arranging the terms in Equation ( ) the following form with the left hand side (LHS) involves function of u (or constants) only, and the right hand side (RHS) consists of function of the variable g(x) (or constants) only.

6 H(u) du = g(x) dxSolution u(x) of Equation ( ) may be obtained by integrating both sides of the above equality and resulting in: cdxxgduuh)()(( )where c is the integration constant to be determined by given specified condition for the the following first order ordinary Differential equation:4)(2 udxxduxBy re arranging the terms in the DE into the separated form with the LHS involving the function u(x) and the RHS with the variable x:Solution:xdxxudu 2)(4followed by integrating both sides to get:cxnuun 2241with c=integration constantThe solution u in the above expression may be re written in the following form: 1'1'2)(44 xcxcxuwith c being another arbitrary constant to be determined by specified conditionTypical formof the equation:0)()()( xuxpdxxduThe solutionu(x) in the above equation is: xFKxu where K = constant to be determined by given condition, and the function F(x) has the form.

7 DxxpexF)()(in which the function p(x) is given in the Differential equation in Equation ( ) of linear, homogeneous Equations ( )( )( ) of linear, homogeneous Equations Cont dSolve the following first order ordinary Differential equation: xxudxxdusin (a)We will first re arrange the terms in Equation (a) in the following way:Solution: 0)(sin)( xuxdxxduBy comparison, we have: p(x) = sin x, which leads to the integral: xdxxpcos and hence have:xdxxpeexFcos)()( from Equation ( )The solution of the Differential equation in Equation (a) is available in Equation ( ), or in the form: xcKexFexucos)()( where K is an arbitrary constant to be determined by an appropriate prescribed of linear Non-homogeneousequations:Typical Differential equation:)()()()(xgxuxpdxxdu ( )The appearance of function g(x)in Equation ( ) makes the DE non-homogeneousThe solution of ODE in Equation ( ) is similar to the solution of homogeneous equation in a little more complex form than that for the homogeneous equation in ( ): )()()()(1)(xFKdxxgxFxFxu( ) dxxpexF)()(where function F(x) can be obtained from Equation ( ) as.

8 ( ) dxxpexF)()(is called the integration factor, where the integral in Equation ( )and K is the integration the following first order non homogeneous Differential equation:xxudxxdux5)(2)(2 Solution:By re arranging the terms, we get:xxuxdxxdu5)(2)(2 (a)xxgandxxp5)(2)(2 By comparison of Equations (a) and ( ), we get:The integration factor in Equation ( ) is xdxxdxxpeeexF22)(2)( The solution of Equation (a), u(x) is obtained by substituting the above integration factor into Equation ( ) and resulted in:xxxxxxKedxexeeKdxxeexFKdxxgxFxFxu2212 222551)()()()(1)( The integration constant K in the above solution of this ODE can be determined by specified conditions, as will be demonstrated in the next example Example the following Differential equation:2)(2)( xudxxduwith a given condition: u(0) = 2 Solution:By comparison of Equation (a) with the typical form in Equation ( ), we will have: p(x) = 2 and g(x) = 2.

9 (a)(b)Thus, from Equation ( ), we get: xdxdxxpeeexF22)()( Following Equation ( ), we have the solution of Equation (a) as xxxxeKeKdxeexFKdxxgxFxFxu2222121)()()()( 1)( (c)The integration constant K may be determined by using the specified condition in Equation (b) with u(x) = 2 at x = 0. Thus by substituting this condition into Equation (c), we will get the relationship:2102 xxeKfrom which we may obtain K = 1 Consequently, we have the complete solution of u(x) in Equation (a) to be:xxeexu22111)( of First Order Differential Equation to Fluid Mechanics AnalysisFundamental Principles of Fluid Mechanics Analysis:FluidsCompressible(Gases)Non-co mpressible(Liquids)-A substance with massbut noshapeMoving of a fluid requires: A conduit, , tubes, pipes, channels, etc.

10 Driving pressure, or by gravitation, , difference in head Fluid flows with a velocityvfrom higher pressure (or elevation) to lower pressure (or elevation) Fluid flows from higher elevation to low elevationtvAQ sec)/(kgvAtQQ sec)/(3mAvQV )(3mtvAtVV The total mass flow,Total mass flow rate, Total volumetric flow rate,Total volumetric flow, Fluid velocity, vCross-sectionalArea, AFluid flowHigher pressure (or elevation) Terminologies in Fluid Mechanics Analysis( ) ( )( )( )in which = mass density of fluid (g/m3), A = cross-sectional area (m2), v = velocity (m/s), and t = duration of fluid flow (s).


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