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Chapter 1MEEN 221-01

Sam Ibekwe, Ph.D., P.E.355 Pinchback Engineering Building

Email: samuel_ibekwe.subr.edu

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Mathematical Modeling and Engineering Problem solving

Chapter 1

• Requires understanding of engineering systems– By observation and experiment– Theoretical analysis and generalization

• Computers are great tools, however, without fundamental understanding of engineering problems, they will be useless.

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Fig. 1.1The engineering

problem-solving process.

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A mathematical model is represented as a functional relationship of the form

Dependent independent forcingVariable = f variables, parameters, functions

Dependent variable: Characteristic that usually reflects the state of the system

Independent variables: Dimensions such as time ans space along which the systems behavior is being determined

Parameters: reflect the system’s properties or compositionForcing functions: external influences acting upon the system

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Newton’s 2nd law of MotionStates that “the time rate change of momentum of a body

is equal to the resulting force acting on it.”The model is formulated as

F = m a (1.2)

F = net force acting on the body (N)

m = mass of the object (kg)

a = its acceleration (m/s2)

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Formulation of Newton’s 2nd law has several characteristics that are typical of mathematical models of the physical world:It describes a natural process or system in

mathematical termsIt represents an idealization and simplification of

realityFinally, it yields reproducible results,

consequently, can be used for predictive purposes.

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Some mathematical models of physical phenomena may be much more complex.

Complex models may not be solved exactly or require more sophisticated mathematical techniques than simple algebra for their solution

Example, modeling of a falling parachutist:

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Figure 1.02.

Schematic diagram of the forces acting on a falling parachutist. – Upward and Downward Forces.


m

cvmg

dt

dv

cvF

mgF

FFF

m

F

dt

dv

U

D

UD

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Expression the acceleration as the time rate of change of the velocity yield Eq. 1.4

The net force includes both downward and upward forces.

FD = force due to gravityFU = force due to air resistanceC = drag coefficient

Net Force is the difference between the downward and upward force.


This is a differential equation and is written in terms of the differential rate of change dv/dt of the variable that we are interested in predicting.

If the parachutist is initially at rest (v=0 at t=0), using calculus

tmcec

gmtv )/(1)(

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vm

cg

dt

dv

Independent variable

Dependent variableParametersForcing function

Simplifying the equation-

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Example 1.1Figure 1.3


Example 1.2 – Numerical SolutionUsing equation Eq. 1.11 to yield to following:

1-13

)()()()( 11 iiiii tttvm

cgtvtv


Section 1.2Conservation Laws and EngineeringConservation laws are the most important and

fundamental laws that are used in engineering.

Change = increases – decreases (1.13)Change implies changes with time (transient). If the

change is nonexistent (steady-state), Eq. 1.13 becomes

Increases =Decreases

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Figure 1.6


For steady-state incompressible fluid flow in pipes:Flow in = Flow out

or100 + 80 = 120 + Flow4

Flow4 = 60

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Fig 1.6


Problems 1.4, 1.5, 1.7


Problem 1.4For a free falling parachutist with linear

drag, assume a first jumper is 70 kg and has a drag coefficient of 12kg/s and a mass of 75 kg, if the second jumper drag coefficient of 15kg/s. how long will it take him to reach the same velocity the first jumper reached in 10 seconds?

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tmcec

gmtv )/(1)(


Problem 1.5

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Compute the velocity of a free falling parachutist using Euler’s method for the case where m= 80 kg and c = 10kgs. Perform the calculation from t = 0 to 20s with a step size of 1 second. Using an initial condition that the parachutist has an upward velocity of 20 m/s at t = 0. At t=10 seconds, assume that the chute is instantaneously deployed sot that the drag coefficient jumps to 50 kg/s


Problem 1.7 Storage tank contains a liquid at depth y where y = 0 when the tank is full.

Figure P1_07.jpg


Problem 1.7 results

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t

Y

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