# CIVL 222 Study Guide - Quiz Guide: Heat Flux, Taylor Series, Advection

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9 Feb 2016

School

Department

Course

Professor

Quiz 1 Review

Part 1: Introduction

A. Numerical Methods

Analytical Methods

Can be used to solve many algebraic expressions as well as some ODEs and PDEs

Advantages: Fast (computer time), exact; no error introduced by the solution method

Disadvantages: limited to certain physical conditions eg: homogeneous material, simple geometry

Numerical Methods

Can be used to solve a greater variety of problems

Advantages: not limited by physical conditions eg: complex geometry, heterogeneous condition

Disadvantages: approximate solution, CPU time can be large

B. Partial Dierential Equations

Order: the order of the highest order partial derivative in the equation

Linearity: is linear if the coefficients depend on the independent variable (x,y) and not the dependent

variable u

∂2u

∂ x2+2xy ∂2u

∂ y2+u=1→linear ∂2u

∂ x2+xu ∂ u

∂ y =x → non−linear

Classication – 2nd Order

The general form of a second order PDE is:

A∂2u

∂ x2+B∂2u

∂ x ∂ y +C∂2u

∂ y2+D∂ u

∂ x +A∂ u

∂ y +Fu+G=0

The values of A, B, C can be used to further classify this PDE based on the sign of the solution to the

equation B2 – 4AC

Elliptic if < 0, typically describe processes that are occurring at steady state (i.e. no time derivative)

Parabolic if = 0, typically describe processes where the dependent variable is changing in time with

smooth changes

qx

qy

qx+x

xz

y

Hyperbolic if > 0, also describe changes in time, but the disturbance propagates with finite speed

where sharp fronts are maintained (i.e. waves do not loose height)

Part 2: Derivation of PDEs

A. Elliptic Equations

Steady state (not changing in time) temperature distribution in 2D, ex. flow of heat through an element:

The steady state energy balance equation would be (with qx flowing through yz, qy through xz):

qx∆ y ∆ z+qy∆ x ∆ z=qx+∆ x ∆ y ∆ z +qy+∆ y ∆ x ∆ z

Can divide through by volume xyz and group the terms:

qx−qx+∆ x

∆ x +qy−qy+∆ y

∆ y =0

Recall that the definition of a derivative is

∂ u

∂ x =u

(

x+∆ x , y

)

−u(x , y)

∆ x →−∂u

∂ x =u

(

x , y

)

−u

(

x+∆ x, y

)

∆ x

Therefore the new expression for the energy balance equation is:

−∂ q

∂ x −∂ q

∂ y =0

We now have an equation for heat flux q, we can proceed further by replacing q with the equation for heat

flux to receive the true equation:

−∂

∂ x

(

−K∂ T

∂ x

)

−∂

∂ y

(

−K∂ T

∂ y

)

=0→∂2T

∂ x2+∂2T

∂ y2=0

qy+yBased on the flow in the x and y direction, we can write:

Accumulation = In – Out +/- Source Sink

Since steady state Accumulation = 0

Assume no source sink, therefore In = Out