c- ( ( a) 0 o) / 6, t ) U I-i Problem 1. (Heat Conduction in a Rod, Ends at Different Temperatures)



Similar documents
1.- L a m e j o r o p c ió n e s c l o na r e l d i s co ( s e e x p li c a r á d es p u é s ).

College of the Holy Cross, Spring 2009 Math 373, Partial Differential Equations Midterm 1 Practice Questions

The two dimensional heat equation

The Heat Equation. Lectures INF2320 p. 1/88

α = u v. In other words, Orthogonal Projection

An Introduction to Partial Differential Equations

Class Meeting # 1: Introduction to PDEs

tegrals as General & Particular Solutions

General Theory of Differential Equations Sections 2.8, , 4.1

FINAL EXAM SOLUTIONS Math 21a, Spring 03

1 Inner Products and Norms on Real Vector Spaces

Solutions to old Exam 1 problems

Module 1 : Conduction. Lecture 5 : 1D conduction example problems. 2D conduction

AB2.5: Surfaces and Surface Integrals. Divergence Theorem of Gauss

Second Order Linear Partial Differential Equations. Part I

Chapter 9 Partial Differential Equations

The Mean Value Theorem

Solutions for Math 311 Assignment #1

MATH 425, PRACTICE FINAL EXAM SOLUTIONS.

OPTIMAl PREMIUM CONTROl IN A NON-liFE INSURANCE BUSINESS

The one dimensional heat equation: Neumann and Robin boundary conditions

Using Predictive Modeling to Reduce Claims Losses in Auto Physical Damage

MATH 31B: MIDTERM 1 REVIEW. 1. Inverses. yx 3y = 1. x = 1 + 3y y 4( 1) + 32 = 1

Høgskolen i Narvik Sivilingeniørutdanningen STE6237 ELEMENTMETODER. Oppgaver

Chapter 8 - Power Density Spectrum

Heat Transfer and Energy

MATH 10550, EXAM 2 SOLUTIONS. x 2 + 2xy y 2 + x = 2

Heat equation examples

H ig h L e v e l O v e r v iew. S te p h a n M a rt in. S e n io r S y s te m A rc h i te ct

W h a t is m e tro e th e rn e t

Solutions to Homework 5

MA107 Precalculus Algebra Exam 2 Review Solutions

Opis przedmiotu zamówienia - zakres czynności Usługi sprzątania obiektów Gdyńskiego Centrum Sportu

i n g S e c u r it y 3 1B# ; u r w e b a p p li c a tio n s f r o m ha c ke r s w ith t his å ] í d : L : g u id e Scanned by CamScanner

Campus Sustainability Assessment and Related Literature

Two-Dimensional Conduction: Shape Factors and Dimensionless Conduction Heat Rates

Higher Order Equations

Intermediate Value Theorem, Rolle s Theorem and Mean Value Theorem

r c cqc\ Cubc no X3*XL 2x p C\QcQQ fl cq (Lk 3±2x Polynomial Functions Monomial: Polynomial: Standard form: Degree of a polynomial: --

1. First-order Ordinary Differential Equations

5.4 The Heat Equation and Convection-Diffusion

1 The 1-D Heat Equation

COMPUTATION OF THREE-DIMENSIONAL ELECTRIC FIELD PROBLEMS BY A BOUNDARY INTEGRAL METHOD AND ITS APPLICATION TO INSULATION DESIGN

A power series about x = a is the series of the form

B rn m e d s rlig e b e h o v... 3 k o n o m i S s k e n d e tils k u d o g k o n o m is k frip la d s... 7 F o r ld re b e ta lin g...

Cl1'15 Charlotte-Mecklenburg Schools

Key Concept. Density Curve

CIRCLE COORDINATE GEOMETRY

Electromagnetism - Lecture 2. Electric Fields

Present. and. RFA No79/2OO5

Chapter 4. Linear Second Order Equations. ay + by + cy = 0, (1) where a, b, c are constants. The associated auxiliary equation is., r 2 = b b 2 4ac 2a

A Brief Review of Elementary Ordinary Differential Equations

A n d r e w S P o m e r a n tz, M D

1. Oblast rozvoj spolků a SU UK 1.1. Zvyšování kvalifikace Školení Zapojení do projektů Poradenství 1.2. Financování

SCO TT G LEA SO N D EM O Z G EB R E-

LINEAR DIFFERENTIAL EQUATIONS IN DISTRIBUTIONS

MATH 381 HOMEWORK 2 SOLUTIONS

ab = c a If the coefficients a,b and c are real then either α and β are real or α and β are complex conjugates

Solving Quadratic & Higher Degree Inequalities

CITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION

i=(1,0), j=(0,1) in R 2 i=(1,0,0), j=(0,1,0), k=(0,0,1) in R 3 e 1 =(1,0,..,0), e 2 =(0,1,,0),,e n =(0,0,,1) in R n.

So far, we have looked at homogeneous equations

Exam 2 Practice Problems Part 1 Solutions

AN EVALUATION OF SHORT TERM TREATMENT PROGRAM FOR PERSONS DRIVING UNDER THE INFLUENCE OF ALCOHOL P. A. V a le s, Ph.D.

Mathematical Finance

THREE DIMENSIONAL GEOMETRY

Chapter 6. Linear Transformation. 6.1 Intro. to Linear Transformation

Section 5.0A Factoring Part 1

Differential Balance Equations (DBE)

Fourth-Order Compact Schemes of a Heat Conduction Problem with Neumann Boundary Conditions

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

AP Calculus BC Exam. The Calculus BC Exam. At a Glance. Section I. SECTION I: Multiple-Choice Questions. Instructions. About Guessing.

NOV /II. 1. Let f(z) = sin z, z C. Then f(z) : 3. Let the sequence {a n } be given. (A) is bounded in the complex plane

An E mpir ical Analysis of Stock and B ond M ar ket Liquidity

Change of Variables in Double Integrals

JANUARY 2016 EXAMINATIONS. Life Insurance I

OPTIMAL SELECTION BASED ON RELATIVE RANK* (the "Secretary Problem")

THE SQUARE PARTIAL SUMS OF THE FOURIER TRANSFORM OF RADIAL FUNCTIONS IN THREE DIMENSIONS

Chapter 17. Review. 1. Vector Fields (Section 17.1)

Application of Fourier Transform to PDE (I) Fourier Sine Transform (application to PDEs defined on a semi-infinite domain)

Derive 5: The Easiest... Just Got Better!

An Introduction to Partial Differential Equations in the Undergraduate Curriculum

Heat Transfer Prof. Dr. Ale Kumar Ghosal Department of Chemical Engineering Indian Institute of Technology, Guwahati

B a rn e y W a r f. U r b a n S tu d ie s, V o l. 3 2, N o. 2, ±3 7 8

SIMPLE LINEAR CORRELATION. r can range from -1 to 1, and is independent of units of measurement. Correlation can be done on two dependent variables.

Probability and Random Variables. Generation of random variables (r.v.)

AP CALCULUS AB 2009 SCORING GUIDELINES

Phasors. Phasors. by Prof. Dr. Osman SEVAİOĞLU Electrical and Electronics Engineering Department. ^ V cos (wt + θ) ^ V sin (wt + θ)

4 Sums of Random Variables

MC78XX/LM78XX. 3-terminal 1A positive voltage regulator. Features. Description. Internal Block Digram.

EXISTENCE AND NON-EXISTENCE RESULTS FOR A NONLINEAR HEAT EQUATION

RAJALAKSHMI ENGINEERING COLLEGE MA 2161 UNIT I - ORDINARY DIFFERENTIAL EQUATIONS PART A

Chapter 22: The Electric Field. Read Chapter 22 Do Ch. 22 Questions 3, 5, 7, 9 Do Ch. 22 Problems 5, 19, 24

Solutions to Homework 10

Binomial lattice model for stock prices

4.2. LINE INTEGRALS ; z = t. ; y = sin

5-58. The two shafts are made of A-36 steel. Each has a diameter of 1 in., and they are supported by bearings at A,

Sect Solving Equations Using the Zero Product Rule

MAT 200, Midterm Exam Solution. a. (5 points) Compute the determinant of the matrix A =

26 Ideals and Quotient Rings

Transcription:

Throughout, k is the mean thermal diffusivity, usually written as Fourier s constant K0 divided by specihe heat c and Problem 1. (Heat Conduction in a Rod, Ends at Different Temperatures) 19 Feb, S2014 Math 3150 Midterm 1 _ I C c ( ( a) 0 o ) U 0 0 + <U () (3 o Ii C) o) 6, t Evaluate all Fourier coefficients. Then display the final answer u(x, t). Solve the rod problem for u(x. t). It is necessary to derive the product solutions. Provide Fourier coefficient formulas. u(x,0) 1!x, 0 <z <1. = kn. 0 <x < 1, t > 0, u(0,t) = 0, t >0, u(1,t) = 1, t >0, (b) [60%] Consider tile heat conduction problem in a laterally insulated rod of length 1 with one end at zero Celsius The answer u(x, t) to this problem is exactly the steady state tenipero.ture. Find tile answer u(x, t) and display a complete answer check. =.r, 0<x<1. u(1,t) = 1, t >0, u(0.t) = 0, t>0, Ut ku,. 011(1 the other end at one Celsius. The initial temperature along the rod is given by function [(c) = 1 + z. a 2., 0 <a; < 1, t> 0, the other end at 1 Celsius. The initial temperature along the rod is given by function [(3;) a;. j (a) [40%] Consider the heat coadiictioii problem in a laterally iiisulated rod of ieugth 1 with 000 011(1 at zero Celsius and moss density per unit voliune p. Name KE

t F 3 3 ii _ ( 0 1,, x r 1 r E r I, C, fr 0

At 4 IlA 4> ], f k [ A (, b) 4 1)d) _;fa ar(, t) ur(df) to obtain v, t) dx. 1 1 5 t ioiiah itito A is ioblem 2. (Total Thermal Energy in a Rod) (lorgy br all I. c Fc 17 ]1 Thç ) (±\, ç i)imaf r?7y K o i;i1 e IdfrAl 1i&ra1 CfM(X 1 v() (c) [20%] Explain the meanilig of the equation in (b) in terms of heat flux and Fourierh ow. = r(,t) Ur(0,t).,. (ii) (i0%1 Differentiate on t across the equation of (a). Simplify the resulting ecluation usilig U) k7trr and v1 = kv1.. A (a) j30%] Explain why J u,(x. t) dx f :::w Au expussom ior the tiuit d peiidtiit total tilernial energy contained in a rod i: = 0 to x =, with mutorni cross T1 i becriuice i tj OYV1e) v& an A (:( drf 11(111 iquitl 1011 l( kit,... Assmniii ( 011(1 f) are coiistaiits. J cpu(xj)adx. 2,.i e bev c Xy iihol ( is flit specific heat, p is the moss density per unit volume and u(x, t) is the rod temperature, satisf ing the I (_,r,.,l.,)...c. I. _c Ø i oi: i (,) (ô, jo Jo v(i I V() vw),) Suppose o(,r 1) and i (x t) are two temperature distributions for the sonic rod which supply the saute total thermal

. Problem 2. (Total rrhei1nal Energy in a Rod) An uxplessioli for the ti111c dpe11deht total tlwriiial energy coiit,aiued in a rod c = 0 to.: =, with uniform cross s 1 n i titi 01(0 A is J I lou ( is f l1( 5l)(Clh( 1iett [) IS tlw 1111155 (lensity per unit volitnie and u(x, t) is the rod teniperature, satisfying the hieiit ( (flliltiofl I( h0,pr. Asslilne ( 011(1 [1 1U0 coiistaiits. Snppose o(r, t) 011(1 cimergy for all 1. e(x. t) are two temperature distributions for the same rod winch supply the same total tliernial J (a) [30%] Explain wily u,(:v. t) dx = v(x, t) (c. (h) [60%] Differentiate on across the equation of (a). Simplify the resulting equatioli using u, to obt,aii r(.t) u = v(,t) lr(0,t). j 1(0t) (c) [20%] Explain the mueanhllg of the equation in (b) in terms of heat flux and Foi.irier s ow. ku audi m = kerr Jc1 ( c.i% t i sj) 11 A i J) A ( J I j J jf.j (i I;, b \I 4,. Jo. i I lj ii! 4 i 104+,, J \ (I i Tkc*cr,H V Ot) 9

1 IC t ) Y. ;:; 7.. ) I I \ Ji 4..c....

I 4 1\ u(m y). f\_ (b) (30%] et f(:r) be the smn of the first three eigciifiiiictioos which is the sum of time first three X answers. Find required threc zero boundary conditions. Ala) ir(.:, p) = 0 for :r = (I tuid,r = 1, n(.r, y) 0 for y = 1, u.(,i, j) = [(a) for p = 0. 70%) Find the product solutions a(.r, y) = X (a)y(). Include a cheek that each product solution satisfies Consider apliwes equation + on the rectangle (I <,t < 1, (1 < p < 1 subject to the boundary conditions d,r ()y ( (: (O) 1 C (o (:Q 6 Y 1: )1 w 3 iv7i o CCX ( f( T 0 y:o ch(: O (,c(: 0 i uj 0 0, () r1 Problem 3. (SteadyState Heat Conduction on a Rectangular Plate) 9c Corr! 5( cv Y( ((: ce 7.z (fl: 0 X (o: X (io 1+ :Q X y

2 (fl.r)7i 1T1t 1 (1) ) UhI1 q (i:)] 4J d J) T1i 3( 3 J i.10 TTTI I T! I 11T..)TiliJTl) it) )).l1 4H.ITJ )II} 1 1))1Il1))i 1 0.T) )J.II X110tfll)T10.) A.IUplIfl()1 1 P!d [%OI () IIo!4llIo 4311p0.Id i i (1i)(;) x = (Ii ) 411p01c Ii ioj (.i)f = (Ii.r) ii Ii.oj o (Ii.r) 1 = piiu n : oj o = (Ii.r) n _1i()._.r() ))UT(IU J.!J1)pI10) II03U1l1).) HTI0i1)ITO.) n])un()q im14 0J 4)Jql0 I > 1 > > > 0 J14 TIC) (wia J1T Tfl3DOH U 110 11OT.)flpt10D WOH spuos) g WOJJ OQr2 o i:ec 9; ),)21i]7 1_i ; 2 josj 11 x2 401 7 (p. O22 K ci o1 (2,3ç)J 4l., }...(1 (i),i 77 P 2 ),) F Q; A c7zj ) ( ) ii ); JI XJ5 (J,,.IJ :()(.2. fr.? )..1 ç J 1,Q :,k (o 9 CQ A)ul, <2 (I 1)ç) C2 0 1 pi),1,nc1 (G() I_JJ <f.

NI dc d1j ii..r(,i:, i) 0 = for :i (I and.r 1, = u(, y) (I for p 1, u(.i, = y) = fcc) for p 0. u(.u, ti). (a) [70%] Find the prodmt solutions u(,r, i) (h) [30%] et ICc) be lie suni of tlit first flirci cigeufunctiuns, winch is the suui of the first three X answers. Find required three zero boundary conditions. Prob em 3. (SteadyState Heat Conduction on a Rectangular Plate) Conshkr aplace s equation + on flit rictongle () < i < 1, (1 < p < 1 sub ject to the boundary conditions = X (.c)y (y). Inelwhlk that each product solution satisfies the 11 C (r \\)Hi k I c_) ). 4 3 S I No (IC3 0 N 00 C c q \E l O1flt 1. tt()9 Co 5c co i, ccc,0 A0 0 CotQ. Fo C t, me. r c( ) 2 1.l( o. o NO._ 1O C çr c c D C1 e e y gc 7?

c I \ II I 0 4, jc J N t I 4 I

4 (5o) f(( (1 (b) [40Yi Calculate u(o, 6) when f(8) = (1 on 0 < 8 <, 1(8) = 50 on 8 < 2. Hint: Tue Poisson integral theorem omit superposition, omit the_series solution and do not develop formulas fur the Fourier coefficients. (a) 1(i0XJ Find the product solutions n = R(r)e(8), then identify tin orthogonal set and the interval, Stop at this stage: mid the Mcmi Value Theorem. I u(2,8) =,f(8), 0<8< 2. 1v(r,8) + v.oo(r,8) = 0, 0 < i <2, 0 <8 <2, 12e M; v f cf!ec. ro, f (p) (: (e.); C) ç ) 7 4 (r,&) fru a), (r, ) Problem 4. (SteadyState Heat Conduction on a Disk) (cushier till stead v state heat conduction problem in polar coordinates 1 9c 7t