Stability investigation for the inverse heat conduction problems Krzysztof Grysa & Artur Maciqg Kielce University of Technology, AL 1000-leciaPP 7,

Similar documents
AP Physics C. Gauss s Law. Free Response Problems

Chapter 24. Gauss s Law

Transactions on Modelling and Simulation vol 8, 1994 WIT Press, ISSN X

Chapter 24. Gauss s Law

Chapter 4: Transient Heat Conduction. Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University

Physics 9 WS E3 (rev. 1.0) Page 1

Please fill in your Student ID number (UIN): IMPORTANT. Read these directions carefully:

Chapter 24. Gauss s Law

Chapter 22. Dr. Armen Kocharian. Gauss s Law Lecture 4

Chapter 23 Term083 Term082

Chapter 15 Appendix Moment of Inertia of a Spherical Shell

E. not enough information given to decide

Counting Dots Kwok-Wai Ng Feb 1, 2007

Chapter 22 Gauss s Law. Copyright 2009 Pearson Education, Inc.

Fokker-Planck Equation with Detailed Balance

Chapter 22 Gauss s Law. Copyright 2009 Pearson Education, Inc.

2 Which of the following represents the electric field due to an infinite charged sheet with a uniform charge distribution σ.

. (70.1) r r. / r. Substituting, we have the following equation for f:

1. Short Answer (25 points total)

Gauss s Law. Name. I. The Law: , where ɛ 0 = C 2 (N?m 2

Chapter 4. Electrostatic Fields in Matter

Physics 208, Spring 2015 Exam #1

Session 5 Heat Conduction in Cylindrical and Spherical Coordinates I

Chapter 22 Gauss s Law

Volumes of Solids of Revolution. We revolve this curve about the x-axis and create a solid of revolution.

Exam 1 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1

Fall 2004 Physics 3 Tu-Th Section

Chapter 2 Gauss Law 1

Chapter (2) Gauss s Law

MURI teleconference 28 May Optical Antimatter. John Pendry and Sebastien Guenneau Imperial College London. 24 May 2004 page 1

Sample Question: A point in empty space is near 3 charges as shown. The distances from the point to each charge are identical.

Gauss s Law. 3.1 Quiz. Conference 3. Physics 102 Conference 3. Physics 102 General Physics II. Monday, February 10th, Problem 3.

free space (vacuum) permittivity [ F/m]

2. Equations of Stellar Structure

Department of Physics Preliminary Exam January 2 5, 2013

Questions Chapter 23 Gauss' Law

Heat and Mass Transfer Prof. S.P. Sukhatme Department of Mechanical Engineering Indian Institute of Technology, Bombay

Phy207 Exam I (Form1) Professor Zuo Fall Semester Signature: Name:

University of Rome Tor Vergata

Notes on multivariable calculus

Angular Momentum. 1. Object. 2. Apparatus. 3. Theory

Homework 4 PHYS 212 Dr. Amir

University Physics (Prof. David Flory) Chapt_24 Sunday, February 03, 2008 Page 1

PHY102 Electricity Topic 3 (Lectures 4 & 5) Gauss s Law

Gauss s Law. Chapter 22. Electric Flux Gauss s Law: Definition. Applications of Gauss s Law

CHAPTER 3 CYLINDRICAL WAVE PROPAGATION

Chapter 10: Steady Heat Conduction

Chapter 24. Gauss s Law

Chapter 21: Gauss s Law

Electricity & Magnetism Qualifier

CHAPTER 4 ANALYSIS AND DESIGN OF THE DUAL INVERTED-F ANTENNA

Gauss Law. Challenge Problems

Flux. Flux = = va. This is the same as asking What is the flux of water through the rectangle? The answer depends on:

Essential University Physics

1 Introduction. Green s function notes 2018

Chapter 23. Gauss Law. Copyright 2014 John Wiley & Sons, Inc. All rights reserved.

1. Electricity and Magnetism (Fall 1995, Part 1) A metal sphere has a radius R and a charge Q.

Electric flux. Electric Fields and Gauss s Law. Electric flux. Flux through an arbitrary surface

Lecture 3. Electric Field Flux, Gauss Law

PHYSICS - CLUTCH 1E CH 12: TORQUE & ROTATIONAL DYNAMICS.

PHYSICS - CLUTCH CH 12: TORQUE & ROTATIONAL DYNAMICS.

TRANSIENT HEAT CONDUCTION

Experiment III Electric Flux

Chapter 21 Chapter 23 Gauss Law. Copyright 2014 John Wiley & Sons, Inc. All rights reserved.

TALLINN UNIVERSITY OF TECHNOLOGY, DIVISION OF PHYSICS 13. STOKES METHOD

The temperature of a body, in general, varies with time as well

CHAPTER 8 CONSERVATION LAWS

A A + B. ra + A + 1. We now want to solve the Einstein equations in the following cases:

Lesson 9: Multiplying Media (Reactors)

PHYS102 EXAM #1 February 17, MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Chapter 21: Gauss law Tuesday September 13 th. Gauss law and conductors Electrostatic potential energy (more likely on Thu.)

Steady and unsteady diffusion

1. Mathematical Tools

Numerical Solution of Integral Equations in Solidification and Melting with Spherical Symmetry

Massachusetts Institute of Technology Department of Materials Science and Engineering

Physics 210: Worksheet 26 Name:

Physics 1302W.500 Lecture 9 Introductory Physics for Scientists and Engineering II

( ) 5. Bearing internal load distribution and displacement. 5.1 Bearing internal load distribution

Phys102 Second Major-181 Zero Version Coordinator: Kunwar, S Monday, November 19, 2018 Page: 1

APPLICATIONS OF GAUSS S LAW

Columbia University Department of Physics QUALIFYING EXAMINATION

Electric flux. You must be able to calculate the electric flux through a surface.

Problem Solving 3: Calculating the Electric Field of Highly Symmetric Distributions of Charge Using Gauss s Law

Separation of Variables in Linear PDE: One-Dimensional Problems

Chapter 5. Effects of Photonic Crystal Band Gap on Rotation and Deformation of Hollow Te Rods in Triangular Lattice

3 The Friedmann-Robertson-Walker metric

Version: A. Earth s gravitational field g = 9.81 N/kg Mass of a Proton m p = kg

Methods in Computer Vision: Introduction to Matrix Lie Groups

Consider a point P on the line joining the two charges, as shown in the given figure.

A cylinder in a magnetic field (Jackson)

PHYSICS GRADUATE SCHOOL QUALIFYING EXAMINATION

Chapter 8. Rotational Equilibrium and Rotational Dynamics. 1. Torque. 2. Torque and Equilibrium. 3. Center of Mass and Center of Gravity

Moment of inertia. Contents. 1 Introduction and simple cases. January 15, Introduction. 1.2 Examples

Gauss s Law. Lecture 3. Chapter Course website:

Sink particle accretion test

week 3 chapter 28 - Gauss s Law

Physics (

PHY103A: Lecture # 4

Electric Flux and Gauss Law

Electricity. Revision Notes. R.D.Pilkington

Transcription:

Stability investigation for the inverse heat conduction problems Krzysztof Grysa & Artur Maciqg Kielce University of Technology, AL 1000-leciaPP 7, Abstract Discretisation with respect to time changes the heat conduction equation into a system of Helmholtz equations. An error introduced by such a simplification is investigated. For the inverse problems the constrains for a time step are considered in the case of two inaccurate internal responses. Nomographs helping to choose a proper time step for inverse problems with two internal responses are presented. 1 Discretisation Consider the heat conduction equation in dimensionless form: with initial and boundary conditions as follows: or r(*,0) = %(x), xeq (2) r(^,f)=7^,f), J%E6Q, fe(0,co) (3a)

94 Advanced Computational Methods in Heat Transfer ~\ <T> (xt,t) = Qt,(x>,t), x>edn, / e (0,«) (3b) CM Approximation of the derivative with respect to time with a first back difference leads to the following formulation of the problem: (x) = /,(4, ^^0, 6 = 1,2,... (4) or where (5a) ^-(xh) = Q,(xi,,kAt), x^dq, k = 1,2,... (5b) CM, A/ - time step, p -, and A/.Y,/X/, A = l,2,... (6) (A-l)Af Solution of such a problem may be expressed in an integral form ^(W/O-W-UW/,), ^50 (7) where $, is a simple layer potential and V^ is a volume potential for the Helmholtz equation. They may be expressed as integrals (8) /r (9) n where h and / stand for densities of the simple layer and volume potentials, respectively. G^ is a fundamental solution of the Helmholtz equation. In the formula (7) the density of the simple layer potential is an unknown function that has to satisfy the boundary condition. Hence, for both boundary conditions, (5a) and (5b), an integral equation has to be solved: for the condition (5a) or S&,p\h) = Tt($-V^,p\h), %e90 (10)

Advanced Computational Methods in Heat Transfer 95 for the condition (5b). The form (7) of the solution is convenient for analytical consideration in the case of body with simple geometry (e.g. flat slab, cylindrical or spherical layer, cylinder, sphere). 2 An inverse problem solution If the boundary condition in the problem (1), (2), (3a) or (3b) is replaced by a condition at an inner surface 9Q*eiQ (called an internal response - IR), then we consider an inverse heat conduction problem (IHCP). An integral form of an approximate solution of the IHCP has the form (7), too, and the simple layer potential density has also to satisfy an equation in form (10) (for the temperature IR) or in form (11) (for the heat flux IR), but in both cases jtedq.the IHCPs belong to the class of ill-posed problems. Besides, the Helmholtz model introduces its own constrains for time step magnitude, in particular respecting the possible IR inaccuracy propagation. Subsequently, the constrains for the time step magnitude are discussed. We confine ourselves to an IHCP in a flat slab. However, the conclusions are widened for a cylindrical and spherical layer. 3 The IHCP solution for aflatslab For ID the fundamental solution of Helmholtz equation reads I GI(X- y,p) = exp(-p\x - y) (12) 2f The solution (7) of the inverse problem in the case of two temperature IRs in a flat slab of unique thickness has a form _/ \ sinh[x*f-j;)]r *

96 Advanced Computational Methods in Heat Transfer s'm\\[pu] (13) where: u = Xg - x^, x e(0,l), k- 1,2,....,x^,jc^ - coordinates of points with IRs, 7^* ^Tgk ~ ^he temperature IRs at Xg,Xj at the moment kat. When mixed IRs are known (in xj - a temperature T, at ^ - a heat flux?*) then we obtain,-x)] pcos\\[pu], A = 1,2,... (14) The case of two heat flux IRs, ><#,Qgk, leads to a likewise formula. 4 Time step choice for the case of two IRs in flat slab It is obvious that the IRs as functions of time are not known precisely. If measured they depend on sensivity of the measuring device; if given they hardly ever are described as functions of time. Hence, they are always inaccurate. If an IR at x^ is given then the greatest inaccuracy of temperature calculation will appear at the boundary x=0, because between the point x^ and the boundary x=\ an IR at Xg is given. Denote the distance from x=0 to x~ as ifi and let t/^=<0, x >. Likewise, 6 <5 u^ is the distance fromx^to jc=l and U^=<xj,\>.

Advanced Computational Methods in Heat Transfer 97 Consider the case of two temperature IRs. With 6^(0) standing for an error of the calculation at x=0 and moment of time equal k&t we find / \ sinh p r / r>\ sinh p- sinh \p[] - u LJ LL g _ (is) where s ^_2 is an average error of the temperature in a slab obtained at ;i( the moment (k-2)kt. It is easy to prove that forp>5 the coefficient at s ^ is greater than the coefficient at s ^_2. Hence, in order to avoid an error propagation in time we require the absolute value of the coefficient at 8 j^_i to be smaller than 1. Finally we arrive to the following inequality: "77 (16) For given u^ the greatest value of p (i.e. the smallest value of the time step A/ denoted as Af %) for which the inequality (16) holds will be found. Likewise, A/ ^ is the smallest value of the time step for a given za Of course, in the case of two temperature IRs A/ ^ = A/ ^. Both, A/ ^ and A/ ^ are called critical time steps. For Af < Af ^ or Af < Af ^ calculations may occur unstable. For mixed IRs and for two heat flux IRs inequalities similar to (16) can be obtained. Then, A/ * and Af ^ can be found. Results concerning the cases of two temperature IRs and mixed IRs are shown on Fig. 1 and Fig. 2, where the nomographs for choosing the time step that ensures no IRerror propagation in time are presented. The hiperbolic lines denote length u of the inverse problem basis (distance from x^ to x ); relation between it, u^ and %/ reads For xj=q is «^=1; «= "n"r T (1?) for x =1 is u^=\. Hence, for each hiperbolic line description is given on the left or lower scale.

98 Advanced Computational Methods in Heat Transfer In the case of mixed IRs the time step to be chosen is the greater one (see Fig. 2). The minimal time step for the same length of u^ (z/) depends on the kind of an IR and it is worth to notice that the heat flux IR gives smaller value of the critical time step than the temperature IR. OOOOOOOOOOOOOOOOOOIT) 888888888888888888S OOOOOOOOOOOOOOOOOOO 1 no - _[ Ii I i n an U.0 U n An R U OA.4U n n on u.zu n nn 1 0.00 2 3 4 5 0.20 6 7 8 9 0.40 10 11 1 0.60 18 17 16 15 14 13 0.80 \ 19 0.000625 0.003000 0.006000 0.010000 0.016000 0.022000 0.030000 0.039000 0.049000 0.059000 0.068000 0.077000 0.085000 0.092000 0.098000 0.103000 0.107000 0.110000 0.112000 U Fig.l. Nomograph for two temperature IRs. Upper and right scales show the critical time step A/ ^ and A/ ^ for z/ and %^, respectively. 1.00

Advanced Computational Methods in Heat Transfer 99 1 00 0 80 060 R U 0 40 0 20 0 00-1 0.00 S 8 8 g 8 g OOOOOOOOOOOOOOOOOOCM - OOOOOOOOOOOOOOOOOOO I I I i! 1 1 1 1 i I! I I 2 ' 4 3 i 5 0.20 6 7 19 18 17 16 15 14 13 12. 11 10 9 8 0.40 U 0.60 0.80 Fig.2. Nomograph for two mixed IRs (T*^ and Q* scales show Af ^ and A/ ^ for i/ and «^, respectively. 1.00 0.000625 0.003000 0.006000 0.010000 0.016000 0.023000 0.031000 0.041000 0.053000 0.068000 0.086000 0.109000 0.134000 0.158000 0.181000 0.200000 0.215000 0.226000 0.232000 Upper and right 5 The IHCP solution for a cylindrical and spherical layers Let Q be a spherical or cylindrical layer with an inner radius r and outer radius r+\ (i.e. layer with a unique thickness). Consider the case of two temperature IRs, at r^ and at r^ r<r^ < ^ < 1 +r. For the spherical layer the IHCP solution has a form similar to that one for a flat slab, therefore we do not quote it here. In the case of cylindrical layer we obtain

100 Advanced Computational Methods in Heat Transfer (18) where (19) 6/z (20) Formulae desribing solution for mixed IRs and for two heat flux IRs for both, spherical and cylindrical layer, are easy to obtain and we do not quote them. An analysis leading to nomographs likewise to those shown at Fig.l and 2 is not difficult for the case of spherical layer; for two temperature IRs results are likewise to those for a flat slab. In the case of cylindrical layer the modified Bessel functions of the first and second kind make an analysis to be rather complicated. 6 Error introduced by the model itself Consider the following direct (initial-boundary) problem for a flat slab of unique thickness: T(x,Qj=Q, T(Q,t) = \QQ, 7/1,0=0. Comparing the exact solution with its approximate value (from the Helmholtz model) we can notice inaccuracies introduced by the model (see Fig.3). The greatest error appears for the first time step and is generated by the difference between first derivative and the first back diffrence of temperature with respect to time (see Fig.4). For an IHCP the effect of the first time step for solution accuracy is even greater and depends on location of a point with IR. In order to diminish the inaccuracy introduced by the first time step the back difference for the first step is modified as follows: dt Q5M At Geometric interpretation of (20) is presented on Fig.4. A%M±%zA) (21)

Advanced Computational Methods in Heat Transfer 101 '00.00 Temp 75.00 50.00 25.00 0.00 0.00 0.50 X 1.00 Fig.3. Exact (solid line) and approximate (dotted line) temperature value in a flat slab after dimensionless time equal to: a) 0.01; b) 0.03; c) 0.05. The parameter 0 can not be too big. For the case shown in Fig.4 0=0.8. For arbitrary jt^e(0,l), x =1 and time step A/, value of 0 can be determined based on analysis of the function sinh[v20p] -100 (22) that describes the difference between exact solution of the direct problem and approximate solution of the IHCP in the case, when the IR is taken from the exact solution. One can prove that if x^ =0 then B(0)=0. Moreover, B(0) is increasing with 0 and for an arbitrary jt^e(0,l) has exactly one solution. However, taking into account physical sense, value of the parameter 0 should be rather close to 0.5 (0e(0,l)).

102 Advanced Computational Methods in Heat Transfer.00 4.00 0.00 Fig.4. Geometric interpretation of the modified first back difference with respect to time (exact solution ybr 7(%,0)=0, 7(0,0=100, 7(l,f)=0 at x=0.13andaf=0.0014). 7 Conclusion The results presented in the paper are obtained in the case of the dimensionless variables. However, in an experiment the IR measurements take place for a real body in time measured in seconds. In order to avoid instable numerical calculation the dimensionless time step should be compared with the results presented in the paper. For instance when considering a steel flat slab of thickness 50 mm, the time step equal to 0,3 s denotes dimensionless time equal to 0.001. At Fig.l one can check that in the case of two temperature IRs the dimensionless distance between points with the IRs (the base of the IHCP) should be at least about equal to 0.97. Moreover, regarding the first time step, it is advisable to find the parameter 0 to minimize thefirsttime step error. Such a procedure seems to be a good way to diminish well known problems with stability of the IHCPs.