One Dimensional Steady State Heat Conduction

1. MODULE 2 One Dimensional Steady State Heat Conduction

2. T(x,y,z) Objectives of conduction analysis To determine the temperature field, T(x,y,z,t), in a body (i.e. how temperature varies with position within the body) T(x,y,z,t) depends on: - boundary conditions - initial condition - material properties (k, cp,  …) - geometry of the body (shape, size) Why we need T(x,y,z,t) ? - to compute heat flux at any location (using Fourier’s eqn.) - compute thermal stresses, expansion, deflection due to temp. etc. - design insulation thickness - chip temperature calculation - heat treatment of metals

3. Unidirectional heat conduction (1D) Area = A x x+x 0 x A qx qx+x = Internal heat generation per unit vol. (W/m3) Solid bar, insulated on all long sides (1D heat conduction)

4. & & & & - + = ( E E ) E E in out gen st ¶ E - + D = q q A ( x ) q & + D x x x ¶ t = r D ¶ E ( A x ) u T = - q kA x ¶ x ¶ ¶ ¶ E u T ¶ = r D = r D q A x A xc = + D x q q x ¶ ¶ ¶ t t t + D x x x ¶ x Unidirectional heat conduction (1D) First Law (energy balance)

5. ¶ ¶ ¶ ¶ ¶ æ ö T T T T - + + D + D = r D ç ÷ kA kA A k x A x q Ac x & ¶ ¶ ¶ ¶ ¶ x x x x t è ø ¶ ¶ ¶ æ ö T T + = r ç ÷ k q c & ¶ ¶ ¶ x x t è ø Thermal inertia Internal heat generation Longitudinal conduction ¶ r ¶ ¶ 2 T q c T 1 T & + = = ¶ ¶ a ¶ 2 x k k t t Unidirectional heat conduction (1D)(contd…) If k is a constant

6. Unidirectional heat conduction (1D)(contd…) • For T to rise, LHS must be positive (heat input is positive) • For a fixed heat input, T rises faster for higher  • In this special case, heat flow is 1D. If sides were not insulated, heat flow could be 2D, 3D.

7. Boundary and Initial conditions: • The objective of deriving the heat diffusion equation is to determine the temperature distribution within the conducting body. • We have set up a differential equation, with T as the dependent variable. The solution will give us T(x,y,z). Solution depends on boundary conditions (BC) and initial conditions (IC).

8. Boundary and Initial conditions (contd…) How many BC’s and IC’s ? - Heat equation is second order in spatial coordinate. Hence, 2 BC’s needed for each coordinate. * 1D problem: 2 BC in x-direction * 2D problem: 2 BC in x-direction, 2 in y-direction * 3D problem: 2 in x-dir., 2 in y-dir., and 2 in z-dir. - Heat equation is first order in time. Hence one IC needed

9. …. . . . . . .. .. .. . . . . .. . .. . . . . .. . . . . . . . . .. .. . . … . . . . . . .. . . . . . . . . . . . . . . . . Cold fluid k Ts,1 .. . . . . . . . . . .. . . . . . . . . . . . . . Ts,2 T∞,2 Hot fluid x=L x=0 1- Dimensional Heat Conduction The Plane Wall : Const. K; solution is:

10. Thermal resistance(electrical analogy) OHM’s LAW :Flow of Electricity V=IR elect Voltage Drop = Current flow×Resistance

11. Thermal Analogy to Ohm’s Law : Temp Drop=Heat Flow×Resistance

12. T∞,1 …. . . . . . .. .. .. . . . . .. . .. . . . . .. . . . . . . . . .. .. . . … . . . . . . .. . . . . . . . . . . . . . . . . Cold fluid k Ts,1 .. . . . . . . . . . .. . . . . . . . . . . . . . Ts,2 T∞,2 Hot fluid x=L x=0 1 L 1 A A A h h k 1 2 1 D Heat Conduction through a Plane Wall T∞,1 Ts,1 Ts,2 T∞,2 qx (Thermal Resistance )

13. Resistance expressions

14. T∞,1 A B C h1 K A K B K C h2 T∞,2 T∞,1 T∞,2 L A L B L C q x 1 1 A A h h 1 2 Composite Walls : Overall heat transfer coefficient

15. TA TB A B Overall Heat transfer Coefficient Contact Resistance :

16. B AB+AC=AA=AD A D K B LB=LC T1 T2 K D K A C K c Series-Parallel :

17. T1 T2 Series-Parallel (contd…) Assumptions : • Face between B and C is insulated. • (2) Uniform temperature at any face normal to X.

18. kI = 20 W/mk AI = 1 m2, L = 1m qx Tf = 100°C h = 1000 W/ m2 k T1 = 0°C kII = 10 W/mk AII = 1 m2, L = 1m Example: Consider a composite plane wall as shown: Develop an approximate solution for the rate of heat transfer through the wall.

19. h1 r1 T∞,1 r2 h2 T∞,2 k1 r3 k2 T∞,1 T∞,2 1 D Conduction(Radial conduction in a composite cylinder)

20. T∞ r0 ri h Ti Critical Insulation Thickness : Insulation Thickness : r o-r i Objective :decrease q , increases Vary r0 ; as r0 increases ,first term increases, second term decreases.

21. Set Critical Insulation Thickness (contd…) Maximum – Minimum problem at Max or Min. ?Take :

22. R t o t good for electrical cables good for steam pipes etc. R c r=k/h r0 Critical Insulation Thickness (contd…) Minimum q at r0 =(k/h)=r c r (critical radius)

23. r2 r1 k T∞,2 Ts,2 Ts,1 T∞,1 1D Conduction in Sphere Inside Solid:

24. & E = q & V = Energy generation per unit volume Conduction with Thermal Energy Generation Applications: * current carrying conductors * chemically reacting systems * nuclear reactors

25. k Ts,1 Ts,2 Assumptions: 1D, steady state, constant k, uniform q & T∞,1 T∞,2 Hot fluid Cold fluid q & x= -L x=+L x=0 Conduction with Thermal Energy Generation The Plane Wall :

26. 2 d T q & + = 0 2 dx k = - = Boundary cond . : x L , T T , 1 s = + = x L , T T , 2 s q & = - + + 2 Solution : T x C x C 1 2 2 k Conduction With Thermal Energy Generation (contd…)

27. Conduction with Thermal Energy Generation (cont..) Use boundary conditions to find C and C 1 2 - + æ ö T T T T 2 2 q L x x & ç ÷ = - + + , 2 , 1 , 2 , 1 s s s s Final solution : T 1 ç ÷ 2 2 k L 2 L 2 è ø Not linear any more dT Derive the expression and show that it is not independent of x any more ¢ ¢ = - Heat flux : q k x dx Hence thermal resistance concept is not correct to use when there is internal heat generation

28. T∞ h ro q & r æ ö 1 d dT q & + = ç ÷ r 0 r dr è dr ø k Ts q & Cylinder with heat source Assumptions: 1D, steady state, constant k, uniform Start with 1D heat equation in cylindrical co-ordinates:

29. = = Boundary cond . : r r , T T 0 s dT = = r 0 , 0 dr æ ö 2 q r & ç ÷ 2 = - + Solution : T ( r ) r 1 T ç ÷ 0 s 2 4 k r è ø 0 Cylinder With Heat Source Ts may not be known. Instead, T and h may be specified. Exercise: Eliminate Ts, using T and h.

30. Cylinder with heat source(contd…) Example: A current of 100A is passed through a stainless steel wire having a thermal conductivity K=25W/mK, diameter 3mm, and electrical resistivity R = 2.0 . The length of the wire is 1m. The wire is submerged in a liquid at 100°C, and the heat transfer coefficient is 10W/m2K. Calculate the centre temperature of the wire at steady state condition.