One Dimensional Steady State Heat Conduction MODULE 2 One Dimensional Steady State Heat Conduction

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 T(x,y,z)

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

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

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

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.

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).

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

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

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

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

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

Resistance expressions

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

Overall Heat transfer Coefficient Contact Resistance : A B TA TB

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

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

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

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

Critical Insulation Thickness : Insulation Thickness : r o-r i Objective : decrease q , increases Vary r0 ; as r0 increases ,first term increases, second term decreases.

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

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

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

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

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

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

Conduction with Thermal Energy Generation (cont..) Use boundary conditions to find C and C 1 2 æ ö q & L 2 x 2 T - T x T + T Final solution : T = ç 1 - ÷ + s , 2 s , 1 + s , 2 s , 1 ç ÷ 2 k L 2 è ø 2 L 2 No more linear dT Heat flux : q ¢ ¢ = - Derive the expression and show that it is no more independent of x k x dx Hence thermal resistance concept is not correct to use when there is internal heat generation

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

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

Cylinder with heat source (contd…) Example: A current of 200A is passed through a stainless steel wire having a thermal conductivity K=19W/mK, diameter 3mm, and electrical resistivity R = 0.99 . The length of the wire is 1m. The wire is submerged in a liquid at 110°C, and the heat transfer coefficient is 4W/m2K. Calculate the centre temperature of the wire at steady state condition. Solution: to be worked out in class