Presentation on theme: "One Dimensional Steady State Heat Conduction"— Presentation transcript:
1 One Dimensional Steady State Heat Conduction MODULE 2One Dimensional Steady State Heat Conduction
2 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 metalsT(x,y,z)
3 Unidirectional heat conduction (1D) Area = Axx+xSolid bar, insulated on all long sides (1D heat conduction)qxqx+xA= Internal heat generation per unit vol. (W/m3)
4 Unidirectional heat conduction (1D) stgenoutinE&=+-)(First Law (energy balance)tEqxA=D+-&)(tTxcAuxED=r)(xqTkAD+=-
5 Longitudinal conduction Unidirectional heat conduction(1D)(contd…)tTcqxkAcAkA=+÷øöçèæD-r&Longitudinal conductionThermal inertiaInternal heat generationtTkcqx=+ar12&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 1- Dimensional Heat Conduction The Plane Wall :……kT∞,2Ts,1Ts,2x=0x=LHot fluidCold fluidConst. K; solution is:
10 Thermal resistance (electrical analogy) OHM’s LAW :Flow of ElectricityV=IR electVoltage Drop = Current flow×Resistance
11 Thermal Analogy to Ohm’s Law : Temp Drop=Heat Flow×Resistance
12 1 D Heat Conduction through a Plane Wall ……kT∞,2Ts,1Ts,2x=0x=LHot fluidCold fluidT∞,1T∞,1Ts,1Ts,2T∞,2qxAh21AkLAh1(Thermal Resistance )
14 Composite Walls : T∞,1 T∞,2 A h Overall heat transfer coefficient h1 BCT∞,1T∞,2h1h2K AK BK CL AL BL Cq xT∞,1T∞,2Ah12Overall heat transfer coefficient
15 Overall Heat transfer Coefficient Contact Resistance :ABTATB
16 Series-Parallel :ABDCT1T2K cK DK AK BAB+AC=AA=ADLB=LC
17 Series-Parallel (contd…) Assumptions :Face between B and C is insulated.(2) Uniform temperature at any face normal to X.
18 Example: qx Consider a composite plane wall as shown: kI = 20 W/mkAI = 1 m2, L = 1mkII = 10 W/mkAII = 1 m2, L = 1mT1 = 0°CTf = 100°Ch = 1000 W/ m2 kqxDevelop an approximate solution for the rate of heat transfer through the wall.
19 1 D Conduction(Radial conduction in a composite cylinder) h1T∞,1k1r1r2k2T∞,2r3h2
20 Critical Insulation Thickness : Insulation Thickness : r o-r iObjective : decrease q , increasesVary r0 ; as r0 increases ,first term increases, second term decreases.
21 Critical Insulation Thickness (contd…) Maximum – Minimum problemSetatMax or Min. ? Take :
22 Critical Insulation Thickness (contd…) Minimum q at r0 =(k/h)=r c r (critical radius)good for steam pipes etc.good for electrical cablesR c r=k/hr0R t o t
23 1D Conduction in Spherer1r2kTs,1Ts,2T∞,1T∞,2Inside Solid:
24 Conduction with Thermal Energy Generation = Energy generation per unit volumeVEq&=Applications: * current carrying conductors* chemically reacting systems* nuclear reactors
25 Conduction with Thermal Energy Generation The Plane Wall :kTs,1Ts,2x=0x=+LHot fluidCold fluidT∞,2T∞,1x= -Lq&Assumptions:1D, steady state, constant k,uniformq&
26 Conduction With Thermal Energy Generation (contd…) xkqTSolutionLcondBoundarydxds+-=2:,.1&
27 Conduction with Thermal Energy Generation (cont..) UseboundaryconditionstofindCandC12æöq&L2x2T-TxT+TFinalsolution:T=ç1-÷+s,2s,1+s,2s,1ç÷2kL2èø2L2Not linear any moredTHeatflux:q=-Derive the expression and show that it is not independent of x any morekxdxHence thermal resistance concept is not correct to use when there is internal heat generation
28 Cylinder with heat source rorTsT∞ hq&Assumptions:1D, steady state, constant k, uniformq&Start with 1D heat equation in cylindrical co-ordinates:kqdrdTrd=+÷øöçèæ1&
29 Cylinder With Heat Source kTcondøè=4,:.sqSolutiondrdTBoundary+÷öçæ-21)(&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.