Chapter 18 – Transient heat conduction

1 1Dr. Primal (850) 410-6323 Chapter 18 Transient heat conductionThermal-FluidsI2 Transient heat conduction In general, The temperature of a body varies with time as well as position. In rectangular co-ordinates this variation is expressed as T(x,y,z,t) x,y,z variations in x,y,zdirections t variation with time The studies in this Chapter is focused on Lumped system analysis Transient heat conduction in large plane walls, long cylinders and spheres with spatial effects Transient heat conduction in semi-infinite solids Transient heat conduction in multi-dimensional systems3 BROAD OBJECTIVE: INVESTIGATE THE PROBLEM OFHOW DO SPHERES COMING OUT OF A OVEN COOL?

2 4 Consider .. An engineer, a psychologist, and a physicist were asked to make recommendations to improve the productivity of an under-producing dairy farm .. Engineer: more technology Psychologist: improve environment Physicist ..5 Consider a spherical cow .. T(t) Great engineers and physicists are able to appropriately simplify problems to extract the physics!6 Lumped system A lumped system is one in which the dependent variables of interest are a function of time alone. In general, this will mean solving a set of ordinary differential equations (ODEs) A distributedsystem is one in which all dependent variables are functions of time andone or more spatial variables.

3 In this case, we will be solving partial differential equations (PDEs) 7 Lumped system Consider a small hot copper ball coming out from an oven. Temperature change with time. Temperature does not change much with position at any given time. Lumped system analysis are applicable to this system Consider a large roast in an oven. Temperature distribution not even. Temperature does change much with position at any given time. Lumped system analysis are notapplicable to this a body of arbitrary shape of mass m, volume V, surface area As, density , and specific heat Cpinitially at a uniform temperature of time t=0, the body is placed into a medium at temperature T heat transfer take place between body and its environmentTemperature of the body change with the time and the temperature of the body at a given time T=T(t) heat transfer into the body at any given time T=T(t)[])(tTThAQs = 10)

4 (TThAQs = heat transfer into the body at temperature T = dtdt time during body the of energy the in increase The period time a during body the intotransfer heat TmcdtTThAps = )(11dTmcdtTThAps= )(Vm =dtVchATTdTps = )( = t0pstTTdtVchATTdTi )()(t0pstTTtVchATTi = )()ln(tVchATTTtTpsi = )(lntVchAipseTTTtT = )(12tVchAipseTTTtT = )(btieTTTtT = )(wheres1unitsVchAbps = Time constant1314 Criteria for lumped system analysisbtieTTTtT = )(s1unitsVchAbps = Characteristic lengthscAVL=Biot number BikhLBic=body the of surface theat resistance Convectionbody the within resistance conduction ==h1kLBic//15body the of surface theat resistance Convectionbody the within resistance conduction ==h1kLBic//Small Bi number indicate low conduction resistance, and therefore small thermal gradient within the body Lumped system is exact when Bi = 0 Generally accepted lumped system analysis when,10Bi.)

5 If Bi < , there is a 5% error or less in estimating temperature throughout body as a single-valued function of time T(t)16 Temperature is uniform throughout Temperature gradients are small inside to conduction within solid much less than resistance to convection across fluid boundary 1stMajor Assumptionbody the of surface theat resistance Convectionbody the within resistance conduction ==h1kLBic//17 Rate of heat energy passing through sphereQ = - h As(Ts- T )(W) = (W/[m2-Ko])(m2)(Ko) heat transfer coefficient isassumed notto be a function of 2ndMajor Assumption18 Problem: Steel balls 12 mmin diameter are annealed by heating to 1150 Kand then slowly cooling to 400 Kin an air environment for which T =325 Kand h=20 W/m2K.

6 Assuming the properties of the steel to be k=40 W/mK, =7800 kg/m3, and cp=600J/kgK, estimate the time required for the cooling number BikhLBic=()3323sc10210363rr4r34 AVL = == == Characteristic length()( ) <= == Therefore, temperature of the steel balls remain approximately uniform: lumped system analysis applicable 20 =TtTTThcLtipc)(ln btieTTTtT = )(pcpscLhVchAb == =TtTTTb1ti)(ln( ) )(ln70418s1122 KmWkgKJmmkg3254003251150206001027800 TtTTThcLt233ipc== = = 21Bi number provide-Measure of temperature drop in solid relative to temperature difference between the surface and the fluid body the of surface theat resistance Convectionbody the within resistance conduction ==h1kLBic// = =TTTTQTTQTTBi2s2s1s2s2s1s,,,,,, Steady state system22 Transient heat conduction in large plane walls.

7 Long cylinders and spheres with spatial effectsIn this section variation of temperature with time and position in one dimensional problems such as those associated with large plane wall, long cylinder and distributedsystem is one in which all dependent variables are functions of time andone or more spatial variables. In this case, we will be solving partial differential equations (PDEs)23 Transient heat conduction in large plane walls, long cylinders and spheres with spatial effects0tatTTi=< 24large plane walls with spatial effectsTransient temperature distribution T(x,t)in a wall results in a partial differential equation, which can be solved using advanced mathematical techniques.

8 The solution however, normally involves infinite series , which are inconvenient and time-consuming to evaluate. Therefore, there is a clear motivation to present the solution in tabular or graphical form. Solution involves so many parameters such as x, L, t, k, , h, Tiand T . In order to reduce the number of parameters, it is defined dimensionless quantities25 Dimensionless parametersDimensionless temperature =TTTtxTtxi),(),( Dimensionless distance from the center LxX=Dimensionless heat transfer coefficient khLBi=(Biot number)Dimensionless time 2Lt =(Fourier number)Nondimensionalization enables us to present temperature data in terms of X, Biand The above defined dimensionless quantities can be used for cylinder or sphere by replacing the variable xby rand Lby cylinder and sphere (not V/A)

9 26 One dimensional Transient heat conduction problem: For above geometries, solutions involve in finite series, which are difficult to deal with. Solutions using one dimensional approximation20 LxeATTTtxTtx11iwall21.),/cos(),(),(>= = Plane wall:20rrJeATTTtrTtr0101icyl21.),/(),(), (>= = Cylinder:20rrrreATTTtrTtr01011isph21.,/) /sin(),(),(>= = Sphere:A1and 1are functions of Bi and their values are listed in Table 18-1 J0is the Zeroth order Bessel function and values are listed in Table 18-22728 Temperature of the center of the plane wall, cylinder and sphere 21eATTTt0Tt01iwall = =),(),(Plane wall: 21eATTTt0Tt01icyl = =),(),(Cylinder: 21eATTTt0Tt01isph = =),(),(Sphere:Once Bi number is known, these relations can be use to determine the temperature anywhere in the medium (interpolations may be required to determine intermediate values).

10 29 Heisler charts Heisler 1947 There are 3 charts associated with each geometry Chart 1: Determine the temperature of the of the center of the geometry (T0) at a given time t. Chart 2: Determine the temperature of another location (T) in terms of (T0) center temperature. Chart 3: Determine the total amount of heat transfer up to the : these plots are valid to > charts -Large plane walls (18-13)(chart 1)Fourier number31 Heisler charts -Large plane walls (18-13)(chart 2)32 Heisler charts -Large plane walls (18-13)(chart 3)Note: Qmax=mcp(T - Ti)Total mount of heat transfer during whole periodQ is amount of heat transfer at finite time period t33 Heisler charts Long cylinder (18-14)Heisler charts - Sphere (18-15)