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

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

2 Objectives When you finish studying this chapter, you should be able to: Assess when the spatial variation of temperature is negligible, and temperature varies nearly uniformly with time, making the simplified lumped system analysis applicable, Obtain analytical solutions for transient one-dimensional conduction problems in rectangular, cylindrical, and spherical geometries using the method of separation of variables, and understand why a one-term solution is usually a reasonable approximation, Solve the transient conduction problem in large mediums using the similarity variable, and predict the variation of temperature with time and distance from the exposed surface, and Construct solutions for multi-dimensional transient conduction problems using the product solution approach.

4 Lumped System Analysis Lumped: 1- Temp. nearly isothermal - It is applicable for small bodies of high K In heat transfer analysis, some bodies are essentially isothermal and can be treated as a lump system. An energy balance of an isothermal solid for the time interval dt can be expressed as A s Heat Transfer into the body during dt = The increase in the energy of the body during dt h SOLID BODY m=mass V=volume ρ=density T i =initial temperature T=T(t) T () Q = has T T t ha s (T -T)dt= mc p dt (4 1)

5 Noting that m=ρv and dt=d(t-t ) since T constant, Eq. 4 1 can be rearranged as d( T T ) has = dt (4 ) T T ρvcp Integrating from time zero (at which T=T i ) to t gives Tt () T has ln = t (4 3) Ti T ρvcp Taking the exponential of both sides and rearranging Tt () T bt has = e ; b= (1/s) T T ρvc i b is a positive quantity whose dimension is (time) -1, and is called the time constant. p (4 4)

6 There are several observations that can be made from this figure and the relation above: 1. Equation 4 4 enables us to determine the temperature T(t) of a body at time t, or alternatively, the time t required for the temperature to reach a specified value T(t).. The temperature of a body approaches the ambient temperature T exponentially. 3. The temperature of the body changes rapidly at the beginning, but rather slowly later on. 4. A large value of b indicates that the body approaches the ambient temperature in a short time.

7 Rate of Convection Heat Transfer The rate of convection heat transfer between the body and the ambient can be determined from Newton s law of cooling Qt () = ha[ ] s Tt () T (W) (4 6) The total heat transfer between the body and the ambient over the time interval 0 to t is simply the change in the energy content of the body: [ ] Q= mcp T( t) T (kj) (4 7) The maximum heat transfer between the body and its surroundings (when the body reaches T ) [ ] Qmax = mcp Ti T (kj) (4 8)

8 Criteria for Lumped System Analysis Assuming lumped system is not always appropriate, the first step in establishing a criterion for the applicability is to define a characteristic length Large Plane Wall (thickness=l): Long Cylinder: Sphere: L c = V A LA Lc = V As = = L A π rl o ro Lc = V As = = π rl o 4 3 ro π = = 3 ro Lc V As = 4r π 3 o Note: if the size of previous shapes are enough small we should take the whole surface area s

9 Criteria for Lumped System Analysis and a Biot number (Bi): It can also be expressed as: Bi hl c = (4 9) k Bi Lc = k Rcond 1 = R = conv h Conduction resistance within the body Convection resistance at the surface of the body T R conv T R cond s h T in

10 Lumped system analysis assumes a uniform temperature distribution throughout the body, which is true only when the thermal resistance of the body to heat conduction is zero. The smaller the Bi number, the more accurate the lumped system analysis. It is generally accepted that lumped system analysis is applicable if Bi 0.1

11 Ex.: The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1.-mm-diameter sphere. The properties of the junction are k = 35 W/m C, ρ= 8500 kg/m 3, and C p = 30 J/kg C, and the heat transfer coefficient between the junction and the gas is h = 65 W/m C. Determine how long it will take for the thermocouple to read 99 percent of the initial temp. difference. Gas h, T Junction D T(t)

12 Analysis The characteristic length of the junction and the Biot number are L c Bi V πd / 6 D m = = = = = m A πd 6 6 surface hlc = = k 3 ( 65 W/m. C )( m) = < 01. ( 35 W/m. C) Since Bi <0.1, the lumped system analysis is applicable. Then the time period for the thermocouple to read 99% of the initial temperature difference is determined from Tt () T b T i T = 001. ha h = = = ρcv ρc L p p c 3 ( 8500 kg / m )( 30 J / kg. C)(0.000 m) 65 W/m. C = s -1 Tt () T T i T bt ( ) t = e 001. = e t = 38.5 s s -1

Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres with Spatial Effects In many transient heat transfer problems the Biot number is larger than 0.

13 Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres with Spatial Effects In many transient heat transfer problems the Biot number is larger than 0.1, and lumped system can not be assumed. In these cases the temperature within the body changes appreciably from point to point as well as with time. It is constructive to first consider the variation of temperature with time and position in one-dimensional problems of rudimentary configurations such as a large plane wall, a long cylinder, and a sphere.

A large Plane Wall A plane wall of thickness L. Initially at a uniform temperature of T i. At time t=0, the wall is immersed in a fluid at temperature T. Constant heat transfer coefficient h.

14 A large Plane Wall A plane wall of thickness L. Initially at a uniform temperature of T i. At time t=0, the wall is immersed in a fluid at temperature T. Constant heat transfer coefficient h. The height and the width of the wall are large relative to its thickness one-dimensional approximation is valid. Constant thermophysical properties. No heat generation. There is thermal symmetry about the midplane passing through x=0.

15 The Heat Conduction Equation One-dimensional transient heat conduction equation problem (0 x L): Differential equation: T x 1 T = α t (4 10a) Boundary conditions: ( 0, t) T = 0 x (4 10b) T( L, t) k = h T( L, t) T x Initial condition: ( ) T x,0 = Ti (4 10c)

17 1- Dimensionless Temp.: Large Wall Long Cylinder Sphere Dimensionless parameters: T ( x, t) T θ ( x, t) = T T T ( r, t) T θ ( rt, ) = Ti T T ( r, t) T θ ( rt, ) = T T - Dimensionless Distance from the center Large Wall Long Cylinder Sphere i i X = x L X = r r o X = r r o

18 3- Dimensionless Heat Transfer Coefficient (Biot Number) Large Wall Long Cylinder Sphere Bi = hl k Bi hro = k Bi hro = k 3- Dimensionless Time (Fourier Number Fo= ) Large Wall Long Cylinder Sphere τ α τ = t L α t α t τ = τ = r r o o

Fourier number ( 1/ L) αt kl ΔT τ = = = L ρc L / t ΔT 3 p The rate at which heat is conducted across L of a body of volume L 3 The rate at which heat is stored in a body of volume

19 Fourier number ( 1/ L) αt kl ΔT τ = = = L ρc L / t ΔT 3 p The rate at which heat is conducted across L of a body of volume L 3 The rate at which heat is stored in a body of volume L 3 The Fourier number is a measure of heat conducted through a body relative to heat stored. A large value of the Fourier number indicates faster propagation of heat through a body.

20 Approximate Analytical and Graphical Solutions The series solutions of Eq. 4-0 and in Table 4 1 converge rapidly with increasing time, and for τ >0., keeping the first term and neglecting all the remaining terms in the series results in an error under percent. Thus for τ >0. the one-term approximation can be used T( x, t) T θwall = = 1 cos λ1 /, τ > 0. T T λτ Plane wall: Ae ( ) 1 x L i λτ Cylinder: Ae ( ) 1 J r r Sphere: Trt (,) T θcyl = = 1 0 λ1 / 0, τ > 0. T T θ sph i Trt (,) T ( λ r r ) sin / λτ = = Ae 1 τ > Ti T λ1r/ r0, 0. (4 3) (4 4) (4 5)

21 The constants A 1 and λ 1 are functions of the Bi number only, and their values are listed in Table 4 against the Bi number for all three geometries. The function J 0 is the zeroth-order Bessel function of the first kind, whose value can be determined from Table 4 3.

22 The solution at the center of a plane wall, cylinder, and sphere: Center of plane wall (x=0): θ Center of cylinder (r=0): Center of sphere (r=0): θ θ T T = = A e 0 0, wall 1 Ti T T T = = A e 0 0, cyl 1 Ti T sph T 0 = = T i T T A e 1 1 λ τ 1 λ τ 1 λ τ (4 6) (4 7) (4 8)

23 Heisler Charts The solution of the transient temperature for a large plane wall, long cylinder, and sphere are also presented in graphical form for τ >0., known as the transient temperature charts (also known as the Heisler Charts). There are three charts associated with each geometry: the temperature T 0 at the center of the geometry at a given time t. the temperature at other locations at the same time in terms of T 0. the total amount of heat transfer up to the time t.

24 Heisler Charts Plane Wall Midplane temperature

25 Heisler Charts Plane Wall Temperature distribution Heat Transfer

26 Heat Transfer The maximum amount of heat that a body can gain (or lose if T i =T ) occurs when the temperature of the body is changes from the initial temperature T i to the ambient temperature ( ) ρ ( ) Qmax = mcp T Ti = Vcp T Ti (kj) (4 30) The amount of heat transfer Q at a finite time t is can be expressed as ( ) Q= ρcp T x, t -Ti dv (4 31) V

27 Assuming constant properties, the ratio of Q/Q max becomes ρc T x, t -T dv ( ) p i Q V 1 = = Q c T T V V ( - ) ( 1 ) V dv (4 3) max ρ p i V The following relations for the fraction of heat transfer in those geometries: Plane wall: Q Q θ sin λ1 λ = 1 0, wall max wall 1 (4 33) Cylinder: Q Q = 1 θ cyl ( λ ) 1 1 0, max λ cyl 1 J (4 34) Q sin λ λ cos λ = 1 3θ sph Q Sphere: , 3 max λ sph 1 (4 35)

28 Remember, the Heisler charts are not generally applicable The Heisler Charts can only be used when: the body is initially at a uniform temperature, the temperature of the medium surrounding the body is constant and uniform. the convection heat transfer coefficient is constant and uniform, and there is no heat generation in the body.

29 Ex.: White potatoes (k = 0.50 W/m C and α=0.13x10-6 m /s) that are initially at a uniform temperature of 5 C and have an average diameter of 6 cm are to be cooled by refrigerated air at C flowing at a velocity of 4 m/s. The average heat transfer coefficient between the potatoes and the air is experimentally determined to be 19 W/m C. Determine how long it will take for the center temperature of the potatoes to drop to 6 C. Also, determine if any part of the potatoes will experience chilling injury during this process.

30 Assumptions: The potatoes are spherical in shape with a radius of r 0 = 3 cm hr W/m. C) m Analysis: First we find the Biot number: 0 ( 19 ( ) Bi = = = 114. k 0.5 W / m. C From Table 4-1 we read, for a sphere, λ 1 = and A 1 = Substituting these values into the one-term solution gives T T λ 6 1 τ (1.635) τ θ = = A e = 1.30e τ=0.753 T T 5 o 0 1 i which is greater than 0. and thus the one-term solution is applicable. Then the cooling time becomes αt τr0 (0.753)(0.03 m) τ = t = = = 513s -6 r m / s 0 α = 1.45 h The lowest temperature during cooling will occur on the surface (r/r 0 = 1), and is determined to be T r T T sin( λ r / r ) T ( r ) T sin( λ r sin( λ r ( ) λ 1 τ = A1e = θ 0 = i T λ1r / r0 Ti T λ1r0 / r0 Ti T λ1r0 / r0 / r ) T T / r )

31 T ( r 0 ) 5 = 6 5 sin(1.635 rad) T( r 0 ) = 4.44 C which is above the temperature range of 3 to 4 C for chilling injury for potatoes. Therefore, no part of the potatoes will experience chilling injury during this cooling process. Alternative solution We could also solve this problem using transient temperature charts as follows: o 1 k 0.50 W/m. C = = = o Bi hr o (19 W/m. C)(0.03m) αt τ = 0.75 (Fig. 4-17a) = To T 6 ro = = Ti T 5 τ r0 t = = α 075)(. 003) 6 m / s = 519s 1.44 h The surface temperature is determined from 0 1 k = = Bi hr T(r) T r T0 T = 1 r0 = 0.6 (Fig. 4-17b) which gives T = T + 06.( T T ) = + 06.( 6 ) = 44.ºC surface o

Transient Heat Conduction in Semi- Infinite Solids A semi-infinite solid is an idealized body that has a single plane surface and extends to infinity in all directions.

32 Transient Heat Conduction in Semi- Infinite Solids A semi-infinite solid is an idealized body that has a single plane surface and extends to infinity in all directions. Assumptions: constant thermophysical properties no internal heat generation uniform thermal conditions on its exposed surface initially a uniform temperature of T i throughout. Heat transfer in this case occurs only in the direction normal to the surface (the x direction) one-dimensional problem.

33 Eq. 4 10a for one-dimensional transient conduction in Cartesian coordinates applies Differential equation: Boundary conditions: Initial condition: T x 1 T = α t ( 0, ) s (, ) T t = T T x t = Ti ( ) T x,0 = Ti (4 10a) (4 37b) (4 10c) The separation of variables technique does not work in this case since the medium is infinite. The partial differential equation can be converted into an ordinary differential equation by combining the two independent variables x and t into a single x variable η, called the similarity variable: η = 4α t

34 Boundary Conditions Case 1) Specified Surface Temperature, T s = constant. Where the complementary error function is defined as: η u u erf ( η) = e du ; erfc( η) 1 e du π = π 0 0 Use Table 4-4 for the complementary error function Case ) Specified Surface Heat Flux η

35 Case 3) Convection on the Surface

36 EX.: A thick wood slab (k = 0.17 W/m C and α=1.8x10-7 m /s) that is initially at a uniform temperature of 5 C is exposed to hot gases at 550 C for a period of 5 minutes. The heat transfer coefficient between the gases and the wood slab is 35 W/m C. If the ignition temperature of the wood is 450 C, determine if the wood will ignite. Hot gases T = 550 C Wood Slab T i = 5 C L=0.3 m 0 x

37 Assumption: The wood slab is treated as a semi-infinite medium subjected to convection at the exposed surface Analysis T ( x, t) T T T i i = erfc x hx h αt + exp erfc t k α k x αt + h αt k where h αt = k h αt = h k (35 W/m αt k. C) = 1.76 ( W/m. C -7 = 1.68 m / s)(5 60 s) = 1.76 Noting that x = 0 at the surface and using Table 4-4 for erfc values, T ( x, t) = erfc(0) exp( ) erfc( ) = 1 (5.0937)(0.077) = T( x, t) = 356 C which is less than the ignition temperature of 450 C. Therefore, the wood will not ignite. Note: One May Use Fig. 4-9 to solve the problem

Transient Heat Conduction in Multidimensional Systems Using a superposition approach called the product solution, the one-dimensional heat conduction solutions can also be used to construct solutions

38 Transient Heat Conduction in Multidimensional Systems Using a superposition approach called the product solution, the one-dimensional heat conduction solutions can also be used to construct solutions for some two-dimensional (and even three-dimensional) transient heat conduction problems. Provided that all surfaces of the solid are subjected to convection to the same fluid at temperature, the same heat transfer coefficient h, and the body involves no heat generation.

39 Example short cylinder Height a and radius r o. Initially uniform temperature T i. No heat generation At time t=0: convection T heat transfer coefficient h The solution: (,, ) (, ) (, ) T r x t T T x t T T r t T = X T T T T T T i Short i plane i infinite Cylinder wall cylinder (4 50)

40 The solution can be generalized as follows: the solution for a multidimensional geometry is the product of the solutions of the one-dimensional geometries whose intersection is the multidimensional body. For convenience, the one-dimensional solutions are denoted by T( x, t) T θ ( xt, ) θ θ wall cyl ( rt, ) semi-inf ( xt, ) = Ti T (, ) T r t T = Ti T (, ) plane wall infinite cylinder T x t T = Ti T semi-infinite solid (4 51)

42 Total Transient Heat Transfer The transient heat transfer for a two dimensional geometry formed by the intersection of two one-dimensional geometries 1 and is: Q Q Q Q = + 1- Qmax Q total, D max Q 1 max Q max 1 (4 53) Transient heat transfer for a three-dimensional (intersection of three one-dimensional bodies 1,, and 3) is: Q Q Q Q = + 1- Qmax Q total, 3D max Q 1 max Q max 1 Q Q Q Qmax Q 3 max Q 1 max (4 54)

EX.: A short brass cylinder ( ρ= 8530 kg/m 3, C p = 0.389 kj/kg C, k = 110 W/m C, and α= 3.39x10-5 m /s) of diameter D = 8 cm and height H = 15 cm is initially at a uniform temperature of T i = 150 C.

43 EX.: A short brass cylinder ( ρ= 8530 kg/m 3, C p = kj/kg C, k = 110 W/m C, and α= 3.39x10-5 m /s) of diameter D = 8 cm and height H = 15 cm is initially at a uniform temperature of T i = 150 C. The cylinder is now placed in atmospheric air at 0 C, where heat transfer takes place by convection with a heat transfer coefficient of h=40 W/m C. Calculate (a) the center temperature of the cylinder, (b) the center temperature of the top surface of the cylinder, and (c) the total heat transfer from the cylinder 15 min after the start of the cooling. D 0 = 8 cm Air T = 0 C x r L = 15 cm Brass cylinder T i = 150 C

44 Assumption: Heat conduction in the short cylinder is twodimensional, and thus the temperature varies in both the axial x- and the radial r- directions Analysis: This short cylinder can physically be formed by the intersection of a long cylinder of radius D/ = 4 cm and a plane wall of thickness L = 15 cm. We measure x from the midplane. (a) The Biot number is calculated for the plane wall to be Bi hl = k = (40 W/m. C)(0.075 m) (110 W/m. C) = The constants corresponding to this Biot number are, from Table 4-, The Fourier number is λ = and A1 1 = αt τ = = L 5 ( m / s)(15 min 60 s / min) ( m) = 544. > 0.

45 dimensionless temperature at the center of the plane wall is T0 T λ1 τ (0.164) (5.44) θo, wall = = A1e = (1.0050) e = Ti T We repeat the same calculations for the long cylinder hr Bi = k 0 = (40 W/m. C)(0.04 m) (110 W/m. C) = λ 1 and A 1 = = αt τ = = r o 5 ( m / s)(15 60 s) (. 004 m) = > 0. θ ocyl, = T T 0 i T T λ1 τ ( ) ( ) 1 = Ae = ( ) e = T (0,0, t) T T i T T (0,0, t) short cylinder = θ o, wall θ o, cyl = T (0,0, t) = 85.1 C = = 0.501

46 (b) The center of the top surface of the cylinder is still at the center of the long cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore, we first need to determine the dimensionless temperature at the surface of the wall. θ ( L, t) wall = T ( x, t) T T i T = A e 1 λ τ (0.164) (5.44) 1 cos( λ L / L) = (1.0050) e cos(0.164) = Then the center temperature of the top surface of the cylinder becomes T ( L,0, t) T T i T T ( L,0, t) short cylinder = θ ( L, t) wall θ o, cyl = T ( L,0, t) = 84. C = = 0.494

47 (c) We first need to determine the maximum heat can be transferred from the cylinder Q m = ρv max = mc = ρπr p ( T i o T L = (8530 kg/m 3 ). [ π(0.04 m) (0.15 m) ] = 6.43 kg ) = (6.43 kg)(0.389 kj/kg. C)(150 0) C = 35 kj Then we determine the dimensionless heat transfer ratios for both geometries as Q Q Q Q max Q Q max wall cyl = 1 θ = 1 θ o, wall o, cyl sin( λ λ J1( λ1) = 1 λ ) sin(0.164) = 1 (0.869) = (0.577) The heat transfer ratio for the short cylinder is max short cylinder Q = Q max plane wall Q + Q max long cylinder Q 1 Q max plane wall = 0.47 = (0.47)( ) = Then the total heat transfer from the short cylinder during the first 15 minutes of cooling becomes Q =.503Q = (0.504)(35 kj) 0 max = 164 kj

Chapter 3: Transient Heat Conduction

Chapter 3: Transient Heat Conduction 3-1 Lumped System Analysis 3- Nondimensional Heat Conduction Equation 3-3 Transient Heat Conduction in Semi-Infinite Solids 3-4 Periodic Heating Y.C. Shih Spring 009 3-1 Lumped System Analysis (1) In heat