Chapter 4: Transient Heat Conduction Yoav PelesDepartment of Mechanical, Aerospace and Nuclear Engineering Rensselaer Polytechnic Institute Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

ObjectivesWhen 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.

Lumped System Analysis• 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 timeinterval dt can be expressed as As SOLID BODYHeat Transfer The increase in the m = mass h V = volume Tinto the body = energy of the body ρ = density Ti = initial temperatureduring dt during dt T = T(t)hAs(T -T)dt= mcpdT (4–1) Q = hAs ⎣⎡T∞ − T (t )⎤⎦

• Noting that m=ρV and dT=d(T-T ) since T constant, Eq. 4–1 can be rearranged as d (T − T∞ ) = hAs dt (4–2) T − T∞ ρVcp• Integrating from time zero (at which T=Ti) to t gives ln T (t) − T∞ = hAs t (4–3) Ti − T∞ ρVcp• Taking the exponential of both sides and rearrangingT (t) − T∞ = e−bt ; b = hAs (1/s) (4–4)Ti − T∞ ρVcp• b is a positive quantity whose dimension is (time)-1, and is called the time constant.

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).2. 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.

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 Q (t) = hAs [T (t) − 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)

Criteria for Lumped System Analysis• Lumped system is not always appropriate, characteristic length Lc = V As• and a Biot number (Bi) as (4–9)• It can also be expressed as Bi = hLc k Lc = Rcond Conduction resistance within the body k Rconv Convection resistance at the surface of the bodyBi = 1 = Rconv Rcond h T Ts Tin h

• 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

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 2L.• Initially at a uniform temperature of Ti.• 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 Æ one-dimensional approximation is valid.• Constant thermophysical properties.• No heat generation.• There is thermal symmetry about the midplane passing through x=0.

The Heat Conduction Equation• One-dimensional transient heat conduction equation problem (0≤ x ≤ L):Differential equation: ∂ 2T =1 ∂T (4–10a) ∂x2 ∂t αBoundary conditions: ⎧ ∂T (0,t ) = 0 (4–10b) ⎪⎪ ⎨ ∂x ) = h ⎡⎣T ( L, t ) − T∞ ⎤⎦ ⎩⎪⎪−k ∂T ( L,t ∂xInitial condition: T ( x, 0) = Ti (4–10c)

Non-dimensional Equation• A dimensionless space variable X=x/L• A dimensionless temperature variable θ(x, t)=[T(x,t)-T ]/[Ti-T ]• The dimensionless time and h/k ratio will be obtained through the analysis given below• Introducing the dimensionless variable into Eq. 4-10a∂θ = ∂θ = Ti L ∂T ; ∂2θ = Ti L2 ∂2T ; ∂θ = 1 ∂T − T∞ ∂x − T∞ ∂x2 Ti − T∞ ∂t∂X ∂(x/ L) ∂X 2 ∂t• Substituting into Eqs. 4–10a and 4–10b and rearranging ∂ 2θ = L2 ∂ 2T ∂θ ; ∂θ (1,t ) = hL θ (1,t ) ; ∂θ (0,t ) = 0 (4–11) ∂x2 ∂X k ∂X ∂X 2 α ∂t

• Therefore, the dimensionless time is τ=αt/L2, which is called the Fourier number (Fo).• hL/k is the Biot number (Bi).• The one-dimensional transient heat conductionproblem in a plane wall can be expressed innondimensional form asDifferential equation: ∂2θ = ∂θ (4–12a) ∂τ ∂X 2Boundary conditions: ⎧ ∂θ (0,τ ) =0 (4–12b) ⎪⎪ ⎨ ∂X = −Biθ (1,τ ) ⎪ ⎪⎩ ∂θ (1,τ ) ∂XInitial condition: θ (X,0) =1 (4–12c)

Exact Solution• Several analytical and numerical techniques can be used to solve Eq. 4-12.• We will use the method of separation of variables.• The dimensionless temperature function θ(X,τ) is expressed as a product of a function of X only and a function of τ only asθ ( X ,τ ) = F ( X )G (τ ) (4–14)• Substituting Eq. 4–14 into Eq. 4–12a and dividing by the product FG gives1 d2F = 1 dG (4–15)F dX 2 G dτ

• Since X and τ can be varied independently, the equality in Eq. 4–15 can hold for any value of X and τ only if Eq. 4–15 is equal to a constant.• It must be a negative constant that we will indicate by -λ2 since a positive constant will cause the function G(τ) to increase indefinitely with time.• Setting Eq. 4–15 equal to -λ2 givesd2F + λ2F = 0 ; dG + λ 2F = 0 (4–16)dX 2 dτ• whose general solutions are⎪⎧F = C1 cos (λ X ) + C2 sin (λ X ) (4–17)⎩⎨⎪G=C3e−λ2τ

θ = FG = C3e−λ2τ ⎡⎣C1 cos (λ X ) + C2 sin (λ X )⎦⎤= e−λ2τ ⎣⎡ Acos (λ X ) + B sin (λ X )⎦⎤ (4–18)• where A=C1C3 and B=C2C3 are arbitrary constants.• Note that we need to determine only A and B toobtain the solution of the problem.• Applying the boundary conditions in Eq. 4–12b gives ∂θ (0,τ ) = 0 → −e−λ2τ ( Aλ sin 0 + Bλ cos 0) = 0 ∂X → B = 0 → θ = Ae−λ2τ cos (λ X )∂θ (1,τ ) = −Biθ (1,τ ) → − Ae−λ2τ λ sin λ = −BiAe−λ2τ cos λ ∂X → λ tan λ = Bi

• But tangent is a periodic function with a period of π, and the equation λtan(λ)=Bi has the root λ1 between 0 and π, the root λ2 between π and 2π, the root λn between (n-1)π and nπ, etc.• To recognize that the transcendental equation λtan(λ)=Bi has an infinite number of roots, it is expressed as λn tan λn = Bi (4–19)• Eq. 4–19 is called the characteristic equation or eigenfunction, andits roots are called the characteristic values or eigenvalues.• It follows that there are an infinite number of solutions of the form θ = Ae−λ2τ cos (λ X ) , and the solution of this linear heat conductionproblem is a linear combination of them,∞ A e−λn2τ n∑ ( )θ = cos λn X (4–20)n=1• The constants An are determined from the initial condition, Eq. 4–12c, ∞θ ( X , 0) = 1 → 1 = ∑ An cos (λn X ) (4–21) n=1

• Multiply both sides of Eq. 4–21 by cos(λmX), and integrating from X=0 to X=1 X =1 X =1 ∞ ∫ cos (λm X ) = ∫ cos (λm X )∑ An cos (λn X ) X =0 X =0 n=1• The right-hand side involves an infinite number of integrals of the form X =1 ∫ cos (λm X ) cos (λn X ) dX X =0• It can be shown that all of these integrals vanish except when n=m, and the coefficient An becomes X =1 X =1 ∫ cos (λn X ) dX = An ∫ cos2 (λn X ) dX X =0 X =0 → An = 2λn 4sin λn ) (4–22) + sin (2λn

• Substituting Eq. 4-22 into Eq. 20a gives=∑θ ∞ 4sin λn e−λn2τ cos (λn X ) n=1 2λn + sin (2λn )• Where λn is obtained from Eq. 4-19.• As demonstrated in Fig. 4–14, the terms in the summation decline rapidly as n and thus λn increases.• Solutions in other geometries such as a long cylinder and a sphere can be determined using the same approach and are given in Table 4-1. FIGURE 4-14

Summary of the Solutions for One-Dimensional Transient Conduction

Approximate Analytical and Graphical Solutions• The series solutions of Eq. 4-20 and in Table 4–1 converge rapidly with increasing time, and for τ >0.2, keeping the first term and neglecting all the remaining terms in the series results in an error under 2 percent.• Thus for τ >0.2 the one-term approximation can be usedPlane wall: θwall = T (x,t) − T∞ = A e−λ12τ cos (λ1x / L), τ > 0.2 (4–23) Ti − T∞ 1 (4–24) (4–25) = T (r,t) − T∞ A1e−λ12τ J0 Ti − T∞ ( )θcylCylinder: = λ1r / r0 , τ > 0.2 ( )θsphSphere: = T (r,t) − T∞ = A e−λ12τ sin λ1r / r0 , τ > 0.2 Ti − T∞ 1 λ1r / r0

• Tonhley,coanndstathnetsirAv1aaluneds λa1rearleistfeudncintioTnasbolef the Bi number Bi 4–2 against the number for all three geometries.• The function J0 is the zeroth-order Bessel function of the first kind, whose value can be determined from Table 4–3.

The solution at the center of a plane wall, cylinder, and sphere:Center of plane wall (x=0): θ0,wall = T0 − T∞ = A e−λ12τ (4–26) Ti − T∞ 1 (4–27) (4–28)Center of cylinder (r=0): θ0,cyl = T0 − T∞ = A e−λ12τCenter of sphere (r=0): Ti − T∞ 1 θ sph = T0 − T∞ = A e−λ12τ Ti − T∞ 1

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>0.2, known as the transient temperature charts (also known as the Heisler Charts).• There are three charts associated with each geometry: – the temperature T0 at the center of the geometry at a given time t. – the temperature at other locations at the same time in terms of T0. – the total amount of heat transfer up to the time t.

Heisler Charts – Plane Wall Midplane temperature

Heisler Charts – Plane WallTemperature distribution

Heat Transfer

Heat Transfer• The maximum amount of heat that a body can gain (or lose if Ti=T ) occurs when the temperature of the body is changes from the initial temperature Ti to the ambient temperatureQmax = mcp (T∞ − Ti ) = ρVcp (T∞ − Ti ) (kJ) (4–30)• The amount of heat transfer Q at a finite time t is can be expressed asQ = ∫ ρcp ⎣⎡T ( x,t ) -Ti ⎤⎦ dV (4–31) V

• Assuming constant properties, the ratio of Q/Qmaxbecomes ρcp ⎣⎡T ( x,t ) -Ti ⎤⎦ dV ρcp (T∞ -Ti )V ∫Q = V = 1 (1−V ) dV (4–32)∫Qmax VV• The following relations for the fraction of heat transferin those geometries:Plane wall: ⎛ Q ⎞ = 1− θ0,wall sin λ1 (4–33)Cylinder: ⎜ Qmax ⎟ λ1 (4–34)Sphere: ⎝ ⎠wall (4–35) ( )⎛ Q ⎜ ⎞ = 1− 2θ0,cyl J1 λ1 ⎝ Qmax ⎟ λ1 ⎠cyl ⎛ Q ⎞ = 1− 3θ0,sph sin λ1 − λ1 cos λ1 ⎜ Qmax ⎟ λ13 ⎝ ⎠sph

Remember, the Heisler charts are not generally applicableThe 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.

Fourier numberτ = αt = kL2 (1/ L) ∆T = The rate at which heat is conducted ∆T across L of a body of volume L3 L2 ρcp L3 / t The rate at which heat is stored in a body of volume L3• 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.

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 Ti throughout.• Heat transfer in this case occurs only in the direction normal to the surface (the x direction) one-dimensional problem.

• Eq. 4–10a for one-dimensional transient conduction in Cartesian coordinates appliesDifferential equation: ∂ 2T =1 ∂T (4–10a) ∂x2 ∂t α ⎧⎪T (0,t ) = Ts (4–37b)Boundary conditions: ⎨ ⎩⎪T ( x → ∞, t ) = TiInitial condition: T ( x, 0) = Ti (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 variable η, called the similarity variable.

Similarity Solution• For transient conduction in a semi-infinite mediumSimilarity variable: η= x 4α t• Assuming T=T(η) (to be verified) and using the chainrule, all derivatives in the heat conduction equationcan be transformed into the new variable (4–39a) ∂ 2T =1 ∂T ∂ 2T = −2η ∂T ∂x2 ∂t α ∂η 2 ∂η

∂ 2T = −2η ∂T (4–39a)∂η 2 ∂η• Noting that η=0 at x=0 and η as x (and alsoat t=0) and substituting into Eqs. 4–37b (BC) give,after simplificationT (0) = Ts ; T (η → ∞) = Ti (4–39b)• Note that the second boundary condition and the initial condition result in the same boundary condition.• Both the transformed equation and the boundary conditions depend on h only and are independent of x and t. Therefore, transformation is successful, and η is indeed a similarity variable.

• To solve the 2nd order ordinary differential equation in Eqs. 4– 39, we define a new variable w as w=dT/dη. This reduces Eq. 4– 39a into a first order differential equation than can be solved by separating variables,dw = −2η w → dw = −2ηdη → ln (w) = −η 2 + C0 wdη• where C1=ln(C0). → w = C1e−η2• Back substituting w=dT/dη and integrating again, η (4–40) ∫T = C1 e−u2 du + C2 0• where u is a dummy integration variable. The boundarycondition at η=0 gives C2=Ts, and the one for η gives ∫∞ π 2 Ti − Ts (4–41)( )Ti = C1 e−u2 du + C2 = C12 π 0 + Ts → C1 =

• Substituting the C1 and C2 expressions into Eq. 4–40 and rearranging,∫T − Ts = 2 η (4–42) e−u2 du = erf (η ) = 1− erfc (η )Ti − Ts π 0• Where∫erf (η ) =2η ; erfc (η ) = 1− ∫2η (4–43) e−u2 du e−u2 du π0 π0• are called the error functionand the complementary errorfunction, respectively, ofargument η.

• Knowing the temperature distribution, the heat flux atthe surface can be determined from the Fourier’s lawto beqs = −k ∂T = −k ∂T ∂η = − kC1e −η 2 1 = k (Ts − Ti ) ∂x 4αt η=0 πα t x=0 ∂η ∂x η =0 (4–44)

Other Boundary Conditions• The solutions in Eqs. 4–42 and 4–44 correspond tothe case when the temperature of the exposed surfaceof the medium is suddenly raised (or lowered) to att=0 and is maintained at that value at all times. Ts• Analytical solutions can be obtained for other boundary conditions on the surface and are given in the book– Specified Surface Temperature, Ts = constant.– Constant and specified surface heat flux.– Convection on the Surface,– Energy Pulse at Surface.

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.

Example short cylinder • Height a and radius ro. • Initially uniform temperature Ti. • No heat generation • At time t=0: – convection T – heat transfer coefficient h • The solution: (r, x,t ) − T∞ ( ) ( )⎛ T r,t − T∞ Ti − T∞ Ti − T∞ =⎜ ⎝⎛ T ⎞ x,t − T∞ ⎞ X ⎛T ⎞ (4–50)⎜ ⎟ Ti − T∞ ⎟ ⎜ ⎟⎝ ⎠ Short ⎠ plane ⎝ ⎠ infinite Cylinder wall cylinder

# Chapter 4: Transient Heat Conduction - nchu.edu.tw

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**Description: ** Lumped System Analysis • In heat transfer analysis, some bodies are essentially isothermal and can be treated as a “lump” system. • An energy balance of an ...

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