TRANSIENT HEAT CONDUCTION

Transcription

1 TRANSIENT HEAT CONDUCTION Many heat conduction problems encountered in engineering applications involve time as in independent variable. This is transient or Unsteady State Heat Conduction. The goal of analysis is to determine the variation of the temperature as a function of time and position T (x, t) within the heat conducting body. In heat transfer analysis, some bodies are observed to behave like a lump whose interior temperature remains essentially uniform at all times during a heat transfer process. The Heat transfer analysis where the temperature of such bodies can be taken to be a function of time only, T(t) is known as lumped system analysis, which provides great simplification in certain classes of heat transfer problems without much sacrifice from accuracy. Consider a small hot copper ball coming out of an oven. Measurements indicate that the temperature of the copper ball changes with time, but it does not change much with position at any given time. Thus the temperature of the ball remains uniform at all times, and we can talk about the temperature of the ball with no reference to a specific location. On the other hand, consider a large roast in an oven. If you have done any roasting, you must have noticed that the temperature distribution within the roast is not even close to being uniform.

2 For instance, consider steady-state conduction through the plane wall of area A as shown in the figure. Although we are assuming steady-state conditions, the following criterion is readily extended to transient heat transfer. One surface is maintained at a temperature T s,1 and the other surface is exposed to a fluid of temperature T <T s,1. The temperature of this surface will be some intermediate value T s,2, for which T s,1.> T s,2 > T. Hence under steady-state conditions the surface energy balance is KKKK LL TT ss,1 TT ss,2 = haa(tt ss,2 TT ) Rearranging this equation, we get TT ss,1 TT ss,2 (TT ss,2 TT ) = hll kk The term hll is called Biot Number. For unsteady state heat transfer, Bi kk is expressed as Bi = hll cc kk = LL cc kk 1/h = CCCCCCCCCCCCCCCCCCCC rrrrrrrrrrrrrrrrrrrr CCCCCCCCCCCCCCCCCCCC rrrrrrrrrrrrrrrrrrrr Where L c is called Characteristic length. Observe from equation 1, that if convection resistance is large (slow convection) and conduction resistance is small (fast conduction) producing a value of Bi < 0.1, the body may be considered to have uniform temperature inside the body as shown in the figure. (1)

3 Consider a body of arbitrary shape of mass m, volume V, surface area A s, density ρ, and specific heat C p initially at a uniform temperature T i. At time t = 0, the body is placed into a medium at temperature T, and heat transfer takes place between the body and its environment, with a heat transfer coefficient h. Assume that T >T i, but the analysis is equally valid for the opposite case. We assume lumped system analysis to be applicable, so that the temperature remains uniform within the body at all times and changes with time only, T = T(t). During a differential time interval dt, the temperature of the body rises by a differential amount dt. An energy balance of the solid for the time interval dt can be expressed as Heat transfer into the body during dt = The increase in the energy of the body during dt h A s (T -T i, ) dt _ m C p dt Noting that m = ρv and dt = d(t T) since T = constant, Above Eq. can be rearranged as

4 dd(tt TT ) (TT TT ) = haa ss ρρρρcc pp dt Integrating from t = 0, at which T = T i, to any time t, at which T = T(t), gives Ln (TT(tt) TT ) (TT ii TT ) = haa ss ρρρρcc pp t Taking the exponential of both sides of the equation gives (TT(tt) TT ) (TT ii TT ) = ee haass ρρρρccpp tt (2) Rearranging haa ss tt ρρρρcc pp = h tt ρρ LL cc CC pp = h LL cc kk tt 2 = h LL cc kk ρρ CC pp LL cc kk kk ρρ CC pp tt LL cc 2 = BBBB αα tt LL cc 2 = Bi*Fo αα tt The term LL 2 is called Fourier Number (Fo), also known as cc θθ dimensionless time and we can write = (TT(tt) TT ) θθ ii (TT ii TT ) = ee BBBB FFFF (3) Additionally if we Write haa ss ρρρρcc pp = bb, we can also write equation 3 as θθ = (TT(tt) TT ) = θθ ii (TT ii TT ) ee bbbb (4) then b will have unit of (time) 1, called time constant. Both Equation 3 and 4 enable 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).

5 The temperature of a body approaches the ambient temperature T exponentially. The temperature of the body changes rapidly at the beginning, but rather slowly later on. A large value of b indicates that the body will approach the environment temperature in a short time. The larger the value of the exponent b, the higher the rate of decay in temperature. Note that b is proportional to the surface area, but inversely proportional to the mass and the specific heat of the body. This the reason why it takes longer to heat or cool a larger mass, especially when it has a large specific heat. Once the temperature T(t) at time t is known from Eq. 3 or 4, the rate of convection heat transfer between the body and its environment at that time can be determined from Newton s law of cooling as QQ (t) = h A s [T(t) T ] Watts The total amount of heat transfer between the body and the surrounding medium over the time interval t = 0 to t is simply the change in the energy content of the body: Q = mc p [T(t) - T i ] kj

6 Lumped System Analysis The lumped system analysis certainly provides great convenience in heat transfer analysis, and we should know when it is appropriate to use it. In order to establish a criterion for the applicability of the lumped system analysis we need to determine two things, Characteristic length L c = V/A s and Biot number Bi = hll cc kk Body Characteristic lengths Slab of thickness 2L L Cylinder of Diameter D o D o /4 Sphere of Diameter D o D o /6 Lumped system analysis therefore assumes a uniform temperature distribution throughout the body, which will be the case only when the thermal resistance of the body to heat conduction (the conduction resistance) is zero. Thus, lumped system analysis is exact when Bi = 0 and approximate when Bi >0. Of course, 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 and the variation of temperature with a location within the body will be slight and can reasonably be approximated as being uniform. If high accuracy is not necessary, lumped system analysis may be used even when the criterion Bi < 0.1 is not satisfied. If we encounter a general situation for which thermal conditions within a solid may be influenced simultaneously by convection, radiation, an applied surface heat flux, and internal energy generation. It is presumed that, initially (t = 0), the temperature of the solid T i differs from that of the fluid T, and the surroundings T sur, and that both surface and volumetric heating (q s and q) are

7 initiated. Applying conservation of energy at any instant t, we can write This Equation is a nonlinear, first-order, nonhomogeneous, ordinary differential equation that cannot be integrated to obtain an exact solution. However, exact solutions may be obtained for simplified versions of the equation. If there is no imposed heat flux or generation and convection is either nonexistent (a vacuum) or negligible relative to radiation, the above Equation reduces to Separating variables and integrating from the initial condition to any time t, a solution can be obtained as Similarly, exact solutions can be obtained for some systems with negligible radiation or only convection and other conditions. As evident from following Examples that these equations can be solved numerically for a wide variety of situations involving variable properties

8 or time-varying boundary conditions, internal energy generation rates, or surface heating or cooling. Consider the thermocouple hangs in a duct with convection and radiation heat transfer from the gas stream and surroundings. If the duct walls are at 400 o C and the emissivity of the thermocouple surface is 0.9, calculate the steady-state temperature of the junction. Also, determine the time for the junction temperature to increase from an initial condition of 25 o C to a temperature that is within 1 o C of its steady-state value. For steady-state conditions, the energy balance on the thermocouple junction has the form E in E out = 0. Recognizing that net radiation to the junction must be balanced by convection from the junction to the gas, the energy balance may be expressed as εεεε TT ssssssss 4 TT 4 h(tt TT ) = 0 Substituting numerical values results in T = o C On the other hand the energy balance for transient conditions is εεεε TT ssssssrr 4 TT 4 h(tt TT ) = ρ C pp dddd dddd The solution to this first-order differential equation can be obtained by numerical integration, giving the result, T(4.9 s) = o C. Hence, the time required to reach a temperature that is within 1 o C of the steadystate value is t = 4.9 s.

9 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 = 320 J/kg C, and the convection heat transfer coefficient between the junction and the gas is h = 210 W/m 2 C. Determine how long it will take for the thermocouple to read 99 percent of the initial temperature difference. Assumptions: The junction is spherical in shape with a diameter of D = m, the thermal properties of the junction and the heat transfer coefficient are constant, Radiation effects are negligible. Analysis: Need to determine the time it takes to register 99 percent of the initial T. The charac length of the junction L c = VV 1 6 = ππdd3 = 1 AA cc ππdd m=1.67*10-4 m Then the Bi number = hl c /k = = << Therefore, lumped system analysis is applicable, and the error involved in this approximation is negligible. In order to read 99 percent of the initial temperature difference T i T between the junction and the gas, we must have TT(tt) TT TT ii TT = 0.01 For example, when T i = 0 C and T = 100 C, a thermocouple is considered to have read 99 percent of this applied temperature difference when its reading indicates T (t ) = 99 C. Considering lump capacitance approach, The value of the b = haa ss ρρρρcc pp = h ρρcc pp LL cc = = Sec-1

10 Substituting these values into Eq. 4 TT(tt) TT TT ii TT Yields t = 10 secs = ee bbbb 0.01 = e * t Therefore, we must wait at least 10 s for the temperature of the thermocouple junction to approach within 1 percent of the initial junction-gas temperature difference. In real practical situations the Biot number is not small, and we must cope with the fact that temperature gradients within the medium are no longer negligible. Use of the lumped capacitance method would yield incorrect results, so alternative approaches, presented in the remainder of this chapter, must be utilized. These problems are mathematically more intensive and often not possible to solve mathematically. We will consider some simpler problems and leave the complex problems for numerical solution techniques. Example Engine valves (k=48 W/m K, C p = 440 J/kgK, and ρ = 7840 kg/m 3 ) are heated to 800 C in the heat treatment section of a valve manufacturing facility. The valves are then quenched in a large oil bath at an average temperature of 45 C. The convection coefficient in the oil bath is 650 W/m 2 K. The valves have a cylindrical stem with a diameter of 8 mm and a length of 10 cm. The valve head and the stem may be assumed to be of equal surface area, and the volume of the valve head can be taken to be 80 percent of the volume of stem. Determine how long will it take

11 for the valve temperature to drop to (a) 400 C, (b) 200 C, and (c) 46 C and (d) the maximum heat transfer from a single valve. (a) Example Case of annealing Bi = hlc/k = 20*(0.012/6)/40 = Since Bi<0.1, we can use lump Approach

12 EXACT SOLUTION OF TRANSIENT CONDUCTION PROBLEM IN A SLAB USING SEPARATION OF VARIABLES (When Bi >0.1) The non-dimensionalized partial differential equation formulated above together with its boundary and initial conditions can be solved using several analytical and numerical techniques, including the Laplace or other transform methods, the method of separation of variables, the finite difference method, and the finite-element method. In the case of transient conduction in a plain wall, for example, the dependent variable is the solution function u(x, t), which is expressed as u(x, t)= F(X)G(t), and the application of the method results in two ordinary differential equation, one in X and the other in t. The method is applicable if (1) the geometry is simple and finite (such as a rectangular block, a cylinder, or a sphere) so that the boundary surfaces can be described by simple mathematical functions, and (2) the differential equation and the boundary and initial conditions in their most simplified form are linear (no terms that involve products of the dependent variable or its derivatives) and involve only one nonhomogeneous term (a term without the dependent variable or its derivatives).

13 TRANSIENT CONDUCTION IN SLAB WITH CONVECTION The dimensionless time τ is usually denoted as Fourier Number Fo seen before. A large value

14 TRANSIENT CONDUCTION IN CYLINDER WITH CONVECTION

15 TRANSIENT CONDUCTION IN SPHERE WITH CONVECTION

16 One-Dimensional Problems Imposed Boundary Temperature in Cartesian Coordinates: A simple but important conduction heat transfer problem consists of determining the temperature history inside a solid flat wall which is quenched from a high temperature. More specifically, consider the homogeneous problem of finding the onedimensional temperature distribution inside a slab of thickness L and thermal diffusivity α, initially at some specified temperature T (x, 0) = f(x) and exposed to heat extraction at its boundaries x = 0 and x = L such that T (0, t) = T (L, t) = 0 (Dirichlet homogeneous conditions), for t > 0. The thermal properties are assumed constant. Convection at the Boundary in Cartesian Coordinates: Same geometry BUT the boundary conditions specify values of the normal derivative of the temperature or when linear combinations of the normal derivative and the temperature itself are used. Consider the homogeneous problem of transient heat conduction in a slab initially at a temperature T = f(x) and subject to convection losses into a medium at T = 0 at x = 0 and x = L. Convection heat transfer coefficients at x = 0 and x = L are, respectively h 1 and h 2. Assume the thermal conductivity of the slab k is constant. Imposed Boundary Temperature and Convection at the Boundary in Cylindrical Coordinates: (i) a long cylinder (radius r = b) initially at T = f(r) whose surface temperature is made equal to zero for t > 0. (ii) A long cylinder (radius r = b) initially at T = f(r) is exposed to a cooling medium which extracts heat uniformly from its surface. Imposed Boundary Temperature and Convection at the Boundary in Spherical Coordinates: (i) quenching problem where a sphere (radius r = b) initially at T = f(r) whose surface temperature is made equal to zero for t > 0. Consider a sphere with initial temperature T (r, 0) = f(r) and dissipating heat by convection into a medium at its surface r = b.

17 Separation of Variables Let us consider the Slab/Convection experiment. Recall that in this case we have: The heat conduction equation in cylindrical or spherical coordinates can be nondimensionalized in a similar way. Note that nondimensionalization reduces the number of independent variables and parameters from 8 to 3 from x, L, t, k, a, h, T i, and T to X, Bi, and Fo. That is, This makes it very practical to conduct parametric studies and to present results in graphical form. Recall that in the case of lumped system analysis, we had u f(bi, Fo) with no space variable.

18 Separation of Variables First, we express the dimensionless temperature function u(x, t) as a product of a function of X only and a function of t only as: Substituting to: we have (C) all the terms that depend on X are on the left-hand side of the equation and all the terms that depend on t are on the r the terms that are function of different variables are separated (and thus the name separation of variables). Considering that both X and t can be varied independently, the equality in Eq. C can hold for any value of X and t only if it is equal to a constant. Further, it must be a negative constant that we will indicate by -λ 2 since a positive constant will cause the function G(t) to increase indefinitely with time (to be infinite), which is unphysical, and a value of zero for the constant means no time dependence, which is again inconsistent with the physical problem. Setting Eq. C equal to -λ 2 gives: whose general solutions are: and

19 Separation of Variables Then it follows that there are an infinite number of solutions of the form, and the solution of this linear heat conduction problem is a linear combination of them, The constants A n are determined from the initial condition This is a Fourier series expansion that expresses a constant in terms of an infinite series of cosine functions. Now we multiply both sides of last eq. by cos(λ m X), and integrate from X=0 to X=1. The right-hand side involves an infinite number of integrals of the form: It can be shown that all of these integrals vanish except when n m, and the coefficient An becomes:

20 Separation of Variables This completes the analysis for the solution of onedimensional transient heat conduction problem in a plane wall. Solutions in other geometries such as a long cylinder and a sphere can be determined using the same approach. The results for all three geometries are summarized in Table. Note that the solution for the plane wall is also applicable for a plane wall of thickness L whose left surface at x =0 is insulated and the right surface at x=l is subjected to convection since this is precisely the mathematical problem we solved.

21 Approximate Analytical Solutions The analytical solutions of transient conduction problems typically involve infinite series, and thus the evaluation of an infinite number of terms to determine the temperature at a specified location and time. However, as demonstrated in Table, the terms in the summation decline rapidly as n and thus λ n increases because of the exponential decay function. This is especially the case when the dimensionless time τ is large. Therefore, the evaluation of the first few terms of the infinite series (in this case just the first term) is usually adequate to determine the dimensionless temperature θ. For example, for τ> 0.2, keeping the first term and neglecting all the remaining terms in the series results in an error under 2 percent. We are usually interested in the solution for times with τ> 0.2, and thus it is very convenient to express the solution using this one-term approximation, given as: where the constants A 1 and λ 1 are functions of the Bi number only, and their values are listed in Table (see next slide) against the Bi number for all three geometries. The function J 0 is the zeroth-order Bessel function of the first kind (see next slide).

22 Useful Tables

23 Approximate Analytical Solutions Noting that cos (0)= J 0 (0)= 1 and the limit of (sin x)/x is also 1, these relations simplify to the next ones at the center of a plane wall, cylinder, or sphere: Comparing the sets of equations above with approximate solution we notice that the dimensionless temperatures anywhere in a plane wall, cylinder, and sphere are related to the center temperature by which shows that time dependence of dimensionless temperature within a given geometry is the same throughout. That is, if the dimensionless center temperature θ 0 drops by 20 percent at a specified time, so does the dimensionless temperature θ 0 anywhere else in the medium at the same time. Once the Bi number is known, these relations can be used to determine the temperature anywhere in the medium.

24 Useful Relationship Again the temperature of the body changes from the initial temperature T i to the temperature of the surroundings T at the end of the transient heat conduction process and the maximum amount of heat that a body can gain (or lose) is simply the change in the energy content of the body: The amount of heat transfer Q at a finite time t is Assuming constant properties, the ratio of Q/Qmax becomes Using the appropriate non-dimensional temperature relations based on the one term approximation for the plane wall, cylinder, and sphere, and performing the indicated integrations, we obtain the following relations for the fraction of heat transfer in those geometries: These Q/Qmax ratio relations based on the one-term approximation are also plotted in Heisler charts, against the variables Bi and h 2 at/k 2 for the large plane wall, long cylinder, and sphere, respectively. Note that once the fraction of heat transfer Q/Qmax has been determined from these charts or equations for the given t, the actual amount of heat transfer by that time can be evaluated by multiplying this fraction by Qmax.

25 Example Consider a steel pipeline (AISI 1010) that is 1 m in diameter and has a wall thickness of 40 mm. The pipe is heavily insulated on the outside, and, before the initiation of flow, the walls of the pipe are at a uniform temperature of 20 o C. With the initiation of flow, hot oil at 60 o C is pumped through the pipe, creating a convective condition corresponding to h =500 W/m 2 K at the inner surface of the pipe. 1. What are the appropriate Biot and Fourier numbers 8 min after the initiation of flow? 2. At t= 8 min, what is the temperature of the exterior pipe surface covered by the insulation? 3. What is the heat flux q (W/m 2 ) to the pipe from the oil at t = 8min? 4. How much energy per meter of pipe length has been transferred from the oil to the pipe at t = 8min? Properties: Table A.1, steel AISI 1010 [T = ( )/2 =300 K]: ρ = 7832 kg/m 3, c = 434 J/kgK, k =63.9 W/mK, _ α = 18.8x10-6 m 2 /s. At t = 8 min, the Biot and Fourier numbers are computed from previous Equations, respectively, with Lc _ L. Hence With Bi = 0.313, use of the lumped capacitance method is inappropriate. However, since Fo = 0.2 and transient conditions in the insulated pipe wall of thickness L correspond to those in a plane wall of thickness 2L experiencing the same surface condition, the desired results may be obtained from the one-term approximation for a plane wall. The midplane temperature can be determined from Equation where, with Bi= 0.313, C 1 =1.047 and ζ 1 = rad from Table. With Fo = 5.64,

26 Hence after 8 min, the temperature of the exterior pipe surface, which corresponds to the midplane temperature of a plane wall, is T(0, 8 min) =T + θο (Ti T ) = ( 20 60) C = 42.9 C Heat transfer to the inner surface at x = L is by convection, and at any time t the heat flux may be obtained from Newton s law of cooling. Hence at t = 480 s, q x (L, 480 s) = q L = h[t(l, 480 s) T ] Using the one-term approximation for the surface temperature, with x*= 1 has the form The heat flux at t = 8 min is then q L = 500 W/m 2 K( ) C = 7400 W/m 2 4. The energy transfer to the pipe wall over the 8-min interval may be obtained from above Equations With it follows that Q = 0.80 ρ c V(T i T ) The minus sign associated with q and Q simply implies that the direction of heat transfer is from the oil to the pipe (into the pipe wall).

27 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, as shown in Fig. This idealized body is used to indicate that the temperature change in the part of the body in which we are interested (the region close to the surface) is due to the thermal conditions on a single surface. Consider a semi-infinite solid that is at a uniform temperature T i. At time t = 0, the surface of the solid at x = 0 is suddenly subjected to convection by a fluid at a constant temperature T, with a heat transfer coefficient h. This problem can be formulated as a partial differential equation, which can be solved analytically for the transient temperature distribution T(x, t). The exact solution of this transient one-dimensional heat conduction problem where the quantity erfc (ζ) is the complementary error function, defined as Despite its simple appearance, the integral that appears in the above relation cannot be performed analytically. Therefore, it is evaluated numerically for different values of ζ, and the results are listed in Table. For the special case of h, the surface temperature T s becomes equal to the fluid temperature T, and above Eq. reduces to

28 Example A semi-infinite aluminum cylinder of diameter D= 20 cm is initially at a uniform temperature T i = 200 C. The cylinder is now placed in water at 15 C where heat transfer takes place by convection, with a heat transfer coefficient of h= 120 W/m 2 K. Determine the temperature at the center of the cylinder 15 cm from the end surface 5 min after the start of the cooling. We will solve this problem using the one-term solution relation for the cylinder and the analytic solution for the semi-infinite medium. First we consider the infinitely long cylinder and evaluate the Biot number:

29 The coefficients λ1 and A1 for a cylinder corresponding to this Bi are determined from Table to be λ1 = and A1= The Fourier number in this case is and thus the one-term approximation is applicable. Substituting these values into Eq. gives The solution for the semi-infinite solid can be determined from First we determine the various quantities in parentheses: Substituting and evaluating the complementary error functions from Table,

30 Now we apply the product solution to get

31 Example A thick wood slab (k = 0.17 W/m C and α = m 2 /s) that is initially at a uniform temperature of 25 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 2 C. If the ignition temperature of the wood is 450 C, determine if the wood will ignite.