CHAPTER
Conduction: 3 One-dimensional transient
and two-dimensional steady state
3.1 Introduction
There are many situations and applications where temperature is a function of time alone, or time and space, for example, in the heat treatment of ball bearings, response of thermocouples, aerodynamic heating of rockets, transients in avionic cooling packages, and so on. In view of this, it is necessary to know the temperature at a location at a given instant of time. One way to do this is to simply measure it. The “analytically” inclined, though, always ask the question, “Can we obtain this information by theory?” The answer is yes, and in this chapter we see how to obtain this information in a few typical cases of engineering interest.
Please recall the general heat conduction equation derived in chapter 2, which is
2T + qv = |
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In Cartesian coordinates, Eq. (3.1) can be written as
∂2 T |
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1 ∂T |
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Now, let us consider a stationary plane wall of thickness 2L initially at Ti that is suddenly immersed in a convection environment with a heat transfer coefficient “h” and an ambient temperature of T∞.
A schematic representation of the plane wall is shown in Fig. 3.1 The following assumptions hold:
1.Ti > T∞
2.One-dimensional, unsteady state heat transfer, that is, T = f (x, t) alone
3.Constant thermophysical properties of the medium
4.No heat generation in the solid, that is, qv = 0
5.(ρ, cp , k) are constant
6.Radiation is negligible
Heat Transfer Engineering. http://dx.doi.org/10.1016/B978-0-12-818503-2.00003-4 |
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Copyright © 2021 Elsevier Inc. All rights reserved.
66 CHAPTER 3 Conduction
FIGURE 3.1
Schematic representation of a one-dimensional slab undergoing transient conduction (Ti >T∞).
The general governing equation reduces to
∂2 T |
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∂x |
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Eq. (3.3) supports two boundary conditions, as it is second order in x and one initial condition, as it is first order in time.
∂T
∂x Boundary condition 2: At x = ±L; −k ∂∂Tx
Initial condition: At time t = 0 s, T = Ti (throughout the slab)
A close look at the above boundary and initial conditions shows that the slab is initially hot and is getting convectively cooled at the two ends. After sufficient time elapses, the entire slab will be at T∞. Hence, there is little interest in the steady state solution to this problem. The challenge is to determine T(x,t) for any (x,t).
We can now carry out a non-dimensionalization of the governing equations and the boundary and initial conditions to get down to the bottom of the problem.
Let ξ, the nondimensional length, be defined as x/L, and φ, the nondimensional
temperature excess, be defined as T −T∞ . Then
Ti −T∞
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3.1 Introduction |
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∂∂Tx = (Ti − T∞ ) ∂∂φx
∂2 T = (Ti − T∞ ) ∂2 φ ∂x2 ∂x2
Now, non-dimensionalizing x as ξ = x/L, we have
∂2 T |
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(T − T ) ∂2 |
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Similarly,
∂T = (Ti − T∞ ) ∂φ ∂t ∂t
Substituting Eqs. (3.6) and (3.7) in Eq. (3.3), we have
(T − T ) ∂2 |
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∂φ |
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∞ |
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∂ξ 2 |
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Eq. (3.9) may also be written as
(3.4)
(3.5)
(3.6)
(3.7)
(3.8)
(3.9)
∂2 φ |
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∂ |
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From the boundary condition 2: At x = ±L; ξ = ±1; −k ∂∂Tx = h(T − T∞ )
−k Ti − T∞ ∂φ = h(Ti − T∞ )φ L ∂ξ
∂∂ξφ = − hLk φ
(3.10)
(3.11)
(3.12)
From the initial condition: At time t = 0 s or αL2t = 0; T = Ti; φ = 1, and from the bound-
ary condition 1: At x = 0 or ξ = 0, ∂∂Tx = 0 or ∂∂ξφ = 0
From the above set of equations, it is clear that we can immediately recognize that all quantities appearing in Eq. (3.10) and Eq. (3.12) are dimensionless.
From the above non-dimensionalization, it is evident that φ can be written as a function of the pertinent dimensionless parameters, as follows
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CHAPTER 3 Conduction |
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Please note that the dependence of φ on ξ and |
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hL |
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dependence on |
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Eq. (3.12). |
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We now introduce, two key dimensionless numbers, namely, Biot number (Bi) |
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and Fourier number (Fo) as: |
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hL |
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Rconduction |
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Biot number = Bi = k |
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convection |
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where A is the area of the slab perpendicular to the plane of the paper. |
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Fourier number = Fo = |
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L |
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The Fourier number can be rewritten as Fo = |
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with |
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having the units of |
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time, and let us denote it as t+, a characteristic time in the problem. Therefore, Fo = tt+ .
In other words, the Fourier number is nothing but a scaled or normalized or dimensionless time.
The Biot number, as shown above, is the ratio of conduction to the convection resistance.
As Rconduction →0, the temperature variation within the solid body approaches zero. The resulting problem at hand is frequently referred to as one with negligible internal temperature gradients. In the parlance of heat transfer, this formulation or simplification is referred to as the lumped capacitance method with the word “lumped” implying that the object is assumed to be spatially isothermal. We now look at this variant of the transient heat conduction problem in a little more detail.
3.2 Lumped capacitance method
Consider a body of arbitrary shape shown in Fig. 3.2 with a mass of m, surface area A, volume V, specific heat cp.
Let Ti be the initial temperature of the body. The body is losing heat to the surroundings, which are at a temperature of T∞ (Ti > T∞ ). Performing an energy balance on the body, we get the following:
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(3.16) |
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Rate of change of internal energy = Ein − Eout |
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mc dt = −Qconvection |
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mcp dT = −hA(T − T∞ ) |
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3.2 Lumped capacitance method |
69 |
FIGURE 3.2
Schematic representation of an irregular body used for lumped capacitance method of solving a transient heat conduction problem.
Let the temperature excess be, θ = T − T∞
mcp ddtθ = −hAθ
The initial condition at t = 0; is T = Ti or θ = θi
dθ = −hAθ dt mcp
dθ = −hA dt θ mcp
Integrating both sides from θ = θi to θ = θ, we have
∫ θ dθ |
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−hA ∫ t |
dt |
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mcp |
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−t |
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(3.19)
(3.20)
(3.21)
(3.22)
(3.23)
(3.24)
(3.25)
(3.26)
70CHAPTER 3 Conduction
where τ is the time constant given by mchAp . Working further, we have
hA |
t = |
hAt |
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ht |
(3.27) |
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The characteristic length, Lc, is given by V/A. The characteristic length, Lc for a transient conduction problem is, L for a plane slab of thickness 2L, R/2 (or D/4) for
a cylinder of radius R (or diameter D) and R/3 (or D/6) for a sphere of radius R (or ht
ρcpV / A
less numbers, Bi and Fo, of interest in this problem, as follows,
ht |
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× |
Lc k |
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hLc |
× αt = BiFo |
ρcpV / A |
ρcp Lc |
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θ = e−BiFo
θi
(3.28)
(3.29)
θ = f (Bi, Fo, dimen-
θi
sionless length scale), with the dependence on the third quantity on the right-hand side vanishing due to the spatial isothermality of the body under consideration.
The temperature time history of the body is an exponential curve and is qualitatively shown in Fig. 3.3.
It would be instructive to calculate the enthalpy lost by the body in the time interval 0 to t.
Enthalpy lost = ∫ t |
Q dt |
(3.30) |
0 |
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hAθ dt |
(3.31) |
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FIGURE 3.3
Schematic representation of the temperature time history of the spatially isothermal body losing heat to cooler surroundings.
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3.2 Lumped capacitance method |
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1− e |
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We can work on eqn. 3.34 to come out with a form which is physically more intuitive to us.
Enthalpy lost = mcpθi ( |
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= mcp (θi − θ ) = mcp (Ti − T )
Eqns. 3.34 and 3.35 are essentially the same with the latter serving as confirmation. The enthalpy lost in the time interval 0 to t is given by the difference between mcpθi and the area under the curve in Fig. 3.3. A quick asymptotic check is in order.
At t = 0, Enthalpy lost is “0” (zero) At t → ∞, Enthalpy lost → mcpθi
These are asymptotically correct and are consistent with the physics of the problem under consideration. The normally accepted criterion for considering a body to be spatially isothermal is that the Biot number (Bi), based on the characteristic length Lc given by the ratio of the volume (V) to the surface area (A), must be less than 0.1.
Example 3.1: A copper sphere 12 mm in diameter is initially 100 °C. It is then immersed in cold surroundings with an ambient temperature of 30 °C, and a natural heat transfer coefficient of h = 7 W/m2 K. The thermal conductivity of copper is 369 W/mk.
1.For this transient conduction problem, determine the Biot number.
2.Can lumped capacitance method be used to solve this problem?
3.If the answer to (2) is yes, determine the temperature at the center of the sphere after 100 s.
4.What is the total enthalpy lost by the copper sphere in the first 50 s? (The thermophysical properties of copper are thermal conductivity = 396 W/mK density = 8960 kg/m3, and specific heat = 385 J/kg k).
Solution:
Given data
The diameter of sphere (D) = 12 mm; initial temperature of the sphere (Ti) = 100 °C; ambient temperature (T∞) = 30 °C; heat transfer coefficient (h) = 7 W/m2K; thermal conductivity (k) = 396 W/mK.
Bi = |
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L = |
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LC = D6
72CHAPTER 3 Conduction
This is a key result. The characteristic length in transient conduction is L for a plain slab of thickness 2L, R/2 or D/4 for a cylinder of radius R and R/3 or D/6 for a sphere of a radius of R, as already mentioned.
1.Bi = 7 ×12 ×10−3 = 35.35×10−6 6 × 396
2.Since Bi < 0.1, the lumped capacitance method is applicable.
3.From the solution for lumped capacitance heat transfer, the temperature excess is given by
T − T |
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ρVc |
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The temperature of the sphere at the center after 100 s (same everywhere in the sphere under the lumped capacitance assumption)
T100s − 30 |
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4. Enthalpy lost = mcp (Ti − T50 s ) = ρVcp (Ti − T50 s )
To be able to calculate this, first we need to obtain the temperature of the sphere at 50 s
T50s − 30 |
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Enthalpy lost = 8960 × 9.04 ×10−7 × 385 × (100 − 96.54) = 10.79 J.
3.3 Semi-infinite approximation
Let us revisit our earlier discussion on transient conduction in an infinitely long one-di- mensional slab of thickness 2L. However, now we are looking at a situation where the Bi > 0.1 and, in view of this, the assumption of spatial isothermality no longer holds.
Consider a plane wall that is semi-infinite in character, that is, the left end is specified (x = 0) while the right end is infinite in extent at an initial temperature of Ti.
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3.3 Semi-infinite approximation |
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FIGURE 3.4
Schematic representation of a semi-infinite plane wall undergoing transient conduction.
Now, at x = 0 (left end) suddenly, the temperature is lowered to T∞ (T∞ < Ti). A schematic is shown in Fig. 3.4. It is intuitive to imagine that the effect of the thermal disturbance at the wall will be initially felt only near the wall, and this region of disturbance, denoted by δ, slowly grows with time along x. Now consider a time interval for which
δL 1
This assumption involves a situation where we do not know the value of δ a priori. However, there will be no flaw in the analysis so long as δ 1. “L” is the
characteristic length scale for the problem. For times when δ L,Llarge portions of the slab would simply not know that a thermal disturbance happened at x = 0. They would still be at T = Ti. We call this the “early regime”. This is the reason why we stated earlier that the body under consideration, a slab in this case, is semi-infinite in extent. In what follows, we will see that the ‘L’ itself does not appear in the final solution, as it should be for a truly semi-infinite approximation.
The δ referred to above is the thermal penetration depth or thermal penetration thickness.
The governing equation for the problem under consideration is
∂2 T |
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The initial condition is for all x: At t = 0 s, T = Ti.
74 CHAPTER 3 Conduction
Boundary conditions: At x = 0, T = T∞ for t > 0. It is intuitively apparent that when x → ∞; T = Ti.
We now perform a scale analysis on this problem, wherein we try to reason out scales or orders of magnitude of the key quantities for the problem under consideration.
Let θ = T–T∞. Eq. (3.36) turns out to be
∂2θ |
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The scale for θ is θi, while the scale for x is δ. Substituting for these in Eq. (3.37), we have
θi |
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As already mentioned,
δL 1
substituting δ from Eq. (3.40), we have
Lαt 1
Hence, αt L for the semi-infinite model to be valid. The key question to be asked now is the motivation to do all this. The answer to this question lies in the fact
that not only for t → 0, T = Ti everywhere, but also for x →∞; that is, when δ 1,
δ L
L 1.” In view of this, one initial and one boundary condition will fuse into just one condition if a
new variable, η, is introduced that takes care of both x and t. If the introduction of η reduces the original partial differential equation (PDE) into an ordinary differential equation (ODE), with the understanding that the latter is a lot easier to solve than the former, then its introduction is well worth the effort. In this effort, the fusing of one initial and one boundary condition will ensure that there is no degeneracy, when the reduction is done. Let us now see if this reduction is possible!
Let η = |
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