FIGURE 3.9
Center-line temperature charts for a long cylinder of radius r0 initially at a uniform temperature of Ti, subject to convection on both sides with a heat transfer coefficient of h and an ambient temperature of T∞.
85 variables of separation of method The 4.3
86 CHAPTER 3 Conduction
FIGURE 3.10
Temperature distribution chart for a long cylinder of radius r0 initially at a uniform temperature of Ti, subjected to convection on both sides with a heat transfer coefficient of h and an ambient temperature of T∞.
FIGURE 3.11
Enthalpy chart for a long cylinder of radius r0 initially at a uniform temperature of Ti, subjected to convection on both sides with a heat transfer coefficient of h and an ambient temperature of T∞.
FIGURE 3.12
Midpoint temperature charts for a sphere of radius r0 initially at a uniform temperature of Ti subject to convection on both sides with a heat transfer coefficient of h and an ambient temperature of T∞.
variables of separation of method The 4.3
87
88 CHAPTER 3 Conduction
FIGURE 3.13
Temperature distribution chart for a sphere of radius r0 initially at a uniform temperature of Ti subjected to convection on both sides with a heat transfer coefficient of h and an ambient temperature of T∞.
FIGURE 3.14
Enthalpy chart for a sphere of radius r0 initially at a uniform temperature of Ti subjected to convection on both sides with a heat transfer coefficient of h and an ambient temperature of T∞.
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3.4 The method of separation of variables |
89 |
well as, Figs. 3.12–3.13, are Heisler’s charts for the plane wall (or plate), cylinder and sphere respectively. For each of the three charts, the first two concern the temperature at the mid-plane followed by the temperature at any other plane. A third chart is usually added to these charts, for all the geometries, and this is known as Grober’s chart. These give the ratio of enthalpy to the initial enthalpy excess of the body in question. In what follows, we present fully worked examples for plane wall. Computer codes based on MATLAB have been developed as a part of this book project to generate Heisler’s charts for the plane wall, cylinder and sphere. These are available through the online support for this book. End of chapter problems 3.7 to 3.9 deal with transient conduction in a cylinder and sphere.
Example 3.2: Consider a thick slab made of a material whose thermal conductivity “k” is unknown. The slab is initially at 100 °C, and one end of the slab is suddenly brought to 30 °C by the use of a temperature bath. A thermocouple, placed 8 mm from the end, exposed to the cold fluid shows 63 °C at 3 min, after the surface is exposed to the cold fluid. Determine the thermal conductivity of the material.The density and specific heat of the material are respectively 1920 kg/m3 and 835 J/kg.K, respectively.
Solution:
We use the semi-infinite approximation to solve this problem.
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63 − 30 |
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100 − 30 |
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0.4714 = erf (η)
From the error function table, η = 0.44 for erf(η) = 0.4714
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αt |
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αt = 8.26 |
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1920 × 835 |
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k = 0.735 W/mK
(3.105)
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(3.108)
(3.109)
(3.110)
(3.111)
(3.112)
90CHAPTER 3 Conduction
Example 3.3: A 1.6 cm thick slab of carbon steel is initially at Ti = 610 °C. This slab is suddenly immersed into a bath of water at T∞ = 25 °C. The heat transfer coefficient h is 104 W/m2 K. The properties of a carbon steel are k = 40 W/m K, and α = 1 × 105 cm2/s.
1.Determine the time t at which the temperature in the mid-plane of the slab
drops to TC = 100 °C. Also determine the average temperature of the slab corresponding to this time.
2.Calculate the corresponding temperature in a plane situated 0.2 cm from one of the ends.
Solution:
The problem is solved in both the methods, i.e., through the analytical method and by using the Heisler’s charts.
Bi = hLk c
Lc = t / 2 for a plate = |
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Bi = 104 × 0.008 = 2 0.1
40
Since Bi 0.1 the lumped capacitance method is not applicable.
Solution through the analytical method:
From the analytical solution, the first term approximation for Eq. (3.92) is given
as
φ = |
4 sin (ξn ) eξn Fo cos(λn x) |
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2ξn + sin (2ξn ) |
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a. For mid-plane x = 0, so the first term approximation for Eq. (3.92) becomes
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From Eq. (3.88)
ξtan ξ = Bi
ξtanξ = 2 tan ξ = ξ2
ξ= tan−1 2
ξ
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3.4 The method of separation of variables |
91 |
Solving for ξ using the successive substitution method‡(i.e., iteratively),
ξ= 1.076
φt , x=0 = T(t ,x=0) − T∞ = 100 − 25 = 0.1282
φi Ti − T∞ 610 − 25
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0.1282 = |
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0.1282 =1.178 ×e−ξ2Fo
e−ξ2Fo = 01..1282178 = 0.1086
−ξ2 Fo = −2.22
Fo = 1.917
αt = 1.917
L2c
1.917 ×(0.008)2
t = 0.1×10−4
t = 12.27 s
From eq. 3.103 we know that
Q
Qmax
Q
Qmax
=1 − L1 A1 e−ξ
=1 − A1e−ξ2Fo
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sin(ξ)
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‡In this method we start with a guess of ξ. Use this value to evaluate ‘ξ’ again using the relation ξ = tan–1(2/ξ). We get a new value of ξ. Use this again to get the next iteration of ξ and continue this till convergence.
92 CHAPTER 3 Conduction
Tavg − T∞
Ti − T∞
Tavg − T∞
Ti − T∞
Tavg
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sin(ξ) |
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= T∞ + 0.1041 × (Ti − T∞ )
=25 + 0.1041 × (610 − 25) = 85.59 °C
Tavg = 85.89 °C
b.Temperature in a plane situated 0.2 cm from the cooled surface of the plate (i.e., temperature in a plane at x = 0.006 m).
φt ,x |
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Fo = 1.917, ξ = 1.076, x = 0.006, t = 12.27 s Lx = 00..006008 = 0.75
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=1.178 0.691 0.108 = 0.0879
φx ,t = T (x, t) − T∞ = 0.0879
φi Ti − T∞
T (x,t) = 76.42 °C
Solution using Heisler charts:
φ= T − T∞ Ti − T∞
φ= 100 − 25 = 0.1282 610 − 25
Bi1 = hLk c = 0.5
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3.4 The method of separation of variables |
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From the Heisler charts the centerline temperature chart for the case of a plane |
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wall (or plate) is given by |
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α2t = 2 |
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t = |
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0.1×10−4 |
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t = 12.8s. |
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φavg = |
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= 98.125 °C |
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The temperature at 0.2 cm from one of the cooled surfaces of the plate is determined as
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0.002 |
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φ = |
T(0.006,t ) −T∞ |
= 0.65 |
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∞ |
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= 73.75 °C
We see that the analytical and graphical solutions are reasonably close. Confirmation of the average temperature of the slab obtained from the analytical solution with that from the Grober’s chart is left as an exercise to the student.
94 CHAPTER 3 Conduction
3.5 Analysis of two-dimensional, steady state systems
Consider a two-dimensional slab of height “W” and length “L.” The depth in the direction perpendicular to the plane of the paper is so large that T varies only with x and y.
Assumptions:
1.Steady state prevails.
2.No internal heat generation (qv = 0).
3.The thermophysical properties such as k, ρ, cp, etc., are constant (Fig. 3.15).
The governing equation for two-dimensional steady state conduction is given by
∂2 T |
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∂2 T |
= 0 |
(3.113) |
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Let us introduce a dimensionless temperature, f, as follows
φ = |
T − TC |
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Therefore, eqn. 3.113 becomes |
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∂2 φ2 + ∂2 φ2 = 0 |
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FIGURE 3.15
Schematic representation of a two-dimensional plane wall with the boundary condition
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T −T |
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C |
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remaining three walls.