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7 E 7 IOXLG

Using equation 2.33 for the temperature distribution and substituting x = L, noting that cosh (0) = 1, implies that 1/ cosh (mL) = 0.05. So, mL = 3.7, which requires that L > 247 mm.

Simple Time Dependent Conduction

The 1-D time-dependent conduction equation is given by Equation 2.10 with no variation in the y or z directions:

2 = (2.43) ∂ x a ∂ t

A full analytical solution to the 1-D conduction equation is relatively complex and requires finding the roots of a transcendental equation and summing an infinite series (the series converges rapidly so usually it is adequate to consider half a dozen terms). There are two alternative possible ways in which a transient conduction analysis may be simplified, depending on the value of the Biot number (Bi = h L /k).

2.3.3 Small Biot Number (Bi << 1): Lumped Mass Approximation

A small value of Bi implies either that the convective resistance 1/h, is large, or the conductive resistance L/k is small. Either of these conditions tends to reduce the temperature gradient in the material. For example there will be very little difference between the two surface temperatures of a heated copper plate (k ≈ 400 W/m K) of say 5 or 10 mm thickness. Whereas for Perspex (k ≈ 0.2 W/m K), there could be

a significant difference. The copper thus behaves as a “lumped mass”. Hence for the purpose of analysis we may treat it as a body with a homogenous temperature. A simple heat balance on a material of mass, m, density ρ, specific heat C, exchanging heat by convection from an area A to surrounds at Ta, gives

4 P & 7 V 7 f K $ 7 V 7 f (2.44) GW

define: Θ = ( T s − T ∞ ) / ( T s , initial − T ∞ , initial ) and : l = A / m C

in forced convection when h ≠ f ( T s − T ∞ ) . This gives the simple solution:

4 O K W RU

In free convection when the heat transfer coefficient depends on the surface to fluid temperature difference, say n h ∝ ( T s − T ∞ ) , then the solution becomes:

− Θ n = 1 + ( n h initial l ) t (2.46)

2.3.4 Large Biot Number (Bi >>1): Semi – Infinite Approximation When Bi is large (Bi >> 1) there are, as explained above, large temperature variations within the material.

For short time periods from the beginning of the transient (or to be more precise for Fo << 1), the boundary away from the surface is unaffected by what is happening at the surface and mathematically may be treated as if it is at infinity.

There are three so called semi-infinite solutions:

N.B. erfc(x) = 1 – erf(x); erf(x) = error function

Given by the series:

erf ( x ) =

Constant surface heat flux

¯ D W ¿ ¼ Constant surface temperature

« ¬ °¯ N © S ¹ °¿ © D W ¹ » ¼ ¬ © N ¹

 x  = erf 

1 / 2  (2.49) ( T initial − T s )

Constant surface heat transfer coefficient

7 [ W 7 V ­ [ ½ ª § K[ K D W ·

HUIF ®

¾ « H[S ¨¨

¸¸ HUIF ®

D W ¾ » (2.50)

N ¿ ¼ As well as being useful in determining the temperature of a body at time, these low Biot number and

7 LQLWLDO 7 V ¯ D W ¿ ¬ © N

large Biot number methods can also be used in the inverse mode. This is the reverse of the above and makes use of the temperature time history to determine the heat transfer coefficient.

Example 2.4

A titanium alloy blade from an axial compressor for which k = 25 W / m K, ρ = 4500 kg / m3 and

C = 520 J/kg K, is initially at 40ºC. Although the blade thickness (from pressure to suction side) varies along the blade, the effective length scale for conduction may be taken as 3mm. When exposed to a hot gas stream at 350ºC, the blade experiences a heat transfer coefficient of 150 W / m2 K. Use the lumped mass approximation to estimate the blade temperature after 50 s.

Firstly, check that Bi << 1 Bi = h L / k = 150 x 0.003 / 25 = 0.018. So the lumped mass method can be used.

Solution

4 O T − T fluid

= exp  −

T initial − T fluid

 

However, the mass m, and surface area A, are not known. It is easy to rephrase the above relationship, since mass = density x volume and volume = area x thickness, where this thickness is the conduction length scale, L. So

T − T fluid

exp( −

T initial − T fluid

From which 

T =  ( T initial − T fluid )  −

 exp( r C L   

t )  + T fluid

T  = 243 . 5 C

2.4 Summary

This chapter has introduced the mechanism of heat transfer known as conduction. In the context of engineering applications, this is more likely to be representative of the behaviour in solids than fluids. Conduction phenomena may be treated as either time-dependent or steady state.

It is relatively easy to derive and apply simple analytical solutions for one-dimensional steady-state conduction in both Cartesian (plates and walls) and cylindrical (pipes and pressure vessels) coordinates. Two-dimensional steady-state solutions are much more complex to derive and apply, so they are considered beyond the scope of this introductory level text.

Fins and extended surfaces are an important engineering application of a one-dimensional conduction analysis. The design engineer will be concerned with calculating the heart flow through the fin, the fin efficiency and effectiveness. A number of relatively simple relations were presented for fins where the surface and cross sectional areas are constant.

Time-dependent conduction has been simplified to the two extreme cases of Bi << 1 and Bi >> 1. For the former, the lumped mass method may be used and in the latter the semi-infinite method. It is worth noting that in both cases these methods are used in practical applications in the inverse mode to measure heat transfer coefficients from a know temperature-time history.

In many cases, the boundary conditions to a conduction analysis are provided in terms of the convective heat transfer coefficient. In this chapter a value has usually been ascribed to this, without explaining how and from where it was obtained. This will be the topic of the next chapter.

2.5 Multiple Choice Assessment

2.5.1 Simple 1-D Conduction

1. Which of these statements is a correct expression of Fourier’s Law G7 w 7 w 7 w 7

G[

2. Which is the correct form of the 2D steady state conduction equation for constant thermal

3. Conductivity, in Cartesian coordinates ? • Which of the following is NOT a boundary condition ?

• Tx=L = 50ºC • qx=L = qinput • –k(dT/dx)x=L = h(Ts – Tf) • Ty=L/2 = T0( 1 – x/L) • k = 16 W/m K

4. The statement Tx=0 = T0, means that: • the temperature at x = 0 is zero • the temperature at x = 0 is constant • the temperature at x = L is zero • the temperature at x = L is constant • the surface at x = 0 is adiabatic

5. The statement –k(dT/dx)x=L = h(Ts – Tf) means that: • the temperature at x = L is constant • the heat flux at x = L is constant • heat transfer by convection is zero at x = L • heat transfer by conduction is zero at x = L • heat transfer by convection equals that by conduction at x = L

6. If Bi << 1, then: • temperature variations in a solid are significant • temperature variations in a solid are insignificant • surface temperature is virtually equal to the fluid temperature • surface temperature is much less than the fluid temperature • surface temperature is much greater than the fluid temperature

7. A large value of heat transfer coefficient is equivalent to: • a large thermal resistance • a small thermal resistance • infinite thermal resistance • zero thermal resistance • it depends on the fluid temperature