One-dimensional heat transfer

To show how numerical models work we start with the one-dimensional heat transfer problem. 

From energy balance (Figure 146), heat gain and heat loss in a volume element with one dimensional heat transfer gives... 

Assuming one-dimensional heat transfer is the mode of the solid bed heating due to the heating of the film by conduction and dissipation, the temperature will only change in the y direction. The same assumption that was made by Tadmor and Klein will be made here that the heat transfer model is a semi-infinite slab moving at a velocity Vsy c (melting velocity) with the boundary conditions T(0) = and j(-oo) = 7 , This assumption is not strictly correct because it will also be proposed that the other four surfaces are melting. The major error will occur at the corners of the solid bed. is the velocity of the solid bed surface adjacent to Film C as it moves toward the center of the solid bed in the y direction. 

In most cases, for composite structures the planar dimensions are sufficiently large compared with the thickness that one-dimensional heat transfer is assumed. In this case,... 

Billig (98) realized this point and used a one-dimensional heat-transfer analysis for the crystal with assumptions about the temperature field in the melt to derive the the following relationship that has been used heavily in qualitative discussions of crystal growth dynamics ... 

Thus, for a small Fourier number, the one-dimensional heat transfer yields the following equation... 

Flfl. 2-1 One-dimensional heat transfer through a composite wall and electrical analog. 

Thegheterogeneous one-dimensional heat transfer parameters and ai can be obtained from (2) ... 

Unlike the radiant loss from an optically thin flame, conductive or convective losses never can be consistent exactly with the plane-flame assumption that has been employed in our development. Loss analyses must consider non-one-dimensional heat transfer and should also take flame shapes into account if high accuracy is to be achieved. This is difficult to accomplish by methods other than numerical integration of partial differential equations. Therefore, extinction formulas that in principle can be used with an accuracy as great as that of equation (21) for radiant loss are unavailable for convective or conductive loss. The most convenient approach in accounting for convective or conductive losses appears to be to employ equation (24) with L(7 ) estimated from an approximate analysis. The accuracy of the extinction prediction then depends mainly on the accuracy of the heat-loss estimate. Rough heat-loss estimates are readily obtained from overall balances. 

Note that in the special case of one-dimensional heat transfer in the x-direction, the derivatives with respect toy and z drop out and the equations above reduce to the ones developed in the previous section for a plane wall (Fig. 2-22). 

The temperature of an exposed surface can usually be measured directly and easily. Therefore, one of the easiest ways to specify the thennal conditions on a surface is to specify the temperature. For one-dimensional heat transfer (hiough a plane wall of thickness L, for example, the specified temperature boundary conditions can be expressed as (Fig. 2-28)... 

For one-dimensional heat transfer in the. v direction in a plate of thickness the convection boundary conditions on both surfaces can be expressed as... 

Consider a plane wall of thickness L whose thermal conductivity varies linearly in a specified temperature range as k T) = kad + PT) where kg and p are constants. The wall surface at x = 0 is maintained at a constant temperature of 7i while the surface at r = (.is maintained at Tj, as shown in Fig. 2-64. Assuming steady one-dimensional heat transfer, obtain a relation for (a) the heat transfer rate through the wall and [b) the temperature distribution 7(x) in the wall. 

Consider a 2-m-hlgh and 0.7-m-wide bronze plate whose thickness is 0.1 m, One side of the plate is maintained at a constant temperature of 600 K vrhile the other side is maintained at 400 K, as shown in Fig. 2-65. The thermal conductivity of the bronze plate can be assumed to vary linearly in that temperature range as k T) = fed + pT) where fe = 38 W/m K and p = 9,21 x lQ-4 ) -i Disregarding the edge effects and assuming steady one-dimensional heat transfer, determine the rate of heal conduction through the plate. 

C How does transient heat transfer differ from steady heat transfer How does one-dimensional heat transfer differ from two-dimensionat heat transfer ... 

A 2-kW resistance heater wire whose thermal conductivity is k = 18 W/m - K has a radius of = 0.15 cm and a length of Z. = 40 cm, and is used for space heating. Assuming Constant thermal conductivity and one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem during steady operation. Do not solve. 

Assuming lerHperalure-dependem ihermal conductivity and one-dimensional heat transfer, express the mathematical formulation (the di feicniial equation and the boundary conditions) of this heat conduction problem during steady operation. Do not solve. 

W/m . Assuming steady one-dimensional heat transfer, (a) express the differential equation and the boundary conditions for heal conduction through the wall, (b) obtain a relation for the variation of temperature in the wall by solving the differential... 

Assuming one-dimensional heat transfer, determine the tem-peratuies at the center of the resistance wire and the wire-plastic layer interface tinder steady conditions. 

Now consider steady one-dimensional heat transfer through a plane wall of thickness L, area A, and thermal conductivity k that is exposed to convection on both sides to fluids at temperatures ro,i and T 2 with heat transfer coefficients /i and hj, respectively, as shown in Fig. 3-6. Assuming T i < < i> variatiqji of temperature will be as shown in the figure. Note that the temperature va es linearly in the wall, and a.symptotically approaches r , and J 2 die fluids we move away from the wall. 

The ihennal resistance concept or the electrical analogy can also be used to solve steady heat transfer problems that involve parallel layers or combined series-parallel arrangements. Although such problems are often two- or even three-dimensional, approximate solutions can be obtained by assuming one-dimensional heat transfer and using the Ihennal resistance network. 

One-dimensional heat transfer through a simple or composite body exposed to convection from both sides to mediums at temperatures Tii and T, can be e.xpressed as... 

A 10-cm long bar with a square cross-section, as shown in Fig. P3-I81, consists of a 1-cm thick copper layer (k = 400 W/m K) and a 1-cm thick epoxy composite layer (k - 0.4 W/m - K). Calculate the rate of heat transfer under a thermal driving force of fiO C, when the direction of steady one dimensional heat transfer is (n) from front to back (i.e. along its length), (h) from left to right, and (c) from lop to bottom. 

SOLUTION Wo have solved this problem in Example 5-1 for the steady case, and here we repeat it for the transient case to demonstrate the application of the transient finite difference methods. Again we assume one-dimensional heat transfer in rectangular coordinates and constant thermal conductivity. The number of nodes is specified to be W = 3, and they arc chosen to be at the two surfaces of the plate and at the middle, as shown in the figure. Then the nodal spacing Ax becomes... 

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