The steady-state temperature pro le in .NET Encode 3 of 9 in .NET The steady-state temperature pro le

The steady-state temperature pro le generate, create barcode code39 none for .net projects Microsoft Windows Official Website Separating variables, we barcode code39 for .NET obtain:. d = 2 2 zdz which may be integrated t .NET 3 of 9 barcode o yield:. ln = 2 z 2 + c = ec e 2 z2. (6.18). The next task is to evaluate the constant of integration, ec . The basal boundary condition The constant of integrati on may be evaluated by using the boundary condition = o on z = 0. In other words, we presume that the temperature gradient at the bed, o , is known or can be estimated. Making these substitutions in Equation (6.

18) yields ec = o . Thus, replacing ec with o and with d /dz in Equation (6.18) yields:.

d 2 2 = o e z dz (6.19). This is a solution for th e temperature gradient as a function of elevation above the bed. The requirement that the temperature gradient in the basal ice be known is fundamentally unavoidable. However, this is not as serious a problem as one might, at rst, expect.

In the steady state, o is adjusted so that all of the heat coming from within the Earth, the geothermal ux, can be conducted upward into the ice. Thus, if the geothermal ux can be estimated, o can be calculated because the constant of proportionality between the two, the thermal conductivity of ice, K, is known. To clarify the physical processes by which o is adjusted, consider a non-steady-state situation in which o is too low.

Some of the geothermal heat would then remain at the ice rock interface where it would warm the ice. Because the temperature decreases upward in the glacier, the ice being colder than the Earth s interior, such warming would increase o until all of the heat could be conducted upward into the ice, thus tending to re-establish the steady state. (For the moment, we neglect basal melting.

) Geothermal heat is produced by radioactive decay in the crustal rocks as well as by residual cooling of the mantle and core. Numerous measurements of the geothermal ux have been made, so we have a fair idea of its magnitude in different geological terranes. Geophysicists use the heat ow unit, or HFU, to describe this ux: 1 HFU is.

Temperature distribution in polar ice sheets Table 6.1. Geothermal uxes in some geological terranes in which glaciers are or were found Heat ux Locality Canadian Shield World average East Antarctica Baf n Bay West Antarctica HFU 0.8 1.2 1.21 1.35 1.41 mW m 2 33 50 50 56 59 Basal gradient K m 1 0.0151 0.0226 0.0226 0.0255 0.0264 Reference Budd et al., 1971 Budd et al., 1971.

Estimated. 1 cal cm 2 s 1 . In glaci ology, however, it is more common to use W m 2 . The world-wide average geothermal ux is 1.

2 HFU or 50 mW m 2 . This corresponds to a temperature gradient in basal ice of 0.0226 K m 1 .

The gradient in the underlying rock will normally be somewhat different as the thermal conductivity of the rock will not be the same as that of the ice. In general, geothermal uxes are highest in volcanic terranes, high in geologically young terranes, and lowest in geologically ancient terranes. A few examples of geothermal uxes in glaciated areas are given in Table 6.

1. In the discussion above, we asserted that knowledge of o was fundamentally unavoidable . It is true, of course, that a boundary value problem such as this could be solved with some other basal boundary condition, such as the basal temperature.

(This will be left as an exercise for the reader.) However, as the basal temperature is one of the quantities that we are particularly eager to determine, and as basal temperatures are much harder to estimate from existing data than are basal temperature gradients, choosing o as the basal boundary condition is the only logical choice in most situations..

The second integration To obtain the actual temp erature distribution, it is necessary to integrate Equation (6.19). Separating variables as before yields:.

s H d = o (h) h 2 z2. (6.20). Here, the integration is from some level, z = h, in the glacier, where the temperature is (h), to the surface at z = H where the temperature. The steady-state temperature pro le is s . (Note that in th is case, rather than solve Equation (6.19) as an inde nite integral and then evaluate a constant of integration by applying a boundary condition, it is more convenient to express the integrals as de nite integrals.

Thus the boundary condition, = s on z = H, is incorporated into the limits of integration. Further discussion of this boundary condition is deferred for the moment.) The integral on the right-hand side of Equation (6.

20) does not have a solution in closed form. However, it occurs frequently, and thus has been tabulated. In addition, many computer statistical packages have solutions.

The challenge is to express it in the terms used in these tables. We rst express the integral on the right-hand side as the difference between integrals over the range 0 H and 0 h, thus:.
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