Newton's And Fourier's Cooling Laws Applied To Heat Flow Boundary Conditions
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Newton's And Fourier's Cooling Laws Applied To Heat Flow Boundary Conditions Many important physical phenomena can be modeled as problems of systems of partial differential equations (PDEs) or ordinary differential equations (ODEs). Usually, the mathematical expressions of the initial conditions (IC) and boundary conditions associated with a particular problem are stated with the PDEs or ODEs. The PDE, BC and IC, together constitute an Initial-Boundary-Value-Problem (IBVP).

Consider the laterally insulated one-dimensional copper rod with length L (figure 14.12(a)), the ends of which are enclosed in containers of liquids at temperatures described by the functions g1(t) and g2(t) respectively.

Use Newton's and Fourier's cooling laws to express the boundary conditions at the ends of the rod in mathematical terms.

The strings:

S7P2A21 (Physical Property)

The math:

Pj Problem of Interest is of type identity. Problems of mathematical modelind are identity problems. The primary interes is to identitify the mathematical structure of the physical problem being modeled. It is in this sense that the Pj Problem of Interest is of type identity. Newton's Law Of Cooling:

Outward flux of heat (at x= 0) = h[u(0,t) - g1(t)]-------(1)
Outward flux of heat (at x= L) = h[u(L,t) - g2(t)]-------(2)

Where, h is a heat-exchange coefficient. It is a measure of the amount of calories that flow across the boundary per unit of temperature difference per second per centimeter.
Outward flux of heat is the amount of calories flowing through the ends of the rod per second. It is positive at either end if the temperature of the rod is greater than the surrounding medium.

Fourier's Law Of Cooling:
Outward flux of heat across a boundary is proportional to the inward normal derivative across the boundary (figure 14.12(b)):

Outward flux of heat (at x= 0) = k∂u(0,t)/∂x----(3)
Outward flux of heat (at x= L) = -k∂u(L,t)/∂x----(4)

where k = thermal conductivity of rod.

Equating equations (1) and (3):
k∂u(0,t)/∂x = h[u(0,t) - g1(t)]
So, ∂u(0,t)/∂x = (h/k)[u(0,t) - g1(t)] -----(5)

Equating equations (2) and (4):

k∂u(L,t)/∂x = h[u(L,t) - g2(t)]
So, ∂u(L,t)/∂x = (h/k)[u(L,t) - g2(t)] -----(6)

Equations (5) and (6) express the boundary conditions in mathematical terms.

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