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Tresset [83]
3 years ago
7

Briefly describe the purpose of specifying boundary conditions.

Engineering
1 answer:
lisov135 [29]3 years ago
3 0

Answer:

Given in the explanation

Explanation:

A boundary condition expresses the behavior of a function on the boundary (border) of its area of definition. An initial condition is like a boundary condition, but then for the time-direction. Not all boundary conditions allow for solutions, but usually the physics suggests what makes sense.  

Boundary value is a condition accompanying a differential equation in the solution of physical problems. In mathematical problems arising from physical situations, there are two considerations involved when finding a solution: (1) the solution and its derivatives must satisfy a differential equation, which describes how the quantity behaves within the region; and (2) the solution and its derivatives must satisfy other auxiliary conditions either describing the influence from outside the region (boundary values) or giving information about the solution at a specified time (initial values), representing a compressed history of the system as it affects its future behaviour.  

The problem is that without additional conditions the arbitrariness in the solutions makes it almost useless (if possible) to write down the general solution. We need additional conditions, that reduce this freedom. In most physical problems these are boundary conditions, that describes how the system behaves on its boundaries (for all times) and initial conditions, that specify the state of the system for an initial time t=0.

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3 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
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Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

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Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

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t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

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Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

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A = \frac{5912500  }{90 \times 341.67} = 192.3 \, m^2

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

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