Steady-State Two-Dimensional Heat Conduction
Previous chapters were devoted to steady-state one-dimensional systems. In this chapter, analytical solution, graphical analysis, method of analogy, and numerical solutions have been presented for two-dimensional steady-state conduction heat flow through
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Steady-State Two-Dimensional Heat Conduction
5.1
Introduction
In the preceding chapters, the cases of one-dimensional steady-state conduction heat flow were analyzed. We consider now, two-dimensional steady-state conduction heat flow through solids without heat sources. The Laplace equation that governs the temperature distribution for two-dimensional heat conduction system is @2t @2t þ 2¼0 2 @x @y
ð5:1Þ
Equation (5.1) is based on the assumption of constant and equal thermal conductivities in both x and y space coordinates. Solution of this equation gives the temperature in the body as a function of coordinates x and y. Knowing the temperature distribution in the body, the heat flow in the x and y directions at a point can be determined from qx ¼ kAx
@t @x
ð5:2aÞ
qy ¼ kAy
@t @y
ð5:2bÞ
The rate of heat flow at the point is the resultant of the components qx and qy, i.e., q ¼ iqx þ jqy
© Springer Science+Business Media Singapore 2017 R. Karwa, Heat and Mass Transfer, DOI 10.1007/978-981-10-1557-1_5
ð5:3Þ
247
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5 Steady-State Two-Dimensional Heat Conduction
It follows that in order to determine the heat flow, the temperature field must be known. Thus, the problem reduces to the solution of Eq. (5.1). The common techniques available for the solution of Eq. (5.1) are (i) (ii) (iii) (iv)
5.2
Mathematical analysis (analytical solution) Graphical analysis Method of analogy Numerical solutions using either a finite-difference or finite-element method.
Analytical Solution of Two-Dimensional Heat Conduction Problems
Consider a rectangular section bar, as shown in Fig. 5.1, which is very long in z-direction. Three lateral sides of the bar are maintained at a constant temperature To. For the fourth side we consider different conditions, case (i) to (iii), as outlined below. Case (i) The fourth side (y = H) has a sinusoidal temperature distribution, T = Tm sin(πx/W), imposed on it. Using θ = T – To, the Laplace equation, Eq. (5.1), is transformed to @2h @2h þ 2 ¼0 @x2 @y
ð5:4Þ
The boundary conditions are ðiÞ h ¼ 0 ðiiiÞ h ¼ 0
at y ¼ 0; ðiiÞ h ¼ 0 at x ¼ 0; at x ¼ W; ðivÞ h ¼ hm sinðpx=WÞ at y ¼ H
ð5:5Þ
where θm = Tm – To is amplitude of the sine function. T = Tm sin(πx/W) is transformed to θ = θm sin(πx/W). Fig. 5.1 A rectangular section bar with given thermal boundary conditions
y T H To
To
x To
W
5.2 Analytical Solution of Two-Dimensional Heat Conduction Problems
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The solution of Eq. (5.4), using the separation of variable technique (based on the assumption that the solution to the differential equation takes a product form), is hðx; yÞ ¼ XY
ð5:6Þ
2 2 1 d X 1 dY ¼ 2 X dx Y dy2
ð5:7Þ
where X = X (x) and Y = Y (y). Substitution in Eq. (5.4) gives
Each side of Eq. (5.7) is independent of the other, since x and y are independent variables. Hence, the left side of Eq. (5.7) can equal the right side only if both the sides have a constant value, greater than zero, say λ2. That is, 2 2 1 d X 1 dY ¼ ¼ k2 2 X dx Y dy2
ð5:8Þ
Thus, we get two ordinary differential equations
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