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Technical Brief

Shape Optimization With Isoperimetric Constraints for Isothermal Pipes Embedded in an Insulated Slab

[+] Author and Article Information
Theodoros Leontiou

General Department,
Frederick University,
Nicosia 1303, Cyprus
e-mail: eng.lt@fit.ac.cy

Marios M. Fyrillas

Department of Mechanical Engineering,
Frederick University,
Nicosia 1303, Cyprus
e-mail: m.fyrillas@frederick.ac.cy

1Corresponding author.

Contributed by the Heat Transfer Division of ASME for publication in the JOURNAL OF HEAT TRANSFER. Manuscript received September 19, 2013; final manuscript received May 28, 2014; published online June 17, 2014. Assoc. Editor: William P. Klinzing.

J. Heat Transfer 136(9), 094502 (Jun 17, 2014) (6 pages) Paper No: HT-13-1494; doi: 10.1115/1.4027780 History: Received September 19, 2013; Revised May 28, 2014

In this paper, we consider the heat transfer from a periodic array of isothermal pipes embedded in a rectangular slab. The upper surface of the slab is sustained at a constant temperature while the lower surface is insulated. The particular configuration is a classical heat conduction problem with a wide range of practical applications. We consider both the classical problem, i.e., estimating the shape factor of a given configuration, and the inverse problem, i.e., calculating the optimum shape that maximizes the heat transfer rate associated with a set of geometrical constraints. The way the present formulation differs from previous formulations is that: (i) the array of pipes does not have to be placed at the midsection of the slab and (ii) we have included an isoperimetric constraint (not changing in perimeter) through which we can control the deviation of the optimum shape from that of a circle. This is very important considering that most of the applications deal with buried pipes and a realistic shape is a practical necessity. The isoperimetric constraint is included through the isoperimetric quotient (IQ), which is the ratio between the area and the perimeter of a closed curve.

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Figures

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Fig. 1

Schematic representation of the problem in the physical domain. A periodic array of isothermal (T1) pipes of arbitrary shape are embedded in a slab whose upper surface is isothermal (T0) and the lower surface is insulated.

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Fig. 2

Schematic representation of the problem in the transformed domain. The physical domain of height H + d and period L has been transformed to a channel of height h∧ and period w∧N. We have assumed that the pipes are symmetric with respect to the vertical axis.

Grahic Jump Location
Fig. 3

Schematic representation of the model problem in the transformed domain along with boundary conditions. Note that we have assumed that the pipes are symmetric with respect to the vertical axis. In addition, the lengths are normalized with wN−3.

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Fig. 4

Analytical versus Numerical for the shape factor associated with a periodic array of isothermal strips embedded in a slab (Fig. 3). The solid curves correspond to numerical results while the dots and circles to boundary element results with ten elements. The circles correspond to the classical collocation method while the dots to desingularized boundary element method [24-27]. The lower set of curve/points correspond to h∧=0.5, the middle curve/points to h∧=1 while the top set of curve/points to h∧=2.

Grahic Jump Location
Fig. 5

Optimum isothermal pipes that minimize heat conduction (shape factor). The results represent the solution to the minimization problem (Eq. (5)) without the last constraint. The distance between two successive pipes is L = 5. The other parameters are as follows: (a) d = 0.01, A = 0.05 and (b) d = 0.05, A = 0.5. There is excellent agreement between numerical (points) and analytical (solid curves) results.

Grahic Jump Location
Fig. 6

Optimum shape of isothermal pipes that maximize the shape factor (S), i.e., heat transfer rate. We consider four different cases (a)–(d), for four different combinations of the period L, perimeter P, and distance d of the pipes from lower (insulated) surface: (a) L = 1, P = 0.5, and d = 0.5; (b) L = 0.5, P = 0.5, and d = 0.5; (c) L = 0.5, P = 1.0, and d = 0.1; and (d) L = 2.0, P = 1.0, and d = 0.1. For all cases, we consider three different values of the IQ: 0.99, 0.8, and the optimum IQ. The optimum value equals to 0.5913, 0.3831, 0.4051, and 0.5610 for (a)–(d), respectively. As expected, values of IQ closer to one result in a more circular shape rather than values closer to zero.

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