Optimization of Heating and Cooling System Locations by Taguchi’s Method to Maximize or Minimize the Natural Convection Heat Transfer Rate in a Room

Abstract

Using Taguchi's method, the best heat transfer locations for the heating and cooling systems are optimized. To investigate the flow field and heat transfer, the double multi-relaxation time lattice Boltzmann method is used. Sixteen different positions for heating and cooling systems are investigated, the full factorial method for the proposed model considers ${4}^{4}=256$ experiments, but using the Taguchi method can significantly reduce the number of experiments. In this study, with respect to the number of parameters and levels, the orthogonal array ${L}_{16}$ of Taguchi's method is proposed which reduces the number of experiments to 16. Various simulations are performed based on the design of the experiments, and the process output changes are measured by a factor called the signal-to-noise ratio. The optimal positions of the heating and cooling systems to maximize and minimize the Nusselt number are found. Also, using analysis of variance (ANOVA), the effective parameters and the percentage of participation of each of them on the Nusselt number are evaluated. In addition, the results show that the height position of the heating system has the greatest effect on the Nusselt number, while the longitudinal position of the cooling system has the least effect.

Appendix

One of the advantages of evaluating the SN ratio in Taguchi's method is the possibility of determining the optimal levels and the effectiveness of each parameter. In SN ratio analysis, the level with the highest signal-to-noise average is equivalent to the best level for each parameter. This section attempts to express the SN ratio calculations as well as the ANOVA for maximizing the Nusselt number with Ra = 10⁴. Table

Table 9 The orthogonal array of ${L}_{16}$

9 shows the Taguchi L₁₆ orthogonal array with the Nusselt number and the SN ratio values for each experiments. The value of the SN ratio for each experiments with respect to maximize the Nusselt number is as follows,

$${\text{SN}}_{{{\text{LB}}}} = - 10\log \left( {\frac{1}{n}\mathop \sum \limits_{i = 1}^{n} \frac{1}{{y_{i}^{2} }}} \right)$$

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in which the values of the Nusselt number will substitute with parameter ${y}_{i}$.

Table

Table 10 Mean values of SN ratio for each parameter at each level

10 shows the SN ratio average for each factor at each level. In the last row of Table 10, the maximum variation of the SN ratio values, Δ, is given for each parameter. Higher values of Δ indicate that the parameter has a greater effect on the output of the problem.

For example, to calculate the value of the SN ratio average for parameter per level 1, the bolded data of Table 9 are used and are equal to,

$$S{N}_{{L}_{H},1}=\frac{\left(2.92+5.25+5.06+3.58\right)}{4}=4.20$$

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The results of the analysis of variance (ANOVA) shown in Table

Table 11 Mean values of SN ratio for each parameter at each level

11 examine the effect of each parameter on the output.

For each of the input parameters in this table, the degree of freedom is defined separately, which equals the number of levels of that parameter minus one. Since four levels are considered for each parameter, the degree of freedom of each parameter is equal to three. Taking the number of experiments and subtracting one, we get the total degree of freedom. Therefore, the value of 15 is considered as the total degree of freedom. The degree of freedom of error is equal to the difference between the degree of total freedom from the sum of the degrees of freedom of the parameters. Therefore, in this study, the degree of freedom of error is equal to three.

Table 11 also shows the sequential sums of squares (Seq SS), adjusted mean squares (Adj MS), F-value test, and also percent contribution of each parameter on the output which is calculated as follows,

$$S_{m} = {\text{Correction}}\,{\text{Factor}} = \frac{{\left( {\sum {\text{SN}}_{i} } \right)^{2} }}{N}$$

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$$S_{T} = {\text{Total}}\,{\text{sum}}\,{\text{of}}\,{\text{squares}} = \sum \left( {{\text{SN}}_{i} } \right)^{2} - S_{m}$$

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$${\text{SS}}_{A} = \frac{{A_{1}^{2} }}{{N_{{A_{1} }} }} + \frac{{A_{2}^{2} }}{{N_{{A_{2} }} }} + \ldots + \frac{{A_{n}^{2} }}{{N_{{A_{n} }} }} - S_{m}$$

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$${\text{MS}}_{i} = \frac{{{\text{SS}}_{i} }}{{{\text{DoF}}\,{\text{of}}\,i}}$$

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$$F_{i} = \frac{{{\text{MS}}_{i} }}{{{\text{MS}}\,{\text{of}}\,{\text{error}}}}$$

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$$\rho \,(\% ) = \frac{{{\text{SS}}_{i} - \left( {{\text{DoF}}\,{\text{of}}\,i} \right)\left( {{\text{MS}}\,{\text{of}}\,{\text{error}}} \right)}}{{S_{T} }} \times 100$$

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in which $N$ is the number of SN ratio, ${A}_{i}$ is the sum of SN ratio values for parameter $A$ at level $i$, and parameter ${N}_{{A}_{i}}$ is the number of SN ratio of parameter $A$ at level $i$. For example, according to the results shown in Table 11, the values of the ANOVA table for the ${H}_{H}$ parameter are calculated as follows,

$$S_{m} = \frac{{\left( {2.92 + 5.01 + 5.06 + 3.69 + \ldots + 5.01 + 3.52} \right)^{2} }}{16} = 393.94$$

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$$S_{T} = \left( {2.92^{2} + 5.01^{2} + 5.06^{2} + \ldots + 5.01^{2} + 3.52^{2} } \right) - 393.94 = 15.81$$

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$$SS_{{H_{H} }} = \frac{{\left( {2.92 + 5.01 + 5.06 + 3.69} \right)^{2} }}{4} + \frac{{\left( {5.25 + 5.93 + 6.11 + 5.71} \right)^{2} }}{4} + \cdots + \frac{{\left( {3.58 + 5.06 + 5.01 + 3.52} \right)^{2} }}{4} - 393.94 = 8.60$$

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$$MS_{{H_{H} }} = \frac{8.60}{{\left( {4 - 1} \right)}} = 2.87$$

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$$F_{{H_{H} }} = \frac{2.87}{{0.00907}} = 316.43$$

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$$\rho \left( \% \right) = \frac{{8.60 - \left( {4 - 1} \right)\left( {0.00907} \right)}}{15.81} \times 100 = 54.22\%$$

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