Rapid evaluation method for thermal performance of phase change material wall
A phase change material and thermal performance technology, applied in the field of building energy conservation, can solve the problems of convenience and economy, time cost challenges, high numerical calculation cost, and achieve the simple, widely applicable, and physical meaning of thermal performance prediction and evaluation. Effect
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Embodiment 1
[0112] The thermal performance orthogonal analysis and corresponding index results of the wall body combined with various factors of the present invention are shown in Table 1.
[0113] Table 1L 18 (3 7 ) orthogonal table and thermal performance calculation results
[0114]
[0115]
[0116] The technical solution of the present invention is described in detail by taking the wall body 13 in Table 1 as a specific example. The main factor level of wall 13 is A 2 B 1 C 3 D. 2 , the specific factor levels are as follows:
[0117] Factor A (level 2): the phase transition temperature is lower than the center temperature of the left boundary, but higher than the temperature of the center of the right boundary;
[0118] Factor B (level 1): The total latent heat of PCM is taken as 56.22kJ;
[0119] Factor C (level 3): The thermal conductivity of the wall material is taken as 1.2W m -1 ·K -1 ;
[0120] Factor D (level 2): The PCM is in the middle of the wall.
[01...
Embodiment 2
[0158] The technical solution of the present invention is described in detail by taking the wall body 11 in Table 1 as a specific embodiment 2. The main factor level of wall 11 is A 1 B 2 C 1 D. 2 , the specific factor levels are as follows:
[0159] Factor A (level 1): the phase transition temperature is equal to the center temperature of the left and right boundaries;
[0160] Factor B (level 2): The total latent heat of PCM is taken as 74.60kJ;
[0161] Factor C (level 1): The thermal conductivity of the wall material is taken as 0.5W m -1 ·K -1 ;
[0162] Factor D (level 2): The PCM is in the middle of the wall.
[0163] Step 1: Score Factor A
[0164] due to T PCM,m =T left,m =T right = 26°C, so factor A score S for wall 11 A = 100;
[0165] Step 2: Score Factor B
[0166] a. According to formula (8), k df =18.14351-689.67702·e -0.08032L =18.14351-689.67702·e -0.08032×74.60 =16.42;
[0167] b. According to formula (9), k tl =0.09768+0.14369L-9.3199...
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