Mathematical modeling of the output power of a mobile power plant with a garland turbine

Introduction. In this study, the factors influencing the use of garland-type microhydroelectric power plants in agricultural electricity supply were selected as follows: water wheel length (x₁), water flow velocity (x₂), and the number of blades on the water wheel (x₃). The variation in the water-contact surface of the wheel depends on its height and length. Because the streams in the foothill regions have relatively shallow water levels, the outer diameter of the water wheel cannot be altered; therefore, only the wheel length was varied. The water flow velocity changes depending on the required output power. The number of blades directly affects both the power output and the rotational speed of the wheel. Based on the results of preliminary experiments, and considering that the response function is nonlinear within the investigated experimental region, the experimental design was developed according to Hartley’s plan [1-7]. Materials and methods.The factors influencing the process of electricity generation from micro hydropower plants and their change intervals and levels (table 1). Table 1. Intervals and levels of factor modification Factors X1, m X2, m/s X3, piece Base level Xi0 (xi0=0) 1.5 2.5 9 Change interval DXi 0.5 0.5 3 Low Level Xip (xip= -1) 1 2 6 High level Xiy (xiy=+1) 2 3 12 The Hartley’s plans spectrum matrix is presented in (table 2). Table 2 shows the results of the experiment (yg1, yg2, yg3) and the average value and variances of m parallel experiments are calculated using the following expressions: m    i gi g y m y (1)        m 2 2 1 1 i g gi g y y m S (2) 05 .0  q We calculate the expression of the experiment for q=0.05 according to the following expression according to the Koxren criterion: 2         S G q   3924 . 0 11 , 2 2326 . 0 66 . 326 76 max g g  v v G S 2 1 1 11 g g This value does not contradict the results of the observation of the response under the hypothesis of the repetition of the experiment. “Фан ва технологиялар тараққиёти” Илмий–техникавий журнал №6/2025 213 Информатика и информационно-коммуникационные системы Table 2. Hartley’s plan matrix and experiment results f matrix of basic functions ^ 2 g S 1 g y 2 g y 3 g y g y g y f0(x) f1(x) f2(x) f3(x) f5(x) f6(x) f8(x) f11(x) f12(x) f13(x) g 1 x1 x2 x3 x1·x2 x1·x3 x2·x3 x1 2 x2 2 x3 2 5 -1 -1 +1 +1 -1 -1 +1 +1 +1 -1 100 104 94 102 25 102 2 +1 -1 -1 -1 -1 +1 +1 +1 +1 +1 200 190 205 198 58 201 3 -1 +1 -1 -1 +1 -1 +1 +1 +1 -1 340 334 337 337 9 338 8 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 361 355 365 360 25 359 9 -1 0 0 0 0 0 +1 0 0 -1 392 390 376 386 76 385 10 +1 0 0 0 0 0 +1 0 0 +1 450 445 447 447 6 449 11 0 -1 0 0 0 0 0 +1 0 0 410 417 420 416 26 417 12 0 +1 0 0 0 0 0 +1 0 0 460 465 463 463 6 462 13 0 0 -1 0 0 0 0 0 1 0 350 354 348 351 9 349 14 0 0 +1 0 0 0 0 0 1 0 300 293 303 299 26 297 15 0 0 0 0 0 0 0 0 0 0 380 385 370 378 58 380 At n=3, the Hartley plan is symmetric, so we determine the moments of the plan by the following expressions:   N ; ... 1 , 1     2 2    n i x N g g i    N ; ; ... 1 , , 1      2 2 3 (3)  j i n j i x x N g g j i      N . ... 1 , 1     4 4 n i x N g g i   Based on the above expression, we calculate plan moments: 54545 .0 4 2    , 45454 .0 3   We determine the constants according to the Hartley plan using the following expressions: 2 2          n n n a (4) 1 1 2 2 3 4   2 1     n n b     (5)   2 2 3 4 1    c (6) 3 1 2 2 3 1        n n d       (7)       2 2 3 1 3 1 (4) – (7) we calculate constants from expressions: 8947 .2  a ; 9705 .0  b ; 9998 . 10  c ; 0734 .3  d . We check the correctness of the calculated values of a, b, c, d using the following expressions: 1 2    nb a (8)   0 2     c nd b (9)     0 1 4 3 2         d c n d b (10) We determine the regression coefficients using the following expressions: “Развитие науки и технологий” Научно–технический журнал №6/2025 214 Информатика ва ахборот-коммуникацион тизимлар         N n N g g y x N b y N a b 1 0 (11) g g g i i 1 1     N 1  (12) g g g i i y x N b 1 2      N 1  (13) g g g j i ij y x x N b 1 3             N n N N g g g i ii y N b y x N d y x N C b 2 (14) g g g i g g i 1 1 1 We determine the value of the regression coefficients through the above expressions: 00714 . 511 0  b ; 5. 30 1  b ; 33 . 74 2  b ; 16 . 21 3   b ; 15 12   b ; 8, 79 13  b ; 33 . 24 23  b ; 387 . 988 11  b ; 8135 . 1020 22  b ; 8. 55 33  b . Coefficient variances and correlation moments are determined using (15):  y S N a b S 2 0 2 ) (  (15)   y S N b S i 2 1 2 2 ) (    (16)   y S N b S ij 2 1 3 2 ) (    (17)   y S N d C b S ii 2 2   (18)    y S N b b b ii 2 0 11 ^ ,    (19)    y S N d b b ij ii 2 11 ^ ,    (20) Determined repeat variance  696 . 29 2  y S and mean variance  898 .9 2  y S taking into account the values, we calculate variances and correlation moments: 6 . 2 ) ( 0 2  b S ;    6498 .1 3 2 2 2 1 2    b S b S b S       9798 .1 23 2 13 2 12 2    b S b S b S ;       133 .7 33 2 22 2 11 2    b S b S b S ^     b b b b b b    ^ ^       8734 . 0 , , , 33 0 11 11 0 11 22 0 11 ^     b b b b b b    ^ ^       7658 .2 , , , 33 22 11 22 11 11 33 11 11 jt - The statistical significance of the regression coefficients in a regression model helps determine whether the independent variables have a statistically significant relationship with the dependent variable. Here’s how to evaluate this: j i b S b t  (21)  j Using the above expression, we calculate the tj criterion value for each coefficient. 166 . 196 0  t ; 4867 . 18 1  t ; 0551 . 45 2  t ; 83 . 12 3  t ; 5765 .7 4  t ; 3071 . 40 5  t ; 2908 . 12 6  t ; 565 . 138 7  t ; 111 . 143 8  t ; 977 . 119 9  t . To check the significance of the regression coefficients, we compare the null hypothesis with the alternative value of the Student t-criterion using the following inequality: )) 1 ( ( 2 1     m N v t t g j (22) In our case 05 .0  q the value of the Student distribution for: 074 . 2 ) 22 2 11 ( 2 1      v t g “Фан ва технологиялар тараққиёти” Илмий–техникавий журнал №6/2025 215 Информатика и информационно-коммуникационные системы Results and discussion. For all coefficients, the condition in expression (22) is fulfilled, which means that all coefficients are significant, and we get the mathematical model in the following form:   2 3 2 2 2 1 3 2 3 1 2 1 3 2 1 ^ 72 .0 1. 15 22 . 33 2917 .1 375 .3 375 .5 87 . 19 83 . 47 63 . 26 23 . 272 , x x x x x x x x x x x x b x y           (23) (21) we present the values of the expression in Table 2. The next stage in processing the experimental results is to test the hypothesis regarding the adequacy of the mathematical model and the response function. Following the regression analysis, this evaluation is performed by comparing the variance of the experimental data with the variance representing model adequacy. The hypothesis concerning the homogeneity of these two variances — that is, whether the model adequately describes the experimental results — is tested using Fisher’s criterion: 2 .  (24)  y S S F resp 2 We calculate the degree of dispersion representation (reproducibility) using the following formula: N N  69697 . 29 11 66667 . 326 2 1 1 2 2           g g S N n N S N y S g g We calculate the average variance as follows: 0 1 1   89899 .9 3 69697 . 29 2 2    m y S y S Selected dispersion 2 . resp S is calculated using the following formula: 2 ^ N       1 y у S g g g 2 . (25) d N resp  Based on the value in the table, we calculate the selected variance: 31087 . 13 2 62173 . 26 2 .   resp S  2 2 . y resp S S  taking into account , we calculate as follows: 2 .    y S S F resp  344669 . 1 89899 . 9 31087 . 13 V1=N-d=11-9=2; V2=N(n-1)=11(3-1)=22 The tabular value of the Fisher criterion at q=0.05 is equal to the following: F= 1.344669 <F1-q(2.22) = 3.44 Therefore, the hypothesis about the compatibility of the mathematical model and the response function does not contradict the observational results[8]. After converting the factors from their coded values to their natural values and making appropriate changes, the mathematical model representing the process of using micro-hydroelectric power plants in agricultural electricity supply looks like this: 9424 . 61 081 .0 4164 . 60 916 . 132 86 .0 5.2 56 . 21 1752 .2 9222 . 181 258 . 483 2 2 2           z v b vz bz bv z v b P Conclusion. In the process of optimizing the parameters of the micro-hydropower plant water wheel intended for agricultural electricity supply, the most influential factors were identified as follows: the water wheel length b = 1.7 m, the water flow velocity v = 2.9 m/s, and the number of blades z = 12. At these parameter values, the output power reached its optimal level of 345.1 W. The factors were studied within the following ranges: 1 ≤ b ≤ 2 m; 2 ≤ v ≤ 3 m/s; 6 ≤ z ≤ 12. “Развитие науки и технологий” Научно–технический журнал №6/2025 216 Информатика ва ахборот-коммуникацион тизимлар The reproducibility of the experimental results was evaluated using the Cochran criterion. The significance of the regression equation coefficients was verified using the Student’s t-test, and the adequacy of the mathematical model with respect to the response function was assessed using Fisher’s criterion