PRELIMINARY STUDY FOR ITSUNUSA AUV BALLAST SYSTEM BASED EXPERIMENTAL IN FRESH WATER
3. RESULT AND DISCUSSION
From the results of the interpolation calculation from the experimental data the detailed depths can be observed when water filling is done. Based on the collected data then calculation of buoyant forces is made in accordance with Archimedes' law to find out the relationship between the experiments using Archimedes's law. The amount of buoyant force is formulated as follows:
Fa = ρcair Vb g
With the data interpolation, the graph can be obtained from each experiment using plastic bottle of different water conditions, namely fresh water and sea water. With this graph, it is easy to see the difference of each calculation data. The intended graphical results are as follows:
Graph of plastic bottle experimental data
Freshwater
Figure 6. The graph of freshwater coming into plastic bottle
From the graph results it can be observed that there is an increase at some point. The increase occurs when the volume of the immersed object or the volume of filled water filled is ¾. To determine the formula of the empirical equation, division is made to each graph in accordance with the difference in the increase in the curve. The determination of the equation is done using the following linear equation formula:
h
From the calculation using the formula above, comes up the equation of each increase in curve for the graph as follows:
𝑦−𝑦1 𝑦2−𝑦1 = 𝑥−𝑥1
𝑥2−𝑥1
From the calculation using the formula above, comes up the equation of each increase in curve as of the graph as follows:
Figure 7. Graph of freshwater entering plastic bottle ≤ 3/4 volume.
From Graph of freshwater entering plastic bottle ≤
¾ volume, the linear equation is as follows:
plastic bottle 0.6 liter y = 0.4x + 1.6716 plastic bottle 1 liter y = 0.56x + 0.1388 plastic bottle 1.5 liter y = 0.44x - 0.3448
Figure 8. Graph of freshwater entering plastic bottle ≥ 3/4 volume.
From Graph of freshwater entering plastic bottle ≥3/4 volume, the linear equation is as follows:
plastic bottle 0.6 liter y = 24.2x - 116.09 plastic bottle 1 liter y = 14.2x - 111.06 plastic bottle 1.5 liter y = 9.3867x - 108.68
From the graphs of the results of experimental data, it results in curves that look relatively the same from each experiment. The changes in depth increase occur at the point ≥ 3/4 of the volume of water is filled into the plastic bottle. From the experiment, this can be used as depth parameter, that is, when the volume of water enters the ballast tank.
From the analysis of experimental data, the results can be used to establish the parameter for active ballast making. These parameters are listed in the table below as follows:
Table 2 List of Standard deviation of experimental data results
No Type of tool
Parameter of active ballast Time
(second)
Volume (m³)
Depth (cm) 1
Plastic Bottle 0.6 liter
78.5 0.00019 18.473 2 Plastic
Bottle 1 liter 127.8 0.000323 18.419 3
Plastic Bottle 1.5 liter
57.1 0.000484 18.417
Table 2, it can be seen that in each experiment there is a standard deviation that functions as a means of finding out the values scattered at each point.
Based on the standard deviation data of the depth, by comparing those of the freshwater medium and those of the sea water one, it shows different results. The resulted standard deviation of the fresh water medium is higher than that of the seawater, indicating that the depth of fresh water has more various increases in value when there is a gradual filling process compared to that of the sea water. In its filling, the time the seawater takes to fill the active ballasts is relatively shorter than that the fresh water takes. Then experimental data was calculated with regression equation.
The regression equation obtained can be written as follows:
𝑌̂ = 𝑏0+ 𝑏1𝑋1+ 𝑏2𝑋2+ ⋯ + 𝑏𝑝𝑋𝑝 (1)
ICMST 2019 August, 1 2019
Figure 9. Results of multiple linear regression from 0.6 Liter freshwater bottle
data of which :
𝑌̂ : estimation of variable Y 𝑏𝑝 : estimation of parameter 𝛽𝑝
𝑋1𝑖 : value of variable 𝑋1 for the observation of i 𝑋𝑝𝑖 : value of variable 𝑋𝑝 for the observation of i 𝑌𝑖 : value of variable Y for the observation of i 𝑛 : number of data
Bottle with a volume of 0,6L in fresh water from the available data obtained :
𝑛 = 46;
∑𝑛𝑖=1𝑋1𝑖= 0,01725;
∑𝑛𝑖=1𝑋2𝑖= 761;
∑𝑛𝑖=1𝑌𝑖= 440,5;
∑𝑛𝑖=1𝑋1𝑖2 = 0,0000072795;
∑𝑛𝑖=1𝑋2𝑖2 = 19268,8617;
∑𝑛𝑖=1𝑋1𝑖𝑌𝑖= 0,2194750;
∑𝑛𝑖=1𝑋2𝑖𝑌𝑖= 13446,26330;
∑𝑛𝑖=1𝑋1𝑖𝑋2𝑖= 0,342255;
Based on the equation (1), the linear regression equation model can be shown as follows:
𝑌 = 𝛽0+ 𝛽1 𝑉𝑜𝑙𝑢𝑚𝑒 + 𝛽2𝑡𝑖𝑚𝑒 + 𝜀 (2)
To get a small residual, the regression equation model can be estimated by using the least squares method. By using the formula in equation (2)), the equation as follows is obtained
46𝑏0+ 0,01725𝑏1+ 761𝑏2= 440,5
0,01725𝑏0+ 0,0000072795𝑏1+ 0,342255𝑏2= 0,2194750
761𝑏0+ 0,342255𝑏1+ 19268,8617𝑏2= 13446,26330 So these are obtained:
𝑏0= −6,99785 𝑏1= 5635,79 𝑏2= 0,87409
Based on equation (2) the regression equation obtained is as follows:
𝑌̂ = −6,99785 + 5635,79 𝑋1+ 0,87409 𝑋2
Whereas by the application of the Minitab software the results as shown in Figure 9 are obtained.
Whereas for 1 liter plastic bottle, to get a small residual, the regression equation model can be estimated by using the least squares method. By using the formula in the equation (2.2) - (2.4), the equation is obtained as follows:
76𝑏0+ 0,04750𝑏1+ 2287,80𝑏2= 742,1 0,04750𝑏0+ 0,0000333450 + 1,7025345𝑏2= 0,6138180
2287,80𝑏0+ 1,7025345𝑏1+ 90662,8950𝑏2= 34890,38020
So as the following are obtained:
𝑏0= −0,495177 𝑏1= −28488,57 𝑏2= 0,9323107
Based on equation (2.5) the regression equation obtained is as follows:
𝑌̂ = −0,495177 − 28488,57 𝑋1+ 0,9323107 𝑋2
And for 1.5 liter plastic bootle, to get a small residual, the regression equation model can be estimated by using the least squares method. By using the formula in the equation (1) the equation is obtained as follows:
1125𝑏0+ 1,054174𝑏1+ 15428,835𝑏2= 10876,098 1,054174𝑏0+ 0,0011065120𝑏1+ 14,724999𝑏2= 13,4370179
15428,835𝑏0+ 14,724999𝑏1+ 214623,4686𝑏2= 166068,78308
So as the following are obtained:
𝑏0= −61,92289
𝑏1= 18413,577 𝑏2= 3,9619473
Based on equation (1) the regression equation obtained is as follows:
𝑌̂ = −61,92289 + 18413,577𝑋1+ 3,9619473𝑋2 4. CONCLUSION
Based the results of the experimental analysis above, it can be concluded that:
1. The submarine is in a sinking state when the volume of water filled is 3/4 of the volume of the active ballast tank.
2. The regression equation of 0.6 liter is 𝑌̂ = −6,99785 + 5635,79 𝑋1+ 0,87409 𝑋2
3. The regression equation of 1 liter is 𝑌̂ = −0,495177 − 28488,57 𝑋1+ 0,9323107 𝑋2 4. The regression equation of 1.5 liter is
𝑌̂ = −61,92289 + 18413,577𝑋1+ 3,9619473𝑋2