Research Article Mathematical Model Based on BP Neural … · 2019. 7. 31. · storage tank and calibration of storage tank chart. Finally, some concluding remarks are drawn in Section

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 923036 13 pageshttpdxdoiorg1011552013923036

Research ArticleMathematical Model Based on BP Neural NetworkAlgorithm for the Deflection Identification of StorageTank and Calibration of Tank Capacity Chart

Caihong Li1 Yali Yuan1 Lulu Song2 Yunjian Tan1 and Guochen Wang3

1 School of Information Science and Engineering Lanzhou University Lanzhou Gansu 730000 China2 College of Tourism Hainan University Haikou Hainan 570228 China3Hefei Rongshida Sanyo Electric Co Ltd Hefei Anhui 230061 China

Correspondence should be addressed to Yali Yuan yuanyali firstfoxmailcom

Received 28 February 2013 Accepted 22 April 2013

Academic Editor Fuding Xie

Copyright copy 2013 Caihong Li et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The tank capacity chart calibration problem of two oil tanks with deflection was studied one of which is an elliptical cylinderstorage tank with two truncated ends and another is a cylinder storage tank with two spherical crowns Firstly the function relationbetween oil reserve and oil height based on the integral method was precisely deduced when the storage tank has longitudinalinclination but has no deflection Secondly the nonlinear optimization model which has both longitudinal inclination parameter 120572and lateral deflection parameter 120573 was constructed using cut-complement method and approximate treatment method Then thedeflection tank capacity chart calibration with a 10 cm oil level height interval was worked out Lastly the tank capacity chart wascorrected by BP neural network algorithm and got proportional error of theoretical and experimental measurements ranges from0 to 000015 Experimental results demonstrated that the proposed method has better performance in terms of tank capacitychart calibration accuracy compared with other existing approaches and has a strongly practical significance

1 Introduction

Tanks for storing oil have been in existence for almost onehundred years Many of the underground storage tanks arehorizontal tanks they are mainly divided into square cylin-drical and elliptic cylindrical and their roofs can be dividedinto flat top conic top ball lacunarity and so on [1 2] In thispaper cylindical storage tank with two spherical crowns wasprimarily discussed which is more practical The oil reservemeasurement of storage tank is a challenging problem espe-cially after a period of time due to foundation deformationand other reasons oil storage tanks tend to be vertical orhorizontal displacement resulting in inaccurate tank volumetables

Although some install the automated measurement sys-tem the measurement accuracy is not high And the price ofthe imported high-precision liquid level instrument is too

highTherefore on the basis of the situation and the develop-ment trend of the current domestic and foreign oil tank liquidlevel measurement technology developing a kind of liquidlevel measurement technology which is suitable for Chinarsquosnational conditions is very important [3] People usually useflow meter and oil level gauge to measure input or outputoil oil height of the tank and other data and to get thechangeable relation of oil height and oil volume by meansof the precalibration tank capacity table (the correspondingrelationship between oil height and oil volume of storagetank) so as to determine whether to add oil or not [4ndash8]

Since the early 1870s some scholars have already triedto solve this problem [9 10] In particular after 2010s manyresearchers have made much study as to how to improve theaccuracy of calibration of tank capacity chart and proposedmany improvedmethodsMost of themadopted pure integraland infinitesimal method to handle tank issues [11ndash13] while

2 Abstract and Applied Analysis

some of them used method of the minimum squares [14 15]However for the methods of error correction few of thepapers use methods to correct the result of the calibration oftank capacity chart [16] And few papers have been presentedon handling tank issues by BP neural network [17] Neverthe-less many issues and problems about calibration have beenaddressed and resolved and got a good result when using BPneural network [18ndash20]

The error between theoretical oil reserve and actual oilreserve results from two main reasons One of the reasons isthe irregular geometry of the actual storage tank and anotheris the volatilization of the oil the thickness and the capillaryabsorption phenomenon of storage tank which leads to a cer-tain deviation between the theoretical oil reserve and actualones As the rule of this kind deviation is relatively fuzzy inorder to further reduce error and improve the accuracy ofthe calibration the BP neural network a method with self-learning ability is adopted in this paper to revise the calibra-tion value [21]

Artificial neural networks (ANNS) are powerful toolsfor prediction of nonlinearities These mathematical modelscomprise individual processing units called neurons thatresemble neural activity [22] After the first simple neu-ral network was developed by McCulloch and Pitts in 1943[23] many types of ANN have been proposed BP neuralnetwork simulates the human nervous system structure andthe neural network model with multilayer perceptron is themost mature widely used model among ANN

Currently the error back-propagation (BP) neural net-work is the most widely used which consists of three layersnamely input hidden and output layers One artificial neu-ron is simple but a lot of artificial neurons can compose com-plicated neural network which can achieve highly nonlinearmapping relation between the input and output through theinteraction of artificial neurons and realize the informationprocessing and storage [24] Due to its highly parallel struc-ture high-speed self-leaning ability self-adaptable processingability arbitrary function mapping ability powerful patternclassification and pattern recognition capabilities for mod-eling complex nonlinear systems [25] BP neural networkalgorithm studies show promising results in calibration andis used in this study

The paper is mainly organized into four sectionsSection 2 describes the model establishment and solution ofsmall elliptic storage tank with deflection identification andcalibration of tank capacity chart It is divided into three partsas follows Section 21 mathematical models of the relationbetween oil reserve and oil height of the small nondeflectionelliptic storage tank Section 22 model 2 for tank capacitychart calibration problem of small elliptic storage tank at aninclination angle 120572 of 41∘ and Section 23 correction modelof calibration based upon BP neural network Section 3describes the establishment and solution of the model withdeflection identification and calibration of tank capacity chartof actual storage tank It also includes three parts Section 31model 3 of actual storage tank with longitudinal inclinationand lateral deflection Section 32 the determination of thedeflection parameter and Section 33 model solving of actual

12m

178m

Figure 1 Cross-section schematic of small elliptic storage tank

119884

119910

119886 119883

2119887

Figure 2 Cross-section diagram of the small elliptical tank withoutdeflection

storage tank and calibration of storage tank chart Finallysome concluding remarks are drawn in Section 4

2 Model Establishment and Solution ofSmall Elliptic Storage Tank withDeflection Identification and Calibrationof Tank Capacity Chart

21 Mathematical Models of the Relation between Oil ReserveandOilHeight of the Small Nondeflection Elliptic Storage TankAccording to the cross-section diagram of the small ellipticalstorage tank as shown in Figure 1 we set up a coordinatesystem as shown in Figure 2 Make the profile nadir of thestorage tank for origin the high for 119910 shaft and the basal leveltangent for 119909 shaft

The cross-section oil surface area of the small ellipticaltank can be calculated according to the application of definiteintegration

1199041(ℎ1) = 2int

ℎ1

0

119909119889119910 (1)

According to the elliptic equation (11990921198862) + (119910 minus 119887)21198872 = 1substitute 119909 = 119886radic1 minus (119910 minus 119887)21198872 into formula (1) and inte-gral which can obtain the following

1199041(ℎ1) =

119886119887

2

120587 minus 119886radic119896ℎ+

119886

119887

ℎ1

radic119896ℎ+ 119886119887 arc sin(minus1 + ℎ1

119887

)

(2)

where 119896ℎis minusℎ1(ℎ1minus 2119887)

Abstract and Applied Analysis 3

Table 1 Testing data and result of model one

Site 1 Site 2OHmm AORL TORL IORL IER OHmm AORL TORL IORL IER15902 31200 32288 31346 047 48689 151200 156474 151108 00617614 36200 37463 36290 025 49895 156200 161649 156120 00519259 41200 42636 41241 010 51097 161200 166824 161133 00420850 46200 47813 46203 001 52295 166200 171998 166146 00322393 51200 52985 51167 006 53490 171200 177173 171159 00223897 56200 358161 56139 011 54682 176200 182346 176171 00225366 61200 63335 61116 014 55872 181200 187519 181184 00126804 66200 68508 66095 016 57061 186200 192695 186200 00028216 71200 73685 71081 017 58248 191200 197868 191212 00129603 76200 78858 76067 017 59435 196200 203043 196227 00130969 81200 84033 81058 018 60622 201200 208220 201244 00232315 86200 89206 86049 018 61809 206200 213395 206259 00333644 91200 94380 91044 017 62996 211200 218567 211271 00334957 96200 99554 96041 017 64185 216200 223743 216286 00436256 101200 104730 101041 016 65375 221200 228916 221299 00437542 106200 109905 106043 015 66567 226200 234089 226310 00538816 111200 115081 111047 014 67763 231200 239267 231327 00640079 116200 120255 116051 013 67854 231583 239661 231709 00541332 121200 125429 121056 012 69053 236583 244837 236723 00642576 126200 130603 126062 011 69082 236706 244962 236845 00643812 131200 135777 131069 010 70285 241706 250140 241860 00645040 136200 140949 136075 009 71491 246706 255311 246869 00746262 141200 146124 141085 008 72703 251706 260488 251882 00747478 146200 151298 146095 007 73919 256706 265659 256889 007Where OH AOR TOR IOR and IER are respectively oil height actual oil reserve theoretical oil reserve improved oil reserve and improved error ratio

We get theoretical oil reserve of small elliptical storagetank according to cylinder volume formula

V1(ℎ1)=119897 (

119886119887

2

120587minus119886radic1199041+

119886

119887

ℎ1radic1199041+ 119886119887 arc sin(minus1 + ℎ1

119887

))

(3)

where 1199041= minusℎ1(ℎ1minus 2119887) 119897 is the length of the storage tank

cylinder and ℎ1is oil height of the tank capacity chart

According to the data provided by the topic A of 2010National Mathematical Contest in Modeling (Table 1) [26]we substitute the known data into formula (3) and thencompare the theory oil reserve with the actual oil reserve Itis obvious that the proportional error is so large as high as34 sim 35 that it is necessary to take an error analysis

211 Error Analysis With the increase of liquid level the partof the pipe submerged in the oil is increasing which makesthe theoretical data larger than the actual data According tothe actual situation the capacity of the pipe in the oil and theprobe will take a linear change Fit the two groups of datanamely the theoretical and actual oil reserve difference valuesand the height of the liquid The results are shown in theFigure 3

It turns out to be that the above two groups of datameet the linear relationship and the curve fitting goes to

1198772

= 09967 from the diagram It also obtains the capacity ofthe pipe in the oil and the probe which is named ΔV

1

ΔV1= 13493ℎ

1minus 12031 (4)

212 Model One (Improved Model 1) The relationshipbetween the capacity and the height of the oil in the tankwithout deflection can be acquired through formula (3) (4)

V1= 119897 (

119886119887

2

120587 minus 119886radicminusℎ1(ℎ1minus 2119887) +

119886

119887

ℎ1

radicminusℎ1(ℎ1minus 2119887)

+119886119887 arc sin(minus1 + ℎ1119887

)) minus ΔV1

(5)

where 119897 is the length of storage tank cylinder and ℎ1is oil

height of the tank capacity chartSubstituting the oil height into formula (5) can obtain the

improved oil reserve and the error ratio is within 047 Itmeans that the precision has been improved by more than 10times compared with the original model The specific testingdata are shown in Table 1

22 Model 2 for Tank Capacity Chart Calibration Problem ofSmall Elliptic Storage Tank at an Inclination Angle 120572 of 41∘When the inclination angle 120572 equals 41∘ take left inclinationfor example as shown in Figure 4


020406080

100120140160

0 20 40 60 80 100 120 140

Erro

r

Oil height

Error fitting curve

Error fitting curveError fitting line

119910 = 13493119909 minus 12031

1198772 = 09967

Figure 3 The fitting result of the difference between the theoreticaland actual oil reserve and the height of the liquid

Oil level probeOil inlet Oil outlet pipe

Oil floater

Oil2119887

Horizontal line120572

11989721198971

Figure 4 Facade schematic of small elliptic storage tank

Considering the relation of mutative oil reserve and oilheight this problem can be divided into three conditions todiscuss as shown in Figure 5

Make the lower left quarter of the storage tank for originthe length of storage tank for 119909 shaft and the high for 119910 shaftThen the coordinate system can be built as shown in Figure 6

Based upon Figures 5 and 6 (1) there is little oil in storagetank that is oil level is under line AB Now 0 le ℎ

2lt 1198972tan120572

(2)There is moderate oil in storage tank that is oil levelshould be between line CD and AB Now 119897

2tan120572 le ℎ

2lt

2119887 minus 1198971tan120572

(3) there is much oil in storage tank that is oil levelshould be over line CD Now 2119887 minus 119897

1tan120572 le ℎ

2le 2119887

For above three situations we build model 2 according tothe relation of oil height and oil reserve The detailed solvingis below

First of all establish equation of the liquid level Obvi-ously the slope of this line is minus tan120572 and

119896 =

ℎ2minus 119887

1198971

= minus tan120572 (6)

where 1198971is the length of OC

The other equation is obtained as follows

119887 = ℎ2+ 1198971tan120572 (7)

So the relation between oil height 119910 and horizontal ordinate119909 is defined as follows

119910 = (1198971minus 119909) tan120572 + ℎ

2 (8)

Make differential on both sides of the function at the sametime which can obtain the following

119889119909 = minuscot120572119889119910 (9)

The theoretical oil reserve of storage tank at longitu-dinal angle 120572 of 41 degrees can be acquired through thestereoscopic volume formula [27] with known parallel cross-section area

V2= int

245

0

119860 (119909) 119889119909 (10)

that is

V2=int

119897

0

(

119886119887

2

120587minus119886radic1199011+

119886

119887

119910radic1199011+ 119886119887 arc sin(minus1 +

119910

119887

))119889119909

(11)

where 1199011= minus119910(119910minus2119887)119860(119909) is the parallel cross-section area

when inclined angle with deflection of 120572 is 41 degrees 119897 isthe length of storage tank cylinder and the value of 119910 is (119897

1minus

119909) tan120572 + ℎ2

As was discussed above the relation model between oilheight ℎ

2and oil reserve of storage tank can be obtained as

shown in the following model

1198812=

10

minus3

int

ℎ2+1198971 tan120572

0

cot120572[1198861198872

120587 minus 119886radicminus119910 (119910 minus 2119887) +

119886

119887

119910radicminus119910 (119910 minus 2119887) + 119886119887 arc sin(minus1 +119910

119887

)] 119889119910 0 le ℎ2lt 1198972tan120572

10

minus3

int

ℎ2+1198971 tan120572

ℎ2minus1198972 tan120572cot120572[1198861198872


119886

119887


119887

)] 119889119910 1198972tan120572 le ℎ

2lt 2119887 minus 119897

1tan120572

V1(2119887) minus 10

minus3

int

ℎ2minus1198972 tan120572

0

cot120572[1198861198872


119886

119887


119887

)] 119889119910 2119887 minus 1198971tan120572 le ℎ

2le 2119887

(12)

According to the data provided by 2010 National Math-ematical Contest in Modeling the title of A [26] (Table 1)

model 2 can be tested by the inclined oil-taking data Whatis more the displayed oil reserve of oil height between


AB

C D

(a) Little oil in storage tank

AB

C D

(b) Moderate oil in storage tank

AB

C D

(c) Much oil in storage tank

Figure 5 Different oil reserve profiles

B

ℎ2

1198971 1198972119900 119888

120572 A

119883

119884

119887

Figure 6 Facade schematic of small elliptic storage tank withlongitudinal inclination

050

100150200250300350400

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

113

120

Erro

r val

ue

Oil height

Displacement curve of error

The displacement error

minus50

Figure 7 Calibrated error curve graph with inclined angle of 41degrees

0 and 120 cm accordingly and theoretical oil reserve ofinclined deflection can be calculated The chart can begenerated as shown in Figure 7 using the calibrated errorvalue and oil height

The original tank capacity chart can no longer reflect thereal oil capacity when the tank inclines As shown in Figure 7when the oil height is more than 90 cm error should besmaller with the increasing oil height

1199091

1199092

119909119895

Input layer

1198821198941198951205791

1205791

12057921205792

120579119894

120579119894

Σ

Σ

Σ

Σ

Σ

Σ

Hidden layer Output layer

1199051

1199052

119905119897

purelin

purelin

purelin

119879119897119894

tan sig

tan sig

tan sig

Figure 8 The topological structure of the BP network model

23 Correction Model of Calibration Based upon BP NeuralNetwork BP neural network is a nonlinear adaptive dynamicsystem consisting of many parallel neurons with learningability memory ability calculation ability and intelligentprocessing ability [28] Commonly a typical BP neuralnetwork model is a full-connected neural network includinginput layer hidden layer and output layer [29 30] Each layerhas multiple neurons and the nodes between two adjacentlayers connect in single direction It has been proved byKolmogorovrsquos theorem a neural network theory theoremthat the fully studied three-layer BP network can approachany function [28]

Some researchers also claim that networks with a singlehidden layer can approximate any continuous function toany desired accuracy and are enough for most forecastingproblems [31ndash33]

In this study a three-layer neural network is appliedin calibration of storage tank chart modeling Whatrsquos morethe network training is actually an unconstrained nonlinearminimization problem and the nonlinear model is used inthis study Therefore it can achieve better effect to processresidual correction of this model by BP neural network

The input node hidden node and output node arehypothesized as 119909

119895 119910119894and 119874

119897 respectively The connection

weight between the input node and the hidden node is 119908119894119895

while the connection weight between the hidden node andthe output node is 119879

119897119894 Giving the maximum iterating times

and error precision Figure 8 is the topological structure ofthe BP neural network model


Begin

Initialize parameters of network

Input data samples

Calculate outputs of hidden layersand output layers

Calculate the output error 119864

Calculate error signal of differentlayers by 119864

Adjust the weights of each layer

Input all the data samples

Storage of network weights

Cycle-index adds one

119864 meets the requirements

Sample number adds one

Y

N

N

Y

End

119864 = 0 get the first sample

Figure 9 Flowchart of back-propagation (BP) neural networks algorithm

Various steps of the BP training procedure are describedin Figure 9

According to Figures 8 and 9 suppose that the expectedoutput value of the output node is 119905

119897 then the BP network

model adopts a learning algorithm for training as followsFirstly Initialize by giving random number between minus1

and 1 to the connection weights 119908119894119895 119879119897119894and threshold values

120579119894 120579119897 choosing a mode and giving network to 119909

119895 119905119897

Secondly the output of the hidden note is 119910119894= 1198911(sum119895

119908119894119895

119909119895minus 120579119894)

The output of the output note is 119874119897= 1198912(sum119894

119879119897119894119910119894minus 120579119897)

Thirdly calculate new connection weights and thresholdvalues The correction value of connection between hiddenand output node is defined as is 119879

119897119894(119896) = 119879

119897119894(119896) + 120578120575

119897119910119894

The correction of the threshold values is 120579119897(119896+1) = 120579

119897(119896)+

120578120575119897


Table 2 The oil reserve of oil-out and analysis of percentage error with BP neural network statistical table

Site 1 Site 2OHmm AORL TORL FENNL PEC OHmm AORL TORL FENNL PEC102065 346474 340510 5961 000 71532 221474 222210 1008 012100773 341474 336180 5315 001 70543 216474 218070 1176 01999432 336474 331590 4864 001 69352 211474 213060 1397 00998096 331474 326940 4520 000 6825 206474 208430 1609 01796710 326474 322010 4205 008 67102 201474 203600 1831 01595601 321474 318010 3959 015 65868 196474 198400 2060 00794154 316474 312720 3634 004 64774 191474 193800 2250 00492969 311474 308320 3360 007 63576 186474 188760 2434 00891644 306474 303340 3045 003 62461 181474 184070 2580 00190414 301474 298660 2746 002 61253 176474 179000 2708 0108919 296474 293950 2447 003 60069 171474 174030 2799 01487923 291474 289030 2139 010 58940 166474 169310 2855 00186899 286474 285020 1896 015 57700 161474 164130 2882 01485513 281474 279550 1581 012 56458 156474 158960 2876 02584402 276474 275120 1345 000 55433 151474 154710 2848 02683164 271474 270160 1107 008 54076 146474 149100 2783 01182047 266474 265650 920 004 52865 141474 144120 2702 00480816 261474 260650 750 003 51719 136474 139420 2609 02579600 256474 255690 624 006 50487 131474 134410 2497 03378504 251474 251190 551 011 49078 126474 128700 2355 01077307 246474 246260 518 012 47806 121474 123590 2220 00976209 241474 241720 530 012 46597 116474 118770 2088 01875081 236474 237030 586 001 45240 111474 113400 1939 00173942 231474 232290 684 006 43998 106474 108530 1804 02472709 226474 227140 832 007 42583 101474 103030 1655 010Where OH AOR TOR FENN and PEC are respectively oil height actual oil reserve theoretical oil reserve fitting error of neural network and proportionalerror with correction

The correction value of connection between input andhidden node is defined as 119908

119894119895(119896 + 1) = 119908

119894119895(119896) + 120578

1015840

120579119894

1015840

119909119895

The correction of the threshold values 120579119894(119896 + 1) = 120579

119894(119896) +

1205781015840

120575119894

1015840Where 120578 and 1205781015840 reflect learning efficiency 120575

119897= (119905119897minus 119874119897) sdot

119874119897sdot (1 minus 119874

119897) 120575119894

1015840

= 119910119894(1 minus 119910

119894) sum119897

120575119897119879119897119894

Lastly select the next input mode and return to step (2)Keep training until the error precision of the network settingsmeets the requirementsThen finish the trainingThus the BPneural network model is established

Calibration correction take the oil-out level height dataof the small longitudinal tilting elliptical tank as input datawhile take the119863-value between the theoretical oil reserve andthe actual measurement of oil as output data Construct a BPneural network model with single input single output andhidden layer with three-node by matlab 2010

Then train the model with inspecting data of the oil-inlevel height data and the practical measurement oil reserveThe training results are shown in Figures 10 and 11

As mentioned above it turns out to be that the accuracyof the results of the correction BP neural network model isvery high Part of the results can be seen in Table 2

As shown in Table 2 the proportional error of theoreticalvalue and experimental measurement value with BP neural

network correction ranges from 000 to 038 Errorreduces a lot more than before Using the correction modelthe calibration of tank capacity chart value can be calculatedwith the internal of oil height for 1 cm after the deflection ofstorage tank as shown in Table 3

3 Establishment and Solution of the Modelwith Deflection Identification andCalibration of Tank Capacity Chart ofActual Storage Tank

31 Model 3 of Actual Storage Tank with LongitudinalInclination and Lateral Deflection

311 Model Establishment of Actual Storage Tank with Longi-tudinal Inclination The graph in Figure 12 clearly shows that

V3= V119862+ V119871+ V119877 (13)

1199101= ℎ + 119897

1tan120572

1199102= ℎ minus (

119897

2

+ 1198972) tan120572

(14)


Table 3 Calibration results of tank capacity chart of the small deflected elliptic storage tank after the correction of BP neural network

Site 1 Site 2 Site 3 Site 4OHcm TORL OHcm TORL OHcm TORL OHcm TORL0 103 31 61549 62 184765 93 3118021 342 32 65015 63 188848 94 3157492 655 33 68536 64 192913 95 3196503 1057 34 72110 65 196962 96 3235084 1559 35 75736 66 200997 97 3273225 2167 36 79410 67 205020 98 3310966 2892 37 83132 68 209035 99 3348437 3739 38 86899 69 213048 100 3385858 4717 39 90708 70 217061 101 3423589 5830 40 94560 71 221080 102 34621410 7085 41 98450 72 225108 103 35020211 8486 42 102378 73 229150 104 35431812 10040 43 106341 74 233210 105 35845913 11751 44 110338 75 237289 106 36246014 13625 45 114366 76 241388 107 36619915 15960 46 118422 77 245508 108 36965716 18085 47 122504 78 249649 109 37287417 20342 48 126610 79 253810 110 37590318 22717 49 130736 80 257986 111 37878119 25201 50 134880 81 262176 112 38153220 27786 51 139038 82 266374 113 38416621 30466 52 143206 83 270577 114 38668222 33236 53 147384 84 274780 115 38908223 36090 54 151564 85 278978 116 39135524 39025 55 155746 86 283166 117 39349025 42038 56 159925 87 287339 118 39435826 45123 57 164096 88 291491 119 39623127 48279 58 168260 89 295619 120 39794028 51502 59 172410 90 29971729 54790 60 176545 91 30378330 58140 61 180664 92 307812Where OH is oil height and TOR is theoretical oil reserve

In the case of no deflection the formula in references [34]can be cited namely

V119862=

119886

119887

119897 [119902ℎ+ 1198872arc sin(ℎ

119887

minus 1) +

1

2

1205871198872

] (15)

where 119902ℎ= (ℎ minus 119887)radicℎ(2119887 minus ℎ)

V119878=

120587119886119888

21198872

[1198872

(ℎ minus 119887) minus

1

3

(ℎ minus 119887)3

+

2

3

1198873

] (16)

where 119888 reflects sagittal of the storage tankCalculate the oil capacity in the tank with longitudinal

angel of 120572 by integration as shown in Figure 12 As bothsides of the storage tank are irregular solid it is difficult tocalculate accurately But the angel 120572 is very tiny accordingto the fact that both longitudinal angle and lateral deflectionangle are small angles so cut-complement method can beadopted to make an approximate disposal The extra volume

of left approximately equals the insufficient volume of theright that is

ΔV119871minus ΔV119877997888rarr 0 (17)

Hence the relation between oil reserve of storage tankand oil height can be defined as follows

V3= V119862+ V119878|ℎ=119910

1

+ V119878|ℎ=119910

2

(18)

The calculations of V119862are as shown in Figure 13

Theboundary of the cylinderrsquos longisection is rectangularAs shown in Figure 13 firstly draw a line perpendicular tothe base through the basersquos midpoint and the line intersectswith the metal line Secondly draw a parallel line to the basethrough the above point of intersection Then the parallelline metal line and two boundaries form two trianglesnamed V1015840 and V10158401015840 both of which are right-angled triangleswith equal vertical angles and horizontal right-angle side


04 05 06 07 08 09 1 110

01

02

03

04

05

06

07

Oil level

119863-v

alue

of t

heor

etic

al o

il re

serv

e and

actu

aloi

l res

erve

Expected outputsPredicted outputs

Figure 10 The outputs of the network prediction

0 5 10 15 20 25 30 350

02

04

06

08

1

12

14

16

18

2

Prop

ortio

nal e

rror

()

times10minus5

Figure 11 The relative error percentage of the BP neural networkprediction

So the two triangles are congruent Apparently their areasare also equal According to the ZuYuan principle theircorresponding volumes in the cylinder are also equal thatis V1015840 = V10158401015840 Therefore it can be acquired through the cut-complement method as follows

V119862=

119886

119887

119897 [119902119910+ 1198872arc sin(

119910119862

119887

minus 1) +

1

2

1205871198872

] (19)

where 119902119910is (119910119862minus 119887)radic119910

119862(2119887 minus 119910

119862)

Suppose that the metal linersquos slope equals 119896 then

119896 =

119910119862minus ℎ3

1198972

= minus tan120572 997904rArr 119910119862= ℎ3minus 1198972tan120572 (20)

RightLeftCenter

1199101

1199102

120572ℎ 119910

1198971 1198972

Δ119871

Δ119877

Figure 12 Facade schematic of actual elliptic storage tankwith angle120572 of longitudinal inclination

1205721199101 119910ℎ

1198971 1198972

1199102

998400998400

998400

Figure 13 Facade schematic of cylinder with inclined angle of 120572

Substituting formula (20) into formula (19)

V119862=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)radic119902

119886

+1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

(21)

where 119902119886is (ℎ3minus 1198972tan120572)(2119887 minus (ℎ

3minus 1198972tan120572))

The spherical crownrsquos metal line of both ends parallelsthe undersurface of the cylindrical after the approximatedisposal which is equivalent to the case of no deflectionFrom formula (16) the calculation of V

119904can be defined as

follows

V119878|ℎ=119910

1

=

120587119886119888

21198872

[1198872

(1199101minus 119887) minus

1

3

(1199101minus 119887)3

+

2

3

1198873

]

V119878|ℎ=119910

2

=

120587119886119888

21198872

[1198872

(1199102minus 119887) minus

1

3

(1199102minus 119887)3

+

2

3

1198873

]

(22)

Substitute formula (14) into formula (22) thus

V119871=

120587119886119888

21198872

[1198872

119905119887minus

1

3

1199053

119887

+

2

3

1198873

] (23)

where 119905119887= ℎ3+ 1198971tan120572 minus 119887

V119877=

120587119886119888

21198872

[1198872

119905119897minus

1

3

1199053

119897

+

2

3

1198873

] (24)


Table 4 The results of calibration of storage tank with deflection of actual storage tank without correction by BP neural network

Site 1 Site 2 Site 3 Site 4OHcm TORL OHcm TORL OHcm TORL OHcm TORL0 358 80 1182788 160 3348474 240 549672710 14031 90 1427104 170 3634002 250 572301320 88588 100 1682393 180 3917791 260 593317430 208466 110 1946814 190 4198345 270 612440240 360150 120 2218649 200 4474133 280 629318050 537210 130 2496273 210 4743573 290 643469160 735389 140 2778126 220 5005013 300 654108570 951461 150 3062689 230 5256696Where OH is oil height and TOR is theoretical oil reserve

where 119905119897is ℎ3minus (1198972 + 119897

2) tan120572 minus 119887

V3= V119862+ V119871+ V119877

=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)

times radic(ℎ3minus 1198972tan120572) (2119887 minus (ℎ

3minus 1198972tan120572))

+ 1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

1199052minus

1

3

1199053

2

+

2

3

1198873

]

+

120587119886119888

21198872

[1198872

1199051minus

1

3

1199053

1

+

2

3

1198873

]

(25)

where 1199051= ℎ3minus (1198972 + 119897

2) tan120572 minus 119887 and 119905

2= ℎ3+ 1198971tan120572 minus 119887

312 Establishment of the Model with Deflection Identificationand Calibration of Tank Capacity Chart of Actual StorageTank As shown in Figure 14

ℎ4= 119903 + 119905

119905 =

119904

cos120573

119904 = ℎ3minus 119903

(26)

Hence the ℎ is defined by

ℎ = (ℎ4minus 119903) cos120573 + 119903 (27)

Substitute formula (27) into formula (25) Then thetheoreticalmodel about120572120573 ℎ of oil reservewith longitudinalinclination and lateral deflection of storage tank is obtainedbelow

V4=

119886

119887

119897 [ (ℎ minus 1198972tan120572 minus 119887)

times radic(ℎ minus 1198972tan120572) (2119887 minus (ℎ minus 119897

2tan120572))

+ 1198872arc sin(ℎ minus 1198972 tan120572

119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

(ℎ + 1198971tan120572 minus 119887)

minus

1

3

(ℎ + 1198971tan120572 minus 119887)3 + 2

3

1198873

]

+

120587119886119888

21198872

[1198872

119903119886minus

1

3

1199033

119886

+

2

3

1198873

] (28)

where ℎ = (ℎ4minus 119903) cos120573 + 119903 119897 is the length of storage tank

cylinder 119903119886is ℎ minus (1198972 + 119897

2) tan120572 minus 119887 and ℎ is the oil height of

the calibration of storage tank chart

32 The Determination of the Deflection Parameter As wasdiscussed in Section 312 and Figure 7 of model 2 the errornearby ℎ

4= 150 is micro even without error For this reason

equations can be established by selecting three contiguousgroups of data near ℎ

4= 150 to ascertain the deflection

parameters 120572 and 120573 The equations can be defined by

V4(151073 120572 120573) minus V

4(150765 120572 120573) = 8676

V4(150765 120572 120573) minus V

4(150106 120572 120573) = 18761

(29)

Substitute the formula (28) into formula (29) Then solvethem by the software of Mathematica [35] and Matlab [36]using quasi-Newton iterative algorithm and the result can begot as follows

120572 = 235920

120573 = 3801270

(30)

The angles are very tiny which is realistic

33 Model Solving of Actual Storage Tank and Calibration ofStorage Tank Chart According to the data provided by 2010NationalMathematical Contest inModeling the title of A [26](Table 2) we substitute 120572 120573 and collected oil height of actualstorage tank into model 3 The calibration of tank capacitychart value can be calculatedwith the internal of oil height for10 cm after the deflection of storage tank as shown in Table 4

Error can be controlled under 2 when testing model 3by actual collected data of storage tank But it is still large forthe volume of this tank Similarly in order to further reduceerror and improve the accuracy of the calibration model 3also uses the BP neural network which is a method with self-learning ability to revise the calibration value


Table 5 The statistics of oil reserve of actual storage tank with oil-out and the percentage of error

Site 1OHcm AORL TORL ATDV FENNL PEC201429 4655267 4513062 142205 142210 0015200374 4627512 4484334 143178 143180 0011199529 4605218 4461273 143945 143950 0006198953 4589988 4445527 144461 144460 0004198562 4579635 4434827 144808 144810 0003197933 4562956 4417595 145361 145360 0002197251 4544837 4398883 145954 145950 0001196931 4536323 4390094 146229 146230 0000196141 4515273 4368370 146903 146900 0004195715 4503903 4356641 147262 147260 0005195130 4488267 4340518 147749 147750 0003194347 4467302 4318907 148395 148390 0006193896 4455206 4306444 148762 148760 0006193405 4442022 4292864 149158 149160 0006192588 4420047 4270239 149808 149810 0005192123 4407520 4257346 150174 150170 0005191098 4379857 4228887 150970 150970 0009189916 4347871 4196006 151865 151860 0006188986 4322643 4170088 152555 152550 0006188442 4307862 4154909 152953 152950 0006187658 4286528 4133010 153518 153520 0005187350 4278136 4124399 153737 153740 0004186244 4247959 4093445 154514 154510 0005185164 4218424 4063169 155255 155250 0005184146 4190525 4034588 155937 155940 0003183685 4177873 4021632 156241 156240 0003182891 4156056 3999299 156757 156760 0002182668 4149923 3993022 156901 156900 0003182043 4132719 3975421 157298 157300 0001181531 4118612 3960991 157621 157620 0003181169 4108629 3950784 157845 157840 0000180790 4098171 3940092 158079 158080 0001180167 4080966 3922507 158459 158460 0001179873 4072840 3914204 158636 158640 0000179075 4050764 3891653 159111 159110 0000178404 4032179 3872676 159503 159500 0001177508 4007331 3847315 160016 160020 0003Where OH AOR TOR ATDV FENN and PEC are respectively oil heightactual oil reserve theoretical oil reserve 119863-value of theoretical and actualoil reserve fitting error of neural network and proportional error withcorrection

By taking the oil-out level height data of the actual storagetank as input data and the 119863-value between the theoreticaloil reserve and the actual measurement of oil as output datathe whole network reflects the function mapping relationbetween input node and output node

Then train the network so it can have a certain ability ofassociation and prediction for this kind of problem

119904

119905

ℎ3119903

ℎ4120573

Figure 14 Cross-section schematic of lateral deflection

0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()


Calibration correction randomly take 150 groups ofthe oil-out level height data of the actual storage tank asinput data corresponding take same groups of the 119863-value between the theoretical oil reserve and the actualmeasurement of oil as output data while 50 groups of dataare taken as testing data A BP neural network model can beconstructed and trained with single input and single outputand three-node hidden layer by matlab 2010 The trainingresult is as shown in Figure 15

As discussed above it turns out to be that the accuracy ofthe results of the correction BP neural network model is veryhigh Part of the results can be seen in Table 5

As shown in Table 5 the proportional error of theoreticalvalue and experimental measurement value with BP neuralnetwork correction ranges from000000 to 000015 Erroris micro Using the correction model the calibration of tankcapacity chart value can be calculated with the internal of oilheight for 10 cm after the deflection of storage tank as shownin Table 6

4 Conclusions

In this paper using geometrical relationship of the storagetank the integral models are established from simple to


Table 6 The results of calibration of storage tank with deflection of actual storage tank with correction by BP neural network


complicated which makes the models simple and easy tounderstand Taking into account many possible oil level con-ditions and giving the common theoretical relation betweenthe oil reserve and oil height with the known deflectionparameters it has strong universality and is easy to popu-larize Cut-complement algorithm is designed to constructthis model according to the special inclined angle And thenonlinear equations are effectively solved by quasi-Newtoniterative method A novel method is applied to calibrate thestorage tank chart which combines the advantages of thepolynomial fitting method and BP neural network Modelsare tested by the known data and the improved models aregot by polynomial fitting method Based upon fuzzificationof systemmeasurement error BP neural network is proposedto correct results Quasi-Newton iterative algorithm is usedto calculate deflection parameters 120572 = 23592∘ 120573 = 380127∘by Mathematica Matlab software However when oil in thestorage tank is approximately full or there is very little oilin it it is unable to get the accurate calibration method somore research efforts should be devoted to validating theseissues Developing better models of solving these problems isthe next step we will undertake

Acknowledgments

This work was supported by the Natural Science Foundationof P R of China (90912003 61073193) the Key Science andTechnology Foundation of Gansu Province (1102FKDA010)Natural Science Foundation of Gansu Province (1107RJZA188) and the Fundamental Research Funds for the CentralUniversities (lzujbky-2012-47 lzujbky-2012-48)

References

[1] J Sun ldquoThe discussion of horizontal tank volume about theproblematic point of the verification and calculationrdquo PetroleumProducts Application Research vol 18 no 5 pp 20ndash24 2000

[2] Z Li ldquoThe calculation of horizontal storage tank volume withelliptic cylinder typerdquoMathematics in Practice and Theory vol2 pp 17ndash26 1997

[3] Y Ji S Song and Y Tu ldquoThe current situation and developmenttendency of liquid level measurement technology of storagetankrdquoPetroleumEngineering Construction vol 32 no 4 pp 1ndash42006

[4] C Li T Liang W zhou and J Lu ldquoThe function relation ofremain liquid volume and oil height of storage tankrdquo Petro-Chemical Equipment no 6 pp 25ndash27 2001

[5] C Li T Liang and M Jin ldquoThe relation of liquid volumeand oil height of storage tank with deformation sectionrdquo Petro-Chemical Equipment no 1 pp 21ndash22 2003

[6] G Si ldquoCalculate the liquid volume by liquid height of tank withdeformation sectionrdquoProcess EquipmentampPiping no 2 pp 63ndash64 2000

[7] B Gao and X Su ldquoThe volume calculation of different liquidheight of horizontal tank with various end enclosurerdquo Petro-Chemical Equipment no 4 pp 1ndash7 1999

[8] C Fu ldquoThe volume calculation of inclined storage tankrdquo Journalof Heilongjiang Bayi Agricultural University no 2 pp 43ndash521981

[9] S Sivaraman andW AThorpe ldquoMeasurement of tank-bottomdeformation reduces volume errorsrdquo Oil and Gas Journal vol84 no 44 pp 69ndash71 1986

[10] F Kelly ldquoShore tank measurementrdquo Quarterly Journal of Tech-nical Papers vol 1 pp 59ndash62 1987

[11] Q Ai J Huang X zhang and M Zhang ldquoModel of theidentification of oil tankrsquos position and the calibration of tankcapacity tablerdquo Light Industry Design vol 4 2011

[12] S Si F Hua X Tian and JWu ldquoThe study of the relation of anyheight and volume in Erect Spherical cap bodyrdquo Automation ampInstrumentation vol 154 pp 15ndash16 2011

[13] Z Li ldquoStudy on tank capacity of tilted oil tank in elliptic cross-sectionrdquo Engineering amp Test vol 51 no 2 pp 38ndash40 2011

[14] Y Chang D Zhou N Ma and Y Lei ldquoThe problem of deflec-tion identification of horizontal storage tank and calibration oftank capacity chartrdquo Oil amp Gas Storage and Transportation vol31 no 2 pp 109ndash113 2012

[15] J Dou Y Mei Z Chen and L Wang ldquoModel of the identifi-cation of oil tankrsquos position and the calibration of tank capacitytablerdquoPure andAppliedMathematics vol 27 no 6 pp 829ndash8402011

[16] Y Ai ldquoThe study of capacity table calibration of storage tankwith deflectionrdquo Business Affection vol 41 pp 170ndash170 2012

[17] S Ou J Wang and S Han ldquoModel of the identification of oiltankrsquos position and the calibration of tank capacity tablerdquo ChinaPetroleum and Chemical Standard and Quality vol 31 no 4 pp25ndash26 2011

[18] Z Wang Y Li and R F Shen ldquoCorrection of soil parametersin calculation of embankment settlement using a BP networkback-analysis modelrdquo Engineering Geology vol 91 pp 168ndash1772007


[19] S Tian Y Zhao HWei and ZWang ldquoNonlinear correction ofsensors based on neural network modelrdquo Optics and PrecisionEngineering vol 14 no 5 pp 896ndash902 2006

[20] W He J H Lan Y X Yin and Z H Zhang ldquoThe neuralnetwork method for non-linear correction of the thermalresistance transducerrdquo Journal of Physics vol 48 no 1 pp 207ndash211 2006

[21] W Liu MATLAB Program Design and Application ChinaHigher Education Press Beijing China 2002

[22] B HM Sadeghi ldquoBP-neural network predictor model for plas-tic injection molding processrdquo Journal of Materials ProcessingTechnology vol 103 no 3 pp 411ndash416 2000

[23] W S McCulloch and W Pitts ldquoA logical calculus of the ideasimmanent in nervous activityrdquo The Bulletin of MathematicalBiophysics vol 5 no 4 pp 115ndash133 1943

[24] L Sun and Y Li ldquoReview of on-line defects detection techniquefor above ground storage tank floor monitoringrdquo in Proceedingsof the 8thWorld Congress on Intelligent Control and Automation(WCICA rsquo10) vol 8 pp 4178ndash4181 July 2010

[25] T Hu P Yuan and J Din ldquoApplications of artificial neuralnetwork to hydrology and water resourcesrdquo Advances in WaterScience vol 11 pp 76ndash81 1995

[26] China Society for Industrial and Applied Mathematics 2010National Mathematical Contest in Modeling the Title of a[DBOL] China Society for Industrial and Applied Mathemat-ics Beijing China 2010

[27] Department of pplied Mathematics of Tongji universityAdvanced Mathematics (New Paris Interiors) Higher EducationPress Beijing China 2006

[28] D Zhang MATLAB Neural Network Application DesignMechanical Industry Press Beijing China 2009

[29] ASCE Task Committee on Application of Artificial Neural Net-works in Hydrology ldquoArtificial neural networks in hydrology Ipreliminary conceptsrdquo Journal of Hydrologic Engineering vol 5no 2 pp 115ndash123 2000

[30] ASCE Task Committee on Application of Artificial Neural Net-works inHydrology ldquoArtificial neural networks in hydrology IIhydrologic applicationsrdquo Journal of Hydrologic Engineering vol5 no 2 pp 124ndash137 2000

[31] G Cybenko ldquoApproximation by superpositions of a sigmoidalfunctionrdquo Mathematics of Control Signals and Systems vol 2no 4 pp 303ndash314 1989

[32] R Hecht-Nielsen Neurocomputing Addison-Wesley MenloPark Calif USA 1990

[33] K Hornik M Stinchcombe and HWhite ldquoMultilayer feedfor-ward networks are universal approximatorsrdquo Neural Networksvol 2 no 5 pp 359ndash366 1989

[34] J Guan and H Zhao ldquoPractical methods of oil volume cali-bration of horizontal storage tankrdquo Metrology amp MeasurementTechnique vol 31 no 3 pp 21ndash36 2004

[35] M Yang and J Lin Mathematica Base and MathematicalSoftware DalianUniversity of Technology Press Dalian China2007

[36] Science and Technology Products Research Center of FeisiNeural Network Theory and the Implementation of Matlab7 pp99ndash108 Electronic Industry Press Beijing China 2005

Submit your manuscripts athttpwwwhindawicom

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MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


some of them used method of the minimum squares [14 15]However for the methods of error correction few of thepapers use methods to correct the result of the calibration oftank capacity chart [16] And few papers have been presentedon handling tank issues by BP neural network [17] Neverthe-less many issues and problems about calibration have beenaddressed and resolved and got a good result when using BPneural network [18ndash20]

The error between theoretical oil reserve and actual oilreserve results from two main reasons One of the reasons isthe irregular geometry of the actual storage tank and anotheris the volatilization of the oil the thickness and the capillaryabsorption phenomenon of storage tank which leads to a cer-tain deviation between the theoretical oil reserve and actualones As the rule of this kind deviation is relatively fuzzy inorder to further reduce error and improve the accuracy ofthe calibration the BP neural network a method with self-learning ability is adopted in this paper to revise the calibra-tion value [21]

Artificial neural networks (ANNS) are powerful toolsfor prediction of nonlinearities These mathematical modelscomprise individual processing units called neurons thatresemble neural activity [22] After the first simple neu-ral network was developed by McCulloch and Pitts in 1943[23] many types of ANN have been proposed BP neuralnetwork simulates the human nervous system structure andthe neural network model with multilayer perceptron is themost mature widely used model among ANN

Currently the error back-propagation (BP) neural net-work is the most widely used which consists of three layersnamely input hidden and output layers One artificial neu-ron is simple but a lot of artificial neurons can compose com-plicated neural network which can achieve highly nonlinearmapping relation between the input and output through theinteraction of artificial neurons and realize the informationprocessing and storage [24] Due to its highly parallel struc-ture high-speed self-leaning ability self-adaptable processingability arbitrary function mapping ability powerful patternclassification and pattern recognition capabilities for mod-eling complex nonlinear systems [25] BP neural networkalgorithm studies show promising results in calibration andis used in this study

The paper is mainly organized into four sectionsSection 2 describes the model establishment and solution ofsmall elliptic storage tank with deflection identification andcalibration of tank capacity chart It is divided into three partsas follows Section 21 mathematical models of the relationbetween oil reserve and oil height of the small nondeflectionelliptic storage tank Section 22 model 2 for tank capacitychart calibration problem of small elliptic storage tank at aninclination angle 120572 of 41∘ and Section 23 correction modelof calibration based upon BP neural network Section 3describes the establishment and solution of the model withdeflection identification and calibration of tank capacity chartof actual storage tank It also includes three parts Section 31model 3 of actual storage tank with longitudinal inclinationand lateral deflection Section 32 the determination of thedeflection parameter and Section 33 model solving of actual

12m

178m

Figure 1 Cross-section schematic of small elliptic storage tank

119884

119910

119886 119883

2119887

Figure 2 Cross-section diagram of the small elliptical tank withoutdeflection

storage tank and calibration of storage tank chart Finallysome concluding remarks are drawn in Section 4

2 Model Establishment and Solution ofSmall Elliptic Storage Tank withDeflection Identification and Calibrationof Tank Capacity Chart

21 Mathematical Models of the Relation between Oil ReserveandOilHeight of the Small Nondeflection Elliptic Storage TankAccording to the cross-section diagram of the small ellipticalstorage tank as shown in Figure 1 we set up a coordinatesystem as shown in Figure 2 Make the profile nadir of thestorage tank for origin the high for 119910 shaft and the basal leveltangent for 119909 shaft

The cross-section oil surface area of the small ellipticaltank can be calculated according to the application of definiteintegration

1199041(ℎ1) = 2int

ℎ1

0

119909119889119910 (1)

According to the elliptic equation (11990921198862) + (119910 minus 119887)21198872 = 1substitute 119909 = 119886radic1 minus (119910 minus 119887)21198872 into formula (1) and inte-gral which can obtain the following

1199041(ℎ1) =

119886119887

2

120587 minus 119886radic119896ℎ+

119886

119887

ℎ1

radic119896ℎ+ 119886119887 arc sin(minus1 + ℎ1

119887

)

(2)

where 119896ℎis minusℎ1(ℎ1minus 2119887)





V1(ℎ1)=119897 (

119886119887

2

120587minus119886radic1199041+

119886

119887


119887

))

(3)






1198772


1

ΔV1= 13493ℎ

1minus 12031 (4)


V1= 119897 (

119886119887

2


119886

119887

ℎ1


+119886119887 arc sin(minus1 + ℎ1119887

)) minus ΔV1

(5)






020406080

100120140160

0 20 40 60 80 100 120 140

Erro

r

Oil height

Error fitting curve


119910 = 13493119909 minus 12031

1198772 = 09967



Oil floater

Oil2119887


11989721198971





2lt 1198972tan120572


2tan120572 le ℎ

2lt

2119887 minus 1198971tan120572


1tan120572 le ℎ

2le 2119887



119896 =

ℎ2minus 119887

1198971

= minus tan120572 (6)



119887 = ℎ2+ 1198971tan120572 (7)


119910 = (1198971minus 119909) tan120572 + ℎ

2 (8)


119889119909 = minuscot120572119889119910 (9)


V2= int

245

0

119860 (119909) 119889119909 (10)

that is

V2=int

119897

0

(

119886119887

2

120587minus119886radic1199011+

119886

119887


119910

119887

))119889119909

(11)



1minus

119909) tan120572 + ℎ2




1198812=

10

minus3

int

ℎ2+1198971 tan120572

0

cot120572[1198861198872


119886

119887


119887

)] 119889119910 0 le ℎ2lt 1198972tan120572

10

minus3

int

ℎ2+1198971 tan120572

ℎ2minus1198972 tan120572cot120572[1198861198872


119886

119887


119887

)] 119889119910 1198972tan120572 le ℎ

2lt 2119887 minus 119897

1tan120572

V1(2119887) minus 10

minus3

int

ℎ2minus1198972 tan120572

0

cot120572[1198861198872


119886

119887


119887

)] 119889119910 2119887 minus 1198971tan120572 le ℎ

2le 2119887

(12)




AB

C D


AB

C D


AB

C D



B

ℎ2

1198971 1198972119900 119888

120572 A

119883

119884

119887


050

100150200250300350400

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

113

120

Erro

r val

ue

Oil height



minus50




1199091

1199092

119909119895

Input layer

1198821198941198951205791

1205791

12057921205792

120579119894

120579119894

Σ

Σ

Σ

Σ

Σ

Σ


1199051

1199052

119905119897

purelin

purelin

purelin

119879119897119894

tan sig

tan sig

tan sig






119895 119910119894and 119874







Begin


Input data samples










Y

N

N

Y

End









119895 119905119897


119908119894119895

119909119895minus 120579119894)


119879119897119894119910119894minus 120579119897)


119897119894(119896) = 119879

119897119894(119896) + 120578120575

119897119910119894


119897(119896)+

120578120575119897





119894119895(119896 + 1) = 119908

119894119895(119896) + 120578

1015840

120579119894

1015840

119909119895


119894(119896) +

1205781015840

120575119894


119897= (119905119897minus 119874119897) sdot

119874119897sdot (1 minus 119874

119897) 120575119894

1015840

= 119910119894(1 minus 119910

119894) sum119897

120575119897119879119897119894










V3= V119862+ V119871+ V119877 (13)

1199101= ℎ + 119897

1tan120572

1199102= ℎ minus (

119897

2

+ 1198972) tan120572

(14)





V119862=

119886

119887

119897 [119902ℎ+ 1198872arc sin(ℎ

119887

minus 1) +

1

2

1205871198872

] (15)


V119878=

120587119886119888

21198872

[1198872


1

3

(ℎ minus 119887)3

+

2

3

1198873

] (16)




ΔV119871minus ΔV119877997888rarr 0 (17)


V3= V119862+ V119878|ℎ=119910

1

+ V119878|ℎ=119910

2

(18)




04 05 06 07 08 09 1 110

01

02

03

04

05

06

07

Oil level

119863-v

alue

of t

heor

etic

al o

il re

serv

e and

actu

aloi

l res

erve



0 5 10 15 20 25 30 350

02

04

06

08

1

12

14

16

18

2

Prop

ortio

nal e

rror

()

times10minus5



V119862=

119886

119887

119897 [119902119910+ 1198872arc sin(

119910119862

119887

minus 1) +

1

2

1205871198872

] (19)


119862(2119887 minus 119910

119862)


119896 =

119910119862minus ℎ3

1198972


RightLeftCenter

1199101

1199102

120572ℎ 119910

1198971 1198972

Δ119871

Δ119877


1205721199101 119910ℎ

1198971 1198972

1199102

998400998400

998400



V119862=

119886

119887


119886

+1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

(21)


3minus 1198972tan120572))



follows

V119878|ℎ=119910

1

=

120587119886119888

21198872

[1198872

(1199101minus 119887) minus

1

3

(1199101minus 119887)3

+

2

3

1198873

]

V119878|ℎ=119910

2

=

120587119886119888

21198872

[1198872

(1199102minus 119887) minus

1

3

(1199102minus 119887)3

+

2

3

1198873

]

(22)


V119871=

120587119886119888

21198872

[1198872

119905119887minus

1

3

1199053

119887

+

2

3

1198873

] (23)

where 119905119887= ℎ3+ 1198971tan120572 minus 119887

V119877=

120587119886119888

21198872

[1198872

119905119897minus

1

3

1199053

119897

+

2

3

1198873

] (24)




where 119905119897is ℎ3minus (1198972 + 119897

2) tan120572 minus 119887

V3= V119862+ V119871+ V119877

=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)


3minus 1198972tan120572))

+ 1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

1199052minus

1

3

1199053

2

+

2

3

1198873

]

+

120587119886119888

21198872

[1198872

1199051minus

1

3

1199053

1

+

2

3

1198873

]

(25)

where 1199051= ℎ3minus (1198972 + 119897

2) tan120572 minus 119887 and 119905

2= ℎ3+ 1198971tan120572 minus 119887


ℎ4= 119903 + 119905

119905 =

119904

cos120573

119904 = ℎ3minus 119903

(26)


ℎ = (ℎ4minus 119903) cos120573 + 119903 (27)


V4=

119886

119887

119897 [ (ℎ minus 1198972tan120572 minus 119887)


2tan120572))


119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

(ℎ + 1198971tan120572 minus 119887)

minus

1

3

(ℎ + 1198971tan120572 minus 119887)3 + 2

3

1198873

]

+

120587119886119888

21198872

[1198872

119903119886minus

1

3

1199033

119886

+

2

3

1198873

] (28)










V4(151073 120572 120573) minus V

4(150765 120572 120573) = 8676

V4(150765 120572 120573) minus V

4(150106 120572 120573) = 18761

(29)


120572 = 235920

120573 = 3801270

(30)









119904

119905

ℎ3119903

ℎ4120573


0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()





4 Conclusions






Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra













V1(ℎ1)=119897 (

119886119887

2

120587minus119886radic1199041+

119886

119887


119887

))

(3)






1198772


1

ΔV1= 13493ℎ

1minus 12031 (4)


V1= 119897 (

119886119887

2


119886

119887

ℎ1


+119886119887 arc sin(minus1 + ℎ1119887

)) minus ΔV1

(5)






020406080

100120140160

0 20 40 60 80 100 120 140

Erro

r

Oil height

Error fitting curve


119910 = 13493119909 minus 12031

1198772 = 09967



Oil floater

Oil2119887


11989721198971





2lt 1198972tan120572


2tan120572 le ℎ

2lt

2119887 minus 1198971tan120572


1tan120572 le ℎ

2le 2119887



119896 =

ℎ2minus 119887

1198971

= minus tan120572 (6)



119887 = ℎ2+ 1198971tan120572 (7)


119910 = (1198971minus 119909) tan120572 + ℎ

2 (8)


119889119909 = minuscot120572119889119910 (9)


V2= int

245

0

119860 (119909) 119889119909 (10)

that is

V2=int

119897

0

(

119886119887

2

120587minus119886radic1199011+

119886

119887


119910

119887

))119889119909

(11)



1minus

119909) tan120572 + ℎ2




1198812=

10

minus3

int

ℎ2+1198971 tan120572

0

cot120572[1198861198872


119886

119887


119887

)] 119889119910 0 le ℎ2lt 1198972tan120572

10

minus3

int

ℎ2+1198971 tan120572

ℎ2minus1198972 tan120572cot120572[1198861198872


119886

119887


119887

)] 119889119910 1198972tan120572 le ℎ

2lt 2119887 minus 119897

1tan120572

V1(2119887) minus 10

minus3

int

ℎ2minus1198972 tan120572

0

cot120572[1198861198872


119886

119887


119887

)] 119889119910 2119887 minus 1198971tan120572 le ℎ

2le 2119887

(12)




AB

C D


AB

C D


AB

C D



B

ℎ2

1198971 1198972119900 119888

120572 A

119883

119884

119887


050

100150200250300350400

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

113

120

Erro

r val

ue

Oil height



minus50




1199091

1199092

119909119895

Input layer

1198821198941198951205791

1205791

12057921205792

120579119894

120579119894

Σ

Σ

Σ

Σ

Σ

Σ


1199051

1199052

119905119897

purelin

purelin

purelin

119879119897119894

tan sig

tan sig

tan sig






119895 119910119894and 119874







Begin


Input data samples










Y

N

N

Y

End









119895 119905119897


119908119894119895

119909119895minus 120579119894)


119879119897119894119910119894minus 120579119897)


119897119894(119896) = 119879

119897119894(119896) + 120578120575

119897119910119894


119897(119896)+

120578120575119897





119894119895(119896 + 1) = 119908

119894119895(119896) + 120578

1015840

120579119894

1015840

119909119895


119894(119896) +

1205781015840

120575119894


119897= (119905119897minus 119874119897) sdot

119874119897sdot (1 minus 119874

119897) 120575119894

1015840

= 119910119894(1 minus 119910

119894) sum119897

120575119897119879119897119894










V3= V119862+ V119871+ V119877 (13)

1199101= ℎ + 119897

1tan120572

1199102= ℎ minus (

119897

2

+ 1198972) tan120572

(14)





V119862=

119886

119887

119897 [119902ℎ+ 1198872arc sin(ℎ

119887

minus 1) +

1

2

1205871198872

] (15)


V119878=

120587119886119888

21198872

[1198872


1

3

(ℎ minus 119887)3

+

2

3

1198873

] (16)




ΔV119871minus ΔV119877997888rarr 0 (17)


V3= V119862+ V119878|ℎ=119910

1

+ V119878|ℎ=119910

2

(18)




04 05 06 07 08 09 1 110

01

02

03

04

05

06

07

Oil level

119863-v

alue

of t

heor

etic

al o

il re

serv

e and

actu

aloi

l res

erve



0 5 10 15 20 25 30 350

02

04

06

08

1

12

14

16

18

2

Prop

ortio

nal e

rror

()

times10minus5



V119862=

119886

119887

119897 [119902119910+ 1198872arc sin(

119910119862

119887

minus 1) +

1

2

1205871198872

] (19)


119862(2119887 minus 119910

119862)


119896 =

119910119862minus ℎ3

1198972


RightLeftCenter

1199101

1199102

120572ℎ 119910

1198971 1198972

Δ119871

Δ119877


1205721199101 119910ℎ

1198971 1198972

1199102

998400998400

998400



V119862=

119886

119887


119886

+1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

(21)


3minus 1198972tan120572))



follows

V119878|ℎ=119910

1

=

120587119886119888

21198872

[1198872

(1199101minus 119887) minus

1

3

(1199101minus 119887)3

+

2

3

1198873

]

V119878|ℎ=119910

2

=

120587119886119888

21198872

[1198872

(1199102minus 119887) minus

1

3

(1199102minus 119887)3

+

2

3

1198873

]

(22)


V119871=

120587119886119888

21198872

[1198872

119905119887minus

1

3

1199053

119887

+

2

3

1198873

] (23)

where 119905119887= ℎ3+ 1198971tan120572 minus 119887

V119877=

120587119886119888

21198872

[1198872

119905119897minus

1

3

1199053

119897

+

2

3

1198873

] (24)




where 119905119897is ℎ3minus (1198972 + 119897

2) tan120572 minus 119887

V3= V119862+ V119871+ V119877

=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)


3minus 1198972tan120572))

+ 1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

1199052minus

1

3

1199053

2

+

2

3

1198873

]

+

120587119886119888

21198872

[1198872

1199051minus

1

3

1199053

1

+

2

3

1198873

]

(25)

where 1199051= ℎ3minus (1198972 + 119897

2) tan120572 minus 119887 and 119905

2= ℎ3+ 1198971tan120572 minus 119887


ℎ4= 119903 + 119905

119905 =

119904

cos120573

119904 = ℎ3minus 119903

(26)


ℎ = (ℎ4minus 119903) cos120573 + 119903 (27)


V4=

119886

119887

119897 [ (ℎ minus 1198972tan120572 minus 119887)


2tan120572))


119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

(ℎ + 1198971tan120572 minus 119887)

minus

1

3

(ℎ + 1198971tan120572 minus 119887)3 + 2

3

1198873

]

+

120587119886119888

21198872

[1198872

119903119886minus

1

3

1199033

119886

+

2

3

1198873

] (28)










V4(151073 120572 120573) minus V

4(150765 120572 120573) = 8676

V4(150765 120572 120573) minus V

4(150106 120572 120573) = 18761

(29)


120572 = 235920

120573 = 3801270

(30)









119904

119905

ℎ3119903

ℎ4120573


0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()





4 Conclusions






Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










020406080

100120140160

0 20 40 60 80 100 120 140

Erro

r

Oil height

Error fitting curve


119910 = 13493119909 minus 12031

1198772 = 09967



Oil floater

Oil2119887


11989721198971





2lt 1198972tan120572


2tan120572 le ℎ

2lt

2119887 minus 1198971tan120572


1tan120572 le ℎ

2le 2119887



119896 =

ℎ2minus 119887

1198971

= minus tan120572 (6)



119887 = ℎ2+ 1198971tan120572 (7)


119910 = (1198971minus 119909) tan120572 + ℎ

2 (8)


119889119909 = minuscot120572119889119910 (9)


V2= int

245

0

119860 (119909) 119889119909 (10)

that is

V2=int

119897

0

(

119886119887

2

120587minus119886radic1199011+

119886

119887


119910

119887

))119889119909

(11)



1minus

119909) tan120572 + ℎ2




1198812=

10

minus3

int

ℎ2+1198971 tan120572

0

cot120572[1198861198872


119886

119887


119887

)] 119889119910 0 le ℎ2lt 1198972tan120572

10

minus3

int

ℎ2+1198971 tan120572

ℎ2minus1198972 tan120572cot120572[1198861198872


119886

119887


119887

)] 119889119910 1198972tan120572 le ℎ

2lt 2119887 minus 119897

1tan120572

V1(2119887) minus 10

minus3

int

ℎ2minus1198972 tan120572

0

cot120572[1198861198872


119886

119887


119887

)] 119889119910 2119887 minus 1198971tan120572 le ℎ

2le 2119887

(12)




AB

C D


AB

C D


AB

C D



B

ℎ2

1198971 1198972119900 119888

120572 A

119883

119884

119887


050

100150200250300350400

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

113

120

Erro

r val

ue

Oil height



minus50




1199091

1199092

119909119895

Input layer

1198821198941198951205791

1205791

12057921205792

120579119894

120579119894

Σ

Σ

Σ

Σ

Σ

Σ


1199051

1199052

119905119897

purelin

purelin

purelin

119879119897119894

tan sig

tan sig

tan sig






119895 119910119894and 119874







Begin


Input data samples










Y

N

N

Y

End









119895 119905119897


119908119894119895

119909119895minus 120579119894)


119879119897119894119910119894minus 120579119897)


119897119894(119896) = 119879

119897119894(119896) + 120578120575

119897119910119894


119897(119896)+

120578120575119897





119894119895(119896 + 1) = 119908

119894119895(119896) + 120578

1015840

120579119894

1015840

119909119895


119894(119896) +

1205781015840

120575119894


119897= (119905119897minus 119874119897) sdot

119874119897sdot (1 minus 119874

119897) 120575119894

1015840

= 119910119894(1 minus 119910

119894) sum119897

120575119897119879119897119894










V3= V119862+ V119871+ V119877 (13)

1199101= ℎ + 119897

1tan120572

1199102= ℎ minus (

119897

2

+ 1198972) tan120572

(14)





V119862=

119886

119887

119897 [119902ℎ+ 1198872arc sin(ℎ

119887

minus 1) +

1

2

1205871198872

] (15)


V119878=

120587119886119888

21198872

[1198872


1

3

(ℎ minus 119887)3

+

2

3

1198873

] (16)




ΔV119871minus ΔV119877997888rarr 0 (17)


V3= V119862+ V119878|ℎ=119910

1

+ V119878|ℎ=119910

2

(18)




04 05 06 07 08 09 1 110

01

02

03

04

05

06

07

Oil level

119863-v

alue

of t

heor

etic

al o

il re

serv

e and

actu

aloi

l res

erve



0 5 10 15 20 25 30 350

02

04

06

08

1

12

14

16

18

2

Prop

ortio

nal e

rror

()

times10minus5



V119862=

119886

119887

119897 [119902119910+ 1198872arc sin(

119910119862

119887

minus 1) +

1

2

1205871198872

] (19)


119862(2119887 minus 119910

119862)


119896 =

119910119862minus ℎ3

1198972


RightLeftCenter

1199101

1199102

120572ℎ 119910

1198971 1198972

Δ119871

Δ119877


1205721199101 119910ℎ

1198971 1198972

1199102

998400998400

998400



V119862=

119886

119887


119886

+1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

(21)


3minus 1198972tan120572))



follows

V119878|ℎ=119910

1

=

120587119886119888

21198872

[1198872

(1199101minus 119887) minus

1

3

(1199101minus 119887)3

+

2

3

1198873

]

V119878|ℎ=119910

2

=

120587119886119888

21198872

[1198872

(1199102minus 119887) minus

1

3

(1199102minus 119887)3

+

2

3

1198873

]

(22)


V119871=

120587119886119888

21198872

[1198872

119905119887minus

1

3

1199053

119887

+

2

3

1198873

] (23)

where 119905119887= ℎ3+ 1198971tan120572 minus 119887

V119877=

120587119886119888

21198872

[1198872

119905119897minus

1

3

1199053

119897

+

2

3

1198873

] (24)




where 119905119897is ℎ3minus (1198972 + 119897

2) tan120572 minus 119887

V3= V119862+ V119871+ V119877

=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)


3minus 1198972tan120572))

+ 1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

1199052minus

1

3

1199053

2

+

2

3

1198873

]

+

120587119886119888

21198872

[1198872

1199051minus

1

3

1199053

1

+

2

3

1198873

]

(25)

where 1199051= ℎ3minus (1198972 + 119897

2) tan120572 minus 119887 and 119905

2= ℎ3+ 1198971tan120572 minus 119887


ℎ4= 119903 + 119905

119905 =

119904

cos120573

119904 = ℎ3minus 119903

(26)


ℎ = (ℎ4minus 119903) cos120573 + 119903 (27)


V4=

119886

119887

119897 [ (ℎ minus 1198972tan120572 minus 119887)


2tan120572))


119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

(ℎ + 1198971tan120572 minus 119887)

minus

1

3

(ℎ + 1198971tan120572 minus 119887)3 + 2

3

1198873

]

+

120587119886119888

21198872

[1198872

119903119886minus

1

3

1199033

119886

+

2

3

1198873

] (28)










V4(151073 120572 120573) minus V

4(150765 120572 120573) = 8676

V4(150765 120572 120573) minus V

4(150106 120572 120573) = 18761

(29)


120572 = 235920

120573 = 3801270

(30)









119904

119905

ℎ3119903

ℎ4120573


0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()





4 Conclusions






Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










AB

C D


AB

C D


AB

C D



B

ℎ2

1198971 1198972119900 119888

120572 A

119883

119884

119887


050

100150200250300350400

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

113

120

Erro

r val

ue

Oil height



minus50




1199091

1199092

119909119895

Input layer

1198821198941198951205791

1205791

12057921205792

120579119894

120579119894

Σ

Σ

Σ

Σ

Σ

Σ


1199051

1199052

119905119897

purelin

purelin

purelin

119879119897119894

tan sig

tan sig

tan sig






119895 119910119894and 119874







Begin


Input data samples










Y

N

N

Y

End









119895 119905119897


119908119894119895

119909119895minus 120579119894)


119879119897119894119910119894minus 120579119897)


119897119894(119896) = 119879

119897119894(119896) + 120578120575

119897119910119894


119897(119896)+

120578120575119897





119894119895(119896 + 1) = 119908

119894119895(119896) + 120578

1015840

120579119894

1015840

119909119895


119894(119896) +

1205781015840

120575119894


119897= (119905119897minus 119874119897) sdot

119874119897sdot (1 minus 119874

119897) 120575119894

1015840

= 119910119894(1 minus 119910

119894) sum119897

120575119897119879119897119894










V3= V119862+ V119871+ V119877 (13)

1199101= ℎ + 119897

1tan120572

1199102= ℎ minus (

119897

2

+ 1198972) tan120572

(14)





V119862=

119886

119887

119897 [119902ℎ+ 1198872arc sin(ℎ

119887

minus 1) +

1

2

1205871198872

] (15)


V119878=

120587119886119888

21198872

[1198872


1

3

(ℎ minus 119887)3

+

2

3

1198873

] (16)




ΔV119871minus ΔV119877997888rarr 0 (17)


V3= V119862+ V119878|ℎ=119910

1

+ V119878|ℎ=119910

2

(18)




04 05 06 07 08 09 1 110

01

02

03

04

05

06

07

Oil level

119863-v

alue

of t

heor

etic

al o

il re

serv

e and

actu

aloi

l res

erve



0 5 10 15 20 25 30 350

02

04

06

08

1

12

14

16

18

2

Prop

ortio

nal e

rror

()

times10minus5



V119862=

119886

119887

119897 [119902119910+ 1198872arc sin(

119910119862

119887

minus 1) +

1

2

1205871198872

] (19)


119862(2119887 minus 119910

119862)


119896 =

119910119862minus ℎ3

1198972


RightLeftCenter

1199101

1199102

120572ℎ 119910

1198971 1198972

Δ119871

Δ119877


1205721199101 119910ℎ

1198971 1198972

1199102

998400998400

998400



V119862=

119886

119887


119886

+1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

(21)


3minus 1198972tan120572))



follows

V119878|ℎ=119910

1

=

120587119886119888

21198872

[1198872

(1199101minus 119887) minus

1

3

(1199101minus 119887)3

+

2

3

1198873

]

V119878|ℎ=119910

2

=

120587119886119888

21198872

[1198872

(1199102minus 119887) minus

1

3

(1199102minus 119887)3

+

2

3

1198873

]

(22)


V119871=

120587119886119888

21198872

[1198872

119905119887minus

1

3

1199053

119887

+

2

3

1198873

] (23)

where 119905119887= ℎ3+ 1198971tan120572 minus 119887

V119877=

120587119886119888

21198872

[1198872

119905119897minus

1

3

1199053

119897

+

2

3

1198873

] (24)




where 119905119897is ℎ3minus (1198972 + 119897

2) tan120572 minus 119887

V3= V119862+ V119871+ V119877

=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)


3minus 1198972tan120572))

+ 1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

1199052minus

1

3

1199053

2

+

2

3

1198873

]

+

120587119886119888

21198872

[1198872

1199051minus

1

3

1199053

1

+

2

3

1198873

]

(25)

where 1199051= ℎ3minus (1198972 + 119897

2) tan120572 minus 119887 and 119905

2= ℎ3+ 1198971tan120572 minus 119887


ℎ4= 119903 + 119905

119905 =

119904

cos120573

119904 = ℎ3minus 119903

(26)


ℎ = (ℎ4minus 119903) cos120573 + 119903 (27)


V4=

119886

119887

119897 [ (ℎ minus 1198972tan120572 minus 119887)


2tan120572))


119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

(ℎ + 1198971tan120572 minus 119887)

minus

1

3

(ℎ + 1198971tan120572 minus 119887)3 + 2

3

1198873

]

+

120587119886119888

21198872

[1198872

119903119886minus

1

3

1199033

119886

+

2

3

1198873

] (28)










V4(151073 120572 120573) minus V

4(150765 120572 120573) = 8676

V4(150765 120572 120573) minus V

4(150106 120572 120573) = 18761

(29)


120572 = 235920

120573 = 3801270

(30)









119904

119905

ℎ3119903

ℎ4120573


0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()





4 Conclusions






Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Begin


Input data samples










Y

N

N

Y

End









119895 119905119897


119908119894119895

119909119895minus 120579119894)


119879119897119894119910119894minus 120579119897)


119897119894(119896) = 119879

119897119894(119896) + 120578120575

119897119910119894


119897(119896)+

120578120575119897





119894119895(119896 + 1) = 119908

119894119895(119896) + 120578

1015840

120579119894

1015840

119909119895


119894(119896) +

1205781015840

120575119894


119897= (119905119897minus 119874119897) sdot

119874119897sdot (1 minus 119874

119897) 120575119894

1015840

= 119910119894(1 minus 119910

119894) sum119897

120575119897119879119897119894










V3= V119862+ V119871+ V119877 (13)

1199101= ℎ + 119897

1tan120572

1199102= ℎ minus (

119897

2

+ 1198972) tan120572

(14)





V119862=

119886

119887

119897 [119902ℎ+ 1198872arc sin(ℎ

119887

minus 1) +

1

2

1205871198872

] (15)


V119878=

120587119886119888

21198872

[1198872


1

3

(ℎ minus 119887)3

+

2

3

1198873

] (16)




ΔV119871minus ΔV119877997888rarr 0 (17)


V3= V119862+ V119878|ℎ=119910

1

+ V119878|ℎ=119910

2

(18)




04 05 06 07 08 09 1 110

01

02

03

04

05

06

07

Oil level

119863-v

alue

of t

heor

etic

al o

il re

serv

e and

actu

aloi

l res

erve



0 5 10 15 20 25 30 350

02

04

06

08

1

12

14

16

18

2

Prop

ortio

nal e

rror

()

times10minus5



V119862=

119886

119887

119897 [119902119910+ 1198872arc sin(

119910119862

119887

minus 1) +

1

2

1205871198872

] (19)


119862(2119887 minus 119910

119862)


119896 =

119910119862minus ℎ3

1198972


RightLeftCenter

1199101

1199102

120572ℎ 119910

1198971 1198972

Δ119871

Δ119877


1205721199101 119910ℎ

1198971 1198972

1199102

998400998400

998400



V119862=

119886

119887


119886

+1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

(21)


3minus 1198972tan120572))



follows

V119878|ℎ=119910

1

=

120587119886119888

21198872

[1198872

(1199101minus 119887) minus

1

3

(1199101minus 119887)3

+

2

3

1198873

]

V119878|ℎ=119910

2

=

120587119886119888

21198872

[1198872

(1199102minus 119887) minus

1

3

(1199102minus 119887)3

+

2

3

1198873

]

(22)


V119871=

120587119886119888

21198872

[1198872

119905119887minus

1

3

1199053

119887

+

2

3

1198873

] (23)

where 119905119887= ℎ3+ 1198971tan120572 minus 119887

V119877=

120587119886119888

21198872

[1198872

119905119897minus

1

3

1199053

119897

+

2

3

1198873

] (24)




where 119905119897is ℎ3minus (1198972 + 119897

2) tan120572 minus 119887

V3= V119862+ V119871+ V119877

=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)


3minus 1198972tan120572))

+ 1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

1199052minus

1

3

1199053

2

+

2

3

1198873

]

+

120587119886119888

21198872

[1198872

1199051minus

1

3

1199053

1

+

2

3

1198873

]

(25)

where 1199051= ℎ3minus (1198972 + 119897

2) tan120572 minus 119887 and 119905

2= ℎ3+ 1198971tan120572 minus 119887


ℎ4= 119903 + 119905

119905 =

119904

cos120573

119904 = ℎ3minus 119903

(26)


ℎ = (ℎ4minus 119903) cos120573 + 119903 (27)


V4=

119886

119887

119897 [ (ℎ minus 1198972tan120572 minus 119887)


2tan120572))


119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

(ℎ + 1198971tan120572 minus 119887)

minus

1

3

(ℎ + 1198971tan120572 minus 119887)3 + 2

3

1198873

]

+

120587119886119888

21198872

[1198872

119903119886minus

1

3

1199033

119886

+

2

3

1198873

] (28)










V4(151073 120572 120573) minus V

4(150765 120572 120573) = 8676

V4(150765 120572 120573) minus V

4(150106 120572 120573) = 18761

(29)


120572 = 235920

120573 = 3801270

(30)









119904

119905

ℎ3119903

ℎ4120573


0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()





4 Conclusions






Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra













119894119895(119896 + 1) = 119908

119894119895(119896) + 120578

1015840

120579119894

1015840

119909119895


119894(119896) +

1205781015840

120575119894


119897= (119905119897minus 119874119897) sdot

119874119897sdot (1 minus 119874

119897) 120575119894

1015840

= 119910119894(1 minus 119910

119894) sum119897

120575119897119879119897119894










V3= V119862+ V119871+ V119877 (13)

1199101= ℎ + 119897

1tan120572

1199102= ℎ minus (

119897

2

+ 1198972) tan120572

(14)





V119862=

119886

119887

119897 [119902ℎ+ 1198872arc sin(ℎ

119887

minus 1) +

1

2

1205871198872

] (15)


V119878=

120587119886119888

21198872

[1198872


1

3

(ℎ minus 119887)3

+

2

3

1198873

] (16)




ΔV119871minus ΔV119877997888rarr 0 (17)


V3= V119862+ V119878|ℎ=119910

1

+ V119878|ℎ=119910

2

(18)




04 05 06 07 08 09 1 110

01

02

03

04

05

06

07

Oil level

119863-v

alue

of t

heor

etic

al o

il re

serv

e and

actu

aloi

l res

erve



0 5 10 15 20 25 30 350

02

04

06

08

1

12

14

16

18

2

Prop

ortio

nal e

rror

()

times10minus5



V119862=

119886

119887

119897 [119902119910+ 1198872arc sin(

119910119862

119887

minus 1) +

1

2

1205871198872

] (19)


119862(2119887 minus 119910

119862)


119896 =

119910119862minus ℎ3

1198972


RightLeftCenter

1199101

1199102

120572ℎ 119910

1198971 1198972

Δ119871

Δ119877


1205721199101 119910ℎ

1198971 1198972

1199102

998400998400

998400



V119862=

119886

119887


119886

+1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

(21)


3minus 1198972tan120572))



follows

V119878|ℎ=119910

1

=

120587119886119888

21198872

[1198872

(1199101minus 119887) minus

1

3

(1199101minus 119887)3

+

2

3

1198873

]

V119878|ℎ=119910

2

=

120587119886119888

21198872

[1198872

(1199102minus 119887) minus

1

3

(1199102minus 119887)3

+

2

3

1198873

]

(22)


V119871=

120587119886119888

21198872

[1198872

119905119887minus

1

3

1199053

119887

+

2

3

1198873

] (23)

where 119905119887= ℎ3+ 1198971tan120572 minus 119887

V119877=

120587119886119888

21198872

[1198872

119905119897minus

1

3

1199053

119897

+

2

3

1198873

] (24)




where 119905119897is ℎ3minus (1198972 + 119897

2) tan120572 minus 119887

V3= V119862+ V119871+ V119877

=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)


3minus 1198972tan120572))

+ 1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

1199052minus

1

3

1199053

2

+

2

3

1198873

]

+

120587119886119888

21198872

[1198872

1199051minus

1

3

1199053

1

+

2

3

1198873

]

(25)

where 1199051= ℎ3minus (1198972 + 119897

2) tan120572 minus 119887 and 119905

2= ℎ3+ 1198971tan120572 minus 119887


ℎ4= 119903 + 119905

119905 =

119904

cos120573

119904 = ℎ3minus 119903

(26)


ℎ = (ℎ4minus 119903) cos120573 + 119903 (27)


V4=

119886

119887

119897 [ (ℎ minus 1198972tan120572 minus 119887)


2tan120572))


119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

(ℎ + 1198971tan120572 minus 119887)

minus

1

3

(ℎ + 1198971tan120572 minus 119887)3 + 2

3

1198873

]

+

120587119886119888

21198872

[1198872

119903119886minus

1

3

1199033

119886

+

2

3

1198873

] (28)










V4(151073 120572 120573) minus V

4(150765 120572 120573) = 8676

V4(150765 120572 120573) minus V

4(150106 120572 120573) = 18761

(29)


120572 = 235920

120573 = 3801270

(30)









119904

119905

ℎ3119903

ℎ4120573


0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()





4 Conclusions






Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra













V119862=

119886

119887

119897 [119902ℎ+ 1198872arc sin(ℎ

119887

minus 1) +

1

2

1205871198872

] (15)


V119878=

120587119886119888

21198872

[1198872


1

3

(ℎ minus 119887)3

+

2

3

1198873

] (16)




ΔV119871minus ΔV119877997888rarr 0 (17)


V3= V119862+ V119878|ℎ=119910

1

+ V119878|ℎ=119910

2

(18)




04 05 06 07 08 09 1 110

01

02

03

04

05

06

07

Oil level

119863-v

alue

of t

heor

etic

al o

il re

serv

e and

actu

aloi

l res

erve



0 5 10 15 20 25 30 350

02

04

06

08

1

12

14

16

18

2

Prop

ortio

nal e

rror

()

times10minus5



V119862=

119886

119887

119897 [119902119910+ 1198872arc sin(

119910119862

119887

minus 1) +

1

2

1205871198872

] (19)


119862(2119887 minus 119910

119862)


119896 =

119910119862minus ℎ3

1198972


RightLeftCenter

1199101

1199102

120572ℎ 119910

1198971 1198972

Δ119871

Δ119877


1205721199101 119910ℎ

1198971 1198972

1199102

998400998400

998400



V119862=

119886

119887


119886

+1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

(21)


3minus 1198972tan120572))



follows

V119878|ℎ=119910

1

=

120587119886119888

21198872

[1198872

(1199101minus 119887) minus

1

3

(1199101minus 119887)3

+

2

3

1198873

]

V119878|ℎ=119910

2

=

120587119886119888

21198872

[1198872

(1199102minus 119887) minus

1

3

(1199102minus 119887)3

+

2

3

1198873

]

(22)


V119871=

120587119886119888

21198872

[1198872

119905119887minus

1

3

1199053

119887

+

2

3

1198873

] (23)

where 119905119887= ℎ3+ 1198971tan120572 minus 119887

V119877=

120587119886119888

21198872

[1198872

119905119897minus

1

3

1199053

119897

+

2

3

1198873

] (24)




where 119905119897is ℎ3minus (1198972 + 119897

2) tan120572 minus 119887

V3= V119862+ V119871+ V119877

=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)


3minus 1198972tan120572))

+ 1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

1199052minus

1

3

1199053

2

+

2

3

1198873

]

+

120587119886119888

21198872

[1198872

1199051minus

1

3

1199053

1

+

2

3

1198873

]

(25)

where 1199051= ℎ3minus (1198972 + 119897

2) tan120572 minus 119887 and 119905

2= ℎ3+ 1198971tan120572 minus 119887


ℎ4= 119903 + 119905

119905 =

119904

cos120573

119904 = ℎ3minus 119903

(26)


ℎ = (ℎ4minus 119903) cos120573 + 119903 (27)


V4=

119886

119887

119897 [ (ℎ minus 1198972tan120572 minus 119887)


2tan120572))


119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

(ℎ + 1198971tan120572 minus 119887)

minus

1

3

(ℎ + 1198971tan120572 minus 119887)3 + 2

3

1198873

]

+

120587119886119888

21198872

[1198872

119903119886minus

1

3

1199033

119886

+

2

3

1198873

] (28)










V4(151073 120572 120573) minus V

4(150765 120572 120573) = 8676

V4(150765 120572 120573) minus V

4(150106 120572 120573) = 18761

(29)


120572 = 235920

120573 = 3801270

(30)









119904

119905

ℎ3119903

ℎ4120573


0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()





4 Conclusions






Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










04 05 06 07 08 09 1 110

01

02

03

04

05

06

07

Oil level

119863-v

alue

of t

heor

etic

al o

il re

serv

e and

actu

aloi

l res

erve



0 5 10 15 20 25 30 350

02

04

06

08

1

12

14

16

18

2

Prop

ortio

nal e

rror

()

times10minus5



V119862=

119886

119887

119897 [119902119910+ 1198872arc sin(

119910119862

119887

minus 1) +

1

2

1205871198872

] (19)


119862(2119887 minus 119910

119862)


119896 =

119910119862minus ℎ3

1198972


RightLeftCenter

1199101

1199102

120572ℎ 119910

1198971 1198972

Δ119871

Δ119877


1205721199101 119910ℎ

1198971 1198972

1199102

998400998400

998400



V119862=

119886

119887


119886

+1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

(21)


3minus 1198972tan120572))



follows

V119878|ℎ=119910

1

=

120587119886119888

21198872

[1198872

(1199101minus 119887) minus

1

3

(1199101minus 119887)3

+

2

3

1198873

]

V119878|ℎ=119910

2

=

120587119886119888

21198872

[1198872

(1199102minus 119887) minus

1

3

(1199102minus 119887)3

+

2

3

1198873

]

(22)


V119871=

120587119886119888

21198872

[1198872

119905119887minus

1

3

1199053

119887

+

2

3

1198873

] (23)

where 119905119887= ℎ3+ 1198971tan120572 minus 119887

V119877=

120587119886119888

21198872

[1198872

119905119897minus

1

3

1199053

119897

+

2

3

1198873

] (24)




where 119905119897is ℎ3minus (1198972 + 119897

2) tan120572 minus 119887

V3= V119862+ V119871+ V119877

=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)


3minus 1198972tan120572))

+ 1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

1199052minus

1

3

1199053

2

+

2

3

1198873

]

+

120587119886119888

21198872

[1198872

1199051minus

1

3

1199053

1

+

2

3

1198873

]

(25)

where 1199051= ℎ3minus (1198972 + 119897

2) tan120572 minus 119887 and 119905

2= ℎ3+ 1198971tan120572 minus 119887


ℎ4= 119903 + 119905

119905 =

119904

cos120573

119904 = ℎ3minus 119903

(26)


ℎ = (ℎ4minus 119903) cos120573 + 119903 (27)


V4=

119886

119887

119897 [ (ℎ minus 1198972tan120572 minus 119887)


2tan120572))


119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

(ℎ + 1198971tan120572 minus 119887)

minus

1

3

(ℎ + 1198971tan120572 minus 119887)3 + 2

3

1198873

]

+

120587119886119888

21198872

[1198872

119903119886minus

1

3

1199033

119886

+

2

3

1198873

] (28)










V4(151073 120572 120573) minus V

4(150765 120572 120573) = 8676

V4(150765 120572 120573) minus V

4(150106 120572 120573) = 18761

(29)


120572 = 235920

120573 = 3801270

(30)









119904

119905

ℎ3119903

ℎ4120573


0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()





4 Conclusions






Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












where 119905119897is ℎ3minus (1198972 + 119897

2) tan120572 minus 119887

V3= V119862+ V119871+ V119877

=

119886

119887

119897 [ (ℎ3minus 1198972tan120572 minus 119887)


3minus 1198972tan120572))

+ 1198872arc sin(

ℎ3minus 1198972tan120572119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

1199052minus

1

3

1199053

2

+

2

3

1198873

]

+

120587119886119888

21198872

[1198872

1199051minus

1

3

1199053

1

+

2

3

1198873

]

(25)

where 1199051= ℎ3minus (1198972 + 119897

2) tan120572 minus 119887 and 119905

2= ℎ3+ 1198971tan120572 minus 119887


ℎ4= 119903 + 119905

119905 =

119904

cos120573

119904 = ℎ3minus 119903

(26)


ℎ = (ℎ4minus 119903) cos120573 + 119903 (27)


V4=

119886

119887

119897 [ (ℎ minus 1198972tan120572 minus 119887)


2tan120572))


119887

minus 1) +

1

2

1205871198872

]

+

120587119886119888

21198872

[1198872

(ℎ + 1198971tan120572 minus 119887)

minus

1

3

(ℎ + 1198971tan120572 minus 119887)3 + 2

3

1198873

]

+

120587119886119888

21198872

[1198872

119903119886minus

1

3

1199033

119886

+

2

3

1198873

] (28)










V4(151073 120572 120573) minus V

4(150765 120572 120573) = 8676

V4(150765 120572 120573) minus V

4(150106 120572 120573) = 18761

(29)


120572 = 235920

120573 = 3801270

(30)









119904

119905

ℎ3119903

ℎ4120573


0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()





4 Conclusions






Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra














119904

119905

ℎ3119903

ℎ4120573


0 5 10 15 20 25 30 350

05

1

15

2

25

3

35times10minus9

Prop

ortio

nal e

rror

()





4 Conclusions






Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra













Acknowledgments


References













































Volume 2014




Journal of











Journal of


Function Spaces






Algebra



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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Research Article Mathematical Model Based on BP Neural … · 2019. 7. 31. · storage tank and calibration of storage tank chart. Finally, some concluding remarks are drawn in Section