CN102011576B - Method for hierarchically diagnosing fault of rod-comprising oil pumping system based on indicator diagram - Google Patents

Abstract

The invention discloses a method for hierarchically diagnosing a fault of a rod-comprising oil pumping system based on an indicator diagram. The method comprises two stages, namely a fault distinguishing stage and a fault recognizing stage, and comprises the following steps of: dividing the indicator diagram into a fault class and a non-fault class according to a statistical law of the normal sample at the fault distinguishing stage; and recognizing a fault type of a fault sample by a search tree method for recognizing the fault of the rod-comprising oil pumping system based on the indicator diagram. By the method, the fault diagnosis performance of a fault diagnosis system is improved, a mechanical model of the rod-comprising oil pumping system does not need establishing or resolving, the problem of a training set is solved, and the basic features of the rod-comprising oil pumping system can be reflected.

Description

Rod pumping system fault based on indicator card is passed the rank diagnostic method

Technical field

The present invention relates to a kind of rod pumping system method for diagnosing faults based on indicator card, relate in particular to a kind of rod pumping system fault based on indicator card and pass the rank diagnostic method, belong to the technical field of rod pumping system method for diagnosing faults.

Background technology

The rod pumping system work under bad environment, the probability broken down is very high, is diagnosed not in time, will cause the waste of the energy, and impact is produced, and causes damage to enterprise, country.

The fault diagnosis of rod pumping system is mainly first according to surface dynamometer card at present, obtain pump merit figure by the rod pumping system mechanical model, using again normal sample and as far as possible the pump merit figure of polymorphic type fault sample as training set, again each sample is carried out to fault diagnosis after training corresponding model.Mainly there is the problem of two aspects in these class methods: the one, and rod pumping system is the Complex Nonlinear System of a mechanical electronic hydraulic coupling, fringe conditions and damped coefficient are difficult to accurately determine, and then are difficult to accurately set up and solve the mechanical model of this system; The 2nd, rod pumping system has that sample size is large, plateau is many, the unbalanced characteristics of classification, make in training set to lack very eurypalynous fault sample, especially the sample of catastrophe failure, and catastrophe failure accurately, real-time diagnosis is significant for rod pumping system.Due to the complexity of rod pumping system, go back so far the fully effective means of neither one to the down-hole fault carry out in real time, fault diagnosis accurately.

Reflection oil pumping system polished rod load is called surface dynamometer card or polished rod indicator card with the figure of its change in displacement rule, is called for short indicator card, and it is the closed curve that indicator measures in suction period of oil pumper.Surface dynamometer card is the firsthand information of pumpingh well oil recovery collection in worksite, realizes that based on surface dynamometer card the real-time diagnosis of rod pumping system is to removing oil well failure, guaranteeing that the normal production of oil well or raising oil well output etc. just seem extremely important and have realistic meaning.

Every group of indicator card data comprise load and displacement, totally 216 pairs of data.Oil field gathers an indicator card in one hour, and 24 groups of data are just arranged in one day, and 168 groups of data are arranged in a week, and within one month, (within 30 days, calculating) has 720 groups of data.So oil well indicator card data are many, sample size is large.And, for oil field, the overwhelming majority is duty stably, comprises two kinds of situations: the one, oil well is normal, and most of indicator card is all normal, and fault sample and fault type are less; The 2nd, some oil wells are with certain fault, as gassiness, shake out, feed flow is not enough, slightly touch certain class faults not too large to Influence of production such as extension, vibration effect is main, occurs that the probability of other fault is lower.Above both of these case does not need to stop production and overhauled immediately, can, when next well workover or in convenient and processing in case of necessity, that is to say that a large amount of oil well indicator card data are all not need to carry out immediately troubleshooting.Once and occur all needing the catastrophe failures such as valve is malfunctioning, rod parting, pump seizure to stop production immediately, take in time to examine pump and well workover measure, to reduce the loss.Oil well well workover measure, particularly to workover or more exchange device can production status and the merit figure of oil well be exerted an influence, the maintenance before and after production status and merit figure often have bigger difference, the training set after maintenance and sample characteristics often need again to form.In addition, in existing oil recovery industry, oil pumper quantity is large, distribution is wide, and the complexity that this has more increased downhole conditions, make each mouthful of well that own characteristic all be arranged, and the fault type that may occur is also fixing.

Summary of the invention

Technical problem: the present invention seeks to the defect existed at present for the rod pumping system fault diagnosis and oil well indicator card sample size is large, plateau is many, the characteristics of classification unbalanced (sample that especially lacks catastrophe failure), using normal (or plateau) sample surface dynamometer card as cutting point, provide a kind of rod pumping system fault based on indicator card to pass the rank diagnosis, at first in the fault stage of differentiating, according to the statistical law of normal (or plateau) sample, indicator card is divided into to failure classes and non-failure classes; Then carry out detailed fault diagnosis identification in the fault diagnosis stage for fault sample.Because most indicator card is all normal (or steadily), first carry out the fault resolution and just saved a large amount of time, thereby improve the efficiency of field management.In addition, the training sample that fault is differentiated comes from normally (or plateau) sample, sample-rich and can reflect self essential characteristic of rod pumping system.

The present invention for achieving the above object, adopts following technical scheme:

The rod pumping system fault that the present invention is based on indicator card is passed the rank diagnostic method and is comprised the steps:

1) fault is differentiated the stage, and the fault stage of differentiating is divided into again two stages:

11) training stage: in the training stage, by the geometric feature on 15 indicator cards of normal or steady sample extraction, according to the characteristic quantity of these samples, obtain ASSOCIATE STATISTICS information:

111) manually choose n normal or steady sample, carry out the extraction of 15 characteristic quantities, wherein n >=50;

112) characteristic quantity extracted is carried out to test of outlier, if abnormal, reject this sample, and re-start test of outlier, until there is no exceptional value;

113) sample after test of outlier is carried out to regularity of distribution check;

114) calculate normal region and fault zone according to regularity of distribution assay;

12) the resolution stage: whether exist knotting point and test sample book whether to have characteristic quantity to fall into fault zone according to test sample book itself and judge whether fault:

121) judging whether test sample book itself exists the knotting point, if exist, is fault sample, jump procedure 2); Otherwise jump procedure 122);

122) extract 15 characteristic quantities of test sample book;

123) whether test sample book has characteristic quantity to fall into fault zone, if having, is fault sample, jump procedure 2); Otherwise jump procedure 124);

124) judge that test sample book, as normal sample, does not need to carry out Fault Identification, finish;

2) the Fault Identification stage: adopt the search tree method of the rod pumping system Fault Identification based on indicator card, fault sample is carried out to the identification of fault type.

Preferably, described 15 characteristic quantities are that bottom dead centre (E point) load, top dead-centre (F point) displacement and load, standing valve are opened the displacement of point (B point) and displacement that load, travelling valve are opened point (D point) and load, area, peak load, minimum load, the total mean change amount of load, the mean change amount of EB, BF, FD, DE section load.

Preferably, step 112) method of described rejecting abnormalities sample adopts t test criterion rejecting abnormalities data method:

21) find out feature value with the normal sample of average phase ratio error maximum as dubious value x in n observation _k, the characteristic quantity that wherein observation is normal sample;

22) to not comprising dubious value x _kin an interior n-1 observation, calculating mean value:

$\stackrel{\‾}{x}=\left(\underset{i=1}{\overset{n}{\underset{i\≠k}{\mathrm{\Σ}}}}{x}_i\right)/(n-1)---\left(1\right)$

And standard deviation estimated value:

$\stackrel{\‾}{s}=\sqrt{\left[\underset{i=1}{\overset{n}{\underset{i\≠k}{\mathrm{\Σ}}}}{(x_i-\stackrel{\‾}{x})}^2\right]/(n-2)}---\left(2\right)$

23) determine relative risk α _t, α _t=0.001, find A from the t distribution table _t(α _t; Then calculate n-2):

$K({\mathrm{\α}}_T,n)=A_T({\mathrm{\α}}_T;n-2)\·\sqrt{\frac{n}{n-1}}$

(3)

24) check x _kif have

$|x_k-\stackrel{\‾}{x}|>K({\mathrm{\α}}_T,n)\·\stackrel{\‾}{s}---\left(4\right)$

Set up, x _kfor exceptional value, should reject jump procedure 25); Otherwise x _knot exceptional value, can not reject, and stop checking;

25) if x _kfor exceptional value, after it is rejected, remaining n-1 observation repeated to above-mentioned steps, until no longer include exceptional value.

Preferably, step 113) describedly sample is carried out to the regularity of distribution verify as sample distribution is being supposed on basis, adopt χ ²whether test of fitness of fot method null hypothesis distribution is consistent with actual distribution:

31) hypothesis H is proposed ₀: X obeys certain alternative distribution pattern; Choose 6 continuity random distribution commonly used (be respectively: exponential distribution, be uniformly distributed, Weibull distribution, normal distribution, rayleigh distributed and gamma distribute) be alternative distribution pattern, carry out respectively hypothesis testing, contain r unknown number in the distribution function of described distribution pattern, r is natural number;

32) real number axis is divided into to k disjoint interval (a ₀, a ₁], (a ₁, a ₂] ..., (a _k-1, a _k], wherein:, when n≤200, k gets respectively 5,7 ... 13; When n>200, k gets respectively 11,13 ... 21; a ₀, a _kpress formula (5) cycle calculations:

a ₀＝x _min-(x _max-x _min)×j×1％，j＝1，2，L 20

(5)

a _k＝x _max+(x _max-x _min)×j×1％，j＝1，2，L 20

33) calculated data falls into each interval frequency n _i, i=1,2,3...k;

34) at H ₀under the condition of setting up, calculate X and fall into each interval Probability p _i:

p _i＝P(a _i-1＜X≤a _i) (6)

And then obtain theoretical frequency np _i(i=1,2, Λ, k);

35) by n _i, np _ithe substitution formula

obtain χ ²value;

36) determine level of significance α _χ, α _χ=0.01, look into χ ²distribution table obtains

37) if

refuse H ₀, otherwise, can accept H ₀.

Preferably, step 114) described characteristic quantity normal region and fault zone computational methods are as follows:

41) each characteristic quantity is chosen to unified probable value α, α is that sample falls into interval, normal region [b ₀, a ₀] probability, α=99.99%, and sample falls into interval, fault zone (∞, b ₀) and the interval (a in fault zone ₀,+∞) probability identical,

P{-∞＜X＜b ₀}+P{b ₀≤X≤a ₀}+P{a ₀＜X＜+∞}＝1 (7)

P{-∞＜X＜b ₀}＝P{a ₀＜X＜+∞} (8)

$P\{-\&infin;<X<b_0\}=P\{a_0<X<+\&infin;\}=\frac{1-P\{b_0\&le;X\&le;a_0\}}{2}=\frac{1-\mathrm{\&alpha;}}{2}---\left(9\right)$

So

$P\{-\&infin;<X<a_0\}=1-\mathrm{\&alpha;}+\frac{1-\mathrm{\&alpha;}}{2}=\frac{3+\mathrm{\&alpha;}}{2}---\left(10\right)$

42) according to distribution pattern and the parameter of formula (9), (10) and this characteristic quantity, calculate a ₀, b ₀, can obtain interval, normal region [b ₀, a ₀] and fault zone interval (∞, b ₀) ∪ (a ₀,+∞); For load mean change amount, significant fault zone is (a ₀,+∞);

43), if the regularity of distribution assay of certain characteristic quantity is to meet a plurality of distributions, calculate the interval, normal region by a plurality of regularities of distribution respectively, and the union of getting them is this interval, characteristic quantity normal region.

44), if the hypothesis testing of certain characteristic quantity be can not determine to its regularity of distribution, getting normal region interval is [μ-5 σ, μ+5 σ], wherein

for sample average,

for the sample standard deviation variance.

45) make b ₁=b ₂λ (a ₂-b ₂), a ₁=a ₂λ (a ₂-b ₂), the threshold value of λ for setting, λ=0.2, wherein b ₂, a ₂be respectively minimum value and the maximum value of sample.Interval [b ₁, a ₁] with step 41) to 44) union in the interval, normal region that calculates is the final normal region interval [b, a] of this characteristic quantity, final interval (∞, b) ∪ (a ,+∞) in fault zone of this characteristic quantity.

Preferably, step 121) describedly judge whether test sample book itself exists the method for knotting point to be:

51) calculate the minimum and maximum value of test sample book displacement;

52) displacement on the test sample book indicator card is divided into to J part, J=10000, j=1,2, L J-1;

53) calculate j bar straight line x _j=x _min+ (x _max-x _minthe intersection point set U{H of) * j/J and indicator card ₁, H ₂;

54) obtain H by interpolation method ₁corresponding load y _j1;

55) obtain H by interpolation method ₂corresponding load y _j2;

56) if | y _j1-y _j2|/y _j1<ε, the threshold value of ε for setting,, there is the knotting point in ε=0.001, finishes;

57) if j<J upgrades j=j+1, jump procedure 53), otherwise judge that there is not the knotting point in test sample book, finish.

Preferably, step 2) described search tree is divided into four large classes: area change class, the class of tiing a knot, vibrate class, touch the extension class, wherein the area change class is divided three classes again: area increases class, the minimum class of area, area is less than normal or normal class.Area class fault search tree comprises for the fault type of identification: standing valve leakage, sucker rod middle and upper part disconnected de-, travelling valve is malfunctioning, oil pipe bottom leakage, Pumping with gushing, sucker rod bottom disconnected de-, gas lock, pump seizure, serious feed flow is not enough, oil pipe leakage, travelling valve leakage, that travelling valve cuts out is slow, plunger is deviate from seating nipple, feed flow deficiency, gases affect, inertial load are large; Knotting class fault search tree comprises for the fault type of identification: under touch pump, secondary vibration, plunger and deviate from seating nipple; Vibration class fault search tree comprises for the fault type of identification: shake out, vibrate excessive; Touching extension class fault search tree comprises for the fault type of identifying: above touch extension.

In the inventive method, step 23) α _t, step 36) α _χ, step 41) α, step 45) λ, step 56) the symbol description of ε, Fig. 5 in ε 1, ε 2, ε 3, ξ be the threshold value of setting, step 32) to 34) k and step 52) J be the initial value of setting, these threshold values and initial value are mainly determined according to required precision and calculating experience, be the invention discloses its preferred numerical value.

Beneficial effect: the rod pumping system fault that the present invention is based on indicator card is passed the rank diagnostic method under the simple prerequisite of maintenance calculating, improved the precision of classification, especially for small sample class target anomalous event, thereby improved the performance of fault diagnosis of fault diagnosis system; It does not need set up and solve the rod pumping system mechanical model, does not have the training set problem yet; First carry out fault and differentiate and can save the plenty of time, improve the efficiency of field management.Example shows that the fault resolution can filter out fault sample from a large amount of data, and the accuracy of resolution is higher; The statistical law of training sample can reflect the production status that oil well is real-time.By example and multiple typical fault indicator card, comprise catastrophe failure, verified that the method can accurately identify single fault and the combined fault of rod pumping system.

The accompanying drawing explanation

Fig. 1: the rod pumping system fault based on indicator card is passed the flow chart of rank diagnostic method

Fig. 2: training stage flow chart;

Fig. 3: normal region and fault zone probability schematic diagram;

Fig. 4: differentiate the stage flow chart;

Fig. 5: area class fault search tree;

The symbol description of Fig. 5 is as follows:

T1-1: area is not more than ε ₁b _area(normal region of establishing area is [b _area, a _area]), ε ₁for being less than 1 little value, ε ₁for the threshold value of setting, ε ₁=0.25

T1-2: area is less than a _areaand be greater than ε ₁b _area

T1-3: area is not less than a _area

T2-1:(peak load-minimum load)/(Z _{peak load}-Z _{minimum load}) be less than ε ₂, ε ₂for being less than 1 little value, ε ₂for the threshold value of setting, ε ₂=0.01

(normal region of establishing peak load is [b _{peak load}, a _{peak load}], the normal region of minimum load is [b _{little load}, a _{minimum load}], Z _{peak load}=0.5 (a _{peak load}-b _{peak load}), Z _{minimum load}=0.5 (a _{minimum load}-b _{minimum load}))

T2-2:(peak load-minimum load)/(Z _{peak load}-Z _{minimum load}) be not less than ε ₂

T3-1: minimum load is not less than ξ a _{minimum load}, ξ is greater than the threshold value of 1, ξ for setting, ξ=1.2

T3-2: minimum load is not more than ε ₃b _{minimum load}, ε ₃for being less than 1 little value, ε ₃for the threshold value of setting, ε ₃=0.8

T3-3: minimum load is less than ξ a _{minimum load}and be greater than ε ₃b _{minimum load}

The T3-4:B point is undesired

The T3-5:B point is normal

T4-1:FD is convex function

T4-2:FD is not convex function

T5-1:B point or F point malposition, the D point is normal

T5-2:D point malposition, B point are normally

T5-3:B, D point malposition

T6-1: indicator card is parallelogram

T6-2: indicator card is not parallelogram

The T7-1:F point is not abnormal

The T7-2:F point is abnormal

T8-1:EB is concave function

T8-2:EB is not concave function

T9-1:BF is concave function

T9-2:BF is not concave function

T10-1:FD is concave function

T10-1:FD is concave function

T11-1:FD is not circular arc

T11-2:FD is circular arc

T12-1: indicator card is parallelogram

T12-2: indicator card is not parallelogram

The T13-1:F point load is not less than b _{the F point load}(normal region of establishing the F point load is [b _{the F point load}, a _{the F point load}])

The T13-2:F point load is less than b _{the F point load}

T14-1:EB is not concave function

T14-2:EB is concave function

T15-1:FD is concave function

T15-2:FD is not concave function

T16-1:FD is circular arc

T16-2:FD is not circular arc

T17-1:FD is concave function

T17-2:FD is not concave function

T18-1:FD is circular arc

T18-2:FD is not circular arc

T19-1:BF is concave function

T19-2:BF is not concave function

T20-1:FD is concave function

T20-2:FD is not concave function

T21-1:FD is circular arc

T21-2:FD is not circular arc

T22-1:FD is concave function

T22-2:FD is not concave function

T23-1:FD is circular arc

T23-2:FD is not circular arc

Fig. 6: vibration class fault search tree;

The symbol description of Fig. 6 is as follows:

ZT1-1:5 load change amount all enters fault zone

ZT1-2:5 load change amount all do not enter fault zone

Vibration is arranged between ZT2-1:BF

Not vibration between ZT2-2:BF

Vibration is arranged between ZT2-3:BF

Not vibration between ZT2-4:BF

Fig. 7: knotting class fault search tree;

The symbol description of Fig. 7 is as follows:

DT1: the displacement of knotting point is less than 1/4 of total displacement

DT2: the displacement of knotting point is greater than the load that 3/4 and F of total displacement order and is greater than the load intermediate value

DT3: the displacement of knotting point is greater than the load that 3/4 and F of total displacement order and is less than the load intermediate value

Fig. 8: touch and hang class fault search tree;

Fig. 8 symbol description is as follows:

P1: peak load is greater than a _{peak load}and near the F point, (normal region of establishing peak load is [b to peak load point _{peak load}, a _{peak load}])

P2: do not meet the P1 rule.

In Fig. 5 to Fig. 8 symbol description, related terms is defined as follows:

Due to the restriction of the conditions such as accuracy of the field test instrument in the mechanical characteristic of sucker rod and oil field, described concave function, circular arc, parallelogram etc. are not definition strict on mathematical meaning, and the present invention adopts and is defined as follows:

Definition 1 slope by 4 lines that on surface dynamometer card, E, B, F, D tetra-point coordinates calculate, if the slope differences of EB and FD, BF and DE, in a minimum scope, claims that this indicator card is parallelogram;

Definition 2 is established f (x) for the curve that formed by multi-point fitting, and the two-end-point of f (x) is connected to a straight line, and the point in curve all at downside, claims that f (x) is convex function apart from distance five points farthest of straight line;

Definition 3 is established f (x) for the curve that formed by multi-point fitting, and the two-end-point of f (x) is connected to a straight line, and the point in curve all at upside, claims that f (x) is concave function apart from distance five points farthest of straight line;

Definition 4 is established f (x) for the curve that formed by multi-point fitting, and two-end-point I, the J of f (x) are connected to a straight line, and apart from this air line distance, point farthest is designated as H to the point in curve, and IH and HJ are convex function, claim that f (x) is circular arc;

Definition 5 fits to straight line by the point between B, F on surface dynamometer card, if B, this straight line of F spacing three points farthest, displacement is to increase progressively successively, and distance is successively decreased successively, claims between BF to have vibration.

The specific embodiment

The present invention is directed to the characteristics of rod pumping system and the problem that fault diagnosis exists at present thereof, provide a kind of rod pumping system fault based on indicator card to pass the rank diagnosis, be divided into fault and differentiate and two stages of Fault Identification.At first in the fault stage of differentiating, according to the statistical law of normal sample, indicator card is divided into to failure classes and non-failure classes; Then statistical theory is combined with search tree, for fault sample, carry out detailed fault type recognition.It does not need set up and solve the rod pumping system mechanical model, does not have the training set problem yet.

The rod pumping system fault resolution method provided, be divided into training and differentiate two stages.In the training stage, after t check rejecting abnormalities data, pass through χ ²random distribution form and the parameter thereof of sample determined in the test of fitness of fot, then, calculates normal region and fault zone; In the resolution stage, according to test sample book, whether there is characteristic quantity to fall into fault zone and judge its classification.Fault is differentiated can filter out fault sample from a large amount of data, and the accuracy of resolution is higher; The statistical law of training sample can reflect the production status that oil well is real-time.

The search tree method of the rod pumping system Fault Identification provided, on the basis of differentiating in fault, the statistical information of normal sample is combined with search tree, set up the search tree of the rod pumping system Fault Identification based on indicator card, fault sample is carried out to the detailed identification of fault type.The method can accurately identify single fault and the combined fault of rod pumping system.

As Fig. 1, the rod pumping system fault that the present invention is based on indicator card is passed the rank diagnostic method and is comprised the steps:

1) fault is differentiated the stage, and the fault stage of differentiating is divided into again two stages:

11) training stage: by the geometric feature on 15 indicator cards of normal or steady sample extraction, according to the characteristic quantity of these samples, obtain ASSOCIATE STATISTICS information in the training stage, as Fig. 2:

111) manually choose n normal or steady sample, carry out the extraction of 15 characteristic quantities, wherein n >=50;

112) characteristic quantity extracted is carried out to test of outlier, if abnormal, reject this sample, and re-start test of outlier, until there is no exceptional value;

113) sample after test of outlier is carried out to regularity of distribution check;

114) calculate normal region and fault zone according to regularity of distribution assay;

12) the resolution stage: according to test sample book itself, whether exist knotting point and test sample book whether to have characteristic quantity to fall into fault zone and judge whether fault, as Fig. 4:

121) judging whether test sample book itself exists the knotting point, if exist, is fault sample, jump procedure 2); Otherwise jump procedure 122);

122) extract 15 characteristic quantities of test sample book;

123) whether test sample book has characteristic quantity to fall into fault zone, if having, is fault sample, jump procedure 2); Otherwise jump procedure 124);

124) judge that test sample book, as normal sample, does not need to carry out Fault Identification, finish;

2) the Fault Identification stage: adopt the search tree method of the rod pumping system Fault Identification based on indicator card, fault sample is carried out to the identification of fault type.

Preferably, step 11), step 11), step 122) described 15 characteristic quantities are that bottom dead centre (E point) load, top dead-centre (F point) displacement and load, standing valve are opened the displacement of point (B point) and displacement that load, travelling valve are opened point (D point) and load, area, peak load, minimum load, the total mean change amount of load, the mean change amount of EB, BF, FD, DE section load.

Preferably, step 112) method of described rejecting abnormalities sample adopts t test criterion rejecting abnormalities data method:

21) find out feature value with the normal sample of average phase ratio error maximum as dubious value x in n observation _k, the characteristic quantity that wherein observation is normal sample;

22) to not comprising dubious value x _kin an interior n-1 observation, calculating mean value:

$\stackrel{\&OverBar;}{x}=\left(\underset{i=1}{\overset{n}{\underset{i\&NotEqual;k}{\mathrm{\&Sigma;}}}}{x}_i\right)/(n-1)---\left(1\right)$

And standard deviation estimated value:

$\stackrel{\&OverBar;}{s}=\sqrt{\left[\underset{i=1}{\overset{n}{\underset{i\&NotEqual;k}{\mathrm{\&Sigma;}}}}{(x_i-\stackrel{\&OverBar;}{x})}^2\right]/(n-2)}---\left(2\right)$

23) determine relative risk α _t, α _t=0.001, find A from the t distribution table _t(α _t; Then calculate n-2):

$K({\mathrm{\&alpha;}}_T,n)=A_T({\mathrm{\&alpha;}}_T;n-2)\&CenterDot;\sqrt{\frac{n}{n-1}}$

(3)

24) check x _kif have

$|x_k-\stackrel{\&OverBar;}{x}|>K({\mathrm{\&alpha;}}_T,n)\&CenterDot;\stackrel{\&OverBar;}{s}---\left(4\right)$

Set up, x _kfor exceptional value, should reject jump procedure 25); Otherwise x _knot exceptional value, can not reject, and stop checking;

25) if x _kfor exceptional value, after it is rejected, remaining n-1 observation repeated to above-mentioned steps, until no longer include exceptional value.

Preferably, step 113) describedly sample is carried out to the regularity of distribution verify as sample distribution is being supposed on basis, adopt χ ²whether test of fitness of fot method null hypothesis distribution is consistent with actual distribution:

31) hypothesis H is proposed ₀: X obeys certain alternative distribution pattern; Choose 6 continuity random distribution commonly used (be respectively: exponential distribution, be uniformly distributed, Weibull distribution, normal distribution, rayleigh distributed and gamma distribute) be alternative distribution pattern, carry out respectively hypothesis testing, contain r unknown number in the distribution function of described distribution pattern, r is natural number;

32) real number axis is divided into to k disjoint interval (a ₀, a ₁], (a ₁, a ₂] ..., (a _k-1, a _k], wherein:, when n≤200, k gets respectively 5,7 ... 13; When n>200, k gets respectively 11,13 ... 21; a ₀, a _kpress formula (5) cycle calculations:

a ₀＝x _min-(x _max-x _min)×j×1％，j＝1，2，L 20

(5)

a _k＝x _max+(x _max-x _min)×j×1％，j＝1，2，L 20

33) calculated data falls into each interval frequency n _i, i=1,2,3...k;

34) at H ₀under the condition of setting up, calculate X and fall into each interval Probability p _i:

p _i＝P(a _i-1＜X≤a _i) (6)

And then obtain theoretical frequency np _i(i=1,2, Λ, k);

35) by n _i, np _ithe substitution formula

obtain χ ²value;

36) determine level of significance α _χ, α _χ=0.01, look into χ ²distribution table obtains

37) if

refuse H ₀, otherwise, can accept H ₀.

Preferably, step 114) described characteristic quantity normal region and fault zone computational methods are as follows, as Fig. 3:

41) each characteristic quantity is chosen to unified probable value α, α is that sample falls into interval, normal region [b ₀, a ₀] probability, α=99.99%, and sample falls into interval, fault zone (∞, b ₀) and the interval (a in fault zone ₀,+∞) probability identical,

P{-∞＜X＜b ₀}+P{b ₀≤X≤a ₀}+P{a ₀＜X＜+∞}＝1 (7)

P{-∞＜X＜b ₀}＝P{a ₀＜X＜+∞} (8)

$P\{-\&infin;<X<b_0\}=P\{a_0<X<+\&infin;\}=\frac{1-P\{b_0\&le;X\&le;a_0\}}{2}=\frac{1-\mathrm{\&alpha;}}{2}---\left(9\right)$

So

$P\{-\&infin;<X<a_0\}=1-\mathrm{\&alpha;}+\frac{1-\mathrm{\&alpha;}}{2}=\frac{3+\mathrm{\&alpha;}}{2}---\left(10\right)$

42) according to distribution pattern and the parameter of formula (9), (10) and this characteristic quantity, calculate a ₀, b ₀, can obtain interval, normal region [b ₀, a ₀] and fault zone interval (∞, b ₀) ∪ (a ₀,+∞); For load mean change amount, significant fault zone is (a ₀,+∞);

43), if the regularity of distribution assay of certain characteristic quantity is to meet a plurality of distributions, calculate the interval, normal region by a plurality of regularities of distribution respectively, and the union of getting them is this interval, characteristic quantity normal region.

44), if the hypothesis testing of certain characteristic quantity be can not determine to its regularity of distribution, getting normal region interval is [μ-5 σ, μ+5 σ], wherein

for sample average,

for the sample standard deviation variance.

According to Chebyshev (Chebyshev) inequality:

$P\left(\right|X-\mathrm{\&mu;}|\&GreaterEqual;\mathrm{\&epsiv;})\&le;\frac{{\mathrm{\&sigma;}}^2}{{\mathrm{\&epsiv;}}^2}---\left(11\right)$

Or $P\left(\right|X-\mathrm{\&mu;}|<\mathrm{\&epsiv;})\&GreaterEqual;1-\frac{{\mathrm{\&sigma;}}^2}{{\mathrm{\&epsiv;}}^2}---\left(12\right)$

To the characteristic quantity regularity of distribution arbitrarily, if the interval, normal region is taken as [μ-3 σ, μ+3 σ], sample falls into the probability α in interval, normal region >=88.89%; If the interval, normal region is taken as [μ-4 σ, μ+4 σ], α >=93.75%; If the interval, normal region is taken as [μ-5 σ, μ+5 σ], α >=96%.In the case, it is far away that the characteristic quantity of considering fault generally departs from the interval, normal region, and it is [μ-5 σ, μ+5 σ] that the present invention gets normal region interval.

45) make b ₁=b ₂λ (a ₂-b ₂), a ₁=a ₂λ (a ₂-b ₂), the threshold value of λ for setting, λ=0.2, wherein b ₂, a ₂be respectively minimum value and the maximum value of sample.Interval [b ₁, a ₁] with step 41) to 44) union in the interval, normal region that calculates is the final normal region interval [b, a] of this characteristic quantity, final interval (∞, b) ∪ (a ,+∞) in fault zone of this characteristic quantity.

Preferably, step 121) describedly judge whether test sample book itself exists the method for knotting point to be:

51) calculate the minimum and maximum value of test sample book displacement;

52) displacement on the test sample book indicator card is divided into to J part, J=10000, j=1,2, L J-1;

53) calculate j bar straight line x _j=x _min+ (x _max-x _minthe intersection point set U{H of) * j/J and indicator card ₁, H ₂;

54) obtain H by interpolation method ₁corresponding load y _j1;

55) obtain H by interpolation method ₂corresponding load y _j2;

56) if | y _j1-y _j2|/y _j1<ε, the threshold value of ε for setting,, there is the knotting point in ε=0.001, finishes;

57) if j<J upgrades j=j+1, jump procedure 53), otherwise judge that there is not the knotting point in test sample book, finish.

Preferably, as Fig. 5-Fig. 8, step 2) described search tree is divided into four large classes: area change class, the class of tiing a knot, vibrate class, touch the extension class, wherein the area change class is divided three classes again: area increases class, the minimum class of area, area is less than normal or normal class.Area class fault search tree comprises for the fault type of identification: standing valve leakage, sucker rod middle and upper part disconnected de-, travelling valve is malfunctioning, oil pipe bottom leakage, Pumping with gushing, sucker rod bottom disconnected de-, gas lock, pump seizure, serious feed flow is not enough, oil pipe leakage, travelling valve leakage, that travelling valve cuts out is slow, plunger is deviate from seating nipple, feed flow deficiency, gases affect, inertial load are large; Knotting class fault search tree comprises for the fault type of identification: under touch pump, secondary vibration, plunger and deviate from seating nipple; Vibration class fault search tree comprises for the fault type of identification: shake out, vibrate excessive; Touching extension class fault search tree comprises for the fault type of identifying: above touch extension.

Because oil is thick, the indicator card feature class of oil well paraffinication, the stifled large three kinds of faults of frictional resistance of pipeline seemingly, in search tree, not by its further classification, need be judged according to the real-time production status of oil well (can referring to corresponding Fault Mechanism Analysis).Travelling valve is malfunctioning, also feature class is seemingly for the indicator card of the leakage of oil pipe bottom, Pumping with gushing, the disconnected de-four kinds of faults in sucker rod bottom, and these four kinds of faults are all extremely serious faults, all need to stop immediately producing and processed, further distinguish and have little significance, thus in search tree also not by its further classification.

Claims (7)

1. the rod pumping system fault based on indicator card is passed the rank diagnostic method, it is characterized in that comprising the steps:

1) fault is differentiated the stage, and the fault stage of differentiating is divided into again two stages:

11) training stage: in the training stage, by 15 geometric features on normal or steady sample extraction indicator card, according to the characteristic quantity of these samples, obtain ASSOCIATE STATISTICS information:

111) manually choose n normal or steady sample, carry out the extraction of 15 geometric features, wherein n >=50;

112) characteristic quantity extracted is carried out to test of outlier, if abnormal, reject this sample, and re-start test of outlier, until there is no exceptional value;

113) sample after test of outlier is carried out to regularity of distribution check;

114) calculate normal region and fault zone according to regularity of distribution assay;

12) the resolution stage: whether exist knotting point and test sample book whether to have characteristic quantity to fall into fault zone according to test sample book itself and judge whether fault:

121) judging whether test sample book itself exists the knotting point, if exist, is fault sample, jump procedure 2); Otherwise jump procedure 122);

122) extract 15 geometric features of test sample book;

123) whether test sample book has characteristic quantity to fall into fault zone, if having, is fault sample, jump procedure 2); Otherwise jump procedure 124);

124) judge that test sample book, as normal sample, does not need to carry out Fault Identification, finish;

2) the Fault Identification stage: adopt the search tree method of the rod pumping system Fault Identification based on indicator card, fault sample is carried out to the identification of fault type.

2. the rod pumping system fault based on indicator card according to claim 1 is passed the rank diagnostic method, it is characterized in that described 15 load that geometric feature is bottom dead centre E, the displacement of top dead-centre F and load, standing valve is opened displacement and the load of a B, and travelling valve is opened displacement and the load of a D, area, peak load, minimum load, the mean change amount that load is total, the mean change amount of EB, BF, FD, DE section load.

3. the rod pumping system fault based on indicator card according to claim 1 is passed the rank diagnostic method, it is characterized in that step 112) method of rejecting abnormalities sample adopts t test criterion rejecting abnormalities data method:

21) find out feature value with the normal sample of average phase ratio error maximum as dubious value x in m observation _k, the characteristic quantity that wherein observation is normal sample;

22) to not comprising dubious value x _kin an interior m-1 observation, calculating mean value:

$\stackrel{\&OverBar;}{x}=\left(\underset{\underset{i=1}{i\&NotEqual;k}}{\overset{m}{\mathrm{\&Sigma;}}}{x}_i\right)/(m-1)---\left(1\right)$ And standard deviation estimated value:

$\stackrel{\&OverBar;}{s}=\sqrt{\left[\underset{\underset{i=1}{i\&NotEqual;k}}{\overset{m}{\mathrm{\&Sigma;}}}{(x_i-\stackrel{\&OverBar;}{x})}^2\right]/(m-2)}---\left(2\right)$

23) determine relative risk α _t, α _t=0.001, find A from the t distribution table _t(α _t; Then calculate m-2):

$K({\mathrm{\&alpha;}}_T,m)=A_T({\mathrm{\&alpha;}}_T;m-2)\&CenterDot;\sqrt{\frac{m}{m-1}}---\left(3\right)$

24) check x _kif have

$|x_k-\stackrel{\&OverBar;}{x}|>K({\mathrm{\&alpha;}}_T,m)\&CenterDot;\stackrel{\&OverBar;}{s}---\left(4\right)$

Set up, x _kfor exceptional value, should reject jump procedure 25); Otherwise x _knot exceptional value, can not reject, and stop checking;

25) if x _kfor exceptional value, after it is rejected, remaining m-1 observation repeated to above-mentioned steps, until no longer include exceptional value.

4. the rod pumping system fault based on indicator card according to claim 1 is passed the rank diagnostic method, it is characterized in that step 113) sample is carried out to the regularity of distribution verify as sample distribution is being supposed on basis, adopt χ ²whether test of fitness of fot method null hypothesis distribution is consistent with actual distribution:

31) H is proposed ₀: X obeys one of alternative distribution pattern; Choose 6 continuity random distribution commonly used: exponential distribution, be uniformly distributed, Weibull distribution, normal distribution, rayleigh distributed and gamma be distributed as alternative distribution pattern, carry out respectively hypothesis testing, contain r unknown number in the distribution function of described distribution pattern, r is natural number;

32) real number axis is divided into to k disjoint interval (a ₀, a ₁], (a ₁, a ₂] ..., (a _k-1, a _k], wherein:, when n≤200, k gets respectively 5,7 ... 13; As n>200 the time, k gets respectively 11,13 ... 21; a ₀, a _kpress formula (5) cycle calculations:

a ₀=x _min-(x _max-x _min)×j×1%,j=1,2,…20(5)

a _k=x _max+(x _max-x _min)×j×1%,j=1,2,…20

33) calculated data falls into each interval frequency n _i, i=1,2,3 ... k;

34) at H ₀under the condition of setting up, calculate X and fall into each interval Probability p _i:

p _i=P(a _i-1<X≤a _i) (6)

And then obtain theoretical frequency np _i(i=1,2 ..., k);

35) by n _i, np _ithe substitution formula

obtain χ ²value;

36) determine level of significance α _χ, α _χ=0.01, look into χ ²distribution table obtains

;

37) if

, refuse H ₀, otherwise, can accept H ₀.

5. the rod pumping system fault based on indicator card according to claim 1 is passed the rank diagnostic method, it is characterized in that step 114) the characteristic quantity normal region and the fault zone computational methods as follows:

41) each characteristic quantity is chosen to unified probable value α, α is that sample falls into interval, normal region [b ₀, a ₀] probability, α=99.99%, and sample falls into interval, fault zone (∞, b ₀) and the interval (a in fault zone ₀,+∞) probability identical,

P{-∞<X<b ₀}+P{b ₀≤X≤a ₀}+P{a ₀<X<+∞}=1 (7)

P{-∞<X<b ₀}=P{a ₀<X<+∞} (8)

$P\{-\&infin;<X<b_0\}=P\{a_0<X<+\&infin;\}=\frac{1-P\{b_0\&le;X\&le;a_0\}}{2}=\frac{1-\mathrm{\&alpha;}}{2}---\left(9\right)$

So $P\{-\&infin;<X<a_0\}=\mathrm{\&alpha;}+\frac{1-\mathrm{\&alpha;}}{2}=\frac{1+\mathrm{\&alpha;}}{2}---\left(10\right)$

42) according to distribution pattern and the parameter of formula (9), (10) and this characteristic quantity, calculate a ₀, b ₀, can obtain interval, normal region [b ₀, a ₀] and fault zone interval (∞, b ₀) ∪ (a ₀,+∞); For load mean change amount, significant fault zone is (a ₀,+∞);

43), if the regularity of distribution assay of certain characteristic quantity is to meet a plurality of distributions, calculate the interval, normal region by a plurality of regularities of distribution respectively, and the union of getting them is this interval, characteristic quantity normal region;

44), if the hypothesis testing of certain characteristic quantity be can not determine to its regularity of distribution, getting normal region interval is [μ-5 σ, μ+5 σ], wherein $\mathrm{\&mu;}=\left(\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{x}_i\right)/n$ For sample average, $\mathrm{\&sigma;}=\sqrt{\left[\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{(x_i-\mathrm{\&mu;})}^2\right]/(n-1)}$ For the sample standard deviation variance;

45) make b ₁=b ₂-λ (a ₂-b ₂), a ₁=a ₂+ λ (a ₂-b ₂), the threshold value of λ for setting, λ=0.2, wherein b ₂, a ₂be respectively minimum value and the maximum value of the training sample after test of outlier; Interval [b ₁, a ₁] with step 41) to 44) union in the interval, normal region that calculates is the final normal region interval [b, a] of this characteristic quantity, final interval (∞, b) ∪ (a ,+∞) in fault zone of this characteristic quantity.

6. the rod pumping system fault based on indicator card according to claim 1 is passed the rank diagnostic method, it is characterized in that step 121) describedly judge whether test sample book itself exists the method for knotting point to be:

51) calculate the minimum and maximum value of test sample book displacement;

52) displacement on the test sample book indicator card is divided into to J part, J=10000, j=1,2 ... J-1;

53) calculate j bar straight line x _j=x _min+ (x _max-x _minthe intersection point set U{H of) * j/J and indicator card ₁, H ₂;

54) obtain H by interpolation method ₁corresponding load y _j1;

55) obtain H by interpolation method ₂corresponding load y _j2;

56) if | y _j1-y _j2|/y _j1<ε, the threshold value of ε for setting,, there is the knotting point in ε=0.001, finishes;

57) if j<J upgrades j=j+1, jump procedure 53), otherwise judge that there is not the knotting point in test sample book, finish.

7. the rod pumping system fault based on indicator card according to claim 1 is passed the rank diagnostic method, it is characterized in that step 2) described search tree is divided into four large classes: area change class, the class of tiing a knot, vibrate class, touch the extension class, wherein the area change class is divided three classes again: area increases class, the minimum class of area, area is less than normal or normal class; Area change class fault search tree comprises for the fault type of identification: standing valve leakage, sucker rod middle and upper part disconnected de-, travelling valve is malfunctioning, oil pipe bottom leakage, Pumping with gushing, sucker rod bottom disconnected de-, gas lock, pump seizure, serious feed flow is not enough, oil pipe leakage, travelling valve leakage, that travelling valve cuts out is slow, plunger is deviate from seating nipple, feed flow deficiency, gases affect, inertial load are large; Knotting class fault search tree comprises for the fault type of identification: under touch pump, secondary vibration, plunger and deviate from seating nipple; Vibration class fault search tree comprises for the fault type of identification: shake out, vibrate excessive; Touching extension class fault search tree comprises for the fault type of identifying: above touch extension.

Images