**2.1. The reasons of the poor accuracy of steam flow rate measurement**

There are many reasons for the poor accuracy of steam measurement data. The most important reasons can be concluded as follows:

1. The working conditions of the instrument deviating from the designed conditions

At present, the mass flow rates are mostly deduced by the volume flow rates and the density. However, the changes of temperature and pressure in the transmission process would lead to the density of steam deviate from the original designed value [3]. The measurement errors would be very large [1]. Any more, some superheated steam would change into a vapor-liquid two-phase medium, which making precision worse.


### **2.2. Determining control limits for fault data detecting**

244 Energy Efficiency – The Innovative Ways for Smart Energy, the Future Towards Modern Utilities

production and energy conservation in iron& steel enterprises.

decreased and the quality of the data would be further improved.

mathematical models of the steam pipe networks is proposed.

Method of pseudonodes (MP), are demonstrated.

**2. Fault data detection and reconstruction** 

important reasons can be concluded as follows:

**Figure 1.** The objective of the work

irrational use of steam, and find the weakness in the management links. Therefore, improving the reliability the steam flow rate measurement data is essential for the normal

The objective of the work may be depicted by figure 1. The real time data (mainly referring to the mass flow rate variables) would be mainly processed by three approaches. By fault data detection and reconstruction, the fault data would be picked out and the real value would be reconstructed or estimated. By gross error detection, the data with gross error would be discovered and re-estimated. By data reconciliation, the random error would be

For fault data detection and reconstruction, the reasons of the low accuracy of steam flow rate data are introduced. By applying the statistical process control theory[3] to determine univariate and multivariate process control limit to monitor the abnormal data online. The approach to calculate the real data (mass flow rates) through the thermal and hydraulic

For the section of gross error detection, the definition of the problem is introduced. And the two basic gross error detection approaches, the Measurement Test (MT) method and the

For the section of data reconciliation, the constrained least-squares problems stated in the section of gross error detection is discussed in detail, including the assumptions for the application,the constraint equations and the selection of the weighted parameter matrix.

The presented approaches of data processing can be programmed for computers to

There are many reasons for the poor accuracy of steam measurement data. The most

1. The working conditions of the instrument deviating from the designed conditions

determine the abnormality and improve the precision of the mass flow rate data.

**2.1. The reasons of the poor accuracy of steam flow rate measurement** 

The abnormal data from certain sensors (including temperature, pressure and flow rate sensors) may be characterized as fluctuation quickly with a large magnitude (induced by the poor contact in the instruments), keeping a certain value without any varying (induced by the failure of the data sampling systems) or being outside the normal variable range. The first two cases are not discussed here because they are easy to be discovered. As for the last case, the statistical process control approaches to determine the control limits for data monitoring is applied. When the values of the controlled variables (or functions) exceed the limits, it shows there are abnormal data except that the process is actually abnormal. The statistical process control includes two types--univariate and multivariate control.

### *2.2.1. Determining the control limits for single variable*

The statistical process control (SPC) and control chats were first proposed by Shewhart in quality prediction. Traditional SPC mainly treated single variables. When the value locates outside the normal range, the system would output an alarming signal to notify the operators to check and make sure whether the state is abnormal. In the work, if the process state is actually normally operating, the values of the variables can be judged as fault data. Reasonably determined control limits can reduce the probability of false alarm. Many researches focus on the problem [4-6]. In the work, empirical distribution function combined with the principle of "3σ"[3]is applied to determine the control limits of different variables.

Denote *<sup>n</sup>* <sup>1</sup> *X R* <sup>×</sup> ∈ as one of the measurement data matrix, n is the number of the samples, each row of X is a measured sample.If 1 2 ( , ,..., )*<sup>T</sup> X xx xn* <sup>=</sup> , the range of the data matrix is:

$$R(X) = \max(\mathbf{x}\_i) - \min(\mathbf{x}\_i), (i \le i \le n) \tag{1}$$

Divide the range into N intervals with the same length, each length of the interval is:

$$\mathcal{L} = \frac{R(X\_1)}{N} \tag{2}$$

Data Processing Approaches for the Measurements of Steam Pipe Networks in Iron and Steel Enterprises 247

However, according the principle of "3σ" and the empirical distribution of X, the control

It shows the corresponding sample value is not appeared in the significance test level range. To determine the limits, the empirical distribution can be linearly extended to outer range or estimated with experience by comparing with the real process

2. Two more crossing points between empirical distribution and the testing level at one side. It shows there are disturbances in the data, which makes the empirical distribution fluctuate at one lateral side. According to principle of "3σ" , find the highest probability interval and search from it to the two laterals for the points first amounting to the

Figure 3-5 show the control limits (4 lines) and monitoring result in a plant. The interval between the two green lines is the "green zone" showing normally working. The intervals outside the red lines are the "red zones" indicating the fault data or process abnormality. The other two intervals are the "yellow zones" which reminds the operators to notice the changes of the data. Figure 4 and 5, in the right side the red and green line are coincident. It

By determining the limits of single variable, the wild values considered as fault data can

The approach of Univariate Statistical Process Control (USPC) can roughly judge fault data.

The shortage of USPC is the operators being prone to "information overload"when the number of the variables is large. Anymore, it's too rough to judge the normalty of data just by monitor alarming. In practice, the variables maybe constrained by certain inherent functions. It

can be explained that the outer equipment constraints the variable freely moving.

easily be discovered. However, two kinds of inherent errors can not be avoided.

*2.2.2. Determining the control limits of multivariate system based on PCA* 

The approach is also appropriate to monitor the temperature and pressure variables.

limits can be deduced reversely by searching the respective probability of limits.

1. No crossing point between empirical distribution and the testing level.

**Figure 2.** The probability density distribution graph of normal distribution

Two kinds of special instances have to be noticed:

performance state.

probability (1-α)/2.

The corresponding intervals are marked as 1 2 , ,..., *<sup>N</sup> cc c* . So each element in X belongs to one of the intervals and these data are divided into N groups. The probability of the sampled variable value locates in each interval can be estimated as:

$$f\_r = \frac{\upsilon\_r}{N} \tag{3}$$

1,2,... , *<sup>i</sup> r Nv* = is the number of the data locate in the ith interval. As the data are independent to each other (usually the hypothesis is reasonable). The probability defined:

$$P\_r\left\{X\_i \in \mathcal{C}\_r\right\} = f\_r \tag{4}$$

The average probability density of each interval can be written as:

$$p\_r\left\{X\_i \in c\_r\right\} = \frac{f\_r}{R(X\_i) / N} = \frac{Nf\_r}{R(X\_i)}\tag{5}$$

The empirical distribution of X can be got with the lengths of intervals as bottom and *<sup>r</sup> p* as the height. When n and N are large enough, the bar graph would be more similar to the global distribution of X. According to the central limit theorem, when the system has only one stable state, the measured data would follow normal distribution as the sample size grows. So the global distribution function of X can be written as normal distribution:

$$f(\mathbf{x}) = \frac{1}{\sqrt{2\pi\sigma}} e^{-\frac{\left(\mathbf{x}-\mu\right)^2}{2\sigma^2}}\tag{6}$$

Figure 2 shows the distribution graph. The probability of data locating in the range of μ±3σis 99.73%. When the value of X is outside the range, it's reasonable to suspect the value being abnormal (otherwise, the process doesn't normally function ). That's the principle of "3σ". In practice, the outside of "3σ"occupying about 0.25% probability is named "red zone", and the range inside μ±2σis named "green zone"(about 95% probability). The intervals between them are "yellow zone". Four control limits are determined by the method.

As the expectation and the deviation of the X global distribution are unknown, and the estimated values with the sample are not accurate especially when the empirical distribution isn't similar to normal distribution, the principle of "3σ" can not be applied directly.

**Figure 2.** The probability density distribution graph of normal distribution

However, according the principle of "3σ" and the empirical distribution of X, the control limits can be deduced reversely by searching the respective probability of limits.

Two kinds of special instances have to be noticed:

246 Energy Efficiency – The Innovative Ways for Smart Energy, the Future Towards Modern Utilities

variable value locates in each interval can be estimated as:

The average probability density of each interval can be written as:

method.

*ri r*

Denote *<sup>n</sup>* <sup>1</sup> *X R* <sup>×</sup> ∈ as one of the measurement data matrix, n is the number of the samples, each row of X is a measured sample.If 1 2 ( , ,..., )*<sup>T</sup> X xx xn* <sup>=</sup> , the range of the data matrix is:

( ) max( ) min( ),( ) *RX x x i i n i i* = − ≤≤ (1)

*R X*( ) *<sup>c</sup>*

The corresponding intervals are marked as 1 2 , ,..., *<sup>N</sup> cc c* . So each element in X belongs to one of the intervals and these data are divided into N groups. The probability of the sampled

*r*

1,2,... , *<sup>i</sup> r Nv* = is the number of the data locate in the ith interval. As the data are independent to each other (usually the hypothesis is reasonable). The probability defined:

{ } ( )/ ( )

The empirical distribution of X can be got with the lengths of intervals as bottom and *<sup>r</sup> p* as the height. When n and N are large enough, the bar graph would be more similar to the global distribution of X. According to the central limit theorem, when the system has only one stable state, the measured data would follow normal distribution as the sample size

grows. So the global distribution function of X can be written as normal distribution:

<sup>2</sup> <sup>1</sup> ( ) 2

πσ

Figure 2 shows the distribution graph. The probability of data locating in the range of μ±3σis 99.73%. When the value of X is outside the range, it's reasonable to suspect the value being abnormal (otherwise, the process doesn't normally function ). That's the principle of "3σ". In practice, the outside of "3σ"occupying about 0.25% probability is named "red zone", and the range inside μ±2σis named "green zone"(about 95% probability). The intervals between them are "yellow zone". Four control limits are determined by the

As the expectation and the deviation of the X global distribution are unknown, and the estimated values with the sample are not accurate especially when the empirical distribution

isn't similar to normal distribution, the principle of "3σ" can not be applied directly.

*fx e*

*f Nf pX c RX N RX*

*r r*

*i i*

2 2 ( )

μ

σ

*x*

<sup>−</sup> <sup>−</sup>

*r*

*<sup>N</sup>* <sup>=</sup> (2)

*<sup>v</sup> <sup>f</sup> N*= (3)

*PX c f ri r r* { ∈ ≈} (4)

∈≈ = (5)

= (6)

Divide the range into N intervals with the same length, each length of the interval is:


Figure 3-5 show the control limits (4 lines) and monitoring result in a plant. The interval between the two green lines is the "green zone" showing normally working. The intervals outside the red lines are the "red zones" indicating the fault data or process abnormality. The other two intervals are the "yellow zones" which reminds the operators to notice the changes of the data. Figure 4 and 5, in the right side the red and green line are coincident. It can be explained that the outer equipment constraints the variable freely moving.

By determining the limits of single variable, the wild values considered as fault data can easily be discovered. However, two kinds of inherent errors can not be avoided.

The approach of Univariate Statistical Process Control (USPC) can roughly judge fault data. The approach is also appropriate to monitor the temperature and pressure variables.

#### *2.2.2. Determining the control limits of multivariate system based on PCA*

The shortage of USPC is the operators being prone to "information overload"when the number of the variables is large. Anymore, it's too rough to judge the normalty of data just by monitor alarming. In practice, the variables maybe constrained by certain inherent functions. It shows the existence of fault data when the functions are unsatisfied. Principal Component Analysis (PCA) is one of the approaches to discover the relationship among the variables by means of statistics. By determining the control limits of two guidelines, Hotelling's T2 and SPE (Squared Predictive Error, Q), the multivariate process can be monitored conveniently.

Data Processing Approaches for the Measurements of Steam Pipe Networks in Iron and Steel Enterprises 249

The proposed approach is to distinguish different stable states of the process by empirical

Denote the measurement matrix as *n m X R* <sup>×</sup> ∈ , n is the size of samples, m is the number

Denote 1 2 ( , ,..., ) *X xx x i i i ni* = (i=1,2,…,m) as the ith column. The empirical distribution of *Xi* can be derived as just mentioned. When the process has two or more stable states,

Figure 6 is the curve of flow rate and the distribution bar graph of a chemical process in a period. The two peaks represent different demands for steam in different states.

Without loss of generality, the two peaks distribution is to be discussed. If the control limits are determined without considering the different states, the sensitivity of detecting fault data will be too low for the multivariate process. Apply two normal distributions to fit the two peak sections. If the averages and deviations are signed as

0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

x 104

[min( ), ] [ ,max( )] *ik i is is ik i s i s i*

γσ

 can be selected in the interval of (1, 3), and the total state region should cover the whole range of the sampled data. Ideally, the state regions should not have folded region or the folded region is narrow and with low probability. Or, the variable have no

By the rule of (7), the samples are separated into two groups. By the same way, the two groups can be further grouped with other variables. However, it's not suggested to divide the sample into many groups with too few elements in each group. Assume

*k n* = 1,2,..., (7)

<sup>0</sup> <sup>5</sup> <sup>10</sup> <sup>15</sup> <sup>20</sup> <sup>25</sup> <sup>0</sup>

x

**Figure 6.** The curve and distribution bar graph of flow rate for a chemical process

*Sx XX*

 ∈ + <sup>=</sup> ∈ −

2 .2 .2

1 .1 .1

γσ

*Sx X X*

necessity for state differentiating, which is not the case discussed here.

, the rule to divide the samples into different groups (different

of the monitored variables. The rows of X are the serial samples in time order.

distribution function, and determine the control limits for each stable state.

the distribution bar graph would characterize as multi peaks.

i. Differentiate the stable states and group the samples

*Xi s*. 1 , *i s*. 1 σ

γ

v alue  , *Xi s*. 2 , *i s*. 2 σ

states) is listed as equation (7):

*S*

0 0.5 1 1.5 2 2.5

the sample number

**Figure 3.** Monitoring the Steam Flow Rate of the Steel Making Process

**Figure 4.** Monitoring the Flow Rate of Start Steam Boiler

**Figure 5.** Monitoring the Steam Flow Rata of CDQ (Coke Dry Quenching)

The proposed approach is to distinguish different stable states of the process by empirical distribution function, and determine the control limits for each stable state.

i. Differentiate the stable states and group the samples

248 Energy Efficiency – The Innovative Ways for Smart Energy, the Future Towards Modern Utilities

**Figure 3.** Monitoring the Steam Flow Rate of the Steel Making Process

0 50 100 150

0

0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.02

0.04

0.06

0.08

0.1

0.12

0.02

0.04

0.06

0.08

0.1

0.12

**Figure 4.** Monitoring the Flow Rate of Start Steam Boiler

50 60 70 80 90 100 110

10 15 20 25 30 35

**Figure 5.** Monitoring the Steam Flow Rata of CDQ (Coke Dry Quenching)

shows the existence of fault data when the functions are unsatisfied. Principal Component Analysis (PCA) is one of the approaches to discover the relationship among the variables by means of statistics. By determining the control limits of two guidelines, Hotelling's T2 and SPE

0

15

65

70

75

80

85

90

95

20

25

30

35

50

100

150

0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2 2.5

4

(Squared Predictive Error, Q), the multivariate process can be monitored conveniently.

Denote the measurement matrix as *n m X R* <sup>×</sup> ∈ , n is the size of samples, m is the number of the monitored variables. The rows of X are the serial samples in time order.

Denote 1 2 ( , ,..., ) *X xx x i i i ni* = (i=1,2,…,m) as the ith column. The empirical distribution of *Xi* can be derived as just mentioned. When the process has two or more stable states, the distribution bar graph would characterize as multi peaks.

Figure 6 is the curve of flow rate and the distribution bar graph of a chemical process in a period. The two peaks represent different demands for steam in different states.

Without loss of generality, the two peaks distribution is to be discussed. If the control limits are determined without considering the different states, the sensitivity of detecting fault data will be too low for the multivariate process. Apply two normal distributions to fit the two peak sections. If the averages and deviations are signed as *Xi s*. 1 , *i s*. 1 σ , *Xi s*. 2 , *i s*. 2 σ , the rule to divide the samples into different groups (different states) is listed as equation (7):

**Figure 6.** The curve and distribution bar graph of flow rate for a chemical process

$$S = \begin{cases} S\_1 & \mathbf{x}\_{i\bar{k}} \in [\min(\mathbf{X}\_i), \mathbf{X}\_{i,s1} + \mathcal{py}\sigma\_{i,s1}] \\ S\_2 & \mathbf{x}\_{i\bar{k}} \in [\mathbf{X}\_{i,s2} - \mathcal{py}\sigma\_{i,s2}, \max(\mathbf{X}\_i)] \end{cases} \quad k = 1, 2, \dots, n \tag{7}$$

γ can be selected in the interval of (1, 3), and the total state region should cover the whole range of the sampled data. Ideally, the state regions should not have folded region or the folded region is narrow and with low probability. Or, the variable have no necessity for state differentiating, which is not the case discussed here.

By the rule of (7), the samples are separated into two groups. By the same way, the two groups can be further grouped with other variables. However, it's not suggested to divide the sample into many groups with too few elements in each group. Assume there are only two groups of samples are derived. They are marked as 1 *n m X R <sup>a</sup>* <sup>×</sup> ∈ and <sup>2</sup> *n m X R <sup>b</sup>* <sup>×</sup> <sup>∈</sup> ( 1 2 *nn n* + ≤ ,some samples may be discarded because they don't belong to any state region).

ii. Determining the control limits of different state with PCA

Through grouping, the sample data is more convergence. First standardize *Xa* , denote:

$$X\_a = (\mathbf{x}\_1, \mathbf{x}\_2, \dots, \mathbf{x}\_m) \tag{8}$$

Data Processing Approaches for the Measurements of Steam Pipe Networks in Iron and Steel Enterprises 251

2 1

2

θ= − , *<sup>c</sup>*

θ

θ

2 1 3

θ θ

0 2

process normally functions, the two statistical variables satisfy the following

1

2

Three variables are considered and some of the data are modified on the original data. The no. 167-172 and no.350-355 samples show the transition procedure of two states. The no.360- 365 samples are modified data which deviate from the normal constraint. Figure 7 shows the

Two experiments are designed to testify the advantage of differentiating states applied in this work. Figure 8 is the case no differentiating states, whilst figure 9 is the differentiating case. As the figures shown, the approach of differentiating states and determining the control limits respectively is more sensitive to the transition of states and fault data.

In the case of MSPC, the fault data can be located and isolated by contribution diagram.

unit matrix. The fault data is judged as the variable with larger contribution to SPE and

<sup>2</sup> ( ( )) *SPE T T Cont I PP x i i* = − ξ

*<sup>i</sup> T* can be written as (22)(23). *<sup>i</sup>*

α

α

2 2 *aj A n*, , *T T*

> <sup>2</sup> *Qaj* α ≤ δ

the second data matrix *Xb* can also be processed to derive the control inequalities:

2 2 *bj A n*, , *T T*

> <sup>2</sup> *Qbj* α ≤ δ

changing curves of the three variables and the scatter diagram in 3-dimention space.

2 2 0 20 0 1 2 1 1 <sup>2</sup> ( 1) (1 ) *<sup>h</sup> c h*

*h*

θ

θ

= = , 1 3

*h h*

α

( 1,2,3)

( ) *<sup>T</sup> T T Q e e X I PP X aj aj aj aj aj* ==− (16)

0

≤ (18)

≤ ′ (20)

′ (21)

ξ

(22)

is the ith column of

(19)

is the threshold value under the

<sup>−</sup> <sup>=</sup> + + (17)

α

of normal distribution.When there's no fault data in the sample or the

SPE(also called statistical variable Q)is defined as:

α

1

= +

iii. The monitoring results for the experimental data

iv. Locate and isolate the fault data

<sup>2</sup> *T* .

The variable contribute to SPE and <sup>2</sup>

 λ*i*

*<sup>m</sup> <sup>i</sup> i j j A*

δ θ

The control limit can be calculated by:

θ

α

In the equation

testing level

inequalities.

In it,, 1 2 , ,..., *<sup>m</sup> xx x* <sup>1</sup> *<sup>n</sup>* <sup>1</sup> *<sup>R</sup>* <sup>×</sup> <sup>∈</sup> . Calculate each average and standard deviation:

$$M\_{\
u} = (m\_1, m\_2, \dots, m\_m) \in \mathbb{R}^{1 \times n\_1}.\tag{9}$$

$$S\_a = (\mathbf{s}\_1, \mathbf{s}\_2, \dots, \mathbf{s}\_m) \in R^{1 \times n\_1} \tag{10}$$

The measured data matrix is standardized as:

$$\tilde{X}\_a = \begin{bmatrix} X\_a - \begin{pmatrix} 1 & 1 & \dots & 1 \end{pmatrix}^T M\_a \end{bmatrix} \text{diag}(\frac{1}{s\_1}, \frac{1}{s\_2}, \dots \frac{1}{s\_m}) \tag{11}$$

Derive the eigenvalues of the covariance matrix <sup>1</sup> 1 *<sup>T</sup> C XXa a <sup>n</sup>* <sup>=</sup> <sup>−</sup> and orthogonalize the matrix, then the model of PCA can be written as:

$$
\hat{X}\_a = \hat{X}\_a + E = T\_a P^T + E \tag{12}
$$

$$T\_a = \tilde{X}\_a P \tag{13}$$

In the equations, *m A P R* <sup>×</sup> ∈ is the load matrix. If arrange the eigenvalues of C in descendent order, the former A elements are noted as 1 2 ( , ,..., ) *a A* λ λλ λ = . The corresponding unitized eigenvectors form the matrix of P, 1 2 ( , ,..., ) *<sup>A</sup> P PP P* = . A is the number of the selected components. <sup>1</sup>*n A <sup>a</sup> T R* <sup>×</sup> <sup>∈</sup> is the scored matrix and E is the error matrix..

The MSPC based on PCA defined two statistical variables: Hotelling's T2 and SPE (Q).

If the jth measurement vector is signed as *Xaj* (row vector), the corresponding scored vector is signed as *aj <sup>T</sup>* (ie. The jth row of matrix *<sup>a</sup> <sup>T</sup>* ), then <sup>2</sup> *aj T* is defined as[7]:

$$T\_{a\circ}^2 = T\_{a\circ} \mathcal{A}\_a^{-1} T\_{a\circ}^T \tag{14}$$

When the testing level is α,its control limit can be calculated according F distribution:

$$T\_{A,n\_1,\alpha}^2 = \frac{A(n\_1 - 1)}{n\_1 - A} F\_{A,n\_1 - 1,\alpha} \tag{15}$$

<sup>1</sup> *A n*, 1, *<sup>F</sup>* <sup>−</sup> α is the critical value corresponding to testing level α, degree of freedom A,n1-1. Data Processing Approaches for the Measurements of Steam Pipe Networks in Iron and Steel Enterprises 251

SPE(also called statistical variable Q)is defined as:

$$Q\_{a\circ} = e\_{a\circ}e\_{a\circ}^T = \tilde{X}\_{a\circ}(I - PP^T)\tilde{X}\_{a\circ}^T \tag{16}$$

The control limit can be calculated by:

250 Energy Efficiency – The Innovative Ways for Smart Energy, the Future Towards Modern Utilities

ii. Determining the control limits of different state with PCA

The measured data matrix is standardized as:

<sup>2</sup> *n m X R <sup>b</sup>*

any state region).

there are only two groups of samples are derived. They are marked as 1 *n m X R <sup>a</sup>*

<sup>×</sup> <sup>∈</sup> ( 1 2 *nn n* + ≤ ,some samples may be discarded because they don't belong to

Through grouping, the sample data is more convergence. First standardize *Xa* , denote:

11 1 [ (1 1 ... 1) ] ( , ,... ) *<sup>T</sup>*

In the equations, *m A P R* <sup>×</sup> ∈ is the load matrix. If arrange the eigenvalues of C in descendent

eigenvectors form the matrix of P, 1 2 ( , ,..., ) *<sup>A</sup> P PP P* = . A is the number of the selected

λ λλ

The MSPC based on PCA defined two statistical variables: Hotelling's T2 and SPE (Q).

2 1 *T aj aj a aj TT T* λ

1 1

, , , 1, 1

( 1) *A n A n A n T F n A*

2 1

α

is the critical value corresponding to testing level

*<sup>a</sup> T R* <sup>×</sup> <sup>∈</sup> is the scored matrix and E is the error matrix..

In it,, 1 2 , ,..., *<sup>m</sup> xx x* <sup>1</sup> *<sup>n</sup>* <sup>1</sup> *<sup>R</sup>* <sup>×</sup> <sup>∈</sup> . Calculate each average and standard deviation:

*a a a*

Derive the eigenvalues of the covariance matrix <sup>1</sup>

order, the former A elements are noted as 1 2 ( , ,..., ) *a A*

vector is signed as *aj <sup>T</sup>* (ie. The jth row of matrix *<sup>a</sup> <sup>T</sup>* ), then <sup>2</sup>

matrix, then the model of PCA can be written as:

If the jth measurement vector is signed as *Xaj*

α

components. <sup>1</sup>*n A*

When the testing level is

<sup>1</sup> *A n*, 1, *<sup>F</sup>* <sup>−</sup> α *X X M diag*

1 2 ( , ,..., ) *X xx x a m* = (8)

1 2 ( , ,..., ) *Ma m* <sup>=</sup> *mm m* <sup>1</sup> <sup>1</sup> . *<sup>n</sup> <sup>R</sup>* <sup>×</sup> <sup>∈</sup> (9)

1 2 ( , ,..., ) *a m S ss s* <sup>=</sup> <sup>1</sup> <sup>1</sup> *<sup>n</sup> <sup>R</sup>* <sup>×</sup> <sup>∈</sup> (10)

*m*

and orthogonalize the

1 2

 λ

,its control limit can be calculated according F distribution:

 α−

1 *<sup>T</sup> C XXa a <sup>n</sup>* <sup>=</sup> <sup>−</sup>

<sup>ˆ</sup> *<sup>T</sup> X X E TP E aa a* = += + (12)

*a a T XP* <sup>=</sup> (13)

= . The corresponding unitized

(row vector), the corresponding scored

<sup>−</sup> = (14)

<sup>−</sup> <sup>=</sup> − (15)

α

*aj T* is defined as[7]:

, degree of freedom A,n1-1.

*ss s* = − (11)

<sup>×</sup> ∈ and

$$\delta\_{\alpha}^{2} = \theta\_{1} (\frac{c\_{\alpha} \sqrt{2 \theta\_{2} h\_{0}}^{2}}{\theta\_{1}} + 1 + \frac{\theta\_{2} h\_{0} (h\_{0} - 1)}{\theta\_{1}^{2}})^{\circ \prime} \tag{17}$$

In the equation 1 ( 1,2,3) *<sup>m</sup> <sup>i</sup> i j j A* θ λ *i* = + = = , 1 3 0 2 2 2 1 3 *h* θ θ θ = − , *<sup>c</sup>*αis the threshold value under the

testing level α of normal distribution.When there's no fault data in the sample or the process normally functions, the two statistical variables satisfy the following inequalities.

$$T\_{a\circ}^2 \le T\_{A, n\_1, \alpha}^2 \tag{18}$$

$$Q\_{a\circ} \le \delta\_\alpha^2 \tag{19}$$

the second data matrix *Xb* can also be processed to derive the control inequalities:

$$T\_{b\circ}^2 \le T\_{A,n\_2,\alpha}^{\prime 2} \tag{20}$$

$$Q\_{b\dagger} \le \delta\_\alpha^{\prime 2} \tag{21}$$

#### iii. The monitoring results for the experimental data

Three variables are considered and some of the data are modified on the original data. The no. 167-172 and no.350-355 samples show the transition procedure of two states. The no.360- 365 samples are modified data which deviate from the normal constraint. Figure 7 shows the changing curves of the three variables and the scatter diagram in 3-dimention space.

Two experiments are designed to testify the advantage of differentiating states applied in this work. Figure 8 is the case no differentiating states, whilst figure 9 is the differentiating case. As the figures shown, the approach of differentiating states and determining the control limits respectively is more sensitive to the transition of states and fault data.

iv. Locate and isolate the fault data

In the case of MSPC, the fault data can be located and isolated by contribution diagram. The variable contribute to SPE and <sup>2</sup> *<sup>i</sup> T* can be written as (22)(23). *<sup>i</sup>* ξ is the ith column of unit matrix. The fault data is judged as the variable with larger contribution to SPE and <sup>2</sup> *T* .

$$\text{Cont}\_{l}^{\text{SPE}} = \left(\xi\_{l}^{T} (I - PP^{T})\mathbf{x}\right)^{2} \tag{22}$$

252 Energy Efficiency – The Innovative Ways for Smart Energy, the Future Towards Modern Utilities

*T TT T* 2 1 *Cont x P P x <sup>i</sup> i i* ξ ξ<sup>−</sup> = Λ (23)

Data Processing Approaches for the Measurements of Steam Pipe Networks in Iron and Steel Enterprises 253

**Figure 9.** The result of monitoring no differentiating states

*2.3.1. The static model of steam pipe network* 

i. The hydraulic and thermal model of single pipe. - The hydraulic model of single pipe

temperatures and pressures according to the steam network model.

3) There is no condensate or secondary steam and the effect is neglected.

**the model** 

**2.3. The steam pipe network modeling and calculating the fault data based on** 

After the fault data are detected, the next step is to estimate the true values of the variables. As the experience shown, the measured temperature and pressure values are usually accurate. So the object of the work is to calculate the flow rate value with the detected

To deduce the model, the real conditions are simplified as: 1) The steam in the pipes is axialdirection one dimensional flow; 2) The network is composed of nodes and branches(pipes).

By the law of momentum conservation the hydraulic model can be written as [8]:

**Figure 7.** The changing curves and scatter diagram of three variables

**Figure 8.** The result of monitoring no differentiating states

**Figure 9.** The result of monitoring no differentiating states

**Figure 7.** The changing curves and scatter diagram of three variables

0 50 100 150 200 250 300 350 400

0 50 100 150 200 250 300 350 400

0 50 100 150 200 250 300 350 400

0

100

0

50

x 3

x 2

50

x 1

100

**Figure 8.** The result of monitoring no differentiating states

*T TT T* 2 1 *Cont x P P x <sup>i</sup> i i*

x 3 ξ ξ

<sup>−</sup> = Λ (23)

20

x1 x2

40

60

80

100

## **2.3. The steam pipe network modeling and calculating the fault data based on the model**

After the fault data are detected, the next step is to estimate the true values of the variables. As the experience shown, the measured temperature and pressure values are usually accurate. So the object of the work is to calculate the flow rate value with the detected temperatures and pressures according to the steam network model.

#### *2.3.1. The static model of steam pipe network*

To deduce the model, the real conditions are simplified as: 1) The steam in the pipes is axialdirection one dimensional flow; 2) The network is composed of nodes and branches(pipes). 3) There is no condensate or secondary steam and the effect is neglected.

	- The hydraulic model of single pipe

By the law of momentum conservation the hydraulic model can be written as [8]:

$$P\_1^2 - P\_2^2 = 1.25 \times 10^8 \frac{\lambda q^2 P\_1 T\_2 Z\_2 L}{D^5 \rho\_m T\_1 Z\_1} \tag{24}$$

Data Processing Approaches for the Measurements of Steam Pipe Networks in Iron and Steel Enterprises 255

calculated with following equations (31),(32),(33). The unclaimed symbols were

ρ ρ, , μ , *<sup>m</sup>*ρ

(30)

(31)

(33)

(34)

(35)

(36)

(37)

= + (32)

can be

, as the input and output steam densities. 1 2

1

<sup>1</sup> ( )*<sup>r</sup> P r r RT* π π

2

<sup>2</sup> ( )*<sup>r</sup> P r r RT* π π

1 2 2 3 3 *<sup>m</sup>* ρ

 ρ

\*

By the law of energy conversation, the static thermal model can be written as

(1 ) (1 ) <sup>1000</sup> <sup>278</sup> <sup>1000</sup> 3600

*c q c q* + +

12 12

*l*

The definitions and units of 1 2 *T T Lq* , , , are the same as eqation (24).(note: changing equation (34) to (35) is to make thermal model has the same pattern with equation

—the appending heat loss coefficient for the appendix of pipe, valve, and

In term of equation IF-97, The unclaimed symbols are defined in the reference[9].

γγ γγ

( ) *<sup>o</sup>*

The amount of heat loss along unit pipe length can be calculated using equation (38)

 γ

γ

*T*

*l l*

*p*

 β

1 0

2 0

π<sup>=</sup> <sup>+</sup>

π<sup>=</sup> <sup>+</sup>

ρ

ρ

ρ

μ = Ψ(,) δτη

*p*

2

*l*

*T*

*p*

β

*<sup>p</sup> c* —specific heat capacity at constant pressure, kJ/kg.K; *<sup>l</sup> q* —the amount of heat loss along unit pipe length, W/m.

278

*qL qL T T*

( )( ) (1 )

<sup>2</sup> 278 (1 ) *p*

*c q q T T CT T*

*q L* = −= − <sup>+</sup>

*c q <sup>C</sup>* β*q L* <sup>=</sup> <sup>+</sup>

support, the value can be 0.15-0.25 depending on pipe laying methods;

2

 =− + τγ

*<sup>p</sup> c R*

β

 − = = ×

Denote 1 2 ρ ρ

defined in reference[9].


1 2

follows:

(26)).

β

1 2 *P P*, —the input and output pressure of the steam in the pipe, Mpa;

1 2 *T T*, —the input and output absolute temperature, K;

1 2 *Z Z*, —compressibility factor of the input and output steam;


Usually, there's little difference between the products of compressibility factors and temperatures in the same pipe. Considering the elbows, reducer extenders and other friction factors, the equivalent coefficient η is added to the equation. Equation (24) changes to (25).

$$P\_1^2 - P\_2^2 = 1.25 \times 10^8 \frac{\lambda q^2 P\_1 (1 + \eta) L}{D^5 \rho\_m} \tag{25}$$

Thus:

$$q = \frac{D^5 \rho\_m}{1.25 \times 10^8 \,\lambda q P\_1 (1 + \eta) L} P\_1^2 - P\_2^2 = C\_p (P\_1^2 - P\_2^2) \tag{26}$$

$$C\_P = \frac{D^5 \rho\_m}{1.25 \times 10^8 \,\lambda q P\_1 (1 + \eta) L} \tag{27}$$

In accordance with Альтщуль equation, frictional factor λ:

$$
\lambda = 0.11 \times (\frac{\Delta}{D} + \frac{68}{\text{Re}})^{0.25} \tag{28}
$$

Δ —the equivalent absolute roughness, mm; Re —the Reynolds number of the steam.

After transferring the units of *D q*, to mm and t/h (ton per hour), Reynolds number is :

$$\text{Re} = \frac{Du\rho\_m}{\mu} = 354 \frac{q}{D\mu} \tag{29}$$

In the equation, *u* —the characteristic velocity of the steam in this pipe, m/s; μ the mean dynamic viscosity coefficient. It's determined by equation (33)[9].

Denote 1 2 ρ ρ, as the input and output steam densities. 1 2 ρ ρ, , μ , *<sup>m</sup>*ρ can be calculated with following equations (31),(32),(33). The unclaimed symbols were defined in reference[9].

$$\rho\_1 = \frac{P\_1}{\pi (r\_\pi^0 + r\_\pi^r)RT\_1} \tag{30}$$

$$\rho\_2 = \frac{P\_2}{\pi (r\_\pi^0 + r\_\pi^r)RT\_2} \tag{31}$$

$$
\rho\_m = \frac{\rho\_1}{3} + \frac{2\rho\_2}{3} \tag{32}
$$

$$
\mu = \Psi(\mathcal{S}, \mathfrak{r}) \mathfrak{q}^"\tag{33}
$$


254 Energy Efficiency – The Innovative Ways for Smart Energy, the Future Towards Modern Utilities

1 2 *T T*, —the input and output absolute temperature, K;

—the weighted mean density of steam, kg/m3.

other friction factors, the equivalent coefficient

5

λ

*D*

*P*

In accordance with Альтщуль equation, frictional factor

λ

1

*<sup>D</sup> <sup>C</sup>*

<sup>Δ</sup>

*m*

ρ

λ

*m*ρ

Thus:

steam.

is :

 —frictional factor; *q* —the mass flow rate, t/h; *L* —the length of the pipe, m;

*D* —the inner diameter of pipe, mm;

Equation (24) changes to (25).

1 2 *Z Z*, —compressibility factor of the input and output steam;

2

λ

Usually, there's little difference between the products of compressibility factors and temperatures in the same pipe. Considering the elbows, reducer extenders and

2

*D*

*m*

22 22

*P*

ρ

 η

+

(1 ) 1.25 10

8 12 12

*qP L*

 η

( ) 1.25 10 (1 )

5 8 <sup>1</sup> 1.25 10 (1 ) *m*

λ

<sup>68</sup> 0.25 0.11 ( ) *<sup>D</sup>* Re

Re 354 *Du <sup>m</sup> <sup>q</sup>*

In the equation, *u* —the characteristic velocity of the steam in this pipe, m/s;

the mean dynamic viscosity coefficient. It's determined by equation (33)[9].

ρ

μ

Δ —the equivalent absolute roughness, mm; Re —the Reynolds number of the

After transferring the units of *D q*, to mm and t/h (ton per hour), Reynolds number

*D*

 μ

<sup>=</sup> × +

ρ

*q P P CP P qP L*

 η<sup>=</sup> −= − × +

λ

22 8 1 1 2 5

*qP L P P*

1 1

−= × (24)

η

−= × (25)

λ:

= ×+ (28)

= = (29)

is added to the equation.

(26)

μ—

(27)

*m*

*D TZ*

ρ

22 8 12 2 1 2 5

*q PT Z L P P*

1.25 10

1 2 *P P*, —the input and output pressure of the steam in the pipe, Mpa;

By the law of energy conversation, the static thermal model can be written as follows:

$$T\_1 - T\_2 = \frac{(1+\mathcal{J})\eta\_l L}{1000c\_p q \times \frac{1000}{3600}} = \frac{(1+\mathcal{J})\eta\_l L}{278c\_p q} \tag{34}$$

$$q = \frac{278c\_pq^2}{(1+\beta)q\_lL}(T\_1 - T\_2) = C\_T(T\_1 - T\_2) \tag{35}$$

$$C\_T = \frac{278c\_pq^2}{(1+\beta)q\_lL} \tag{36}$$

The definitions and units of 1 2 *T T Lq* , , , are the same as eqation (24).(note: changing equation (34) to (35) is to make thermal model has the same pattern with equation (26)).

β —the appending heat loss coefficient for the appendix of pipe, valve, and support, the value can be 0.15-0.25 depending on pipe laying methods;

*<sup>p</sup> c* —specific heat capacity at constant pressure, kJ/kg.K;

*<sup>l</sup> q* —the amount of heat loss along unit pipe length, W/m.

In term of equation IF-97, The unclaimed symbols are defined in the reference[9].

$$\mathcal{L}\_p = -\mathcal{R}\pi^2(\mathcal{Y}\_{\mathcal{W}}^\rho + \mathcal{Y}\_{\mathcal{W}}^\rho) \tag{37}$$

The amount of heat loss along unit pipe length can be calculated using equation (38)

$$q\_l = \frac{T\_0 - T\_a}{\frac{1}{2\pi\mathcal{E}} \ln\frac{D\_o}{D\_i} + \frac{1}{\pi D\_0 a\_w}}\tag{38}$$

Data Processing Approaches for the Measurements of Steam Pipe Networks in Iron and Steel Enterprises 257

According to the hydraulic and thermal equations, for the pipe network equations:

\* \* *T T*

Equations (26), (27), (35), (36), (42), (43), (44) comprise the static model.

Industrial networks are equipped with temperature, pressure and flow meters except intermediate nodes. The proposed algorithm tries to calculate the flow rate of each pipe with the given condition. The flow rate meter readings are applied to evaluate of the algorithm.

Set figure 10 as an example. the index of the nodes and pipes are shown in the figure:

<sup>−</sup> <sup>=</sup> <sup>−</sup>

<sup>−</sup>

−

**Step 1.** Determine the elements of matrices A, Q as the preliminarily defined.

−

*A*

Substitute (42) into (41):

**Figure 10.** Layout of steam pipe network

*2.3.2. Hydraulic and thermal calculation based on searching* 

*Aq Q*+ = <sup>0</sup> (41)

*P T q CAP CAT* = = (42)

\* <sup>0</sup> *<sup>T</sup> AC A P Q <sup>P</sup>* + = (43)

\* <sup>0</sup> *<sup>T</sup> AC A T Q <sup>T</sup>* + = (44)

(45)

<sup>0</sup> , *<sup>a</sup> T T* —the out surface temperature of the pipe and the environment temperature, K; <sup>0</sup> , *D Di* —the outer and inner diameters of the steam pipe, mm;

ε—the coefficient of thermal conductivity of the heat insulated material, W/m.K;

In equation (38), *wa* is determined by:

$$a\_w = 11.6 + 7\sqrt{v} \tag{39}$$

*v* —the velocity of environmental air, m/s;

	- Incidence matrix of pipe network

Draw out pipe network map, number the nodes in the order of steam sources, users, three-way nodes, and number the pipes in the order of leaf pipes, branch pipes. Suppose there are *m*1 intermediate nodes and *m*2 outer nodes, there are total *mm m* 1 2 = + nodes and *p m*= − 1 pipes. Set the element of the ith row and the jth column in *m p A R* <sup>×</sup> <sup>∈</sup> as *ij <sup>a</sup>* , it's defined as:

$$a\_{ij} = \begin{cases} 1 & \text{(flow } \text{ in } \text{ from } node \text{ i)} \\ -1 \text{(flow } \text{ out } \text{ from } node \text{ i)} \\ 0 & \text{(unculated)} \end{cases} \tag{40}$$


Denote: 22 2 1 2 ( , ,..., )*<sup>T</sup> <sup>m</sup> P PP P* <sup>=</sup> , <sup>2</sup> ,( 1,..., ) *<sup>i</sup> Pi m* <sup>=</sup> the square of the ith node's pressure, MPa2;

1 2 ( , ,..., )*<sup>T</sup> <sup>m</sup> T TT T* = , ,( 1,..., ) *<sup>i</sup> Ti m* = the ith node temperature, °C; 1 2 ( , ,..., )*<sup>T</sup> <sup>p</sup> q qq q* <sup>=</sup> , ,( 1,..., ) *<sup>j</sup> qj p* <sup>=</sup> the flow rate of the jth pipe, t/h;

1 2 ( , ,..., )*<sup>T</sup> Q QQ Qm* <sup>=</sup> , ,( 1,..., ) *Qi m <sup>i</sup>* <sup>=</sup> the flow rate of the ith node, unit t/h; For steam souce the value is minus *<sup>j</sup> q* , and for the user the value is positive *<sup>j</sup> q* , otherwise, the value is 0.

$$\mathbf{C}\_{p}^{\*} = \text{diag}(\mathbf{C}\_{p1}, \mathbf{C}\_{p2}, \dots, \mathbf{C}\_{p}), \mathbf{C}\_{pj} (j = 1, \dots, p)\text{ parameters are determined by equation (27):}$$

\* 1 2 ( , ,..., ), ,( 1,..., ) *C diag C C C C j p <sup>T</sup> T T Tp Tj* = = the parameter are determined by equation(36).

In term of mass conservation law, the total flow rate of each node should be 0, that is:

Data Processing Approaches for the Measurements of Steam Pipe Networks in Iron and Steel Enterprises 257

$$A\vec{q} + Q = 0\tag{41}$$

According to the hydraulic and thermal equations, for the pipe network equations:

$$\vec{q} = \mathbf{C}\_P^\* A^T P = \mathbf{C}\_T^\* A^T T \tag{42}$$

Substitute (42) into (41):

256 Energy Efficiency – The Innovative Ways for Smart Energy, the Future Towards Modern Utilities

*l*

In equation (38), *wa* is determined by:

ii. Steam pipe network synthetic model

Denote: 22 2

1 2 ( , ,..., )*<sup>T</sup>*

1 2 ( , ,..., )*<sup>T</sup>*

the value is 0.

equation(36).

\*

(27); \*

is:

MPa2;


*v* —the velocity of environmental air, m/s;

column in *m p A R* <sup>×</sup> <sup>∈</sup> as *ij <sup>a</sup>* , it's defined as:

*ij*


1 2 ( , ,..., )*<sup>T</sup>*

 = − 

ε

2

<sup>0</sup> , *D Di* —the outer and inner diameters of the steam pipe, mm;

*<sup>q</sup> <sup>D</sup>* πε

<sup>−</sup> <sup>=</sup>

0

1 1 ln

*o*

*T T*

*a*

 π*D Da*

<sup>0</sup> , *<sup>a</sup> T T* —the out surface temperature of the pipe and the environment temperature, K;

—the coefficient of thermal conductivity of the heat insulated material, W/m.K;

Draw out pipe network map, number the nodes in the order of steam sources, users, three-way nodes, and number the pipes in the order of leaf pipes, branch pipes. Suppose there are *m*1 intermediate nodes and *m*2 outer nodes, there are total *mm m* 1 2 = + nodes and *p m*= − 1 pipes. Set the element of the ith row and the jth

> 1 ( ) 1( )

*a flow out from node i unrelated*

*flow in from node i*

*<sup>m</sup> P PP P* <sup>=</sup> , <sup>2</sup> ,( 1,..., ) *<sup>i</sup> Pi m* <sup>=</sup> the square of the ith node's pressure,

1 2 ( , ,..., )*<sup>T</sup> Q QQ Qm* <sup>=</sup> , ,( 1,..., ) *Qi m <sup>i</sup>* <sup>=</sup> the flow rate of the ith node, unit t/h; For steam souce the value is minus *<sup>j</sup> q* , and for the user the value is positive *<sup>j</sup> q* , otherwise,

1 2 ( , ,..., ), ,( 1,..., ) *C diag C C C C j p <sup>p</sup> P P Pp Pj* = = parameters are determined by equation

1 2 ( , ,..., ), ,( 1,..., ) *C diag C C C C j p <sup>T</sup> T T Tp Tj* = = the parameter are determined by

In term of mass conservation law, the total flow rate of each node should be 0, that

0( )

*<sup>m</sup> T TT T* = , ,( 1,..., ) *<sup>i</sup> Ti m* = the ith node temperature, °C;

*<sup>p</sup> q qq q* <sup>=</sup> , ,( 1,..., ) *<sup>j</sup> qj p* <sup>=</sup> the flow rate of the jth pipe, t/h;

+

0

(38)

(40)

11.6 7 *wa v* = + (39)

*i w*

$$A\mathcal{C}\_P^\* A^T P + \mathcal{Q} = 0 \tag{43}$$

$$A\stackrel{\ast}{C}\_T^\ast A^T T + \mathcal{Q} = 0\tag{44}$$

Equations (26), (27), (35), (36), (42), (43), (44) comprise the static model.

#### *2.3.2. Hydraulic and thermal calculation based on searching*

Industrial networks are equipped with temperature, pressure and flow meters except intermediate nodes. The proposed algorithm tries to calculate the flow rate of each pipe with the given condition. The flow rate meter readings are applied to evaluate of the algorithm.

**Figure 10.** Layout of steam pipe network

Set figure 10 as an example. the index of the nodes and pipes are shown in the figure:

**Step 1.** Determine the elements of matrices A, Q as the preliminarily defined.

$$A = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 \\ -1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & -1 \end{pmatrix} \tag{45}$$

$$Q = \begin{pmatrix} -q\_{1'}, q\_{2'}, q\_{3'}, q\_{4'}, 0, 0 \end{pmatrix}^T \tag{46}$$

Data Processing Approaches for the Measurements of Steam Pipe Networks in Iron and Steel Enterprises 259

presume the intermediate pressure values as the result of step 3, reduce step size and go

errors and calculation stability. The calculation flow chart is depicted with figure 11.

The specifications of each pipe are shown in table 1. The parameters in the model and

β= 0.15 ,

Substitute the specification data in Table 1 and temperature and pressure data in Table 2 to the algorithm calculating flow rate values.The comparision results are listed in Table 2. It shows the largest difference is less than 6%. Actually, the neglected factors in model, the

. If it is satisfied, the calculation ends with the result of step 3. If not,

ε

Outer Diameter (mm)

> Measured q(t/h)

Calculated q (t/h)

the relative difference

, , can be set as 0.05-0.3 depending on the demand for

 = 0.035 , 1,2,3 ξ

= 0.3 .

Heat Insulation Layer Thickness (mm)

\*

*<sup>T</sup> AC A T Q <sup>T</sup>* + ≤

algorithm are initially set as:

3

In the algorithm, the value 123

Pipe No. Length(m) Inner Diameter

**Table 1.** The Specifications of the Pipes

No. Pressure(MPa) Temperature

**Table 2.** The Comparisons of the Measured Data and the Calculated Data.

Outer node

to step 2 and start another calculation cycle.

*2.3.3. Comparisons of calculation results and measurements* 

η

parameter errors and measurement errors add to the difference.

ξξξ

= 0.2 , Δ = 0.2 ,

(mm)

1 600 700 720 110 2 2300 600 630 105 3 700 400 426 95 4 1300 500 529 100 5 1200 600 630 105

(°C)

1 0.705 286 117 121 3.4% 2 0.679 195 19 20 5.2%

3 0.563 246 19 19 0

4 0.445 256 79 82 3.8%

ξ

As the previous definitions for equation (41)(42), <sup>5656</sup> *PPTTq* , ,,, are unknown. From equation (27), it's easy to discover friction factor and density are relative with <sup>5656</sup> *PPTT* , , , . Only by presetting *q* can equations (42) (43) be applied for hydraulic iterative calculation. However, in thermal calculation only *<sup>p</sup> c* relates with <sup>5656</sup> *PPTT* , , , . Changing equation (34) to (35), *q* can be calculated directly.

$$q\_{i} = \frac{(1+\beta)q\_{li}L\_{i}}{278c\_{p}\Delta T\_{i}} , (i = 1,2,...,5) \tag{47}$$

In equation (47), *i* —the pipe index, *<sup>i</sup>* Δ*T* —the temperature difference of input and output. **Step 2.** Thermal calculation

It can be inferred by the flow direction of steam that:

$$P\_2 < P\_5 < P\_{1'} \max(P\_{3'}P\_4) < P\_6 < P\_5 \tag{48}$$

$$T\_2 < T\_5 < T\_{1'} \max(T\_{3'}T\_4) < T\_6 < T\_5 \tag{49}$$

Presume:

$$P\_5 = \frac{1}{2}(P\_2 + P\_1); P\_6 = \frac{1}{2}(P\_5 + \max(P\_3, P\_4))\tag{50}$$

Set the searching start value for <sup>5</sup> *T* , <sup>6</sup> *T* as 1*T* and max( , ) 3 4 *T T* respectively, and set the searching step size as h then begin searching <sup>5</sup> *T* , <sup>6</sup> *T* in the range (49). For each searching step size, calculate *<sup>p</sup> c* (equation (37)) and flow rate of each pipe (equation (49),(50)), and test whether *Aq Q* <sup>1</sup> + ≤ ξ is satisfied. If it is not satisfied, update 5 *<sup>T</sup>* , <sup>6</sup> *<sup>T</sup>* with one step to continue calculation, or retain the temperature value and the corresponding flow rates.

#### **Step 3.** Hydraulic calculation

Just as the thermal calculation, first presume the temperature of the intermediate nodes as the value searched by the former step, and set the start point and step size for the two intermediate pressures. Then begin searching in the region determined by inequality (48) with the step size to calculate the density (equation (32)), frictional factor ((28), (29), (33)), and flow rate ((26), (42)) of each pipe. When calculating the flow rate, presetting the initial values as the result of step 2 will reduce the iterative calculation time. Test whether *Aq Q* <sup>2</sup> + ≤ ξ is satisfied, If the inequality is not satisfied, update <sup>5</sup>*<sup>P</sup>* , <sup>6</sup>*P* with one step size to continue calculation, or retain the pressure values and the corresponding flow rates.

#### **Step 4.** Thermal model Verification

Substitute the intermediate nodes' temperature (determined by step 2), pressure and flow rate values (determined by step3) into thermal model to verify the inequality

\* 3 *<sup>T</sup> AC A T Q <sup>T</sup>* + ≤ ξ. If it is satisfied, the calculation ends with the result of step 3. If not,

presume the intermediate pressure values as the result of step 3, reduce step size and go to step 2 and start another calculation cycle.

In the algorithm, the value 123 ξξξ , , can be set as 0.05-0.3 depending on the demand for errors and calculation stability. The calculation flow chart is depicted with figure 11.

#### *2.3.3. Comparisons of calculation results and measurements*

258 Energy Efficiency – The Innovative Ways for Smart Energy, the Future Towards Modern Utilities

*i*

It can be inferred by the flow direction of steam that:

ξ

<sup>5656</sup> *PPTT* , , , . Only by presetting *q*

Changing equation (34) to (35), *q*

**Step 2.** Thermal calculation

test whether *Aq Q* <sup>1</sup> + ≤

**Step 3.** Hydraulic calculation

*Aq Q* <sup>2</sup> + ≤ ξ

**Step 4.** Thermal model Verification

Presume:

As the previous definitions for equation (41)(42), <sup>5656</sup> *PPTTq* , ,,, are unknown. From equation (27), it's easy to discover friction factor and density are relative with

iterative calculation. However, in thermal calculation only *<sup>p</sup> c* relates with <sup>5656</sup> *PPTT* , , , .

(1 ) ,( 1,2,...,5) <sup>278</sup> *li i*

In equation (47), *i* —the pipe index, *<sup>i</sup>* Δ*T* —the temperature difference of input and output.

5 2 1 6 5 34 1 1 ( ); ( max( , ))

Set the searching start value for <sup>5</sup> *T* , <sup>6</sup> *T* as 1*T* and max( , ) 3 4 *T T* respectively, and set the searching step size as h then begin searching <sup>5</sup> *T* , <sup>6</sup> *T* in the range (49). For each searching step size, calculate *<sup>p</sup> c* (equation (37)) and flow rate of each pipe (equation (49),(50)), and

continue calculation, or retain the temperature value and the corresponding flow rates.

Just as the thermal calculation, first presume the temperature of the intermediate nodes as the value searched by the former step, and set the start point and step size for the two intermediate pressures. Then begin searching in the region determined by inequality (48) with the step size to calculate the density (equation (32)), frictional factor ((28), (29), (33)), and flow rate ((26), (42)) of each pipe. When calculating the flow rate, presetting the initial values as the result of step 2 will reduce the iterative calculation time. Test whether

is satisfied, If the inequality is not satisfied, update <sup>5</sup>*<sup>P</sup>* , <sup>6</sup>*P* with one step size

to continue calculation, or retain the pressure values and the corresponding flow rates.

Substitute the intermediate nodes' temperature (determined by step 2), pressure and flow rate values (determined by step3) into thermal model to verify the inequality

2 2

*p i q L q i c T* + β

can be calculated directly.

<sup>1234</sup> ( , , , ,0,0)*<sup>T</sup> Q qqqq* = − (46)

can equations (42) (43) be applied for hydraulic

= = Δ (47)

2 5 1 34 6 5 *P P P PP P P* < < ,max( , ) < < (48)

2 5 1 34 6 5 *T T T TT T T* < < ,max( , ) < < (49)

*P P P P P PP* =+ =+ (50)

is satisfied. If it is not satisfied, update 5 *<sup>T</sup>* , <sup>6</sup> *<sup>T</sup>* with one step to

The specifications of each pipe are shown in table 1. The parameters in the model and algorithm are initially set as:η = 0.2 , Δ = 0.2 , β = 0.15 , ε = 0.035 , 1,2,3 ξ= 0.3 .

Substitute the specification data in Table 1 and temperature and pressure data in Table 2 to the algorithm calculating flow rate values.The comparision results are listed in Table 2. It shows the largest difference is less than 6%. Actually, the neglected factors in model, the parameter errors and measurement errors add to the difference.


**Table 1.** The Specifications of the Pipes


**Table 2.** The Comparisons of the Measured Data and the Calculated Data.

Data Processing Approaches for the Measurements of Steam Pipe Networks in Iron and Steel Enterprises 261

The results show the validation of the model and effectiveness of the algorithm, and the proposed model and algorithm can be applied to simulate the running of static steam pipe

As for the larger scale steam network with more than 3 intermediate nodes, it's difficult to apply the algorithm directly. However, the pipe network can be divided into several smaller

A section of the steam network named "S2" for an iron &steel plant in China is shown in figure 12. In the figure, N1-N7 represent the different production processes. The arrows point to the direction of steam flow, the variables Xi or Xij on behalf of the real steam mass flow. The electric valves are remotely controlled by the operators. Many industrial steam

If all of the variables above have been measured, suppose that the electric valves are fixed at a certain position, and the pipeline leakage and the amount of condensate water can be neglected, the constraint equations can be written out on the basis of the mass balance.

network, reconstruct the discovered fault data of the mass flow rates.

networks from the nodes with known temperature and pressure.

systems are similar in structure to this system, but the scale is much larger.

**3. Gross errors detection** 

**Figure 12.** A section diagram of steam network

**3.1. Problem definition** 

**Figure 11.** The flow chat of flow rate calculation

The results show the validation of the model and effectiveness of the algorithm, and the proposed model and algorithm can be applied to simulate the running of static steam pipe network, reconstruct the discovered fault data of the mass flow rates.

As for the larger scale steam network with more than 3 intermediate nodes, it's difficult to apply the algorithm directly. However, the pipe network can be divided into several smaller networks from the nodes with known temperature and pressure.
