**7.3 Practical example**

of added units and their cost are reduced if the two system are interconnected

A 4 12.63 5.652 2 9.44 1.054 B 2 16.42 4.852 1 8.75 2.045

Therefore, it can be concluded from the above analysis that both systems will benefit from the interconnection. The reliability of both systems can be improved, and consequently the cost of service will be reduced through interconnection and reserve sharing. However, this is not the overall saving because the systems must be linked together in order to create an integrated system. The next stage must, therefore, assess the economic worth that may result from either interconnection or

Since embarking on the national industrial development and the industrials program in the Kingdom of Saudi Arabia, the Ministry of Energy, Industry and Mineral Resources launched two solar PV projects with a combined generation capacity of 1.51 GW enough to power 226,500 households. These projects will be tendered by mid-2019 to attract a total investment of \$1.51 billion Saudi Riyals creating over 4500 jobs during construction, operations, and maintenance [13]. The program will be phased and rolled out in a systematic and transparent way to ensure that the Kingdom benefits from the cost-competitive nature of renewable energy. The National Renewable Energy Program aims to substantially increase the share of renewable energy in the total energy mix, targeting the generation of 27.3 gigawatts (GW) of renewable energy by 2024 and 58.7 GW by 2030. This initiative sets out an organized and specific road map to diversify local energy sources, stimulate economic development, and provide sustainable economic stability to the Kingdom in light of the goals set for Vision 2030, which include establishing the renewable energy industry and supporting the advancement of this promising sector.

As a result of the continuous subsidy and generous support of the government for the electricity sector, the ministry has been able to accomplish many electrical projects in both urban and rural areas, resulting in electric services that can reach remote areas and sparsely populated areas, over rough roads and rugged terrain. In fact, electric services require large sums of money to finance, build, operate, safeguard, and sustain. Another important component that must be considered along with the continuous operation and maintenance expenditures is fuel costs. Therefore, constant maintenance measures ought to be implemented to ensure the level and continuity of the flow of electrical energy without fluctuation, decline, or

**No. of unit Cost (MSR)** *ϵENS* **(MWh) No. of unit Cost (MSR)** *ϵENS* **(MWh)**

**System Isolated Interconnected**

increasing generating capacity individually and independently.

**7. Transmission and distribution reliability evaluation**

**7.2 Role of the government in the electricity sector**

rather than being isolated.

**7.1 Introduction**

interruption.

**156**

System costs as isolated and interconnected:

*Reliability and Maintenance - An Overview of Cases*

One practical example demonstrating the evolving of industry of electric sector in the Kingdom of Saudi Arabia will be shown in this section. The availability of network can be analyzed in a similar manner to that used in generating capacity evaluation (Section 3.1). Therefore, the probability of failing to satisfy the criterion of service adequacy and continuity can be evaluated. Provided the appropriate component reliability indices are known, it is relatively simple to evaluate the expected failure rate (*λ*) of the system, the average duration of the outage I, and the unavailability or annual outage time (*U*). To do this, the values of *λ*, *r*, and *U* are required for each component of the system.

### *7.3.1 State probabilities*

The state-space transition diagram for a two-component system is shown in **Figure 10**.

The probability of a component being in the up state is . Also, the probability of a component being in the down state is *<sup>λ</sup> λ*þ*μ*.

$$\text{Probability of being in state } 1 = \frac{\mu\_1}{\mu\_1 + \lambda\_1} \cdot \frac{\mu\_2}{\mu\_2 + \lambda\_2}$$

$$\text{Probability of being in state } 2 = \frac{\lambda\_1}{\mu\_1 + \lambda\_1} \cdot \frac{\mu\_2}{\mu\_2 + \lambda\_2}$$

$$\text{Probability of being in state } 3 = \frac{\mu\_1}{\mu\_1 + \lambda\_1} \cdot \frac{\lambda\_2}{\mu\_2 + \lambda\_2}$$

$$\text{Probability of being in state } 4 = \frac{\lambda\_1}{\mu\_1 + \lambda\_1} \cdot \frac{\lambda\_2}{\mu\_2 + \lambda\_2} \tag{10}$$

The most accurate method for analyzing networks including weather states is to use the Markov modeling. However, this becomes impractical for all except the simplest system. Instead, therefore, an approximate method is used based upon simple rules of probability.

**Figure 10.**

*State-space diagram for two-component system, where λ is the failure rate and μ is the repair rate* <sup>¼</sup> <sup>1</sup> *<sup>r</sup>* ð Þ *r* ¼ *repair time :*

#### *7.3.2 Series components*

The requirement is to find the reliability indices of a single component that is equivalent to a set of series-connected components as shown in **Figure 11**.

If the components are in series from a reliability point of view, both must operate, i.e., be in upstate, for the system to be successful, i.e., the upstate of a series system is state 1 of the state-space diagram shown in **Figure 11**.

From the above equation (state 1), the probability of being in this upstate is In addition, since , the above equation becomes

$$r\_s = \frac{\lambda\_1 r\_1 + \lambda\_2 r\_2 + \lambda\_1 \lambda\_2 r\_1 r\_2}{\lambda\_s} \tag{11}$$

$$=\frac{\lambda\_1 r\_1 + \lambda\_2 r\_2}{\lambda\_2} \tag{12}$$

*λ<sup>s</sup>* ¼ *λ*<sup>1</sup> þ *λ*<sup>2</sup> ¼ ∑

*rs* <sup>¼</sup> <sup>∑</sup>*<sup>n</sup>*

In particular, the order of evaluation is usually *λs*ð Þ ¼ ∑*λ , Us*ð Þ ¼ ∑*λ r* and

Although these equations were derived from the assumption of exponential distribution, they are expected or average values and can be shown to be valid

Many systems consist of both series and parallel connections. These systems can

In this case, the requirement is to find the indices of a single component that is

If the components are in parallel from a reliability point of view, both must fail for resulting in a system failure, i.e., the down state of a parallel system is state 4 of the state-space diagram shown in **Figure 10**. From (10), the probability of being in

Also, the rate of transition from state 4 of the two-component state-space dia-

<sup>¼</sup> *<sup>λ</sup><sup>p</sup> λ<sup>p</sup>* þ *μ<sup>p</sup>*

*λ*1*λ*<sup>2</sup> ð Þ *μ*<sup>1</sup> þ *λ*<sup>1</sup> ð Þ *μ*<sup>2</sup> þ *λ*<sup>2</sup>

be seen in transmission lines, in combinations of transformers, cables, feeders, relays, protection and control devices, etc. As an example, **Figure 12** displays two parallel lines that are both connected in series with another line. In these situations, and from a reliability point of view, it is essential to consequently reduce the network in order to estimate its overall reliability. This is accomplished by repeatedly combining sets of parallel and series components into equivalent network components until a single component remains. The reliability of the last component

is equal to the reliability of the original system (**Figure 12**).

equivalent to two parallel components as shown in **Figure 12**.

*rs*ð Þ ¼ *Us=λ<sup>s</sup>* .

*7.3.3 Parallel components*

this downstate is

gram is . Therefore

**Figure 12.**

**159**

*State-space diagram for a two-component system.*

irrespective of the distribution assumption.

*Reliability Evaluation of Power Systems DOI: http://dx.doi.org/10.5772/intechopen.85571*

Thus, the unavailability for series systems (*US*Þ can be expressed as

*n i*¼1

*<sup>i</sup>*¼1*λ<sup>i</sup> ri* ∑*<sup>n</sup> <sup>i</sup>*¼1*λ<sup>i</sup>*

*λ<sup>i</sup>* (14)

*US* ¼ *λ<sup>s</sup> rs* (16) ¼ ∑*λ r* (17)

(15)

(18)

$$\dot{r} = \frac{\sum\_{i=1}^{n} \lambda\_i \ r\_i}{\lambda\_i} \tag{13}$$

Also, the rate of transition from state 1 of the two-component state-space diagram is , therefore

**Figure 11.** *State-space diagram for two-component system.*

*Reliability Evaluation of Power Systems DOI: http://dx.doi.org/10.5772/intechopen.85571*

$$
\lambda\_t = \lambda\_1 + \lambda\_2 = \sum\_{i=1}^n \lambda\_i \tag{14}
$$

$$r\_s = \frac{\sum\_{i=1}^{n} \lambda\_i \ r\_i}{\sum\_{i=1}^{n} \lambda\_i} \tag{15}$$

Thus, the unavailability for series systems (*US*Þ can be expressed as

$$U\_S = \lambda\_s \ r\_s \tag{16}$$

$$= \sum \lambda r \tag{17}$$

In particular, the order of evaluation is usually *λs*ð Þ ¼ ∑*λ , Us*ð Þ ¼ ∑*λ r* and *rs*ð Þ ¼ *Us=λ<sup>s</sup>* .

Although these equations were derived from the assumption of exponential distribution, they are expected or average values and can be shown to be valid irrespective of the distribution assumption.

### *7.3.3 Parallel components*

*7.3.2 Series components*

*<sup>r</sup>* ð Þ *r* ¼ *repair time :*

*Reliability and Maintenance - An Overview of Cases*

**Figure 10.**

*repair rate* <sup>¼</sup> <sup>1</sup>

diagram is , therefore

*State-space diagram for two-component system.*

**Figure 11.**

**158**

The requirement is to find the reliability indices of a single component that is

From the above equation (state 1), the probability of being in this upstate is

*rs* <sup>¼</sup> *<sup>λ</sup>*1*r*<sup>1</sup> <sup>þ</sup> *<sup>λ</sup>*2*r*<sup>2</sup> <sup>þ</sup> *<sup>λ</sup>*1*λ*2*r*1*r*<sup>2</sup> *λs*

> <sup>¼</sup> *<sup>λ</sup>*1*r*<sup>1</sup> <sup>þ</sup> *<sup>λ</sup>*2*r*<sup>2</sup> *λs*

> > *<sup>i</sup>*¼<sup>1</sup>*λ<sup>i</sup> ri λs*

<sup>¼</sup> <sup>∑</sup>*<sup>n</sup>*

Also, the rate of transition from state 1 of the two-component state-space

(11)

(12)

(13)

equivalent to a set of series-connected components as shown in **Figure 11**. If the components are in series from a reliability point of view, both must operate, i.e., be in upstate, for the system to be successful, i.e., the upstate of a series

system is state 1 of the state-space diagram shown in **Figure 11**.

*State-space diagram for two-component system, where λ is the failure rate and μ is the*

In addition, since , the above equation becomes

Many systems consist of both series and parallel connections. These systems can be seen in transmission lines, in combinations of transformers, cables, feeders, relays, protection and control devices, etc. As an example, **Figure 12** displays two parallel lines that are both connected in series with another line. In these situations, and from a reliability point of view, it is essential to consequently reduce the network in order to estimate its overall reliability. This is accomplished by repeatedly combining sets of parallel and series components into equivalent network components until a single component remains. The reliability of the last component is equal to the reliability of the original system (**Figure 12**).

In this case, the requirement is to find the indices of a single component that is equivalent to two parallel components as shown in **Figure 12**.

If the components are in parallel from a reliability point of view, both must fail for resulting in a system failure, i.e., the down state of a parallel system is state 4 of the state-space diagram shown in **Figure 10**. From (10), the probability of being in this downstate is

$$\frac{\lambda\_1 \lambda\_2}{(\mu\_1 + \lambda\_1)(\mu\_2 + \lambda\_2)} = \frac{\lambda\_p}{\lambda\_p + \mu\_p} \tag{18}$$

Also, the rate of transition from state 4 of the two-component state-space diagram is .

Therefore

**Figure 12.** *State-space diagram for a two-component system.*

$$\frac{1}{r\_p} = \frac{1}{r\_1} + \frac{1}{r\_2} \tag{19}$$

where 8760 is the total number of hours in a year, using Eq. (20)

<sup>¼</sup> <sup>5</sup> � <sup>5</sup>

<sup>¼</sup> <sup>2</sup>*:*<sup>854</sup> � <sup>10</sup>�<sup>4</sup> � <sup>2</sup>*:*5*=*<sup>8760</sup>

<sup>¼</sup> <sup>7</sup>*:*<sup>135</sup> � <sup>10</sup>�<sup>4</sup> <sup>h</sup>*=*yrs*:*

*λ<sup>C</sup>* ¼ *λ<sup>B</sup>* þ *λ*<sup>3</sup>

*rC* <sup>¼</sup> *UB* <sup>þ</sup> *<sup>λ</sup>*3*r*<sup>3</sup> *λC*

¼ 9*:*977 h*:*

*UC* ¼ *λCrC*

events are very unreliable, the opposite effect may occur.

applying the previous series equations to these overlapping outages.

*7.3.4 Network reduction for failure mode analysis*

above) and then combined with line 3 in series, using Eq. (14):

<sup>5</sup> <sup>þ</sup> <sup>5</sup> <sup>¼</sup> <sup>2</sup>*:*5 h*:*

<sup>¼</sup> <sup>8</sup>*:*<sup>145</sup> � <sup>10</sup>�<sup>8</sup> yrs*=*yrs <sup>¼</sup> *probability*

To find indices at busbar C, lines 1 and 2 must be combined in parallel (as done

<sup>¼</sup> <sup>2</sup>*:*<sup>854</sup> � <sup>10</sup>�<sup>4</sup> <sup>þ</sup> <sup>0</sup>*:*<sup>1</sup> <sup>¼</sup> <sup>1</sup>*:*<sup>003</sup> � <sup>10</sup>�<sup>1</sup> *<sup>f</sup> <sup>=</sup>yr*

*rC* <sup>¼</sup> <sup>7</sup>*:*<sup>135</sup> � <sup>10</sup>�<sup>4</sup> <sup>þ</sup> <sup>0</sup>*:*<sup>1</sup> � <sup>10</sup> <sup>1</sup>*:*<sup>003</sup> � <sup>10</sup>�<sup>1</sup>

<sup>∴</sup>*UC* <sup>¼</sup> <sup>1</sup>*:*<sup>003</sup> � <sup>10</sup>�<sup>1</sup> � <sup>9</sup>*:*<sup>977</sup> ¼ 1*:*001 h*=*yrs*:*

In this case, it is seen that the indices of busbar C are dominated by the indices of line 3. This is clearly expected since busbar C will be lost if either line 3 or lines 1 and 2 simultaneously fail. Consequently, loss of line 3 is a first-order event, and loss of lines 1 and 2 are a second-order event. It must be stressed that this is only true if the reliability indices of the components are comparable; if the component forming the low-order event is very reliable and the components forming the higher order

In some cases, some critical or unreliable areas become absorbed into equivalent elements and become impossible to identify. The alternative is to impact the system and compose a list of failure nodes, i.e., component outages that must overlap to cause a system outage. These overlapping outages are effectively parallel elements and can be combined using the equations for parallel components. Any one of these overlapping outages will cause system failure and therefore, from a reliability point of view, are effectively in series. The system indices can therefore be evaluated by

*rB* <sup>¼</sup> *<sup>r</sup>*<sup>1</sup> *<sup>r</sup>*<sup>2</sup> *r*<sup>1</sup> þ *r*<sup>2</sup>

*Reliability Evaluation of Power Systems DOI: http://dx.doi.org/10.5772/intechopen.85571*

*UB* ¼ *λBrB*

Using Eq. (23)

**161**

or

$$r\_p = \frac{r\_1 r\_2}{r\_1 + r\_2} \tag{20}$$

From the above equations, it yields that

$$
\lambda\_p = \frac{\lambda\_1 \lambda\_2 (r\_1 + r\_2)}{1 + \lambda\_1 r\_1 + \lambda\_2 r\_2} \tag{21}
$$

$$=\lambda\_1 \lambda\_2 (r\_1 + r\_2) \tag{22}$$

Thus, the unavailability for parallel systems *Up* can be expressed as

$$U\_p = \lambda\_p r\_p \tag{23}$$

In practice, the order of evaluation is usually

$$
\lambda\_s = \sum \lambda \qquad \qquad \nu\_s = \sum \lambda \cdot r \qquad \qquad \text{and} \qquad \qquad r\_s = \frac{\upsilon\_s}{\lambda\_s}
$$

Although these equations were derived from the assumption of exponential distribution, they are expected or average values and can be shown to be valid irrespective of the distributional assumption.

**Example (series/parallel):** To illustrate the applications of these techniques, let us consider the transmission lines supplying the newly large industrial park constructed near Riyadh city (the capital of the KSA) within what is called "industrial cities" in the main cities of the KSA. The transmission lines with their data load points are given below (see **Figure 13**). It is required to evaluate the load point (busbar) reliability indices at busbars B and C.

To find the indices at busbar B, lines 1 and 2 must be combined in parallel using Eq. (22):

$$\begin{aligned} \dot{\lambda}\_{\mathcal{B}} &= \dot{\lambda}\_1 \dot{\lambda}\_2 \, (r\_1 + r\_2) \\ &= \mathbf{0}. \mathbf{5} \times \mathbf{0}. \mathbf{5} (\mathbf{5} + \mathbf{5}) / \mathbf{8760} \\ &= \mathbf{2.854} \times \mathbf{10}^{-4} f / \mathbf{y} \end{aligned}$$

**Figure 13.** *Transmission lines configuration with data load points.*

1 *rp* ¼ 1 *r*1 þ 1 *r*2

From the above equations, it yields that

*Reliability and Maintenance - An Overview of Cases*

Thus, the unavailability for parallel systems *Up*

In practice, the order of evaluation is usually

irrespective of the distributional assumption.

(busbar) reliability indices at busbars B and C.

*Transmission lines configuration with data load points.*

*rp* <sup>¼</sup> *<sup>r</sup>*<sup>1</sup> *<sup>r</sup>*<sup>2</sup> *r*<sup>1</sup> þ *r*<sup>2</sup>

*<sup>λ</sup><sup>p</sup>* <sup>¼</sup> *<sup>λ</sup>*<sup>1</sup> *<sup>λ</sup>*2ð Þ *<sup>r</sup>*<sup>1</sup> <sup>þ</sup> *<sup>r</sup>*<sup>2</sup> 1 þ *λ*1*r*<sup>1</sup> þ *λ*2*r*<sup>2</sup>

Although these equations were derived from the assumption of exponential distribution, they are expected or average values and can be shown to be valid

us consider the transmission lines supplying the newly large industrial park constructed near Riyadh city (the capital of the KSA) within what is called "industrial cities" in the main cities of the KSA. The transmission lines with their data load points are given below (see **Figure 13**). It is required to evaluate the load point

**Example (series/parallel):** To illustrate the applications of these techniques, let

To find the indices at busbar B, lines 1 and 2 must be combined in parallel using

*λ<sup>B</sup>* ¼ *λ*<sup>1</sup> *λ*<sup>2</sup> ð Þ *r*<sup>1</sup> þ *r*<sup>2</sup>

¼ 0*:*5 � 0*:*5 5ð Þ þ 5 *=*8760

<sup>¼</sup> <sup>2</sup>*:*<sup>854</sup> � <sup>10</sup>�<sup>4</sup>*<sup>f</sup> <sup>=</sup><sup>y</sup>*

¼ *λ*1*λ*2ð Þ *r*<sup>1</sup> þ *r*<sup>2</sup> (22)

can be expressed as

*Up* ¼ *λprp* (23)

or

Eq. (22):

**Figure 13.**

**160**

(19)

(20)

(21)

where 8760 is the total number of hours in a year, using Eq. (20)

$$r\_B = \frac{r\_1 r\_2}{r\_1 + r\_2} = \frac{5 \times 5}{5 + 5} = 2.5 \text{ h}.$$

$$U\_B = \lambda\_B r\_B$$

$$= 2.854 \times 10^{-4} \times 2.5 / 8760$$

$$= 8.145 \times 10^{-8} \text{ yrs/yrs} = 
probability$$

$$= 7.135 \times 10^{-4} \text{ h/yrs}.$$

To find indices at busbar C, lines 1 and 2 must be combined in parallel (as done above) and then combined with line 3 in series, using Eq. (14):

$$\begin{aligned} \dot{\lambda}\_C &= \dot{\lambda}\_B + \dot{\lambda}\_3\\ &= 2.854 \times 10^{-4} + 0.1\\ &= 1.003 \times 10^{-1} f/yr\\ r\_C &= \frac{U\_B + \dot{\lambda}\_3 r\_3}{\dot{\lambda}\_C} \\ r\_C &= \frac{7.135 \times 10^{-4} + 0.1 \times 10}{1.003 \times 10^{-1}} \\ &= 9.977 \text{ h} \end{aligned}$$

Using Eq. (23)

$$\begin{aligned} U\_C &= \lambda\_C r\_C \\ \vdots U\_C &= \mathbf{1.003} \times \mathbf{10^{-1}} \times \mathbf{9.977} \\ &= \mathbf{1.001} \,\mathrm{h/yrs} .\end{aligned}$$

In this case, it is seen that the indices of busbar C are dominated by the indices of line 3. This is clearly expected since busbar C will be lost if either line 3 or lines 1 and 2 simultaneously fail. Consequently, loss of line 3 is a first-order event, and loss of lines 1 and 2 are a second-order event. It must be stressed that this is only true if the reliability indices of the components are comparable; if the component forming the low-order event is very reliable and the components forming the higher order events are very unreliable, the opposite effect may occur.

#### *7.3.4 Network reduction for failure mode analysis*

In some cases, some critical or unreliable areas become absorbed into equivalent elements and become impossible to identify. The alternative is to impact the system and compose a list of failure nodes, i.e., component outages that must overlap to cause a system outage. These overlapping outages are effectively parallel elements and can be combined using the equations for parallel components. Any one of these overlapping outages will cause system failure and therefore, from a reliability point of view, are effectively in series. The system indices can therefore be evaluated by applying the previous series equations to these overlapping outages.

The following case study showcases the existing tie-line interconnecting the eastern region (ER) with the central region (CR) (400 km apart) in the Kingdom of Saudi Arabia (KSA). The ER is actually the incubator of the oil industry and all its refineries and infrastructures. Riyadh is located in the CR, which is the domicile of the Saudi Electric Company (SEC). The latter is envisioning tremendous expansion with vast increasing industrial future loads. Therefore, a huge bulk of electric power is transferred from the ER to the CR via the interconnecting tie-line. Therefore, to evaluate its reliability using the concepts and methodology stated above, the tie-line (see **Figure 14**) is considered bearing the following data:
