## Power flow in radial network

### Symmetrical load

In three phase system the load for each phase should be the same. We call symmetrical load **when the load of each phase is equal**.

In symmetrical circuit each phase has same configuration, so in drawing we can represent only one phase and omit 2 other phases. The diagram that shows only one line is called single line diagram.

The load of different types has different representations. In the power system the best presentation is power load **S = (P, Q)**.

The point to where the load is connected is called node. Ordinary this is may be a feeder, a derivation of MV network, a transformer post. The tension at this point we call node voltage.

### Electrical scheme, single diagram

Suppose the loads are symmetrical. So for **phase A** the scheme is the same as for **phase B** or **phase C**. To show one phase is sufficient.

**At the beginning node of a branch:**

- Node voltage by
**U1**, - Active power flow from the beginning by
**P’**and - Reactive power flow by
**Q’**

**At the end node of a branch:**

- Node voltage by
**U2**, - Active power flow to the end node by
**P”**and - Reactive power flow by
**Q”**

**Voltage drop has 2 components, a longitudinal and a transversal component.**

**DU2**= (P” r + Q”x)/U2 (longitudinal)**dU2**= (P” x – Q” r)/U2 (transversal)**DU1**= (P’ r + Q’ x)/U1**dU1**= (P’ x – Q’ x)/U1

In general when the power factor is about **0.8** the transversal component of voltage drop is small compare to the longitudinal one. **dU2 can be drop out in calculation.**

**Node voltage:**

U1 = √((U2 + DU2)^{2} + dU2^{2})

U2 =√((U1-DU1)^{2} + dU1^{2})

**In normal case, when power factor is greater than 0.8:**

U1 ≈ U2 + DU2

U2 ≈ U1 – DU1

**Voltage loss:**

DU = U1 – U2

**Power loss:**

DP = ( P”^{2} + Q”^{2}) r/U2^{2} DQ = (P”2 + Q”^{2}) x/U2^{2}

DP = ( P’^{2} + Q’^{2}) r/U1^{2} DQ = (P’2 + Q’^{2}) x/U1^{2}

**does not contain any loop (tree like configuration)**.

#### Principles of calculation (an example)

Beginning | Ending | Branch | Comment |

A | B | AB | |

B | C | BC | |

C | D | CD | |

D | E | DE | E is last node |

B | F | BF | F is last node |

D | G | DG | |

G | H | GH | H is last node |

**To find out the path from the first node to last node:**

- Write down branch of last node, ex:
**DE** - Get the previous node to the node D:
**C so we have CDE** - Repeat this process until the first node

**BCDE**

**ABCDE**

For the last node F: **ABF**

For the last node H: **ABCDGH**

**Compute a radial network (approximately):**

- Find the path to each last node
- Compute the power flow for each branch
- From the first to the last node compute all node voltage, voltage drop in each branch and power loss.

#### Example of radial network calculation

**1.** The path is {0, 1, 2}

**2.** From node 2 the power flows are:

**P12** = P2 + 0 = P2 = 1 MW

**Q12** = Q2 + 0 = Q2 = 0.6 MVar

**From node 1 the power flows are:**

**P01** = P1 + P12 = 0.8 + 1 = 1.8 MW

**Q01** = Q1 + Q12 = 0.3 + 0.6 = 0.9 MVar

**3.** Compute voltage drop and power loss:

**From node 0 to 1 (branch { 0, 1})**

DU1(01) = (P01.r01 + Q01.x01)/U0 = (1.8 x 1+0.9 x 1)/22 = **0.123 kV**

U1 = U0 – DU1 = **21.877 kV**

**From node 1 to node 2 (branch {1, 2}) **

**DU1(12)** = (P12.r12+Q12.x12)/U1 = (1 x 1.5 + 0.6 x 1)/21.877 = **0.096 kV**

**U2** = U1 – DU1(12) = 21.877 – 0.096 = **21.781 kV**

**Total voltage drop =**22 – 21.781 = 0.219 kv that means

**1%**.

**Reference //** Handbook for installation of medium voltage lines by Mr. Ky Chanthan, Mr. Theng Marith