Power flow in MV radial network and calculation of voltage drop and loss

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.

Load in three phase system
Figure 1 – Load in three phase system

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.

Power flow in MV radial network, calculation of voltage drop and loss
Power flow in MV radial network, calculation of voltage drop and loss (photo credit: MAMA Z via Flickr)

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

The node where the load is connected
Figure 2 – The node where the load is connected

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.

Power flow around a node
Figure 3 – Power flow around a node

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”
Representation of a branch
Figure 4 – Representation of a branch

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 + dU22)
U2 =√((U1-DU1)2 + dU12)

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/U22 DQ = (P”2 + Q”2) x/U22
DP = ( P’2 + Q’2) r/U12 DQ = (P’2 + Q’2) x/U12

Radial network is a network whose configuration does not contain any loop (tree like configuration).

Principles of calculation (an example)

Radial network
Figure 5 – Radial network

DEDEE is last node
BFBFF is last node
GHGHH is last node

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

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

For the last node F: ABF
For the last node H: ABCDGH

Compute a radial network (approximately):

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

Example of radial network calculation

Example of radial network calculation
Figure 6 – 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

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About Author //


Edvard Csanyi

Edvard - Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry fascilities. Professional in AutoCAD programming. Present on


  1. Shandukani Vhulondo
    Jul 02, 2017

    Thank You very much for sharing such valuable information

  2. José Jamal
    Jul 01, 2017

    Thanks for sharing such a wonderful issue. It is likely to be forgoten with time.

  3. Alfred Gjylameti
    Nov 02, 2016

    Hi Mr. Edvart

    I need a design program for 20kV overhead transmission lines. May you help me?

    Best Regards

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