2.3 DERIVATION OF THE TRANSMISSION LOSS FORMULA

This section presents an easier method for derivation of the transmission loss formula. Fig. 2.1 represents two generating units 1 and 2 feeding a number of loads, totaling to I_L through a large transmission network. Out of the many branches within the network, consider a branch (any three-phase transmission line) designated as b.

Let the switch in Fig.2.1(a) be open. Then the total load current I_L is supplied by generator 1 only. With generator 1 acting alone, the current in the line b is I_b1 whereas the generator current is I₁.

Now define current distribution factor a_b1 as:

Similarly, let the switch in Fig.2.1(b) be open and the total load current I_L be supplied by generator 2 only. With generator 2 acting alone, the current in the line b is I_b2 whereas the generator current is I₂.

Now define current distribution factor a_b2 as:

When both the generators are connected to supply the load current, the current through the branch b i.e. I_b can be obtained by applying the principle of superposition as:

 

The current distribution factors are taken as real numbers owing to the following assumptions:

1. The phase angle of all load currents are the same.
2. The ratio X/R for all transmission lines in the transmission network is the same.

Where θ₁, θ₂ are the phase angles with respect to a common reference.

 

 

From Fq.(2.12).

 

Now, let

P₁, P₂ = 3-Phase real power outputs by generators 1 and 2 respectively.

I₁ = (P₁/ × V₁cosθ₁); I₂ = (P₂/ × V₂cosθ₂)

cosθ₁, cosθ₂ = power factors at generator ends 1 and 2 respectively

V₁, V₂ = bus voltages at generator ends 1 and 2 respectively

R_b is the resistance of branch b

Let k be the number of branches in the transmission network.

Now, the total transmission loss in the entire network is given by

Substitute I²_b from Eq. (2.14) in Eq.(2.15)

Substituting I₁ and I₂

The above loss equation in terms of B-coefficients can be written as:

 

where,

The terms B₁₁, B₁₂ and B₂₂ are called loss coefficients or B-Coefficients. The units of B-coefficients are in MW^–1 when the voltages are in KV and branch resistances in ohms.

B-coefficients for multi-machine power systems.

Let Eq. (2.16) be extended for a power system with n_g number of generators. Total transmission losses in the system are:

Where B_mn are elements of the S-Matrix.

Diagonal terms in the B-Matrix:

Off-Diagonal terms in B-Matrix:

Expanding Eq.(2.17).

 

In the matrix form of Eq.(2.20),

In the condensed form, Eq.(2.21) can be written as

 

Example 2.1

A four-bus power system network is shown in Fig 2.2(a).

Currents flowing in the lines are marked and are given as:

 

 

The branch impedances are:

 

 

Take voltage at bus 1 as p.u and base MVA as 100. Compute B-coefficients of the network.

Solution:

 

The load currents at bus 2 and bus 4 are:

 

 

Determination of current distribution factors:

Step-1: Let us assume that the total load be supplied by generator G₁ alone.

I_L = Total load current = I_L1 + I_L2 = 9 – j 2.25 p.u referring to Fig.2.2(b), the current distribution factors can be determined as below.

Step-2: Let the total load be supplied by generator G₂ alone.

Referring to Fig 2.2(c), the current distribution factors are computed as shown below:

With the help of Fig. 2.1(a), the bus voltages can be determined as follows:

The generator currents and voltages are:

The phase angle at the Generator-1 terminal Φ₁ = 0° – (–14.03°) = 14.03°

The phase angle at the Generator-2 terminal Φ₂ = –6.115 – (–14.03°) = 7.915°

The plant on generator power factors are:

 

 

The current phase angles at G₁ and G₂ terminals are:

Therefore, cos(δ₁ – δ₂) = cos 0° = 1 (i.e. I₁ and I₂ are in phase)

On a 100MVA base the loss coefficients can be obtained by dividing the p.u values by 100

Note:

The following assumptions are made for simplifying the procedure of computing B-coefficients.

1. All load currents have the same phase angle with respect to a common reference. It can be seen in Example 2.1, that

   I_L1 = 6 – j 1.5 = 6.18 p.u and

   I_L2 = 3–j0.75 = 3.092 p.u. The phase angle of both the load currents is the same i.e., 14°.

2. The X/R ratio for all branches of the network are the same. It can be seen in Example 2.1 that the X/R ratio of the branches x, y and z are the same and its value is 4.

   The above two assumptions are essential to make the current distribution factor as real numbers, rather than complex numbers.

3. Voltage magnitudes at the generator buses remain constant.
4. The phase angles between generator bus voltages and currents for all the generating units are the same. In other words, power factor at each plant remains constant.

   It can be seen in Example 2.1 that the Generator-1 current I₁ = 4.123 and Generator-2 current I₂ = 5.153 That is, δ₁ = δ₂ = – 14.03°.

Example 2.2

Consider the power system shown in Fig.2.3.

The power loss equation is represented as P_L = B₁₁P²₁ + 2B₁₂P₁P₂ + B₂₂P₂². All the terms in the equation for P_L are standard notations. The resistances of lines are marked in the figure. Compute the B-coefficients in terms of resistances.

Solution:

 

From the figure,

 

where,

The total power loss in the system is:

 

Simplying the above,

Representing the above in the standard form,

 

where

and

Example 2.3

The power system in Example 2.2 is modified as shown in the Fig. 2.4. Compute the B-coefficients

Solution:

 

Comparing Fig.2.3 and Fig.2.4, it can be understood that R₃ = 0 in this case. Substituting R₃ = 0,

The modified loss equation is

 

Example 2.4

The power system in the Example 2.2 is modified as shown in the Fig 2.5. Compute the B-coefficients.

Solution:

 

In this case, the load is connected near Generator-2.

By substituting R₂ = R₃ = 0, the B-coefficients are:

The modified loss equation is:

Any amount of power supplied by G₂ will not produce losses.