Products

Holooly Rewards

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

Holooly Ads. Manager

Advertise your business, and reach millions of students around the world.

Holooly Tables

All the data tables that you may search for.

Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Holooly Sources

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Holooly Help Desk

Need Help? We got you covered.

Chapter 21

Q. 21.1

A load of 30 MW, 45 MVAR is connected to a line where X to R ratio is 5 and the short circuit capacity of the load bus is 250 MVA. The supply voltage is 11 kV and the load is star connected. Determine the load bus voltage.

Step-by-Step

Verified Solution

The load per phase is (10+j 15) MVA and SCC. \frac{250}{3} MVA, voltage L-N= \frac{11}{\sqrt{3}} \mathrm{kV}.

The equivalent short circuit impedance \frac{(11 / \sqrt{3})^{2}}{250 / 3}

=\frac{121}{250}=0.484  \Omega

 

\phi_{s c}=\tan ^{-1} 5=78.69^{\circ}

\therefore \quad R = 0.0949 and X = 0.4746 Ω

\left(\frac{11}{\sqrt{3}}\right)^{2}=\left(V \cos \phi+\frac{P}{V} R\right)^{2}+\left(V \sin \phi+\frac{Q}{V} X\right)^{2}

 

=\left(V \times 0.5547+\frac{10 \times 10^{6}}{V} \times 0.0949\right)^{2}+\left(V \times 0.832+\frac{15 \times 10^{6}}{V \times 10^{3}} \times 0.4746\right)^{2}

 

\frac{121}{3}=\left(0.5547  V+\frac{10}{V} \times 0.0949\right)^{2}+\left(0.832  V+\frac{15 \times 0.4746}{V}\right)^{2}

 

=\left(0.5547   V+\frac{0.949}{V}\right)^{2}+\left(0.832  V+\frac{7.119}{V}\right)^{2}

 

40.33  V^{2}=\left(0.5547  V^{2}+0.949\right)^{2}+\left(0.832  V^{2}+7.119\right)^{2}

 

40.33  V^{2}=0.3077  V^{4}+0.9+1.053  V^{2}+0.69  V^{4}+50.68+11.846  V^{2}

 

or \quad 0.9977  V^{4}-27.43  V^{2}+51.59=0

 

V^{4}-27.5  V^{2}+51.7=0

 

V^{2}=\frac{27.5 \pm \sqrt{756.25-206.32}}{2}=\frac{27.5+23.45}{2}=25.47

or V = 5.047 kV

The drop in volts is 6.350 – 5.047 = 1.303 kV

\therefore pu drop 0.2583.

Now using the expression

\begin{aligned} \frac{\Delta V}{V} &=\frac{1}{S_{s c}}\left(P \cos \phi_{s c}+Q \sin \phi_{s c}\right)+j \frac{1}{S_{s c}}\left(P \sin \phi_{s c}-Q \cos \phi_{s c}\right) \\ &=\frac{1}{250}(30 \times 0.196+45 \times 0.98)+j \frac{1}{250}(30 \times 0.98-45 \times 0.196) \\ &=\frac{1}{250}\{(5.88+44)+j(29.4-8.82)\} \\ &=\frac{1}{250}(49.88+j 20.58) \\ &=\frac{53.96}{250}=0.2158 \end{aligned}

The difference in the two values, one obtained through the exact calculation and the other through approximate calculation, is because of the fact that E ≠ V. In fact, the real problem lies in the amount of power being transmitted. For 11 kV system, it is not desirable to transmit such a large amount of power. Also X : R assumed as 5 is high and is not true in actual system designs. Overloading of the system results in larger voltage sags which leads to lower power transfer capability of the system and hence lower margin for stable operation of the system.

Stability and voltage levels of the system are the basic requirements of power system. By stability is meant the tendency of the power system to continue to operate in the desired mode (voltage and frequency to remain within desired limits) even when the system is subjected to a fault, sudden change of load or any other external condition which may have tendency to make the system unstable. Like any other system, it is the inherent property of the system and is a measure of developing restoring forces within the system so as to counter the disturbing forces e.g., short circuits or increase in loads etc. During disturbances there is exchange of power or energy amongst various components of power system. Various synchronous machines can run and deliver or absorb energy only at synchronous speeds (under normal operation) or around synchronous speed (during a short time of disturbance). Transmission line is one of the most important components of power system through which exchange of energy takes place amongst various synchronous machines.

Like any other physical system, the stable or unstable operation of power system depends mainly on two factors (i) initial operating condition (ii) the level of perturbation i.e., small perturbation or large perturbation i.e., a small increase in load or a large increase in load or a short circuit.

It is possible (even though not desirable) that in a particular operating condition all the equipments are at their maximum limits and even a small perturbation might lead to unstable operation of the system. On the other hand there could be another initial operating condition for which even a large perturbation of specific amount might retain the system in stable operating condition. Therefore, larger the margin between the initial operating condition and the limiting operating condition of the system, the greater are the chances of the system to operate under stable operating condition for a specific disturbance.

As mentioned earlier, the exchange of energy amongst various synchronous machines takes place through transmission lines. One of the limitations of the line is that for a given length of the line the stability tends to become less as the initial transmitted power is increased. That is, if the initial transmission angle δ is large the smaller is the stability margin. This is why normally the transmission lines are loaded equivalent to 30° angle. A still smaller angle of transmission is desirable from stability point of view but it will be under utilisation of the power transfer capability of the line. The capital investment in transmission lines is very large these days and these should be used optimally.

As the power transfer through the line is increased a limit is reached when an attempt to increase power results actually reduction in power. Thus the two synchronous machines at the two ends of the line would lose synchronism. This limit of power transfer is known as steady state stability limit because it is the maximum steady power that can be transferred stably. The steady state stability limit of a link can be increased by increasing the excitation level of the synchronous machines at the two ends, by interconnecting another line between the two ends, by the use of series capacitors etc.

A transmission line has three critical loadings (i) Natural loading; (ii) Steady state stability limit; (iii) Thermal limit loading. Out of the three for a compensated line, the Natural Loading is the lowest and before thermal limit loading is reached, steady state stability limit is achieved.