Voltage Drop Calculation in Substation Design

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An electrical substation is mainly designed to perform voltage transformation and power flow control within the electrical power system, and it is supposed that voltage within the substation must remain within the prescribed limits. This is very important to ensure equipment efficiency and reliability.

A voltage drop can be defined as the reduction in voltage with the flow of electric current through conductors, transformers, or switchgear components used in an electrical substation.

If the voltage drop exceeds a certain limit in the substation, then it can result in the following consequences −

• Poor performance of electrical equipment
• Inaccurate operation of protective relays
• Higher energy losses
• Deviation from prescribed voltage limits, etc.

Therefore, it is very important to perform accurate voltage drop calculation in substation to make the substation reliable and efficient.

In this chapter of the substation design tutorial, we will learn about voltage drop calculation in substation design. Here, we will cover some important topics such as theory of voltage drop, acceptable voltage-drop limits, and step-by-step voltage drop calculation with example.

Theory of Voltage Drop

Voltage drop is nothing but a reduction in voltage with the flow of electric current through a conductor or electrical component. This is mainly due to electrical resistance (R) and reactance (X).

In electrical substation, the voltage drop in a cable or a line is generally expressed by

$$\mathrm{\Delta V \:=\: I \:\times\: (R\cos\phi \:+\: X\sin\phi)}$$

Here, ΔV is the voltage drop, I is the current flowing through line or cable, R is the resistance, X is the reactance, and φ is the power factor angle of the load.

The voltage drop can also be expressed as a percentage of the rated voltage as follows −

$$\mathrm{\%\Delta V \:=\: \frac{\Delta V}{V_{rated}} \:\times\: 100}$$

Permissible Voltage Drop Limits in Substations

The following table provides acceptable voltage drop limits in various systems of an electrical substation −

Voltage Drop Calculation in Substation Design

The step-by-step calculation of voltage drop in substation design is explained below −

Step 1 – Determine the load current

$$\mathrm{I \:=\: \frac{P}{\sqrt{3} \:\times\: V \:\times\: \cos\phi}}$$

Step 2 – Calculate total line impedance for the given line length

$$\mathrm{R_{T} \:=\: R\:per\:km \:\times\: line\:length}$$

$$\mathrm{X_{T} \:=\: X\:per\:km \:\times\: line\:length}$$

Step 3 – Calculate the voltage drop in line

$$\mathrm{\Delta V \:=\: \sqrt{3} \:\times\: I \:\times\: (R\cos\phi \:+\: X\sin\phi)}$$

Step 4 – Calculate the percentage voltage drop

$$\mathrm{\%\Delta V \:=\: \frac{\Delta V}{V_{rated}} \:\times\: 100}$$

The value of %ΔV must be within the acceptable limits for the given substation.

Numerical Example

A 33 kV electrical substation is supplying an electrical load of 5 MW at a power factor of 0.87 lagging located at a distance of 5 km away. If the line conductor used is ACSR Wolf with a resistance of 0.275 Ω/km and line reactance of 0.35 Ω/km. Determine the voltage drop within the feeder of the substation.

Solution– Given data,

Voltage = 33 kV = 33000 V

Load power, P = 5 MW = 5 × 10⁶ W

Power factor of load, cosφ = 0.87 lag

Line length l = 5 km

Line resistance R = 0.275 Ω/km

Line reactance X = 0.35 Ω/km

Then, load current carried by the feeder will be,

$$\mathrm{I \:=\: \frac{P}{\sqrt{3} \:\times\: V \:\times\: \cos\phi} \:=\: \frac{5 \:\times\: 10^{6}}{\sqrt{3} \:\times\: 33000 \:\times\: 0.87}}$$

$$\mathrm{\therefore\: I \:=\: 100.55\: A}$$

Calculating the impedance of the feeder as,

$$\mathrm{R_{T} \:=\: 0.275 \:\times\: 5 \:=\: 1.38\: \Omega}$$

$$\mathrm{X_{T} \:=\: 0.35 \:\times\: 5 \:=\: 1.75\:\Omega}$$

Now, calculating the voltage drop in the feeder,

$$\mathrm{\Delta V \:=\: \sqrt{3} \:\times\: I \:\times\: (R\cos\phi \:+\: X\sin\phi)}$$

$$\mathrm{\Rightarrow\: \Delta V \:=\: \sqrt{3} \:\times\: 100.55 \:\times\: (1.38 \:\times\: 0.87 \:+\: 1.75 \:\times\: 0.49)}$$

$$\mathrm{\Rightarrow\: \Delta V \:=\: 358.42\:V}$$

The percentage voltage drop will be,

$$\mathrm{\%\Delta V \:=\: \frac{358.42}{33000} \:\times\: 100 \:=\: 1.086\%}$$

This voltage drop is within the acceptable limits for the 33 kV distribution feeder which is 3 to 5%.

Factors Affecting Voltage Drop in Substation Design

The following are some key factors that significantly affect the voltage drop in a conductor or cable in substation design −

• Conductor Material and Size − Conductors with low resistance material and large cross-sectional area results in lower voltage drop.
• Length of Conductor − The voltage drop increases with increasing length of the conductor.
• Load Power Factor − A lagging load power factor results in higher voltage drop.
• Operating Temperature − Increase in temperature of conductor results in higher voltage drop.

Voltage Drop in Substation Busbars

In electrical substations, voltage drop across busbars must also be checked to ensure uniform distribution of voltage across outgoing feeders.

But in the case of busbars, the length is too short and hence the reactance is minimal and negligible. Therefore, busbars have voltage drop due to resistance only which is given by,

$$\mathrm{\Delta V_{bus} \:=\: I \:\times\: R_{bus}}$$

Voltage Drop in Substation Control Circuits

In substation control and protection circuits, the voltage drop can significantly affect the operation of protective relays and trip coils.

As per standards, it is recommended that for a 110 V DC system, the voltage across terminals of the control or protection device must not fall below 90% of the nominal voltage.

Conclusion

Voltage drop calculation in substation design plays an important role in reliable and efficient operation of electrical substations. In this chapter, we explained in detail the concept of voltage drop and its calculation in substation design with step-by-step procedure and solved numerical example.