Single line diagram
This technical article explains how to calculate and draw a single line diagram of the three-phase, 60-Hz system power system with generators, motors, transformers and lines.
The following components comprise a simplified version of a power system, listed in sequential physical order from the generator location to the load:
- Two steam-electric generators, each at 13.2 kV
- Two step-up transformers, 13.2/66 kV
- Sending-end, high-voltage bus at 66 kV
- One long transmission line at 66 kV
- Receiving-end bus at 66 kV
- A second 66 kV transmission line with a center-tap bus
- Step-down transformer at receiving-end bus, 66/12 kV, supplying four 12 kV motors in parallel and
- A step-down transformer, 66/7.2 kV, off the center-tap bus, supplying a 7.2 kV motor
1. Identify the Appropriate Symbols
For electric power networks an appropriate selection of graphic symbols is shown in Figure 1 (common power symbols used in single line diagrams):
2. Draw the Required System
The system described in the problem is shown in Figure 2. The oil circuit breakers are added at the appropriate points for proper isolation of equipment.
It is the general procedure to use single line diagrams for representing three-phase systems. When analysis is done using symmetrical components, different diagrams may be drawn that will represent the electric circuitry for positive, negative, and zero-sequence components.
This type of notation is shown in Figure 3.
Per-unit method of solving of 3-phase problems
For the system shown in Figure 4, draw the electric circuit or reactance diagram, with all reactances marked in per-unit (p.u.) values, and find the generator terminal voltage assuming both motors operating at 12 kV, three-quarters load, and unity power factor.
|Motor A||Motor B||Transmission|
|13.8kV||25,000 kVA||15,000 kVA||10,000 kVA||–|
|25,000 kVA 3-phase||13.2/69 kV||13.0 kV||13.0 kV||–|
|X” = 15 percent||XL = 15 percent||X” = 15 percent||X” = 15 percent||X = 65 Ω|
Calculation Procedure in 8 steps
1. Establish Base Voltage through the System
By observation of the magnitude of the components in the system, a base value of apparent power S is chosen. It should be of the general magnitude of the components, and the choice is arbitrary. In this problem, 25,000 kVA is chosen as the base S, and simultaneously, at the generator end 13.8 kV is selected as a base voltage Vbase.
(13.8 kV)(69 kV / 13.2 kV) = 72.136 kV
The base voltage of the motors is determined likewise but with the 72.136 kV value, thus:
(72.136 kV)(13.2 kV / 69 kV) = 13.8 kV
The selected base S value remains constant throughout the system, but the base voltage is 13.8 kV at the generator and at the motors, and 72.136 kV on the transmission line.
2. Calculate the Generator Reactance
No calculation is necessary for correcting the value of the generator reactance because it is given as 0.15 p.u. (15 percent), based on 25,000 kVA and 13.8 kV. If a different S base were used in this problem, then a correction would be necessary as shown for the transmission line, electric motors, and power transformers.
3. Calculate the Transformer Reactance
It is necessary to make a correction when the transformer nameplate reactance is used because the calculated operation is at a different voltage, 13.8 kV / 72.136 kV instead of 13.2 kV / 69 kV.
(nameplate per-unit reactance) (base kVA/nameplate kVA) (nameplate kV/base kV)2 =
(0.11) (25,000/25,000) (13.2/13.8)2 = 0.101 p.u.
This applies to each transformer.
4. Calculate the Transmission-Line Reactance
Use the equation:
- Xper unit = (ohms reactance)(base kVA)/(1000)(base kV)2 =
- Xper unit = (65) (25,000)/(1000)(72.1)2 = 0.313 p.u.
5. Calculate the Reactance of the Motors
Corrections need to be made in the nameplate ratings of both motors because of differences of ratings in kVA and kV as compared with those selected for calculations in this problem. Use the correcting equation from Step 3, above.
X”A = (0.15 p.u.) (25,000 kVA / 15,000 kVA) (13.0 kV / 13.8 kV)2 = 0.222 p.u.
For motor B:
X”B = (0.15 p.u.)(25,000 kVA /10,000 kVA)(13.0 kV / 13.8 kV)2 = 0.333 p.u.
6. Draw the Reactance Diagram
The completed reactance diagram is shown in Figure 5:
7. Calculate Operating Conditions of the Motors
If the motors are operating at 12 kV, this represents 12 kV/13.8 kV = 0.87 per-unit voltage. At unity power factor, the load is given as three-quarters or 0.75 p.u.
Iper unit = per-unit power/per-unit voltage = 0.75/0.87 = 0.862 ∠0° p.u.
8. Calculate the Generator Terminal Voltage
The voltage at the generator terminals is:
- VG = Vmotor + the voltage drop through transformers and transmission line
- VG = 0.87∠0° + 0.862 ∠0°(j0.101 + j0.313 + j0.101)
- VG = 0.87 + j0.444 = 0.977 ∠27.03° p.u.
In order to obtain the actual voltage, multiply the per-unit voltage by the base voltage at the generator. Thus,
- VG = (0.977 ∠27.03°) (13.8 kV) = 13.48 ∠27.03° kV
In the solution of these problems, the selection of base voltage and apparent power are arbitrary. However, the base voltage in each section of the circuit must be related in accordance with transformer turns ratios.
Zbase = (base kV)2 (1000) / (base kVA).
For the transmission line section in this problem, Zbase = (72.136)2 (1000) / (25,000) = 208.1
Thus the per-unit reactance of the transmission line equals (actual ohms) / (base ohms) = 65 / 208.1 = 0.313 p.u.
66kv substation control room introduction
Reference // Handbook of el. power calculations by H. Wayne Beaty (Get hardcover from Amazon)