Practical Power Plant Engineering. Zark Bedalov
when the faulty transformer is brought back follows a similar procedure in reverse order. The incomer B is switched ON, on its HV side. This initiates the tie breaker T to open first, followed by closing the incomer breaker B. During this restoration to normal operation, Bus B again passes through a “dead” transition, and its feeder breakers and contactors are turned OFF during a brief transition. Latched feeder breakers are restored immediately, while the other loads must be restored by the plant control system by reviewing the permissive logic of each circuit. If the switchgear breaker cell control switches are left on Loc/Rem, they must be turned to Remote to allow the automatic reintegration of the lost services to proceed.
2.7.10.2 Switching from Control Room
Automatic switching from the control room is possible only if the breaker Loc/Rem switches are held on “Remote.” All the breakers and controllers are provided with means of operating from a local and remote position. The breakers that are left on Loc position (most likely intentionally) will not be restored. The means of communication from the plant control system to the switchgear is by Ethernet (see Chapter 17).
Breaker Loc/Rem switch operation:
Loc (Local) means manual.
Rem (Remote) means automatic from control room.
2.7.11 Plant Transformation
Every effort shall be made to minimize the number of transformations from the grid to the loads. Unnecessary transformers add to the voltage drops as well as to the cost of the plant in the transformers and additional associated switching and protection equipment.
For this plant, we can use two or maximum three levels of transformation, as follows:
Grid → 230 kV–T1–13.8 kV → 13.2 kV–T2–4.16 kV → 4.0 kV Motors, or
Grid → 230 kV–T1–13.8 kV → 13.2 kV–T2–0.48 kV → 0.46 kV Motors and feeders
Grid → 230 kV–T1–13.8 kV → 13.2 kV–T2–4.16 kV → 4.16 kV–T3–0.48 kV → 0.46 kV Loads
Grid → 230 kV–T1–13.8 kV → 13.2 kV–T2–0.48 kV → 0.48 kV–T3–0.48 kV → 0.48 kV Ltg. Panels1
2.7.12 Voltage Regulation
2.7.12.1 Voltage Regulation ΔV
The full range of load regulation of a power transformer is the change in secondary voltage Vs, expressed in percentage of rated Vs for a specified power factor. This occurs when the rated MVA output at a specified power factor is reduced to zero, with Vp maintained constant. One can use these approximate formulae to calculate ΔV at any operating load and power factor:
(2.6)
For instance: Transformer: 30 MVA as a MVAb = 1 pu, having Z = 9% (0.09 pu) impedance
on 30 MVA base, efficiency 99.5% at rated load.
Total loss = (1 − 0.995) × 30 MVA = 150 kW. The total loss includes no load and load losses.
With this information, we can calculate transformer resistance R, followed by calculating the reactance X. After that, we calculate the voltage regulation.
Transformer resistance R is calculated from the load losses. Since we do not have that figure we will assume that the load loss is equal to 80% of total loss. The rest is the no load loss.
Assume operating load of 25 MVA (MVA load pu = 0.833 pu) at power factor pf = 0.85.
R or Load loss in % = 0.8 × 100 = 0.4%. Z is given as 9%
Pload = MVApu × cos ϕ = 0.833 × 0.85 = 0.71 pu
Qload = MVApu × sin ϕ = 0.833 × 0.52 = 0.43 pu
ΔVpu = (0.004 × 0.71 + 0.0899 × 0.43) = 0.0414 pu or 4.1 % . From Eq. (2.6).
2.7.12.2 Motor Start Voltage Drop
The plant will include a number of large motors, for which cable sizing must be verified with respect to the voltage drop during its start. During the motor start, the voltage at the motor terminals must be >85% of the motor nominal voltage. Here is a quick check without going through a computer study.
For selecting the power cable for the motor we use National Electrical Code (NEC) for ampacities of Cu and Al cables [1]. Most of the engineers would have this booklet (Code) on their desks and use it for the various engineering activities ranging from the switchgear and cable selection to fire protection regulations. A similar Code is also available in Canada.
Let us calculate voltage drop for a motor rated as follows:
Motor: 100 kW, 0.85 pf, 480 V
Motor sub‐transient Impedance: Zm = 17%,
Power cable length: 100 m.
Select the cable. Calculate motor nominal current:
Select the cable from NEC Ampacity table.
We look for a cable size for: 1.25 × 125 A = 155 A (∼25% margin was added).
Cable selected from the code: 3c # 1/0, Cu, 90 °C, capable of carrying 170 A.
We calculate the voltage drop on the motor kW base (kWb): 100 kW = 1 pu.
Motor impedance on the motor base if not known can be assumed as: Zm = 0.17 pu as per ANSI.
Cable impedance ZΩ for cable 3c #1/0 AWG = 0.035 Ω/100 m. Value was taken from relevant tables.
Calculate impedance Zc pu on per unit (pu) value for 100 m cable:
Cable impedance in pu: .
Calculate voltage drop ratio corresponding to the motor/cable impedance ratio:
Therefore, the voltage % on motor terminals during the motor start is equals to 91.8% OK!For motor cable 100 m long: ΔV = 8.2% < 15% allowed.
This calculation assumed the 480 V bus is operating at 100% voltage at the time the motor is initiated to start.
2.7.12.3 Conclusion
This motor operating with the proposed cable 3c #1/0 AWG is acceptable for the actual cable length of 100 m. If the distance is doubled to 200 m, a new heavier cable would have to be selected as the ΔV = 16.4% > 15% is over the operating limit.
2.7.13 Overhead Distribution Lines
The major part of the power distribution between the main substation and the remote areas and the main process plant will be by overhead distribution lines. The lines will follow the most direct routes or along the roads to facilitate ease of construction and maintenance. The 13.8 kV feeders will be laid out either as radial feeders or ring main loop feeders to connect several unit substations and other major loads in a loop fed from two different sources.
The MV distribution lines will use wood poles and wood cross‐arms (see