This document discusses different power distribution systems including radial, ring, and network systems. It provides details on: - The permissible voltage ranges for low-voltage customers - Causes and calculations for voltage drop and power losses - The reliability and efficiency advantages of ring and network systems over radial systems - Selection criteria for single-phase versus three-phase distribution based on economic considerations - Calculating motor performance values like speed and torque given specifications like voltage, frequency, resistance

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TASK 1
INTRODUCTION
As a Sarawak Energy technician working in one of the distribution Substation (Pending) and transmission
Substation ( Sejingkat Power Plant ), I was assigned to control and monitor the operating range (Lower,
Normal or Over) of electrical power layout (topology) system. Provided, the daily data report of the data
collected over a month period interval.
SOLUTION

2. PERMISSIBILITY
The permissible voltage of the terminal voltage of low-voltage customers is considered within 220±13V.
VOLTAGE DROP / SAG AND POWER LOSSES
The NEC allows a maximum 5 percent voltage drop in the combination of a branch and a feeder circuit;
however, a 5 percent voltage reduction represents a significant power loss in a circuit. We can calculate
power loss due to voltage drop as V2/R, where V2 is the voltage drop of the circuit, and R is the resistance
of the conductors of the circuit.
One of the main elements of a substation is the transformers. In order to decrease the amount of magnetic
loss, transformer windings are wound around iron cores. Iron cores concentrate the magnetic lines of
force, so that better coupling between the primary and secondary windings is accomplished. Two types of
transformer cores are illustrated in. These cores are made of laminated iron to reduce undesirable eddy
currents, which are induced into the core material. These eddy currents cause power losses.
RADIAL POWER DISTRIBUTION SYSTEM
Radial systems are the least reliable in terms of continuous service, since there is no back-up distribution
system connected to the single power source. If any power line opens, one or more loads are interrupted.
There is more likelihood of power outages. However, the radial system is the least expensive. This system
is used in remote areas where other distribution systems are not economically feasible.
RELIABILITY
The least reliable in terms of continuous service, since there is no distribution system connected to the
The NEC allows a maximum 5 percent voltage drop in the combination of a branch and a feeder circuit;
however, a 5 percent voltage reduction represents a significant power loss in a circuit. We can calculate
power loss due to voltage drop as V2/R, where V2 is the voltage drop of the circuit, and R is the resistance
of the conductors of the circuit.
POWER EFFICIENCY
CONSISTENCY
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RING POWER DISTRIBUTION SYSTEM
Ring distribution systems are used in heavily populated areas. The distribution lines encircle the service
area. Power is delivered from one or more power sources into substations near the service area.
The power is then distributed from the substations through the radial power lines. When a power line is
opened, no interruption to other loads occurs. The ring system provides a more continuous service than
the radial system. Additional power lines and a greater circuit complexity make the ring system more
expensive.
RELIABILITY
Quite reliable in heavily populated area and when a power line is opened, no interruption to other loads
occurs. The ring system provides a more continuous service than the radial system.
POWER EFFICIENCY
CONSISTENCY
NETWORK POWER DISTRIBUTION SYSTEM
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4. Network distribution systems are a combination of the radial and ring systems. They usually result when
one of the other systems is expanded. Most of the distribution systems in the United States are network
systems. This system is more complex, but it provides very reliable service to consumers. With a network
system, each load is fed by two or more circuits.
RELIABILITY
Very reliable service to the customer because whenever there is an opened power line, the load will fetch
another power line connected to actively provide power to the load.
POWER EFFICIENCY
CONSISTENCY
TASK 2
INTRODUCTION
I am responsible to select single or three-phase based on economic consideration in the distribution area
and convincing the potential client (industrial related) on the advantages and of using three-phase
induction motors, instead of ac single phase or universal dc machines or motor.
SOLUTION
Are three-phases better than single phase? (You need to produce the answer of this question by
comparing the economics of the single-phase and three-phase distribution/transmission substations?)
There are advantages and disadvantages of both three-phases and single phase system.
ADVANTAGES
1. Efficiency drives up to 80% compared to single phase electronics (for example a single-phase
floor cleaner machine which took about 8 hours to finish 1000 sq. ft. room, whereas three-phases
floor cleaner machine only took 2 hours to finish the job)
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DISADVANTAGES
TASK 3
INTRODUCTION
Huge construction project (e.g. Kuching International Airport) need electrical power or economic
viewpoint/ study because total cost of electrical estimated work is 40% of the total cost of the project.
SOLUTION
1. 3.1 You are required to provide the reasons and advantages from the viewpoint basis of economic
considerations why a huge facility (e.g. Kuching International Airport) need to choose 3-Φ induction
motors for required load drives in efficiency factor.
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6. 2. A 3-Φ, wound-rotor induction motor, 4-poles, has a rating of 110 kilo Watt, 1700 rev/min, 2.3
kV, 60 Hz (USA Standard). Three external resistors of 2Ω are connected in wyes across the rotor
slip-rings. Under these conditions the motor develops a torque of 300 Newton/m. at a speed of
1000 rev/min.
Poles, p: 4 Power, P: 110 kilowatt Speed, S: 1700 rev/min
Voltage: 2.3 kilovolt Frequency, f: 60Hz
External Resistance, R external: 2Ω x 3 = 6Ω
2.1. Calculate the speed for a torque of 400 Newton/m.
Solution:
2.2. Calculate the value of the external resistors so that the motor develops 10kW at 200
rev/min
Solution:
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