THE NACABS POLYTECHNIC AKWANGA

SCHOOL OF ENGINEERING
TECHNOLOGY
DEPARTMENT OF ELECTRICAL
ENGINEERING
Frist Semester Examination 2024/2025 Academic Session
Course: ELECTRICAL POWER SYSTEM II
Course Code: EEC232 Examiner: Mr luka Jacob
Level: ND2 Time Allow: 2 30 Hours
Instruction: Attempt Four questions in all.

Question1a. Under scheme of connection distribution systems are further classified as radial
system, ring main system or interconnected system explain?(10mark)

Answer: i) I Radial System – In this system feeders branch out radially from a common source
and feed the distributors at one end only. In this type of system if a feeder fails due to a fault,
the supply to the consumer is interrupted until repairs are done. It is the simplest distribution
circuit and has the lowest initial cost but has some drawbacks such as,

• Any fault on the feeder or distributor cuts off supply to the consumers on the side of the fault
away from the substation as they are dependent on a single feeder and distributor.

• The consumer at the farthest end of the distributor would be subject to voltage fluctuations
when the load on the distributor changes.

ii) Ring Main System – In the ring main distribution system the feeder branches out in the form
of a loop or ring. The loop circuit starts from the substation bus bars makes loop through the
area to be served and returns to the substation. This makes a complete loop and has isolating
switches provided at the poles at strategic points for isolating a particular section in case of a
fault. Thus failure of one interconnecting feeder does not interrupt the supply.

Question1b. D.C. supply can be obtained as either 2-wire system or 3-wire system for
distribution explain?(7.5mark)

Answer: 1) 2-wire D.C. System

This system has 2 wires; the outgoing or positive wire and the return or negative wire. Due to
its low efficiency, it is not used for transmission purposes but for distribution of D.C. power.

2) 3-wire D.C. System

This system has 3 wires, the middle wire which is the neutral is earthed. The voltage between
either of the outer wires and neutral is half that between the negative and positive wire making
two voltages available at the consumer terminal.

Question2a. Explain the three functions of circuit breakers?(6mark)

Answer: A circuit breaker is a piece of equipment which can

(i) Make or break a circuit either manually or by remote control under normal
conditions.

(ii) Break a circuit automatically under fault conditions

(ii) Make a circuit either manually (or by remote control) as well automatic control for
switching functions. The latter control employs relays and operates only under fault
conditions.

Question2b.Write five (5) properties of insulators? (10mark)

Answer: The five properties of insulator are:

i. High mechanical strength in order to withstand conductor load, wind load.

ii. High electrical resistance of insulator material in order to avoid leakage currents to
earth.
iii. High relative permittivity of insulator material in order that dielectric strength is
high.

iv. The insulator material should be non-porous; free from impurities and cracks otherwise
the permittivity will be lowered.

iv. High ratio of puncture strength to flashover.

Question2C.What are the three requirements of a distribution system? (1.5mark)

Answer: The three requirements of a distribution system are:

PROPER VOLTAGE

AVAILABILITY OF POWER ON DEMAND

RELIABILITY

Question3a.What the five advantage of suspension types insulators? (10mark)

Answer:

i. Suspension type insulators are cheaper than pin type insulators for voltages beyond
33KV.
ii. If any one disc is damaged, the whole string does not become useless because the
damaged disc can be replaced.
iii. Suspension arrangement provides greater flexibility to the line.

iv. Suspension type insulators are generally used with steel towers.

iv. Each unit or disc of suspension type insulator is designed for low voltage usually 11KV.

Question3b.Mechanical energy is supplied to a d.c. generator at the rate of 4200 J/s. The
generator delivers 32·2 A at 120 V. (i) What is the percentage efficiency of the generator ? (ii)
How much energy is lost per minute of operation ? (7.5mark)

Solution. (i) Input power, Pi = 4200 J/s = 4200 W

Output power, Po = EI = 120 × 32·2 = 3864 W

∴ Efficiency, η = P P o i ×= × 100 3864 4200 100 = 92 %

(iii) Power lost, PL = Pi − Po = 4200 − 3864 = 336 W

∴ Energy lost per minute (= 60 s) of operation = PL × t = 336 × 60 = 20160 J

Note that efficiency is always less than 1 (or 100 %). In other words, every system is less
than 100 % efficient.

Question4a. Mention the six (6) stages of steam power generation? (6mark)

1. Coal and ash handling arrangement

2. Steam generating plant

3. Steam turbine

4. Alternator

5. Feed water
6. Cooling arrangement

Question4b. A thermal station has the following data : Max. demand = 20,000 kW ;

Load factor = 40% Boiler efficiency = 85% ;

Turbine efficiency = 90%

Coal consumption = 0·9 kg/kWh ;

Cost of 1 ton of coal = Rs. 300

Determine (i) thermal efficiency and (ii) coal bill per annum. (5mark)

Question4c. A hydro-electric generating station is supplied from a reservoir of capacity 5 ×

106 cubic metres at a head of 200 metres.

Find the total energy available in kWh if the overall efficiency is 75% (6.5mark).

Question5a.Define load curve and with graphical example explain daily load curve?
(7.5mark).

Answer: The curve showing the variation of load on the power station with respect to
(w.r.t) time is known as a load curve.

The load on a power station is never constant; it varies from time to time. These load
variations during the whole day (i.e., 24 hours) are recorded half-hourly or hourly and are
plotted against time on the graph. The curve thus obtained is known as daily load curve as it
shows the variations of load w.r.t. time during the day. Fig. 3.2. shows a typical daily load
curve of a power station.

Question5b.How can you represent Per-unit system for power, voltage, current and
impedance in single phase and three phase in power system?

the per unit quantity may be obtained by dividing by the respective base of that quantity as
follows: For single phase

For three phase

Question6a.Define the following with mathematical example Connected load, Maximum
demand, Demand factor, Average load, Load factor, Diversity factor, Plant capacity factor and
Plant use factor?(17.5mark)

Answer:

(i) Connected load. It is the sum of continuous ratings of all the equipment connected
to supply system.
(ii) Maximum demand : It is the greatest demand of load on the power station during a
given period.
(iii) Demand factor. It is the ratio of maximum demand on the power station to its
connected load i.e., Demand factor = Maximum demand /Connected load
(iv) Average load. The average of loads occurring on the power station in a given period
(day or month or year) is known as average load or average demand.
Daily average load = No. of units (kWh) generated in a day/ 24 hours
Monthly average load = No. of units (kWh) generated in a month/ Number of hours
in a month
Yearly average load = No. of units (kWh) generated in a year/ 8760 hours
(v) Load factor. The ratio of average load to the maximum demand during a given
period is known as load factor i.e., Load factor = Average load /Max. demand
(vi) Diversity factor. The ratio of the sum of individual maximum demands to the
maximum demand on power station is known as diversity factor i.e.,
Diversity factor = Sum of individual max. demands /Max. demand on power station
(vii) Plant capacity factor. It is the ratio of actual energy produced to the maximum
possible energy that could have been produced during a given period i.e.,

(viii) Plant use factor. It is ratio of kWh generated to the product of plant capacity and the
number of hours for which the plant was in operation i.e.
Plant use factor = Station output in kWh /Plant capacity . Hours of us × e
 THE NACABS POLYTECHNIC AKWANGA
SCHOOL OF ENGINEERING
TECHNOLOGY
DEPARTMENT OF ELECTRICAL
ENGINEERING
Frist Semester Examination 2024/2025 Academic Session
Course: ELECTRICAL POWER SYSTEM II
Course Code: EEC232 Examiner: Mr luka Jacob
Level: ND2 Time Allow: 2 30 Hours
Instruction: Attempt Four questions in all.

Question1a. Under scheme of connection distribution systems are further classified as radial
system, ring main system or interconnected system explain?(10mark)

Question1b. D.C. supply can be obtained as either 2-wire system or 3-wire system for
distribution explain?(7.5mark)
Question2a. Explain the three functions of circuit breakers?(6mark)

Question2b.Write five (5) properties of insulators? (10mark)

Question2b.What are the three requirements of a distribution system? (1.5mark)

Question3a.What the five advantage of suspension types insulators? (10mark)

Question3b.Mechanical energy is supplied to a d.c. generator at the rate of 4200 J/s. The
generator delivers 32·2 A at 120 V. (i) What is the percentage efficiency of the generator ? (ii)
How much energy is lost per minute of operation ? (7.5mark)

Question4a. Mention the six (6) stages of steam power generation? (6mark)

Question4b. A thermal station has the following data : Max. demand = 20,000 kW ;

Load factor = 40% Boiler efficiency = 85% ;

Turbine efficiency = 90%

Coal consumption = 0·9 kg/kWh ;

Cost of 1 ton of coal = Rs. 300

Determine (i) thermal efficiency and (ii) coal bill per annum. (5mark)

Question4c. A hydro-electric generating station is supplied from a reservoir of capacity 5 ×

106 cubic metres at a head of 200 metres.

Find the total energy available in kWh if the overall efficiency is 75% (6.5mark).

Question5a.Define load curve and with graphical example explain daily load curve?
(7.5mark).

Question5b.How can you represent Per-unit system for power, voltage, current and
impedance in single phase and three phase in power system?

Question5a.Define the following with mathematical example Connected load, Maximum

demand, Demand factor, Average load, Load factor, Diversity factor, Plant capacity factor and
Plant use factor?(17.5mark)