Uploaded by Lu Casthe

toaz.info-power-plant-lecture-for-me-review-pr c47fac4622c388c1c765f9d36448814c

advertisement
POWER PLANT REVIEW LECTURE
by:
ENGR. JERMHEL M. SOLIS, REE, RME
Power Plant Engineering
Art of designing and installing generating plant that will result in maximum return on investment
(profit) over the expected life of the equipment. And also operating to achieve reliable, continuous and cheap
power service.
Power Plant – A station or establishment which houses the prime – movers, electric generators and auxiliaries,
for conversion of mechanical, chemical, and/or nuclear energy into electrical energy.
I. Types of Power Plants (As to source of energy)
1. Thermal Power Plant – an electric generating station using heat as source of energy.
a) Oil – fired Steam Plant – makes use of heavy fuel oil, light fuel oil or bunker oil for production of
energy.
b) Coal – fired Thermal Plant – makes use of pulverized coal as fuel
c) Dendro – Thermal Plant – makes use of wood (ipil – ipil)
d) Nuclear – Steam Plant – makes use of steam generated in a reactor by heat from fission process
of nuclear fuel (Uranium 325, Uranium – 238)
e) Gas Turbine Plant – makes use of combustible gasses as fuel from gas turbine engine prime –
mover.
f) Geothermal Plant – makes use of generated hear from the inherent steam from the Earth’s
magma fuel.
g) Solar Steam Plant – makes use of steam generated from solar radiation.
2. Hydro – Electric Power Plant – an electric generating station using flowing water to drive theprime –
movers (hydraulic turbines) either impulse of reaction type.
Two Types of Hydraulic Turbines
a) Impulse Type – use for high – heads and low volume, example is the “Pelton Wheel”.
b) Reaction type –use for low – heads and high volume, examples are “Francis and Kaplan”
Turbine
Types of Hydro – Electric Plant
a) Run-of-river – using pondage of steam flow as it occurs; more power can be generated in a rainy
season than in dry season.
b) Plant with storage capacity – associated with a large water reservoir. This permits regulated
supply of water o that the power output is constant throughout the year.
c) Pump – storage plant – where energy is generated during periods of high system demand using
water which has been pumped into a reservoir usually during periods of relatively low system
demands.
3. Diesel Power Plant – a plant of internal combustion engine (ICE) prime – mover using diesel as fuel in
producing energy.
4. Windmill Plant – using a series of windmills as prime – mover.
5. Sea waves / Ocean Tides Plant – makes use of the natural rising of tide to simulate flowing
water.
1
6. Solar or Photo – Voltaic Plant – chemical conversion of radiant energy of the sun to electric (dc).
Types of Power Plant (as to use)
1) Base – Load Plant – plant that assumed load requirements under normal conditions.
2) Peaking Plant – plant that is normally operated to provide only during peak load periods.
3) Regulating Plant – plant capable of carrying load for the time interval either during off – peak or peak
periods and usually responds to changes in system frequency.
4) Reserve or Stand – Plant – for peak or system deficiencies.
Types of Plant Reserves
a) Cold reserve – portion of the installed reserve kept in operable condition and available for service but
not for immediate loading.
b) Operating reserve – refers t capacity in actual service in excess of peak load.
c) Hot reserve – refers to unit available, maintained at operating temperature and ready for service
although not in operation.
d) Spinning reserve – generating capacity connected to bus ready to take load.
STEAM POWER PLANT - operates in the principle of “Rankine Cycle”.
Block Diagram of Steam Power Plant (1 kg steam with single extraction)
Regeneration – The process of extracting or bleeding – off steam from the turbine to open – heater and back to
the boiler for steam production.
2
Thermal Efficiency (et)
e t =Heat equivalent of mechanical energy transmitted ¿ the turbine
et =
W net Q A−Q R
Q
T
=
=1− R =1− COLD
QA
QA
QA
T HOT
¿
Heat of fuel combustion
Where: W net =W t – ΕW p
ΕW p=∑ of all pumpwork
Overall Efficiency (eo)
eo = heat equivalent of electrical output
eo = e 1 x eg
where: eg = electrical efficiency = alternator efficiency
HEAT RATE (HR)
1
Hr= ( K ) ; where: K = constant conversion factor
et
Values of k based from the following Conversion Factor
1 kWh = 3600 KJ
1 Hp – hr = 2545 BTU
1 kWh = 3413 BTU
1 kWh = 860 Kcal
HYDRO – ELECTRIC POWER PLANT (SCHEMATIC ARRANGEMENT)
3
IMPORTANT TERMS & PLANT FACTORS
1. Connected Load (CL) – It is the sum of continuous rating of all utilization equipment in the consumer
premises connected to the supply system.
2. Maximum Demand or Peak Load (MD) – It is the greatest demand of load on the power station during a
given period (say 15 – 30 minutes)
3. Average Load (AL) – It is the average load occurring on the power station in a given period (say day,
months or year), sometimes called “average demand”.
4. Demand Factor (DF) – It is the maximum ratio demand on the power station to its connected load.
DF =
MD
<1.0
CL
5. Diversity Factor (Div F) – the ratio of the sum of individual maximum demand to the simultaneous
maximum demand on the power station.
¿.F=
∑ MD ' s
Simultaneous MD
≥ 1.0
6. Coincidence Factor (CO. F) – reciprocal of diversity factor
Co . F=
Simultaneous MD
1
=
≤1.0
¿
.F
∑ MD ' s
7. Load Factor (LF) – is the ratio of average load to the maximum demand during a given period.
LF=
AL
TEP Where:
=
MD MD x Sr
Sr = service hours or operating
hours
8. Capacity Factor (CF) – it is the ratio of the actual energy produced to the maximum possible energy that
could have been produced during a given period.
CF =
TEP
IC x Pr
Where:
IC = installed capacity or rated capacity of the
machine as an individual
9. Name plate or Capacity rating – ability to carry load under condition specify by the manufacturer.
10. Capacity Rating – maximum output of equipment obtained by test under specific operating condition
(not specified by manufacturer)
11. Plant Use (or Output) – it is the ratio of energy generated to the product of plant capacity and the
number of hours for which the plant was in operation.
PUF=
TEP
IC x Sr
12. Utilization Factor (UF) – it is the ratio of maximum generator demand to the generator capacity.
UF =
MD
IC
4
13. Operating Plant Factor (OPF)– it is the ratio of average load for a given period to the operating capacity
in acual service only (for multi – set plant).
Where:
TEP
OPF=
OPC x Sr
OPC = operating plant capacity
14. Operating Load Factor (OLF) – is the ratio of the average load for the given period of the time to the
average of daily maximum demand during the same period. (say for 30 days period)
OLF=
TEP/30 (24)
MD1 + MD2 +...+ MD30
30
15. Operating (or Service Factor) – is the ratio between capacity factor (CF) to the use factor (UF), or it is
the ratio of service hours (Sr) to the period hours (Pr).
OF =
CF
Sr
=
PUF Pr
16. Plant Ratio (PR) – it is the per annum measure of the plant’s capacity factor, it is the ratio of Net KWH
output of the year to the name plate capacity of the plant times hours of the years.
PR=
TEP per year
IC x Pr( ¿ 8760)
ECONOMICS ON POWER GENERATION
The art of determining the per unit (or per KWh) cost of production of electrical energy.
Characteristics of an Electric rate (Tarrif)
a) Fair
b) Simple
c) Reasonable
The following Elements enters into the costs of Electric Energy to consumers:
1. Fixed Elements (FE) – to start the enterprise, includes the following:
a) power-plant, lands, building structures, etc.
b) primary distribution lines and subtraction including supports
c) managements, administrative staff
d) depreciation
e) General maintenance required whether the plant is in operation or not.
Note: FE is dependent on the size of enterprise.
2. Energy Elements (EE) – Sometimes called running cost, it is operating expenses to generate power /
energy which includes the following.
a) Fuels
b) Labor of operation
c) Facilities, water, oil supplies
d) Maintenance required when the plant is in operation only.
3. Customer Elements (CE) – cost to bring power to the users / consumers which includes the following:
5
a)
b)
c)
d)
Secondary distribution system, including service drops & meters.
Personnel for lines, meter readers, collection of bills.
Advertisement
Franchise (50 year period)
4. Investors Profit / Return-on-Investments (IP) – controlled by the government.
RATE (or Tariff) – rate at which electric energy is supplied to a consumer.
TYPES OF TARIFF
1. Uniform rate tariff – there is a fixed rate per KWH consumed.
2. Flat rate tariff – when different types of consumers are charged at different uniform rates.
3. Block rate tariff – also called “room rates” when a block of energy is charged at a specified rate and the
succeeding blocks of energy are charged at progressively reduced rates.
4. Two – parts tariff – also called “two – charge rates” charge on the basis of maximum demand of the
consumer and the energy consumed.
Total charges = P b x KW (of MD) + P c x KWh (consumed)
Where: P b = charge per rate KW of MD
P c = charge per KWH of energy consumed
5. Maximum demand tariff–power factor of consumer’s load is taken into consideration.
Billing KWH = (Meter Reading) x K
K = 1.0 if p.f. is from 80.1 % to 85.0 %
K > 1.0 if p.f. is lower than 80.1 % (k is from 1.01 to 1.09)
K < 1.0 if p.f. is higher than 85.0 % (k is from 0.95 to 0.99)
Economical Limit of power factor
Bi 2
p . f .= 1−
=most economical p . f .
100 A
√
(
)
Where:
A = charge per annum KWa maximum demand
i = interest and depreciation charges
B = cost of synchronous condenser per KVAR
6. Three – Part tariff – also called “three – charge rate” or “Doherty Rate” total split into three.
Total charge = P a + P b x KW (of MD) + P c x KWH (consumed)
Where:
Pa = fixed charge made during each billing period, includes interest, depreciation on the
cost of secondary distribution and labor cost.
KELVIN’S ECONOMIC LAW
6
The most economical cross – section of a conductor is that which makes the interest on the capital
outlay plus depreciation due to the conductor in the cable to the annual cost of energy loss.
POWER PLANT REVEW PROBLEMS
1. Heating value of coal depends largely on ___________.
a) moisture content
c) volatile matter
b) size of coal particles
d) ash content
2. To produce one KW-hr, a power plant burns 0.9 lb of coal with a heating value of 13,000 per lb. What is the
heat rate of the plant?
a) 6,250 BTU/kW-hr
c) 9,500 BTU/kW-hr
b) 11,700 BTU/kW-hr
d) 8,700 BTU/kW-hr
3. A 100 MW power plant has a heat rate of 2.88 x 10 6 calories per KW-hr. It is a base load plant and runs at
full load 24 hrs a day. How many tons of coal is needed per day? (EE Board Prob. Oct ’98).
a) 250
b) 625
c) 830
d) 960
4. Determine the thermal efficiency of a steam power station given that a coal consumption of 0.54 kg/kWh
and calorific value of coal of 6,400 kcal/kg.
a) 25%
b) 32%
c) 28%
d) 35%
5. An effective remover of dusts, carbon particles and others from the flue gas of the power plant is
_____________. (EE Board Prob. Apr. ’97).
a) Electrostatic precipitator
c) Mechanical collector
b) Soot blower & collector
d) dust scrubber
6. Which of the following element is radioactive? (EE Board Prob. Oct ’99)
a) Uranium
c) Californium
b) Cobalt
d) Plutonium
7. In the list below, which is not a type of present day nuclear power plant? (EE Board Prob. Mar. ’98)
a) boiling water
c) fusion – fission reactor
b) low pressure reactor
d) fast breeder reactor
8. A gas turbine works on ________ cycle.
a) Carnot
b) Otto
c) Brayton
d) Reversed Carnot
9. Power output in a reaction – turbine type hydroelectric plant can be controlled by the following, EXCEPT:
a) movable wicked gate
c) servo-motor via wicked gate
b) governor
d) movable need;e type nozzle
10. How does the output of hydraulic turbine vary with the diameter of the blade? (EE Board Prob. Oct. ’96).
a) It varies directly as the square of the diameter of the blade.
b) It varies directly as the two – thirds power of the diameter of the blade.
c) It varies directly as the diameter of the blade.
d) It varies directly as the cube of the diameter of the blade.
11. A ten – year investigation of the river’s potential gave an average water flow of 25 cm per second at the
bottom and 90 cm per second at the surface. The average cross – section at the same location is 80 m 2.
What is the average flow rate in cubic meter per hour? (EE Board Prob. Oct. ’98).
a) 165,000
c) 46,500
b) 57,500
d) 5,760
7
12. A power plant gets water from a dam from a height of 122.45 meters at the rate of 1,000 cubic meters per
minute. If the output of the plant is 15,000 kW, what is the efficiency? (EE Board Prob. Mar. ’98).
a) 80%
b) 70%
c) 75%
d) 65%
13. A natural waterfalls, 50 meters high, consistently discharges 1.3 m 3 per second. A mini hydroelectric plant
is to be constructed at the bottom of the waterfalls. Calculate the electric generator KW rating, assuming 90
% mechanical to electrical conversion efficiency and water turbine design efficiency of 70 %.
a. 401.72
b. 410.55
c. 420.42
d. 425.10
14. A pump, driven by 440 V, 3 – phase induction motor lifts 1,000 cubic feet of water per minute against a total
head of 100 ft of water. The efficiency of the pump and that of its motor are 0.75 and 0.92 respectively. The
power factor of the motor is 0.9. Calculate the cost of operating pump for 24 hours a day duty when power cost
P 0.30 per KW – hr.
a. P 1,723
b. P 2,035
c. P 1,619
d. P 1,905
15. A farmer has a small stream on his property, which he thinks might supply enough power to light his farm
load. The stream was found to be flowing at 0.5 m 3/sec. The available fall is 5 m. A small hydraulic turbine
and generator will have a combined efficiency of 75%. Assuming a transmission line efficiency of 82%, how
many 50 – W lamps can be served by the generator simultaneously?
a. 28 lamps
b. 32 lamps
c. 36 lamps
d. 30 lamps
16. A small wind generator is designed to generate 50 kW of power at a wind velocity of 25 mph. What is the
approximate windmill diameter in feet? And if the wind velocity actually varies between 20 kph and 50
kph. What is the average output power available?
a) 32 ft. and 40.9 kW
c) 36 ft. and 51.2 kW
b) 39 ft. and 60.9 kW
d) 34 ft. and 31.2 kW
17. The maximum tidal head available for a proposed tidal – power station is 6m. What must be the area of the
tidal bay to generate an average of 1000 MW of power?
a) 126.8 sq. km.
c) 186.2 sq. km.
b) 168.2 sq. km.
d) 286.1 sq. km.
18. A diversity factor of 2.5 gives a savings of ____ percent in the generating equipment.
a) 60
b) 40
c) 50
d) 25
19. The current loads for four circuits are as follows:
Circuit No. 1 = 25 amperes
Circuit No. 3 = 18 amperes
Circuit No. 2 = 38 amperes
Circuit No. 4 = 45 amperes
What is the minimum ampacity of the feeder conductor?
a) 126 amperes
c) 84 amperes
b) 189 amperes
d) 152 amperes
20. A new colony is being established in a city. There will be 200 quarters each having an average connected
load of 2kW. The business center will have a total connected load of 180 kW. What is the increase in peak
demand of the city sub – station? Given the following data:
Demand Factor
Group Diversity
Peak Diversity
Residential load
45%
3.5
1.4
Business load
60%
1.5
1.1
a) 123.4 kW
c) 102.3 kW
b) 288 kW
d) 580 kW
21. The metering of a powercustomer gave the following data:
kW-hr = 200,000
kVAR-hr = 180,000
maximum demand = 380 kW
billing days = 30
What is the load factor? (EE Board Prob. Oct. ’98)
8
a) 68%
b) 80%
c) 73%
d) 62%
22. The power customer has four circuits of 220 volts, three – phase. The circuits have maximum demands
follows:
Circuit No. 1 = 35 amperes
Circuit No. 3 = 46 amperes
Circuit No. 2 = 72 amperes
Circuit No. 4 = 57 amperes
The diversity factor is 1.5, the load factor is 50% and the power factor is 80%. What is the approximate kW
demand of the customer?
a) 15 kW
c) 43 kW
b) 21 kW
d) 53 kW
23. If a generating station has annual load factor of 30.5% and the maximum loads of the substations were
7,500 kW, 5,000 kW, 3,400 kW, 4600 kW and 2,800 kW and diversity factor among substations is 1.3. What
is the energy generated annually for the station?
a) 48,900 MWh
c) 47,900 MWh
b) 49,700 MWh
d) 46,700 MWh
24. A 100MW coal fired power plant has average heat rate of 9,500 BTU/kWh. The plant load factor is 75%.
The heat value of coal is 12,000 BTU/lb. Calculate the amount of coal usage for one day. (EE Board Prob.
Oct. ’97).
a) 1.425 x 106 lbs
c) 2.235 x 106 lbs
5
b) 2.235 x 10 lbs
d) 1.826 x 106 lbs
25. Which of the following reaction is incorrect?
a) capacity factor = utilization factor x load factor
b) demand factor x connected load = maximum demand
c) load factor x maximum load = average load
d) none of these
26. The annual peak load on a 15,000 kW power plant is 10,500 kW. Two substations are supplied by this
plant. Annual energy dispatched through substation A is 27,500,000 kWh with a peak at 8900 kW, while
16,500,000 are sent through B with a peak at 6,650 kW. Neglect line losses. What is the capacity factor of
the power plant?
a) 33.35%
b) 35.3%
c) 47.8%
d) 48.7%
27. A power plant has a maximum demand of 15 MW. The annual load factor and capacity factor are 50% and
40% respectively. Determine the reserve capacity of the plant.
a) 7530 kW
c) 5730 kW
b) 3750 kW
d) 3075 kW
28. A power plant having a three turbo – alternators rated 25 MVA, 0.9 pf, 10 MVA, unity pf and 20 MVA, 0.85
pf has a maximum demand of 40 MW. What is the utilization factor of this plant?
a) 0.88
b) 0.90
c) 0.81
d) 0.73
29. A powerplant has an annual factor as follows: load 58.5% capacity 40.9% and use 45.2%. The reserve
carried over the above peak load is 8,900 kW. The hours per year not in service and installed capacity if this
plant is __________ respectively.
a) 792 hrs& 20,346 kW
c) 833 hrs& 39,147 kW
b) 933 hrs& 49,417 kW
d) 822 hrs& 30,463 kW
30. Three towns A, B and C are situated at the corner of an equilateral triangle of 10 km sides. It is proposed to
supply the power to the extent of 5,000 kW, 3,000 kW & 2,000 kW at A, B and C respectively from the
substation. Where should the substation be located?
a) at 5.5 km horizontally from C & 4.33 vertically from C
9
b) at 4.5 km horizontally from B & 5.33 vertically from C
c) at 5,0 km horizontally from C & 3.33 vertically from C
d) none of these
31. A coal thermal power plant has a boiler, turbine, and alternator efficiencies of 35%, 86% and 93%
respectively. Coal with heating value of 12,000 BTU per lb and cost P 1.50 per lb. What is the fuel cost in
producing one kW-hr? (EE Board Prob. Oct. ’98).
a) P 1.900 per KWh
c) P1.75 per KWh
b) P 1.52 per KWh
d) P 1.62 per KWh
32. A certain coal fire power plant has a heat rate of 2.88 x 106. Calories per KW-hr. Coal costs P 2,500 per ton.
How much is the fuel costs component of producing one KW-hr? (EE Board Prob. Oct. ’98).
a) P 2.50
b) P 1.00
c) P 1.75
d) P 1.25
33. A diesel generator set burns diesel with a heating value of 18,000 BTU per lb. The diesel engine has an
efficiency of 30% and the alternator has an efficiency of 95%. Determine the fuel cost component of
producing one KW-hr if diesel costs P 2.80 per lb. (EE Board Prob. Oct. ’98).
a) P 0.15
c) P 3.28
b) P 2.15
d) P 1.86
34. A 100 MW power plant has a heat rate of 2.88x106 cal/kwhr. It is a base load plant and runs at full load 24
hrs a day. How many tons of coal is needed per day? ( Board prob- Oct.’98 )
a. 250
b. 625
c. 830
d. 960
35. A generating station has an over-all efficiency of 25% and 0.8 kg of coal is burned per kwhr oitput of the
station. Determine the calorific value of coal in kcal/kg.
a. 7300
b. 4300
c. 5300
d. 5530
36. A power generating station has 2-100 MW units each running for 8200 hrs in a year and 1-50 MW unit
running for 1500 hrs in a year. The station output is 1050x106 kwhr/year. What is the plant use factor?
a. 72.3%
b. 73.2%
c. 61.2%
d. 62.3%
37. If the power plant has a peak demand of 15 MW, capacity factor of 52.5%, use factor of 85% and 70% load
factor, find the maximum energy that could be produced daily when the plant operation is fully loaded. (EE
Board Problem May 2009)
a. 290.3 MWHR
b. 309.1 MWHR
c. 305.7 MWHR
d. 296.5 MWHR
38. A generating station supplies the ff. Loads: 15 MW, 12 MW, 8.5 MW, 6 MW and 0.45 MW. The station has a
maximum demand of 22 MW. Determine the diversity factor.
a. 1.25
b. 1.3
c. 1.45
d. 1.9
10
POWER PLANT SUPPLEMENTARY PROBLEMS
1. The feeder is usually heated before it is injected to the boiler for the following reasons EXCEPT one. Which
one is this? (EE Board Prob. Oct. ’96)
a) Improves the overall performance of the plant
b) Recover part of the flue gas heat which is otherwise lost to the atmosphere
c) Dearating types of preheaters permit the removal of dissolves gases from the feedwater.
d) Increase theefficiency of the boiler
2. The commonly used material of condenser tubes is __________.
a) aluminum
c) cast iron
b) admiralty brass
d) mild steel
3. Which one is essential for combustion of fuel?
a) oxygen to support combustion
b) proper ignition temperature
c) correct fuel air ratio
d) all of these
4. The following are the essential parts of a hydroelectric power station EXCEPT:
a) spiral case
c) draft tube
b) impeller
d) throttle valve
5. A diesel power station has a fuel consumption of 0.37 lb/KWh, the calorific value of fuel being 30,000
BTU/lb. Determine the engine efficiency if the alternator efficiency is 95%. Neglect brake power on the
engine.
a) 32%
c) 36%
b) 29%
d) 34%
6. A 55 MW steam power plant uses coal of heating value of 12,000 BTU/lb. The fuel consumption when
supplying full rated load is 32,550 lbs/hr. What is the overall efficiency of the station?
a) 32%
c) 48%
b) 38%
d) 53%
7. In fission, energy is produced when _______. (EE Board Prob. Mar. ’98).
a) chemical compound splits into its constituents
b) nuclear particles combine
c) nuclear particle splits
d) molecules combine
8. Which of the following is/or not a secondary nuclear fuels?
a) Uranium – 235
c) Plutonium – 239
b) Uranium – 233
d) both a and b
9. Reflecting mirrors used for exploiting solar energy in a solar power plant is ________.
a) mantle
c) ponds
b) diffuser
d) heliostats
10. The power customer has fur circuits of 220 Volts, three – phase. The circuits have the maximum demand as
follows: Circuit No. 1 = 35 amp; Circuit No. 2 = 46 amp; Circuit No. 3 = 72 amp; Circuit No. 4 = 57 amp. The
diversity factor is 1.5, the load factor is 50%. Determine the maximum KVA demand on the customer? (EE
Board Prob. Oct. ’98).
a) 46 KVA
c) 80 KVA
b) 26.5 KVA
d) 53 KVA
11
11. The annual load duration curve of a certain power station can be considered as a straight line from 20 MW
to 4 MW. To meet this load, three turbo alternator units, two rated at 10 MW each and one rated 5 MW are
installed. Determine the plant use factor.
a) 48 %
c) 80 %
b) 60 %
d) 38 %
12. A power plant gets water from a dam a height of 122.45 m at the rate of 1000 cubic meters per minute. If
output of the plant 15,000 kW, what is the plant efficiency?
a.80%
b. 70%
c. 75%
d. 65%
13. A100 MW thermal plant has an overall efficiency of 34%. If the coal used has a heating value of 10,800 Btu
per pound, calculate the coal consumption of the plant per KW – hr output.
a. 0.84 lb
b. 0.96 lb
c. 0.89 lb
d. 0.93 lb
14. A certain coal – fired power plant has a heat rate of 2.88 x 10 6 calories per KW – hr. Coal costs P 2,500 per
ton. How much is the fuel cost component of producing one KW – hr? Assume the heating value of coal used
equal to 13,000 Btu/lb.
a. P 2.50
b. P 1.75
c. P 1.00
d. P 1.25
15. A power plant has a fuel consumption of 1 lb per KW – hr generated. Determine the heating value of the
fuel used in Btu per pound. The overall efficiency of the plant is 36%.
a. 10,000
b. 9,500
c. 9,800
d. 10,050
16. A power plant consumes 100,000 pounds of coal per hour. The heating value of the coal is 12,000 BTU per
pound. The overall efficiency of the plant is 30%. What is the KW output of the plant?
a. 105,500 kW
b. 205,000 kW
c. 142,500 KW
d. 175,000 kW
17. A plant has a total operating capacity of 800 kW. The coal consumption is 1,900 lbs per hour. Heating value
of coal is 9,500 BTU per pound. What approximate percent of the heat in the coal is converted into useful
energy?
a. 12.3
b. 15.0
c. 8.7
d. 17.5
18. A power plant has an overall efficiency of 30%. If this plant can consumed 4,200 kilograms of coal per hour,
estimate the total electric energy produced in one day. Assume the calorific value of the coal being used is
5,000 kcal per kilogram.
a. 183 MW – hr
b. 168 MW – hr
c. 155 MW – hr
d. 176 MW – hr
19. A coal power plant has an overall plant efficiency of 28%. Coal with a heating value of 12,000 Btu per
pound cost P 1.50 per pound. What is the fuel cost of producing one kW – hr?
a. P 1.90
b. P 1.52
c. P 1.75
d. P 1.62
20. A diesel electric generating unit supplies a load of 70 kW. The heating value of the oil used is 12,000 kcal
per kg. If the overall efficiency of the unit is 40%, determine the mass of oil required per hour.
a. 11.8 kg
b. 10.3 kg
c. 12.5 kg
d. 9.67 kg
21. A diesel generator set burs diesel with a heating value of 18,000 BTU per pound. The diesel engine has an
efficiency of 30% and the alternator has an efficiency of 95%. Determine the fuel cost component of
producing one kW – hr if diesel costs P 2.80 per pound.
a. P 0.15
b. P 3.28
c. P 2.15
d. P 1.86
22. To produce one kW – hr, a power plant burns 0.9 lb of coal with a heating value of 13,000 BTU. What is the
heat rate of the plant?
a. 6,250 BTU/kW-hr
b. 9,550 BTU/kW-hr c. 11,700 BTU/kW-hr d. 8,700 BTU/kW-hr
12
23. A substation transformer is to serve the following loads:
Classification
Total Load
Demand Factor
Lighting
300 kW
60%
Power
1200 kW
80%
Heating
500 kW
90%
If the diversity factor among the load types is 1.5, determine the maximum demand on the transformer.
a. 2000 kW
b. 1060 kW
c. 1225 kW
d. 1180 kW
24. The power customer has four feeder circuits of 220 V, three – phase. The circuits have maximum demand
follows:
Circuit No. 1 = 35 A
Circuit No. 3 = 72 A
Circuit No. 2 = 46 A
Circuit No. 4 = 57 A
The diversity factor is 1.5; determine the maximum kVA demand of the customer.
a. 46
b. 26.5
c. 80
d. 53
25. The current loads of four circuits are as follows:
Circuit No. 1 = 25 A
Circuit No. 3 = 18 A
Circuit No. 2 = 38 A
Circuit No. 4 = 45 A
If the diversity factor is 1.5, what is the minimum ampacity of the feeder conductor?
a. 126 A
b. 189 A
c. 84 A
d. 152 A
26. A power plant has two 75 MW sets each operating at 7200 hours per year and a 50 MW set operating at
3000 hours per year. If the plant output is 800 x 106 kW – hr per year, solve for the plant use factor.
a. 63.4%
b. 65%
c. 68.35
d. 70.6%
27. An industrial plant has an aggregated load of 100 kW. The demand factor is 60%. If the average total energy
consumption of the plant in one year is 186 MW – hr, what is the yearly load factor of the plant?
a. 32.45%
b. 42.25%
c. 38.67%
d. 35.39%
28. The power customer has four circuits of 220 V, three – phase. The circuits having maximum demand are as
follows:
Circuit No. 1 = 35 A
Circuit No. 3 = 72 A
Circuit No. 2 = 46 A
Circuit No. 4 = 57 A
The diversity factor is 1.5, the load factor is 50% and the power factor is 80%. What is the approximate kW
demand of the customer?
a.15 kW
b. 43 kW
c. 21 kW
d. 53 kW
29. The metering of a power customer was read and gave the following data:
kW – hr
=
200,000
kVAR – hr
=
180,000
Maximum demand =
380 kW
Billing days
=
30
What is the load factor?
a. 68%
b. 73%
c. 80%
d. 38.28%
30. A thermal power plant has two generating sets rated 600 kW and 800 kW respectively being operated at
rated capacity. The total coal consumption I 680 kg per hour. If the coal has a heating value of 20,000 joules
per gram, what fraction of heat produced from the coal is converted into useful electrical energy?
a. 37.06%
b. 40.20%
c. 33.67%
d. 38.28%
31. A 55,000 kW thermal plant of National Power Corporation delivers an annual output of 238,000,000 kW –
hr with a peak load of 44 MW. What is the capacity factor?
a. 49.4%
b. 42.4%
c. 48.2%
d. 44.6%
13
32. A central generating station has an annual factor report as follows:
Plant capacity factor
=
50%
Load factor
=
60%
Use factor
=
45%
Reserved
=
10,500 kW
Determine the rated capacity of the station.
a. 63,000 kW
b. 60,000 kW
c. 65,000 kW
d. 70,000 kW
33. A certain power plant has a reserve capacity above the peak load of 10 MW. The annual factors are load
factor = 59%, capacity factor = 41% and use factor = 46%. Determine hour per year not in service.
a. 952 hours
b. 965 hours
c. 924 hours
d. 937 hours
34. A certain power plant has a reserve capacity above the peak load of 10 MW. The annual factors are load
factor = 59%, capacity factor = 41% and use factor = 46%. Determine installed capacity.
a. 33.45 MW
b. 31.85 MW
c. 32. 77 MW
d. 30.47 MW
35. The average weekly energy generated by a thermal plant is 7,000,000 kW – hr. The peak load of the plant is
75,000 kW during the week. If the plant has an installed capacity of 100 MW, solve for the load factor of the
plant.
a. 48.92%
b. 56.34%
c. 50.44%
d. 55.55%
36. A power plant has two 75 MW sets each operating at 7,200 hours per year and a 50 MW set operating at
3000 hours per year. If the plant output is 800 x 106 kW – hr per year, solve for the plant use factor.
a. 63.4%
b. 65.0%
c. 68.3%
d. 70.6%
37. An industrial plant has an aggregate load of 100 kW. The demand factor is 60%. If the average total energy
consumption of the plant in one year is 186 MW – hr, what is the yearly load factor of the plant?
a. 32.45%
b. 42.25%
c. 38.67%
d. 35.39%
14
Download