Economies of Power Generation MCQ Quiz - Objective Question with Answer for Economies of Power Generation - Download Free PDF
Last updated on May 14, 2025
Latest Economies of Power Generation MCQ Objective Questions
Economies of Power Generation Question 1:
Which of the following types of engines is most commonly adapted to run on liquid biomass fuels?
Answer (Detailed Solution Below)
Economies of Power Generation Question 1 Detailed Solution
Internal combustion engine
Internal combustion engines (ICEs) are the most commonly adapted to run on liquid biomass fuels, such as ethanol, biodiesel, and other biofuels. These fuels can be used in modified gasoline or diesel engines with relative ease compared to other engine types listed.
Reasons to use an internal combustion engine:
- ICEs are designed to burn liquid fuels, like gasoline and diesel, which makes them naturally suited for biofuels such as ethanol and biodiesel.
- ICEs can be optimized to improve efficiency and emissions with biofuels.
- Governments promote the use of biofuels in ICEs to reduce dependence on fossil fuels and cut emissions, making them an attractive option for adaptation.
Economies of Power Generation Question 2:
Which of the following options best describes the role of a gasifier in an agro-chemicalbased power plant?
Answer (Detailed Solution Below)
Economies of Power Generation Question 2 Detailed Solution
Agro-chemical-based power plant
- An "agrochemical-based power plant" typically refers to a facility that generates electricity using agricultural waste or biomass, often in conjunction with or as a byproduct of agrochemical manufacturing processes.
- These plants may use technologies like combined heat and power (CHP) or cogeneration to produce steam or electricity, which can then be used for in-house operations or sold to the grid.
Function of the gasifier:
A gasifier is a key component in biomass-based or agrochemical-based power plants. Its role is to convert solid biomass (like wood chips, crop waste, etc.) into a combustible gas mixture (called syngas or producer gas) using a thermochemical process involving partial oxidation (limited oxygen).
Economies of Power Generation Question 3:
If a residential area has an annual energy consumption of 28,908,000 kWh, what is its average demand?
Answer (Detailed Solution Below)
Economies of Power Generation Question 3 Detailed Solution
Concept
The average demand for a residential area is given by:
\(Average \space demand={Total \space energy \space consumption\over Time \space Period}\)
Calculation
Given, Annual energy consumption = 28,908,000 kWh
Time Period = 365 × 24 = 8760 hours
\(Average \space demand={28908000 \over 8760}\)
Average demand = 3300 kWh
Economies of Power Generation Question 4:
An apartment complex consists of 15 apartments, each with a peak demand of 8 kW. The maximum system demand recorded is 60 kW. What is the diversity factor?
Answer (Detailed Solution Below)
Economies of Power Generation Question 4 Detailed Solution
Concept
The diversity factor is given by:
\(Diversity \space factor={ Sum \space of \space \space individual \space maximum \space demands \over Maximum \space demand \space of \space the \space system}\)
Calculation
Given, Number of apartments = 15
Peak demand per apartment = 8 kW
System maximum demand = 60 kW
Sum of individual demands = 15 × 8 = 120 kW
\(Diversity \space factor={120\over 60}=2\)
Economies of Power Generation Question 5:
The difference between Capacity Factor and Load Factor represents __________.
Answer (Detailed Solution Below)
Economies of Power Generation Question 5 Detailed Solution
Load factor
Load factor measures the efficiency with which electrical energy is used.
It is defined as the ratio of the average load over a given period to the maximum load during that same period.
\(Load \space factor={Average \space load\over Maximum\space load}\)
The value of the load factor is always less than 1.
Capacity factor
It measures how often a plant is running at maximum power.
\(Capacity \space factor={Average \space demand\over Plant\space capacity}\)
The difference between both the two factors shows how much of the plant’s total capacity is not being utilized to meet current demand, thus representing potential reserve for future growth.
Top Economies of Power Generation MCQ Objective Questions
A consumer consumes 600 kWh per day at a load factor of 0.5. Without increasing the maximum demand, if the consumer increases the load factor to 0.8, the consumption of energy in kWh would be:
Answer (Detailed Solution Below)
Economies of Power Generation Question 6 Detailed Solution
Download Solution PDFConcept:
Load factor:
The load factor is the ratio of average energy consumed to maximum demand.
Load factor = average energy consumed / maximimum energy consumed
Calculation:
Given load factor = 0.5
Average energy consumed at 0.5 load factor = 600 kWh
Maximum energy consumed = \(\frac{{600}}{{0.5}}\) = 1200 kWh
Now maximum energy consumed is constant and load factor is increased to 0.8
Average energy consumed = load factor × maximum energy consumed
= 0.8 × 1200
= 960 kWhWhat is the reserve capacity, if the maximum demand of generation of power is 50 MW, the load factor of the plant is 60% and the plant capacity factor is 50%?
Answer (Detailed Solution Below)
Economies of Power Generation Question 7 Detailed Solution
Download Solution PDFLoad factor \(=\frac{average~demand}{maximum~demand}\)
Average demand = (50) (0.6) = 30 MW
Plant capacity factor \(=\frac{average~demand~}{plant~capacity}\)
Plant capacity \(=\frac{30}{0.5}=60~MW\)
Reserve capacity = Plant Capacity – Maximum Demand = 60 - 50 = 10 MWThe diversity factor between transformers for residential lighting is:
Answer (Detailed Solution Below)
Economies of Power Generation Question 8 Detailed Solution
Download Solution PDFConcept:
Diversity factor: The ratio of the sum of individual maximum demands to the maximum demand on the power station is known as a diversity factor.
\(Diversity\;factor = \frac{{Sum\;of\;individual\;maximum\;demands}}{{Maximum\;demand\;on\;power\;station}}\)
A power station supplies load to various types of consumers whose maximum demands generally do not occur at the same time. Therefore, maximum demand on the power station is always less than the sum of the individual maximum demands of the consumer. Hence diversity factor is always greater than 1.
The knowledge of diversity factor is vital in determining the capacity of the plant equipment.
The greater the diversity factor, the lesser is the cost of generation of power. Because greater diversity factor means lesser maximum demand. Now, lower maximum demand means a lower capacity of the plant which reduces the cost of the plant.
Explanation:
Diversity factor can be used to estimate the total load required for a facility or to size the transformer.
Diversity factors have been developed for main feeders supplying a number of feeders, and the typical values are given below.
- Residence consumer - 1.2 to 1.3
- Commercial load - 1.1 to 1.2
- Power and lighting loads - 1.50 to 2.00
An industrial consumer has a load pattern of 2000 kW at 0.8 lagging for 12 hours and 1000 kW at unity power factor for 12 hours. Its load factor is
Answer (Detailed Solution Below)
Economies of Power Generation Question 9 Detailed Solution
Download Solution PDFConcept:
Load factor: The ratio of average load (AL) to the maximum demand (MD) during a given period is known as the load factor.
\(Load factor =\dfrac{AL}{MD}\)
If the plant is in the operation of T hours
\(Load factor = \dfrac{{AL\times T}}{{MD \times T}}\)
Note: To find Load factor, Demand Factor, Diversity Factor, etc, we used unit of Power in kW or W, and unit of Energy in kWh or Wh, not in kVA or kVAh
Application:
Given,
P1 = 2000 kW for 12 hr,
P2 = 1000 kW for 12 hr,
Since, maximum power is P1 hence,
MD = 2000 kW
Now, Average load (AL) can be calculated by the ratio of total energy consumed in kWh to the total time
\(AL=\dfrac{(2000\ kW\ \times\ 12\ hr)+(1000\ kW \times\ 12\ hr)}{24\ hr}=1500\ kW\)
From above concept,
\(Load \ factor=\dfrac{AL}{MD}=\dfrac{1500}{2000}\)
Load Factor = 0.75
The load factor of domestic load is
Answer (Detailed Solution Below)
Economies of Power Generation Question 10 Detailed Solution
Download Solution PDF
Type of load |
Demand factor |
Diversity factor |
Load factor |
Domestic |
70-100% |
1.2-1.3 |
10-15% |
Commercial |
90-100% |
1.1-1.2 |
25-30% |
Industrial |
70-80% |
1.3-1.6 |
60-65% |
Agricultural |
90-100% |
1-1.5 |
15-25% |
A power station has 4 consumers with their maximum demand as 40 MW, 20 MW, 30 MW and 50 MW. The maximum demand of the station is 100 MW. The diversity factor of the plant is ______.
Answer (Detailed Solution Below)
Economies of Power Generation Question 11 Detailed Solution
Download Solution PDFDiversity factor:
The ratio of the sum of individual maximum demands to the maximum demand on the power station is known as a diversity factor.
\(Diversity\;factor = \frac{{Sum\;of\;individual\;maximum\;demands}}{{Maximum\;demand\;on\;power\;station}}\)
A power station supplies load to various types of consumers whose maximum demands generally do not occur at the same time. Therefore, the maximum demand for the power station is always less than the sum of the individual maximum demands of the consumer. Hence diversity factor is always greater than 1.
The knowledge of the diversity factor is vital in determining the capacity of the plant equipment.
The greater the diversity factor, the lesser is the cost of generation of power. Because greater diversity factor means lesser maximum demand. Now, lower maximum demand means a lower capacity of the plant which reduces the cost of the plant.
Calculation:
Sum of maximum individual demand = 40 + 20 + 30 + 50 = 140 MW
Maximum demand of the station = 100 MW
Diversity factor \( = \frac{{140}}{{100}} = 1.4\)
The annual load duration curve is a straight line from 20 MW to 6 MW. Find the load factor.
Answer (Detailed Solution Below)
Economies of Power Generation Question 12 Detailed Solution
Download Solution PDFLoad Factor
Load factor is defined as the ratio of the average load over a given period to the maximum demand (peak load) occurring in that period.
\(Load \space factor={Average \space demand \over Maximum \space demand}\)
The average demand is given by:
\(Average \space demand={Area \space under \space curve \over Total \space time(in\space hrs)}\)
Calculation
\(Average \space demand={(6\times 8760)+(1/2\space \times \space 8760 \space \times 14 ) \over 8760}\)
Average demand = 13 MW
\(Load \space factor={13\over 20}=0.65\)
Load factor = 65%
Which of the following expressions depicts the Utilization Factor?
Answer (Detailed Solution Below)
Economies of Power Generation Question 13 Detailed Solution
Download Solution PDFUtilization factor: It is the ratio of maximum demand on the power station to the rated capacity of the power station.
Utilization factor = maximum demand / rated capacity
Important Points
Diversity factor: The ratio of the sum of individual maximum demands to the maximum demand on the power station is known as a diversity factor.
\(Diversity\;factor = \frac{{Sum\;of\;individual\;maximum\;demands}}{{Maximum\;demand\;on\;power\;station}}\)
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.
\(Plant\;capacity\;factor = \frac{{{\rm{Actual\;energy\;produced}}}}{{Maximum\;energy\;that\;could\;have\;been\;produced}}\)
\( = \frac{{Average\;demand \times T}}{{Plant\;capacity \times 100}}\)
\(= \frac{{Average\;demand}}{{Plant\;capacity}}\)
\(Annual\;plant\;capacity\;factor = \frac{{{\rm{Annual\;kWh\;output}}}}{{Plant\;capacity \times 8760}}\)
Demand factor: It is the ratio of maximum demand on the power station to its connected load.
Demand factor = Maximum Demand / Connected load
Utilization factor of a power station is the ratio of:
Answer (Detailed Solution Below)
Economies of Power Generation Question 14 Detailed Solution
Download Solution PDFUtilization factor: It is the ratio of maximum demand on the power station to the rated capacity of the power station.
Utilization factor = maximum demand / rated capacity
Demand factor: It is the ratio of maximum demand on the power station to its connected load.
Demand factor = Maximum Demand / Connected load
Diversity factor: The ratio of the sum of individual maximum demands to the maximum demand on the power station is known as a diversity factor.
\(Diversity\;factor = \frac{{Sum\;of\;individual\;maximum\;demands}}{{Maximum\;demand\;on\;power\;station}}\)
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.
\(Plant\;capacity\;factor = \frac{{{\rm{Actual\;energy\;produced}}}}{{Maximum\;energy\;that\;could\;have\;been\;produced}}\)
\( = \frac{{Average\;demand \times T}}{{Plant\;capacity \times 100}}\)
\(= \frac{{Average\;demand}}{{Plant\;capacity}}\)
\(Annual\;plant\;capacity\;factor = \frac{{{\rm{Annual\;kWh\;output}}}}{{Plant\;capacity \times 8760}}\)
Determine the average load from the load curve.
Answer (Detailed Solution Below)
Economies of Power Generation Question 15 Detailed Solution
Download Solution PDFConcept:
- The curve which shows the variation of load on the electrical power station with respect to time is known as the load variation curve or load curve.
- The Daily load curve gives information about the load on the power station during different running hours of the day.
- The area under the daily load curve gives the total units of electrical energy generated.
- Units Generated/day = Area under daily load curve (kW)
- The maximum demand of the station on that day is found from the highest point of the daily load curve.
- Average Load = Area under the daily Load Curve (kWh)/24 hrs
Calculation:
From the given daily load curve, the load distribution is as follows:
(0 – 6) hours – 40 MW for 6 hours
(6 to 10) hours – 50 MW for 4 hours
(10 to 12) hours – 60 MW for 2 hours
(12 to 16) hours – 50 MW for 4 hours
(16 to 20) hours – 70 MW for 4 hours
(20 to 24) hours – 40 MW for 4 hours
Total load = (40 × 6) + (50 × 4) + (60 × 2) + (50 × 4) + (70 × 4) + (40 × 4)
= 1200 MW
Total number hours = 24
Average load = 1200/24 = 50 MW