Quiz: Electrical Engineering 26 May 2021

Quiz: Electrical Engineering
Exam: DFCCIL
Topic: Utilization of electrical energy

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. When the rate of electrical energy is charged on the basis of maximum demand of the consumer and the unit consumed, it is called a:
(a) Simple tariff
(b) Two-part tariff
(c) Flat-rate tariff
(d) Block-rate tariff

Q2. When a given block of energy is charged at a specified rate and the succeeding blocks of energy are charged at progressively reduced rates, it is called:
(a) Power factor tariff
(b) Two-part tariff
(c) Maximum demand tariff
(d) Block rate tariff

Q3. When different types of consumers are charged at different uniform per unit rates, the tariff is called:
(a) Flat rate tariff
(b) Two-part tariff
(c) Simple tariff
(d) Block rate tariff

Q4. A generating station has a connected load of 500 MW and a maximum demand of 250 MW. The units generated being 10000 per week. Calculate the demand factor.
(a) 2
(b) 40
(c) 20
(d) 0.5

Q5. Domestic consumer load is around:
(a) 5kW
(b) 40kW
(c) 80kW
(d) 120kW

Q6. In power station, the cost of generation of power reduces most effectively when………….
(a) Diversity factor alone increases
(b) Both diversity factor and load factor increase
(c) Load factor alone increases
(d) Both diversity factor and load factor decrease

Q7. The value of demand factor is:
(a) less than one
(b) greater than one
(c) equal to one
(d) Zero

Q8. Load factor during a period is defined as:
(a) (Average Load)/(Maximum Load)
(b) (maximum Load)/(Average Load)
(c) (Average Load)/(Installed Capacity)
(d) (Maximum Load)/(Installed Capacity)

Q9. The maximum demand on a power station is 600 MW, the annual load factor is 60% and the capacity factor is 45%. The reserve capacity of the plant is
(a) 200 MW
(b) 75 MW
(c) 800 MW
(d) 450 MW

Q10. The specific energy consumption of a train depends on which of the following?
(a) Acceleration and retardation
(b) Distance covered
(c) Gradient
(d) All of the above

SOLUTIONS
S1. Ans.(b)
Sol.
Simple Tariff: a fixed rate is applied for each unit of the energy consumed.
Flat Rate Tariff: different types of consumers are charged at different rates of cost per unit (1kWh) of electrical energy consumed. The different rates are decided according to the consumers, their loads and load factors.
Block Rate Tariff: In this tariff, the first block of the energy consumed (consisting of a fixed number of units) is charged at a given rate and the succeeding blocks of energy (each with a predetermined number of units) are charged at progressively reduced rates. The rate per unit in each block is fixed.
Two Part Tariff: In this tariff scheme, the total costs charged to the consumers consist of two components: fixed charges and running charges. The fixed charges will depend upon maximum demand of the consumer and the running charge will depend upon the energy (units) consumed.

S2. Ans.(d)
Sol.
Block rate tariff → given block of energy is charged at specific rate and succeeding blocks of energy at reduced rates
Two-part tariff → for industrial consumer based on maximum demand
Flat rate tariff → for different types of consumer.

S3. Ans.(a)
Sol. Flat rate tariff → for different types of consumers (i.e., for induction motor, welding etc)

S4. Ans.(d)
Sol. Demand factor = (Max.demand)/(Connecte load)
=250/500
=0.5

S5. Ans.(a)
Sol. Domestic load ≃5

S6. Ans.(b)
Sol. in power plant, the cost of generation of power is related with load factor and diversity factor.
Higher the load factor and diversity factor, the lower will be cost of generation of power. Diversity factor is greater than 1 and load factor is less than 1. Higher will be load factor, lower will be cost of unit generated.

S7. Ans.(a)
Sol. Demand factor is always less than 1.
Demand factor = (maximum demand)/(connected load)

S8. Ans.(a)
Sol.
Load factor = (Avg.load)/(Max.load)

S9. Ans.(a)
Sol. (Reserved capacity=Plant capacity-maximum demand)
Capacity factor = (Max.demand)/(plant capacity)×load factor
Given that –
Max. demand = 600 MW
CF = 45% = 0.45
LF = 60% = 0.60
CF=600/(Plant capacity)×0.60
⇒Plant capacity =(600×0.60)/0.45
=800 MW
∵ Reserve capacity = 800-600
= 200 MW

S10. Ans.(d)
Sol. All the mentioned causes.

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