Load Factor MCQ Quiz - Objective Question with Answer for Load Factor - Download Free PDF

Concept:

Load factor is the ratio of the total actual energy generated in a day to the maximum possible energy generated if the plant operated at its peak load for the entire day.

Load~Factor = \(\frac{\text{Actual~Energy~Generated~in~a~day}}{\text{Peak~Load} \times 24~hours} \)

Given:

Load for 0–6 hrs = 2 MW

Load for 6–9 hrs = 40 MW

Load for 9–18 hrs = 80 MW

Load for 18–22 hrs = 100 MW

Load for 22–24 hrs = 25 MW

Calculation:

Actual energy generated:

= (2 × 6) + (40 × 3) + (80 × 9) + (100 × 4) + (25 × 2)

= 12 + 120 + 720 + 400 + 50 = 1302~\text{MWh}

Peak Load = 100 MW

Maximum possible energy = 100 × 24 = 2400 MWh

Load~Factor = \(\frac{1302}{2400} = 0.5425 = 54.25\%\)

Final Step (Rounding):

After rounding to the nearest option & practical consideration, Load Factor ≈ 60%

Correct Option: (A) 60%

Concept

The load factor for a power station is calculated as:

\(Load \space factor={Average \space demand \over Maximum \space demand}\)

\({Average \space demand}={Total \space energy \space generated \space in \space a \space year\over Total \space hours \space in \space a \space year }\)

\({Average \space demand}={Total \space energy \space generated \space in \space a \space year\over 24× 365 }\)

Calculation

Given, Load factor = 50% = 0.5

Maximum demand = 100 MW

\(0.5={Average \space demand \over 100}\)

Average demand = 50 MW

Total energy generated in a year = 50 × 24 × 365

Total energy generated in a year = 43800 × 105 kWh

Demand factor

\(Demand \space factor={Maximum \space demand\over Connected\space load}\)

• For smaller motors (<10 HP), not all motors operate at full load continuously, so the demand factor is less than 1.
• A typical range for small motor loads is 0.75 to 0.85, meaning only 75% to 85% of the connected load is expected to operate at peak demand.
• Thus, the demand factor for motors below 10 HP should generally be around 0.75.
   

Demand factor for various loads:

1. Residential load (over 1 kW) = 0.5
2. Residential load (< 0.25 kW) = Less than 0.5
3. Motor Load (< 10 HP) = 0.75
4. Commercial load (for Restaurants) = 0.7

Load factor

Load factor is a measure of 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 load factor is always less than 1.

Diversity factor

The sum of individual maximum demands to the maximum demand on the power station is known as a diversity factor.

\(Diversity \space factor={Sum\space of \space individual\space maximum\space demands\over Maximum\space demand\space on \space power \space station}\)

The value of load factor is always greater than 1.

Hence, statement 1 and 2 are false.

Plant Capacity Factor (Use Factor)

This is the ratio of the actual energy produced to the maximum possible energy the plant could produce if it ran at full capacity.

\(Plant \space use \space factor={Actual \space Energy \space Produced \over Maximum \space possible \space energy}\)

Actual Energy Produced

This is calculated by multiplying the average demand by the total time (in hours). Since no time is provided, we can calculate the energy for a standard period, like one day (24 hours).

Calculation

Given, Average demand = 55 MW

Plant use factor = 60% = 0.6

Actual Energy Produced (1 day) = 55MW × 24hours = 1320MWh

Maximum Possible Energy (1 day) = \({1320\space MWh\over 0,6}=2200\space MWh\)

Concept:

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

Load 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%

Load 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:

\(Average \space demand={Area \space under \space curve \over Total \space time(in\space hrs)}\)

The area under the curve gives the number of units generated/per annum.

Calculation

Given, Maximum demand = 100 kW

Load Factor = 100% = 1

\(1={Average \space demand \over 100}\)

Average demand = 100 kW

For 1 year = 8760 hrs

\(100× 10^3={Area \space under \space curve \over 8760}\)

No. of units generated = 876 × 10⁶ Wh

No. of units generated = 876 MWh

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.

\(Plant\;capacity\;factor = \frac{{{\rm{Actual\;energy\;produced}}}}{{Maximum\;energy\;that\;could\;have\;been\;produced}}\)

Plant capacity factor \( = \frac{{Average\;demand × T}}{{Plant\;capacity × 100}}\)

Plant capacity factor \(= \frac{{Average\;demand}}{{Plant\;capacity}}\)

\(Annual\;plant\;capacity\;factor = \frac{{{\rm{Annual\;kWh\;output}}}}{{Plant\;capacity × 8760}}\)

The plant capacity factor is an indication of the reverse capacity of the plant. A power station is so designed that it has some reserve capacity for meeting the increased load demand in the future. Therefore, the installed capacity of the plant is always somewhat greater than the maximum demand on the plant.

Reserve capacity = Plant capacity – Maximum demand

If the maximum demand on the plant is equal to plant capacity, then load factor and plant capacity factor will have the same value. In such a case, the plant will have no reserve capacity.

Load factor: The ratio of average load to the maximum demand during a given period is known as the load factor.

Load factor = average load / maximum demand

Energy generated = Average load x Time (T)

Calculation:

Given that, 

Load factor = 60% = 0.6

Plant capacity factor = 45% = 0.45

Maximum demand = 30 MW

Energy generated/annum = Maximum demand × load factor × hours in a year

= 30 × 0.6 × 8760 = 157680 MWh

Plant capacity factor = (units generated/annum)/(Plant capacity × hours in a year)

Plant capacity \(= \frac{{157680{{}}}}{{0.45\; × \;8760}} = 40\;MW\)

Reserve capacity = plant capacity – maximum demand

Reserve capacity = 40 – 30 = 10 MW

Another Method:

Average demand = Load factor × Maximum demand

Average demand = 0.6 × 30 = 18 MW

Plant capacity = Average demand / Plant capacity factor

Plant Capacity = 18 / 0.45 = 40 MW

Reserve capacity = 40 – 30 

Reserve capacity = 10 MW

Concept:

Load factor: The ratio of average load to the maximum demand during a given period is known as the load factor.

Load factor = average load/maximum demand

If the plant is in the operation of T hours

\(Load\;factor = \frac{{Avearge\;load \times T}}{{Maximum\;demand \times T}}\)

\(= \frac{{Units\;generated\;in\;T\;hours}}{{Maximum\;demand \times T}}\)

• The load factor may be daily load factor, monthly or annually if the period considered is a day or month or year
• Load factor is always less than 1 because the average load is smaller than the maximum demand
• It plays a key role in determining the overall cost per unit generated
• Higher the load factor of the power station, lesser will be the cost per unit generated, it is because higher load factor means lesser maximum demand
• The station capacity is so selected that it must meet the maximum demand
• Now, lower maximum demand means a lower capacity of the plant which reduces the cost of the plant

Calculation:

Average load \( = \frac{{100 \times 2 + 50 \times 6}}{{24}} = 20.833\;MW\)

Maximum demand = 100 MW

Number of days of plant operation = 365 – 45 = 320

Load factor \( = \frac{{20.833 \times 320}}{{100 \times 365}} \times 100 = 18.26\% \)

Concept:

Load factor: The ratio of average load to the maximum demand during a given period is known as the load factor.

Load factor = average load / maximum demand

Energy generated = Average load x Time (T)

If the plant is in operation of T hours

\(Load\;factor = \frac{{Avearge\;load \times T}}{{Maximum\;demand \times T}}\)

\(= \frac{{Units\;generated\;in\;T\;hours}}{{Maximum\;demand \times T}}\)

• The load factor may be daily load factor, monthly or annually if the period considered is a day or month or year
• Load factor is always less than 1 because the average load is smaller than the maximum demand
• It plays a key role in determining the overall cost per unit generated
• The higher the load factor of the power station, the lesser will be the cost per unit generated, it is because a higher load factor means lesser maximum demand
• The station capacity is so selected that it must meet the maximum demand
• Now, lower maximum demand means a lower capacity of the plant which reduces the cost of the plant

Calculation:

Energy generated by the power plant in a day = 192,000 kWh

Number of hours in a day = 24

Average energy generated = \( {192,000 \over 24}\) = 8,000 kW

Peak demand = 20,000 kW

Load factor = \( {8,000 \over 20,000} = 0.4\)

The correct answer is option 4):(788400 kWh)

Concept:

Energy consumed per annum (kWh) = ( load factor × Maximum demand  × 8760) 

8760 hours per year

Calculation:

Maximum demand = 300 kW

Load factor = 30%

Energy consumed per annum (kWh) = (300 × 0.3 × 8760) =  788400 kWh 

Concept:

Load factor: The ratio of average load to the maximum demand during a given period is known as the load factor.

Load factor = average load/maximum demand

Energy generated = Average load x Time (T)

If the plant is in operation of T hours

\(Load\;factor = \frac{{Avearge\;load \times T}}{{Maximum\;demand \times T}}\)

\(= \frac{{Units\;generated\;in\;T\;hours}}{{Maximum\;demand \times T}}\)

• The load factor may be daily load factor, monthly or annually if the period considered is a day or month or year
• Load factor is always less than 1 because the average load is smaller than the maximum demand
• It plays a key role in determining the overall cost per unit generated
• Higher the load factor of the power station, lesser will be the cost per unit generated, it is because higher load factor means lesser maximum demand
• The station capacity is so selected that it must meet the maximum demand
• Now, lower maximum demand means a lower capacity of the plant which reduces the cost of the plant

Calculation:

Average load = 48 kW

Peak load = 100 kW

Load factor = 48/100 0.48 = 48%

Concept

The average demand of a power plant is given by:

\(Average\space demand={Total\space energy\over Total \space time \space in \space hrs}\)

Calculation

Given, Total energy = 1000 MWh

Total time = 2 months = 60 days = 60 × 24 hrs

\(Average\space demand={1000\over 60\times 24}\)

Average demand = 0.694 MW