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A factory takes a steady load of 200 kW at a lagging power factor of 0.8. The tariff is Rs. 100 per kVA of maximum demand per annum plus 5 paise per kWh. The phase advancing plant costs Rs. 500 per kVAR and the annual interest and depreciation together amount to 10%. The most economical p.f. is
  • a)
    0.707 lag
  • b)
    0.955 lag
  • c)
    0.625 lag
  • d)
    0.866 lag
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A factory takes a steady load of 200 kW at a lagging power factor of 0...
Peak load of factory, P = 200 kW
Original power factor, cosϕ1=0.8lagging
Max. demand charges, x = Rs. 100/kVA/annum
Charges on phase advancing plant are
y=Rs.500×0.1=Rs.50/kVAR/annum
Most economical p.f.,
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Most Upvoted Answer
A factory takes a steady load of 200 kW at a lagging power factor of 0...
Most economical power factor for the factory is 0.866 lag.

Given data:
- Load = 200 kW
- Power factor (lagging) = 0.8
- Tariff = Rs. 100 per kVA of maximum demand per annum + 5 paise per kWh
- Cost of phase advancing plant = Rs. 500 per kVAR
- Annual interest and depreciation = 10%

To determine the most economical power factor, we need to compare the costs associated with different power factors. The total cost can be divided into two parts: the cost of maximum demand and the cost of energy consumption.

1. Cost of maximum demand:
The cost of maximum demand is given by the tariff, which is Rs. 100 per kVA of maximum demand per annum. The maximum demand is calculated using the formula:
Maximum demand (kVA) = Load (kW) / Power factor

Let's calculate the cost of maximum demand for the given power factor of 0.8:
Maximum demand (kVA) = 200 kW / 0.8 = 250 kVA
Cost of maximum demand = 250 kVA * Rs. 100/kVA = Rs. 25,000

Now, let's calculate the cost of maximum demand for a power factor of 0.866 lag:
Maximum demand (kVA) = 200 kW / 0.866 = 230.94 kVA
Cost of maximum demand = 230.94 kVA * Rs. 100/kVA = Rs. 23,094

2. Cost of energy consumption:
The cost of energy consumption is given by the tariff, which is 5 paise per kWh. The energy consumption can be calculated using the formula:
Energy consumption (kWh) = Load (kW) * Time

Let's assume the time is 1 year. The cost of energy consumption for the given power factor of 0.8 can be calculated as:
Energy consumption (kWh) = 200 kW * 1 year = 200 kWh
Cost of energy consumption = 200 kWh * 5 paise/kWh = Rs. 10

Now, let's calculate the cost of energy consumption for a power factor of 0.866 lag:
Energy consumption (kWh) = 200 kW * 1 year = 200 kWh
Cost of energy consumption = 200 kWh * 5 paise/kWh = Rs. 10

3. Total cost:
The total cost is the sum of the cost of maximum demand and the cost of energy consumption.

For the power factor of 0.8:
Total cost = Cost of maximum demand + Cost of energy consumption
= Rs. 25,000 + Rs. 10
= Rs. 25,010

For the power factor of 0.866 lag:
Total cost = Cost of maximum demand + Cost of energy consumption
= Rs. 23,094 + Rs. 10
= Rs. 23,104

Comparing the total costs, we can see that the most economical power factor is 0.866 lag (Option D) with a total cost of Rs. 23,104.
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A factory takes a steady load of 200 kW at a lagging power factor of 0.8. The tariff is Rs. 100 per kVA of maximum demand per annum plus 5 paise per kWh. The phase advancing plant costs Rs. 500 per kVAR and the annual interest and depreciation together amount to 10%. The most economical p.f. isa)0.707 lagb)0.955 lagc)0.625 lagd)0.866 lagCorrect answer is option 'D'. Can you explain this answer?
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A factory takes a steady load of 200 kW at a lagging power factor of 0.8. The tariff is Rs. 100 per kVA of maximum demand per annum plus 5 paise per kWh. The phase advancing plant costs Rs. 500 per kVAR and the annual interest and depreciation together amount to 10%. The most economical p.f. isa)0.707 lagb)0.955 lagc)0.625 lagd)0.866 lagCorrect answer is option 'D'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A factory takes a steady load of 200 kW at a lagging power factor of 0.8. The tariff is Rs. 100 per kVA of maximum demand per annum plus 5 paise per kWh. The phase advancing plant costs Rs. 500 per kVAR and the annual interest and depreciation together amount to 10%. The most economical p.f. isa)0.707 lagb)0.955 lagc)0.625 lagd)0.866 lagCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A factory takes a steady load of 200 kW at a lagging power factor of 0.8. The tariff is Rs. 100 per kVA of maximum demand per annum plus 5 paise per kWh. The phase advancing plant costs Rs. 500 per kVAR and the annual interest and depreciation together amount to 10%. The most economical p.f. isa)0.707 lagb)0.955 lagc)0.625 lagd)0.866 lagCorrect answer is option 'D'. Can you explain this answer?.
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