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The load on an installation is 600 kW, 0.8 lagging p.f. which works for 3000 hours per annum. The tariff is Rs. 100 per KVA plus 20 Paise per kWh. If the power factor is improved to 0.9 lagging by means of loss of free capacitors costing Rs. 60 per kVAR, calculate the annual cost saving (in Rupees). Allow 10% per annum for interest and depreciation on capacitors.
Correct answer is between '7300,7500'. Can you explain this answer?
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The load on an installation is 600 kW, 0.8 lagging p.f. which works fo...
D = 600 kW
Cosϕ1 = 0.8
Cosϕ2 = 0.9
Loading KVAR taken by the capacitors.
= P (tanϕ1 – tanϕ2)
= 600 (tan (cos-1(0.8)) – tan (cos-1 (0.9)))
= 159.41 KVAR
Annual cost before p.f. correction
Maximum KVA demand = 600/0.8=750
KVA demand charges = 750 × 100 = 75,000
Units consumed per annum = 600 × 3000 = 1800000 kWh
Energy charges per year = 0.2 × 1800000 = 360000
Total annual cost = 75000 + 360000 = 435000
Annual cost after p.f. correction
Maximum KVA demand =600/0.9 = 666.67
KVA demand charges = 666.67 × 100 = 66, 667
Energy charges = 360000
Capital cost of capacitors = 60 × 159.41 = 9564.6
Annual interest and depreciation = 0.1 × 9564.6 = 956.46
Total annual cost = 66667 + 360000 + 956.46 = 427623. 46
Annual saving = 435000 – 427623.46 = 7376.54
Most Upvoted Answer
The load on an installation is 600 kW, 0.8 lagging p.f. which works fo...
Given:
Load = 600 kW
Power factor = 0.8 lagging
Working hours = 3000 hours per annum
Tariff = Rs. 100 per KVA + 20 Paise per kWh
Capacitor cost = Rs. 60 per kVAR
Interest and depreciation = 10% per annum
To find:
Annual cost saving after improving the power factor to 0.9 lagging by means of loss of free capacitors
Solution:
1. Calculation of KVA:
Apparent power (S) = Real power (P) / Power factor (PF)
S = 600 / 0.8 = 750 kVA
2. Calculation of active power:
Active power (P) = Real power (P) x Power factor (PF)
P = 600 x 0.8 = 480 kW
3. Calculation of reactive power:
Reactive power (Q) = √(Apparent power (S)^2 - Active power (P)^2)
Q = √(750^2 - 600^2) = 450 kVAR
4. Calculation of present billing:
KVA billing = 750 x Rs. 100 = Rs. 75,000
Energy billing = 600 x 3000 x 20 / 100 = Rs. 3,60,000
Total billing = Rs. 4,35,000
5. Calculation of improved power factor:
New apparent power (S) = Real power (P) / Improved power factor (PF)
S = 600 / 0.9 = 666.67 kVA
6. Calculation of new reactive power:
New reactive power (Q) = √(New apparent power (S)^2 - Active power (P)^2)
Q = √(666.67^2 - 600^2) = 240.83 kVAR
7. Calculation of required capacitors:
Capacitive kVAR required = Present reactive power (Q) - New reactive power (Q)
Capacitive kVAR required = 450 - 240.83 = 209.17 kVAR
8. Calculation of capacitor cost:
Capacitor cost = Capacitive kVAR required x Rs. 60 per kVAR
Capacitor cost = 209.17 x Rs. 60 = Rs. 12,550
9. Calculation of interest and depreciation on capacitors:
Interest and depreciation = Capacitor cost x 10% per annum
Interest and depreciation = Rs. 12,550 x 10 / 100 = Rs. 1,255
10. Calculation of new billing:
KVA billing = 750 x Rs. 100 = Rs. 75,000
Energy billing = 600 x 3000 x 20 / 100 = Rs. 3,60,000
Total billing = Rs. 4,35,000
11. Calculation of cost saving:
Cost saving = Present billing - New billing - Interest and depreciation on capacitors
Cost saving = Rs. 4,35,000 - Rs. 4,32,745 - Rs. 1,255 = Rs. 7,000
Therefore, the annual cost saving after improving the power factor to 0.9 lagging by means of loss of free capacitors is Rs. 7,000 (approximately).
Load = 600 kW
Power factor = 0.8 lagging
Working hours = 3000 hours per annum
Tariff = Rs. 100 per KVA + 20 Paise per kWh
Capacitor cost = Rs. 60 per kVAR
Interest and depreciation = 10% per annum
To find:
Annual cost saving after improving the power factor to 0.9 lagging by means of loss of free capacitors
Solution:
1. Calculation of KVA:
Apparent power (S) = Real power (P) / Power factor (PF)
S = 600 / 0.8 = 750 kVA
2. Calculation of active power:
Active power (P) = Real power (P) x Power factor (PF)
P = 600 x 0.8 = 480 kW
3. Calculation of reactive power:
Reactive power (Q) = √(Apparent power (S)^2 - Active power (P)^2)
Q = √(750^2 - 600^2) = 450 kVAR
4. Calculation of present billing:
KVA billing = 750 x Rs. 100 = Rs. 75,000
Energy billing = 600 x 3000 x 20 / 100 = Rs. 3,60,000
Total billing = Rs. 4,35,000
5. Calculation of improved power factor:
New apparent power (S) = Real power (P) / Improved power factor (PF)
S = 600 / 0.9 = 666.67 kVA
6. Calculation of new reactive power:
New reactive power (Q) = √(New apparent power (S)^2 - Active power (P)^2)
Q = √(666.67^2 - 600^2) = 240.83 kVAR
7. Calculation of required capacitors:
Capacitive kVAR required = Present reactive power (Q) - New reactive power (Q)
Capacitive kVAR required = 450 - 240.83 = 209.17 kVAR
8. Calculation of capacitor cost:
Capacitor cost = Capacitive kVAR required x Rs. 60 per kVAR
Capacitor cost = 209.17 x Rs. 60 = Rs. 12,550
9. Calculation of interest and depreciation on capacitors:
Interest and depreciation = Capacitor cost x 10% per annum
Interest and depreciation = Rs. 12,550 x 10 / 100 = Rs. 1,255
10. Calculation of new billing:
KVA billing = 750 x Rs. 100 = Rs. 75,000
Energy billing = 600 x 3000 x 20 / 100 = Rs. 3,60,000
Total billing = Rs. 4,35,000
11. Calculation of cost saving:
Cost saving = Present billing - New billing - Interest and depreciation on capacitors
Cost saving = Rs. 4,35,000 - Rs. 4,32,745 - Rs. 1,255 = Rs. 7,000
Therefore, the annual cost saving after improving the power factor to 0.9 lagging by means of loss of free capacitors is Rs. 7,000 (approximately).
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The load on an installation is 600 kW, 0.8 lagging p.f. which works for 3000 hours per annum. The tariff is Rs. 100 per KVA plus 20 Paise per kWh. If the power factor is improved to 0.9 lagging by means of loss of free capacitors costing Rs. 60 per kVAR, calculate the annual cost saving (in Rupees). Allow 10% per annum for interest and depreciation on capacitors.Correct answer is between '7300,7500'. Can you explain this answer?
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The load on an installation is 600 kW, 0.8 lagging p.f. which works for 3000 hours per annum. The tariff is Rs. 100 per KVA plus 20 Paise per kWh. If the power factor is improved to 0.9 lagging by means of loss of free capacitors costing Rs. 60 per kVAR, calculate the annual cost saving (in Rupees). Allow 10% per annum for interest and depreciation on capacitors.Correct answer is between '7300,7500'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The load on an installation is 600 kW, 0.8 lagging p.f. which works for 3000 hours per annum. The tariff is Rs. 100 per KVA plus 20 Paise per kWh. If the power factor is improved to 0.9 lagging by means of loss of free capacitors costing Rs. 60 per kVAR, calculate the annual cost saving (in Rupees). Allow 10% per annum for interest and depreciation on capacitors.Correct answer is between '7300,7500'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The load on an installation is 600 kW, 0.8 lagging p.f. which works for 3000 hours per annum. The tariff is Rs. 100 per KVA plus 20 Paise per kWh. If the power factor is improved to 0.9 lagging by means of loss of free capacitors costing Rs. 60 per kVAR, calculate the annual cost saving (in Rupees). Allow 10% per annum for interest and depreciation on capacitors.Correct answer is between '7300,7500'. Can you explain this answer?.
The load on an installation is 600 kW, 0.8 lagging p.f. which works for 3000 hours per annum. The tariff is Rs. 100 per KVA plus 20 Paise per kWh. If the power factor is improved to 0.9 lagging by means of loss of free capacitors costing Rs. 60 per kVAR, calculate the annual cost saving (in Rupees). Allow 10% per annum for interest and depreciation on capacitors.Correct answer is between '7300,7500'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The load on an installation is 600 kW, 0.8 lagging p.f. which works for 3000 hours per annum. The tariff is Rs. 100 per KVA plus 20 Paise per kWh. If the power factor is improved to 0.9 lagging by means of loss of free capacitors costing Rs. 60 per kVAR, calculate the annual cost saving (in Rupees). Allow 10% per annum for interest and depreciation on capacitors.Correct answer is between '7300,7500'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The load on an installation is 600 kW, 0.8 lagging p.f. which works for 3000 hours per annum. The tariff is Rs. 100 per KVA plus 20 Paise per kWh. If the power factor is improved to 0.9 lagging by means of loss of free capacitors costing Rs. 60 per kVAR, calculate the annual cost saving (in Rupees). Allow 10% per annum for interest and depreciation on capacitors.Correct answer is between '7300,7500'. Can you explain this answer?.
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