This document contains 16 questions and answers related to power systems. Some example questions include calculating diversity factors, load duration curves, power factor correction, and comparing tariff structures. The answers provide numerical calculations and formulas used to solve problems related to electrical power distribution, generation, and billing.

1. CHAPTER TWO QUESTIONS WITH ANSWERS
Q1. A substation supplies power by four feeders to its consumers. Feeder no 1 supplies six
consumers whose individual demands are 70 kW, 90 kW, 20 kW, 50 kW, 10 kW and 20 kW, while
maximum demand on the feeder is 200kW. Feeder no 2 supplies four consumers whose daily
maximum demands are 60 kW, 40 kW, 70 kW and 30 kW while the maximum demand on feeder no
2 is 160 kW. Feeders no 3 and 4 have daily maximum demands of 150 kW and 200 kW respectively,
while the maximum demand on the station is 600 kW.
Determine the diversity factors for the consumers of feeder no 1 and feeder no 2 and for the four
feeders.
Ans: 1.3; 1.25 and 1.183
Q2. A generating station has the following daily loads:
0 to 6 hr 4000 kW
6 to 8 hr 3000kW
8 to 12 hr 8000kW
12 to 14 hr 2000kW
14 to 18 hr 7500kW
18 to 20 hr 3000kW
20 to 24 hr 4500kW
Sketch load duration curve and determine the load factor and plant capacity factor assuming
capacity of the plant is 11000 kW.
Ans: 0.625, 0.4545
Q3. An industrial consumer having a maximum demand of 100 kW, maintains a load factor 60%. The
tariff rates are Rs. 1500 per KVA of maximum demand per annum plus paise 30 per kWh of energy
consumed. If the average pf is 0.8 lagging, calculate the total energy consumed per annum and
annual bill.
Q4. Compute the cost of electrical energy and average unit cost for consuming 450 kWh under block
rate quoted as below:
First 50 kWh at Rs. 3.80 per kWh
Next 50 kWh at Rs. 3.20 per kWh
Next 50 kWh at Rs. 2.80 per kWh

2. Next 50 kWh at Rs. 2.40 per kWh
Excess over 200 kWh at Rs. 2.10 per kWh.
Q5. A synchronous motor having a power consumption of 50 kW is connected in parallel with a load
of 200 kW having pf 0.8 lagging. The excitation of motor is adjusted until the combined pf becomes
0.9 lagging. Calculate the kVA input to the motor.
Ans: 57.76 kVA
Q6. It is desired to correct the power factor 0.95 by means of static condensers connected across
each phase of a 3-phase, 400 V, 50 Hz motor installation having a maximum load of 50 kVA at pf of
0.75 lagging. What must be the capacity of each delta – connected capacitor?
Ans: 244 uF
Q7.A factory having a load of 5 MW at 0.8 lagging for 2000 hours per annum buys energy on tariff of
Rs 500 per kVA plus Rs 2.5 per kWh consumed. If the power factor is improved to 0.95 lagging by
means of capacitors costing Rs 1350 per kVA and having a power loss of 200 watts per kVA, calculate
the annual charge for electrical energy (a) before and (b) after the installation of capacitors and (c)
the saving effected by their use. Allow 8% per annum for interest and depreciation on the
capacitors.
Q8. A consumer has an annual consumption of 70080 kWh. The charge is Rs 100 per kW of
maximum demand plus 5 paise per kWh.
(a) Find the annual bill and the overall costs per kWh is the load factor is 40%.
(b) What is the overall cost per kWh if (i) consumption was reduced by 25% with the same load
factor (ii) consumption remains the same as in (a) but the load factor is reduced to 30%?
Ans. (a) Rs 5504; 7.85p (b) (i) 7.85 p (ii) 8.80 p
Q9. A consumer takes a steady load of 1000 kW at a power factor of 0.707 lagging and pays Rs 80
per annum per kVA of maximum demand. Phase advancing plant costs Rs 128 per kVAr. Determine
the capacity of the phase advancing plant required for minimum overall annual expenditure. Interest
and depreciation total 10%. What will be the value of the new power factor of the supply? Deduce
and formula used.
Ans. 838 kVAr; 0.987 (lagging)
Q10 Two 50hp (metric) induction motors are each supplied on a tow part tariff at Rs 80 per kVA
of maximum demand plus 5 paise per kWh. The service needed is 3000 hours per annum at full load.
Motor A has efficiency 0.878 and p.f. 0.83 plus a capacitor costing Rs 75 per kVA to raise the p.f. to
0.91. Motor B with an efficiency 0.888 and p.f. 0.91 costs 1000 more than motor A. Assuming
interest and depreciation to be 12.5 % per annum for each motor and 8% for the capacitor, compare
the annual charges for two arrangements.
Ans. Net annual saving with motor B is Rs 36.80.

3. Q11 A factory takes a load of 200 kW at 0.85 p.f. lagging for 2500 hours per annum and buys
energy on a tariff of Rs 100 per kVA plus 5 paise per kWh consumed. If the power factor is improved
to 0.90 lagging by means of capacitors costing Rs 350 per kVA and having a power loss of 100 watts
per kVA calculate the annual charge for electrical energy (a) before and (b) after the installation of
capacitors and (c) the saving effected by their use. Allow 8% per annum for interest and depreciation
on the capacitors.
Ans. (a) Rs 48530 (b) Rs 48270 (c) Rs 260
Q12. A generating station supplies the following loads:
15 MW, 12 MW, 8.5 MW, 6 MW and 0.45 MW. The annual load factor of the power station is 48%.
Calculate (i) the number of units supplied annually (ii) diversity factor and (iii) demand factor, if the
station has a maximum demand of 22 MW.
Ans. (i) 92.5 X 10e6 kWh (ii) 1.907 (iii) 52.5%
Q13. A power station has a maximum demand of 200 MW, a plant load factor of 0.6, a plant
capacity factor of 0.5 and a plant use factor of 0.8. Find (i) the daily energy consumed (ii) the reserve
capacity of the plant and (iii) maximum energy that can be produced daily.
Ans. (i) 2880 MWh (ii) 40 MW (iii) 5706 MWh
Q14. The maximum demand of a power supply system is 30 MW. The annual factors as follows:
Load factor = 60%; Capacity factor = 50%; Use factor = 75%.
Determine (i) the annual energy production (ii) reserve capacity and (iii) the hours during which the
plant is not in service per year.
Ans. (i) 157.68 X 10e6 kWh (ii) 6 MW (iii) 2920 hours.
Q15. The maximum demand of a consumer is 1 MW and the units consumed per annum are
500000 kWh. Calculate the reduction in cost if the power factor is raised from 0.5 to 0.8. The tariff is
Rs 900 per annum per kW demand plus Rs 1.50 per kWh plus Re 0.30 per kVAh reactive. Calculate
the flat rate tariff for a unity power factor load.
Ans. Rs 147310.00; Rs 3.30 per kWh
Q16. A consumer, whose connected load is 400 kVA, consumes 1 X 10e6 kWh per annum and the
maximum demand registered is 250 kVA. He is offered the following two alternative tariffs:
Tariff A: Rs 1000 per annum per kVA of connected load + first 150 kWh per KVA of connected load at
Rs 3.0/kWh, next 50000 kWh at Rs 2.00 per kWh and the rest at Re 1.0/kWh.
Tariff B: Rs 1200 per annum per kVA of maximum demand + first 150 kWh per kVA of maximum
demand at Rs 3.0 per kWh; next 50000 kWh at Rs 2.0/kWh and the rest at Re 1.0/kWh.
Which tariff is economical and what is the annual gain?
Ans. Tariff B is cheaper; Annual gain = Rs 14500000