116 Principles of Power System
Energy consumed/year = 288·4 × 8760 = 25,26384 kWh
Annual energy charges = Rs 0·06 × 25,26,384 = Rs 1,51,583
Total annual bill = Rs (37,440 + 1,51,583) = Rs 1,89,023
(b) When synchronous motor runs at p.f. 0·8 leading. As the synchronous motor runs at
leading p.f. of 0·8 (instead of 0·8 p.f. lagging), therefore, it takes now 59·25 leading kVAR. The
lagging kVAR taken by induction motors are the same as before i.e., 179·4.
∴ Net lagging kVAR = 179·4 − 59·25 = 120·15
Total active power = Same as before i.e., 288·4 kW
∴ Total kVA =
12015 288 4
2
.
afaf
+⋅
= 312·4
Annual kVA demand charges = Rs 100 × 312·4 = Rs 31,240
Annual energy charges = Same as before i.e., Rs 1,51,583
Total annual bill = Rs (31,240 + 1,51,583) = Rs 1,82,823
∴ Annual saving = Rs (1,89,023 − 1,82,823) = Rs 6200
TUTORIAL PROBLEMSTUTORIAL PROBLEMS
TUTORIAL PROBLEMSTUTORIAL PROBLEMS
TUTORIAL PROBLEMS
1. What should be the kVA rating of a capacitor which would raise the power factor of load of 100 kW from
0·5 lagging to 0·9 lagging ? [125 kVA]
2. A 3-phase, 50 Hz, 3300 V star connected induction motor develops 250 H.P. (186·5 kW), the power
factor being 0·707 lagging and the efficiency 0·86. Three capacitors in delta are connected across the
supply terminals and power factor raised to 0·9 lagging. Calculate :
(i) the kVAR rating of the capacitor bank.
(ii) the capacitance of each unit. [(i) 111·8 kVAR (ii) 10·9
µµ
µµ
µF]
3. A 3-phase, 50 Hz, 3000 V motor develops 600 H.P. (447·6 kW), the power factor being 0·75 lagging and
the efficiency 0·93. A bank of capacitors is connected in delta across the supply terminals and power
factor raised to 0·95 lagging. Each of the capacitance units is built of five similar 600-V capacitors.
Determine the capacitance of each capacitor. [156
µµ
µµ
µF]
4. A factory takes a load of 800 kW at 0·8 p.f. (lagging) for 3000 hours per annum and buys energy on tariff
of Rs 100 per kVA plus 10 paise per kWh. If the power factor is improved to 0·9 lagging by means of
capacitors costing Rs 60 per kVAR and having a power loss of 100 W per kVA, calculate the annual
saving effected by their use. Allow 10% per annum for interest and depreciation on the capacitors.
[Rs 3972]
5. A station supplies 250 kVA at a lagging power factor of 0·8. A synchronous motor is connected in
parallel with the load. If the combined load is 250 kW with a lagging p.f. of 0.9, determine :
(i) the leading kVAR taken by the motor.
(ii) kVA rating of the motor.
(iii) p.f. at which the motor operates. [(i) 28·9 kVAR (ii) 57·75 kVA (iii) 0·866 lead]
6. A generating station supplies power to the following :
(i) a lighting load of 100 kW;
(ii) an induction motor 800 h.p. (596·8 kW) p.f. 0·8 lagging, efficiency 92%;
(iii) a rotary converter giving 150 A at 400 V at an efficiency of 0·95.
What must be the power factor of the rotary convertor in order that power factor of the supply station may
become unity ? [0·128 leading]
7. A 3-phase, 400 V synchronous motor having a power consumption of 50 kW is connected in parallel
with an induction motor which takes 200 kW at a power factor of 0·8 lagging.
(i) Calculate the current drawn from the mains when the power factor of the synchronous motor is
unity.