TO MEASURE THE ELECTRICAL ENERGY IN A GIVEN A.C. CIRCUITS ANDCALCULATE POWER FACTOR”

Equipments used:- Energy Meter (meter constant =1200rev/kwh) ,voltmeter, ammeter, lamp-board (one 100W and three 200 W bulbs), inductor , capacitor and variac.

Meter calibration check with resistive load:-

ECA=V.I. time taken in one revolution (in hours)

KM= 1/ECA ;

1. Calculated meter constant is not exactly same in all the cases and is different from 1200as specified in Meter .This difference because of heating of Aluminum disc and othereffects like friction etc.

2. Energy consumption as recorded by the meter is different from ECA . The meter wouldread incorrectly at low voltages when eddy currents will be weak.

Energy consumption with inductive load :-

Load:- 3x200 W and 1x 100W , Volt=230 volt

I(A) = 0.68 Amp , Time for 5 rev. = 750 sec.

No. of revolution per hour = (5/750)x3600

=24

VS = 6.7 Volt , VL = 224.7 volt

Apparent power (V*I) = 156.4 VA

tan β= VL/Vs = 33.54 ;

Serialno.

V Load I(current) (inamp.)

Timefor 5rev.(insec.)

Timeperrev.(insec)

Power=V.I

EnergyECA

Meterconstant(KM)

1. 230 1x200 W 0.8 82 16.4 184 0.00083 1193

2. 230 2x200 W 1.58 41 8.2 363.4 0.000827

1208

3. 115 1x200 W 0.54 275 55 62.1 0.000948

1054

4. 115 1x200 W1x100 W

0.84 175 35 96.6 0.000939

1064

5. 115 3x200 W1x100 W

1.96 73 14.6 225.4 0.000914

1094

β= 88.29 Degree

Load power factor ( cos β) =0.029

V = (VS2 + Vl

2)1/2

V = (6.72 +224.72)1/2 = 224.8 Volt

Energy consumed in KWH = (no of rev per hour/meter constant)

= (24 / 1208)

= 0.0198 KWh

Improvement of power factor :--

Load = 3 x 200 W and 1x 100W , Volt=230 volt

I (A) = 0.6 Amp , Time for 5 rev. = 752 sec.

No. of revolution per hour = (5/752) x 3600

= 23.9

Vs = 6.7 Volt , VL= 224.7h Volt

Energy consumed in KWH = (no of rev per hour/meter constant)

= 23.9/1208

= 0.0198 KWh

Load power factor = (Real power)/ (apparent power)

= (230 x 0.68 x 0.029)/ (230 x 0.6)

=0.032

As we can see energy consumption in both the cases is same because meter shows realpower consumed, and real power is the power consumed by the resistance only andpotential difference across the resistance is same in both cases.

In second case we have attached a capacitor in circuit so reactive power will decreasebecause reactive power of capacitor is negative. Therefore apparent power will

decrease( as we can see from above result) since it is sum of the real and reactive powerand real power is fixed so power factor will increase.

Precautions:-

1. Any connection shouldn’t be loose.2. The circuit should be checked properly before switching on the supply.