For a5A-230Venergymeter,the no. of revolutions per kwhis 480. If in a test at full load, unity power factor, the disc makes 6 revolution in 30 seconds. Calculate the error if any.
Answers
The percentage error is 32%.
Explanation:
Correct statement:
5A - 230V energy meter,the no. of revolutions per kwhis 4800. If in a test at full load, unity power factor, the disc makes 60 revolution in 30 seconds. Calculate the error if any.
Given data:
- Current = 5 A
- Energy = 230 V
- Number of revolutions per KW = 4800
- Number of revolutions per second = 60
- Percentage error = ?
Solution:
The total energy consumed = VIt (cos Φ) Wh
The total energy consumed = 230 x 5 x (60 / 3600) (1.0)
The total energy consumed = 1150 x 0.016 x 1
The total energy consumed = 18.4 Wh
The indicated reading IR (energy) = Revolutions made / KE in rev / KWh
IR (energy) = (60 / 4800 )10^3
IR (energy) = 0.0125 x 1000
IR (energy) = 12.5 Wh
Thus energy error = TR (energy) - IR (energy)
Energy error = 18.4 - 12.5
Energy error = 5.9
Percentage error = (-106.6 / 18.4) x 100
Percentage error = 0.320 x 100 = 32%
Thus the percentage error is 32%.