User's Manual

905U-K.MAN Rev. 1.1 January 2002
 ELPRO Technologies Pty Ltd Page 17
Total messages for digital input = 4 + 364 + 32 = 400
Pulse input Update time 1 day Sensitivity 50
Average pulse rate is 1 pulse per hour, with peak rate of 20 per hour, for 10 hours, three times per year.
No. of change messages (normal rate) = 0 (time for 50 pulses is more than the update time)
No. of change messages (peak rate)= 3 (three per year) * 200 (20 per hr for 10 hrs) / 50
= 12
No. of update messages per year = 363 (approx)
Total messages for pulse input = 12 + 363 = 375
Analogue input Sample time 1 hour Warm-up time 5 secs
Sensitivity 3% Update time 1 day
Average changes of >3% is twice per day
No. of change messages per year = 2 (twice per day) * 365
= 730
No. of update messages per year = 0 (always be a change message each 1 day)
Total messages for analogue input = 730
Total input messages per year = 400 + 375 + 730 = 1505
Power consumed in transmissions = 350 * 1505 * 2 (2 per message)
= 1,053,500 µA-hour per year
Power for analogue loop supply (assume average loop current is 12mA)
No. of analogue measurements per year = 365 days * 24 hours * 1(sample time)
= 8,760
Power for analogue loop supply = 17 (from above table) * 5 (warm-up time) * 8,760
= 744 600 µA-hour per year
Power for pulse input
Average pulse rate is 1 pulse per hour (0.0003Hz), so power required = 60 x 0.0003 per day
= 7 µA-hour per year
Quiescent power
Power for quiescent current = 3600 per day * 365
= 1 314 000 µA-hour per year
Total power consumption per year= 1053500 + 744 600 + 7 + 1 314 000
= 3,112,107 µA-hour