User manual

User Manual
man_905K_1.10.doc Page 21
No. of “on” update messages per year = 2 (twice per year) * 16 (4 hours @ 15 min update)
= 32
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
Analog 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 analog input = 730
Total input messages per year = 400 + 375 + 730 = 1505
Power consumed in transmissions = 0.005 * 1505 * 2 (2 transmissions per message)
= 15 mAHr per year
Power for analog loop supply (assume average loop current is 12mA)
No. of analog measurements per year = 365 days * 24 hours * 1(sample time)
= 8,760
Power for analog loop supply = 0.012 (from above table) * 5 (warm-up time) * 8,760
= 526 mAHr per year
Power for pulse input
Average pulse rate is 1 pulse per hour (0.0003Hz), so power required = 0.06 x 0.0003 per day
= zero
Quiescent power
Power for quiescent current = 3.4 per day * 365
= 1241 mAHr per year
Total power consumption per year = 15 + 526 + 0 + 1241