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• VRV Systems • Network Solution
40
10 Explanations of Power Proportional Distribution
10 - 2 Count Accuracy
10 - 2 - 1Cause of error
System example
<Case of arising error>
쪧 + ≠ Count conclusive total for indoor unit #1~#8 → Refer to the next page
쪨 ≠ Count conclusive total for indoor unit #1~#5
≠ Count conclusive total for indoor unit #6~#8 → Refer to the next page
쪧 ≠ ≠ Count conclusive total for indoor unit #1~#8*: The reason to get and the error size.
• REASON 1
iPU counts every one hour’s power consumption.
Though fraction in case of computation occurs at this time, it is computed after leaving off a 1-W figure to avoid the risk for the owners. As a result, the
error by the leaving-off occurs by 0.5W/ hour in average value of all indoor units.
Calculation example
(1) Count for errors in 8-day
Tenant A + B : 0.5 (Wh) × 24 hours × 8 days × 5 units = + 0.480 kWh
Tenant C : 0.5 (Wh) × 24 hours × 8 days × 3 units = + 0.288 kWh
total = + 0.768 kWh
(2) Assuming that the reads on watthour meters are as follows:
W1: read on watthour meter = 490 kWh
W2: read on watthour meter = 200 kWh
total = 690 kWh
(3) Finally it is concluded as total error = 0.768/690 × 100 = 0.11%
Watthour meter
(with oscillator)
Power supply
3 phase
#1 #2 #3 #4 #5 #6
#: Indoor unit's address
#7 #8
W2
W1
Outdoor unit
Tenant A Tenant B Tenant C
Legend
: Read on wattmeter
Power supply
single phase
iPU
To other iPU
PC
HUB
W1 W2
W1
W2
W1 W2
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