•• VRV Systems • Network Solution4010 Explanations of Power Proportional Distribution10 - 2 Count Accuracy10 - 2 - 1Cause of errorSystem example + ≠ Count conclusive total for indoor unit #1~#8 →Refer to the next page ≠ Count conclusive total for indoor unit #1~#5≠ Count conclusive total for indoor unit #6~#8 →Refer to the next page ≠ ≠ Count conclusive total for indoor unit #1~#8*: The reason to get and the error size.• REASON 1iPU 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, theerror by the leaving-off occurs by 0.5W/ hour in average value of all indoor units.Calculation example(1) Count for errors in 8-dayTenant A + B : 0.5 (Wh) × 24 hours × 8 days × 5 units = + 0.480 kWhTenant C : 0.5 (Wh) × 24 hours × 8 days × 3 units = + 0.288 kWhtotal = + 0.768 kWh(2) Assuming that the reads on watthour meters are as follows:W1: read on watthour meter = 490 kWhW2: read on watthour meter = 200 kWhtotal = 690 kWh(3) Finally it is concluded as total error = 0.768/690 × 100 = 0.11%Watthour meter(with oscillator)Power supply3 phase#1 #2 #3 #4 #5 #6#: Indoor unit's address#7 #8W2W1Outdoor unitTenant A Tenant B Tenant CLegend : Read on wattmeterPower supplysingle phaseiPUTo other iPUPCHUBW1 W2W1W2W1 W2