Question

Water Pumping Power Electrical Architecture

2. Water pumping power architecture The diagram of the figure shows the architecture of a water pumping system that can be powered either from a Photovoltaic (solar panel, from the grid, or from an energy storage element (battery). Assume all efficiencies are constant regardless the power flow direction. n = 85% P/ (max)= lin (max)= 12 = 90% P2 (max)= lina (max)= out/ Vin=110V (2) 200V ph-rated =110V Trated =0.5 Nm Orated=377 rad/sec DO Ipk Inc. Tous DC Ppv (max)=305W V Pmax=36.3V IPmax A | DCD 12-15 V 55 Ah 5.5A max Loch 13 = 95% 14 = 95% Pout2 (max)= Tout2 (max)= pamu? PA(max)= Ime (max)=

1) Scenario 1: Assume the pump is working at maximum power, the batteries are being charged at maximum power, and there is no PV generation. What is the power consumed from the grid?
2) Scenario 2: The conditions from Scenario A are maintained, but now the PV is outputting maximum power. What is the net power consumed from the grid in this case?
3) Scenario 3: at sunset the batteries were fully charged (15V) and the pump continues running on grid power. There is a failure during the night, the grid becomes unavailable, and the pump continues to run on battery power. Assuming the battery voltage varies linearly as the battery discharges, how long can the pump be operated at rated conditions until the battery is depleted and reaches 12V?
4) Complete all maximum ratings indicated in the figure assuming the worst-case scenario for each converter.

Answer 1

given system Iout! 700 Vin - 11 OV At 2=881.. n₂=90% Ven-rated = 110 Irated=0.5 wyate -377 Yadda 20SW (Max) to A toc IP (max) TISU SSA S.SA Mar Na=g5t. nu=g5t. As from the the capacity of . motor operated , maximum rated power of motor is given by Pmerated) : Trated Wrated : 0.5 X 377 188.5 Watt So, maximum current Ipn = Pmcrated) X Ven X0.90 drawn by motor - Assume prof 1 mator 0.99 In 188.5 3xloxo.g 1986.5 { 1,9. Ame? 40.63317 / 1. 9 Amp 0.633 A

i Now efficiency - 902 of converter (2) - Input power to converter Pi (max) = pm Crated). 0.g 189.5 og (max): 209 4 W is Voltage 200 v across so, converte a maximum ih put current Iinel max ) = R(max) = 200 209.4 200 (merou7 Amp so, . so converter / has input Vol power -P2 (mux) 209.4 h = 085% to converter 0.85 of voltage 110 Pit = 0.15 (Assume, with input 1 2364) the 2u6. 3 2 lin(max) = 110x0.95

Now PV hey Yating Pou (max)305 W Vp (max) = 36. 30 so, maximum current to PV Por cmax) = Volmax) Ipl max] Ipemer) - 305 -/9.047 36.3 o converter 3 has h = o.g5 so, 04+PUL power of converter 3 Pout 2 (max) = Peu imaxe.eus 305 X0.95 Pout 2 : 289.75 output of converter 3 7 input of converter 4 connected to input of converter 2, which has Voltage of 2000 , means that the DC link between 3 P u also has Voltage of 2000 Voat2 = 200V 312.75 200 Town (max) 5 Poutz (max), - Youth (max) = 1.61 AMP

Now has - hatter s batter has max maximum voltage is Current is 5.5 A. T is To maximum power to stored by battery is Pbattery = ISX 5.5 Poatter : 82.5 watt converter 4 has etticiency of 95%, so so Pu (max) = Pbattery 87.5. 0.95 nu - Pual max) = 86.847 Iin (u) max = Pul max, 86.84 200 200 0.434 Amp upto Here we can Here atings of we the calculated maximum converter.

scenario 1: Assume Pump working at maximum power battery Charged at maximum Power there is no PV. Power consumed from grid given so, by at maximum power required If pump working Power than the from grid Picmur) = 246.3 7 Calculated before At at power same time battery maximum power. to convertu also Charged So input u . is Pu ( max) : 86.84 watt Now 95%. converter So, 1 hos Power efficiency & required is Picmax) 5 06-84 10216 w/ 0.85 I so grid P power taken from = P, (max) tp cmax) s 246.3 + 702.16 Ps 3u8.00 watt

I scenario 2 : PV power is Now in outputing which is this case the muximum 35 289.75 watt Pout = . at And our converter requirement ③ D (W) for is Peodd : Pacmax) + Pu (max) 209.4 t 86.8u Pload = 296. 24 Watt Yequired output power from grid = 296.24 -289.75 - = 6.09 watt efficiency of converter 1 is 85). so, Power Fed by grid - 8.09 0.85 - 77.63 watt 7. 63 watt

scenario :3 it hey + battery is fully charged & capacity of " 55 Ah . So, at 55 Ah s 15u I can supply the power It in لاط| کر = 55 x 15 1825 here I watt hy - Now If is down cet 55 Ah available the batter to Ru capacity power. voltage that It has 5 12x55 sooo who * So to during k... power of down i t voltage can from supply 16v

? 825-680 925-660 - 165 watt 825 her watt ke so to supply it take hr 825 watt in 165 watt in Ihr 1x165 825 - 0.2 hy - 0.2 X 60 minute ninute
Note: If you find any problem in solution ask in comment box below thank you?? ?