Question

Household with 3 people has following load. Four 20-watt lights for 5 hours per night and one 50-watt TV for 3 hours per night. All of the loads are DC and the site has a peak sun hour of 5 for the lowest month of the year. Find the minimum rating of array to meet the household's demand based on the data provided in table 1. Table 1 PSH 25 50 100 Loads DC only Daily Load Energy That system can supply Size of Array (watts) 150 200 250 300 350 310 413 516 620 723 354 472 590 708 826 398 531 664 797 929 400 450 3.5 52 103 207 826 929 59 118 236 944 1062 4.5 66 133 266 1062 1195 1328 5 74 148 295 443 590 738 885 1033 1180

Answer 1

Total Load demand for the entire house can be calculated per night based on the data given in the question. From the questiion we have following data :

Total power consumption of bulbs=4*20=80Watts and they operate for 5 hours hence total energy per night consumed by light bulbs is 80*5=400Whr

Total energy consumption for a TV of 50W for 3 hours=1*50*3=150Wh

Therfore total energy consumption per day/night=550Wh.

Now since the site has peak sun hour of 5 in the lowest month which means that there are 5 hours per day which receives total radiatons of 1000 watts/hr.

Now to calculate the no of solar panel considering the 5 PSH and 550Wh as total power consumptions

Total Output from Solar panel required= 550/PSH= 550/5= 110Watts.

Hence choosing 3 no of solar panels with 50W connected in series to give an output of 150W which will adequate for supplying the entire load. 50W as mentioned on the solar panel is rated output which can not be achieved under any operating conditions.

Hence the rating of individual solar panel shall be 50W and qty 3 and connected to give an output of 150W.