Question

ILO 3 - 10 Marks PV string based on daily household energy consumption. [10 Marks) Part A. 16 Marks The standard derating (correction) factor estimating a household energy consumption is about 30 percent to account for the power losses. A certain household has the following items as seen in Table I. Table 1. Household items Quantity Power (Watt) Time day (Hours) Light bulbs TV Fans Fridge 1. Calculate the total power consumption of the appliances 12.marks! 2. Compute the total energy demand from the PV panels marks 3. Suppose the total energy demand, including losses, from a certain type of PV system, is 8.540 Wh The suggested sun-hour for this particular location is 6.6 hours. Also, the system is powered by 24VDC, 230Wp solar panels. Caculate the total watt-peak (Wp) capacity of the PV system in an hour 12 marks Part B. 14 Marks/ 1. Suppose the total watt peak (Wp) capacity of the PV system in an hour is 1.100 Wp. What is the total number of PV panels needed for the system? Given: The system is powered by 24 VDC, 230 Wp solar panels [2 marks 2. Consider the same household items and energy variables as given in Table 1. Also given are: battery efficiency is 85%, 2-days of autonomy, a Dob of 0.6 and 24V nominal battery voltage. Compute the required battery bank capacity in Ampere-hour (Ah) 12 marks Instructions: Answer all the questions in Parts A and B then write your steps and real answer in the box given below:

Answer 1

Answes Here given, Tirie Iday fowe Chours) Quantity power Ccoatt ) Conoumplia It ero 6x3ox5- 900 30 hight bulbs /x/50 5 = 750 150 TV 1 3 x 20 004 = 240 fans 200 IosLox 12= 6000 Fuidge 500 10050 WHI/.day Total 2) Golal eneupy demand from A panel conoléeainp 30%0 dos. lolse Power regdued = Tolas Powej vconsumoptiso (cday + host Powen, To bat powes aued loos0 t Clooso o 80)/00) lo lou Power leajured - 100so + 30.15 = 18065 WAHI|.coy 3) Watl peat capacily coj sydeno ño Ha. hin = 6540 Jun hous = 6.6hhours . 2 6.6 x365 = 2409 haluday - 2409 Whlugear. watt peak = 8540/2409 > 3.5 Wp,