Family Cartman has a small house in the country side which is not connected to the grid. The place enjoys 4 equivalent sun hours. Therefore, Mr Cartman has decided to install an off-grid PV system in the house to supply the required electicity in the house.
The house electricity needs are summarized in the table below.
Load Quantity Power per item (W) Time of use (h) Type
Light bulb 10 20 2 DC
TV 1 80 2 AC
DVD 1 40 2 AC
Fridge 1 100 4 AC
a) When all loads are connected, what is the total DC power load in W?
b) When all loads are connected, what is the total AC power load in W?
c) Assuming that the overall efficiency for the cables, batteries and charge controller is 85% and the efficiency of the inverter is 95%, how much energy in Watt-hours will the PV panels have to produce to cover the electricity demand?
4. STANDALONE PV SYSTEM II (1 point possible)
The available panels that Mr. Cartman is looking at have the following specifications:
Power output = 120Wp
Vmpp = 20V
Impp = 6A
Voc = 22V
Isc = 7A
d) Assume the panels work under MPP conditions. How many panels will be needed to produce that energy?
4. STANDALONE PV SYSTEM III (2 points possible)
Since the system is disconneced form the grid, the family Cartman will also need some energy storage. Mr. Cartman has specified that 3 days of autonomy will be sufficient. In this case, batteries with the following specifications will be used:
Battery capacity = 105Ah
Battery voltage = 12V
Maximum allowed depth of discharge = 60%
e) The system is designed to use 24V as operating voltage. What will be the minimum battery capacity needed in Ah?
f) How many of the batteries specified will be needed?
ans of c part is 1287.92
a:200
b:220
c:??
d:3
e:??
f:??
e) 260.10
f) 6
please answer the e and f ????
thank u
Thanks a lot...
Thank u so much u helped me alot
what is the reference of this exercise?? its from a book?
thanks all
To answer these questions, we need to calculate the power requirements and energy demand for the loads in the house and then determine the number of PV panels and batteries required.
a) To calculate the total DC power load, we need to multiply the quantity of each load by its power and sum them up.
Light bulb power = 10 * 20 W = 200 W (DC)
TV power = 1 * 80 W = 80 W (AC)
DVD power = 1 * 40 W = 40 W (AC)
Fridge power = 1 * 100 W = 100 W (AC)
Total DC power load = 200 W
b) To calculate the total AC power load, we need to consider the efficiency of the inverter.
Total AC power load = (80 W + 40 W + 100 W) / 0.95 = 227.37 W (approx.)
c) To calculate the energy demand, we need to multiply the power of each load by its time of use and sum them up. We also need to consider the overall efficiency.
Total energy demand = (200 W * 2 h + 227.37 W * 2 h) / (0.85 * 0.95) = 538.06 Wh (approx.)
d) To determine the number of panels needed, we need to divide the energy demand by the power output of each panel. Since we want to work under MPP conditions, we'll consider the maximum power point (MPP) at Vmpp and Impp.
Number of panels needed = 538.06 Wh / (120 Wp * 20 V / 20 V * 6 A / 7 A) = 1.5 (approx.)
Since we cannot have fractional panels, we round it up to 2 panels.
e) To calculate the minimum battery capacity needed, we need to consider the energy demand, system voltage, and depth of discharge (DOD).
Minimum battery capacity = Energy demand / (System voltage * DOD) = 538.06 Wh / (24 V * 0.6) = 37.42 Ah (approx.)
f) To determine the number of batteries needed, we divide the minimum battery capacity by the capacity of each battery.
Number of batteries needed = 37.42 Ah / 105 Ah = 0.36 (approx.)
Since we cannot have fractional batteries, we round it up to 1 battery.
Therefore, the answers are:
a) The total DC power load is 200 W.
b) The total AC power load is 227.37 W (approx.).
c) The PV panels need to produce 538.06 Wh of energy.
d) The number of panels needed is 2.
e) The minimum battery capacity needed is 37.42 Ah (approx.).
f) The number of batteries needed is 1.