I have made some progress since last night. basically worked out the battery and the PV module. would appreciate it if you take a look.
Battery: Need to choose what device needed to be charged first in order to choose the appropriate battery size and PV panel respectively. I assume, load (the device needed to charged is a phone) is 3500 mAh with 3.7v. PV panel: To calculate the energy content of the battery:
E=V*I*T
3.7*3500mAh=12.95Wh ~ 13Wh
Considering the Li-ion battery loss which is close to 20%:
E= 13Wh/0.8=16.25 Wh~ 17Wh
The Lithium ion battery is the most suitable option here as it has a higher rated power, more compact and also Peukart effect is negligible in this type of batteries.
Candide component: Lithium Ion Battery pack 4400 mAh 2 x 2200 mAh 3.7V akku
(**broken link removed**)
PV module: There are three main factors to consider in order to choose a solar panel:
How much energy will your appliance(s) use over a period?
How much energy can your battery store?
How much energy can a Solar panel generate over a period?
The amount of energy the load uses over a period has been already addressed in previous section. Consequently, a battery which can facilitate the same amount of energy and current has been chosen. Therefore, the PV module has to produce at least 17Wh power.
PV module sizing: The fact that different PV module sizes generate different amount of power, must be taken into consideration. The peak Watt produced (Wp) not only depends on the size but also the climate of site location (Radiation angel). In other words, the Total Watt-Peak-Rating (Wp) would be on its maximum when the sunlight and absorbing surface (radiation incident) are preopercular.
The Relationship between S module and S horizontal is as followed:
SMODULE=├ Shorizontal*sin(α+β) ┤/Sinα
As mentioned earlier solar electricity generation depends on many factors such as Incident angel, sun elevation, longitude and latitude of the site, etc.
Below two tables which demonstrate annual estimated solar electricity generated in (Birmingham, UK), have been compared to each other. Two tables below, one with the incident angel of 39˚(optimum) and the other one 90˚ (window mounted) have been compared to each other.
Ed = panel generation factor.
Referring to information provided in the table on the left (optimum, south facing). Watt-peak rating needed for PV module would be the total Watt-hours per day needed from the PV modules divided by daily solar radiation:
Wp=needed PV module energy/ Ed
Wp=17Wh/1.03=16.50Wh
Now to calculate the number of PV modules:
number of PV modules=Wp/(PV module rated Wp)
number of PV modules=16.50Wh/21Wh=0.78 ~ 1
However, this result only represents the outdoor use, as the results have been calculated using the optimum solar radiation angel (39˚). By going through a similar procedure, the unit’s feasibility in a different environment would be tested. In order to replicate a window mounted stationary PV module (indoor use, 90˚, south facing), the table on the right has been used.
Wp=17Wh/1.14Wh=14.91Wh
number of PV modules=14.91Wh/21Wh=0.71 ~ 1
The above calculations examined the daily solar radiation in December which is the lowest in the year (worst case scenario). The results show that even under mentioned circumstances only one PV panel is needed in both situations.
Calculations above show that only one 21Wh foldable PV panel would satisfy the project needs in all conditions.
Candid Component: 21W, 5V Foldable Solar Panel.
I am not sure about the solar incident radiation. The PFG I used was in kWh/m2/day. I simply ignored the k and used it. in formula otherwise it wouldn't have worked!