Need help!! Please provide a step-by-step solution along with a detailed explana
ID: 2248940 • Letter: N
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Need help!! Please provide a step-by-step solution along with a detailed explanation! This is very URGENT!!! PLEASE reply as soon as possible!! Thank you!!
(2) (30 points) Problem 3.14 (Messenger and Abtahi) A battery charge controller incorporates an MPPT to optimize the charging of the batteries. Assume the maximum power voltage (Vmp) of a PV array is 99V and the bulk charge voltage level for a 48V vented lead-acid battery bank is 56V. If the conversion efficiency of the MPPT is 98%, estimate the percentage increase in charge delivered to the batteries under the conditions over that which would be delivered by a controller that causes the array to operate at 56V rather than 99V You may assume the PV array consists of three modules connected in series, that Vmp is independent of irradiance levels, and each module has Voc 42V, Isc 6.25A, Vmp = 33V, and Imp = 5.76AExplanation / Answer
The power delivered to rhe batteries using the MPPT is given by (0.98X99VX5.76A) = 558.8 W.
If the battery is operated without MPPT, and assuming that the current at 56 volts is approximately equal to the short circuit current, we get (56V)(6.25A) = 350 W.
The electrical perfrmance of PV modules is rated at different conditions like under Normal Operating Conditions(NOC), Standard Test Conditions etc.
Irradiance is generally in the range of 800 to 1000 W/m2 .
Given that PV array consists of 3 modules.Assume them as similar PV devices.
we have Vseries = V1* n
where n = number of PV devices
therefore we get Vseries = 3V1 = 3*33 = 99V
and Iseries = I1 = I2 = ......In
Iseries = 5.76 Amps
We know that the Power and Current output of a PV device are proportional to Solar irradiance.
E2/E1 = I2/I1 = P2/P1
and also we have Conversion Efficiency = Pmp / (E*A)
where E = solar Irradiance(W/m2) and A = Area of cross section
To get the percentage increase or the percentage gain in charge delivered to batteries can be found out by
(558.8W)/(350W) = 1.60. (The current delivered at 56 v is about 10 Amps when using the MPPT.)
therefore 60% increase in charge delivered to the batteries under the above conditions.
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