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Assignment 2: Design a Off Grid Solar System For Your Home

KISSAN EDUCATION www.kissanedu.com You are going to install a solar power system in our home for a total load of 800W where the required backup time of the battery is 3 hours. Optimize it. If you are living in Kolkata, West Bangla. Find out Charging Current for Battery . Take Solar Inverter, efficiency = 95% ( Luminious Inverter) Battery, efficiency = 85 % DoD = 80 % Solution :  Since , Total Wattage  = 800 W According to saftey Factor = 800 * 1.25 = 1000 Watt Step 1 : Size of inverter =  1 KW Step 2 : Size of Battery: Backup = 3 hour Back up = Total energy / Total wattage Total input energy in battery = 3 * 800 = 2400 wh Output of Battery = 2400 * efficiency = 2400 * 0.80 = 1920 wh if Voltage = 24v, Ah = 1920 / 24 = 80 Ah Charging current of battery = 10 % of Ah = 16 Amp(ideal value)  = 20 amp(real value) 24V, 80V, Step 3 : Solar irradiance in kolkata = 4.12 kwh panel size after losses = 2400 / 0.70 = 3428...

Design of Solar Water Pump by Realistic Techniques

KISSAN EDUCATION www.kissanedu.com 1 2 3 4 5 6 7 8. 9. According to Dc pump rating , controller would = 96V, 15   Amp 10. 

Design of Solar Water Pump Lite version Techniques

KISSAN EDUCATION www.kissanedu.com 1 2 3. Note : This is only a lite technique to estimate design of Solar plant. It has some limitation , TDH should between 10 m to 12 m  Q . Design a PV water Pump System, which is required to draw 250000 l or 25 Cubic meter of water every day. ANS :   Known Data with Value : a. Amount  of water  to be pump per day : 25m3 b. Total vertical lift = 12 m (by bottle method or sound detector) c. Water Density = 1000kh/m3 d. Acceleration due to gravity = 9.8m/s2 e. Solar module used  = 75Wp (according to users) f. Operating factor = 0.75 g. Pump Efficiency = 30 % h. Mismatch factor = 0.85      // if MPPT is not used 1 is MPPT is used Dummy Module..

How to optimize off Grid solar System ?

Take inverter efficiency , according to "Essen Solar Off Grid Inverter" or "Sun grow inverter" Solution : First of all solve This problem according to Question given data Step 1 = Power = 840 Watt  Total Wattage = 2900 Wh According to safety Factor = Watt = 840 * 1.25 = 1050 = 1000 W Wh = 2900 * 1.25 = 3625 Wh Step 2 : Size of Battery Output energy of battery = 3625 / 0.90 =  4028 Wh   let us take DoD = 90% Ah =  4028 / 0.90 =  4476 Wh Since system is 1000 W , so it is not compatible to take 12 V battery, For this 24 V is taken A h = 186 Ah Step 3 : Calculate detail about module, for 250 Wp , because it is given or demanded by customer. Total panel wattage needed =  Battery output / Efficiency =  4740 Wh Total solar radiation in Delhi , Nirman Vihaar in winter = 4740/4.4 =  1077 Watt . =   1 kw Peak Sunshine hour in winter =  Dec - March Summer Season Month =  June to Sept For 250 Wp ...