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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 wh
size of panel = 3428/4.12 = 832 watt
Let us take Size of Panel 270 watt


Capacity = 1000 Va Batte…

Design of Solar Water Pump by Realistic Techniques

KISSAN EDUCATION www.kissanedu.com
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9. According to Dc pump rating , controller would = 96V, 15Amp
10. 









Design of Solar Water Pump Lite version Techniques

KISSAN EDUCATION www.kissanedu.com


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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


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
  total number of module = 1000/250  = 4

For 250 Wp  Open Circuit Voltage (Voc) in Vo…

Assignment : How to Design a Solar Power System for home

KISSAN EDUCATION www.kissanedu.com

Suppose,
Living District = Una, Himachal
 Inverter's Efficiency = 95%
Battery's Efficiency = 85%
Converter Efficiency = 92%
Take Loom Solar Panel
Inverter = Microtech
consider only night backup for battery

Solution:
Total Day time consumption = 22800 Wh
Total Night-Time Consumption = 17280 Wh
Total Energy Consumption =  40080 Wh

Since Size of Plant is very high, so we have to consider only the night backup plan for optimizing

Step 1 :
Size of Inverter = Total Wattage of Appliances = Wattage * quantity = 5440 watt = 6 KW inverter

Step 2 :
Size of Battery:

Output Energy of Battery = Input energy of Inverter =(output of inverter) / (efficiency of the inverter)
                                          = 17280 /0.95  = 18190 Wh                                     

Energy Storage of Battery =  Output Energy of Battery / DoD of Battery

                                           = 18190 / 0.80 = 22738 Wh

Note DoD of Lead-acid Battery = 50 %
Dod of …

How Much Power Emitted by "the SUN" ?

KISSAN EDUCATION(www.kissanedu.com)
In early class,we students used to study "SUN is Major sources of Energy" . Here i try to clear every one what is actual meaning of this sentence.


Power emits from sun = 3.8 * 1030 W Power Density= 6.25 * 1011 W/m2 Solar Perihelion = 1414 W/m2    3 Jan(longest distance from sun) Aphelion = 1322 W/m2              3 July(Shortest distance from Sun)



Calculation to find out Solar Power density at outer atmosphere over year
St = S[1 + 0.033Sin(360n/365)] Jan 1, n = 1
Jan 2, n = 2
Jan 31, n = 31
Dec 31, n = 365
Suppose,If you want to find out Solar radiation in upper atmosphere in 31 march.




Thus you can calculate value of Solar Constant For any particular day.
In next Blog, how i would Discuss CAN WE USE  SOLAR PV FOR REQUIREMENT OF WORLD POWER ,

FOUR EXTRA Bills that you pay along your "ELECTRICITY BILLS" read it.

KISSAN EDUCATION
(www.Kissanedu.com)

Q. Are you Paying only electricity that you used?
Ans : No, There are Five components that you pay along your Electricity uses:
1. Electricity Charges per Unit 2. Fixed Charges 3. F.A.C 4. Wheeling Charges 5. Electricity Duty ( Tax on Electricity)



1. Electricity Charges Per Unit : This Charges are main charges of electricity that you used,This charges varies State to State. Here Rate of charges considered 5.73 per unit. Hence, Total unit Consumed by Consumer is 319 Units Total Electricity Charges = 5.75 *319 =  Rs 1830.30
2. Fixed Charges : Such Charges are  fixed either you use electricity or not .This charges varies State to State. Fixed Charges =  Rs 60
3. F.A.C : It stands for Fuel Adjustment Charges,  Such charges varies according to demand and supply or availability of Fossil fuel. Availability of fossil fuel becomes not constant. This charges varies State to State. Here , F.A.C is 0.297 per unit.
F.A.C = 0.297 *319 = Rs 93.36
4. Wheeling Charges = S…