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A Presentation On
Design of PV system
• Presented by
• Keshar Rawal
• Suman Shrestha
objective
• To design a PV system for a house.
• To familiar with the practical application of
solar power system.
• Market survey and economical analysis of
component and comparison between solar
system and diesel system
Solar cell
• convert light into a small electrical output -
milli watts output.
• need a bank/array of cells for useful output.
• efficiency of cells is up to 23%/ improving.
Solar panel
• are situated on roof of building.
• absorb heat in the form of radiation from sun.
• basically system is like a domestic central
heating radiator painted black/insulated.
Solar pv
Calculation of number of lamp
• Calculation of lamp sued in rooms is as follows
• Number of lamp= E*A/(O*CU*ME)
• Where E=require luminance
• For room 1, N=300*12*15/(3.3*3.3*6500*0.7*0.8)=2
• For room 4, N=300*14*10/(3.3*3.3*2800*0.7*0.8)=3
• For room 5, N=300(*16*6+10*6)/(3.3*3.3*2100*0.7*0.8)=3
• For room 6, N=200*13*10/(3.3*3.3*3000*0.7*0.8)=2
• For room 7, N=300*10*5.5/(3.3*3.3*1100*0.7*0.8)=3
• For room 14, N=300*12*15/(3.3*3.3*6500*0.7*0.8)=2
• For room 15, N=300*14*10/(3.3*3.3*1100*0.7*0.8)=3
• For room 16, N=300(*16*6+10*6)/(3.3*3.3*2100*0.7*0.8)=3
• For room 17, N=300*13*10/(3.3*3.3*2100*0.7*0.8)=3
• Total wattage-hours per day demand of appliances= 67492.6 watt
• 2.2 Solar PV sizing
• Ampere hours needed by appliances is= total energy needed per
day/(power factor*system voltage)
• =67492.6/(0.9*48)
• =1562.32
• Array current Iarray= Total load in AH/(peak sun*detraction
factor*columbic efficiency )
• =1562.32/(0.9*0.95*5)
• =365.4549Ah
• Number of solar panel needed in parallel = 365.45/6.85=53.35=54
panel
Solar panel selection
Solar panel selection
Battery sizing
• Capacity(Ah)= total load in Ah * day of
autonomy/(DOD*eff)
• 1562*3/(0.8*0.8)=7321.575Ah
• Total number of battery require in parallel=
7321.5/200 = 37
• Hence 37 battery having voltage rating 48 and
wattage rating of 200Ah is selected.
Battery selection
Inverter sizing
• Total AC power consumed
by AC loads= 12268.1watt
• Considering each AC
appliances take 3 time of
the current rating at son
time.
• Power rating of inverter is =
(AC power(W)*serge
current factor)/p.f=
12268.1*3/0.8= 46KVA
• A 50 KVA pure sine wave
inverter fulfill our needs
Feature of selected inverter
Solar charge controller selection
• Total maximum charge
current(Imax)= string* short
circuit current of string
•
=54*7.41= 400.14AH
• Hence a CR of 400AH meet
our requirement.
• Maximum possible load
current =72912.5/48=1519AH
• Lets take a charge controller of
50AH.
• Hence total number of CR =
54*7.41*1.3/50 =11
• Hence each 5 array consist of 1
CR.
Feature of selected charge controller
Economic Analysis
Cost Estimation For Solar System
S.N Particulars Price/piece(NRs) Numbers Total price(NRs)
1 OME 48 Volt mono solar panel 7060 54 381240
2 Inverter 980000 1 980000
3 Charge Regulator 25578 11 281358
4 48 volt battery 166306 37 6153322
5 Transportation Cost 45000
6 Installation Cost 20000
Total 7860920
Total cost of installation = Rs.7860920/-
Total annual expenditure = Rs.2500/-
Net present worth for the PV system is calculated as:
NPW (PV system) = -7860920-6153322(P/F, 4%, 10) - 6153322 (P/F,
4%, 20) - 2500(P/A, 4%, 30)
[Since, batteries have to be replaced in
every ten years.]
= -7860920-4156963-2808295-43230
= NRs.-14869408/-
Cost estimation for Diesel System
• Size of plant required: 4kVA
• From data observed, it was found that the generator at full load (3500W)
requires 2ltr of diesel.
• At 75% of load 1.5ltr is required.
• At 50% of load 1ltr is required.
• At 25% of load 0.5ltr is required.
• The diesel-generator is required to supply full load for 4 hours, 75% load
for 4 hours, 50% load for 6 hours, and 25% of load for 10 hours.
• Therefore, total diesel required = (2x4 + 1.5x4 + 1x6 + 0.5x10) = 25ltrs
daily.
• Total annual cost of operation = Rs(25x109x365)
• =NRs.994625/-
• Initial Cost of diesel-generator = NRs.138000/-
• Total annual expenditure = NRs.1500/-
•
• NPW (Diesel-generator) = -138000-138000(P/F, 4%, 10)
- 138000(P/F, 4%, 20) - 994625 (P/A, 4%, 30) –
1500(P/A, 4%, 30)
• [Since, diesel generator works properly for about
10 years, we need 3 generators during 30 years of
period.]
• = -138000-93227-62981-17199088-25938
= NRs.17519234/-
• The NPW of diesel generator is greater than that of PV
system. Hence, the PV system is financially feasible
compared to diesel generator.
Wire sizing
• Solar pane are going to be placed in the top of the celling and the load cater at
stair.
• Array to load center (allowable voltage drop 3%)= 6m(S1)
• Load center to battery(allowable voltage drop 1%)= 1m(s2)
• Load center to inverter (allowable voltage drop 3%)= 1m(s3)
• Inverter to energy meter= 2m(s4)
• Energy meter to distribution box=0.5m(s5)
• DB to SDB1=0.5m(s6)
• DB to SDB2=3m(s7)
• S1=(0.3*6*400)/3=240sq mm
• S2=(0.3*1*400)/3=40sq mm
• S3=(0.3*1*400)/3=40sq mm
• S4=(0.3*2*400)/3=80sq mm
• S5=(0.3*0.5*400)/3=20sq mm
• S6=(0.3*0.5*400)/3=20sq mm
• S7=(0.3*3*400)/3=120sq mm
Load connection
Load connection
Distribution box
Earthling

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PV system design for house

  • 1. A Presentation On Design of PV system • Presented by • Keshar Rawal • Suman Shrestha
  • 2. objective • To design a PV system for a house. • To familiar with the practical application of solar power system. • Market survey and economical analysis of component and comparison between solar system and diesel system
  • 3. Solar cell • convert light into a small electrical output - milli watts output. • need a bank/array of cells for useful output. • efficiency of cells is up to 23%/ improving.
  • 4. Solar panel • are situated on roof of building. • absorb heat in the form of radiation from sun. • basically system is like a domestic central heating radiator painted black/insulated.
  • 6.
  • 7.
  • 8.
  • 9.
  • 10.
  • 11.
  • 12. Calculation of number of lamp • Calculation of lamp sued in rooms is as follows • Number of lamp= E*A/(O*CU*ME) • Where E=require luminance • For room 1, N=300*12*15/(3.3*3.3*6500*0.7*0.8)=2 • For room 4, N=300*14*10/(3.3*3.3*2800*0.7*0.8)=3 • For room 5, N=300(*16*6+10*6)/(3.3*3.3*2100*0.7*0.8)=3 • For room 6, N=200*13*10/(3.3*3.3*3000*0.7*0.8)=2 • For room 7, N=300*10*5.5/(3.3*3.3*1100*0.7*0.8)=3 • For room 14, N=300*12*15/(3.3*3.3*6500*0.7*0.8)=2 • For room 15, N=300*14*10/(3.3*3.3*1100*0.7*0.8)=3 • For room 16, N=300(*16*6+10*6)/(3.3*3.3*2100*0.7*0.8)=3 • For room 17, N=300*13*10/(3.3*3.3*2100*0.7*0.8)=3
  • 13. • Total wattage-hours per day demand of appliances= 67492.6 watt • 2.2 Solar PV sizing • Ampere hours needed by appliances is= total energy needed per day/(power factor*system voltage) • =67492.6/(0.9*48) • =1562.32 • Array current Iarray= Total load in AH/(peak sun*detraction factor*columbic efficiency ) • =1562.32/(0.9*0.95*5) • =365.4549Ah • Number of solar panel needed in parallel = 365.45/6.85=53.35=54 panel
  • 16. Battery sizing • Capacity(Ah)= total load in Ah * day of autonomy/(DOD*eff) • 1562*3/(0.8*0.8)=7321.575Ah • Total number of battery require in parallel= 7321.5/200 = 37 • Hence 37 battery having voltage rating 48 and wattage rating of 200Ah is selected.
  • 18. Inverter sizing • Total AC power consumed by AC loads= 12268.1watt • Considering each AC appliances take 3 time of the current rating at son time. • Power rating of inverter is = (AC power(W)*serge current factor)/p.f= 12268.1*3/0.8= 46KVA • A 50 KVA pure sine wave inverter fulfill our needs
  • 20. Solar charge controller selection • Total maximum charge current(Imax)= string* short circuit current of string • =54*7.41= 400.14AH • Hence a CR of 400AH meet our requirement. • Maximum possible load current =72912.5/48=1519AH • Lets take a charge controller of 50AH. • Hence total number of CR = 54*7.41*1.3/50 =11 • Hence each 5 array consist of 1 CR.
  • 21. Feature of selected charge controller
  • 22. Economic Analysis Cost Estimation For Solar System S.N Particulars Price/piece(NRs) Numbers Total price(NRs) 1 OME 48 Volt mono solar panel 7060 54 381240 2 Inverter 980000 1 980000 3 Charge Regulator 25578 11 281358 4 48 volt battery 166306 37 6153322 5 Transportation Cost 45000 6 Installation Cost 20000 Total 7860920 Total cost of installation = Rs.7860920/- Total annual expenditure = Rs.2500/- Net present worth for the PV system is calculated as: NPW (PV system) = -7860920-6153322(P/F, 4%, 10) - 6153322 (P/F, 4%, 20) - 2500(P/A, 4%, 30) [Since, batteries have to be replaced in every ten years.] = -7860920-4156963-2808295-43230 = NRs.-14869408/-
  • 23. Cost estimation for Diesel System • Size of plant required: 4kVA • From data observed, it was found that the generator at full load (3500W) requires 2ltr of diesel. • At 75% of load 1.5ltr is required. • At 50% of load 1ltr is required. • At 25% of load 0.5ltr is required. • The diesel-generator is required to supply full load for 4 hours, 75% load for 4 hours, 50% load for 6 hours, and 25% of load for 10 hours. • Therefore, total diesel required = (2x4 + 1.5x4 + 1x6 + 0.5x10) = 25ltrs daily. • Total annual cost of operation = Rs(25x109x365) • =NRs.994625/- • Initial Cost of diesel-generator = NRs.138000/- • Total annual expenditure = NRs.1500/- •
  • 24. • NPW (Diesel-generator) = -138000-138000(P/F, 4%, 10) - 138000(P/F, 4%, 20) - 994625 (P/A, 4%, 30) – 1500(P/A, 4%, 30) • [Since, diesel generator works properly for about 10 years, we need 3 generators during 30 years of period.] • = -138000-93227-62981-17199088-25938 = NRs.17519234/- • The NPW of diesel generator is greater than that of PV system. Hence, the PV system is financially feasible compared to diesel generator.
  • 25. Wire sizing • Solar pane are going to be placed in the top of the celling and the load cater at stair. • Array to load center (allowable voltage drop 3%)= 6m(S1) • Load center to battery(allowable voltage drop 1%)= 1m(s2) • Load center to inverter (allowable voltage drop 3%)= 1m(s3) • Inverter to energy meter= 2m(s4) • Energy meter to distribution box=0.5m(s5) • DB to SDB1=0.5m(s6) • DB to SDB2=3m(s7) • S1=(0.3*6*400)/3=240sq mm • S2=(0.3*1*400)/3=40sq mm • S3=(0.3*1*400)/3=40sq mm • S4=(0.3*2*400)/3=80sq mm • S5=(0.3*0.5*400)/3=20sq mm • S6=(0.3*0.5*400)/3=20sq mm • S7=(0.3*3*400)/3=120sq mm