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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
PV system design for house
PV system design for house
PV system design for house
PV system design for house
PV system design for house
PV system design for house
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.
  • 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