Najah National Najah National University University Faculty of Electrical Faculty of Electrical engineering engineering Introduction to Graduation Project " Techno-economic feasibility of using photovoltaic generators instead of diesel motor for water pumping ” Supervisor Prof .Dr. Marwan M . Mahmoud Prepared by Sabreen Staiti Heba Asedah Yasmeen Salah 1 1
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Najah National University Faculty of Electrical engineering Introduction to Graduation Project "Techno-economic feasibility of using photovoltaic generators.
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Najah National University Najah National University Faculty of Electrical Faculty of Electrical
engineeringengineeringIntroduction to Graduation Project
"Techno-economic feasibility of using photovoltaic generators instead of diesel
motor for water pumping”Supervisor
Prof .Dr. Marwan M . Mahmoud
Prepared by Sabreen Staiti Heba Asedah
Yasmeen Salah11
ContentsContents
contentcontent pagepage
Chapter 1Chapter 1
introductionintroduction
Chapter 2Chapter 2
The content of pv and diesel systemsThe content of pv and diesel systems
Chapter 3Chapter 3
The data of well neededThe data of well needed
Chapter 4Chapter 4
The calculation of wells capital cost The calculation of wells capital cost for pv and diesel systemsfor pv and diesel systems
Chapter 5Chapter 5
The calculation of average value of The calculation of average value of yearly pumping wateryearly pumping water
Chapter 6Chapter 6
Economical evaluation of energy Economical evaluation of energy supply systemssupply systems
Chapter 7Chapter 7
Net present value ( NPV )Net present value ( NPV )
Chapter 8Chapter 8
Life cycle costLife cycle cost
Chapter 9Chapter 9
22
Aim of projectAim of project1 -knowledge the component of
photovoltaic water pumping system and diesel pumping
2 -Advantages and disadvantages of using photovoltaic generator instead of diesel motor for water pumping
3 -Decide which one of them the most economical
33
Chapter 2Chapter 2
The content of photovoltaic and diesel water The content of photovoltaic and diesel water pumping systempumping system
44
Components of Photovoltaic water pumping system
55
Components of Water Components of Water pumping system driven by a pumping system driven by a
diesel electric motordiesel electric motor
66
Chapter 3Chapter 3 Data of well neededData of well needed
77
Well depth.Swl (static water level)
Dwl (dynamic water level)Compute of the well (m3/
hour ) ,(hour / day) . Total pumping head (m)
Total water capacity needed per day (m3/ hour )
88
99
عويص رائد
Well depth=70m.
Swl = 20m.
Dwl = 60m.
Pipe = 2inch.
Compute of the well = 3 m3/ hour,
12 hour / day .
Total pumping head (ht) = 60 +(6m) = 66m .
Total water capacity needed per day =3 ×12=36 m3Eh =0.002725×V × hT
= 0.002725×36× 66 =6.47kwh / dayE p in =11.76kwh /day.
1010
E asm in = 14.34 Kwh / dayE inv in =15.42 kwh / dayPsh =5.4 kwh /day.
1kwp →5.4 kwh /day.
→ ? 15.42 kwh / day.
PPV= 2.85 kwpPPV= 2.85 kwp. .
Pv → 1 wp = 2.5Pv → 1 wp = 2.5$ $
The cost of Pv cells :- 2850* 2.5 = 7125The cost of Pv cells :- 2850* 2.5 = 7125 $$
Inverter cost = 1800Inverter cost = 1800 $ $
Motor pump set = 1000Motor pump set = 1000$ $
piping and accessories = 400piping and accessories = 400$ $
The capital cost = 7125 The capital cost = 7125 +1800+1000+400+1800+1000+400
Output of the well = 3 m3/ hour for 5 hour / day .
Total pumping head (hT) = 40 +(6m) = 46m .
Total water capacity needed per day =3 ×5=15 m3Eh =0.002725×V × hT
= 0.002725×15×46 =1.88kwh / dayE p in =3.42 kwh /day .
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E ASM in = 4.17 Kwh / day. E inv in = =4.48 kwh / day. Psh =5.4 kwh /day.
1kwp →5.4 kwh /day . → ? 4.48 kwh / day.
PPV= .829 kwp.
The cost of Pv cellsThe cost of Pv cells = = 829829 * *2.52.5==2072.52072.5 $ $
Inverter cost = 1400Inverter cost = 1400 $ $
Motor pump set = 800Motor pump set = 800 $ $
Piping and Piping and accessories = 300accessories = 300 The capital cost = 2072.5 +1400 The capital cost = 2072.5 +1400 +800 +3000+800 +3000
= = 4572.54572.5$ $
1313
.. غانم غانم خليل خليلWell depth=60mWell depth=60m..
Swl = 30mSwl = 30m..Dwl = 50mDwl = 50m..
Pipe = 2inchPipe = 2inch.. Output of the well = 4 m3/ hour for 12 hour / dayOutput of the well = 4 m3/ hour for 12 hour / day . .
Total pumping head (hT) = 50 +(6m) = 56mTotal pumping head (hT) = 50 +(6m) = 56m . .Total water capacity needed per day =4 ×12=48 m3Total water capacity needed per day =4 ×12=48 m3
Pipe = 2inchPipe = 2inch.. Output of the well = 3 m3/ hour for 12 hour / dayOutput of the well = 3 m3/ hour for 12 hour / day . .
Total pumping head (h T) = 56 + (6m) = 62 mTotal pumping head (h T) = 56 + (6m) = 62 m . .Total water capacity needed per day =3 ×12=36 m3Total water capacity needed per day =3 ×12=36 m3
Eh =0.002725×V × h TEh =0.002725×V × h T = = 0.0027250.002725××3636××6262 = =6.08kwh / day6.08kwh / day
E p in =11.05 k w h /dayE p in =11.05 k w h /day. .
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E ASM in = 13.48 K w h / dayE ASM in = 13.48 K w h / day. .
E inv in = =14.49 k w h / dayE inv in = =14.49 k w h / day. . P sh =5.4 k w h /dayP sh =5.4 k w h /day..
The cost of Pv cells :- 2068*2.5 =5170The cost of Pv cells :- 2068*2.5 =5170 $ $ Inverter cost = 1800Inverter cost = 1800 $ $
Motor pump set = 1000Motor pump set = 1000$ $ Piping and accessories = 400Piping and accessories = 400$ $
The capital cost = 1800 +1000 +400 +5170The capital cost = 1800 +1000 +400 +5170 = = 83708370$ $
1717
أنسمساد أنسمساد ..Well depth=Well depth=440m0m
Swl = 12mSwl = 12m..Dwl = 35mDwl = 35m..
Pipe = 2inchPipe = 2inch.. Output of the well = 2.5 m3/ hour for 3 hour / dayOutput of the well = 2.5 m3/ hour for 3 hour / day . .
Total pumping head (h T) = 35 + (6m) = 41 mTotal pumping head (h T) = 35 + (6m) = 41 m . .Total water capacity needed per day =2.5 ×3=7.5 m3/ dayTotal water capacity needed per day =2.5 ×3=7.5 m3/ day. .
Eh =0.002725×V × h TEh =0.002725×V × h T = = 0.0027250.002725××7.57.5××4141 = =0.84kwh / day0.84kwh / day
E p in =1. E p in =1. 53 k w h /day53 k w h /day. . E ASM in = 1 .86 K w h / dayE ASM in = 1 .86 K w h / day. .
E inv in = =2 k w h / dayE inv in = =2 k w h / day. . P sh =5.4 k w h /dayP sh =5.4 k w h /day..
Pipe = 2inchPipe = 2inch.. Output of the well = 3 m3/ hour for 12 hour / dayOutput of the well = 3 m3/ hour for 12 hour / day . .
Total pumping head (h T) = 60 + (6m) = 66 mTotal pumping head (h T) = 60 + (6m) = 66 m . .Total water capacity needed per day =12 ×3=Total water capacity needed per day =12 ×3=3636 m3/ day m3/ day. .
Eh =0.002725×V × h TEh =0.002725×V × h T = = 0.0027250.002725××3636××6666 = =66..4747 kwh / daykwh / day
E p in =11. E p in =11. 76 k w h /day76 k w h /day. . E ASM in = 14 .34 K w h / dayE ASM in = 14 .34 K w h / day. .
E inv in = =15.42 k w h / dayE inv in = =15.42 k w h / day. . P sh =5.4 k w h /dayP sh =5.4 k w h /day..
Pipe = 2 inchPipe = 2 inch.. Output of the well = 50 m3/ hour for 4 hour / dayOutput of the well = 50 m3/ hour for 4 hour / day . .
Total pumping head (h T) = 32 + (6m) = 38 mTotal pumping head (h T) = 32 + (6m) = 38 m . .Total water capacity needed per day =50 ×4=200 m3/ dayTotal water capacity needed per day =50 ×4=200 m3/ day. .
Eh =0.002725×V × h TEh =0.002725×V × h T = = 0.0027250.002725××200200××3838 = =20.7120.71 kwh / daykwh / day
E p in =20.710/ .55 = 37.65 k w h /dayE p in =20.710/ .55 = 37.65 k w h /day. . E ASM in =37.65 / .82 = 45.9 K w h / dayE ASM in =37.65 / .82 = 45.9 K w h / day. .
E inv in =45.9 / .92 =49.37 k w h / dayE inv in =45.9 / .92 =49.37 k w h / day. . P sh =5.4 k w h /dayP sh =5.4 k w h /day..
The cost of Pv cells :- 9140 * 2.5 =22850The cost of Pv cells :- 9140 * 2.5 =22850 $ $
Inverter cost = 4500$Inverter cost = 4500$ Motor pump set = 2000Motor pump set = 2000$ $
Piping and accessories = 500Piping and accessories = 500$ $ The capital cost = 22850 +4500 +2000 The capital cost = 22850 +4500 +2000
+500+500 = = 2985029850$ $
2323
4-24-2 the calculation of capital cost for the calculation of capital cost for diesel systemdiesel system
عويص عويص رائد رائدWell depth=70mWell depth=70m..
Swl = 20mSwl = 20m..Dwl = 60mDwl = 60m..
Pipe = 2inchPipe = 2inch..Compute of the well = 3 m3/ hour ,12 hour / dayCompute of the well = 3 m3/ hour ,12 hour / day . .
Total pumping head (hT ) = 60 +(6m) = 66mTotal pumping head (hT ) = 60 +(6m) = 66m . .Total water capacity needed per day =3 ×12=36 m3Total water capacity needed per day =3 ×12=36 m3
COST OF EQUIPMENTCOST OF EQUIPMENTDiesel and engine and S.G = 3125Diesel and engine and S.G = 3125$ $
Motor pump set = 1000Motor pump set = 1000$ $ Piping and accessories = 400Piping and accessories = 400$ $
Capital cost = 4525Capital cost = 4525$ $
2525
مساد مساد جميل جميلWell depth=45mWell depth=45m..Swl = 12mSwl = 12m..Dwl = 40mDwl = 40m..Pipe = 2inchPipe = 2inch.. Output of the well = 3 m3/ hour for 5 hour / dayOutput of the well = 3 m3/ hour for 5 hour / day . . Total pumping head (hT) = 40 +(6m) = 46mTotal pumping head (hT) = 40 +(6m) = 46m . .Total water capacity needed per day =3 ×5=15 m3Total water capacity needed per day =3 ×5=15 m3Eh =0.002725×V × hTEh =0.002725×V × hT = = 0.0027250.002725××1515××4646 = =1.88kwh / day1.88kwh / dayE p in =3.42 kwh /dayE p in =3.42 kwh /day. . E ASM in = 4.17 Kwh / dayE ASM in = 4.17 Kwh / day. . E S.G inE S.G in = 4.17/0.75 = 5.56 kwh = 4.17/0.75 = 5.56 kwh E DIESEL inE DIESEL in =5.56/0.30 = 18.53 kwh =5.56/0.30 = 18.53 kwh Number of liters diesel needed daily =Number of liters diesel needed daily = 18.53/10.5 = 1.76 18.53/10.5 = 1.76Diesel cost = 6* 6 NIS =10.56 NIS/dayDiesel cost = 6* 6 NIS =10.56 NIS/day
2626
COST OF EQUIPMENTCOST OF EQUIPMENT: : Diesel and engine and S.G = 3125Diesel and engine and S.G = 3125$ $ Motor pump set = 800Motor pump set = 800$ $ Piping and accessories = 300Piping and accessories = 300$ $ Capital cost = 4225Capital cost = 4225$ $
2727
****************************************************************************************************** غانم غانم خليل خليلWell depth=60mWell depth=60m..Swl = 30mSwl = 30m..Dwl = 50mDwl = 50m..Pipe = 2inchPipe = 2inch.. Output of the well = 4 m3/ hour for 12 hour / dayOutput of the well = 4 m3/ hour for 12 hour / day . . Total pumping head (hT) = 50 +(6m) = 56mTotal pumping head (hT) = 50 +(6m) = 56m . .Total water capacity needed per day =4 ×12=48 m3Total water capacity needed per day =4 ×12=48 m3Eh =0.002725×V × hTEh =0.002725×V × hT = = 0.0027250.002725××4848××5656 = =7.32kwh / day7.32kwh / dayE p in =13.3 kwh /dayE p in =13.3 kwh /day. . E ASM in = 16.23 Kwh / dayE ASM in = 16.23 Kwh / day. . E S.G inE S.G in = 16.23/0.75 = 21.64 kwh = 16.23/0.75 = 21.64 kwh E DIESEL inE DIESEL in =21.64/0.30 = 72.13 kwh =21.64/0.30 = 72.13 kwh Number of liters diesel needed daily =Number of liters diesel needed daily = 72.13/10.5 = 6.87 72.13/10.5 = 6.87Diesel cost = 6* 6.87 NIS =41.22NIS/dayDiesel cost = 6* 6.87 NIS =41.22NIS/day
2828
COST OF EQUIPMENTCOST OF EQUIPMENT: : Diesel and engine and S.G = 3125Diesel and engine and S.G = 3125$ $ Motor pump set = 1200Motor pump set = 1200$ $ Piping and accessories = 420Piping and accessories = 420$ $ Capital cost = 4745Capital cost = 4745$ $
2929
Well depth=60mWell depth=60m..Swl = 20mSwl = 20m..Dwl = 56mDwl = 56m..Pipe = 2inchPipe = 2inch.. Output of the well = 3 m3/ hour for 12 hour / dayOutput of the well = 3 m3/ hour for 12 hour / day . . Total pumping head (h T) = 56 + (6m) = 62 mTotal pumping head (h T) = 56 + (6m) = 62 m . .Total water capacity needed per day =3 ×12=36 m3Total water capacity needed per day =3 ×12=36 m3Eh =0.002725×V × h TEh =0.002725×V × h T = = 0.0027250.002725××3636××6262 = =6.08kwh / day6.08kwh / dayE p in =11.05 k w h /dayE p in =11.05 k w h /day. . E ASM in = 13.48 K w h / dayE ASM in = 13.48 K w h / day. . E S.G inE S.G in = 13.48/0.75 = 17.97 kwh = 13.48/0.75 = 17.97 kwh E DIESEL inE DIESEL in =17.97/0.30 = 59.9 kwh =17.97/0.30 = 59.9 kwh Number of liters diesel needed daily =Number of liters diesel needed daily = 59.9/10.5 = 5.7 59.9/10.5 = 5.7Diesel cost = 6* 5.7 NIS =34.2 NIS/dayDiesel cost = 6* 5.7 NIS =34.2 NIS/day
3030
COST OF EQUIPMENTCOST OF EQUIPMENT ::Diesel and engine and S.G = 3125Diesel and engine and S.G = 3125$ $ Motor pump set = 1000Motor pump set = 1000$ $ Piping and accessories = 400Piping and accessories = 400$ $ Capital cost = 4525Capital cost = 4525$ $
3131
أنسمساد أنسمساد ..Well depth=40mWell depth=40mSwl = 12mSwl = 12m..Dwl = 35mDwl = 35m..Pipe = 2inchPipe = 2inch.. Output of the well = 2.5 m3/ hour for 3 hour / dayOutput of the well = 2.5 m3/ hour for 3 hour / day . . Total pumping head (h T) = 35 + (6m) = 41 mTotal pumping head (h T) = 35 + (6m) = 41 m . .Total water capacity needed per day =2.5 ×3=7.5 m3/ dayTotal water capacity needed per day =2.5 ×3=7.5 m3/ day. . Eh =0.002725×V × h TEh =0.002725×V × h T = = 0.0027250.002725××7.57.5××4141 = =0.84kwh / day0.84kwh / dayE p in =1. E p in =1. 53 k w h /day53 k w h /day. . E ASM in = 1 .86 K w h / dayE ASM in = 1 .86 K w h / day. . E S.G inE S.G in = 1.86/0.75 = 2.48 kwh = 1.86/0.75 = 2.48 kwh E DIESEL inE DIESEL in =2.48/0.30 = 8.2 kwh =2.48/0.30 = 8.2 kwh Number of liters diesel needed daily =Number of liters diesel needed daily = 8.2/10.5 =.78 8.2/10.5 =.78Diesel cost = 6* .78 NIS =4.68 NIS/dayDiesel cost = 6* .78 NIS =4.68 NIS/day
3232
COST OF EQUIPMENTCOST OF EQUIPMENT: : Diesel and engine and S.G = 3125Diesel and engine and S.G = 3125$ $ Motor pump set = 700Motor pump set = 700$ $ Piping and accessories = 300Piping and accessories = 300$ $ Capital cost = 4125Capital cost = 4125$ $
3333
************************************************************************************************************** العامودي العامودي محمود محمودWell depth=70mWell depth=70mSwl = 30mSwl = 30m..Dwl = 60mDwl = 60m..Pipe = 2inchPipe = 2inch.. Output of the well = 3 m3/ hour for 12 hour / dayOutput of the well = 3 m3/ hour for 12 hour / day . . Total pumping head (h T) = 60 + (6m) = 66 mTotal pumping head (h T) = 60 + (6m) = 66 m . .Total water capacity needed per day =12 ×3=Total water capacity needed per day =12 ×3=3636 m3/ day m3/ day. . Eh =0.002725×V × h TEh =0.002725×V × h T = = 0.0027250.002725××3636××6666 = =6.476.47 kwh / daykwh / dayE p in =11. E p in =11. 76 k w h /day76 k w h /day. . E ASM in = 14 .34 K w h / dayE ASM in = 14 .34 K w h / day. . E S.G inE S.G in = 14.34/0.75 = 19.12 kwh = 14.34/0.75 = 19.12 kwh E DIESEL inE DIESEL in =19.12/0.30 = 63.73 kwh =19.12/0.30 = 63.73 kwh Number of liters diesel needed daily =Number of liters diesel needed daily = 63.73/10.5 =6.06 63.73/10.5 =6.06Diesel cost = 6* 6.06 NIS =36.36 NIS/dayDiesel cost = 6* 6.06 NIS =36.36 NIS/day
3434
COST OF EQUIPMENTCOST OF EQUIPMENT: : Diesel and engine and S.G = 3125Diesel and engine and S.G = 3125$ $ Motor pump set = 1000Motor pump set = 1000$ $ Piping and accessories = 400Piping and accessories = 400$ $ Capital cost = 4525Capital cost = 4525$ $
3535
عباس ابراهيم عمر عباس كامل ابراهيم عمر كاملWell depth=50mWell depth=50mSwl = 18mSwl = 18m..Dwl = 32mDwl = 32m..Pipe = 2inchPipe = 2inch.. Output of the well = 50 m3/ hour for 4 hour / dayOutput of the well = 50 m3/ hour for 4 hour / day . . Total pumping head (h T) = 32+ (6m) = 32 mTotal pumping head (h T) = 32+ (6m) = 32 m . .Total water capacity needed per day =50×4=200m3/ dayTotal water capacity needed per day =50×4=200m3/ day. . Eh =0.002725×V × h TEh =0.002725×V × h T = = 0.0027250.002725××200200××3232 = =20.7120.71 kwh / daykwh / dayE p in = 20.71 / .55 =37.65 k w h /dayE p in = 20.71 / .55 =37.65 k w h /day. . E ASM in = 37.65 / 0.82 = 45.9 K w h / dayE ASM in = 37.65 / 0.82 = 45.9 K w h / day. . E S.G inE S.G in = 45.9/0.75 = 61.2 kwh = 45.9/0.75 = 61.2 kwh E DIESEL inE DIESEL in =61.2/0.30 = 204 kwh =61.2/0.30 = 204 kwh Number of liters diesel needed daily =Number of liters diesel needed daily = 204/10.5 =19.43204/10.5 =19.43$ $ Diesel cost = 6* 19.43 NIS =116.58 NIS/dayDiesel cost = 6* 19.43 NIS =116.58 NIS/day
3636
COST OF EQUIPMENTCOST OF EQUIPMENT: : Diesel and engine and S.G = 3125Diesel and engine and S.G = 3125$ $ Motor pump set = 2000Motor pump set = 2000$ $ Piping and accessories = 500Piping and accessories = 500$ $ Capital cost = 5625Capital cost = 5625$ $
3737
Chapter 5Chapter 5The calculation of average value of yearly The calculation of average value of yearly
pumping waterpumping water((m3 / daym3 / day))
3838
عويص عويصJanuaryJanuary
E E pv outpv out = 2.85 ×2.85 =8.12 kwh = 2.85 ×2.85 =8.12 kwh
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
8.128.12 kwh / daykwh / day ? →? →Average value of daily pumping water dnring JanuaryAverage value of daily pumping water dnring January
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
9.129.12 kwh / daykwh / day ? →? →Average value of daily pumping water during February =21.29 Average value of daily pumping water during February =21.29
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
14.8214.82 kwh / daykwh / day ? →? →Average value of daily pumping water duringAverage value of daily pumping water during MarchMarch =34.59 m =34.59 m33
The cost of unit generated (1 KWh) = total The cost of unit generated (1 KWh) = total annual cost / total energy outputannual cost / total energy output..
The cost of unit generated (1 The cost of unit generated (1 KWh)=1290.625/2361.55 = 0.546 KWh)=1290.625/2361.55 = 0.546 $/kwh$/kwh..
Evaluation of the cost of 1 mEvaluation of the cost of 1 m33: :
Total annual cost = 1290.625Total annual cost = 1290.625.$ .$
Q=13757.1 mQ=13757.1 m33/ year/ year.. The cost of 1 mThe cost of 1 m33 =1290.625/13757.1= =1290.625/13757.1=
.0938 .0938 m m33/ year/ year5757
The cost of 1 Kwh in$
Diesel systemPV systemSystem typeWells
2.546Raad well
2.9.83Jameel well
2.547Khaleel well
2.557Ahmad well
3.51.3Anas well
2.547Mahmoud well
5858
The cost of 1 m3 in$
Diesel systemPV systemSystem typeWells
.373.0938Raad well
.369.o58Jameel well
.308.0914Khaleel well
.34.093Ahmad well
.533.145Anas well
.373.097Mahmoud well
5959
Chapter 7Chapter 7Net present value ( NPV )Net present value ( NPV )
6060
Net present value ( NPV )Net present value ( NPV )
The NPV of an investment project at time t=0 is the sum of the present The NPV of an investment project at time t=0 is the sum of the present values of all cash inflows and outflows linked to the investmentvalues of all cash inflows and outflows linked to the investment: :
NPV = -INPV = -I0 0 ++ (R (Rt t –I–Itt)q)q-t -t + L+ LTTqq-t-t
qq-t -t = (1+= (1+ ))-t-t
where Iwhere I0 0 is the investment cost at the beginning (t=0) ,T is the life time of is the investment cost at the beginning (t=0) ,T is the life time of
project in years, Rproject in years, Rtt is the return in time period t , I is the return in time period t , I t t is the investment in time is the investment in time
period t , qperiod t , q-t -t is discounting factor , i is the discount rate and L is discounting factor , i is the discount rate and LTT is the salvage is the salvage
valuevalue. . A project is profitable when NPV 0 and the greater the NPV the more A project is profitable when NPV 0 and the greater the NPV the more
profitableprofitable..Negative NPV indicates that minimum interest rate will not be metNegative NPV indicates that minimum interest rate will not be met..
6161
Net Present ValueNet Present Value For Raad wellFor Raad well
Evaluation results and conclusionsEvaluation results and conclusions Based on above results the following Based on above results the following
conclusion can be madeconclusion can be made: : The annuity and the production cost of energy The annuity and the production cost of energy
uint (KWh) of PV-system are less than the uint (KWh) of PV-system are less than the diesel systemdiesel system..
The net present value ( NPV)of the PV-system The net present value ( NPV)of the PV-system is much higher than the NPV of diesel systemis much higher than the NPV of diesel system..
The life cycle cost of the PV-system is less than The life cycle cost of the PV-system is less than the diesel systemthe diesel system..
Therefore , utitizing of PV-system is more Therefore , utitizing of PV-system is more economic feasible for pumping water .in economic feasible for pumping water .in additional the PV-system do not pollute the additional the PV-system do not pollute the environment as the case of using diesel environment as the case of using diesel generatorgenerator
7575
6060 mm33/day/dayWell depth=60mWell depth=60m..
Swl = 30mSwl = 30m..Dwl = 50mDwl = 50m..
Pipe = 2 inchPipe = 2 inch.. Output of the well = 5mOutput of the well = 5m33/ hour for 12 hour / day/ hour for 12 hour / day . .
Total pumping head (hTotal pumping head (hTT) = 50 +(6m) = 56m) = 50 +(6m) = 56m . .Total water capacity needed per day =5*12 =60Total water capacity needed per day =5*12 =60
EEhh =0.002725×V × h =0.002725×V × hTT
= = 0.0027250.002725××6060××5656 = =9.156kwh / day9.156kwh / day
E E p in p in =(9.156 kwh /day)/.55 =16.65 kwh/day=(9.156 kwh /day)/.55 =16.65 kwh/day..
E E ASM inASM in = (16.65 Kwh / day ) /.82= 20.3 kwh /day = (16.65 Kwh / day ) /.82= 20.3 kwh /day..