najah national university faculty of electrical engineering introduction to graduation project...
TRANSCRIPT
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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
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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
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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
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Chapter 2Chapter 2
The content of photovoltaic and diesel water The content of photovoltaic and diesel water pumping systempumping system
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Components of Photovoltaic water pumping system
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Components of Water Components of Water pumping system driven by a pumping system driven by a
diesel electric motordiesel electric motor
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Chapter 3Chapter 3 Data of well neededData of well needed
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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 )
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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.
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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
= = 1032510325$ $
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مساد جميلWell depth=45m.Swl = 12m.Dwl = 40m.Pipe = 2inch.
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$ $
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.. 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
Eh =0.002725×V × hTEh =0.002725×V × hT = = 0.0027250.002725××4848××5656 = =7.32kwh / day7.32kwh / day
E p in =13.3 kwh /dayE p in =13.3 kwh /day. . 1414
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E ASM in = 16.23 Kwh / dayE ASM in = 16.23 Kwh / day. .
E inv in = =17.45 kwh / dayE inv in = =17.45 kwh / day. . Psh =5.4 kwh /dayPsh =5.4 kwh /day..
1kwp →5.4 kwh /day1kwp →5.4 kwh /day . . → ? → ? 17.4517.45 kwh / daykwh / day..
PPV= 3.23 kwpPPV= 3.23 kwp . . The cost of Pv cells :- 3230 *2.5 =8075The cost of Pv cells :- 3230 *2.5 =8075
$ $Inverter cost = 2000Inverter cost = 2000 $ $
Motor pump set = 1200Motor pump set = 1200$ $ Piping and accessories = 420Piping and accessories = 420$ $
The capital cost = 8075 +2000 +1200 +420The capital cost = 8075 +2000 +1200 +420 = = 1169511695$ $ 1515
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عامودي عامودي احمد احمد
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 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..
1kwp →5.4 kwh /day1kwp →5.4 kwh /day . . → ? → ? 14.4914.49 kwh / daykwh / day..
PPV= 2.68 kwpPPV= 2.68 kwp . . ..
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$ $
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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..
1kwp →5.4 kwh /day1kwp →5.4 kwh /day . . → ? → ? 22 kwh / daykwh / day..
PPV= .37 kwpPPV= .37 kwp . . 1818
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The cost of Pv cells :- 370 * 2.5 =925The cost of Pv cells :- 370 * 2.5 =925 $ $
Inverter cost = 1300Inverter cost = 1300 $ $ Motor pump set = 700Motor pump set = 700$ $
Piping and accessories = 300Piping and accessories = 300$ $ The capital cost = 925 +1300 +700 +300The capital cost = 925 +1300 +700 +300
= = 32253225$ $
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العامودي العامودي محمود محمود
Well depth=Well depth=770m0mSwl = 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 = =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..
1kwp →5.4 kwh /day1kwp →5.4 kwh /day . . → ? → ? 15.4215.42 kwh / daykwh / day..
PPV= 2.86 kwpPPV= 2.86 kwp . . 2020
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The cost of Pv cells :- 2860 * 2.5 =7150The cost of Pv cells :- 2860 * 2.5 =7150 $ $
Inverter cost = 1800Inverter cost = 1800 $ $ Motor pump set = 1000Motor pump set = 1000$ $
Piping and accessories = 400Piping and accessories = 400$ $ The capital cost = 7150 +1800 +1000 +400The capital cost = 7150 +1800 +1000 +400
= = 3035030350$ $
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عباس ابراهيم عمر عباس كامل ابراهيم عمر كامل
Well depth=50 mWell depth=50 mSwl = 18 mSwl = 18 m..Dwl = 32mDwl = 32m..
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..
1kwp →5.4 kwh /day1kwp →5.4 kwh /day . . → ? → ? 49.349.3 kwh / daykwh / day..
PPV= 9.14 kwpPPV= 9.14 kwp . . 2222
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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$ $
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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
Eh =0.002725×V × hTEh =0.002725×V × hT = = 0.0027250.002725××3636× × 6666 = =6.47kwh / day6.47kwh / day
E p in =11.76kwh /dayE p in =11.76kwh /day. . E ASM in = 14.34 Kwh / dayE ASM in = 14.34 Kwh / 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 = 63.7 kwh =19.12/0.30 = 63.7 kwh
Number of liters diesel needed daily =Number of liters diesel needed daily = 63.7/10.5 = 6 63.7/10.5 = 6Diesel cost = 6* 6 NIS =36 NIS/dayDiesel cost = 6* 6 NIS =36 NIS/day
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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$ $
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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
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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$ $
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****************************************************************************************************** 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
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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$ $
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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
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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$ $
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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
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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$ $
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************************************************************************************************************** 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
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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$ $
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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
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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$ $
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Chapter 5Chapter 5The calculation of average value of yearly The calculation of average value of yearly
pumping waterpumping water((m3 / daym3 / day))
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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
→ →18.918.9 mm33 / day / day..FebruaryFebruary
EEpv outpv out = 2.85 ×3.2 =9.12 kwh = 2.85 ×3.2 =9.12 kwh
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
mm33 / day / day..MarchMarch
EEpv outpv out = 2.85 ×5.2 =14.82 kwh = 2.85 ×5.2 =14.82 kwh
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
/ day/ day..
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AprilApril
EEpv outpv out = 2.85 ×6.1 =17.38 kwh = 2.85 ×6.1 =17.38 kwh
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
17.3817.38 kwh / daykwh / day ? → ? →
Average value of daily pumping water during April =40.57 mAverage value of daily pumping water during April =40.57 m33 / day / day..
MayMay
EEpv outpv out = 2.85 ×7.6 =21.66 kwh = 2.85 ×7.6 =21.66 kwh
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
21.6621.66 kwh / daykwh / day ? → ? →
Average value of daily pumping water during May =50.56 mAverage value of daily pumping water during May =50.56 m33 / day / day . .
JuneJune
EEpv outpv out = 2.85 ×8.2 =23.37 kwh = 2.85 ×8.2 =23.37 kwh
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
23.3723.37 kwh / daykwh / day ? → ? →
Average value of daily pumping water during june =54.56 mAverage value of daily pumping water during june =54.56 m33 / day / day.. 4040
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JulyJuly
EEpv outpv out = 2.85 ×8.1 =23.08 kwh = 2.85 ×8.1 =23.08 kwh
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
23.0823.08 kwh / daykwh / day ? → ? →
Average value of daily pumping water during July =53.88 mAverage value of daily pumping water during July =53.88 m33 / day / day . .
AugustAugust
EEpv outpv out = 2.85 ×8 =22.8 kwh = 2.85 ×8 =22.8 kwh
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
22.822.8 kwh / daykwh / day ? → ? →
Average value of daily pumping water during August =53.23 mAverage value of daily pumping water during August =53.23 m33 / day / day . .
SeptemberSeptember
EEpv outpv out = 2.85 ×6.2 =17.67 kwh = 2.85 ×6.2 =17.67 kwh
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
17.6717.67 kwh / daykwh / day ? → ? →
Average value of daily pumping water during September=41.25 mAverage value of daily pumping water during September=41.25 m33 / day / day . .
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OctoberOctober
EEpv outpv out = 2.85 ×4.7 =13.39 kwh = 2.85 ×4.7 =13.39 kwh
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
13.3913.39 kwh / daykwh / day ? → ? →
Average value of daily pumping water during October =31.26 mAverage value of daily pumping water during October =31.26 m33 / day / day . .
NovemberNovember
EEpv outpv out = 2.85 ×3.65 =10.4 kwh = 2.85 ×3.65 =10.4 kwh
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
10.410.4 kwh / daykwh / day ? → ? →
Average value of daily pumping water during November =24.28 mAverage value of daily pumping water during November =24.28 m33 / / dayday..
DecemberDecember
EEpv outpv out = 2.85 ×2.9=8.26 kwh = 2.85 ×2.9=8.26 kwh
15.4215.42 kwh / daykwh / day → 36 m → 36 m33 / day / day
8.268.26 kwh / daykwh / day ? → ? →
Average value of daily pumping water during December =19.28 mAverage value of daily pumping water during December =19.28 m33 / / dayday . .
Q Q year year = =(19.28+24.28+31.26 +41.25+34.59+40.57+53.23+53.88= =(19.28+24.28+31.26 +41.25+34.59+40.57+53.23+53.88
+ + 54.56+50.56+14.82+21.29+18.954.56+50.56+14.82+21.29+18.9×)×)3030
= = 13757.113757.1 mm33 / year / year..4242
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monthEpv outm3 / day
January8.1218.9
February9.1221.29
march14.8234.59
April17.3840.57
May21.6650.56
June23.8754.56
July23.0853.88
August22.853.23
September17.6741.25
October13.3931.26
November10.424.28
December8.2619.28
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مسا مسا جميل ددجميلQ Q year year = =
=(8.04+10.13+16.9+13.83+8.87+7.9+21.09+13.04+17.21=(8.04+10.13+16.9+13.83+8.87+7.9+21.09+13.04+17.21
+ + 22.2+22.43+22.7322.2+22.43+22.73×)×)3030==9856.119856.11 mm33 / ye / yearar
4545
monthEpv outm3 / day
January2.367.9
February 8.8713.83
march4.3113.83
April 5.0516.9
May6.321.09
June 6.7922.73
July 6.722.43
August 6.6322.2
September5.1417.21
October3.8913.04
November3.0210.13
December2.48.04
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غانم غانم خليل خليلQ Q year year = =
=(25.77+54.19+46.18+28.4+25.3+32.43+67.53+72.83=(25.77+54.19+46.18+28.4+25.3+32.43+67.53+72.83+++ + 71.96+71.07+55.04+41.7571.96+71.07+55.04+41.75×)×)3030==15998.115998.1 mm33 / /
yearyear..
4747
MonthEpv outm3 / day
January9.225.3
February 10.3328.4
march16.7946.18
April 19.754.19
May 24.5567.53
June 26.4872.83
July 26.1671.96
August 25.8411.84
September20.0355.09
October15.1841.75
November11.7932.43
December9.3725.77
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عامودي عامودي احمد احمدQ Q year year = =
=(19.3+24.29+31.28+18.98+21.3+34.63+40.62+50.61=(19.3+24.29+31.28+18.98+21.3+34.63+40.62+50.61+++ + 54.58+53.9+53.27+41.2454.58+53.9+53.27+41.24×)×)3030==1332013320 mm33 / year / year..
4949
monthEpv outm3 / day
January7.6418.98
February 8.57621.3
march13.9434.63
April 16.3540.62
May 20.3750.61
June 21.9754.58
July 21.753.9
August 21.4453.27
September16.641.24
October12.5931.28
November9.7824.29
December7.7719.3
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أنسمساد أنسمساد..
Q Q year year = =(4.01+5.06+6.52+3.9+4.44+7.2+8.48+10.45 = =(4.01+5.06+6.52+3.9+4.44+7.2+8.48+10.45
+11.36+11.25+11.1+8.63)×30=2774.7 m+11.36+11.25+11.1+8.63)×30=2774.7 m33 / year / year..
5151
monthEpv outm3 / day
January1.053.9
February 1.1844.44
march1.927.2
April 2.268.84
May 2.8110.54
June 3.0311.36
July 311.25
August 2.9611.1
September2.38.63
October1.746.52
November1.355.06
December1.074.01
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monthEpv outm3 / day
January8.1519.03
February 9.1521.36
march14.8734.72
April 17.4540.74
May 21.7450.75
June 23.4554.75
July 23.254.08
August 22.8853.42
September17.7341.39
October13.4431.38
November10.4424.37
December8.2919.35
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Chapter 6Chapter 6
Economical evaluation of Economical evaluation of energy supply systemsenergy supply systems
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6-1 The economic evaluation of Pv power system for wells:
عوي عوي رائد صصرائدEvaluation of the cost of 1 KWhEvaluation of the cost of 1 KWh
Total energy out from the pv syste=2361.55 Total energy out from the pv syste=2361.55 KWh/yearKWh/year..
The total fixed cost of this system = 10325The total fixed cost of this system = 10325$ $
Depreciation factor = 10Depreciation factor = 10. % . %
Maintenance cost = 2.5 % fixed costMaintenance cost = 2.5 % fixed cost. .
Annual fixed cost = depreciation factor * total fixed Annual fixed cost = depreciation factor * total fixed costcost
Annual fixed cost = 0.1*10325 = 1032.5$Annual fixed cost = 0.1*10325 = 1032.5$. .
Annual running cost = maintenance costAnnual running cost = maintenance cost
Annual running cost = 0.025 * 10325 = 258.125Annual running cost = 0.025 * 10325 = 258.125 .$ .$
Total annual cost =total Annual fixed cost + total Total annual cost =total Annual fixed cost + total Annual running costAnnual running cost..
Total annual cost = 1032.5 + 258.125 = 1290.625Total annual cost = 1032.5 + 258.125 = 1290.625$ $ ..
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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
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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
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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
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Chapter 7Chapter 7Net present value ( NPV )Net present value ( NPV )
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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..
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Net Present ValueNet Present Value For Raad wellFor Raad well
NPV= 520(P/F,8%,20) +4400 (P/A,8%,20) -NPV= 520(P/F,8%,20) +4400 (P/A,8%,20) -250(P/A,8%,20) -10325 =30473.3250(P/A,8%,20) -10325 =30473.3$ $
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For jameel wellFor jameel well
NPV= 230(P/F,8%,20) +1500 (P/A,8%,20) -NPV= 230(P/F,8%,20) +1500 (P/A,8%,20) -110(P/A,8%,20) -4572.5=9123.85110(P/A,8%,20) -4572.5=9123.85$ $
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For khaleel wellFor khaleel well
NPV= 600(P/F,8%,20) +4900 (P/A,8%,20) -NPV= 600(P/F,8%,20) +4900 (P/A,8%,20) -290(P/A,8%,20) -11695=33694.6290(P/A,8%,20) -11695=33694.6$ $
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For Ahmad wellFor Ahmad well
NPV= 500(P/F,8%,20) +4000 (P/A,8%,20) -NPV= 500(P/F,8%,20) +4000 (P/A,8%,20) -250(P/A,8%,20) -9900=27024.75250(P/A,8%,20) -9900=27024.75 $ $
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For AnasFor Anas
NPV = 160(P/F,8%,20) +1000 (P/A,8%,20) -NPV = 160(P/F,8%,20) +1000 (P/A,8%,20) -80(P/A,8%,20) -3225=5841.8880(P/A,8%,20) -3225=5841.88$ $
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for Mahmood wellfor Mahmood well
NPV = 510(P/F,8%,20) +4400 (P/A,8%,20) -NPV = 510(P/F,8%,20) +4400 (P/A,8%,20) -260(P/A,8%,20) -10350=30405.9$260(P/A,8%,20) -10350=30405.9$
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Chapter 8Chapter 8Life cycle costLife cycle cost
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Life cycle costLife cycle costFor Raad wellFor Raad well
PV systemPV system
PW = 520(P/F,8%,20) -260(P/A,8%,20) -10325= -12766.14PW = 520(P/F,8%,20) -260(P/A,8%,20) -10325= -12766.14 $$
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Diesel systemDiesel system
PW = 230(P/F,8%,20) - 4400(P/A,8%,20) -PW = 230(P/F,8%,20) - 4400(P/A,8%,20) -4522-4292(P/F,8%,10)= -49659.94522-4292(P/F,8%,10)= -49659.9$ $
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For Jameel wellFor Jameel well
PV systemPV system
pw= 230(P/F,8%,20) -110(P/A, 8%,20) -4572= pw= 230(P/F,8%,20) -110(P/A, 8%,20) -4572= -5602,645-5602,645$ $
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Diesel systemDiesel system
PW = 220(P/F,8%,20) - 1500(P/A,8%,20) -PW = 220(P/F,8%,20) - 1500(P/A,8%,20) -4225-4005(P/F,8%,10)= -20759.94225-4005(P/F,8%,10)= -20759.9$ $
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For Ahmad wellFor Ahmad well
PV systemPV system
PW = 500(P/F,8%,20) -250(P/A, 8%,20) -PW = 500(P/F,8%,20) -250(P/A, 8%,20) -9900= -122479900= -12247$ $
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Diesel systemDiesel system
PW = 230(P/F,8%,20) - 4400(P/A,8%,20) -PW = 230(P/F,8%,20) - 4400(P/A,8%,20) -4522-4292(P/F,8%,10)= -49659.94522-4292(P/F,8%,10)= -49659.9$ $
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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
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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..
EEinv ininv in=21.82=21.82
11 kWkWpp ----------- ----------- 5.4 kwh/day 5.4 kwh/day----------------- ??----------------- ?? 21.82 kwh/day 21.82 kwh/day
21.8221.82 kwh/day --------------kwh/day --------------60 m60 m33/day/dayPpv = 4.04Ppv = 4.04
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22229.122229.1 mm33 / year / year ==Q Q yearyear
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average value of daily pumping watermonths
31.66January
35.55February
57.76March
67.77April
84.43May
91.09June
89.98July
88.87August
68.88September
52.21October
40.55November
32.22December
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Total energy from pv systemTotal energy from pv system EEhh*365=3341.94 kwh/year*365=3341.94 kwh/year
11 wwp p -------------- --------------2.5$2.5$
4040w4040wpp ------------ ------------??10100$10100$= = cost of cellscost of cells
Capital cost = cost cells+ cost inv+cost motor pupm + Capital cost = cost cells+ cost inv+cost motor pupm + accessories costaccessories cost
Capital cost= 10100+2200+1400+450 = 14150$Capital cost= 10100+2200+1400+450 = 14150$Annual fixed cost = .1 *capital cost = .1* 14150 =1415$Annual fixed cost = .1 *capital cost = .1* 14150 =1415$Maintenance cost = .025* capital cost = .025* 14150= 353.75 Maintenance cost = .025* capital cost = .025* 14150= 353.75
$/year$/yearTotal annual cost= total fixed cost + total running costTotal annual cost= total fixed cost + total running cost
= =1415+353.751415+353.75==1768.751768.75/$ /$ yearyearCost of (1kwh)= total annual cost /total energy Cost of (1kwh)= total annual cost /total energy
=1768.75/3341.94=.53 $/year=1768.75/3341.94=.53 $/year/$/$mm33 = total annual cost/ total pumping rate = 1768.75/22229.1 = total annual cost/ total pumping rate = 1768.75/22229.1
=.0796 $/m=.0796 $/m33
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3535 mm33/day/dayWell depth=60mWell depth=60m..
Swl = 30mSwl = 30m..Dwl = 50mDwl = 50m..
Pipe = 2inchPipe = 2inch..Compute of the well = mCompute of the well = m33/ hour ,12 hour / / hour ,12 hour /
dayday . . Total pumping head (ht) = 50 +(6m) = 56mTotal pumping head (ht) = 50 +(6m) = 56m . .
Total water capacity needed per day =3 ×12=36 Total water capacity needed per day =3 ×12=36 mm33
EEhh =0.002725×V × h =0.002725×V × hTT
= = 0.0027250.002725××3535× × 5656 = =5.3415.341 kwh / daykwh / day
E E p in p in =5.341 /.55 =9.7 kwh /=5.341 /.55 =9.7 kwh /dayday. .
E E asm inasm in =9.7 / .82 = 11.83 Kwh / day =9.7 / .82 = 11.83 Kwh / day. .
E E inv ininv in =11.83 /.92 =12.86 kwh / day =11.83 /.92 =12.86 kwh / day
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PPshsh =5.4 kwh /day =5.4 kwh /day..
1kw1kwp p → 5.4 kwh /day → 5.4 kwh /day.. → ? → ? 12.8612.86 kwh / daykwh / day..
PPPVPV= 2.38 kwp= 2.38 kwp. . Pv Pv → → 1 wp 1 wp = 2.5 = 2.5$ $
Q Q year year =12948.9 m=12948.9 m33 / year / year..
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average value of daily pumping watermonths
18.48January
20.7February
33.86March
39.5April
48.98May
53June
52.46July
51.82August
40September
30.44October
23.64November
18.78December
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Total energy from pv systemTotal energy from pv system EEhh*365=1949.465 kwh/year*365=1949.465 kwh/year
11 wwp p -------------- --------------2.5$2.5$2380w2380wpp ------------ ------------??59505950= $ = $ cost of cellscost of cells
Capital cost = cost cells+ cost inv+cost Capital cost = cost cells+ cost inv+cost motor pupm + accessories costmotor pupm + accessories cost
Capital cost=5950+1780+1000+400 Capital cost=5950+1780+1000+400 =9130$=9130$
Annual fixed cost = .1 *capital cost Annual fixed cost = .1 *capital cost = .1*9130=9138$= .1*9130=9138$
Maintenance cost = .025* capital cost Maintenance cost = .025* capital cost = .025* 9130= 228.25$/year= .025* 9130= 228.25$/year 8181
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Total annual cost= total fixed cost + Total annual cost= total fixed cost + total running costtotal running cost
= =913+228.25913+228.25==1141.251141.25/$ /$ yearyearCost of (1kwh)= total annual cost /total Cost of (1kwh)= total annual cost /total
energy =1141.25/1949.465=.585$/yearenergy =1141.25/1949.465=.585$/year/$/$mm33 = total annual cost/ total pumping = total annual cost/ total pumping
rate = 1141.25/12948.9 =.088 $/mrate = 1141.25/12948.9 =.088 $/m33
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