najah national university faculty of electrical engineering

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

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

= = 1032510325$ $

1111

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 .

1212

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

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

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

عامودي عامودي احمد احمد

Well depth=60mWell depth=60m..Swl = 20mSwl = 20m..Dwl = 56mDwl = 56m..


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. .

1616

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..


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 $ $


The capital cost = 1800 +1000 +400 +5170The capital cost = 1800 +1000 +400 +5170 = = 83708370$ $

1717

أنسمساد أنسمساد ..Well depth=Well depth=440m0m


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

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$ $

1919

العامودي العامودي محمود محمود

Well depth=Well depth=770m0mSwl = 30mSwl = 30m..Dwl = 60mDwl = 60m..


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..


PPV= 2.86 kwpPPV= 2.86 kwp . . 2020


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$ $

2121

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..


PPV= 9.14 kwpPPV= 9.14 kwp . . 2222


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..


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

2424

COST OF EQUIPMENTCOST OF EQUIPMENTDiesel and engine and S.G = 3125Diesel and engine and S.G = 3125$ $


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


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


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


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


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

→ →18.918.9 mm33 / day / day..FebruaryFebruary

EEpv outpv out = 2.85 ×3.2 =9.12 kwh = 2.85 ×3.2 =9.12 kwh


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



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..

3939

AprilApril



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



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



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

JulyJuly



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


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



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 . .

4141

OctoberOctober



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



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


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

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

4343

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


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

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

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


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

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


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

5353


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

Chapter 6Chapter 6

Economical evaluation of Economical evaluation of energy supply systemsenergy supply systems

5555

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$ $ ..

5656

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

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$ $

6262

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$ $

6363

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$ $

6464

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 $ $

6565

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$ $

6666

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$260(P/A,8%,20) -10350=30405.9$

6767

Chapter 8Chapter 8Life cycle costLife cycle cost

6868

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 $$

6969

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$ $

7070

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$ $

7171



7272

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$ $

7373



7474

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..


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

7676

22229.122229.1 mm33 / year / year ==Q Q yearyear

7777

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

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

7878

3535 mm33/day/dayWell depth=60mWell depth=60m..


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

7979

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..

8080

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

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

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

8282

najah national university faculty of electrical engineering

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