Photovoltaic Systems Engineeringprofsite.um.ac.ir/~karimpor/pv/lec10_Economic.pdf · Photovoltaic Systems Engineering Third Edition CRC Roger Messenger, Jerry Ventre Ali Karimpour

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Photovoltaic Systems Engineering

Ali KarimpourAssistant ProfessorAssistant Professor

Ferdowsi University of Mashhad

Reference for this lecture:Photovoltaic Systems Engineering Third Edition CRC Roger Messenger, Jerry Ventre

Ali Karimpour May 2012

Engineering Economy Tenth Edition Prentice Hall, E. Paul DeGarmo, William G. Sullivan, James A. Bontadelli, and Elin M. Wicks.

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Lecture 10Lecture 10

Economic ConsiderationsIntroduction

* Consumer price index * Inflation and deflation ratep

* Discount or Interest rate * Cash flow

Life Cycle Costing (LCC)

Annual Life Cycle Costing (ALCC)Annual Life Cycle Costing (ALCC)

Capital Cost

Unit Electricity Cost(Levelized Cost)


2Payback period and rate of return

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IntroductionIntroduction

Consumer Price Index(CPI)

Consumer Price Index: CPI is a composite price index thatp pmeasures price changes in food, shelter, medical care, transportation, apparel, andtransportation, apparel, and other selected goods andservices used by individuals and familiesand families.


3http://www.cbi.ir/category/1624.aspx

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Inflation and deflation rate

General price inflation: General price inflation is defined as anpincrease in the price paid forgoods and services bringing about a reduction power ofabout a reduction power of purchasing power of the monetary unit.

100)(

)()(

1

1 ×−

=−

−

k

kk

CPICPICPIi

100)(

)()(

1387

13871388 ×−

=CPI

CPICPIi


47.10%1003.183

3.183203=×

−=

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Example 1: Suppose one want to change a battery after 5 p pp g yyears and the present price is100$. Find the price ofbattery, if the inflation rate be 15%.

5432 )1()1()1()1()1( iCiCiCiCiCC +=+=+=+=+= 012345 )1()1()1()1()1( iCiCiCiCiCC +=+=+=+=+=

$201)15.01(100 55 =+=C


5

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Discount or Interest rateThe amount charged, expressed as a percentage of principal, by a lender to a borrower for the use of assets. Interest rates are typically noted on an annual basis known as the annual percentagetypically noted on an annual basis, known as the annual percentage rate (APR).

)1(1 dCC kk += )1(1 dCC kk ++

d is discount or interest rate.

Ck is the real value of an asset in the kth period.


6

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Example 2: Suppose one diposit1000$ for a period of 4 p pp p pyears with a discount rate 17%, what is the value of his money after 4 years?

40

31

2234 )1()1()1()1( dCdCdCdCC +=+=+=+=

$1874)17.01(1000 44 =+=C

Example 3: Suppose one diposit10,000,000 Rials for a HajExample 3: Suppose one diposit10,000,000 Rials for a Hajprogram in a discount rate of 16%, what is the value of his money after 80 months?

RialsC 690,852,28)1217.01(000,000,10 80

80 =+=


7

12

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Cash flow

Cash flow is the movement of money into or out of a business, project, or financial product. It is usually measured during a specified, finite

i d f iperiod of time.


8

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Borrowing Money

Suppose we borrow C0

)( 10201 ACiACiA −+=+ 12 )1( AiA +=⇒ 11 )1()1( AiAiA nnn +=+= −

Balance:

( ) 012

121 )1(...)1()1(1... CiiiAAAA nn =+++++++=+++ −

i )1( + ni


91)1(01 −+= ni

iCA1)1(

)1(010 −+

+=+= n

n

iiiCAiCPMTANN

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IntroductionExample 4: Suppose we borrow 70 000 000 Rials in 14% interest rate

d i 5 i d

Introduction

and in a 5 years period. a) Determine the monthly payment.

liChln

86281)1( +

01167.0%167.11214%,000,000,700 ==== iRialsC

Rialsi

iiCPMTMonthly n 778,628,11)1(

)1(0 =

−++

=

b) Determine the settlement payment after 2.5 years.

?)...( 30210 =++− AAAC1)1(01 −+

= niiCA

1)1()1()1(1)1(

1)1())1(...)1(1(

30

0

30

0029

10 −++−+

=−+

−+−=+++++− n

n

n iiiC

ii

iiCCiiAC


10835,028,411)1(

)1()1(paymentSettlement 60

3060

0 =−++−+

=i

iiC

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Levelized Cost of Electricity

Capital Cost(Levelized Cost)



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Life Cycle Costing (LCC)y g

LCC is a sensible means of evaluating any purchase

Acquisition costs

LCC is a sensible means of evaluating any purchase option when a decision has been made to purchase.

q

Maintenance costsAt the start andduring operation

Replacement costs

during operation

p

At the endSalvage value

At the endDecommissioning cost

d i i hi d hiAli Karimpour May 2012

12And… Loan, incentives, third party ownership, externalities, …

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13

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14

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The time value of money

let Inflation rate (i ) and Discount or interest rate of (d):niCnC )1()( += iCnC )1()( 0 +=

ndNN )1()(

C0 is the present value of an asset, C(n) is the value of asset after n years. ndNnN )1()( 0 +=

N0 is the present money at hand, N(n) is the money after n years.

001 CPCiPW r

n

=⎠⎞

⎜⎝⎛ +

=Present worth

001 d r⎠

⎜⎝ +

Present money needed to buy and asset n years later in


15Inflation rate of i and discount or interest rate of d

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Life Cycle Costing (LCC)Life Cycle Costing (LCC)

Example 5: Suppose one want to change a battery after 5 p pp g yyears and the present price is100$. a) Find the price of battery, if the inflation rate be 15%.b) Fi d h l f h d d b hb) Find the present value of the money needed to bye the

battery.(i=15% and d=20%)

$201)1501(100 5+C $201)15.01(100 55 =+=C

5

$83.801002.0115.01 5

0 =⎟⎠⎞

⎜⎝⎛

++

== CPPW r


16

⎠⎝

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Cumulative Present Worth

0

1

0

2

00 11...

11

11 C

diC

diC

diCPW

n−

⎟⎠⎞

⎜⎝⎛++

++⎟⎠⎞

⎜⎝⎛++

+⎟⎠⎞

⎜⎝⎛++

+=111 ddd ⎠⎝ +⎠⎝ +⎠⎝ +

( ) 012

0 11...1 CxxxxCPW

nn −

=++++= −( ) 00 1 x−

P

PW

aP

Cumulative Present Worth FactorPW

( ) 02

01... CxxxxxCPW

nn −=+++=


17

( ) 00 1 x−

aa xPP =1

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Annual Life Cycle Costing (ALCC)y g

LCC (Life Cycle Costing):

ALCC (Annual Life Cycle Costing):ALCC (Annual Life Cycle Costing):


19

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Levelized Cost of Electricity (LCOE)y

ALCC (Annual Life Cycle Costing):

Unit Electricity Cost (Levelized Cost of Energy (LCOE) )Unit Electricity Cost (Levelized Cost of Energy (LCOE) )

The LCOE is the minimum price at which energy must be ld f j h l f

)/$/($ALCCLCOE KWhorMWh=

sold for an energy project to have present value of zero.


20

)/$/($Energy ProducedYearly

LCOE KWhorMWh=

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LCC

T=0

Capital CostT=0T 0

CostCapatal


21)/$/($

CapacityCostCapatalCost(NCC)CapatalNormalized KWorMW=

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Example 6: Buying a suitable refrigerator

Refrigerator A:

Cost: 600 $ (800 W) 150 kWh/month no repair for 10 years

Refrigerator B:

Cost: 800 $(650 W) 120 kWh/month no repair for 10 years

Suppose discount rate is 10% and inflation rate is 3%.

Which one is better of the cost of electricity is 0.07$/kWh?


x= (1+i)/(1+d)=1 03/1 1=0 9364 suppose:PW


22

x= (1+i)/(1+d)=1.03/1.1=0.9364, suppose:

Pa=7.57

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Example 6: Buying a suitable refrigeratorp y g g


600 600 800 800600 600 800 800

126 954 100.8 763

Which one is better of the cost of electricity is 0.15$/kWh?1554 1563

y $

600 600 800 800

270 2045 216 1636


232645 2436

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Example 6: Buying a suitable refrigeratorp y g g

LCC, Capital Cost, NCC, ALCC and LCOE (ele. Price. 0.07$/kWh)

1554 1563

600 800

No meaning No meaningNo meaning No meaning

205 206


24No meaning No meaning

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Example 7: Choosing suitable light bulb for home

Suppose we need 18 light bulb each uses 3 h/days averagely.

Light A (Incandescent):60 W, 1000 hours(life time) , 400 T(price)

Light B (Energy saving): 11 W, 8000 hours(life time) , 3400 T(price)

Li ht C (LED) 3 W 30000 h (lif ti ) 12000 T( i )Light C (LED): 3 W, 30000 hours(life time) , 12000 T(price)

Electricity price is 120 T/kwh

Suppose discount rate is 18% and inflation rate is 15%.

x= (1+i)/(1+d)=1.15/1.18=0.9746


25n=27, Pa=19.715

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Example 7: Choosing suitable light bulb for home

Which light bulb is better ?

First PW First PW First PW

Bulb A Bulb B Bulb C

year year year

Purchase Price (T)

18×400 7200 18×3400 61200 18×12000 216000Price (T)

Replace Price (T) 157720 2.5×3400 167577 - -20×400Price (T)

Electrical Price (T) 141912 26017

0.162×365×120 7095

3.24×365×120

0.594×365×120Price (T)

LCC (T)

141912 26017 ×120 7095×120 ×120

306832 254794 223095


( )

26

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Example 8: PV or gasoline generator to power a warning sign

We need a 24 hours/day warning sign with 2 kWh/day and 20 year lifetime20-year lifetime.

981.011 15.7% of rateinflation and 18% of ratediscount Suppose =++

=dix

2 kWh/day 83 W system

Option #1 Option #2

500 W array 4 $/watty

900 $ storage battery. 500 W gasoline generator 250$

2 kWh/gal 365 gal/yearCh i b tt i h 5

300 $ Charge controller.

2 kWh/gal 365 gal/year

2 $/gal

Changing batteries each 5 years.

100 $/year maintenance. 400 $/year maintenance.


27Changing generator each 5 years.

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4516177161Pr981011 =

−==

−===

+=

xxPxPxixnn

n 45.161

,77.161

,Pr,981.01

1 =−

==−

===+

=x

xPx

Pxd

x aa

2000 2502000 2502000300900900

250

250250

2000300900818

250

2272065x

5x10x900

900900100

250250730400

7436751645

06187

122426708

x

1aP

10x15x

15xaP

aP

198207081


28

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Determine LCC, Capital Cost, NCC, ALCC and LCOE

198207081

2503200

250/0.5=5003200/0.5=6400

1182422


29

1182422

1182/730=1.61422/730=0.58

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30Payback period

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Payback Period and Rate of Return

Payback Periody)costCapital(0C

nAAAA 321

0min 0 ≥⎠

⎞⎜⎝

⎛−∑

θ

θ k CA1

0⎠⎝

∑=

θ kk

Discounted Payback Period ?


31

y

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Payback Period

Example 9: A capital investment of 1500 MT can be made on a biomass l t S 10 lif ti f th l t d 10 MT (fi d)plant. Suppose a 10 years life time for the plant and 10 MT (fixed)

expense as annual O&M and 50 MT as a annual revenue(fixed revenue) . Suppose the expense of overhaul after 5 years is 300 MT(fixed). Let i=15% and d=20%.

a) Draw the cash flow of project. b) Find LCC of project.

c) Is the project financially profitable? d) Calculate payback period?

1500

300

10


32550

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Payback Period

a) Draw the cash flow of project. b) Find LCC of project.

c) Is the project financially profitable? d) Calculate payback period?

15001500

30010

55010

MTPxLCC a 1662103001500 15 =++=19.4

11,833.0

2.011

10

1 =−−

==+

=x

xxPx a

5642550Revenue × LCCP 78201500540min =→≥⎞

⎜⎛∑ θ

θ


33

5.642550Revenue 1 =−×= LCCPa 78.201500540min1

=→≥⎠

⎜⎝

−∑=

θθ k

k

What about repair price?

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Payback Period

Example 10: Consider the biggest solar plant of Mashhad.

Detail of project(25 years life time):216 panel (200 W) each 630 KT totally 136 MT.18 T k h 6 5 MT t t l 117MT18 Tracker each 6.5 MT total 117MT6 inverter each 7KW each 7MT totally 42 MTFoundation 25 MT------------------------320 MT in 43.2 KW no maintenance and let i=15% and d=20%.

Suppose yearly average PSH is 5 hours/day and each kWh is 150 TSuppose yearly average PSH is 5 hours/day and each kWh is 150 Ta) Find LCC of the project. b) Draw the cash flow of project.

) I h j fi i ll fi bl ? d) C l l b k i d?c) Is the project financially profitable? d) Calculate payback period?

LCC=Capital cost is 320 MT Yearly energy=43.2×5×365=78840 kWh


34Yearly income=150×78840=11.8 MT

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Payback Period

Example 10: Consider the biggest solar plant of Mashhad.

a) Find LCC of the project. b) Draw the cash flow of project.

) I th j t fi i ll fit bl ? d) Calc late pa back period?c) Is the project financially profitable? d) Calculate payback period?

LCC=Capital cost is 320 MT Yearly energy=43.2×5×365=78840 kWh

Yearly income=150×78840=11.8 MT

320 11501 25+ x32006.15

11,958.0

2.0115.01 1 =

−−

==++

=x

xxPx a

MTLCCP 142320177811Revenue 1 −=−=−×= MTLCCPa 1423201778.11Revenue 1×

moneyback pay No


358.11…..

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Photovoltaic Systems Engineeringprofsite.um.ac.ir/~karimpor/pv/lec10_Economic.pdf · Photovoltaic Systems Engineering Third Edition CRC Roger Messenger, Jerry Ventre Ali Karimpour