Idealized Power Curve Cut in windspeed, rated windspeed, cut-out - - PowerPoint PPT Presentation

Idealized Power Curve

Cut –in windspeed, rated windspeed, cut-out windspeed

Figure 6.32

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Idealized Power Curve

• Before the cut-in windspeed, no net power is generated
• Then, power rises like the cube of windspeed
• After the rated windspeed is reached, the wind turbine
• perates at rated power (sheds excess wind)
• Three common approaches to shed excess wind
• Pitch control – physically adjust blade pitch to reduce

angle of attack

• Stall control (passive) – blades are designed to

automatically reduce efficiency in high winds

• Active stall control – physically adjust blade pitch to

create stall

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Idealized Power Curve

• Above cut-out or furling windspeed, the wind is too

strong to operate the turbine safely, machine is shut down, output power is zero

• “Furling” –refers to folding up the sails when winds

are too strong in sailing

• Rotor can be stopped by rotating the blades to

purposely create a stall

• Once the rotor is stopped, a mechanical brake locks

the rotor shaft in place

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Average Power in the Wind

• How much energy can we expect from a wind turbine?
• To figure out average power in the wind, we need to know

the average value of the cube of velocity:

• This is why we can’t use average windspeed vavg to find the

average power in the wind

 

3 3

1 1 (6.29) 2 2

avg avg avg

P Av A v          

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Example Windspeed Site Data

Figure 6.22

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Wind Probability Density Functions

Windspeed probability density function (p.d.f) – between 0 and 1, area under the curve is equal to 1

Figure 6.23

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Temperature Correction for Air Density When wind power data are presented, it is often assumed that the air density is 1.225 kg/m3; that is, it is assumed that air temperature is 15◦C (59◦ F) and pressure is 1 atmosphere

Altitude Correction for Air Density

IMPACT OF TOWER HEIGHT

An anemometer mounted at a height of 10 m above a surface with crops, hedges, and shrubs shows a windspeed of 5 m/s. Estimate the windspeed and the specific power in the wind at a height of 50 m. Assume 15◦C and 1 atm of pressure.

• Not all of the power in the wind is retained - the rotor spills high-speed winds and low-speed winds are too slow to overcome losses
• Depends on rotor, gearbox, generator, tower, controls, terrain, and the wind
• Overall conversion efficiency (Cp·ηg) is around 30%

Estimates of Wind Turbine Energy

W

P

B

P

E

P

Power in the Wind Power Extracted by Blades Power to Electricity

P

C

Rotor Gearbox & Generator

g



Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Power Mass flow rate

Maximum Rotor Efficiency

Rotor Tip=Speed Ratio

A 40-m, three bladed wind turbine produces 600 kW at a wind speed of 14 m/s. Air density is the standard 1.225 kg/m3. Under these conditions,

• a. At what rpm does the rotor turn when it operates with a TSR of 4.0?
• b. What is the tip speed of the rotor?
• c. If the generator needs to turn at 1800 rpm, what gear ratio is needed to match the rotor speed to the

generator speed?

• d. What is the efficiency of the complete wind turbine (blades, gear box,generator) under these conditions?

Economies of Scale

• Presently large wind farms produce electricity more economically

than small operations

• Factors that contribute to lower costs are
• Wind power is proportional to the area covered by the blade

(square of diameter) while tower costs vary with a value less than the square of the diameter

• Larger blades are higher, permitting access to faster winds
• Fixed costs associated with construction (permitting,

management) are spread over more MWs of capacity

• Efficiencies in managing larger wind farms typically result in

lower O&M costs (on-site staff reduces travel costs)

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

• The economic evaluation of a renewable energy resource

requires a meaningful quantification of cost elements

• fixed costs
• variable costs
• We use engineering economics notions for this purpose

since they provide the means to compare on a consistent basis

• two different projects; or,
• the costs with and without a given project

Energy Economic Concepts

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

• Basic notion: a dollar today is not the same

as a dollar in a year

• Would you rather have $10 now or $50 in five

years?

• What would a $50,000 purchase you’ll make in

10 years be worth today?

• The convention we use is that payments
• ccur at the end of each period

Time Value of Money

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Time Value of Money – Principle and Interest

• Principle – the initial sum
• Interest – productivity of money over time, money

today vs. money tomorrow

• Simple interest – not compounded, interest is only paid on

the principle amount

• Compound interest - when interest is also paid on the

interest (what we consider)

• Difference between the two is greater when: the interest

rate is higher, compounding is more frequent, duration of payments is longer

P = principal i = interest value

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Compound Interest

e.o.p. amount owed interest for next period amount owed for next period P Pi P + Pi = P(1+i ) 1 P(1+i ) P(1+i ) i P(1+i ) + P(1+i ) i = P(1+i ) 2 2 P(1+i ) 2 P(1+i ) 2 i P(1+i ) 2 + P(1+i ) 2 i = P(1+i ) 3 3 P(1+i ) 3 P(1+i ) 3 i P(1+i ) 3 + P(1+i ) 3 i = P(1+i ) 4 n-1 P(1+i ) n-1 P(1+i ) n-1 i P(1+i ) n-1 + P(1+i ) n-1 i = P(1+i ) n n P(1+i ) n

The value in the last column for the e.o.p. (k-1) provides the value in the first column for the e.o.p. k (e.o.p. is end of period)

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Terminology

• We call (1 + i) n the single payment compound amount

factor

• We define

and is the single payment present worth factor

• F is called the future worth; P is called the present worth
• r present value at interest i of a future sum F

 

1

1 i 





 

1

n n

i 



 

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Loan P for 1 year year 0 P repay P + iP = P ( 1 + i ) at the end of 1 year Loan P for n years year 0 P year 1 ( 1 + i ) P repay/reborrow year 2 ( 1 + i )2 P repay/reborrow year 3 ( 1 + i )3 P repay/reborrow

.

year n ( 1 + i )n P repay

Example: Loan amount P, interest i

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Cash Flows

• A cash flow is a transfer of an amount A t from one

entity to another at e.o.p. time t

• Each cash flow has (1) amount, (2) time, and (3) sign

I take out a loan I make equal repayments for 4 years 1 2 3 4 Ex.

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Cash Flows Diagrams - Overview

1 2 3 4 Present End of year 1

Incoming cash flows

Initial purchase Payments made Take out a loan Revenue collected

Ex.

Ex.

Outgoing cash flows

Convention for cash flows  inflow  outflow

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Discount Rate

• The interest rate i is typically referred to as the

discount rate and is denoted by d

• In converting a future amount F to a present worth

P we can view the discount rate as the interest rate that can be earned from the best investment alternative

• A postulated savings of $ 10,000 in a project in 5

years is worth at present

 

5 5 5

10,000 1 P F d 



  

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Discount Rate

• For d = 0.1, P = $ 6,201,

while for d = 0.2, P = $ 4,019

• In general, the lower the discount factor, the

higher the present worth

• The present worth of a set of costs under a given

discount rate is called the life-cycle costs

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Equivalence

• We represent the time value of money by the standard

approach of discounted cash flows

• “Discounting” refers to moving cash flows to obtain an

equivalent amount in another year

• Two cash-flow sets

under a given discount rate d are said to be equivalent cash-flow sets if their worths, discounted to any point in time, are identical.

   

: 0,1,2,..., : 0,1,2,...,

a b t t

A t n A t n   and

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Equivalence, Example

• Are these cash-flow sets are equivalent with d = 7%?

1 2 3 a 4 5 6 7

2000 2000 2000 2000 2000

1 2 b

8,200.40

 

a t

A

 

b t

A

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Equivalence, cont.

• Let’s move each cash flow set to year 2
• Therefore, are equivalent cash flow sets

under d = 7%

   

a b t t

A A and

1 2 2 3 4 5

2000(1 ) 2000(1 ) 2000(1 ) 2000(1 ) 2000(1 ) = 8200.40 F i i i i i

    

         

Cash flow set a Cash flow set b

2

8200.40 F 

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Present and Future Value, Example

• Consider the set of cash flows illustrated below

1 2 3 4 5 6 7 8 $ 300 $ 300 $ 200 $ 400 $ 200 d = 6%

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Example, cont.

• We compute F 8 at t = 8 for d = 6%
• We next compute P
• We check that for d = 6%

       

7 5 8 4 2

300 1 .06 300 1 .06 200 1 .06 400 1 .06 200 951.56 F $          

         

1 3 4 6 8

300 1 .06 300 1 .06 200 1 .06 400 1 .06 200 1 .06 597.04 P $

    

          

 

8 8

597.04 1 .06 951.56 F $   

Future Value Present Value

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

• A capital investment, such as a renewable energy

project, requires funds, either borrowed from a bank, or obtained from investors, or taken from the

• wner’s own accounts
• Conceptually, we may view the investment as a loan

with interest rate i that converts the investment costs into a series of equal annual payments to pay back the loan with the interest

Annualized In Investment

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Annual Payments, Example

1 2 3 4 5 A $ 2000 i = 6% A A A A What value must A have to make these cash flows equivalent?

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Cash Flows, cont.

• We define the present worth P of the cash flow set as

 

1

n n t t t t t 0 t 0

P A A i 

  

  

 

 

1 1

1 0.06

n n t t t t

P A A 

  

  

 

1

(1 ) 1 (1 )

n n t n t

d d d 



   



(1 ) 2000 (A|P,6%,5) (1 ) 1

n n

d d A P d       $474.79 A 

Annualized Value

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Annualized Investment

• Then, the equal annual payments are given by
• The capital recovery factor, CRF(i,n), is the inverse of

the present value function PVF

• CRF measures the speed with which the initial

investment is repaid

• Capital recovery function in Microsoft Excel:

PMT(rate,nper,pv)

1

n

d A P   

Capital Recovery Factor

CRF( , ) (5.20) A P i n  

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Mortgage payment example

• What is the monthly payment for a 100K, 15 year mortgage

with a monthly interest rate of 0.5%?

• = PMT(0.005,180,100000)
• =$843.86 per month
• If terms are changed to 20 years payment goes to $716/month
• Assume a 100K investment in a PV installation with a 15

year life, monthly interest rate of 0.5%, and no O&M

• expenses. What is monthly income needed to cover the

loan?

• Solution is the same as above

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Present and Future Worth - Equivalence

 

1

n

F P i  

compound interest Lump sum repayment at the end of n

• periods. F is

called the future worth, while P is called the present worth Need not be integer-valued

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Infinite Horizon Cash-Flow Sets

• Consider a uniform cash-flow set with
• Then,

For an infinite horizon uniform cash-flow set

 

: , 1, 2, ...

t

A A t   n  

 

1 1

n

P A A n d d      A d P 

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Infinite Horizon Cash-Flow Sets, cont.

• We may view d as the capital recovery factor with the

following interpretation: For an initial investment of P, dP = A is the annual amount recovered in terms of returns on investment

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Internal Rate of Return

• We consider a cash-flow set
• The value of d for which

is called the internal rate of return (IRR)

• The IRR is a measure of how fast we recover an

investment or stated differently, the speed with which the returns recover an investment

 

: , 1, 2, ...

t

A A t  

n t t t 0

P A 



 



Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Internal Rate of Return Example

• Consider the following cash-flow set

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Internal Rate of Return

• The present value

has the (non-obvious) solution of d equal to about 12%.

• The interpretation is that under a 12% discount rate, the

present value of the cash flow set is 0 and so 12% is the IRR for the given cash- flow set

• The investment makes sense as long as other investments yield less

than 12%.

8

1 30,000 6,000 P d        

1 1 d   

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Internal Rate of Return

• Consider an infinite horizon simple investment
• Therefore
• For I = $ 1,000 and A = $ 200, d = 20% and we interpret

that the returns capture 20% of the investment each year or equivalently that we have a simple payback period of 5 years

I

A d I 

ratio of annual return to initial investment

A A A

1 2

. . .

n

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Efficient Refrigerator Example

• A more efficient refrigerator incurs an investment of

additional $ 1,000 but provides $ 200 of energy savings annually

• For a lifetime of 10 years, the IRR is computed from the

solution of

• r

10

1 1,000 200 d     

10

1 5 d   

The solution of this equation requires either an iterative approach or a value looked up from a table

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Efficient Refrigerator Example, cont.

• IRR tables show that

and so the IRR is approximately 15% If the refrigerator has an expected lifetime of 15 years this value becomes

10

15

1 5.02

d %

d 



 

15

18.4

1 5.00

d %

d 



 

As was mentioned earlier, the value is 20% if it lasts forever

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Impacts of Inflation

• Inflation is a general increase in the level of prices in

an economy; equivalently, we may view inflation as a general decline in the value of the purchasing power

• f money
• Inflation is measured using prices: different products

may have distinct escalation rates

• Typically, indices such as the CPI – the consumer

price index – use a market basket of goods and services as a proxy for the entire U.S. economy

• reference basis is the year 1967 with the price of $ 100 for

the basket (L 0); in the year 1990, the same basket cost $ 374 (L 23)

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Inflation (Escalation) Rate

• With escalation, an amount worth $1 in year zero becomes $(1+e) in

year 1, etc., so becomes

• We can compare terms to find an equivalent discount rate d’:

   

2

1 1 1 PVF( , ) + ... (5.8) 1+ 1+ 1+

n

d n d d d   

       

2 2

1+ 1+ 1+ PVF( , , ) + ... (5.13) 1+ 1+ 1+

n n

e e e d e n d d d    1+ 1 (5.14) 1+ 1+ ' e d d 

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Equivalent Discount Rate d’

• From
• We solve for d’ and obtain the following identities
• Now, inflation can be incorporated into all of the

present value relationships just by using d’ in lieu of d.

1+ 1 (5.14) 1+ 1+ ' e d d 

' (5.15) 1+ d e d e   ' 1+ ' d d e d  

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Inflation Example, 5.7

• What is the net present value of a premium motor that

costs an extra $500 initially and saves $192/yr (at current electricity prices) for 20 years if interest is 10% and inflation is 5%?

 

20 20

0.1 0.05 0.04762 1 0.05 1.04762 1 PVF 12.72 yrs 0.04762 1.04762 NPV PVF $192 / yrs 12.72yrs -$500 $1942 d A P                

Compare this to $1135 without fuel escalation In Excel - PV(0.04762,20,1)

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Cash-Flows Incorporating Inflation

• Cash-flows can be expressed either in terms of dollars that

take into account inflation (current, inflated, after inflation), or in terms of dollars that do not take into account inflation (constant, inflation free, before inflation).

• We’ll define the set of constant (inflation free) currency flows
• We’ll define the set of current (inflated) currency flows

 

: 1,2, ... ,

t

A t 0 , n 

 

: ,1,2, ... ,

t

W t n 

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Cash-Flows Incorporation Inflation

• We use the relationship for inflated (current) dollars
• r equivalently

with W t expressed in reference year 0 (today’s) dollars, and e giving the rate of inflation

 

1

t t t

A W e  

 

1

t t t

W A e



 

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Cash-Flows Incorporating Inflation

• Then, we have
• Therefore, the real interest rate d’ is used to discount the

constant (inflation free) flows while the buyer’s discount rate d is used for the inflated flows.

n t

P 



 

t t

A

         

1 1 1 1 1

n t t t t n t t n t t

e d e d d

    

        

  

t t t

W W W

 

• 1

1 Recall that 1 and 1 1 '

t t

e d d d       

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Cash-Flows Incorporating Inflation

• Whenever inflation is taken into account, it is

convenient to carry out the analysis in present worth rather than future worth or on a cash – flow basis

• Under inflation, e > 0, it follows that a uniform set
• f cash flows implies a real

decline in the cash flows

 

: 1,2, ... ,

t

A A t n  

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Inflation Calculation Example

• Consider an annual inflation rate, e = 4 %, and

assume the cost for a piece of equipment is constant for the next 3 years in terms of today’s $

• The corresponding cash flows in current $ are

 

1

1,000 1,000 1 .04 1,040 A $ A = $   

1 2 3

1,000 W W W W $    

   

2 2 3 3

1,000 1 .04 1,081.60 1,000 1 .04 1,124.86 A = $ A = $    

The interpretation is that $1,125 in three years has the same value as $1,000 today.

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

• Ex. 5.8: IRR for HVAC Retrofit with Inflation
• An energy efficiency retrofit of a commercial site

reduces the annual HVAC load consumption from 2.3 GWh to 0.8 GWh and the peak demand by 0.15 MW

• Electricity costs are 60 $/MWh and demand charges

are 7,000 $/(MWmo) and these prices escalate at an annual rate of e  5 %

• The retrofit requires a $ 500,000 investment today and

is planned to have a 15 – year lifetime

• Evaluate the IRR for this project

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

• Ex. 5.8: IRR for HVAC Retrofit with

Inflation

• The annual savings are
• The IRR0 (without fuel inflation) is the value of d’ that

results in a net present value of zero

      

: 2.3 0.8 60 / 90,000 : .15 7000 / ( ) 12 12,600 : 90,000 12,600 102,600 energy GWh $ MWh $ demand MW $ MWh mo mo $ total $      

500,000 102,600 (PVF( ',15)) d    ' 19% IRR d  

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

• Ex. 5.8: IRR for HVAC Retrofit with Inflation
• With inflation factored in, we have
• Can also use

  

1+ 1+ 1+ ' (5.14) (1.05)(1.19) 1.25 d e d   

25%

E

IRR d  

0(1

) (5.19)

E

IRR IRR e e   

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Levelized Bus-Bar Costs

• Various alternatives must be compared on a

consistent basis taking into account

• inflation impacts
• fixed investment costs
• variable costs
• The customary approach for cost valuation consists of

the following steps:

• present worthing of all the cash-flow
• determining the equal amount of an equivalent annual

uniform cash-flow set

• determination of the yearly total generation
• The ratio of the equivalent annual cost to annual

electricity generated is the levelized bus – bar cost

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Levelized Bus-Bar Costs

• Present value of escalating annual costs
• Now find an equivalent annual cost

This is called the levelized annual cost, and the levelizing factor is

annual costs

PV =A PVF( ', ) d n 

A,L annual costs

C =PV CRF( , ) d n  LF=PVF( ', ) CRF( , ) d n d n 

A,L

C =PVF( ', ) CRF( , ) A d n d n  

LF = 1 means no inflation

Professor O. A. Mohammed, EEL5285 Lecture Notes, Spring 2018

Wind Power Probability Density Functions

Weibull and Rayleigh Statistics

where k is called the shape parameter, and c is called the scale parameter. When the shape parameter k is equal to 2, the p.d.f. is called Rayleigh

Average Power in the Wind with Rayleigh Statistics

Example: Estimate the average power in the wind at a height of 50 m when the windspeed at 10 m averages 6 m/s. Assume Rayleigh statistics, a standard friction coefficient α = 1/7, and standard air density ρ = 1.225 kg/m3.

OR

HW 3 Problems : 6.1 6.2 6.4 6.5