Wind EngineeringModule 6.1: Cost and Weight Models
Lakshmi N. Sankar
lsankar@ae.gatech.edu
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Overview
•In this module, we will briefly examine modelsfor estimating the cost of energy (in cents perKWhr) that the operator needs to charge.
•We will look at two approaches
–Engineering models based on weight and cost(This module 6.1)
–Models suitable for hybrid power systems(Module 6.2)
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Some Definitions
•Debt: Money the operator borrows to financea wind turbine project
•Interest on debt: Interest charged per year byfinance institution (expressed in percentage)
•Equity: Funds the operator raises by issuingstocks
•Return on equity: Return the share-holdersexpect on their investments (expressed inpercentage per $1 invested).
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Definitions, continued..
•AWCC: Average weighted cost of capital
•Example:
–20% equity
–13% return on equity
–80% loan
–6.94% interest on loan
•AWCC for this example is (0.20*13+0.80*6.94) =8.15%=0.0815
•Inflation-adjusted AWCC = (AWCC-Inflation)/(1+Inflation).
•For example if inflation is 3%, the inflation adjusted AWCCis (0.0815-0.03)/(1.03) = 0.05=5%
•This is sometimes called discount rate.
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Cost of EnergySource: NREL /TP-500-40566
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Definition
•FCR: fixed charge rate. It includes
–AWCC (payment to the bank loan and equity holders)
–Depreciation
–Income tax
–Property tax
–Insurance
–Other finance fees
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Initial Capital Cost
Sum of turbine system cost for elements listed below + balance of station costs
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Initial capital Cost (Continued..)
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Annual operating Expenses
•Include land lease, operation andmaintenance, cost of replacing or overhaulingparts.
•Expressed in dollars per KWh.
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Net Average Energy Production (AEP)Overview
•Units are in KWh
•We may view this as power production integrated over time for a wholeyear.
•Here is a very crude description of how this is computed.
–Power production depends on how hard wind blows and how often
–It is assumed that the wind speed at a particular site has a Weibulldistribution.
–This distribution gives the probability that the wind is blowing at a given speed
–With some knowledge of the wind turbine power characteristics (rated power,peak Cp, tip speed ratio at which peak Cp occurs, etc), power production atdifferent wind speeds is estimated.
–This is multiplied by the Weibull probability that wind is blowing at that speed.
–Summation is done over all the wind speeds.
–The result is multiplied by 365 days x 24 hours/day
•Capacity Factor = AEP / (Rated Power x 365 x 24) may also be computed.
•See weibull_betz5_lswt_baseline.xls for example calculations.
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Example: Turbine Capital CostNREL Report
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Blade Cost
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Example continued..We compare baseline and projected
Rating (kWs)
1500
1500
Baseline
Projected
Component
Component
Component
Costs $1000
Costs $1000
Foundations
49
49
Transportation
51
51
Roads, civil works
79
79
Assembly & installation
51
51
Elect interfc/connect
127
127
Permits, engineering
33
33
BALANCE OF STATION COST (BOS)
388
388
Project Uncertainty
162
162
Turbine cost from previous slide
921
921
Initial capital cost (ICC)
1,472
1,472
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Other costs
Baseline
In $1000
Projectedin$1000
LEVELIZED REPLACEMENT COSTS (LRC)($10.7 per kW)
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16
O&M $20/kW/Yr (O&M)
30
30
Land ($/year/turbine)
5
5
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Example, continued..
•We next computeprobability of windblowing at a particularspeed.
•Weibull probabilityfunction is used.
•This depends on aparameter called Kfactor, and wind speedat the hub.
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Weibull Distribution
•K: Shape factor
•Changing k shiftsprobability to theleft or right.
• : Scale parameter
•In our example, k= 2
• = Wind Speed atthe hub
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Efficiency of the Turbine
•We next compute efficiencyof the turbine when itoperates at power other thanrated power.
•If field data is available, it isused.
•Otherwise a simple logic isused:
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Hub Power
•If wind velocity is less thancut-in speed, hub power iszero.
•If wind velocity less thanrates speed it is found from
•At higher than rated speeds,rated power is used.
•At greater than cutoutspeeds, power is zero.
•The hub power, whenmultiplied by Weibullprobability and efficiency ,gives turbine energy outputat that speed.
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Annual Energy Production
• Other losses may include electrical system losses
• We divide by 4 because the wind speeds are binned (or grouped by
¼ m/sec increments.
• We will find power, for example at 2, 2.25, 2.50, and 2.75 m/sec and take
the average.
• 365 x 24 is 8760
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Cost of Energy
•Once all theinformation isavailable, we canfind the cost ofenergy per KWh.
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