World's Largest Machines

Copyright ©2014 by Paul Niquette. All rights reserved.

Technical Factors

As noted in the puzzle formulation, a few references on the web deal superficially with the elementary blade-count question.  One factor mentioned is rotating speed, which can be summarily dismissed, inasmuch as the drive train between the external rotor and the alternator inside the nacelle assures compliance with the line frequency of the power grid.  For the solution to the World's Largest Machines puzzle, we can take the typical rotating speed of 18 revolutions per minute as a constant – whatever the blade count. 
Likewise such published factors as auditory noise, blade thickness, cyclical loading, drive train, gear box, peak torques, reliability, and solidity can be regarded merely as dependent variables that must be accommodated in the design of the turbine, thus they are not discriminating factors with respect to blade count.  The puzzle also suggested that solvers conduct an investigation into aerodynamic interferences between successive blades (see Technical Factors: Aerodynamic Interferences).

IllHaving dismissed blade-to-blade interference, we turn now to Technical Factors that require ascertaining the amount of power that will be captured by the wind turbine under operating conditions wherein wind-speed varies.  Our Business Model for that calculation must apply a distinguished form of Return on Investment, so to speak, for each blade on a rotor and will be addressed in the Economic Issues below. 
Nota bene, the conventional objective of maximizing ‘efficiency’ is not well suited to the evaluation of wind turbine designs, given that wind has zero unit cost.
Maximizing the ratio of electrical power output to wind power input is trumped by simply maximizing the power output captured from the wind by a given wind turbine configuration,  while minimizing losses in the drive train within the nacelle.  As we shall see, on that basis Return on Investment actually benefits from not trying to capture 100% of the wind energy that passes through the area swept by the blades.

Economic Issues

Given the present reality – that wind turbines with three blades dominate the market – it makes sense to adopt the three-blade design as the base case for our business model.  That enables us to treat the three-blade configuration as intrinsic to what engineers call a Design Center, a suite of specifications for technical parameters.
For example, a certain three-bladed wind turbine operates over a specified range of local wind speed vLW up to its Nameplate Limit. The pitch-control system automatically makes adjustments that effectively decline to accommodate some of the power offered by the wind, when wind speed becomes exceptionally high. 
A four-blade configuration will doubtless produce more power for a given wind speed.  However, without changing the Design Center to call for an increase in the Nameplate Power, that fourth blade will merely assure reaching that maximum at a lower wind speed.  Accordingly our model will need to answer this question…
Will the increased power justify the incremental investment in the fourth blade?
IllLet us take as our hypothesis that the answer is no.   If confirmed by the business model, we have the solution to our puzzle and no need to study configurations with more than four blades.  For completeness, of course, we also need to use the same Design Center to confirm that the cost of three blades are indeed justified relative to a configuration with only two blades. 

Operating Environment

For a wind turbine, the most relevant feature of the operating environment is, of course, wind speed.  Here is typical distribution of wind speeds measured over the period of one year (8,760 hours) and presented in the form of a histogram or ‘frequency distribution’.  Periods of calm air (vLW = 0) at the far left, total about 80 hours per year.  Winds as high as 20 – 25 mps (44 – 55 mph)  are experienced for a total of six hours per year.  The most frequently occurring wind speed is vLW = 6 mps (13 mph) accounting for 500 hours per year.


Design Center

For our Base Case business model, let us postulate a wind turbine with three blades, each have a length of 60 meters (195 ft) operating at a hub-height of 100 meters (325 ft).  Let the Nameplate Power pNP be specified as five megawatts (5,000 kW).  The turbine is not required to operate below its Cut-in Wind Speed vCI 4 mps (9 mph). 

IllAvailable power output from the turbine increases with local wind speed vLW up to some intensity vNP at which the design limit, Nameplate Power, is reached, pNP = 5,000 kW.  For vLW > vNP, the pitch-control system will automatically maintain the output of the wind turbine at the Nameplate Power pNP = 5,000 kW.  For its own protection, the wind turbine will feather its blades and stop hub rotation at or above the Cut-off Wind Speed, vLW > vCO, specified at 23 mps (52 mph).

Operating Parameters

For all wind speeds, the hub rotates at a constant rate of 18 rpm producing a tip speed tangential to the Plane of Blade Rotation vTS = 113 mps (250 mph).  The wind turbine nacelle is automatically positioned by the yaw-control system to maintain the Plane of Blade Rotation normal to the local wind vLW.  The consequent Relative Wind vRW is given by the Theorem of Pythagoras...
 vRW  = (vTS 2 + vLW 2 )1/2
…and the Relative Wind makes an angle to the Plane of Blade Rotation uRW given by…
uRW = arc tan (vLW / vTS)
Observe that both vRW and uRW are independent of the pitch angle of the blade.  Moreover, for estimating aerodynamic forces on the blade in the operating range of the wind turbine, Relative Wind speed vRW and the Relative Wind angle are bounded as indicated...
At the Cut-in Wind Speed vCI, vRW = 113.1 mps (250 mph) and uRW = 2o ;
At the Cut-out Wind Speed vCO, vRW = 115.3 mps (255 mph) and uRW =
The airfoil chord of the turbine blade is positioned by the pitch-control system at an angle uAB to the Plane of Blade Rotation, which controls the Drive Force fDF on the blade and therefore the amount of power being generated pPGHere is the expression for uAB in the model…
uAB = uA + uB + uRW
...where uA = Angle-of-Attack ( α ) for appropriate Lift-to-Drag ratio, uB = the Angle-of-Incidence ( β ) chosen to control the power being captured, and uRW = angle made by the Relative Wind with the Plane of Blade Rotation. 
At the Cut-in Wind Speed vCI, uAB = 7o.  Operating at Nameplate power pNP, uAB = 12o ; and at the Cut-out Wind Speed vCO, uAB = 17o.  As the wind turbine cuts out, the pitch-control system automatically feathers the blades to uAB = 90o , enabling the rotor to be braked to a stop.
Drive Force

Solvers are invited to review aerodynamic forces for deriving the Drive Force fDF.

The blade as a whole drives the turbine.  Our business model works with only one element of the blade, an airfoil located near the tip with its chord positioned by the pitch-control system to maintain an angle uAB with the Plane of Blade Rotation.  The blade itself is designed with a ‘twist’, which increases the effective value of uAB with decreasing radius toward the hub so that the local Angle-of-Attack uA assures appropriate Lift-to-Drag ratio.
The ‘torque’ delivered by the blade to the drive shaft at the hub varies with radius to the airfoil element.  Values of these blade parameters are assumed in the business model to be appropriately integrated along the length of the blade such that one lumped parameter for Drive Force fDF can be used to characterize the blade.
IllAccordingly, the model derives a measure of the power produced by the turbine correlated with the local Wind Speed vLW starting with an estimate of the Drive Force fDF acting on the blade from the two aerodynamic forces, Lift fAL and Drag fAD.  With comparisons as the objective to our ‘reverse engineering’ exercise, it is unnecessary to determine the explicit value of fDF, only to apply principles to relate aerodynamic forces to vRL and uRL as developed above.
Let us define consolidated aerodynamic coefficients cAL and cAD, which resulting in dimensional units of force divided by velocity squared, thus…
  fAL = cAL vRW2  and  fAD = cAD vRW2 
…and for the base case business model, let the lift-to-drag ratio cAL / cAD = 10.
The model calculates fDF using a coefficient cDF in units of force divided by units of velocity squared, thereby consolidating the two aerodynamic coefficients cAL and cAD by virtue of this trigonometric expression… 
cDF = cAL sin (uAB) – cAD cos (uAB)
 …such that the Drive Force fDF acting tangentially on the turbine blade is given by…
fDF = cDF vRW2
Bending Force

The Drive Force fDF acts tangentially and produces electrical power.  A second force acts on each blade normal to the Plane of Blade Rotation -- the Bending Force fBF given by…
fBF = cBF vRW2 where cBF = cAL cos (uAB) + cAD sin (uAB)
  …and fBF is much greater than the Drive Force fDF
For example, with a lift-to-drag ratio cAL / cAD = 10,  fBF ~ 10 fDF. 
The bending force on each blade does not participate in the production of electrical power but is passed through to the tower structure of the wind turbine via the hub and ultimately is manifested as a cantilever-load at the foundation.  Accordingly, increasing from three blades to four blades, the Bending Force fBF imposes incremental capital costs as a form of ‘overhead’, which must be added to the costs attributable to the blade itself. 
Accordingly, for a given Nameplate Power pNP, adding a fourth turbine blade calls for changes in the Design Center from that of a wind turbine with three blades.  The proportionality (4/3) applies to the whole structure not just to the blade assembly.
Power Generation

The business model has applied the principle that aerodynamic forces increase according as the square of Relative Wind speed vRW.  Thus the Drive Force fDF likewise increases according as the square of the Relative Wind speed vRW.  Since fDF acts tangentially in the Plane of Blade Rotation, it produces electrical power according as the product of fDF and the tangential speed at the blade tip vTS. 

Let a modeling coefficient for Power Generation cPG be adopted with the dimensions kilowatts per unit of force per unit of velocity, such that Power Generated pPG is given by 
pPG = cPG  fDF vTS in kilowatts.
As formulated, with constant vTS, the Local Wind velocity vLW is the only independent variable.  Using the Nameplate power in the model above, we have one point on a curve...
for  vLW = 14 mps (32 mph),  pPG = 5,000 kilowatts.
  …which facilitates the piecewise tabulation of power pPG, scaled against local Wind Speed vLW for the Base Case.  That tabulation is graphed below for all values of vLW.

Base Case Formulation

The graph here summarizes the Base Case business model.  The Design Center Limits for the three-blade wind turbine are superimposed upon the Annual Wind Speed Distribution in the Operating Environment. 

IllThroughout a typical year (8,760 hours), whenever vLW < vCI or vLW > vCO the wind turbine is turned off, generating zero power.  Those extremes prevail for about 440 hours per year (5%).  Power Generation is seen to increase with wind speed from about 1,000 kilowatts to the Nameplate power pNP = 5,000 kilowatts, where it is maintained constant by the pitch-control system.

Power Generation: Base Case

The business model applies straightforward piecewise tabulations to estimate the power generated by the wind turbine.  As derived above, pPG = cPG  fDF vTS in kilowatts, which is shown as a cumulative distribution based on wind speed.

As noted on the graph, the total annual energy produced by the Design Center wind turbine is 18,000 megawatt-hours, about half being generated at wind speeds below 10 mps (22 mph).

Power Generation: Comparison Cases

For brevity, let b = number of blades in a wind turbine configuration being analyzed by the Business Model.  In the Base Case, b = 3.  To solve the World’s Largest Machines puzzle, only two Comparison Cases are needed, b = 4 and b = 2.  An elementary modeling coefficient cPG(b) is introduced here for Comparison Cases:  pPG(b) = cPG(b) pPG, where…
cPG(4) = 4/3 = 1.333
cPG(3) = 1.0 Base Case
= 2/3 = 0.666

Here are the results from the business model in the form of Available Power Generated pPG(b) as a function of Local Wind Speed vLW for each of the three configurations.

For vLW < vCI and for vLW > C0,  pPG(b) = 0.
For vLW = vCI, the respective power levels generated will be given by…
pPG(4) = 1,333 kW
= 1,000 kW Base Case
=  666 kW
For vCI < vLW <vNP, Available Power pPG(b) increases on separate curves, reaching Nameplate Power pNP = 5,000 kW at different wind speeds vNP(b).
Given that the wind turbine will operate at pNP for differing amounts of time, in hours per year tNP(b), the expression 100 tNP(b) / 8,760% might be considered a 'figure of merit' for the wind turbine configuration.  See Duty Cycle.

Energy Pays the Bills

The economic value of a wind turbine is determined by the amount of electrical energy it produces from the wind over the period of a year.  As studied in the solution to Erg and Ugh, the electrical unit for electrical energy is watt-hour or multiples thereof  (kWh, mWh, gWh).  

The Business Model used for solving the World’s Largest Machines puzzle provides the curves in this graph, which features Cumulative Energy as a function of Wind Speed ranges, in the form of piecewise products ΣpPG(b).

Accordingly, the business model provides the following annual estimates… 
ΣpPG(4) = 22,575 mwh
ΣpPG(3) = 18,032 mwh
IllTaking capital costs for a wind turbine to be a function of the number of blades  CC(b), we have reasoned above that CC(b) scales directly with the number of blades b, such that, flexing the comparison from the base case (b = 3), we conclude that… 
CC(4) /  CC(3) = 1.33 and ΣpPG(4) / ΣpPG(3) = 1.25
Accordingly, based on the Operating Environment in the Base Case, four blades are not justified economically over three blades, which confirms our hypothesis and solves the World's Largest Machines puzzle. 

Economic Breakeven

The criteria for the requisite financial investment in a wind turbine would include breakeven, so let us assume that the estimate for the total construction cost of a 5-megawatt wind turbine is $1 million.  A breakeven calculation based on our model above indicates that the machine will produce 18,000 mwh per year of electrical energy.  One needs an estimate for the likely revenue produced by the wind turbine.
A certain puzzle-master living in France paid €600 last year for 2,868 kwh of electrical power.  The conversion rate on the date of this calculation is 1.366 $/€, which means that he paid 28.6 ¢/kwh.  A reasonable mark-up from wholesale to retail is 2-to-1, and if 40% of wholesale can be used to amortized the construction cost of a nearby wind turbine, the breakeven time for that investment is 1 year.
That estimate applied a half-dozen numerical assumptions, which are highlighted in blue.  The point is not what the actual payback interval amounts to; it's the fact that Economic Breakeven can be computed for wind turbines and presumably for other energy technologies. 

Notice that the 'feed-stock' for a wind turbine is -- well, it's wind.  Economic breakeven for a fossil-fueled power plant must take into account estimates for the cost of fuel fuel consumed over time, giving particular emphasis to electrical energy 'yield'.  For a further analysis of the effect of yield see Energy Breakeven.


Some solvers may be interested in an offbeat history of the
World's Largest Machines puzzle.

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