# Are Larger Wind Turbines Paid for Faster?

2015-04-29

And what is the financial pay-back period for large wind turbines? Common questions!

Note that we're considering financial pay-back, not energy pay-back.

*Fifty-seven 15 kW wind turbines and one 850 kW wind turbine ... approximately to scale. The nacelle on the 850 kW machine is a bit bigger than it would be in real life! Which is cheaper? A person would be about as big as this asterisk * ...*

The answer to the question in the title is that for given environmental and economic conditions, bigger is usually better: the reasons often cited are:

- The power output of a wind turbine depends on the swept area, i.e. usually on the
*square*of rotor radius: power increases rapidly with size; - Manufacturing costs do not increase as rapidly, resulting in a lower relative cost of manufacture;
- Large wind turbines are mostly more aerodynamically
*efficient*than small ones; - The wind speeds are higher, the higher up you go, and the power output of a wind turbine depends on the
*cube*of the wind speed.

These are all true. The first three reasons, advantages of size and efficiency, can all be lumped into the turbine's euro cost per rated kilowatt installed, i.e. how much it will cost to have a wind turbine running at your site.

As it turns out, the pay-back period depends directly on this cost per kilowatt (kW) installed (*T _{cost}*), on the

*cube*of the rated wind speed of the turbine (

*U*), but inversely on both the cube of the average wind speed at the proposed site (

*U*) and the income per kilowatt-hour (kWh) of electricity produced (c

_{avg}*)*, i.e.

*Pay-back (years) ∝ T _{cost} * (U^{3}/ U_{avg}^{3}) * (1/c)
*

With an assumption or two, you can write this as:

*Pay-back (years) = (π/6) * T _{cost} * (U^{3}/ U_{avg}^{3}) * (1/8760 c)
*

(The derivation of this rough approximation is at the bottom of the page. All this neglects maintenance and other expenses that can arise. It's taken for granted that the wind turbine will perform as expected, i.e. the cost per kW installed depends on a real kW rating.)

As a sanity check, taking a machine that I know something about, the Vestas V52 850 kW machine at DkIT, a college here in Ireland, cost about € 1500 per kW installed and reaching rated power at a wind speed of 13 m/s. Much of the turbine's power is used to offset grid purchased power and the rest is sold. Assuming that *all* the power is used to offset grid power, purchased at commercial rates, the income per kWh might be something around €0.12. The average wind speed at hub height is something close to 6 m/s. Putting these values into the expression above, the pay-back period will be around 8 years. This isn't so far off the actual period of around ten years.

It follows that wind turbines, *a* and *b*, of different sizes will have pay-back ratio of:

*Pay-back ratio = (T _{cost a} / T _{cost b}) * (c_{b}/c_{a})* (U_{a} * U_{b avg})^{3}/ (U_{b} * U_{a avg})^{3}
*

It's important that the installed cost is calculated based on an accurate power curve, i.e. that the rated wind speed corresponds to the point that rated power is practically achieved.

At present, small- and medium-sized wind turbines - those with rotor diameters below 36 m or ratings below 500 kW - are relatively expensive, with costs per kW installed of around €5,000, more or less. Large wind turbines have variable costs per kW installed but these can be up to about €1,500. You can find numbers in text books and papers. See chapter 11 in Wind Energy Explained by Manwell, McGowan and Rogers, for example.

Using these values, and considering the payback ratio of the small wind turbine to the large wind turbine, we can write:

*Pay-back ratio = (5000/1500)*(c _{b}/c_{a})* (U_{a} * U_{b avg})^{3}/ (U_{b} * U_{a avg})^{3}
*

*= (3.33)*(c _{b}/c_{a})* (U_{a} * U_{b avg})^{3}/ (U_{b} * U_{a avg})^{3}*

You can see from this that if everything else is equal, i.e. the rated wind speeds are the same, the average wind speeds at the site are the same and the value of the energy produced is the same, then the small wind turbine payback depends only on the ratio of the installed costs.

Large and small wind turbines are usually rated at different wind speeds, somewhere around 11 m/s for small wind and around 15 m/s for large wind turbines. Using these numbers,

*Pay-back ratio = (3.33)*(c _{b}/c_{a})* (10* U_{b avg})^{3}/ (15 * U_{a avg})^{3}
*

*= (3.33)*(0.394)*(c _{b}/c_{a})* (U_{b avg})^{3}/ (U_{a avg})^{3}
*

*= 1.3*(c _{b}/c_{a})* (U_{b avg})^{3}/ (U_{a avg})^{3}*

The pay-back ratio depends on the *cube* of the ratio of the rated speeds. If the small wind turbine is rated at 10 m/s rather than 11 m/s, the factor 1.3 reduces to 0.98. To be conservative, we'll stick with 11 m/s for now.

If we assume that the value of the energy from the wind turbines is the same, which can be true on grid connected systems, then *c_b/c_a = *1 and:

*Pay-back ratio = 1.3* (U _{b avg})^{3}/ (U_{a avg})^{3}*

*V = V _{0}*(h/h_{0})^{α}*

Where *V *is the wind speed at the higher altitude *h*, *V _{0}* is the wind speed at the reference, lower altitude

*h*, and

_{0}*a*is a number that depends on the how cluttered the landscape is. At a perfect site, i.e. with little clutter close to the surface,

*a*might have a value of around 0.1. This means that a small wind turbine close to the ground at

*h*will be as well placed as it's possible to be, assuming that it's in a windy part of the world. If we imagine a small wind turbine on a 15 m tower and a large wind turbine on an 80 m tower in such a site, the ratio of V/

_{0}*V*will be 1.18. The ratio of the cube of these speeds will then be 1.652, i.e. 1.18

_{0}^{3}.

Assuming that this is the case,

*Pay-back ratio = 1.3* 1.652= 2.15
*

In this case, all else equal, the small wind turbine will take more than twice as long to recoup the initial investment as the large wind turbine.

In a cluttered location, for *α* = 0.25, a turbine on an 80 m tower and a cost per kW of €1,500 will have a pay-back period that is 4.6 times smaller than that of the small wind turbine costing €1,500 per kW installed on a 15 m tower, assuming the same value of electricity produced.

So, a small wind turbine on 15 m tower, generating power at a given site, may have a pay-back period that is between 2 and 5 times longer than a large wind turbine at the same site.

This is why, in general, the trend is for larger wind turbines. Simple pay-back is often faster.

At the same time, while illustrative, there are quite a few assumptions and simplifications in this!

There are many situations where smaller wind turbines can make sense. I assumed a constant value for the power produced. There are remote locations where the cost of providing grid power drives the value for *c* upwards, reducing the simple pay-back period for a well-matched small system.

Feed-in tariffs also affect *c. *Until recently, the UK's generous micro-generation FIT made smaller systems cost-effective when compared with larger installations.

In some installations, it may not make sense to have a machine larger than necessary.

Additionally, turbine manufacturers world-wide work constantly to reduce the cost per kW installed of small- and medium-sized systems. Evidently, if this can be reduced while maintaining safety and quality, then the pay-back period reduces.

Finally, finding high wind speeds make a very large difference.

All this points to three important targets for wind turbine manufacture and installation at all scales:

1. To push down the cost of manufacture and installation for high quality, safe machines.

2. To better understand the wind resource and to put turbines in suitable locations.

3. To understand the value of the power generated - this will depend on where, when and how it's used. Solar photovoltaic panels, batteries and re-built grids are re-shaping the economics of energy. The cost of electricity will, to a large extent, determine the future of wind energy.

### Links

- NREL has a nice pay-back calculator - with a more detailed model. Note the assumption that all the energy is used on-site.

### Derivation

The most important equation in wind energy is:

*P = (1/2) ρ C _{p} A U^{3}*

where *P *is the power output from the wind turbine in watts, *ρ* is the local air density, *C _{p}* is the efficiency,

*A*is the swept area, i.e.

*ρ R*

^{2}where

*R*is the radius of the swept area, and

*U*is the wind speed at hub height.

The cost of a wind turbine, installed, can be given as a multiple of its rated power. This is a bit tricky as manufacturers tend to rate at different wind speeds. However, most turbines are operating close to rated in a wind speed of 10 m/s.

So, the cost of installing a wind turbine can be given as:

*T _{cost} = P * T_{cost per kW} = (((1/2) ρ C_{p} A U^{3}) / 1000) * T_{cost per kW}
*

After Carlin, the average power output from the wind turbine at a particular site can be estimated using,

*P _{avg} = (3/4)*ρ*C_{p}*(D^{2})*(U_{avg})^{3}*

where *P _{avg}* is the average power output,

*ρ*is the air density,

*D*is the rotor diameter and

*U*is the average wind speed at the site. The mechanical losses are assumed to be wrapped up in

_{avg}*C*(See p64 of Wind Energy Explained by Manwell, McGowan and Rogers, for more.)

_{p. }Using this, we can say that the total amount of energy generated in a year by a wind turbine at a particular site is,

* E = (P _{avg} * 8760)/1000*

This is in kilowatt-hours. The total annual revenue from this wind turbine is *I _{Total} = E * c*, where

*c*is the average price paid in euro for each kWh of electricity, i.e.

*I _{Total}* =

*(P*= (

_{avg}* 8760 * I)/1000*(3/4)*ρ**

*C*euro

_{p}*(D)^{2})*(U_{avg})^{3}* 8760 * c)/1000
(Of course, *c* is complicated and will change over the lifetime of the wind turbine but we will use a nominal value close to the current amount paid. You can easily experiment with different values).

Finally, the installed cost of the wind turbine divided by the annual income from the wind turbine gives the simple pay-back, i.e.

* Pay-back (years) = *

* (((1/2) ρ C _{p} π (D/2)^{2} U^{3}) / 1000) * T_{cost per kW}
*

---------------------------------------------

(*(3/4)*ρ***C _{p}*(D)^{2})*(U_{avg})^{3} * 8760 * c)/1000*

Simplifying this, you end up with:

*Pay-back (years) = (π/6) * T _{cost} * (U^{3}/ U_{avg}^{3}) * (1/8760 c)*

*Q.E.D.*