# Illustrating Efficiency

### Simple Calculations & a self-test Quiz

To accompany this lesson, there’s another simple online calculator (click below and it’ll open in a new window):

CALCULATOR: Efficiency & Energy

Make sure you read the “READ THIS” tab of the calculator before you use it, in order to be clear about the various inputs and outputs. Then, to illustrate some outcomes around efficiency and energy resources, use the calculator and try the self-test quiz below.

Q1.

In a sunny country, where the solar panels will receive 2,000kWh on each square metre each year, and 300,000 MWh of electricity needs to be generated each year, at minimum how many more square km of land will be needed, if 15% efficient panels are chosen over 20% efficient ones.

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For this, you are working with the right-hand column on the calculator.

With 20% efficient panels, a solar collector area of 750,000 sq. metres is required, as compared to 1,000,000 sq. metres for 15% efficient panels, a difference of 250,000 sq. metres, or 0.25 sq. km.

Of course this is only the collector area, so this captures the minimum amount of extra land required. In practice the land area will be quite a bit larger than this, as there will need to be spaces between the additional rows of panels to allow access for cleaning and maintenance. It will also depend on the orientation of the collectors (what angle they are tilted at), whether they are on tracking systems and so on.

Q2.

For the same energy target (300,000 MWh) and 15% efficient panels, what is the minimum difference in land requirements between an area receiving primary solar irradiation of 2,000 kWh on each square metre over a year and one receiving only 1,200 kWh?

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0.67 km2

With the 15% efficient panels in a low solar resource area, a solar collector area of 1.67 sq. km is required, as compared to 1 sq. km in the “sunny” case (with the same notes as regards total land use as in the previous answer).

Q3.

Looking again at the previous calculations: using 15% efficient panels, how has the reduction in primary solar resource from 2,000 to 1,200 kWh/m2/yr impacted the cost of building a project capable of producing 300,000 MWh of energy per year, assuming an installed cost of \$2 per Watt?

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\$170 million

A decrease in the primary energy resource means more capacity needs to be built in order to meet the specified energy generation target. With 2,000 kWh/m2/yr available, a capacity of 128 MW needs to be built; with 1,200 kWh/m2/yr, the required capacity is 213 MW, a difference of 85 MW, at \$2 per W.

In fact the efficiency of the panels is irrelevant here. Lower efficiency means more land is needed for each unit of capacity, but it is the availability of the energy resource which determines the capacity factor, and hence the capacity required to generate a specified energy amount.

Q4.

In the low resource area (1,200 kWh/m2/yr) a project using 15% efficiency panels can be built for \$1.25 per W, whereas one with 20% efficient panels would cost \$1.75 per W. Assume the same 300,000 MWh energy target as before, an actual land use area that is three times that of the collector area, and a land purchase cost of \$10m per km2. What is the difference in investment required between the two projects?

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\$93.9m (low efficiency project is cheaper)

As we’ve already seen, with that low resource and 15% efficient panels, we require 1.67 km2 of collector area (needing a total 5 km2 of land if land is 3x collector area) comprising 213 MW of capacity. At the lower cost of \$1.25 per W of capacity, we therefore need to spend \$265.6m on plant and \$50m on land. A total of £315.5m.

Compare that with 20% efficient panels at \$1.75 per W, where the required plant investment is \$371.9m and land for the (smaller) collector area (3.75 km2) is \$37.5m. A total of \$409.4m.

It’s an example showing that low efficiency is not always bad economics! Even though using low efficiency panels means more land is required to meet the energy target (because more resource needs to be collected), the increased land costs are vastly outweighed by the fact that the installed costs of the plant are much cheaper if low efficiency panels are chosen. In practice, it’s also likely land would be leased rather than purchased, so that cost would fall into operational costs (opex) rather than up-front investment.

Q5.

A 100MW open-cycle gas power plant operates only at peak times (achieving a capacity factor of 10%) and with an average efficiency of just 35%. It cost \$0.6 per W to build and can sell its peak-time electricity at \$100/MWh. If natural gas costs \$5/MMBtu, how many years will it take for its sales revenue to pay back its investment and fuel costs? (ignoring any other costs or financial considerations).

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8 years

For this, you are working with the left-hand column of the calculator. Since you haven’t been given an energy target, but have been given the installed capacity and capacity factor, the first thing to do is to calculate the energy output that would result from a 100MW plant operating at 10% capacity factor (= 100MW x 10% x 8760 hours = 87,600 MWh). [You could also get close to this by simply adjusting the energy input until the calculator produces the stated 100MW capacity as a result].

With the other inputs in place, the ratio in the last row of the calculator is 8. This is calculated by dividing the initial investment cost by the net cash available each year to pay it back (i.e. the revenue generated minus the cost of fuel); so it provides a very simple representation of payback time.

Of course a calculation of actual payback will be much more complex, incorporating other costs such as financing, maintenance and much more; plus consideration of other adjustments such as discounting.

Q6.

Taking the previous scenario*, how much longer would the simple payback period be if the plant was built in a market where gas cost closer to \$10/MMBtu?

(*A 100MW plant, capacity factor of 10%, average efficiency of 35%, cost \$0.6 per W to build, electricity price at \$100/MWh, natural gas costs \$5/MMBtu)

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1.6 years

This is simply the same calculation with a different gas price.

You might find it useful to play around with the other variables to see what scale of changes (e.g. to capacity factor or electricity price) produce the same impact as changes in gas price.

Q7.

With this \$10/MMBtu gas price, the plant modelled previously has an opportunity to generate more often, beyond the very peak times, achieving a capacity factor of 25%. In doing so, it will be able to operate more efficiently – at 40%. However in selling outside the very peak times, the average sale price of its electricity will drop to \$55/MWh. Should it take this opportunity?

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Yes.

Starting once again by calculating the energy output of a 100MW plant with 25% capacity factor, gives a value of 219,000 MWh. So there is much more energy to sell, than when operating at a capacity factor of just 10%. But more fuel is burned.

At the much lower price, if it still operated at 35% efficiency, the simple “payback” ratio would be slightly worse (10.4 rather than 9.6 in the previous calculation). So the extra fuel costs are outweighing the extra energy sales.

However with the gain in efficiency included (for example because of less regular and rapid ramps up and down in output), fuel consumption falls and the simple payback ratio drops to 9.1. So at least on this – very simplistic! – basis, the change in operation would make sense.

The key takeaways are:

1. How a number of variables – in this case output, efficiency, fuel cost and electricity price – have combined to determine the economics of a power project.

2. How some of these variables (efficiency and capacity factor, or capacity factor and price, for example) are potentially related to each other.

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