San José State University
Department of Economics

applet-magic.com
Thayer Watkins
Silicon Valley
USA

The Delusion that Fuel-free Energy
is Necessarily Cheaper than Energy
Requiring Fuel:The Role of Capital Cost

Wind turbines, solar panels and hydroelectric dams are notable not only because they do not require fuel but also because of the their relative high capital costs. The annual capital cost for equipment is the annual payment required to pay its initial cost in its economic lifetime plus any annual maintenance and security costs. The computation of the annual payment required to pay off the initial cost in the lifetime of the equipment necessarily involves an interest rate. That interest rate reflects the scarcity value of the capital investment.

Consider this simplified statement of the problem. Let M be the initial capital investment required to produce 1 kilowatt-hour (kwh) of energy per year. This must take into account the proportion of the time the equipment is operating, the so-called load factor. The lifetime in years of the equipment and structure is denoted as T.

If r is the interest rate the annual payment p required to pay off an amount M in T years is

#### p = Mr/(1−1/(1+r)T)

The derivation of this formula is given in the appendix to this paper.

Let f and s be the fuel and maintenance-security cost, respectively, per kwh of annual production. Then the cost c per kilowatt-hour of energy production is

#### c = f + s + Mr/(1−1/(1+r)T)

Now consider two methods of energy production, one with f=0 and the other with f>0. Let the data for the first method be subscripted 0 and the other subscripted 1. Then the cost for the one using fuel, c1, will be less than the cost of the fuel-free one, c0, if and only if

#### f1 + s1 + M1r/(1−1/(1+r)T1) < s0 + M0r(1−1/(1+r)T0)

Clearly there are values of M0 such that the above holds true; i.e.,

#### M0 > [1−1/(1+r)T0][f1 + Δs1 + M1r/(1−1/(1+r))]T1)

Of course, the fuel-free method could be the cheapest, but the point is that it is not necessarily the cheapest. For an illustration of this problem in the Soviet Union see Soviet hydropower.

## Appendix

The payments of p are made at the end of the years starting with the first year and ending with the T-th year. The present value P of these payments is

#### P = p/(1=r) + p/(1+r)2 + … + p/(1+r)T

If P is divided by (1+r) and the result subtracted from the expression for P, as shown below

#### P = p/(1+r) + p/(1+r)2 + … + p/(1+r)T P/(1+r) =       + p/(1+r)2 + … + p/(1+r)T + p/(1+r)T+1 P[1 − 1/(1+r)] = p/(1+r) − p/(1+r)T+1which reduces to Pr/(1+r) = (1/(1+r))p[1 − 1/(1+r)T] and further Pr = p[1 − 1/(1+r)T]

If P is to be equal to the initial investment M then