## HOMER Energy Support

### Marginal diesel generation cost

Last Updated: Jul 20, 2016 04:15PM MDT

I am a little perplexed at HOMER’s output calculation of the “Marginal generation cost” under the tab “generator 1” in the simulation results. I would assume that marginal generation cost would be the \$/kWh not including capital costs (i.e. just fuel at the diesel’s average efficiency and O&M). However, the number that Homer always displays here is much too low to be this marginal cost. If I calculate this cost, neglecting O&M, I would calculate it as:

Fuel cost [\$/l] /  ( HHV [kWh/kg] x fuel density [kg/l] x ave efficiency ) = \$/kWh

Good question about the generator’s marginal generation cost.  I haven’t yet defined that in the help file.  Your equation is almost correct, except that the relevant efficiency is the marginal efficiency, not the average efficiency.  Also, you need to use the LHV, since HOMER’s definition of efficiency is based on LHV.

The marginal generation cost is the cost of producing one more kWh, once the generator is already running.  Since the generator is already running, you are already paying the fixed cost associated with O&M, replacement, and the no-load fuel consumption.  When you ask for one more kWh from the generator, the only thing that changes is the fuel consumption, and its rate of change is the fuel curve slope.  So the marginal generation cost [\$/kWh] is equal to the fuel price [\$/L] times the fuel curve slope [L/kWh].

By the way, the marginal efficiency is equal to one divided by the product of the fuel curve slope [L/kWh], the LHV [kWh/kg], and the density [kg/L].  So you could write the equation for the marginal generation cost as the fuel cost [\$/L] divided by the product of the LHV [kWh/kg], the density [kg/L], and the marginal efficiency.  That’s the same as your equation, except for the two things I mentioned before, the LHV instead of the HHV, and the marginal efficiency instead of the average efficiency.

At the default fuel curve slope of 0.25 L/kWh and using the figures for diesel fuel, whose LHV is 43.2 MJ/kg = 12 kWh/kg = 9.84 kWh/L, the marginal efficiency is 1 / (0.25 * 9.84) = 41%.

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