Variable Load and Power Plant Economics

A power plant almost never runs at its rated output; demand rises and falls all day. This lesson is about how engineers describe that varying load and how they cost the power a plant produces. The math is mostly ratios and simple economics, so once you keep the factor definitions straight, these are quick and reliable points on the Industrial and Power Plant Engineering paper. Work in SI: power in kW or MW, energy in kWh or MWh.

The load curve and load-duration curve

A load curve plots the demand on a plant against time over a day. Two quantities read straight off it:

• The peak (maximum) load is the highest point on the curve.
• The average load is the total energy generated divided by the time: average load = energy / hours.

Rearranging the load curve so the loads are stacked from highest to lowest gives the load-duration curve, which shows for how many hours the load equalled or exceeded each value. The area under either curve is the same: it is the total energy generated in the period. The load-duration curve makes it easy to see how much of the year the plant runs near its peak versus near its base.

Load factor and capacity factor

The load factor measures how evenly a plant is used:

load factor = average load / peak load

Worked example: a station generates 4,800 MWh in a day with a peak load of 300 MW. The average load is 4,800/24 = 200 MW, so the load factor is 200/300 = 0.667, or 66.7%. A high load factor means the plant runs steadily near its average, which is cheaper per kWh.

The capacity factor (plant capacity factor) compares the energy actually produced with what the plant could produce at its full rated capacity:

capacity factor = actual energy generated / (rated capacity x hours)

Worked example: a 50 MW plant generates 219,000 MWh in a year. The maximum possible is 50 x 8,760 = 438,000 MWh, so the capacity factor is 219,000/438,000 = 0.50, or 50%. If the peak load equals the rated capacity, the capacity factor equals the load factor; otherwise the capacity factor is lower.

Demand, diversity, and plant-use factors

Four more ratios round out the vocabulary:

• Demand factor = maximum demand / connected load. It is at most 1, because not every connected device runs at once. If a plant serves a connected load of 1,500 kW with a maximum demand of 900 kW, the demand factor is 900/1,500 = 0.60.
• Diversity factor = sum of individual maximum demands / maximum demand of the system. It is greater than 1, because different consumers peak at different times. If the individual peaks sum to 1,200 kW but the system peaks at 800 kW, the diversity factor is 1,200/800 = 1.5. High diversity lets a utility serve many customers with less generating capacity.
• Plant-use factor = energy generated / (plant capacity x hours the plant was in operation). It differs from the capacity factor by counting only the hours the plant actually ran. A plant generating 200,000 MWh in 6,000 operating hours at 40 MW capacity has a use factor of 200,000/(40 x 6,000) = 200,000/240,000 = 0.833.
• Reserve capacity = plant capacity minus maximum demand: the spare margin kept for growth and outages.

Keep the two "greater or less than 1" facts straight: the demand factor is at most 1, while the diversity factor is at least 1.

The cost of generated power

The cost of power splits into three parts:

• Fixed (capital) cost: interest and depreciation on the plant, independent of how much energy is produced.
• Operating (running) cost: fuel, water, and consumables, roughly proportional to energy generated.
• Semi-fixed cost: labor, supervision, and maintenance, partly fixed and partly variable.

The cost per kWh is the total annual cost divided by the annual energy generated. Because much of the cost is fixed, a plant with a high load factor spreads that fixed cost over more kWh and so delivers cheaper energy. This is the economic reason to run base-load plants steadily and near capacity.

Plant depreciation

Depreciation spreads a plant's capital cost, less its salvage value, over its useful life. Two standard methods appear on the MELE:

• Straight-line method: an equal amount each year.

  annual depreciation = (first cost minus salvage value) / life in years
  

  Worked example: a plant costs 5,000,000 with a salvage value of 500,000 and a 20-year life. Straight-line depreciation is (5,000,000 minus 500,000)/20 = 4,500,000/20 = 225,000 per year.

• Sinking-fund method: a smaller yearly deposit that grows with interest to equal the depreciable amount at the end of life.

  annual deposit = (first cost minus salvage) x i / [(1 + i)^n minus 1]
  

  Worked example: same plant at i = 8% and n = 20 years. Here (1.08)^20 = 4.661, so the deposit is 4,500,000 x 0.08 / (4.661 minus 1) = 360,000/3.661 = about 98,300 per year. The sinking-fund deposit is smaller than the straight-line charge because interest does part of the work.

Exam-day strategy

• Load factor = average/peak; average load = energy/hours. These two solve most load-curve items in one step.
• Capacity factor uses rated capacity and all the hours in the period; plant-use factor uses only the hours the plant ran.
• Demand factor is at most 1 (max demand over connected load); diversity factor is at least 1 (sum of peaks over system peak).
• The area under both the load curve and the load-duration curve is the same total energy.
• Straight-line depreciation is (cost minus salvage)/life; the sinking-fund deposit is smaller because it earns interest.