Load duration curve for a small utility

Chapter 9 Questions

  1. Consider the following very-simplified load duration curve for a small utility:
    a. How many hours per year is the load less than 200 MW?
    b. How many hours per year is the load between 200 MW and 600 MW?
    c. If the utility has 600 MW of base-load coal plants, what would their average capacity
    factor be?
    d. Find the energy (MWh/yr) delivered by the coal plants.
  2. The economics of a given power plant depend on a number of factors, including how
    much of the time it operates (i.e. its capacity factor). After accounting for its capital
    costs as well as its fuel and maintenance costs, a “revenue required” curve ($/yr per kW
    of generating capacity) can be drawn such as the one below.

a. Suppose these plants run as peakers with a capacity factor of 0.1. Find their
levelized cost of electricity (LCOE, ¢/kWh).
b. Compare your result if instead this plant is able to run more regularly, with a 60%
capacity factor.

  1. Suppose the utility in Problem 1 has 400 MW of combustion turbines operated as
    peaking power plants.
    a. What would be their average capacity factor?
    b. How much energy will these turbines deliver (MWh/yr)?
    c. If these peakers have the “revenue required” curve shown in Problem 2, what would
    the LCOE from these plants (¢/kWh) need to be?
  2. Estimating the levelized cost of electricity (LCOE) for a fossil-fuel fired power plant is
    complicated by the likelihood that fuel costs will increase in the future. To simplify that
    analysis, we have introduced a fuel Levelization Factor (LF), which depends on an
    estimated annual fuel price escalation rate and an appropriate owner discount factor (d)
    that brings those future costs back to a present value.
    The following figure combines cost escalations and discount rates to help us estimate
    the levelizing factor
    Source: Masters, G., Renewable and Efficient Electric Power Systems,
    Wiley, 2013.
    a. What LF would result from an owner’s discount rate of 15% if they are estimating a
    6%/yr annual fuel cost escalation rate?
    b. Solution Box 9.1 used a Fuel-Levelization Factor of 1.4. If that owner was using a
    10% discount rate, roughly what must have been their estimated fuel-escalation rate?
    c. If the cost analysis in Solution Box 9.1 had ignored future fuel price increases, what
    would its LCOE have been?
    Chapter 10 Questions
  3. The rate structure for a small office building includes an energy charge of $0.12/kWh
    and a summer demand charge of $15/mo per peak kW. As a money saving strategy,
    during the 21 days per month that the office is occupied, it has been decided it can
    easily shed 100 kW of summer air-conditioning load for one hour by simply turning the
    thermostat up by a degree or two during the periods of peak demand.
    a. How much energy (kWh/mo) will be saved?
    b. By what amount will their bill be reduced ($/mo) during those months?
    c. What is the value of the energy saved in ¢/kWh. Compare it to the simple 12¢/kWh.
  4. Suppose a small business can elect to use either the time-of-use (TOU) rate
    schedule shown below or the rate structure involving a demand charge.

Their energy use is 30,000 kWh/mo, 20,000 of which is used during the peak demand
period. Their peak demand is 100 kW. Which rate schedule would have the lowest
bills?

  1. An Ice Bear air-conditioning unit freezes 450 gallons of water at night, when utility
    rates are low, and then blows room warm air over the ice for daytime cooling when rates
    are high.
    a. How many Btu’s of heat does it take to melt that ice (you may want to review Section
    4.4.3)?
    b. The ratio of Btus removed by an air conditioner per watt-hour of electricity is called
    the Seasonal Energy Efficiency Ratio (SEER). If this unit has an SEER of 12 Btu/Wh,
    how many kWh would it take to provide the amount of cooling that you found in (a)?
    c. If this unit can shift the above kWh from a 15¢/kWh daytime rate to cheaper 5¢/kWh
    night rates, how many $/day would be saved by this unit?
  2. Make a simple economic comparison of a fuel-cell system with and without waste
    heat recovery. Begin with a 10-kW fuel cell costing $20,000 that converts natural gas
    costing $1.20/therm (100,000 Btu/therm) to electricity with 40% efficiency. We expect it
    to have an 80% capacity factor.

a. Suppose we amortize the capital cost with the equivalent of a 6%, 20-yr loan.
Ignoring complications associated with fuel-cost escalation and potential tax-deductible
interest, estimate the cost per kWh without any waste heat recovery.
b. Now let’s include the value of waste heat recovery. Suppose 70% of that fuel cell’s
waste heat can be used to displace heat that would have been delivered from an 85%
efficient natural gas boiler. What now is the cost of electricity from the fuel cell?

Sample Solution

ACED ESSAYS