Energy Economics, Cogeneration

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ECE 333 Green Electric Energy

• Lecture 15
• Energy Economics, Cogeneration
• Professor Tom Overbye
• Department of Electrical andComputer Engineering

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Announcements

• Be reading Chapter 5
• Homework 6 is due Oct 22. It is 4.9, 5.2, 5.4,
  5.6, 5.11
• Wind Farm field trip will be on Nov 5 from about
  8am to 4pm turn in forms to sign up.
• Campus Wind Turbine Project Let your voice be
  heard! If you would like to email your student
  senators with your wind turbine views some folks
  to talk with are Bradley Tran (president-iss_at_illin
  ois.edu), Bobby Gregg (bobby.gregg_at_gmail.com),
  Melanie Cornell (cornell4_at_illinois.edu)

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Infinite Horizon Cash-Flow Sets

• Consider a uniform cash-flow set with
• Then,
• For an infinite horizon uniform cash-flow set

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Infinite Horizon Cash-Flow Sets, cont.

• We may view d as the capital recovery factor with
  the following interpretation
• For an initial investment of P,
• dP A
• is the annual amount recovered in terms of
  returns on investment

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Internal Rate of Return

• We consider a cash-flow set
• The value of d for which
• is called the internal rate of return (IRR)
• The IRR is a measure of how fast we recover an
  investment or stated differently, the speed with
  which the returns recover an investment

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Internal Rate of Return Example

• Consider the following cash-flow set

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Internal Rate of Return

• The present value
• has the (non-obvious) solution of d equal to
  about 12.
• The interpretation is that under a 12 discount
  rate, the present value of the cash flow set is 0
  and so 12 is the IRR for the given cash- flow
  set
• The investment makes sense as long as other
  investments yield less than 12.

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Internal Rate of Return

• Consider an infinite horizon simple investment
• Therefore
• For I 1,000 and A 200, d 20 and we
  interpret that the returns capture 20 of the
  investment each year or equivalently that we have
  a simple payback period of 5 years

I
ratio of annual return to initial investment
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Efficient Refrigerator Example

• A more efficient refrigerator incurs an
  investment of additional 1,000 but provides
  200 of energy savings annually
• For a lifetime of 10 years, the IRR is computed
  from the solution of
• or

The solution of this equation requires either an
iterative approach or a value looked up from a
table
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Efficient Refrigerator Example, cont.

• IRR tables show that
• and so the IRR is approximately 15
• If the refrigerator has an expected lifetime of
  15 years this value becomes

As was mentioned earlier, the value is 20 if it
lasts forever
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Impacts of Inflation

• Inflation is a general increase in the level of
  prices in an economy equivalently, we may view
  inflation as a general decline in the value of
  the purchasing power of money
• Inflation is measured using prices different
  products may have distinct escalation rates
• Typically, indices such as the CPI the consumer
  price index use a market basket of goods and
  services as a proxy for the entire U.S. economy
• reference basis is the year 1967 with the price
  of 100 for the basket (L 0) in the year 1990,
  the same basket cost 374 (L 23)

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Figuring Average Rate of Inflation

• Calculate average inflation rate e from 1967 to
  1990

Current(1/2009)basketvalue is about 632.
Source http//en.wikipedia.org/wiki/FileUS_Histo
rical_Inflation.svg
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Inflation (Escalation) Rate

• Recall the Present Value Function definition
• Find present value (P) given annual values (A)
  after n years and with discount rate d
• Present value function in Microsoft Excel
  PV(rate,nper,pmt)

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Inflation (Escalation) Rate

• With escalation, an amount worth 1 in year zero
  becomes (1e) in year 1, etc., so
• becomes
• We can compare terms to find an equivalent
  discount rate d

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Equivalent Discount Rate d

• From
• We solve for d and obtain the following
  identities
• Now, inflation can be incorporated into all of
  the present value relationships just by using d
  in lieu of d.

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Inflation Example, 5.7

• What is the net present value of a premium motor
  that costs an extra 500 initially and saves
  192/yr (at current electricity prices) for 20
  years if interest is 10 and inflation is 5?

In Excel - PV(0.04762,20,1)
Compare this to 1135 without fuel escalation
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Cash-Flows Incorporating Inflation

• Cash-flows can be expressed either in terms of
  dollars that take into account inflation
  (current, inflated, after inflation), or in terms
  of dollars that do not take into account
  inflation (constant, inflation free, before
  inflation).
• Well define the set of constant (inflation free)
  currency flows
• Well define the set of current (inflated)
  currency flows

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Cash-Flows Incorporation Inflation

• We use the relationship for inflated (current)
  dollars
• or equivalently
• with W t expressed in reference year 0 (todays)
  dollars, and e giving the rate of inflation

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Cash-Flows Incorporating Inflation

• Then, we have
• Therefore, the real interest rate d is used to
  discount the constant (inflation free) flows
  while the buyers discount rate d is used for the
  inflated flows.

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Cash-Flows Incorporating Inflation

• Whenever inflation is taken into account, it is
  convenient to carry out the analysis in present
  worth rather than future worth or on a cash
  flow basis
• Under inflation, e gt 0, it follows that a uniform
  set of cash flows
  implies a real decline in the cash flows

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Inflation Calculation Example

• Consider an annual inflation rate, e 4 , and
  assume the cost for a piece of equipment is
  constant for the next 3 years in terms of todays
  
• The corresponding cash flows in current are

The interpretation is that 1,125 in three years
has the same value as 1,000 today.
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Ex. 5.8 IRR for HVAC Retrofit with
Inflation

• An energy efficiency retrofit of a commercial
  site reduces the annual HVAC load consumption
  from 2.3 GWh to 0.8 GWh and the peak demand by
  0.15 MW
• Electricity costs are 60 /MWh and demand
  charges are 7,000 /(MW-mo) and these prices
  escalate at an annual rate of e 5
• The retrofit requires a 500,000 investment
  today and is planned to have a 15 year lifetime
• Evaluate the IRR for this project

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Ex. 5.8 IRR for HVAC Retrofit with
Inflation

• The annual savings are
• The IRR0 (without fuel inflation) is the value of
  d that results in a net present value of zero

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Ex. 5.8 IRR for HVAC Retrofit with
Inflation

• With inflation factored in, we have
• Can also use

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Annualized Investment

• A capital investment, such as a renewable energy
  project, requires funds, either borrowed from a
  bank, or obtained from investors, or taken from
  the owners own accounts
• Conceptually, we may view the investment as a
  loan with interest rate i that converts the
  investment costs into a series of equal annual
  payments to pay back the loan with the interest

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Annualized Investment

• For this purpose, we use a uniform cash flow set
  and the relation

present worth
equal payment term
equal payment series present worth factor PVF(d,n)

• Just rewrite the above equation to solve for A

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Annualized Investment

• Then, the equal annual payments are given by
• The capital recovery factor, CRF(i,n), is the
  inverse of the present value function PVF
• CRF measures the speed with which the initial
  investment is repaid
• Capital recovery function in Microsoft Excel
  PMT(rate,nper,pv)

capital recovery factor
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Ex. 5.9, Efficient Air Conditioner

• An efficiency upgrade of an air conditioner
  incurs a 1,000 investment and saves 200/yr
• The 1,000 is a 10-year loan repaid at 7
  interest
• Find the annual savings
• The annual payments are
• Total annual savings are
• The benefits/cost ratio is

In Excel -PMT(0.07,10,1)
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Ex. 5.10, PV System

• Consider a 3 kW PV system with CF 0.25
• The initial cost of 10,000 is paid for with a
  20-year 6 loan
• Assume no other annual costs
• Find the cost of electricity (/kWh)

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Ex. 5.10, PV System

• The annual loan repayments are
• The annual energy generated is
• The cost of electricity is

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Levelized Bus-Bar Costs

• Various alternatives must be compared on a
  consistent basis taking into account
• inflation impacts
• fixed investment costs
• variable costs
• The customary approach for cost valuation
  consists of the following steps
• present worthing of all the cash-flow
• determining the equal amount of an equivalent
  annual uniform cash-flow set
• determination of the yearly total generation
• The ratio of the equivalent annual cost to annual
  electricity generated is the levelized bus bar
  cost

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Levelized Bus-Bar Costs

• Present value of escalating annual costs
• Now find an equivalent annual cost
• This is called the levelized annual cost, and
  the levelizing factor is

LF 1 means no inflation
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Ex. 5.11, Microturbine Engine
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Ex. 5.11, Microturbine Engine

• For the microturbine with the characteristics
  given in the table, find the levelized (/kWh)
  cost over a 20-year lifetime
• The levelized fixed costs CF,L from (5.27) are
• The annual cost of fuel and OM in the initial
  year (not including inflation) is

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Ex. 5.11, Microturbine Engine

• We need to levelize A0 to account for inflation
• The levelizing factor (LF) is
• Compute d and LF

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Ex. 5.11 Microturbine Engine

• Then, the levelized annual cost CA,L is
• Add levelized fixed cost CF,L and levelized
  annual cost CF,L
• Get the levelized bus-bar cost CBB,L

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Combined Heat and Power (CHP)

• Some DG technologies such as microturbines and
  fuel cells produce usable waste heat
• Higher temperature waste heat is more versatile
• Using waste heat can displace the need to buy
  electricity or an expensive fuel such as propane
• One challenge is to appropriately time the use of
  the electric and thermal energy

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CHP Regional Application Centers

• Project sponsored by US Department of Energy
• Goal is to double the amount of installed CHP
  capacity by 2010 for 46,000 MW of new capacity
• Include internal combustion engines,
  microturbines, and fuel cells
• Recovered thermal energy used for cooling,
  heating, controlling humidity

http//www.chpcentermw.org/home.html
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CHP Efficiency Simple approach

• Difficult to quantify the value of a unit of
  electricity is much higher than unit of thermal
  energy
• Simplest approach
• Doesnt distinguish between the two types of
  energy

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Evaluating CHP Efficiency

• Compare CHP to separate heat and power
• Better approach
• Drawback - need to estimate efficiency of the
  boiler

Electricity
Heat
Cogeneration Plant
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Evaluating CHP Efficiency
60 units waste heat
22 units waste heat
90 units input heat
33 efficient grid
100 units input heat
30 units electricity
30 units electricity
Electricity
12 units waste heat
60 units input heat
48 units thermal
Cogeneration Plant
80 efficient boiler
48 units thermal
Heat
100 units input -gt 78 units output
150 units input -gt 78 units output
Figure 5.7
Figure 5.8
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Evaluating CHP Efficiency
Combined Heat and Power
Separate Electricity and Heat
100 units input -gt 78 units output
150 units input -gt 78 units output

• Without CHP, we need more input for the same
  output

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CHP Overall Energy Savings
Combined Heat and Power
Separate Electricity and Heat
100 units input -gt 78 units output
150 units input -gt 78 units output
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Energy Chargeable to Power (ECP)

• Same units as heat rate (Bth/kWh or kJ/kWh)
• Displaced thermal input based on efficiency of
  the replaced boiler
• Describes the extra thermal input needed to
  generate electricity with cogeneration
• Assumes the heat is needed anyway

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Example 5.13

• 30kW microturbine
• 29 electrical efficiency
• 47 usable heat recovered
• The waste heat offsets the fuel needed by a 75
  efficient boiler
• Find ECP

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Example 5.13
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Another ECP Equation

• Start with the first equation
• Suppose the total thermal input is 1 Btu, then

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Another ECP equation

• In terms of ?P (CHP electrical efficiency), ?H
  (CHP useful heat efficiency), and ?B (displaced
  boiler efficiency)
• Try it for Ex. 5.13, ?P 29, ?H 47, ?B
  75

Same answer!
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Cost Chargeable to Power (CCP)

• ECP is modified to account for the cost of fuel
• A measure of the added cost of electricity from
  cogeneration
• Operating cost chargeable to power (CCP)
• Units are /kWh

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Cost of Electricity vs Fuel-to-Heat Efficiency
Figure 5. 10 Cost of electricity from a fuel
cell. CF 0.9, ?B 0.85, fuel 8/106Btu,
plant cost 3/W, loan 8, 20 yrs
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Design of CHP Systems

• Inherently challenging
• Power-to-heat-ratio of equipment may be constant,
  but demand of power and heat changes

Figure 5. 11
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Cooling, Heating, and Cogeneration

• P/H ratio of buildings varies greatly, and we
  want to smooth it out
• Heat Pumps - Use electricity instead of heat in
  the winter
• Absorption Cooling - Use heat instead of
  electricity in the summer
• Cooling is a large part of the load, so it is
  important to talk about