ELG 4126 – DGDSustainable Electrical Power Systems

Winter 2015

DGD Introduction

• TA: Viktar Tatsiankou (PhD student)

• e-mail: viktar.tatsiankou@gmail.com

• Objectives of the DGD:– to teach students the economics of the distributed

energy generation

– to assist students with the background information for case studies

– to learn!

• The DGD is worth 15% of the course

• There will be 2 or 3 quizzes based on the DGD material, unannounced!

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Review from last DGD

• Distributed resources– Distributed generation

– Grid resources

– Demand-side resources

• Electric utility rate structure

– Standard residential rates

– Residential time-of-use (TOU) rates

– Demand charges

– Demand charges with a ratchet adjustment

– Load factor

– Real-time pricing

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Distributed generation

• Generation of electric power by a variety of small-scale producers (up to 50 MW) located close to the point of use

• Examples:

– Wind turbines

– Solar farms

– Fuel cells

– Mini-hydro

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Standard residential rates

• Example of standard residential electric rate

• This is an inverted block rate structure (designed to discourage excessive consumption)

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Power is measured in W or J/sEnergy is measured in J or W·s

1 J = 2.78 x 10-7 kW·h1 kW·h = 3.60 x 106 J

Table 1. Example of the standard residential rates

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Residential time-of-use rates

• Example of residential time-of-use (TOU) rates

• Implemented to shift the customers loads away from the daily energy peak

• A solar installation generation well coincides with the summer energy demand peak

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Table 2. Example of the time of use rates

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Demand charges

• Usually applies to industrial or commercial entities

• Based on the highest (typically monthly) power drawn (averaged over a 15 min period)

• For example, if the peak power draw is 100 kW over 15 min in June, demand charges are $900 for that month.

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Table 3. Example of the demand charges

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Demand charges with a ratchet adjustment

• Problem with demand charges

– Monetary significant for one month of the year

– No sufficient for the utility to pay for the peaking power plant they had to build to supply the load

• Solution is to implement a ratchet adjustment into the demand charges

• Benefits

– Huge penalties for customers who add a few kWh to their load right at their annual peak

– Gives incentive for customers to reduce their annual peak energy demand

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Load factor

• Ratio of a customer’s average power demand to its peak demand

• Useful way for utilities to characterize the cost of providing power to the customer

• For example, a customer with a hourly peak demand of 200 kW that uses 876,000 kWh/yr would have a load factor for 50%.

1 year = 365 days x 24 hr = 8760 hr/yr

Average power = 876,000 kWh/yr / 8760 hr/yr = 100 kW

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Real-time pricing (RTP)

• Time-of-use (TOU) rates are crude mechanisms that attempt to capture the true cost of utility service.

• TOU rates use large time blocks and have only two seasons

• RTP on other hand offers pricing schemes as:

– one-day-ahead

– hour-by-hour

– real-time pricing

• RTP allows utilities to charge customers the true cost of energy generation and distribution.

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Energy Economics – DGD 2

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Outline for today’s DGD

• Energy economics– Simple payback period

– Initial rate of return

– Net present value

– Internal rate of return

– Net present value and internal rate of return with fuel escalation

– Annualizing the investment

– Levelized bus-bar costs

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Simple payback period

• The simplest way to evaluate the economic value of a project.

• For example, an energy-efficient air conditioner that costs an extra $1000 and saves $200/year in electricity, would have a simple payback of 5 years.

• But what if the air conditioner malfunctions in 3 years?

• Rarely used because it does not include any information about the longevity of a project.

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Initial (simple) rate of return

• The initial rate of return is the inverse of the simple payback period.

• For example, if an efficiency investment with a 20% initial rate of return, which sounds very good, lasts only 5 years, then just as the device finally pays for itself, it dies and the investor has earned nothing.

• Usually used as a “minimum threshold” indicator.

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Net present value (NPV) 1

• In general, $1 today is not the same as $1 in 10 years. Why?

• So if an investment P earns an interest i, after one year, you will have F = P + i×P or P×(1+i). Analogously, after two years, you will have F = P×(1+i) 2 and so on F = P×(1+i) n .

• Typically, in the context of energy economics an interest irefers to a discount rate d.

• It can be thought of as the interest rate that you could have earned if you would have invested the money into an alternative investment.

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NPV 2

• Typically, an energy efficiency investment results in a stream of annual cash flows A, for n years, at some discount rate d.

• What is the present worth P of an energy efficiency investment based on this information?

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NPV 3

• For example, say you save $100/yr by installing the energy efficient lighting in your home. In 10 years this is equivalent to $1000, future worth. The question is “What is the present worth of this investment?” with discount d.

• If d = 1%, P = $100/yr × 9.47 yr = $947

• If d = 30%, P = $100/yr × 3.09 yr = $309

• So if d = 30%, would it make sense to spend $500 on installing the energy efficient lighting?

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Example 1

Two 100-hp electric motors are being considered—call them

“good” and “premium.” The good motor draws 79 kW and costs

$2400; the premium motor draws 77.5 kW and costs $2900. The

motors run 1600 hours per year with electricity costing

$0.08/kWh. Over a 20-year life, find the net present value of the

cheaper alternative when a discount rate of 10% is assumed.

We know that

• d = 10%

• n = 20 years

• A = Power consumption × Operation hrs × Electricity cost

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Example 1 (Solution)

The annual electricity cost for the two motors is

A(good) = 79 kW × 1600 hr/yr × $0.08/kWh = $10,112/yr

A(premium) = 77.5 kW× 1600 hr/yr × $0.08/kWh = $9920/yr

The present value factor for these 20-year cash flows with a 10%

discount rate is

The present value of the two motors, including first cost and

annual costs, is therefore

P(good) = $2400 + 8.5136 yr × $10,112/yr = $88,489

P(premium) = $2900 + 8.5136 yr × $9920/yr = $87,354

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Example 1 (Solution)

The premium motor is the better investment with a net present

value of

NPV = $88,489 − $87,354 = $1,135

or alternative solution is:

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Internal rate of return (IRR) 1

• IRR is the discount rate d which makes the NPV equal to zero

• IRR is the most persuasive measure of the value of an energy-efficiency project

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Internal rate of return (IRR) 2

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Table 1. Present value functions to help estimate the IRR

Internal rate of return (IRR) 3

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Internal rate of return (IRR) 4

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Table 1. Present value functions to help estimate the IRR

NPV and IRR with fuel escalation 1

• We must take into the account the rising cost of fuel.

• Add the 1+e term to the PVF function as shown, where e is the fuel escalation rate.

• $1 of fuel today, costs $1×(1+e) in 1 year, costs $1×(1+e)×(1+e) in year 2, costs $1×(1+e)n in year n.

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NPV and IRR with fuel escalation 2

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Example 1

The premium motor in Example 1 costs an extra $500 and

saves $192/yr at today’s price of electricity. If electricity

rises at an annual rate of 5%, find the net present value

of the premium motor if the best alternative

investment earns 10%.

Solution:

The equivalent discount rate with fuel escalation is

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Example 1 (Solution)

The present value function for 20 years of escalating

savings is

The net present value is

Without fuel escalation, the net present value of the

premium motor was only $1135 (from Example 1).

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Annualizing the investment

• In most cases the extra capital required for an energy-efficiency project must be borrowed.

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Example 2

A 3-kW photovoltaic system, which operates with a

capacity factor (CF) of 0.25, costs $10,000 to install.

There are no annual costs associated with the system

other than the payments on a 6%, 20-year loan. Find the

cost of electricity generated by the system (¢/kWh).

Solution:

The capital recovery factor in this case is 0.0872/yr.

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Example 2 (Solution)

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Table 2. Capital recovery factors as a function of interest rate and load

Example 2 (Solution)

The annual payments for the loan are

The actual energy generated by the PV systems is

The cost PV electricity is

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Levelized bus-bar costs 1

• There are two key components to the cost of electricity for a power plant:

– Up-front fixed cost to build the power plant

– Assortment of costs incurred in the future

• Operation and maintenance (O&M)

• Fuel escalation

• The ratio of the equivalent annual cost ($/yr) to the annual electricity generated (kWh/yr) is called the levelized, bus-bar cost of power:

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Levelized bus-bar costs 2

• Converts all of the costs into a series of equal annual amounts.

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Levelized bus-bar costs 3

• Converts all of the costs into a series of equal annual amounts.

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Levelized bus-bar costs 4

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Levelized bus-bar costs 5

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+

= Levelized Bus-Bar Cost

Example 3

A microturbine has the following characteristics:

– Plant cost = $850/kW

– Heat rate = 12,500 Btu/kWh

– Capacity factor = 0.70

– Initial fuel cost = $4.00/106 Btu

– Variable O&M cost = $0.002/kWh

– Fixed charge rate = 0.12/yr

– Owner discount rate = 0.10/yr

– Annual cost escalation rate = 0.06/yr

Find its levelized ($/kWh) cost of electricity over a 20-

year lifetime.

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Example 4 (Solution)

The levelized fixed cost is

The initial annual cost for fuel and O&M is

This needs to be levelized to account for inflation. The

inflation adjusted discount rate d would be

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Example 4 (Solution)

The levelizing factor for annual costs is

The levelized annual cost is therefore

This needs to be levelized to account for inflation. The

The levelized fixed plus annual cost is

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Questions

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