assignment of tpp

8/12/2019 Assignment of Tpp

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A

ASSINGMENT REPORT

ON

STUDIE OF ECONOMICS OF THERMAL POWER PLANTDURING COURSE OF ACTION OF THERMAL POWER

PLANT ENGINEERING (TPP)

Submitted in partial fulfillment for Course Work of Thermal Power PlantEngineering

RAJASTHAN TECHNICAL UNIVERSITY

KOTA

SESSION 2010-2011

Guided by : Submitted by:

Mr. Dharmendra Hariyani Vikram Singh

Deptt. of Mechanical Engg. M.Tech. IInd

Sem

SKIT, Jaipur Thermal Engg.

DEPARTMENT OF MECHANICAL ENGINEERING

SWAMI KESHVANAND INSTITUTE OF TECNOLOGY

MANAGEMENT & GRAMOTHAN

JAGATPURA, JAIPUR-302017

PH. 0141 2758253-56


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Introduction

With the varied and fast changing global power market, the complexity of turbine

system economics has increased dramatically. In the past, power plants were primarily

government regulated and base loaded. Dispatch and electricity pricing was relatively

predictable. In todays market, with IPPs, there are endless variationsin the way

power is produced, provided, regulated, and purchased. OEMs and power producers

need to understand methods to quantify and compare parameters, and to understand

the drivers and uncertainties to properly evaluate decisions and their potential for profi

tability in this constantly changing marketplace.

The Power Market Drivers

To understand the power market, one must keep in mind the key differences

between this market and others:

Electric power can not be economically stored.

Unlike other commodities, electric power cannot be easily or economically stored. For

the most part, it must be produced on demand. While there are some efforts to retainenergy generated during off peak hours using technologies such as pumped storage, fl

ywheels, and/or superconductors, the cost is high and the effi ciency and reliability of

these methods is low.

The demand for electric power is constantly fluctuating.

The fluctuating demand for electric power is clear. Demand varies during the day,

with a morning and evening peak and varies over the year with a winter and summer

peak. Some of this fl uctuation can be predicted based on historical information, such

as the typical change in consumption over a day, and the typicalseasonal variations,

but the fl uctuations can shift signifi cantly from the norm due to uncontrollable events

like periods of severe weather.

Utilities have a high capital investment cost.


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The initial required investment for a power plant varies based on the type of power

plant. Typically, utilities have very high fi xed costs, spending almost fi ve times the

initial investment per dollar than other manufacturing endeavors1. These fi xed costs,

which are typically between $475/kW and $1430/kW2, include equipment for

generation, transmission, distribution, and permitting.

Power plants have a relatively long life cycle.

Power plants do require signifi cant initial investment, but they compensate

with very long life. Power plants operate for decades, with units operating 30 years

or more.Fuel prices are subject to negotiation and electricity prices are constantly

varying.

These factors impact the ability to compete effectively in the deregulated

market.There is an unfl inching expectation that required electric power is always and

immediately available.

Electricity has become a critical and integral part of the economy and there

is no tolerance for an inadequate supply no matter what the circumstances. This is

refl ected in the fact that electric consumption is generally accepted as one of the lead

economic indicators.

5.0

Turbine System Economics

@siemens.com

These unique features of the power market create a complex situation to evaluate and

choose effective strategies for power generation.The nature of this constantly

fluctuating demand for a commodity that must be essentially produced on demanddrives a varied supplier base, which can be broken into three basic types of operating

modesbase loaded plants, intermediate loaded plants, and peakers. Base load plants

operate continuously for long periods of time. They are typically large plants (>

200MW), which are economical and reliable to operate. These plants often do not

have the ability to change load quickly and take advantage of spot market peak

pricing.These units operate year round and all day with an overall economy, which

allows them to compete and operate profitably, even at low demand times. Base load


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plants provide the core of the power grid, and act to regulate and maintain the grid

frequency.Intermediate load supplier includes plants that operate to meet normal

fluctuating demands in the morning and evening hours and typically operate for 10 to

14 hours per day.Peak load suppliers include plants which can start up quickly and

supply power to meet high demands during periods of high or

low temperature when the combined base and intermediate load capacity is not

adequate. These plants are typically more expensive to operate, but offer operational

flexibility. Electricity supplied during a peak demand is sold at a premium, making

this the most profitable

time to generate.The US market is currently a hybrid market consisting of regulated

market regions and deregulated market regions, with each having different economic

drivers. The regulated markets have contracts that are cost based. Rates are negotiated,

fixed, and

renegotiated allowing for a set return on investment for the power producer. The

deregulated market is an auction market, with suppliers bidding into the market,

offering power to the grid. A controller ranks the bids and purchases power as needed

from the lowest cost supplier on up in price until the demand is met.The traditional

power market is regulated to meet local demand for power. In this scenario, the cost of

electricity is typically locked in by regulation and varies little. A network of base load

units and expensive peakers is put in place to enable the supply to meet

the fluctuating demand for power. Profits are based on a cost plus regulated model.

For a power producer, when choosing an OEM (Original Equipment Manufacturer) or

AE (Architect Engineer), the primary economic variables to consider are first time

plant cost,service costs, fuel costs, and plant availability and reliability. Fuel costs faroutweigh the other factors and the driver in this case for the OEM industry is

efficiency.

The power market has changed significantly with deregulation and different variables

must be considered to appropriately assess the potential from a customer and supplier

point of view. In the deregulated market power prices fluctuate drastically over time.

In different areas and different countries, the rules vary, but it is common to have

power generators bidding to supply to the grid and in some cases, there are penalties


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for promising power and then not being able to deliver. This market, which seems as

harried and volatile as the stock exchange floor, bids the operation of units which vary

in their ability to come up to full power, to sustain partial load and to assure delivery

of power at a precise time.

RAM and Economics

When a power plant is off line for maintenance, there is no income stream. When

power is offered on the spot market and the plant does not start, there may be

penalties. These scenarios drive a need to consider RAM. RAM is an acronym for

Reliability, Availability.

Reliability is used to express and quantify the unplanned maintenance needs of a

power plant. Reliability is a measure of how often a plant is available in comparison to

the total number of hours the plant would be available with no unexpected

maintenance. An ideal power plant has a reliability is 100%.Availability is a measure

of how often a unit is capable of providing service. The availability can be quantified

as the ratio of the total number of hours the unit is actually available in comparison to

the total number of hours. Availability considers both scheduled

and unscheduled maintenance and compares that to an ideal situation with no

maintenance outages at all. An ideal power plant has an availability that is less than

100%. Units with less frequent and shorter maintenance intervals have higher

availabilities.

Maintainability is used to express the cost of maintenance. This includes the cost for

parts, and the cost of the servicing.Maintainability can be used to compare plants that

require frequent, lower cost servicing, with plants that require less frequent, highercost servicing.

Combining these considerations, RAM looks at how often you can use the equipment

and how much it costs to keep it in operating condition. The concept of RAM is used

to consider the trade off between higher technology immature technologies, and less

advanced, but more reliable operation.

Modern Power Plant Economics


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Throughout the US market exists a hybrid market in which the trend has been to move

towards deregulation. The change from regulation to deregulation changed the

motivation and strategy of power producers. The old descriptions of the power

industry as stable and constant were traded in for adjectives like competitive, flexible,

dynamiceven risky. This volatile market favors plants that are highly fuel efficient

for base load and plants with quick, push button starting capability to meet daily

peak demands. As a power producer, the ideal situation is to be efficient enough to

compete in the base loaded market or flexible enough to change load quickly to

compete in the peak market. For an OEM to have access to the largest market share,

this means providing engines that can do both.

This need to have it all has driven the technology of power plants. Equipment

manufacturers are competing to deliver plants with the highest efficiency and the most

flexibility. This demand for more performance results in pushing the technology

envelope.There have been positive and negative consequences to this change in the

industry. The driver for the change was to increase

competition in the marketplace, and consequently to reduce utility bills. A discussion

of the result in that regard is out of the scope of this paper, but significant secondary

consequences occurred which are relevant. The technology of gas turbines has

advanced very quickly

in this market. Efficiencies have increased and emissions have reduced. The

introduction of new technologies has increased in volume and scope, resulting in

issues in reliability and availability for less mature engines. A power producer must

decide between buying the most advanced technology with the highest efficiency, or

buying a model that is more mature, and more reliable, but not as efficient.Utilities are businesses. Regulated utilities are mainly concerned with maintaining the

lowest life cycle costs for their units.For deregulated enterprises profitability is what is

important. Prior to deregulation, sound economic decisions were important, and with

regulation, these calculations were fairly simple and reliable. Subsequent to

deregulation, sound economic decisions became absolutely critical to survival, while

becoming far more complex. Utilities must understand how to make appropriate

purchasing decisions, and OEMs must understand how these choices are made to


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compete effectively. The economics and the technology are intimately tied together.

While it may be clear which technical decision advances technology the fastest, it is

not always as clear which technical decision makes the best business case. Power plant

economics explores the cost of a decision over time.

Basic Power Plant Economics

Economic evaluation of a power plant can be explored in a number of ways. All

methods account for the cost of a decision over time. There are several accepted

methods to examine cost over time. A common approach is to evaluate net present

value (NPV).

The net present value method looks at the value of the project over time by converting

all income and expenditures into equivalent values at the current time and subtracting

the initial investment. To do this, the future interest rate and the rate of inflation must

be estimated and expressed as a discount rate, r. While these estimates are somewhat

inaccurate, the sensitivity to the assumption

can be explored to understand the implication of one choice over another.Where;In:

initial investment

r: discount rate or weighted average capitol cost for a given company

t: time

CashFlow: Incomeexpenses

The basic rule of thumb is that the NPV should be greater than zero for an investment,

though most investors will strive to get a hurdle rate that is higher.

To calculate the NPV of a power plant, the following parameters are needed:

1) Capital investments,

2) projected price of electricity,


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3) size of the plant,

4) capacity factors (how many hours the plant will operate in a given year),

5) dispatch payments

6) projected fuel costs,

7) operating and maintenance costs,

8) start up time and costs,

9) regulating costs, and

10) the discount rate (the cost of money).

For a power producer, the initial investment is the cost of the plant and initial costs for

support equipment and hiring staff.The income is calculated per annum, and is income

from the electricity sold to the grid. This is the price of electricity ($/MWh)multiplied

by number of hours the facility is producing power (MWh) in a typical year. In a

deregulated market, the price of electricity is determined by the market and for

calculation, should be evaluated based on historical data and forecasted fuel and

electricity prices.

In some markets, dispatch payments are another source of income. Dispatch payments

are fees paid to producers that are capable of providing power to the grid with a very

short lead time, typically in the range of 10 minutes from demand to supply (spinning

and nonspinning

reserve). These payments are for the assurance of capability and are paid regardless of

whether the capability is leveraged.

This provides incentive to suppliers to develop and maintain a fast start supply, so the

grid can adequately respond to unplanned peak needs.Expenses are determined on a

per annum basis and include the cost of fuel, and operating and maintenance costs.

The fuel cost is a variable cost and must be estimated. For the calculation, fuel cost

must be converted to annual cost in dollars by multiplying the fuel price times

cumulative fuel consumed per annum. The operating and maintenance costs include


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personnel costs, and the costs for scheduled and unscheduled maintenance. Operating

and maintenance costs are impacted directly by the mode of operation of a plant.The

frequency of maintenance is influenced by both operating hours and number of starts.

A single start for a combined cycle facility can result in a significant incremental

increase in maintenance costs, with one producer estimating $20,000 in incremental

costs for one combined cycle start3.Expenses may also include regulating costs.

Regulating costs are government instituted economic consequence to encourage

industry to make decisions that have been determined to be for the good of the people.

These costs include taxes and the cost of complying with government regulations.

Taxes may be implemented to influence companies to choose preferred technologies

or to impact the local job market. Technologies may be politically preferable due to

environmental or safety issues as viewed by the regulating government. Certain fuel

sources may be preferred to enhance energy independence or to promote local industry

and employment.Regulating costs also include costs for adhering to regulations that

are put in place to assure the safety of a facility, such as OSHA requirements.

Regulations and taxes are dependent on the current political climate and are subject to

frequent changes, often resulting in

the changing position of a certain facility in the market place. An example of this kind

of influence is environmental regulations where emissions credits are traded. Over

time a facility built to meet a certain regulation may become covered by a regulation

with a more

aggressive limit. This may mean that additional operating costs are incurrent to

purchase additional emissions credits, thus influencing the economics of the power

producer.The calculation complexity increases further when looking over the life of the power

plant. Power plants have a long life and many changes occur over the life of the plant.

Some of these risks can be hedged by investing in futures to fix the future price of

commodities such as fuels, or to insure against adverse business conditions, such as

long periods of mild weather.

Some gas turbines have the capability to operate on alternate fuels. Some units are

purchased with the flexibility to switch between gas and oil, and various qualities of


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oil can be considered. As the price for oils and gases fluctuate, a plant with this

capability can be more competitive, but at a price. This option requires more capital

investment, both in the unit and in the supporting auxiliaries,and potentially in

licensing and permitting fees.

Units desire to remain competitive over time by being highly efficient and as a result

of this are also driven to increase capital investment over time. New technology is

regularly introduced and can be purchased as upgrades to improve efficiency. To

develop a symbiotic relationship with customers, some OEMs offer access to upgrades

to customers who purchase long term service agreements, thereby integrating the need

for consistent high quality service with the need for continually competitive

technology.

To understand the drivers for profitability, and the importance of RAM, the sensitivity

of the calculation can be explored to further understand the uncertainty of the

calculation. The variables can be examined in terms of controllable variable and

uncontrollable variables.

Operating Strategies and Options

For a base loaded plant, low margins are compensated by long operating times. Long,

uninterrupted operating times are supported by reliability and maintainability. Gas

turbine technology has been in service for many decades, and most units have

reliabilities of greater than 95%. However, the base line efficiency of a competitive

unit is constantly increasing. The old robust engine with learned out technology is

simply not efficient enough.

At the other end of the operating spectrum are peakers that are looking to leverage thehigh costs of electricity during peak needs. In June 25 of 1998, the price per

megawatt-hour of electricity in parts of the Midwest soared briefly from $40 to

$70004. Though the higher end of this scale is the exception and not the norm, the

implication is clear that the investment costs and fuel costs pale in the face of this

return and the only significant factor is how much the plant can generate. In this

market the goal is to be ready to run when the prices increase. Here again RAM is the


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driver since for the most part, a window of high potential for peak need can be

identified,

and so owners can schedule planned maintenance outside these windows. However,

availability and reliability are extremely important because if an owner pushes the

start button and does not get power, a competitor will quickly jump in and take over

that share of the market. If a plant is inoperable due to unplanned maintenance (low

reliability) then the opportunity to compete during this need will not even be possible.

Operators in the intermediate load business are balancing all of these needs. They

want to be chosen for operation, so efficiency is important, and they need to ready to

operate. Reliability, availability, and maintainability are all equally important.

Gas turbines operate in all three operating regimes. Simple cycle gas turbines are

installed to meet peak demands. They are extremely flexible, relatively low in initial

investment cost, and quick to install. While internal machine efficiencies are quite

high, gas turbine simple cycles exhaust at roughly 1000F, wasting a significant

amount of energy and resulting in a rather low cycle efficiency,and limiting the

application to peak markets.

For intermediate and base load applications, gas turbines are used in combined cycles.

In this arrangement, the gas turbine waste heat is used in a heat recovery steam

generator to power a bottoming steam cycle. The combined efficiency of such power

plantsare quite competitive (> 55%), easily fitting into the intermediate load market,

and capable of competing as a base loaded units.

In both scenarios there are economic complexities. The base loaded plants must

maintain very low costs. The business cases are based on low margins and highvolume. Large plants with high availability and low maintenance needs have the

advantage in this market. The business cases for load following plants are based on a

low volume high profit model, which requires them to be available when the need is

present. The economics improve when an unexpected peak in demand occurs. Those

who can meet these peaks quickly and reliably can reap the benefit of selling when the

market is at its highest. For these reason, RAMreliability, availability and

maintainability become significant driving factors in the power market.


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Controlling RAM While Increasing Technology Level

To maintain competitiveness, OEMs are aware that new technology must be

introduced at a lower risk level. Steps are taken to control risk during design, prior to

implementation, and during operation. During design, risk analysis and management

methods are used which allow a quantified assessment of the probability of a

particular failure and the consequences. Results of these analyses are used to mitigate

risks by either changing the design, or altering the consequential impact. Prior to

implementation, new technologies are being tested more thoroughly. In the past, most

of the testing was done with similar technologies in small engines, or aircraft engine

test beds. These approaches are similar, but not the same as the IGT (Industrial Gas

Turbine) application. To further reduce risk, some OEMs such as Siemens, have built

IGT test beds, so fully scaled new technology can be fully instrumented and tested in a

controlled

environment. For further reaching changes, an entire test site is constructed through

cooperation between an OEM and a power producer to validate a new design. In some

cases, new technologies are introduced to a single customer site with an agreement to

test out the

technology prior to release to a larger fleet. Subsequent to implementation, engines

can be fitted with improved monitors and sensors, condition monitoring, and better

controls. OEMs are now offering producers the option to purchase a monitoring

contract, where the OEM constantly monitors plant operation. Monitoring a fleet of

engines allows the OEM to develop probabilistic indicators of potential failures. Whensymptoms occurs, the owner is informed, and corrective actions can be taken in a

controlled manner, at a convenient time, greatly reducing the likelihood of unexpected

failure during a peak need. All of these features combine to reduce the risk of

increased RAM costs.

Since deregulation the focus on efficiency so over shadowed other needs that the

technical envelope was pushed very hard, very fast. The result was an improvement in


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the efficiency of gas turbines, but there was a partnering risk when leveraging

immature technologies. Using RAM in business models allows appropriate evaluation

of the benefit and risk of immature technologies and allows user to apply these

technologies intelligently. Today, OEMs and operators are both cognizant of the need

to make sound economic evaluations of technology options and to consider technology

maturity and the resultant RAM into their calculations. Additional actions (further

testing) are taken and additional products (such as online monitoring) are being

offered to reduce and control RAM. With these considerations, good choices can be

made by power producers that will provide for profit for the company and reliable

power for the

communities served.

Bonnie Marini

LOAD CURVES

A load curve is a plot showing the variation of load with respect to time. Load curve

of a locality

indicates cyclic variation, as human activity in general is cyclic. This results in load

curve of a day does not vary much from the previous day.

Load curves are useful for generation planning and enable station engineers to study

the pattern of variation of demand. They help to select size & number of generating

units and to create operating schedule of the power plant

Load curves used in power stations may be:

Daily load curve: -- Load variations during the day (24Hrs), may be half-hourly

(recorded once in 30 minutes) or hourly.Weekly load curve: -- Load variations during the week at different times of the day

plotted against No. of days. Weekend load can be seen to be significantly lower than

that during other working days for known reasons.

Yearly load curve: -- Load variations during the Year, which is derived from monthly

load curves of a particular year.

Information obtained from load curves:

Area under the load curve = Units generated,


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Highest point of the curve = Maximum Demand

(Area under curve) / (by total hours) = Average load

Load Curve For a Locality is usually obtained as given below

Load is divided into number of categories like private, public, Commercial,

Entertainment,

Hospitals, Transport, Industrial, Waterworks, Street Light etc. After preparing the load

sheet for a locality indicating the total load in each category (each category may have

different types of loads such as light, fan, refrigerator, heater, pump etc) load curve is

plotted for each category over a day (every hour or every 30 minutes) and then the

final load curve for the locality is obtained by summing them. This is Daily Load

Curve for that locality.

Maximum demand ( MD ):

The greatest demand of load on the power station during a given period. The highest

peak on the

power station load curve.

Demand Factor ( D.F ):

Ratio of maximum demand to connected load. This is usually less than 1. D.F = M.D /

C.L

Average load:

This is the average of loads on the power station in a given period.

Daily average load:

Average of loads on a power station in 1 day (24Hrs).

= total number of units (KWHr) / 24 Hrs

Monthly average load:Average of loads on a power station in 1 month (24Hrs x No. of days).

= Total number of units (KWHrs) / (24Hrs x No. of days)

Yearly average load:

Average of loads on a power station in 1 year (8760Hrs).

= total number of units / 8760 Hrs

Load factor ( LF ):


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The ratio of average load to maximum demand. typical less than 1. This is the

measure of the

effective use of the power station.

L.F = Average Load / Maximum Demand

= Annual Output (in KWHr) / (Installed Capacity X 8760 Hrs)

High LF Low cost per unit generated.

Diversity factor ( DF ):

The ratio of the sum of all individual maximum demands on the power station to the

Maximum

demand on the station. Consumer maximum demands donot occur at the same time

thus maximum demand on power station will always be less than the sum of

individual demands.

DF = Individual Maximum Demand / Total Station Maximum Demand

High DF Low MD Low plant capacity Low investment capital required.

Plant capacity factor ( PCF ):

The ratio of actual energy produced to the maximum possible energy that can be

produced on a

given period. This indicates the reserve capacity of a plant.

Economics

The success of any engineering undertaking depends on adequate financial planning to

ensure that the proceeds of the activity will exceed the costs. The construction of a

new power plant or the upgrading of an old one involves a major financial investment

for any energy company. Financial planning therefore starts long before ground is

broken, detailed design is begun, and orders are placed for equipment. Cost analysis

and fiscal control activities continue throughout the construction project and the

operating life of the plant. This section briefly introduces fundamentals of engineering

economics, with a slant toward power plant cost analysis as well as issues of

maintenance and equipment replacement.


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The cost to construct a power plant, waterworks, dam, bridge, factory, or other major

engineering work is called its capital cost. It is common to discuss the capital cost of

building a power plant in terms of dollars per kilowatt of plant power output. A plant

may cost $1100 per kilowatt of installed power generation capacity, for instance.

In addition to the cost of building the plant, there are many additional expenditures

required to sustain its operation. These are called operating costs. They may be

occasional, or they may occur regularly and continue throughout the life of the plant.

Often these costs are periodic, or are taken to be periodic for convenience of analysis.

There are, for instance, annual fuel costs, salary expenses, and administrative and

maintenance costs that are not associated with the initial cost of the plant but are the

continuing costs of generating and selling power. Operating costs are sometimes

related to the amount of electrical energy sold. Usually they are expressed in cents per

kilowatt-hour of energy distributed to customers.

Thus the expenses associated with power generation and other business endeavors

may be thought of as two types:

(1) initial costs usually associated with the purchase of land, building site preparation,

construction, and the purchase of plant equipment; and

(2) recurring operating costs of a periodic or cyclic nature.

It is frequently desirable to express all costs on a common basis. The company and its

investors may wish to know what annual sum of money is equivalent to both the

capital and operating costs. The company may, for example, borrow money to finance

the capital cost of the plant and then pay the resulting debt over the expected usefullife of the plant, say, 30 or 40 years. On the other hand, they may wish to know what

present sum would be required to ensure the payment of all future expenses of the

enterprise.

It is clear that $100 in hand today is not the same as $100 in hand ten years from now.

One difference is that money can earn interest. One hundred dollars invested today at

8% annual compound interest will become $215.89 in ten years. Clearly, an important

aspect of engineering economics is the time value of money.


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Compound Interest

If Alice lends Betty $500, who agrees to pay $50 each year for five years for the use of

the money, together with the original $500, then at the end of the fifth year, Alice will

have earned $250 insimple interest and receive a total of $750 in return. The annual

interest rate is

i = Annual interest / Capital = 50 / 500 = 0.1

or (0.1)(100) = 10% rate of return.

If, however, Betty keeps the interest instead of paying it to Alice annually, and

eventually pays 10% on both the retained interest and the capital, the deal involves

compound interest. The total sum to be returned to Alice after 5 years is computed as

follows: At the end of the first year Alice has earned $50 in interest. The interest for

the 161 next year should be paid on the original sum and on the $50 interest earned in

the first year, or $550. The interest on this sum for the second year is 0.1 $550 =

$55. The following table shows the calculation of the annual debt for the five-year

loan of $500 at 10% interest:

At the End of: The Accumulated Debt is: This Sum

First year $500 + 0.1 x $500 = $550.00

Second year $550 + 0.1 x $550 = $605.00

Third year $605 + 0.1 x $605 = $665.50Fourth year $665.50 + 0.1 x $665.50 = $732.05

Fifth year $732.05 + 0.1 x $732.05 = $805.26

It is evident that the interest earned on the preceding interest accumulation causes

the annual indebtedness to grow at a increasing rate. It can be shown that the future

sum, S, is given by


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S =P (1 + i)n

whereP is theprincipal, the initial sum invested; i is the interest rate, and n is the

number of investment periods, in this case the number of years. Here the factor

multiplying the principal,

S / P = (1 + i)n

is called the compound amount factor, CAF. The difference between simple and

compound interest may not be spectacular for short investment periods but it is very

impressive for long periods of time such as the operating life of a power plant. For our

example, the CAF is (1 + 0.1)5 = 1.6105, and S = 500(1.6105) = $805.26. Now,

consider the following closely related problem.

EXAMPLE

What sum is required now, at 8% interest compunded annually, to produce one million

dollars in 25 years?

Solution

Thefuture sum is

S =P (1 + i)n = 1,000,000 =P (1 + 0.08)25

Solving forP, thepresent sum is 1,000,000/(1.08)25 = $146,017.90. Thus, compound

interest brings a return of almost over seven times the original investment here. The

162same present sum invested at 8%simple interest for twenty-five years would produce

a

future sum of less than half a million dollars.

_____________________________________________________________________

In the example, the inverse of the CAF was used to determine thepresent worth of a

future sum. The inverse of the CAF is called thepresent-worth factor, ( PWF):


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PWF = P/ S = 1/ (1 +I)n

Thus we see that the time value of money is related to the compound interest that can

be earned, and that taking compounding into account can be important. To recklessly

adapt an old adage, .A dollar in the hand is worth two (or more) in the future (if

invested wisely)..

Capital Recovery

Another important aspect of compound interest is the relationship between a present

sum of money and a regular series of uniform payments. Consider a series of five

annual payments ofR dollars each, when the interest rate is i. What is the present

dollar equivalent of these payments? Applying the CAF as in the preceding

example,withR as the future sum, the present sum associated with the first payment is

R/(1 + i).

The present sum associated with the second payment isR/(1 + i)2. Thus the present

worth of the five payments is

P = R [ (1 + i) . 1 + (1 + i) . 2 + (1 + i) . 3 + (1 + i) . 4 + (1 + i) . 5 ]

It may be shown that this expression can be written as

P = R [(1 + i)5 . 1]/[i(1 + i)5].

The factor multiplying the annual sum R is called theseries present-worth factor,

SPWF, which for n years is:

SPWF =P/ R = [(1 + i)n . 1]/[i(1 + i)n]

Solving forR, we obtain an expression for the regular annual payment for n years

needed to fund a present expenditure ofP dollars at an interest rate i. The resulting

factor is called the capital recovery factor, CRF, which is the reciprocal of the series

present worth factor:

CRF =R / P = i(1 + i) n / [(1 + i) n . 1]


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EXAMPLE

What uniform annual payments are required for forty years at 12% interest to retire the

debt associated with the purchase of a $500,000,000 power plant?

Solution

Using equation (4.2), we get

R = Pi (1 + i)/[(1 + i) n . 1] = 5108(0.12)(1 + 0.12) 40/(1.1240 . 1)

= $60,651,813

This sum may be regarded as part of the annual operating expense of the plant. It must

be recovered annually by the returns from the sale of power.

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A Preliminary Design Analysis of a 500-MW Plant

Consider the design of a 500-megawatt steam power plant with a heat rate of 10,000

Btu/kW-hr and a water-cooled condenser with a 20F cooling-water temperature rise

produced by heat transfer from the condensing steam. The plant uses coal with a

heating value of 10,000 Btu/lbm. Let us estimate the magnitude of some of the

parameters that characterize the design of the plant. The reader should verify carefully

each of the following calculations.

A 500-megawatt plant operating at full load produces 500,000 kW and an annual

electrical energy generation of

500,000 _ 365 _ 24 = 4.38 109 kW-hr

With a heat rate of 10,000 Btu/kW-hr, this requires a heat addition rate of

500,000 _ 10,000 = 5 109 Btu/hr

Coal with an assumed heating value of 10,000 Btu/lbm must therefore be supplied at a

rate of 5 109 / 104 = 500,000 lbm/hr or 500,000 / 2000 = 250 tons/hr.


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A dedicated coal car carries about 100 tons. Hence the plant requires 250 /100 = 2.5

cars per hour of continuous operation. A coal unit train typically has about 100 cars.

Then the plant

Needs

2.5 _ 24 / 100 = 0.6 unit trains per day, or a unit train roughly every two days.If coal

costs $30 per ton, the annual cost of fuel will be

30 _ 250 _ 24 _ 365 = $65,700,000

The cost of fuel alone per kW-hr, based on 100% annual plant capacity, will be

65,700,000/(500,000 _ 365 _ 24) = $0.015/kW-hr _ 1.5 cents/kW-hr

The annual plant factor, or annual capacity factor, expressed as a decimal fraction , is

the ratio of the actual annual generation to the annual generation at 100 % capacity. If

the coal has 10% ash, the plant will produce 250 _ 0.1 = 25 tons of ash per hour.Under

some circumstances the ash may be used in the production of cement or other paving

materials. If it is not marketable, it is stabilized and stored in nearby ash ponds until it

can be moved to a permanent disposal site.

Similarly, if 2% of the coal is sulfur and half of it is removed from the combustion

products, 2.5 tons per hour is produced for disposal. If the sulfur is of sufficient

purity,it may be sold as an industrial chemical.With an air-fuel ratio of 14, an air flow

rate of 14 _ 500,000 = 7,000,000 lbm/hr is required for combustion. This information

is important in determining the size of the

induced- and forced-draft fans, that of their driving motors or turbines, and of the

plant.s gas path flow passages.

The heat rate of 10,000 Btu/kW-hr corresponds to a thermal efficiency of 3413/10,000= 0.3413 or 34.13%. If we approximate the heat of vaporization of water as 1000

Btu/lbm, the throttle steam flow rate, with no superheat, would be about 10,000 _

500,000 / 1000 = 5,000,000 lbm/hr

This determines the required capacity of the feedwater pumps and is important in

sizing the passages for the water path. The above thermal efficiency implies that about

65% of the energy of the fuel is rejected into the environment, mostly through the


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condenser and the exiting stack-gas energy. As an upper limit, assume that all of the

heat is rejected in the condenser. Thus

(1 . 0.3413)(5 109) = 3.29 109 Btu/hr must be rejected to condenser cooling water.

With 20 water temperature rise in the condenser, this rate of cooling requires a

cooling-water flow rate to the condenser of 3.29 109/(1.0 20) = 1.65 108 lbm /

hr

assuming a water heat capacity of 1.0 Btu/lbm-R. This gives information relevant to

the design sizing of cooling-water lines, cooling towers, and water pump capacities.

These back-of-the-envelope calculations should not be regarded as precise, but they

are reasonable estimates of the magnitudes of important power plant parameters. Such

estimates are useful in establishing a conceptual framework of the relationships among

design factors and of the magnitude of the design problem.

EXAMPLE

Relating to the above rough design of a 500-MW plant, and assuming the capital cost

information of Example 4.3, determine the capital cost per kW of generation capacity

and estimate the minimum cost of generation for the plant if it is predicted to have an

annual plant factor of 80% and maintenance and administrative costs of $0.007 /kW-

hr.

Solution

The unit cost of the power plant is

$500,000,000/(500 _1000) = $1000 per kW-hr

of capacity. The capital cost part of the annual cost of power generation is(60,651,813 _100)/(365 _ 24 _ 0.8 _500,000) = 1.73 cents per kW-hr

The cost of coal was determned to be 1.5 cents/kW-hr. The minimum cost of

producing electricity is then

1.73 + 1.5 + 0.7 = 3.93 cents per kW-hr

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assignment of tpp

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