Life Cycle Costing

Engineering economy is the application of economic factors and

criteria to evaluate alternatives, considering the time value of
money.
The engineering economy study involves computing a specific
economic measure of worth for estimated cash flows over a
specific period of time.
The terms interest, interest period and interest rate are useful in
calculating equivalent sums of money for an interest period.
Interest is the manifestation of the time value of money.
Simple Interest
Simple interest is calculated using the principal only, ignoring
any interest accrued in preceding interest periods.
The total simple interest over several periods is computed as:
Simple Interest = (Principal)x(Number of Periods)x(Interest Rate)
Here the interest rate is expressed in decimal form. The total sum
accrued at the end of n interest periods is given by:
S = Sum accrued at the end of interest periods (also called Future
Worth)
P = Principal (also called Present Worth)
n = Number of interest periods (normally one year is taken as one
interest period)
i = Interest rate (normally annual interest rate)
( ) i n P S + = 1
Compound Interest
For compound interest, the interest accrued for each interest period is
calculated on the principal plus the total amount of interest
accumulated in all previous periods.
Compound interest reflects the effect of the time value of money on
the interest also. The interest for one period is calculated as:
Compound Interest =(Principal + All accrued Interest)x (Interest Rate)
The total sum accrued after a number of interest periods can be
calculated from the following expression:
S
n
= Sum accrued at the end of n interest periods
P = Principal
i = Interest rate expressed in decimal form (annual interest rate)
n = Number of interest periods (number of years)
( )
n
n
i P S + = 1
Compound Interest-1
We can see from the above two expressions that the sum accrued
at the end of first year would be same for both simple interest and
compound interest calculations .
However, for interest periods greater than one year, the sum
accrued for compound interest would be larger.
What happens if the interest is compounded more than once in a
year?
We need to modify equation (2) and is given by:
nm
n
m
i
P S

+ = 1
Compound Interest-2
m = Number of periods the interest is compounded in one year
i = Annual interest rate in decimal form
n = Number of years
We can extend equation (3) to calculate the sum accrued if the
interest is compounded continuously. Here m tends to .
Taking the limits such that m goes to infinity, we get the
following expression:
in
e P S =
For all practical purposes, equation (2) is used for interest
calculations and repeated here for convenience:
( )
n
n
i P S + = 1
Compound Interest-3
Here,
S
n
= Future Worth of money .
P = Present Worth of the money.
(1+i)n = Future Worth Factor.
Given the present worth, annual interest rate and number of
years, we can calculate the future worth.
There may be situations when the future worth of money is given
and we need to find the present worth of the money.
The above equation can be re-arranged to calculate the present
worth, given by:
( )
n
n
i
S
P
+
=
1
Here,
( )
n
i + 1
1
= Present Worth Factor
Compound Interest-4
To carry out calculations, it is convenient to draw what is called
as cash flow diagram.
The following figure gives one such cash flow diagram:
The cash flow diagram helps in analyzing the problem better.
Equations (2) and (5) are used in problems concerning single payment.
In todays world we deal with problems that involve annual/monthly
equal payments such as home mortgage payments, vehicle loans or
loans for consumer electronic goods.
Compound Interest-5
The following relationships hold good for problems involving
such uniform series:
( )
( )

+
+
=
n
n
i i
i
A P
1
1 1
P = Present worth
A = Uniform Annual amount (installments)
( )

+
=
i
i
A S
n
n
1 ) 1
S
n
= Future worth
From these equations, we can calculate present worth or future
worth given uniform annual amounts
Compound Interest-6
We can also calculate the uniform annual amounts given either
present worth or the future worth.
A typical example would be person borrowing money from a
financial institute for buying a vehicle.
Knowing the interest rate and number of installments, the person
can calculate the uniform equal amounts he or she has to pay
depending on the amount borrowed.
A typical cash flow diagram would look as follows:

P
A
The up-arrow indicates the amount coming in such as borrowing
and the down arrow indicates the amount going out such re-payments
towards the borrowing.
Inflation
In all the above equations, we had assumed that there is no
inflation .
Inflation is an increase in the amount of money necessary to
obtain the same amount of product before the inflated price was
present.
Inflation occurs due to downward change in the value of the
currency.
If C is the cash in hand today for buying a product, f is the
inflation rate, then the amount we need to pay for the same
product after n years would be C(1 + f)
n
, assuming uniform
inflation over the years.
The present worth of such money with interest component added
is given by:
Inflation-1
P
f
= Present worth with inflation taken into account.
If i = f, no change in worth, year after year.
If i > f, save and do not buy the product now.
If i < f, buy the product now and do not save.
An important relationship between the present worth and the
uniform annual amount taking inflation into account is given by
the following equation:
( )
( )
n
n
f
i
f
C P
+
+
=
1
1

+
+

+
=
n
i
f
f i
f
A P
1
1
1
1
for i f . If i = f, then we get the following relationship:
n A P =
Life Cycle Cost
Life cycle costing or LCC is an important factor for comparing
the alternatives and deciding on a particular process for
completing a project.
The different components taken into account for calculating
LCC are:
Here, Capital is the present worth. Replacement cost that may
occur at a later years need to converted to present worth.
Maintenance cost is annual maintenance cost and needs to be
converted to present worth and so is the energy cost.
LCC = Capital + Replacement cost + Maintenance cost + Energy
cost Salvage
Life Cycle Cost-1
Salvage is the money that is obtained while disposing the
machinery at the end of life cycle period.
Even this amount has to be converted to present worth for
calculating LCC.
Once we have the LCC value, we can easily find the Annual Life
Cycle Costing using the following equation:

+
+

+
=
n
i
f
f i
f
LCC
ALCC
1
1
1
) 1
Comparison of Alternative Energy
Systems using Life Cycle Cost Analysis
Electricity is a major secondary energy carrier and is
predominantly produced from fossil fuels .
Challenging concerns of the fossil fuel based power generation
are depletion of fossil fuels and global warming caused by
greenhouse gases (GHG) from the combustion of fossil fuels.
To achieve the goal of environmental sustainability in the power
sector, a major action would be to reduce the high reliance on
fossil fuels by resorting to the use of clean/renewable sources and
efficient generation/use of electricity.
Comparison of Alternative Energy
Systems using Life Cycle Cost Analysis
In order to consider the long-term implications of power
generation, a life cycle concept is adopted, which is a cradle-to-
grave approach to analyse an energy system in its entire life
cycle.
Life cycle assessment (LCA) is an effective tool to pin point the
environmental implications.
Life cycle cost analysis (LCCA) provides effective evaluation
to pinpoint cost effective alternatives.
LCA and LCCA should be combined to identify cost effective
power generation alternative scheme.
SOLAR PV
To calculate the life cycle cost per KWh the basic components
of a PV system are considered as follows.
PV panels
Batteries
Inverters
Charge controllers
SOLAR PV-1
SOLAR PV-2
We will ignore adding in the cost of the charge controller, since
this is only a few hundred dollars (whereas the whole system cost
will be in the thousands of dollars).
The user specified variables will be:
Peak power required to power appliances
Total energy produced/consumed per day
Hours of sunshine (average)
Cost of inverter as function of peak power required:
The amount of peak power the system can deliver will be
determined by the size of the system inverter .
The inverter being the device which converts the dc battery power
to ac.
SOLAR PV-3
P
peak
,
usage
= P
peak,
,
inverter
As determined by surveying current market prices for
inverters, the costs of an inverter are about Rs.50 per watt, or
(multiplying by 1000):
Cost
inverter
= Rs. 50000/kilowatt
Thus, the cost of the inverter, as a function of the peak power
used, is therefore:
Cost
inverter (Ppeak, usage)
=
Ppeak, usage
x Cost
inverter
Cost
inverter
= P
peak, usage
x Rs. 50000/kilowatt
SOLAR PV-4
Cost of solar panels as a function of energy usage:
The peak power produced by the solar panels is determined by
the type and number of solar panels one uses:
P
peak panels
= number of panels x power per panels
But peak usage is not necessarily equal to peak panel power.
This is because the power generated by the solar panels is stored
up over time by batteries.
so more peak power (but not energy) can be delivered by the
inverter than is produced by the panels.
E
produced
= E
used
SOLAR PV-5
Also, we need to know how long the sun shines each day on
average. Let this be denoted by T
sun
,
T
sun
= Hours of Sunshine on average.
Using the formula for power and energy (Power = Energy / Time),
we have,
P
peak panels
= E
used
/ T
sun
.
As determined from a survey of current market prices
Cost
panels
= Rs. 4 Lakhs /Kilowatt.
Cost
panels
= (Eused / Tsun ) x Rs. 4 Lakhs/kilo-watt
SOLAR PV-6
Cost of batteries as a function of energy usage:
The amount of energy stored (by batteries) determines how much
energy can be used after dark, or on a rainy day.
The lifetimes of deep cycle batteries are fairly short (3 - 10 years),
and depend on how well they are maintained (for example, one
needs to avoid overcharging, and overdrawing.
We will assume, in order not to discharge the battery more than
50%, that the batteries will be able to store twice the amount of
energy we use:
E
stored
= 2 x E
used
SOLAR PV-7
Presently, the cost of batteries is about Rs. 5000 per kilowatt-
hour of storage:
Cost
batteries
= Rs. 5000/kilowatt-hour
The cost of batteries, therefore, as a function of energy used, is
Cost
batteries
= 2 x E
used
x $100/kilowatt-hour
Calculation of life cycle cost per KWh:
Todays solar panels are estimated to last atleast 25 years. We
will therefore use 25 years as our life time.
SOLAR PV-8
Let us say i=10%
Capital Cost = 4Lakhs+ 50000+ 10000=Rs. 460000
Replacement Cost =
1 PV panel Rs. 4 Lakhs/KW 25 Years
2 Inverter Rs. 50000/KW 25 Years
3 Battery Rs. 10000/KWh 5 Years
20 15 10 5
) 1 (
10000
) 1 (
10000
) 1 (
10000
) 1 (
10000
i i i i +
+
+
+
+
+
+
=Rs. 13945/KWh
Maintenance Cost:As we are considering only from generating
point of view maintenance cost is negligible part.
Energy Cost: It does not require any external energy (because the
system uses sun energy) to produce the electrical energy.
SOLAR PV-9
CASE STUDY:
For a large home let us say
Life Cycle Cost =
20 13945 10000 20 400000
6
20
50000 5 + + +
=Rs. 2062233.333

Total KWh
used
= 25 Years
365 Days E
used
= 9125 E
used
Therefore Cost per KWh = Life Cycle Cost / (9125 E
used
)
=Rs. 11.3 per KWh
SOLAR THERMAL PLANT
The major components of this system to be considered in
calculating life cycle cost are:
Heat energy Collectors
Boiler
Steam turbine
electric generator
SOLAR THERMAL PLANT-1
The costs of the above mentioned components are listed in the
following table.
Item Cost in Rs/KW Life period
Heat energy
collectors
25000 20years
Boiler+ steam
turbine
13900 10years
Electric
generator
5500 10years
Accessories,
tools
1000 5years
SOLAR THERMAL PLANT-2
Now let us say interest rate i=10%
Then the life cycle cost per KW is calculated as follows:
Capital Cost = Cost of (heat energy collectors+boiler +steam turbine+
electric generator+accessories)
= 25000+13900+5500+1000
=Rs.45400
Replacement Cost =
15 10 5 10 10
) 1 . 1 (
1000
) 1 . 1 (
1000
) 1 . 1 (
1000
) 1 . 1 (
5500
) 1 . 1 (
13900
+
+
+
+
+
+
+
+
+
= Rs.8725.4
SOLAR THERMAL PLANT-3
Maintenance cost = 1% of total capital cost per year
= Rs. 3865.15
Therefore,
Life cycle cost per KW = 45400+8725.4+3865.15
=Rs.57990.55
MICRO HYDEL PLANT
These plants are used for power requirements less than 100KW.
The costs of all major components are specified in the table, the
costs of civil works and permits are not included.
1. Approximate Micro Hydel system Costs (Battery based):
Component 100W (flow rate 4lps and
head at 5m.
lifetime
Penstock Rs 32500 10years
Turbine-generator Rs 125000
Controller Rs 20000
MICRO HYDEL PLANT-1
Batteries Rs 26000 5years
Inverter Rs 60000
Power house Rs 10000
Miscellaneous Rs 10000
Maintenance Rs 2000per year
Estimates provided by Energy Alternatives Ltd.
MICRO HYDEL PLANT-2
Now let us say ,
Interest rate i=10%
Life period =20years
Then,
Capital Cost(C) = Cost of (Penstock + Turbine-Gen + Controller
+ Batteries + Inverter + (Power house+ Miscellaneous)
= Rs.283500
Replacement Cost(R):
Penstock
Battery 32392
) 1 . 1 (
26000
) 1 . 1 (
26000
) 1 . 1 (
26000
15 10 5
=
+
+
+
+
+
=
MICRO HYDEL PLANT-3
Therefore total replacement cost = Rs.44922
Maintenance Cost (M)
17027 }
) 1 . 1 (
1
1 {
1 . 0
20000
20
=
+
=
Life Cycle Cost (LCC) = C+R+M
= Rs.332919.
MICRO HYDEL PLANT-4
Approximate Micro Hydel system Costs (AC-Direct system):
Component 3.5 KW (flow rate at 14lps
and head at 50m)
Life time
Penstock Rs.80000 15years
Turbine-Gen Rs.165000
Controller Rs.95000
Power house Rs.50000
Miscellaneous Rs.82500
Installations Rs.100000
Maintenance Rs.1000 per year
Estimates provided by Thomson and Howe Energy Systems Inc.
MICRO HYDEL PLANT-5
Now for a life period of 30 years, at an interest rate of 10%,
Capital Cost (C):
= Cost of (Penstock + Turbine - Gen + Powerhouse + Mis. +
Installations)
= Rs.572500
Replacement Cost(R):
15
) 1 . 1 (
80000
+
=
= Rs.19151
Maintenance Cost (M):
= Rs.9426
Therefore,
Life Cycle Cost = C+R+M
= Rs.601077
BIOMASS PLANT
The major components of Biogas plant are listed as follows.
Gassifier
Piping
Sand filter
Diesel engine
Electric Generator
BIOMASS PLANT-1
The costs of different components of Biogas plant are specified
in the following table.
Item Cost in Rs.for 5KW
plant
Life
period
Biogas plant 127700 20years
Piping 8300 10years
Sand filter 4150 10years
7hp diesel engine 37700 15years
5KVA generator 78150 15years
Accessories, Tools 20750 10years
Engine room 16550 20years
BIOMASS PLANT-2
Now let us say interest rate i=10%
Then the life cycle cost is calculated as follows:
Capital Cost =127700+8300+4150+37700+78150+20750+16550
=Rs.293300
Replacement Cost =
10 15 15 10 10
) 1 . 1 (
20750
) 1 . 1 (
78150
) 1 . 1 (
37700
) 1 . 1 (
4150
) 1 . 1 (
8300
+
+
+
+
+
+
+
+
+
= Rs40533.6
Maintenance cost = 1% 0f total capital cost per year
= Rs. 24970.28
Therefore,
Life cycle cost =293300+40533.6+24970.58 = Rs.358803.8
WIND ENERGY SYSTEM
The major components of a wind energy system are:
Wind mill
Gear box
Controller
Wind turbine
Electric generator
WIND ENERGY SYSTEM-1
The costs of the above mentioned components are listed in the
following table.
Item Cost in Rs per KW Life period
Wind mill 25000 20years
Gearbox 2500 10years
Controller 2000 10years
Wind turbine 10500 15years
Electric generator 5500 15years
Accessories 1000 5years
WIND ENERGY SYSTEM-2
Now let us say interest rate i=10%
Then the life cycle cost per KW is calculated as follows:
Capital Cost = 25000+2500+2000+10500+5500+1000
=Rs. 46500
Replacement Cost =
15 10 5 15 15 10 10
) 1 . 1 (
1000
) 1 . 1 (
1000
) 1 . 1 (
1000
) 1 . 1 (
5500
) 1 . 1 (
10500
) 1 . 1 (
2000
) 1 . 1 (
2500
+
+
+
+
+
+
+
+
+
+
+
+
+
= Rs.6811
Maintenance cost = 1% of total capital cost per year
= Rs. 3958.8
Therefore,
Life cycle cost per KW = 46500+6811+3958.8 =Rs57269.8
Remarks
Different renewable sources are considered for Life Cycle Cost
analysis.
It was found that every source has its own advantage and
disadvantage, depending upon their location of installation and
in terms of various costs like capital, maintenance etc .
The microhydel plant was found to be having less installation
cost when compared to other plants.
The capital cost required for Wind and Wave energy system
found to be more.