2DG cost evaluationCost is arguably the most important factor when evaluating feasibility of DG applications.Evaluation should be performed for the entire system life; i.e. total cost of ownership. Some of the factors affecting this calculation include:Capital costInterestsInstallationOperation costMaintenance costDepreciationDowntime cost (influenced by system availability)Other costs (taxes, emissions financial impact)Total cost of ownership is compared with traditional solutions (e.g. grid powered system) to determine system feasibility.
3FundamentalsMoney is a dynamic variable. I.e. money value changes in time.When money is invested, its amount changes with the return.When money is transferred into assets, its value change due to depreciation and/or inflation.Discounted cash flow is a technique that allow us to evaluate values as they change over time.Let’s define how we call things first:P is the present value.i is the interest for a given period of time, i.e. the cost of moneyduring the period under study.F is the future value.n is the number of periods of time we consider.Fundamental concept: “time is money.” Hence, as time ticks forward, so does the value of money.
4FundamentalsA dollar bill today worth the same as a dollar bill a year ago? Well, not really. The value of the same dollar bill today is not the same as a year ago…If a year ago I had deposited the dollar bill in a savings account in the bank, today that dollar bill would worth more.Conversely, if a year ago I bought a chocolate bar for $1, today I would need that same dollar plus inflation to buy the same chocolate bar (of course, if I consider that that precisely chocolate bar is as good as a year ago).Hence, in most simple cases, present value and future value are related byF = (1+i)nPThe factor (1+i)n is the compound interest.
5FundamentalsThe changing value of money and transactions over time can be represented with cash flow diagrams.What is the value of a chocolate bar that I bought a year ago for $1 if the annual inflation rate was 5%?$ 1.051t$ 1The values are obtained using F = (1+i)nP by performing the following calculation:$1.05 = (1+0.05)1($1)
6FundamentalsNow consider a more interesting example. Consider that you need to determine how to power a load that will double after 3 years of the initial operation. You have the option of buying enough DG units to power the initial load and the load increase in 3 years for $ 2M or you can buy enough for the initial load by $1M and buy the rest of the DG units in 5 years. Consider that the annual inflation rate is 3 %. You are financing your capital investment with a loan. Loans interest rate are a couple of points above a savings account interest rate which, in turn, is a couple of points above the inflation rate. So the savings account rate is 5 % and the loan’s rate is 7 %. What should you do? You have 3 options:Get a loan today for $ 2 M for all the capacity including the load increase in year 3.Get a loan for $ 1M now to power the initial load and after 3 years get more money for the additional load.Get a loan today for $ 2M, use only what it is needed initially ($1M) and invest the rest until using it in year 3
7Fundamentals First, a couple of notes about inflation. Inflation creates an additional cost for money. Circulating money in an economy can come from different sources. One are loans (i.e government bonds). Another is just printing bills without any real treasury income of equivalent value. Hence, in this last case there are more bills without any income to back them up, each of them worth less. So inflation can be seen as an additional hidden tax that affects money value.Over time all money regardless the source needs to be adjusted by inflation. Indexed cash flows do not consider inflation.The real interest rate for a loan is (Fisher equation):(1 + i) = (1+ r)(1 + d)Hence,(1 + r) = (1 + i)/(1 + d)For common small interest rate values the above equations can be replaced by rates sum or rates difference, respectively.
8FundamentalsCase #1: Get a loan today for $ 2 M for all the capacity including the load increase in year 3.$ 2.25 M123t$ 2 MSince the loan’s interest rate is 7 % - 3 % of inflation, the cost of that money (or the equipment that I bought with it without considering depreciation) is:$2.25M = (1+0.04)3($2M)Note: I could have considered the effect of equipment depreciation with a higher interest rate. Other methods can also be used.
9FundamentalsCase #2: Get a loan for $ 1M now to power the initial load and after 3 years get more money for the additional load.$ M123t$ 1 M$ MNow I need to consider the cost of the initial loan plus the cost of the additional investment adjusted by inflation.$2.217M = (1+0.04)3($1M) + (1+0.03)3($1M)Note: As with any thing, money can have different values.
10FundamentalsCase #3: Get a loan today for $ 2M, use only what it is needed initially ($1M) and invest the rest until using it in year 3.$ 2.28 M123t$ M$ 1.06 M$ 1 M$ 1 MNow I also need to consider the cost of the initial loan plus the cost of the additional investment adjusted by inflation. But after 3 years the investment to buy the equipment to power the additional load comes from the $1 M I initially deposited in the bank. Hence$2.28 M = (1+0.04)3($2M) + [(1+0.03)3($1M) - (1+0.02)3($1M)]
11Fundamentals Observations from the previous examples: I compare the three cases at the same instant (i.e. year 3)Same money (e.g. dollars) have different values.ConclusionsInflation affects the value of moneyInterest rate indicates how expensive or inexpensive is a given amount of money.Since the most expensive money is the one coming from loans it is better to get it as later as possible. But with inflation, once we borrow money it is better to spend it right away.Scalable systems are better because they allow for a better financial result over time.
12FundamentalsDG systems are often compared with the option of powering from the grid.Utility rates are divided among;Types of customersEnergy consumption (by layers or bands)Time-of-UseIndustrial and commercial customers also usually receive an extra charge, called demand charge, based on the highest amount of power consumed by the facility. Hence, DG may reduce electric bills even when there is not enough capacity for the entire facility by reducing the demand charge.Peak power consumption can be characterized with the load factorReal-time pricing is a rate structure in which prices are set based on the different energy cost during periods every day
13Economic Feasibility Evaluation One of the methods is based on the cost of electricity (COE) compared with the average electricity price (http://www.energy.ca.gov/distgen/economics/decision.html)Total cost of electricity in $/kWh isCOE = C&I + O&M + Fwhere C&I is the capital and installation cost, O&M is the operation and maintenance cost, and F is fuel cost.To simplify the analysis let’s assume that there is no inflation and the cost of money is zero. The analysis can include cost of money and inflation by preparing a cash flow diagram and calculating the net present value.Also to simplify the analysis, only some basic costs are considered. For example, downtime cost is not included. This cost affects both the COE and the average utility price. Downtime cost can be included by considering the cost of a down minute (or second or hour) in combination with the mean-time between failures and the mean down time of both the DG system under study and the grid.
14Economic Feasibility Evaluation C&I cost calculation:where TIC is the total installed cost, FCR is the fixed charge rate and CF is the capacity factor.FCR is calculated withwhere Ay is the amortization period in years. This calculation yields 1/Ay only because we assumed that the cost of money is zero.The capacity factor CF represents how much is the system utilized and it is calculated with
15Economic Feasibility Evaluation O&M cost per kWh calculation includes both variable and fixed costs (e.g. space).Cost of fuel implicitly includes energy inefficiency costs. It is calculated withwhere FP is the fuel price (this value can be changed in time according to a cash flow diagram) and HR is the heat rate.The heat rate is a measurement on how efficiently DG units can convert the energy in the fuel into the electric energy; i.e. it is the thermal input required to produce 1kWh of electric energy. It can be calculated withThe analysis implies comparing the COE with the electricity cost at the facility under study.
16Economic Feasibility Evaluation One simple way to compare whether or not it is convenient to pay some extra cost for a good is to calculate the simple payback period:Some disadvantages are that the calculation does not consider any cost of money, or non-uniform cash flows, or how long the compare goods are expected to last .A related way of comparing economic feasibility is by using the initial (simple) rate of return, which is the inverse of the simple payback period.An alternative that considers different cash flows is to calculate the value of the different cash flows being evaluated at the same time (e.g. present value) and then choose the option with the lowest cost. The present value of all project’s costs is called the life-cycle cost. The difference between the present cost of two projects is called the net present value.
17Economic Feasibility Evaluation Internal rate of return (IRR) is, arguably, the most widely used method for comparison.IRR allows to compare the return that could be obtained from competing options.In any cash flow process there are positive and negative influxes. The IRR is the discount rate that makes the net present value for a given cash flow equal to zero.Hence, IRR calculation implies considering some particular money values and some time horizon.The project with the highest IRR is usually the one selected (there are exceptions).The advantage of this method is that simplifies complex cash flows. However, solving the equation to find the IRR is not easy.
18Economic Feasibility Evaluation IRR calculationConsider the cash flow belowThe IRR is the discount rate r that makes the NPV of the above cash flow zero. Hence, it requires solving the next equation with r as the unknown.A1A23ATt12TA3A0
19Economic Feasibility Evaluation IRR example: Suppose that you want to determine whether or not to power a facility from a microgrid. The initial cost of the microgrid is $ 1M but it yields the net savings indicated below when compared with the case of powering the system from the grid. Should you install a microgrid, or should you better borrow the $1M and put it in an investment that yields 10% of return over a 10 year period? Assume that the microgrid components life is 10 years, too.
20Average annual savings Economic Feasibility EvaluationIRR example: One way of solving the problem is building an Excel spreadsheet and the iterate by trial an error to find the discount rate.Since the IRR is over 10%, it makes sense to use the microgrid option.This is the discount rate r cell. I change it until the bottom cell is close to zero.Initial investmentAnnual savingsAverage annual savings
21Economic Feasibility Evaluation Calculations of the NPV and the IRR can be adjusted by inflation. For example, fuel costs in the cash flow can be adjusted with a fuel escalation factor e.From slide 7, a nominal rate i translates into a real rate r when adjusted with an escalation factor e:(1 + r) = (1 + i) / (1 + e)With standard rates the relationship can be approximated tor = i - eSo the NPV isAnd the IRR is
22Economic Feasibility Evaluation When financing the capital investment with a loan, it is usually necessary to prepare a cash flow analysis that represent the principal borrowed and the regular payments.If the annual payments are represented by A, the principal is P, the loan nominal interest rate is i and the loan term is N, thenA = P.(CRF)where the capital recovery factor CRF isIf the payments are made on monthly basis then
23Economic Feasibility Evaluation One useful comparison index is called the levelized bus-bar cost (LBBC).The LBBC is the ratio between the equivalent annual cost to the annual electricity generated.Levelized costs refers to an equivalent uniform cash flow adjusted by an escalation factor.It is possible to find many additional ways to evaluate the economical feasibility of a project. Many of those try to capture some particular situation that affects an accurate evaluation.$Actual costsLevelized coststime
24DG Technologies C&I Costs Cost of many of the technologies are expected to go down significantly during the next few years.Some approximate capital costs are:Microturbines: 650 $/kWReciprocating engines: 200 $/kW (kW) to 1500 $/kW (small units)Small wind turbines: 1000 to 2000 $/kWPV systems: 4000 $/kW to 30 $/kWFuel cells:PEMFC: 4000 $/kWPAFC: 3000 to 3500 $/kWMCFC: 800 to 2000 $/kWSOFC: 1300 to 2000 $/kWCheck the presentations on energy storage for their approximate costs
25DG Technologies C&I Costs Typical C&I distribution in a microgrid.In traditional backup systems energy storage accounts for 40% of the total cost. In microgrids that 40% is divided into 30% for power generation and 15 % for energy storage with power delivery profile.“Other costs” include primarily financial and space costs.Microsources contribution decreases as system capacity increases.For residential applications with no energy storage contributions are divided equally between microsources, installation, and electronic interfaces.
26Combined-heat and power (CHP) Combined heat and power (CHP) refers to on-site generation of both electric power and heat.The generated heat can be used to warm or cool a facility, to drive some industrial process, or for other applications.In many microgrids, implementation of CHP is essential to achieve economic feasibility.In standard systems power is generated from electricity. Cooling is also achieved from electricity. Heating needs are usually meet by separate processes, usually by burning natural gas.In CHP, heating and cooling needs are met by using the waste heat generated by some types of microsources, such as microturbines and some fuel cells.
27Combined-heat and power (CHP) Usually, it is simple to understand how waste heat can be used to meet heating needs in a facility.Understanding how waste heat can be used to cool facilities requires a brief explanation on how air conditioning systems work.In standard air conditioning systems that energy comes from the electric grid. In CHP systems the energy comes from waste heat. Absorption chillers are the air conditioning equipment that removes heat by using waste heat.Take the example of a telecom or IT facility on the right.Cooling is achieved by removing heat from the place where heat is generated.The word “remove” implies that a work is needed. Hence, energy needs to be provided to the air conditioning system to remove the heat.
28Economic Feasibility Evaluation Energy efficiency evaluation :Energy Efficiency Ratio (EER):Coefficient of performance (COP):Discontinuous operation reduces the efficiency up to a halfEPA 430-F
29Economic Feasibility Evaluation The economic value of using CHP can be understood by remembering that the cost of fuel is given bywhere the heat rate is given byWithout reusing the waste heat typical thermal efficiencies are in the range of 20% to 40%. When the heat rate is used, efficiencies increase by 30 % to 40 %.CHP issues:Lack of modularity (systems are not scalable).Absorption chillers efficiency is relatively low.CHP may not meet heating/cooling requirements because design is done for the electric portion and waste heat is just a byproduct.CHP design and planning is usually “custom” made for each site.Without a good systems design CHP may have lower reliability than traditional cooling/heating solutions.
30Downtime costDowntime cost may have a significant impact on the total cost of ownership. Here there are some downtime cost examples:Hence, it it very important to understand system reliability