LIFE-CYCLE CALCULATION OF BUILDING ENERGY INVESTMENS. Kai Sirén AALTO UNIVERSITY

Transcription

1 LIFE-CYCLE CALCULATION OF BUILDING ENERGY INVESTMENS Kai Sirén AALTO UNIVERSITY October 2016

2 CONTENT 1. Introduction Moving the incomes and expences in time The basic problem Compounding and discounting of single entries Compounding and discounting of repeated entries Superimposing discounting factors for repeated entries LIFE-CYCLE CALCULATION METHODS Simple payback period Discounted payback period Net present value Internal rate of return Examples of using the life cycle calculation methods INTEREST RATES Nominal interest rate Inflation, real interest rate Inflation of energy prices - escalation OTHER APPLICATIONS OF THE COST FUNCTION Price for the saved or produced energy Maximum investment Further applications of the cost function SELECTION OF THE CALCULATION PARAMETERS Investment cost Energy price Interest rates Energy price escalation Calculation period LITERATURE

3 1. INTRODUCTION The building stock forms a remarkable national property in most developed countries. This property needs continuous care and renovation. The stock is also been constantly renewed when new buildings are built and old ones removed. Maintenance of existing buildings and building of new ones need resources and investments. Energy is one important resource needed during the buildings lifetime. The need for energy is largely determined in the construction phase of the building. Later on the energy efficiency can be improved by renovation measures. The energy related solutions, like the insulation levels, windows, heating, cooling and ventilation systems and so forth, determine the investment needed but also the energy costs for the building lifetime. For this reason the proper approach in estimating the profitability of any building energy related investment is the life-cycle approach. Every engineer working in the building energy sector should be familiar with the basic principles of life-cycle cost calculation (LCC). And for those working especially with buildings' life-cycle issues this is a part of core competencies. The following document is written in a compact form, focusing in the fundamental principles of the life-cycle cost calculation applied to building energy related issues. The reader is assumed to have previous knowledge of building HVAC and other energy related systems, their operating principles and features for which reason these systems are not described in this context. Also many aspects or methods completing the basic LCC principles like uncertainties and risk assessment, decision processes, tax and accounting aspects, financing as well as optimization of the investments are beyond the scope of this document. 2. MOVING THE INCOMES AND EXPENCES IN TIME 2.1. The basic problem Since money has a cost in terms of interest payable, it follows that a euro today is worth more than a euro in the future. When estimating the profitability of an energy economical investment, it is not enough to just to regard what happens in the present but the whole life span of the investment must be considered. In such a case the investments, expenses and possible incomes take place during different moments in time, Fig 2.1. The value of the different entries depends on the time as an interest rate, inflation or deflation is influencing their value, they cannot be added up just like that. The basic principle is to move all entries (investments, expenses, incomes), using appropriate interest rates, into the same moment of time. Then they can be added together to see the whether the sum is positive or negative. What is this moment in time is in principle indifferent and can be chosen. The move can be done into the present time or any other time during the lifetime of the investment. Fig. 2.1 shows a general presentation of the lifetime of an investment project. The first investment I 0 (the investment is treated as a negative entry) takes place at the moment t = 0 or the present time. A complementing investment I 3 is done in the end of the third year. The project returns yearly an amount of T k euro, starting from the end of the first year. At the same time the yearly expenses are K k euro also starting from the end of the first year. The duration of the whole project is n years after which the investments still have a final value S which can be either positive or negative 3

4 Figure 2.1 Expenses and earnings taking place in different times Now the question is whether this investment scheme is profitable or not when a yearly nominal interest rate of i [%/a] is chosen? The most straightforward approach is to move all entries into the present time and add them up to see whether the sum is positive or negative. Let's name this sum the net Present value or the cost function of the investment scheme and designate on a general level (2.1) Slightly simplifying one can say that if the cost function is a positive number the investment is profitable and if the cost function is a negative number the investment is not profitable. To do these movements in the time, interest and discounting factors are needed Compounding and discounting of single entries The interest rate is increasing the nominal value of capital when moving forward in time. When using a nominal yearly rate I, the compounded value of the principal amount P is after one year P + Pi = P(1+i). After two years it is P(1+i) + P(1+i)i = P(1+i) 2 and so on. At the end of year k the value has increased to 1 (2.2) where c sk = (1+i) k is the annual compounding factor for a single entry over k years. Correspondingly, when moving capital backwards in time from future towards present the 4

5 process is opposite. One year shift of the amount F yields a value F/(1+i), two years shift a value of F/(1+i) 2 and a k years shift /1 (2.3) where a sk = 1/(1+i) k is the annual discount factor for a single entry over k years. This factor is used to move an amount of cash flow from the end of year k to the present time or between any two moments in time being k years apart from each other. The influence of the time and interest rate on the discount factor is shown in Fig An important conclusion based on the figure is that the higher the interest rate, the lower is the impact of expenses or incomes taking place in the distant future when they are moved backwards to the present time. A high interest rate is emphasizing the present and the near future and ignoring what happens during a longer time span and a low interest rate is conversely appreciating all entries more equally during the lifetime of the investment. Figure 2.2 Annual discount factor for a single entry a sk = 1/(1+i) k. By applying now the above discounting principles onto the investment case in Figure 1 for a time span of n years the general cost function (2.1) takes the following form: *! " * & " * ' " ( # #$% " # #$% " #$% " #$% ) (2.4) 5

6 where T k is the income at the end of the year k, K k is the yearly expanses, I k is the investment term and S is the final (residual) value. The summation is done for the investment terms starting from the present time k = 0 to the end of the last year k = n. Other terms are equal to zero during the first year and the summation goes from k = 1 to k = n. In case there is no income or cost or investment during some year, the corresponding term in the summation is set equal to zero Compounding and discounting of repeated entries If during the calculating period the income is every year equal T k = T and it is realized at the end of each year, the cumulative sum of the incomes is developing according to Table 2.1. It can be noticed that the cumulative values at the end of each year is the sum of a geometric series where the first term is T and the common ratio between successive term is q = (1+i). As is well known, the sum of a geometric series having n terms is S n = b (1-q n ) / (1-q) where b is the first term of the series. Table 2.1 Development of cumulative income during n years. year income during the year amount in beginning of year growth during the year cumulative value at end of year 1 T 0 2 T T Ti T[(1+i)+1] 3 T T[(1+i)+1] T[(1+i)+1]i T[(i+1) 2 +(i+1)+1].... n T T[(i+1) n-2 + (i+1) n-3 + +(i+1)+1] T[(i+1) n-2 + (i+1) n-3 + +(i+1)+1]i T[(i+1) n-1 + (i+1) n-2 + +(i+1)+1] By applying the sum of the geometric series to calculate the cumulative sum of repeated incomes during n years we get the future value of the sum as * * # +1 + #,#$%) + #$%),# + #,#$% % -. (2.5) The compounding factor c rk is now the factor for repeated entries of equal size over n years. This kind of analysis is naturally not restricted to incomes only but the entry T in Eqn. (2.5) can be interpreted as any fixed income, expense or net income (income expense) which has a constant value during the calculation time. The same reasoning can also be done by moving the income T realized at the end of each year forward in time to year n using the compounding factors for a single entry, Fig Now the income of the first year is shifted n-1 years forwards with the factor (1+i) n-1, the income of the second year is shifted with the factor (1+i) n-2 and so forth until the shift from the end of the second last year to the end of the last year is done using the factor (1+i). Finally at the end of the last year the final entry of T is realized and we have all factors of the sum shown in table 2.1 together. 6

7 Figure 2.3 Shifting repeated entries forwards in time to year n. In a similar way can be derived the sum of repeated entries shifted backwards in time from year n to the present. Shifting from the end of the first year is done using the discounting factor 1/(1+i), shifting from the end of the second year using 1/(1+i) 2 and so forth until shifting from the end of year n is done using 1/(1+i) n. The first term of the series is now T/(1+i) and the common ratio is q = 1/(1+i). By applying the equation for the sum of a geometric series we get *! #,#$%.) #$% #,#$% #,#$%.) + +./ *. (2.6) % The discounting factor in the above equation a n for repeated entries over n years can be found in few different forms * #,#$%.) % #$%),# %#$% ). (2.7) Because the above discounting factor is used frequently in the following, it is worth to show its dependence on time and interest rate, Fig One conclusion drawn from the figure is that the discounting factor has a maximum value which is the inverse of the interest rate. Using a high rate the growth of the discounting factor is saturated quite rapidly. As a consequence a high interest rate is weighting entries that realize in the present and near future and ignoring what happens during a longer time span. And vice versa a low interest rate is appreciating all entries during the calculation time more equally. This behavior is similar to what we saw earlier in the case of discounting a single entry. This is in close connection with choosing the interest rate for the calculation and good to remember. 7

8 Figure 2.4 Discounting factor for repeated equal entries a n = (1-(1+i) -n )/i. If we now assume that in the example case of Fig. 1 all incomes remain constant T k = T and all expenses remain constant as well K k = K, the cost function in Eqn. (2.4) can be written in the following form * + * 0 1 ' 2 #$% 2 ( #$% ) (2.8) where now the summations are in a simpler form. It is underlined that this kind of treatment requires fixed size yearly repeated entries. If the incomes and expenses vary during the time span, Equation (2.4) operating with discounting of single entries should be used instead. In some cases there is a need to shift all entries related to an investment, incomes, expenses and also investments, as equally large yearly sums along the calculation time span, Fig These equal annuities are created by shifting all entries first into the present time and then by distributing this sum to equally sized entries for each year. 8

9 Figure 2.5 Converting incomes, expenses and investments to equal annuities. Because converting into annuities is an inverse process of discounting, it is done using an annuity factor which is the inverse of the discounting factor in Eqn. (2.8) c n = 1/a n = I / [1- (1+i) -n ]. In Table 2.2 there are shown all compounding factors, discounting factors and annuity factors related to repeated entries with an interest of i and a time span of n years. With these factors all shifts in time forwards and backwards as well as spreading into equal annuities can be handled. Table 2.2 Compounding, discounting and annuity factors with a yearly interest rate of i and a time span of n years. Result Initial value Present value P Value at the end F Annuity A Present value P 1 1 * 11,* Value at the end F 1 1 * 1 1 * 1 Yearly entry A 11,* 1 * Superimposing discounting factors for repeated entries Sometimes a situation occurs when during the time span under consideration there are few periods with a yearly constant income or cost, Fig

10 Figure 2.6 An example of two fixed incomes during the time under consideration. The most straightforward approach is to apply Eqn. (2.4) to discount each separate entry and then carry out the summation as follows 5! / *! 3 * & #$% " 4 " 1 #$% " #$% " ( # #. #$% ) (2.9) On the other hand, the approach of discounting factors for repeated entries can be used as well if one just discovers that these discounting factors can be added up and especially subtracted. Because the discounting factor in Eqn. (2.7) always covers the time from present (k = 0) to year n (k = n), it is not possible to discount the period 2 in Fig. 2.6 starting later than year zero, using this factor. But we have to think that period 2 (years 4 n) is a remainder of period 1+2 (years 0 n) and period 1 (years 0 3). In this way the discounting factor for years 4 n is 4,* * 5 (2.10) where a n is the discounting factor between years 0 n and a 3 is the discounting factor between years 0 3. In this way the cost function in Eqn. (2.9) can be written as follows 5 + # * * 0 1 ( #$% (2.11) ) where the expenses K are further assumed to be constant during the calculation period. Equation (2.11) works naturally also more general and serves here just to illustrate the superimposing principle. 10

11 3. LIFE-CYCLE CALCULATION METHODS 3.1. Simple payback period In the (simple) payback period method the idea is to compute that time span after which the cumulative incomes generated by the investment are equal to the investment itself. The simple payback does not account for the interest rate. Starting from the present value or cost function (2.4) and ignoring the residual value we get 8 7 " # 1 #$% " (3.1) where A k is the net income (gross income minus expenses) of year k and N is the payback period in years. Setting the interest rate equal to zero ignores the influence of the discounting as 1/(1+0) k = 1. Further assuming the net income to be fixed A k = A and setting the cost function equal to zero P = 0 the simple payback period becomes 9 1 /:. (3.2) This simply means that we divide the investment with the annual net income to get the payback period. The profitability of a project can now be estimated by comparing the computed payback period with a goal. If the computed payback period is no longer than the target, the project is regarded feasible. The target for the payback is in turn a subjective time set by the decision maker and is influenced by many factors like other competing investments, liquidity, type of investment, personal preferences etc. Often for example for energy saving measures the target is lower than for energy production projects because for example in energy production the primary business is not to save energy but to produce and sell energy. A typical target for the payback time is between 2 and 10 years. In some occasions like investments in renewable energies it could be even longer. The payback time can also be compared with the estimated lifetime of the investment. If the computed payback is longer than the lifetime, the investment is certainly not feasible Discounted payback period The simple payback period does not account for the time value of money, but the discounted payback period does. This means that the influence of the interest rate cannot be ignored. Starting from equation (3.1) and assuming again a constant value for the net income and using the notation (2.7) for the discounting factor we get : #,#$%.; % 1 0 (3.3) and solving further for N 11

12 9 = >? A B % >? / /CD. (3.4) For a long time period the maximum value of the discounting factor for repeated entries is a n = 1/i, Fig As a consequence N d has an upper limit because the year-by-year shrinking discounted cumulative income does not cover the investment anymore. For the same reason Eqn. (3.4) can be used for values i < A/I 0 only. When the interest is set equal to zero (i = 0) the discounted payback period is naturally same as the simple payback period. In case we cannot accept the simplifications used above, the discounted payback period needs to be calculated based on Eqn. (2.4) by numerically finding that number of years which gives the value zero (P = 0) for the cost function. The payback period method is simple to understand and illustrative. On the other hand, it has one severe weakness. It does not account in any way the incomes after the payback period. The lifetime of the investment can be much longer than the calculated payback period and the influence of this is fully ignored by this method. For this reason the payback method should be regarded as a directional method which can be used to do a rapid first assessment of profitability. However, it should always be complemented with some more advanced method to gain a better understanding Net present value Net present value (NPV) of an investment is the sum of all into the present time discounted incomes, expenses and investments, over a period of time. In a mathematical form this principle was previously shown as equation (2.4) *! " * & " * ' " ( # #$% " # #$% ". #$% " #$% ) (3.5) The profitability of an investment can now be estimated based on the value of the sum (3.5) in a following way: P < 0 The required rate of return set by the interest rate is not realized. The investment can be unprofitable. However, it can also be profitable but with a smaller rate. P = 0 The investment is profitable and the rate of return is realized exactly as demanded. P > 0 The investment is profitable. The required rate of return is exceeding the demand which was set by the interest rate. The absolute limit of profitability can be examined by setting the interest rate equal to zero and computing the net present value. If P < 0 in this case, the investment is generating loss and thus is definitely unprofitable. In case the incomes and expenses have fixed values, the net present value can be formulated using the discounting factors for repeated entries 12

13 * + * 01 ( #$% ) (3.6) where a nt and a nk are discounting factors for incomes and expenses respectively and which often are not equal but have deviant values, like we later on find out. The net present method is one of the most frequently used ways to estimate the profitability of an investment or to compare several investments. It takes into account the time value of money and it can be used to find out whether the demanded rate of return is realized or not Internal rate of return The internal rate of return (IRR) of an investment is the interest rate that makes the cost function (net present value) equal to zero. To find out the value for the internal rate of return we can pose the question: what interest rate makes the equation *! " * * & " ' " # #$% " # #$% " 0. #$% ) (3.7) #$% " ( to be fulfilled? The result is the profit from the investment expressed in form of an interest rate. In a general case the rate cannot be solved in a closed form but has to be found using numerical methods. Let us investigate a simplified case where the incomes and expenses have fixed values, there is only one investment I 0 and the residual value is S=0. Thus, using the discounting factors for repeated entries Eqn. (3.7) turns into * :1 0 (3.8) where A is the yearly net income or gross income minus gross expenses. Based on above equation and the definition of the discounting factor (2.7) the interest rate can be solved as #,#$%.). (3.9) ' A /7 From this equation the interest rate must be iterated, but the expression is simple and the iteration usually converges fast. It can also be noticed that the denominator in Eqn. (3.9) is the previous simple payback time. The internal rate of return is very useful because its value reveals the profitability of the investment and especially because it can be compared with the return rate of any competing investment. In this way energy saving or energy production investment can be compared for example with a bank deposit or a stock investment. 13

14 3.5. Examples of using the life cycle calculation methods In the following the usage of life cycle calculation methods is illustrated with few simple examples. EXAMPLE 3.1 The owner of a detached house is considering a transformation of the old mechanical exhaust ventilation system to a mechanical supply and exhaust system with heat recovery. To estimate the economic profitability of the investment following facts are available: Ventilation air flow is q v = 70 l/s, yearly efficiency of heat recovery η = 0.80 and the heating degree hour for the location S 17 = 4200 o Cday. Because of a new supply fan, the increase of fan power is P = 70 W compared to the old ventilation system. The ventilation is running 8760 hours per year. The investment, including supply air ducting, a new air handling unit as well as the installation is Price for electricity is 89 /MWh, price for district heat is 55 /MWh and the expected lifetime of the investment is n = 15 years. The task is to estimate the profitability using different life cycle calculation methods when the building is heated a) with district heating b) with direct electrical heating. Let's first compute the yearly saved energy by heat recovery. Assuming ρ = 1,2 kg/m 3 for air density and c p = 1 kj/kgk for air specific heat capacity we get for the annual saved energy E hr = ρ c p q v S 17 η = 6774 kwh = MWh. The annual additional electricity need for the supply fan is E sf = P sf t = 70W 8760h = 0.61 MWh. The annual net income is A 1 = 6.77 MWh 55 /MWh 0.61 MWh 89 /MWh = 318 /a A 2 = 6.77 MWh 89 /MWh 0.61 MWh 89 /MWh = 548 /a for district heating for electrical heating. a) Simple payback period According to the definition of the simple payback period N 1 = I 0 / A 1 = 3200 / 318 /a = 10.1 years for district heating N 2 = I 0 / A 2 = 3200 / 548 /a = 5.8 years for electrical heating. So the payback period for both cases is clearly shorter than the estimated lifetime of the investment. For the electrical heating case the payback is shorter as the saved energy is more expensive. b) Net present value Because the annual net income has a fixed value we can utilize the discounting factor for repeated entries. Setting the interest rate to i = 12 % (which is quite high) the discounting factor is a n = [(1-(1+0.12) -15 ] / 0.12 = 6.81 a. By applying Eqn. (2.8) or (3.8) we get for the net present value P 1 = 6.81 a 318 /a 3200 = for district heating P 2 = 6.81 a 548 /a 3200 = +532 for electrical heating. For district heating the rate of return that was set is the form of the interest rate is not realized because NPV is negative. On the other hand for the electrical heating case the rate of return is surpassed. To further investigate whether the district heating case is absolutely unprofitable we set i = 0 and write according to Eqn. (3.5) * #

15 and substituting the numeric values we get P = 15 a 318 /a 3200 = This reveals that the district heating case is not unprofitable but it just does not fulfil the demand of an interest of 12 %. Here actually the influence of inflation should be taken into account as well, but we come back to it later. c) Internal rate of return First we need the discounting factors which are solved from Eqn. (3.8) for each case separately a n = I 0 / A = 3200 / 318 = 10.1 for district heating a n = I 0 / A = 3200 / 699 = 5.8 for electrical heating. We notice that the discounting factor in this case is equal to the simple payback period. Next we find for the interest rate such a value that the discounting factor above is fulfilled. Using Eqn. (3.9) for iteration of the interest rate we get: i = 0,053 = 5.3 % for district heating i= 0,152 = 15.2 % for electrical heating. The computed internal rates support the conclusions drawn earlier. The electrical heating case is more profitable as the saved energy is more expensive. However, the interest rate for the district heating case shows that a 5.3 % yearly return is to be expected even though it does not reach the 12 % demand set for the NPV case. In the example above all incomes and expenses had fixed values over the whole calculation time. In such a case the discounting factor for repeated entries can be used for discounting the entries of the whole time span. If the incomes and/or expenses change from year to year, the discounting must also be done year by year for example according to Eqn. (3.5). This kind year by year approach is also called cash flow calculation. Cash flow shows how much money has come in and how much is paid out from the "cashbox" during the year. In this context the inflows and outflows are restricted to the influence of the investment under consideration. EXAMPLE 3.2 The owner of a commercial building decides to make a lighting renovation. The investment is done in two stages. Half of the lamps, one thousand pieces, are substituted with new ones immediately. In the beginning of the third year the remaining one thousand lamps are replaced. The electrical power of the old lamps is 100 W each and the new lamps 40 W each. The lighting is estimated to be in use 2500 hours per year. The price for one lamp replacement work is 2 per lamp and the price for the new lamps is 15 each. The price for electricity is 80 /MWh. Estimate the NPV for the renovation investment in case the interest rate is set to i = 10 % annual and the lifetime of the whole project is estimated to be 4 years from the first replacement. Let's apply Eqn. (3.5) for the net present value, calculating the incomes and expenses as a cash flow calculation. During years 1-2 the saved energy of the first 1000 lamps is E 1-2 = 1000 (100-40)W 2500 h = 150 MWh and its monetary value is T 1-2 = 150 MWh 80 /MWh = Analogously for years 3-4 when the remaining 1000 lamps are replaced E 3-4 = 2000 (100-40)W 2500 h = 300 MWh and further T 3-4 = 300 MWh 80 /MWh =

16 Additional to the lamp investment also the labor cost has to be taken into account. The entries of each year are discounted to the present moment using the discounting factor a k = 1 / (1+ i) k where the index k is referring to the end of the year (end of 1 st year = beginning of 2 nd year etc.). Finally and all discounted values are summed together. The cash flow calculation looks as follows: Discounted Cash Discounting cash Year Incomes Expenses Investment flow factor flow , , , , , Sum We state that the net present value of the investment is well positive, so the investment is profitable and the 10 % return is exceeded. EXAMPLE 3.3 Let's calculate the NPV of the investment of the previous example again but now using the superposition principle of the discounting factors. The cost function or NPV can be written for this case following the principle of Eqn. (2.11): 6 + #, ,4 0 1 & 3$' 3 #$% 3. The discounting factors for repeated entries are a 2 = (1-(1+0.1) -2 )/0.1 = a 4 = (1-(1+0.1) -4 /0.1 = And the net present value is P = ( ) / (1+0.1) 2 = The result is exactly the same as in the previous example, as it should be. If the calculation periods are long and there are just few changes in the repeated entries, the superposition principle can be a more convenient way than the yearly cash flow calculation. 16

17 4. INTEREST RATES 4.1. Nominal interest rate The interest rate is used to define the time value of money. Money available at the present time is worth more than the same amount in the future. A high interest rate is emphasizing the incomes and the expenses at the present and the near future. With a low interest rate the differences between present time and the future are evened. For a loan the interest is the price for the borrowed money. The nominal interest rate is indicating the absolute rate of return which is demanded from the investment. In principle the decision maker is placing this demand on some chosen level and having done this, makes the decision of the investment based on the profitability and some other often non-monetary arguments. Sometimes other than energy economic things can be dominating the decision. For example the renovation of the building façade can be realized despite the fact that increasing the insulation thickness as such is unprofitable. Then other arguments, like the value of the real estate, visual appearance, indoor conditions etc. Thus additional insulation is often done as a part of a larger renovation package and no profitability of the insulation itself is demanded. The profitability of an energy related investment can also be compared with alternative measures like an equity investment. In business the determination of the interest rate can be done based on different arguments. It can be the rate of a loan with a risk premium or without, the return of an alternative investment or just based on a subjective assessment. Often a higher return is required from an energy saving investment than from a productional investment. When looking at a project having just costs but no incomes, the interest rate cannot be interpreted as a demand for the return but it must be chosen on another basis. Loan interest rates, cost growth rate or GDP growth rate are some starting points to be used for this purpose. In energy related investments the choice of the interest rate is also affected by risk, which is mainly caused by the rise in future energy prices, which in the long is difficult to predict. The higher the risk is the higher is the demand for return. For example in selecting the heat energy source for a building, a local bio-based alternative is likely to be more stable and secure alternative than imported energy which price and availability is strongly affected by the global economic and political situation Inflation, real interest rate The overall cost increase is called inflation. Inflation reduces the purchasing power of money. Over time, the same amount of money will always be of less value. Inflation is affected by the balance of demand and supply of products and for example the rise in wages. Wage growth is easily transferred to product prices. General inflation is measured by the consumer price index and its changes. In investment calculations the effect of inflation must be taken into account. As the interest rate increases the nominal amount of money when moving forward in time, inflation is reducing purchasing power of the same amount. When combining the effect of interest rate and inflation into the same expression, a capital P at the present time is reaching the following value at the end of year k 17

18 #$%" #$E " (4.1) where i is the nominal interest rate and f the rate of inflation. The nominal interest rate in the numerator increases the value of the expression but at the same time the rate of inflation in the denominator is reducing it, assuming that f > 0 (f < 0 means deflation which is quite unusual). Often the effect of the nominal interest rate and inflation rate are combined into a single factor called real interest rate. The impact of real interest rate on the value of capital when moving forward or backward in time is the same as the nominal interest rate and inflation together. This textual definition can be presented as an equation: #$%" #$E " 1F (4.2) where P is the capital at the present time and r is the real interest rate. From the above equation we get for the real interest rate F %,E #$E. (4.3) In the calculations shown in the previous chapters the nominal interest rate can now be simply substituted by the real interest rate to take the inflation into account. Thus for example equation (2.7) of the discounting factor for repeated entries is transformed into * #,#$-.) - (4.4) which now is accounting for both nominal interest rate and general inflation Inflation of energy prices - escalation The general inflation rate and the consumer price index are reflecting the average increase in the price of a selected group of products with time. The price evolution of individual products or commodities can clearly deviate from the level of general inflation. This is called specific inflation or escalation. Energy is one such commodity. Its price may change quickly and differently from the general price development as a result of changes for example in the global economic or political situation. Therefore the price of energy should not, in economic calculations, be tied on the general price development but should be considered separately. The real interest rate for energy, which takes into account the inflation (escalation) of the energy price is calculated exactly in the same way as the real interest rate due to the general inflation. The only difference is that instead the general inflation rate the escalation for energy price is used. According to Eqn. () we get for the real interest rate for energy 18

19 F H %,E I #$E I (4.5) where f e is the escalation for the energy price in question. Further, the discounting factor for the energy related entries is now * #,#$- I.) - I (4.6) where two apostrophes mean that the influence of escalation is included in the factor. 19

20 5. OTHER APPLICATIONS OF THE COST FUNCTION 5.1. Price for the saved or produced energy The price of saved or produced energy is a good indicator which can be used to estimate the profitability of an energy investment. The saved energy is replacing purchased energy. If the price for the saved energy is lower than the price for the purchased energy, the investment is worthwhile. The price for the saved energy can be calculated from the previously presented cost function. In general, the cost function for an energy saving investment is +0 1 ( #$- ) (5.1) where T is the annual income and K the annual expenses which both are discounted to present value using an appropriate discounting factor. In this case the revenue is the value of the saved energy. Expenses may include maintenance costs, energy costs and other running costs of the invested system. The residual value has to be evaluated case by case but quite often it is set equal to zero. When the income is set equal to the annual saved energy times the energy price and the expenses are the sum of maintenance costs and energy costs and the residual value is ignored, the cost function becomes J K L J L L 1 (5.2) where E s is the annual energy savings, p s is the price for the saved energy, M the annual maintenance costs, E a the annual auxiliary energy and p a the price for the auxiliary energy. By notating now P = 0 meaning that the demand for return which is embedded in the interest rate is fulfilled, the price for the saved energy can be solved as LM N$L O MM P O Q O $' A L R SSP R. (5.3) EXAMPLE 5.1 An industrial hall has huge heat losses through ventilation. A heat recovery (HR) device is estimated to save 500 MWh of purchased heat annually. The investment cost of the device is Because of the pressure loss caused by the HR, the annual energy demand of the fans is increased by 5 MWh. The annual maintenance cost of the HR device is The price for purchased electricity is 80 /MWh. What is the price for the saved energy when the nominal interest rate is set to 10 %, the inflation is 2 %, the escalation for heat is 4 % and for electricity 2 % and the lifetime of the investment is assumed to be 15? The real interest rate accounting for the general inflation is r = ( ) / (1+0.02) = = 7.84 %. The real interest rate for the saved heat is r s = ( ) / (1+0.04) = = 5.77 %. and the real interest rate for the auxiliary electricity is r a = ( ) / (1+0.02) = = 7.84 %. 20

21 The corresponding discounting factors for 15 years are a = (1-( ) -15 ) / = 8.64 a a s ' = (1-( ) -15 ) / = 9.86 a a a ' = (1-( ) -15 ) / = 8.64 a. By substituting the numeric values into Eqn. (5.3) we get p s = (8.64 a 2000 /a a 5 MWh/a 80 /MWh ) / (9.86 a 500 MWh/a) p s = /MWh. This price can now be compared with the price of the purchased heat and if the price for the saved energy is lower than the price for the purchased energy the investment is attractive. Applying the same principle, the profitability of an energy production investment or any kind combined saving and production investment can be evaluated. For this purpose we can make a variant of Eqn. (5.2) taking into account necessary entries ' " #$- " SS J SS H J H H S K SS L J L L # 1 (5.4) where subscript s is for the saved energy (savings and/or local use of own production), subscript e is for export (exported part of own production), a is for purchased auxiliary energy and the sum term accounts for all supplementary investments during the life cycle apart from the initial investment I 0. The price for the saved/produced energy is now * LM N$L O MM P O Q O $ " "U/ /CT " $' A,L MM I P I Q I L MM. (5.5) R P R If the cost for energy saved/produced by an invested device or process is more affordable than the cost for purchased energy, the investment is worth realizing. EXAMPLE 5.2 A house owner wants to know would it be profitable to invest in PV. He looks for an alternative an energy providing company is offering. The total price for 20 m 2 PV system including all components and installation is A subsidy reducing the investment with 900 is available. The estimated annual electricity production is 2500 kwh. Because of the mismatch between production and demand the house owner can utilize on-site only 30 % of the produced energy and he has to sell the surplus to the utility for a price of 40 /MWh. The estimated annual maintenance cost is 100 including one renewal of the inverter. The lifetime of the system is assumed to be 25 years. The annual nominal interest rate is set on a level of 3 %, the general inflation is 1 % and the estimated escalation of electricity 2 %. Calculate the price for the electricity produced with the PV system and make a conclusion of the profitability when the price for grid electricity is 120 /MWh. The real interest rate accounting for the general inflation is r = ( ) / (1+0.01) = = 1.98 % 21

22 and the real interest rate for the electricity is r e = ( ) / (1+0.02) = = 0.98 %. The corresponding discounting factors for 25 years are a = (1-( ) -25 ) / = a a s ' = a e ' = (1-( ) -25 ) / = a. By applying Eqn. (5.3) we get p s = (19.57a 100 /a + ( ) a MWh/a 40 /MWh ) / (22.08a MWh/a) p s = ( ) / MWh = 423 /MWh. This is often called the levelized cost of electricity (LCOE). As LCOE is much higher than the grid electricity it seems that from a purely economic point of view the investment is not lucrative. We can also calculate the simple payback time for the investment. The initial net investment is I 0 = The annual savings from the on-site used PV electricity is 90 and the income from the sold electricity is 70. According to Eqn. (3.2) the payback is N = 6600 / ( ) = 41 years Maximum investment From the cost function the investment term can be solved when all other terms are known. In this way the highest investment price maintaining the project still profitable can be found. Starting from Eqn. (5.2) and setting again P = 0 the investment is 1 J K L J L L (5.6) so the investment just is the difference between the discounted incomes and discounted expenses. Let's again illustrate the method with an example. EXAMPLE 5.3 The profitability of additional roof insulation for an old house in under investigation. The roof area is 200 m 2 and U-value with the existing insulation is U 0 = 0.40 W/m2K. In the attic there is space for a 300 mm layer of blow wool which has a heat conductivity of λ = 0,045 W/mK. The building is located in a place where the heating degree day is S = 4620 Cday. How much at most can the blow wool insulation cost when the price for heating energy is 40 /MWh, the nominal interest rate is set to 10 %, escalation for the heating energy is 4 % and the calculation time period is set to 30 years? Let's first calculate the rates and the discounting factors: r s = (0,10-0,04)/(1+0,04) = 0,0577 a s ' = (1-( ) -30 ) / = a. The new U-value after the renovation would be U 1 = 1 / (1/0.40 W/m 2 K m / W/mK) = W/m 2 K. 22

23 Annual energy savings are E s = S (U 0 - U 1 ) = 4620 o Cday ( ) W/m 2 K = 32.3 kwh/m 2. As there are no running or maintenance costs for the insulation the maximum investment is according to (5.6) I 0 = 14.11a 32.3 kwh/m 2 a 0.04 /kwh = /m 2 and the maximum investment for a 200 m 2 roof is Calculated per volume of blow wool this is /m Further applications of the cost function The cost function further gives the opportunity, from the economic profitability perspective, to set targets for technical features of systems. In general, we need just to know all other terms in the cost function except the one we want to solve by setting P = 0. EXAMPLE 5.4 The planner of the HVAC systems of a swimming hall has been given the task to find out whether heat recovery (HR) from the shower water is profitable or not. Annually some 4500 m 3 of water from the showers is going into the sewage having an average temperature of 30 C. The mean temperature of incoming cold water is 7 C and heating it up using district heat costs 55 /MWh. The investment price for the proposed HR system is and its annual cleaning and other maintenance costs are What is the minimum HR efficiency for the investment to be profitable? The nominal interest rate is set to 10 %, inflation is 2 %, escalation for district heat is 4 % and the estimated lifetime of the equipment 15 years. The annual amount of recovered heat is E s = ρ V c p dt = ρ V c p η (T 1 T 2 ), where ρ is the density of water, V is the volume of water, cp is the specific heat capacity of water, η is the average efficiency of the heat recovery and T 1 and T 2 are the temperatures of the extracted warm and incoming cold water respectively. By substituting this into Eqn. (5.2) we get SS V W Q X + # + 6 S K 1 0 The above equation now contains the asked efficiency term which can be solved as X L M N$' A L R MM Y Z [ \! /,! 3 Q R. Let's compute the numeric values. First the interest rates and the discounting factors: r = ( ) / (1+0.02) = r s = ( ) / (1+0.04)= 0,0577 a' = (1- (1+0,0784) -15 ) / 0,0784 = 8.64 a a s '' = (1- (1+0,0577) -15 ) / 0,0577 = 9.86 a. The efficiency then is 23

24 η = (8.64 a 2000 /a ) / (9.86 a kg/a 4,2 kj/kgk (30-7)K 55 /MWh) η = / = = 99.7 %. This means that the HR efficiency should be practically one hundred percent which from a technical point of view is impossible. If the same equipment could be used in a case where the annual volume of water would be three times as much, the required efficiency would be η = 33 % which could be technically achievable. 24

25 6. SELECTION OF THE CALCULATION PARAMETERS 6.1. Investment cost The investment cost is a key parameter and input value in the estimation of energy economic profitability. The investment cost for a simple, fixed priced equipment procurement is usually easy to find out. A time switch, which is connected to control lights or some electrical device could be an example. The device is fixed priced, and there is no need to purchase work for bringing it to function. On the other hand, getting a good estimate of the expenses of a larger investment containing plenty of work and materials is often difficult and laborious. In general, no single real value for the investment does even exist, but there are several estimates from different sources that may vary greatly depending upon the provider's own work situation and order book. The price of the equipment and the materials contained depend not only on where they are acquired but also who acquires them. Various contractors and professionals have their own supply channels and agreements with equipment suppliers. The final cost for an energy investment can often be obtained only through calls for tender. However, the magnitude can certainly be evaluated by using lighter procedural ways Price /m External wall - wooden frame with board cladding External wall - wooden frame with brick cladding External wall - concrete frame with board cladding External wall - concrete frame with brick cladding Roof - hollow-core slab Roof - roof truss construction Roof - roof truss construction and blow wool Roof - self supporting construction Floor - woden frame with crawling space Floor - concrete frame with crawling space Floor - slab on ground ,1 0,2 0,3 U-value W/m 2 K Figure 6.1 Investment prices for some building structures (VAT 0%) as a function of the U- value, price level

26 Using optimization will introduce an additional dimension to the evaluation of the investment cost. If the energy economic profitability calculation is applied in combination with energy optimization, the costs of different variants for the same investment must be known. For example, when computing the optimal insulation thickness for some structure like a wall, a roof etc. one must know the price for the structure as a function of the insulation thickness or U-value. Here the problem is that the price of the structure is not necessarily proportional to the amount of insulation. More work might be needed and when the structure gets thicker also some constructional changes might be necessary, which are driving up the price in an irregular way. Similarly, for example when searching for the most economical option for ventilation heat recovery, one should know prices for various device sizes and options. For optimization purposes the prices may be presented as a function of the design parameter as shown in Figure Energy price The price of energy at present time is information that is readily available. The price of electricity, district heating, fuel etc. are public information that can be found in e.g. on the web sites of energy companies and from numerous other sources. It should be noted that there are several prices, depending on the buyer, the seller and numerous other factors. Different size consumers, small properties, large buildings, industry etc. have different prices. In addition, various energy companies have different price ranges of products. For example, the price of electricity has myriad of variations, depending on whether it is universal electricity, seasonal electricity, a spot market price, or something else. Further, the price also depends on duration of the contract and how the energy is produced. In addition, it is worth noting that the price of energy itself forms only a part of the total price. For electricity in the tariff is comprised of three parts: energy, transfer and tax, which are roughly of same size each. In the price of district heat there is the energy price, a fixed basic fee depending on the power needed and the tax. For a small consumer the fixed part can be a significant amount of the total expense Interest rates The choice of interest rates describing the time value of money is often done on a rather vague basis. This is understandable, since the question is of subjective views as well as predicting the future and no single truth in this matter exists. The yield requirement, i.e. the nominal interest rate or real interest rate is basically freely selectable. Certain benchmarks such as loan interest rates, GDP growth, etc. are existing. One way to determine the real interest rate is to weigh the rate proportional to the equity and liabilities and to take the inflation into account as previously shown in the following way F ] I% I $]^%^,E #$E (6.1) where i e and i l are the interests of the equity and debt, ω e and ω l are the corresponding relative proportions and f is the annual growth of inflation. Determination of the interest rate for a loan is relatively simple based on the lenders loan terms or through financial markets. 26

27 Rate of return for equity is perhaps a more subjective choice, even though there are methods based on the company's performance. This is, however, beyond the scope of this treatment and will not be addressed in more detail. The rate of return is convenient to express directly by the real interest rate, ignoring inflation. This is a good way in the sense that the demand for return is indicated using a "net interest" where the influence of inflation is integrated. When working with building energy investments the real interest rates typically are in a range from 3% to 6%. The lower interest rate reflects a "macro-economic" approach in which profit maximization is not the main issue but the low rate of interest is encouraging for energy efficiency enhancing and environmentally friendly investments. The higher interest rates reflect a commercially oriented approach putting more emphasis on the economic viability but will lead more easily to a decision where the investment is estimated unprofitable and will not be realized. Figure 6.2 Consumer price index 2005=100 (Statistics Finland). Information on the general inflation can be received by looking at the history of the consumer price index, Fig The Finnish consumer price index describes the price development of goods and services purchased by households in Finland. The consumer price index is calculated by a method in which the prices of different commodities are weighted according to their shares of consumption. The building cost index, in turn, illustrates the evolution in construction costs over the years, Fig