District heating investment planning

Transcription

1 Energiatekniikan laitos District heating investment planning Risto Lahdelma Professor, Energy technology for communities Tel: Energy technology, School of Engineering, Aalto University Otakaari 4, ESPOO, Finland R.Lahdelma 1 Investment planning The goal is to determine if the facility is profitable or not The idea is to add up all revenues and expenses over the facility lifetime Initial investment Future operation and maintenance costs Revenues from selling the production Possible decomission costs or end-value after lifetime Cash flows in the future are discounted to present value using specified interest rate Future revenues and costs can be estimated using a simulation/optimization model based on forecasts for demand and price level R.Lahdelma 2 1

2 Methods for evaluating profitability of investments Net present value (NPV) method Discount all future costs and reveues to present and compute the net present value of investment Chosen interest rate affects results Internal rate of return (IRR) method Compute the interest rate that makes the NPV = 0 A priori interest rate is not required Annuity method Distribute initial investment into equal annual payments and combine with other annual costs and revenues Use real interest rate R.Lahdelma 3 Annuity method for investments Idea: Assume that investment of size I is financed by taking a loan for n years with effective interest rate r The loan is amortized (paid back) with fixed constant annual amounts P (to be calculated as follows) The net present value (NPV) of a payment after i years is P(1+r) -i To determine the annual payment, the net present value of payments is set equal the investment I = i=1n (1+r) -i P P = I / i=1n (1+r) -i = I/a The annuity factor a is obtained by computing the sum of the geometric series: a = (1- (1+r) -n )/r R.Lahdelma 4 2

3 For example 15 year annuity with 5% interest rate gives annuity factor To amortize 1000 in 15 years requires 1000/ = payment each year Annuity factors Interest Years 1% 2% 3% 4% 5% 6% R.Lahdelma 5 District Heat Demand District heat demand depends on The season The weekday / special holidays Time of the day The variable demand makes it beneficial to apply different technologies to produce base load, middle load, and peak load In addition sufficient spare capacity should be maintained In Finland, district heating is used both for heating space and for producing hot tap water This means that the DH system is in operation also in the summer R.Lahdelma 6 3

4 Annual consumption of district heat (City of Turku) Load R.Lahdelma Outdoor temperature during a year 40.0 Temp R.Lahdelma 4

5 Weekly consumption of district heat Load R.Lahdelma Outdoor temperature during a week 2.0 Temp R.Lahdelma 5

6 Detecting dependencies from history data: heat demand vs outdoor temp A scatter chart reveals if two quantities have significant dependency Also the type of dependency can be seen (linear/nonlinear etc) Correlation = linear dependency = -95% Outdoors temperatures over 17 o C do not affect heat demand anymore S17 = heating degree days for temperature below 17 o C R.Lahdelma 11 Kulutus (MW) Kaukolämmön kulutus ja regressiosuora Ulkoilman 0 lämpötila District Heat Demand Base load Continuous use Low operating costs ( /MWh) High usability Technologies: CHP-plants and solid fuel boilers Medium load Continuous use Can run also on partial load Fairly low operating and investment costs ( /MWh, /MW) Technologies: solid fuel boilers, natural gas boilers Peak load and spare capacity Low investment costs ( /MW) Easy and fast start-up Technologies: Oil boilers, natural gas boilers R.Lahdelma 12 6

7 Optimal capacity division between base and peak load A historical or forecasted heat load curve can be used for running a production simulation & optimization Optimal capacity limits can be determined by varying the capacity limits in the simulation In presence of dynamic dependencies this is the only way In absence of dynamic dependencies, the capacity division can be determined analytically from the load duration curve R.Lahdelma 13 Heat load curves Q' Qload,b Heat load Peak heating period Peak shaving heat production Heat load Peak heating period Q' Peak shaving heat production Qload,b Basic district heat production Basic district heat production 0 t1 Operation time of basic heat producer t2 d n Time 0 Operation time of basic heat producer n Duration time (a) heat load curve in chronological sequence (b) heat load duration curve R.Lahdelma and H.C. Wang 14 7

8 Optimal capacity division between base and peak load The load duration curve can be expressed as function = f(h) = heat load h = hours in year when heat load is The inverse function is h = f -1 ( ) = g( ) The annual energy is obtained by integrating over f or g E = 0 hmax f(h)dh = 0 max g( )d The hourly load is satisfied by base & peak plants The annual investment costs (annuity, /MW) satisfy I base > I peak The operating costs (mostly fuel costs, /MWh) satisfy C base < C peak R.Lahdelma 15 Optimal capacity division between base and peak load Determine the capacity of base load base to minimize total annual costs Min C = I base base + I peak peak + C base E base + C peak E peak E base = 0 base g( )d E peak = E E base peak = max base Substituting expressions for E base, E peak, peak gives C = I base base + I peak ( max base ) + C base 0 base g( )d + C peak (E 0 base g( )d ) Forming derivative dc/d base and setting it = 0 gives optimum dc/d base = I base I peak + (C base C peak )g( base ) = 0 g( base ) = (I base I peak )/(C peak C base ) // hours base = f((i base I peak )/(C peak C base )) R.Lahdelma 16 8

9 Optimal capacity division between base and peak load Example Heat pump for base load and electric boiler for peak load Heat pump has COP-factor 3, and investment cost /MW Electric boiler has investment cost /MW Power price is 80 /MWh Assume 12 year life time for investment and 4% interest rate Annuity factor = What is the optimal size of the heat pump? Optimal solution Peak operating hours ( )/( ) = 672 h/a Assume piecewise linear load duration curve f(1)=10 MW, f(800)=2 MW f(8760)=0 Interpolation gives pump size = 3.3 MW R.Lahdelma 17 9