Semester Project 4: Projection of Expansion: Assens to Ejby

PROJECTION OF EXPANSION:
ASSENS TO EJBY
19. May 2016
Maibrit Vester Hansen Lars Bo Jensen
64967718 410749
Klaus Christian Jespersen Martin Grunnow Vinstrup
73474570 412964
Søren Andreas Aagaard Christian Nissen Ahrenkilde
409215 412673
4. Semester-Project
Group: 7
Field of study: Energy-Technology
University of Southern Denmark
Supervisor: Maryamsadat Tahavori
Semester: Spring 2016

University of Southern Denmark
Faculty of Engineering
Campusvej 55
5230 Odense M
Www.sdu.dk
Title:
Projection of expansion: Assens to Ejby
Project:
4th Semester project
Project period:
February 2016 - May 2016
Project group:
7
Participants:
Maibrit Vester Hansen
Lars Bo Jensen
Klaus Christian Jespersen
Martin Grunnow Vinstrup
Søren Andreas Aagaard
Christian Nissen Ahrenkilde
Supervisor:
Maryamsadat Tahavori
Copy: 3
Pages: 82
Standard pages: 39
Appendix: 11
Finshed 19-05-2016
Synopsis:
This project is a projection of a possible
expansion of the district heating
infrastructure from Assens to Ejby.
It revolves around both the dimensioning
of the pipeline itself and the production
plan for the two cities to go with it.
Linear optimization are being used to
calculate the optimal production of the
two cities, with the transmission line
as a decision variable subject to the
production and demand in the cities.
The applicable legislation will be
provided and explained to help grasping
the legal aspect of expanding a district
heating infrastructure.
Diﬀerent scenarios will be investigated to
determine the resilience of the model and
to ﬁgure out if this expansion is a viable
option should the subsidies change.
The content of the rapport is freely accessibly, but publication (with sources acknowledged) is only allowed with
permission from the authors.

Preface
We would like to thank our supervisor Maryamsadat Tahavori for being helpful throughout the
project. We would also like to thank the employees of Assens Fjernvarme A.m.b.a, especially
Marc Roar Hintze and Brian Kjær Ottosen, for allowing us to come visit them and providing
us with data necessary for the completion of the report. Lastly we would like to thank all of
our lecturers on the fourth semester for guidance in the project.
Reading guide
The target audience of this report are people with an engineering background, or above
average technical understanding.
The following project consist of different chapters and sections. Before each chapter or
section, a small introduction to the following section is written in italic text. After sections
of significant importance, a partial conclusion is written in italic too. Any source needed in the
project is referenced to in the project by using square brackets with a number inside. Notice
that dot are used as thousands separator and comma are used as decimal separator.
If reading the project as a PDF: The references to the sources is a hyperlink which can be
used to jump to the bibliography in the back of the project. In the bibliography, the internet
sources are hyperlinks. Also all references made in the project to other sections, chapters,
figures or tables are hyperlinks.
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Abstract
1
This report applies linear optimization to the district heating production of Assens Fjernvarme
A.m.b.a, and its theoretical expansion to the city of Ejby, in order to evaluate whether this
expansion would be socioeconomically sound, and as such, whether it could be approved
according to the law.
The motivation for such expansion, is to reach more consumers and thereby enabling the
possibility of producing more electricity. Since the production is primarily based on biomass,
electricity can be sold at a fixed price of 0,60 DKK/kWh. Therefore an expansion would
enable the company to produce more heat, and as electricity is a byproduct of this production,
enabling a higher level of the profitable electricity production.
Based on the result of the optimization, the report calculates the optimal transmission line
diameter, in regards to pump costs and heat losses, both of which are calculated and valuated.
It also offers suggestions as to the placement of said pipes, as well as theoretical pump
placements. It does not offer any insight or suggestions as to the distribution infrastructure,
as one already exists, and is assumed to be functional.
In order to contextualize the model, the development of Assens Fjernvarme A.m.b.a and
district heating in general, as well as the legal framework surrounding the production is
reviewed briefly.
The report concludes that the expansion is only socioeconomically viable as long as the
company’s current agreement regarding electricity prices is upheld. If the electricity prices are
changed from a fixed 0,60 DKK/kWh to be based on the spot price, plus a 0,15 DKK/kWh,
the project ceases to be viable, for either the company, or the city of Ejby itself.
The report also offers analysis of several scenarios where the usage of several storage tanks
is included in the optimization. It concludes that the storage tanks have limited usage as
anything else than buffers for peak load production as long as the fixed price for electricity is
upheld. It is only in the case of a transition to a spot price based agreement that the storage
come into its own. In this case it enables the company to move production of electricity to
times of high electricity prices, instead of being dictated by the high and lows of consumption.
It also concludes that if further expansions are made, then full capacity of the company’s wood
chip boiler will be reached in the winter months, necessitating the startup of the wood pellet
boilers and as a consequence, a higher production cost, removing the very incentive for the
expansion in the first place.
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Group 7 Energy Technology SDU
1.1 Symbol explanation
Symbol Explanation Unit
β Parameter describing the insulation of a pipe [-]
Equivalent roughness for the pipe [mm]
ηElec Electricity effectiveness [%]
λi Thermal conductivity of isolation used in pipes [ W
mK ]
λg Thermal conductivity of soil [ W
mK ]
µ Dynamic viscosity [ N
s·m2 ]
ρ Density [Kg
m3 ]
Φ Heat consumption [kJs]
Lossheat,pump Expenses due to loss [DKK]
A Annuity, yearly payment [DKK]
CMn Slope counter pressure line for heat unit n [-]
cn Heat production cost for a given heat production unit n [DKK
MWh ]
cp,1 Specific heat capacity for supply [ KJ
kg·K ]
cp,2 Specific heat capacity for return [ KJ
kg·K ]
D Diameter of the pipe [m]
D1 Half the distance between the centers of the two pipes [m]
f Friction factor [-]
FV Future value [DKK]
g Gravitational acceleration [m
s2 ]
Gradient The heat demands increases over a month in percentage [%]
∆h Change in height [m]
H The distance between the surface of the ground and the center of a pipe [m]
ha Heat loss factor of the anti-symmetrical problem [-]
HDAssens,t Head demand for the town of Assens in a given hour [MWh]
HDfebraury,hr Head demand for the month of February in a given hour [MWh]
HDfebraury,total Head demand for the month of February in total [MWh]
HDEjby,t Head demand for the town of Ejby in a given hour [MWh]
HL,major Pressure loss due to friction and turbulence [mWc]
HL,minor Pressure loss due to bends [mWc]
hs Heat loss factor of the symmetrical problem [-]
i Interest rate [%]
KL Pressure loss coefficient [-]
L Length of the pipe [m]
˙m Mass flow [kg
s ]
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m Index for storage units [-]
N Time variable in question [year]
∆p Pressure difference [mWc]
P Pressure [KPa]
PV Present value [DKK]
pk,t Electricity revenue for a given hour [-]
q1 Heat loss, supply pipe [W
m ]
q2 Heat loss, return pipe [W
m ]
qa Heat loss in the anti-symmetrical problem [W
m ]
qs Heat loss in the symmetrical problem [W
m ]
qtotal Total heat loss [W
m ]
Q Volume flow [m3
s ]
Qprod Heat produced in a day [MWh]
Qt
prod Heat produced in a given hour [MWh]
ri Inner radius of a pipe [m]
ro Outer radius of a pipe [m]
Re Reynolds number [-]
SAssens,t Sum of discharge or charge of storage for all units in Assens for a given hour [MWh]
Scapmax,m Maximum storage capacity for a given storage unit m [MWh]
Scapmin,m Minimum storage capacity for a given storage unit m [MWh]
Scurrent,m Capacity in storage m before the optimization start / start current capacity in storage m [MW]
Scurrentcap,m Capacity in storage when starting the optimization for a given storage unit m [MWh]
Sflow,m Maximum flow out / into storage per hour for a given storage unit m [MWh]
Sm,t Discharge or charge for a given hour, for a given storage unit [MWh]
SPcorr The sales price of electricity in a day [DKK
MWh ]
SPt
corr The sales price of electricity per hour [DKK
MWh ]
t Time dependent variable, t=1 for the first hour in the simulated year, t=8760 for the last [hr]
T The time period chosen for calculation of degree-days [hr]
T1 Temperature of the water in pipe 1 – the supply pipe [°C]
T2 Temperature of the water in pipe 2 – the return pipe [°C]
T0 Surface Temperature of soil [°C]
Ta Temperature used in the anti-symmetrical problem [°C]
Ts Temperature used in the symmetrical problem [°C]
Ti,Base Indoor base temperature [°C]
To,Mean Mean temperature for the chosen time period [°C]
V Velocity of the pipe [m
s ]
Wpump Work done by pump [W]
Z Production cost for 24 hour production [DKK]
Zunit,cost The total unit cost [DKK]
∆X Heat transport capacity between Assens and Ejby [MW]
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∆Xmax Heat transport max capacity between Assens and Ejby [MW]
∆Xavg Average heat transport capacity between Assens and Ejby [MW]
XAssens,t The sum of heat production located in Assens [MWh]
XEjby,t The sum of heat production located in Ejby [MWh]
xmin,n Minimum heat production capacity for a given heat production unit n [MWh]
xmax,n Maximum heat production capacity for a given heat production unit n [MWh]
xn,t Heat production given for a speciﬁc hour [MWh]
zk,t Electricity production for a speciﬁc hour [MWh]
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1.2 Abbreviation
Abb Explanation
BSF Basic feasible solution
CHP Combined heat and power
CPF Corner point feasible solution
DKK Danish crowns
e.g. Example given
LHS Left hand side
MW Mega watt (106 W)
MWh Mega watt hours (106 Wh)
NPV Net Present Value
O&M Operation and maintenance
Re Reynolds number
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Contents
1 Abstract 4
1.1 Symbol explanation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Abbreviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2 Introduction 11
2.1 Prediction of feasibility to expand to Ejby . . . . . . . . . . . . . . . . . . . . . . 13
3 General District heating 14
3.1 District Heating in Denmark . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.1.1 The historical aspects of district heating . . . . . . . . . . . . . . . . . . . 14
3.1.2 Pipeline infrastructure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.1.3 CO2 emission due to district heating . . . . . . . . . . . . . . . . . . . . . 15
3.1.4 Power plant vs. combined heat and power plant . . . . . . . . . . . . . . . 15
3.1.5 Beneﬁts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.2 Assens Fjernvarme A.m.b.a. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.3 Legal aspect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.3.1 Prices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.3.2 Establishment of district heating . . . . . . . . . . . . . . . . . . . . . . . 18
4 Model development 19
4.1 Optimization in general . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.1.1 Linear optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2 Method choice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.2.1 Linear optimization beneﬁt and limitations . . . . . . . . . . . . . . . . . 20
4.2.2 Chosen optimization method . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.2.3 Software choice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.3 Simplex method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.4 Pipe dimensioning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.5 Degree-day correction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.6 Termis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.6.1 Designing the transmission line . . . . . . . . . . . . . . . . . . . . . . . . 34
4.7 Optimization of electricity and heat production . . . . . . . . . . . . . . . . . . . 36
4.7.1 Delimitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4.7.2 - Without optimization of storage (Matlab model 1 year) . . . . . . . . . 36
4.7.3 - With storage (Excel model 24 hours) . . . . . . . . . . . . . . . . . . . . 43
5 Results 47
5.1 Validation of model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
5.2 Validation of optimization with storage . . . . . . . . . . . . . . . . . . . . . . . . 49
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5.3 Result presentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
5.3.1 Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
5.3.2 Optimal pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
5.3.3 Pressure and heat loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
5.4 Net Present Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
6 What-if analysis 58
6.1 What-if analysis - 1 year (Matlab model) . . . . . . . . . . . . . . . . . . . . . . . 58
6.2 What-if analysis - 24 hours (Excel model) . . . . . . . . . . . . . . . . . . . . . . 59
6.2.1 Description of scenarios . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
6.2.2 Scenario 1 – Fixed settlement price: 2016-2018 . . . . . . . . . . . . . . . 60
6.2.3 Scenario 2 - Spot price settlement: 2019- . . . . . . . . . . . . . . . . . . . 63
7 Conclusion 68
8 Appendix 70
8.1 Time schedule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
8.2 Group contract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
8.3 Group proces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
8.3.1 Team Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
8.3.2 Issues and their solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
8.3.3 Expectations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
8.4 Data used to optimize production without storage: . . . . . . . . . . . . . . . . . 76
8.5 Subject in report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
8.6 Excel model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
8.7 Transmission line model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
Bibliography 81
Webpage Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
Other Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
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Introduction
2
This report will apply linear optimization to the district heating company Assens Fjernvarme
A.m.b.a and their potential expansion to Ejby. Based on the results of this optimization, a
suggestion will be made, as to what the optimal dimension of the transmission line will be, for
said expansion.
At the beginning of 2016, the authors of this report approached the company Assens
Fjernvarme A.m.b.a, a district heating company, with the intent of writing a semester project
about their recent expansion of operations to Ebberup. After a meeting with representatives
of the company however, it was made clear that they themselves would be more interested in a
project revolving their potential future expansion to Ejby.
The company produces heat primarily based on biomass. This enables them, not only to
sell their district heating at a profit, but also electricity with a surcharge. The electricity
has become so profitable, that it has become the primary motivator for the production
optimization
It is assumed the relationship between electricity production and heat production remains
constant, and the net amount of heat production to be directly tied to the net amount
costumer consumption. The natural consequence of this is, that in order to increase company
revenue, either the effectiveness of the production must be increased, or an increase in demand
must occur – which in reality means an expansion of operation to new costumers.
As such, an expansion of their operations would enable them, not only to reach a greater
number of district heating costumers, but also heighten their production of electricity, with
the following increase in revenue and lowered district heating prices. However this expansion
cannot occur, if it does not represent a socioeconomic improvement, as the company will then
not be permitted to expand.
This leads to the following question: If Assens Fjernvarme A.m.b.a was to expand their
operations to the city of Ejby, would it be social economically sound?
The structure of the report is as follows:
In the first part of the report, in chapter 3, the project is contextualized – the concept and
development of district heating is paraphrased and legal framework guiding it, is established.
Then, in chapter 4, based on the framework established in the first part, the method of the
optimization, and two models to analyse the production cost will be made. One model without
optimization of the storage, the current sitatuon at Assens Fjernvarme A.m.b.a., and one
model where storage is optimized
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In Chapter 5, the results of the optimization will be presented including the optimal pipe
dimensions between Assens and Ejby and additional pump loss, heat loss, production cost and
income by expanding to Ejby. This result will be used to valuated the entire project, in order
to establish the potential socioeconomic gains.
Finally the optimization will be put through its paces in chapter 6, by exploring various
scenarios, in which the conditions for the optimization will be changed in order to explore
the eﬀect on the result. Also the potential usage of heat storage is explored in – what do they
represent in terms of operations, and to what extent can the potential be exploited?
In the end, in chapter 7, the report will conclude, based on the above, whether the expansion
to Ejby is socioeconomically sound, and provide recommendations as to the operations of
Fjernvarme Fyn A.m.b.a, based on the calculations.
First however, a small assessment into the viability of the project, in order to estimate its
potential, see section 2.1.
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2.1 Prediction of feasibility to expand to Ejby
This section will make a rough calculation of whether an expansion from Assens to Ejby is
economically attractive to determine whether a more detail analysis and the project is relevant
to complete.
The cost per MWh district heating for the consumers in Ejby has to decrease according to
the law whereby there is a socioeconomic gain for the Ejby heat consumers. In order for
the investment to be viable for the heat consumer in Ejby, the total savings made by the
transmission has to exceed the cost of the pipeline. The total savings is equal to the heat price
in Ejby subtracted by the heat price in Assens.
The assumptions include:
• Discount rate on 2%
• Investment rate on 4%
• Investment write-oﬀ period on 50 years for the pipeline
• Cost of 2000 DKK per meter pipeline, with a total length of 19.304 meter of pipe.
Figure 2.1. A visualisation of the NPV calculation
The net present value based on the above assumption results in around 17,4 million DKK in
revenue. The positive result indicates a viable investment and therefore a socioeconomic gain.
Because the investment probably is viable the project will continue, where the production cost
and proﬁt will be calculated to make a more accurate result. This more detailed calculation is
located in section 5.4, where the formulas used in this section also is described.
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General District heating
3
3.1 District Heating in Denmark
In this section district heating will be described in general. Both about the development,
pipelines, benefits and downsides about district heating.
3.1.1 The historical aspects of district heating
District heating is an old way of producing heat. Back in Pompeii’s time, water from
the towns heat sources were sent to the houses and used to Romans baths. The first real
decentralized district heating plant was established in USA in 1878-79. The first district
heating plant in Denmark was established in 1903 at Frederiksberg. The first application of
district heating was to convert trash in the town to heat and electricity, which the hospital
and the poorhouse were the first consumer of district heating. Since then the technology have
been developed, and thereby spread to almost all parts of Denmark.[1] In Denmark district
heating has become the typical heating form. Currently, 64% of all the Danish households
have district heating which corresponds to about 3,2 million Danes. It is in particular cities
that have district heating. But the network is developed to other areas including many natural
gas areas and houses with oil boilers are being converted to district heating. This is due to
the economy, policy and the environment, in connection to the time when the oil boilers was
phasing out.[2]
3.1.2 Pipeline infrastructure
The consumers are provided with district heating through pipelines. Here the consumers are
paying for maintenance of the pipelines through the fixed tax on the heating bill. The network
is a system of supply-pipelines and return-pipelines, this can be seen in figure 3.1 where
the red lines are supply-pipelines and the blue are return-pipelines. The network consists of
transmission pipelines and distribution pipelines. The transmission pipelines lead the water
from the producer to the distribution pipelines and back again. The distribution pipelines
lead the water to the consumers and back again. This cycle is a closed system. The water
used as district heating is not tap water, as this would cause the pipelines to corrode from
the inside, so therefore the water is filtered. Furthermore hot water is used instead of, the
previously used, steam. The reason water is used as opposed to any other liquid is because
water is able to store a lot of heat pr. unit volume (High Cp-value), and water is not bad
for the environment. In addition, water is accessible in most part of the world and water is
also relatively cheap.[3] Furthermore the district heating companies only deliver heat and not
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domestic water, therefore the consumers need to have a hot water tank or heat exchanger to
transfer the heat to the domestic water.
Figure 3.1. The network of a district heating system with supply- and return-pipelines. [4]
3.1.3 CO2 emission due to district heating
The total use of energy have not increased in Denmark even though there have been a growth
in the Danish economy. One of the reason is the increase of district heating especially after
1990. The production of district heating has increased by 42% since 1990. In the same period
the emission of greenhouse gasses from district heating is reduced by 23%.[5] This is mostly
due to the change in fuel and optimized heat production.[6]
3.1.4 Power plant vs. combined heat and power plant [7]
There are some benefits about choosing a CHP plant instead of a power plant. When only
producing electricity, merely around 40% of the energy is consumed. This is because the heat,
which the electricity production has as a surplus, is not used. Instead, if both electricity and
heat are produced together at a CHP plant, as muxh as 90% of the energy is used because the
surplus of heat is used as district heating. Most CHP plants are using coal as primary fuel but
the renewable sources are increasing. Today 80% of the district heating production comes from
surplus from electricity production.
3.1.5 Benefits
There are some benefits about district heating being the right solution for Denmark.[8]
• In Denmark district heating decreases the import of oil and other fossil fuels.
• District heating is primarily produced combined with electricity, which reduces the need
of fossil fuel.
• It is also possible to produce district heating at local energy sources e.g. solar,
geothermal, wind and biomass.
• District heating production is always able to adapt to other fuel types, hence the
infrastructure is independent. E.g. many gas-fired plants are converting to biomass.
A benefit is the low maintenance the consumers have by using district heating. In some
cases district heating comes from surplus heat, which otherwise would be wasted, hereby the
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environmental impact is being held down. In other cases, fossil fuels are used for production
and heat are cooled to produce more electricity.[9] There are also some downsides to district
heating. E.g. many places around the country there is obligatory connection, which means it is
not possible to determine the heating you want in your house.[10] Furthermore, if a radiator or
floor heating is leaking, the water damage gets bigger because the water would keep flowing,
this is when it is direct district heating. When it is an indirect district heating system the
pressure cannot be used as power in the heating of the house, so there will be an extra cost to
run a circulating pump.[11] When it is an indirect district heating system the district heating
is physically separated from the domestic water, by a heat exchanger, whereby a circulation
pump in the house installation is necessary
3.2 Assens Fjernvarme A.m.b.a. [12]
In this section some general information about Assens Fjernvarme A.m.b.a. will be described.
The development from Assens Fjernvarme A.m.b.a was established until today is described.
Assens Fjernvarme A.m.b.a. was established in 1961 as a limited responsibility cooperative
(A.m.b.a.). The company was located on Hardersvej in Assens and had a boiler fueled by
light oil. The task was to supply the center of the inner city with district heating. The boilers
were continuously replaced along with an increased demand by the expanding the distribution
network. In the 1970s a transition to fuel oil was completed.
In 1982 Fuglebakken was supplied by district heating from Assens Fjernvarme A.m.b.a.. Many
of the house owners started to insulate their houses, during this period. As a result, there
was a surplus in capacity, which allowed Assens Fjernvarme A.m.b.a. to supply a larger area
using the existing boilers and a few small assist stations for the wintertime. In 1982 Assens
Fjernvarme A.m.b.a. was the most expensive district heating company on Funen due to a
poorly maintained plant and distribution network.
In 1983 Assens Fjernvarme A.m.b.a. replaced fuel oil with Tall-oil as it was a significantly
cheaper fuel type including oil taxes.
Tall-oil got prohibited as a source of fuel in 1984 and Assens Fjernvarme A.m.b.a. installed
two coal boilers located at Stejlebjergvej in Assens. The boilers had a capacity of 6,3 MW
each. Coal was significantly cheaper than oil in particular because of the low taxes. At this
time, there were only a few district-heating companies transitioning to coal. As a consequence,
Assens Fjernvarme A.m.b.a. went from the most expensive heating plant to the cheapest
heating plant on Funen.
At the end of the 1980s many district heating companies were transitioned to coal and the
State lost a lot of tax revenue on oil. To compensate for the loss the tax on coal was raised
over several times.
In 1989 Assens Fjernvarme A.m.b.a. was the first producer of energy in Denmark to gradually
convert to wood pellets. The production capacity of the two boilers on Stejlebjergvej were only
5 MW on wood pellets but they got improved and after a few years they were able to produce
6 MW per boiler.
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In 1990, the business strategy of Assens Fjernvarme A.m.b.a. had evolved to supply all of
Assens city with district heating. A third boiler was build on Stejlebjergvej, engineered to use
dry wood chips. The boiler at Hardersvej was considered a reserve.
In 1992 the Danish Energy Regulatory Authority imposed that all district heating plants
should produce electricity with natural gas as the likely choice of fuel. It was possible to avoid
gas if the plant could produce electricity using renewable energy. Since there is only one gas
supplier in the market, Assens Fjernvarme A.m.b.a. decided to explore the use of woodchips.
In February 1999, the first Danish decentralized thermal power plant based on renewable
energy was ready in Assens. Its capacity was 4,6 MW electricity and 13,5 MW heat. From
the beginning many had seen this project as unfeasible, but a strong connection to the Danish
Energy Agency, who needed a success with renewable energy, resulted in lucrative subsidies
from the sale of electricity and the economy in Assens Fjernvarme A.m.b.a. were very good.
This placed Assens Fjernvarme A.m.b.a. as the cheapest district heating plant on Funen and
between the 10% cheapest in the country.
To increase the production during summer, Nyhuse and Torøhuse are getting supplied by
district heating. Further a cooperation with Glamsbjerg and Hårby was attempted but
without success. Instead the possibility of supplying Sønderby/Ebberup was explored.
In 2011 Kærum, Sønderby and Ebberup were supplied by district heating. To have enough
capacity during winter one or more boilers from Stejlebjergvej would supply the thermal power
plant. In 2014 Saltofte was also supplied by district heating. Thereby district heating from
Assens Fjernvarme A.m.b.a. is getting even cheaper.
3.3 Legal aspect
The following section revolves around the relevant judicial framework regarding the production
of district heating and the associated electricity. It will describe laws that may affect the result
of the project, and discuss the relevance. It will not summarize the laws to their full extent, but
merely focus on the aspects that are found to be relevant to this project.
First order of business is to determine what laws have been considered relevant for the
purpose of this project. The problem statement is focused on the production and transmission
of district heating, produced by a collective power plant based on renewable fuels – more
specifically woodchips and – pellets, which constitutes biomass. As such, laws which may
affect various aspects of the production of district heating, such as the cost of production and
the price paid by consumers are considered relevant. Furthermore, any law work that may
affect the establishment of the planned transmission line is also considered relevant.
3.3.1 Prices
An important aspect of the production optimization would naturally be the income aspect and
as such, the conditions regarding the heat and electricity prices need to be established. Section
20 of the heat supply act.[13] states that collective power plants may include any production
based costs into their price – so that any expenses associated with production may be covered.
Furthermore, according to section 20-b the district heating company Assens FJV may include
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a profit in the price because they base their production on biomass. Finally section 20 point 5
states that collective power plants may give different groups of costumers different prices. In
conclusion, Assens FJV can include all expenses from production in their prices, may profit
from the district heating provided, and can give different prices to costumers in Assens from
the costumers in Ejby. Common practice stipulates that each consumer pays his proportionate
share of the investment.[14, p. 334] Furthermore the company may profit from the electricity
it produces – for now Assens FJV sells electricity at 0,60 DKK/kW,[15] as per section 45 point
3, according to sustainable energy act.[16] This is in place until the 31st of December 2018,
after which the company may sell the electricity at the price which the market dictates, while
receiving 0,15 DKK/kWh as a surcharge, in accordance to section 45. It is known that the
subsides change, this will be analyzed further in scenarios, placed in section 6.1.
3.3.2 Establishment of district heating
The approval of suggested establishment of, and major changes to the district heating plants,
is up to the local municipality council, as per Bekendtgørelse af lov om varmeforsyning,
section 4.[17] The council ought to dertermine whether the project is the most social
economically beneficent option, as per Bekendtgørelse om godkendelse af projekter for
kollektive varmeforsyningsanlæg, section 6.[18] Furthermore the council may approve projects
that ensure the change in supplier from natural gas to biomass based central heating, if
compensation is payed for each property affected by the change (section 8).
Finally, in order for the project to be accepted, the council needs to make and energy-,
environmental- and social economical assessment of the project (section 26).
It should be noted that there is a chance that the expansion of operations to the neighbouring
city of Ejby will attract the attention and ire of competing gas companies – as was the
case when Assens FJV suggested the establishment of a transmission line to Ebberup.[19]
According to the documented decision to the complaint filed by Naturgas Fyn Distribution
A/S, the plaintiff first claimed that the social economic situation was not against maintaining
the status quo (page 2, line 25), and as such the suggested transmission line should be
rejected. After a revised plan the plaintiff claimed that it would result in losses of 31.000 DKK
per property by the year 2025 (page 5, line 12) – a substantial increase from the 6.500 DKK
per property offered by Assens FJV (page 4, line 23).
While the complaint was ruled to be in favor of Assens FJV in that particular case, this does
not exclude the possibility of further judicial actions in the case of further expansions – and
as such, a similar chain of events could occur. While this should not spell the end on the
suggested project, as the legal practice has been established to be seemingly in favor of such
a project, it may lead to delays and legal expenses.
Assens FJV can impose the consumer in Ejby to pay the extra cost of an expansion to Ejby.
In order for an expansion to Ejby to be approved, the project needs to give a socioeconomic
gain for the consumers in Ejby. The electricity price for the woodchip boiler is 0,6 DKK/kWh
until 31st December 2018, after which they may claim the spot price and a surcharge of 0,15
DKK/kWh.
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Model development
4
4.1 Optimization in general
In this section the fundamental understanding of optimization will be clarified to ensure the
understanding of the subsequent model development.
Optimization is a process of making something as a system, as fully perfect or effective as
possible. Hereby a mathematical optimization is the selection of the best elements, with
regards to some criteria, from some set of alternatives. In the simplest case, an optimization
problem consists of maximizing or minimizing a function by systematically choosing input
values within the feasible region and computing the value of the function. The maximum
and minimum of a function are the largest and smallest value of the function, either within
a given range of on the entire domain of a function. Furthermore unbounded infinite set, such
as the set of real numbers, have no minimum or maximum. The goal of optimization is to find
maximum and minimum of the function. A method of finding a global maximum or minimum
is to look at all the local maximum or minimum in the interior, also looking at the values on
the boundary and then take the largest or smallest value.
Generally, optimization includes finding the best available values of an objective function given
a defined set of constraints, including a variety of different types of objective functions and
different feasible regions. This object function provides a quantitative measure of the wanted
solution. Hereby an optimization will be to find a combination of variables, which maximize
or minimize the object function. These variables will be assigned restrictions, which are given
by the type of problem. Any restrictions on the values that can be assigned to the decision
variables, which are represented as n related quantifiable decisions to be made (x0, x2, ..., xn),
are also expressed mathematically, typically by means of inequalities or equations (e.g. P =
3x1 + 2x2 + ... + 5xn). Such mathematical expressions for the restrictions are called constraints.
4.1.1 Linear optimization
Linear optimization is a method to achieve the best outcome in a mathematical model whose
requirements are represented by linear relationships. Linear optimization is a technique for
the optimization of a linear objective function, subject to linear equality and linear inequality
constraints. The feasible region is the collection of all feasible solutions, which is a solution,
where all the constraints are satisfied. The solution can also be an infeasible solution, which is
a solution for which a least one constraint is violated. Furthermore, a feasible solution that
has the most favourable value of the objective function is an optimal solution. The most
favourable is the largest value if the objective function is to be maximized, whereas it is the
smallest value if the objective function is to be minimized. When finding a solution of a linear
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optimization a corner point feasible solution, hereafter referred to as CPF solution, can be
found, which is a solution that lies at a corner of the feasible region.
Generally the greatest difference between linear and non-linear optimization is that the linear
optimization works with the CPF solution and the non-linear optimization does not. Here it is
also possible that the optimal solution is inside the boundary of the feasible region.
4.2 Method choice
In this section, the optimization method and software to solve for the optimal production in
Ejby and Assens, will be chosen.
4.2.1 Linear optimization benefit and limitations
Linear optimization has the benefits, that with Simplex,described in section 4.3 it can find the
optimal solution very fast and the model only need to consider a CPF solution which is not
the case for nonlinear optimization. This is useful in this particular case, due to the fact that
the model will be faster.
Linear optimization requires that functional constraints and object function are proportional.
As such it is not possible to include startup costs in the optimization in this project. Startup
costs can be included if mixed integer programming is chosen instead. Furthermore, the
parameter in the model is not allowed to change with the decision variable; thereby production
unit efficiency needs to be assumed constant.
4.2.2 Chosen optimization method
Linear optimization will be used in this project, as this has been the focus of the current
semester. A better choice will be to include especially startup costs to ensure that production
units will not start up in very short periods. On the other hand, linear optimization is very
reliable and less computer-consuming.
4.2.3 Software choice
A lot of software packages to solve linear problems exist, some of them are shown in the table
below with their advantages and disadvantages.
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Advantages
Excel solver Matlab Mathematical modelling
programming languages
The model is easy to set up Good to process and
analyses large amount of
data
Often possible to automatically
create restrictions which are
similar (because of 8 heat
production units – there will
be a lot of these situations)
Easy to repeat calculations
and restrictions
Scalable
The group has experience
with Excel solver
The group has experience
with Matlab
The program is so widely
used, that it easily can be
adopted by third party
Table 4.1. Advantages by use of different solvers
Disadvantages
Excel solver Matlab Mathematical modelling
programming languages
Not possible to simulate one
year because Excel solver
is restricted to 200 decision
variables (our model has
around 240 decision variables
beyond a lot of constraints)
– Open Solver is a solution
because it can handle a lot
more decision variables and
constraints.
This model requires
more time to set up in
comparison to excel.
The method is complex and
resource intensive.
Need to program a macro
that runs the optimization
for a whole year – nobody in
the group have worked with
Visual Basic.
Data processing is more
difficult than for other tools
like Matlab.
Table 4.2. Disadvantages by the use of different solvers
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Due to the complexity of mathematical modelling programming languages, they have been
rejected for use is this optimization problem, as their complexity seems to go beyond the
boundaries of the standard set by the course lectures. Furthermore, what benefits could be
reaped from its use, does not seem to justify the time necessary to learn it. Matlab is chosen
instead due to its easy scalability. On the other hand, Excel with OpenSolver will be used as a
prototyping tool to ensure the constraints are correct before implementation in Matlab because
of the possibility of fast setup of a small optimization problem.
4.3 Simplex method [20]
In this section the simplex method is explained, which can solve linear problems. This could
potentially be the way to solve the production optimization, but because it included a lot of
decision variables and restrictions a light example is provided here to give an idea of the
concept. In reality the problem is not solved by the hand but instead a computer solver like
Matlab or Excel Solver are used.
One application of this equation, could be to maximize the profit, where the decision variable
is the amount sold and the coefficient is the sales price.
Maximize: Z = 20x1 + 10x2
Subject to:
x1 − x2 ≤ 1 (4.1)
3x1 + x2 ≤ 7 (4.2)
x1 ≥ 0, x2 ≥ 0 (4.3)
To solve a model like this example the simplex method needs a system of equations, where the
nonnegative slack variables (named x3, x4) are introduced, so the model becomes:
Z − 20x1 − 10x2 = 0 (4.4)
x1 − x2 + x3 = 1 (4.5)
3x1 + x2 + x4 = 7 (4.6)
x3 ≥ 0, x4 ≥ 0 (4.7)
Notice: This example is a maximization problem, where the production optimization in this
project is a minimization problem. In practice, the minimization problem can be rewritten as
a maximization problem, by multiplying the object function by -1
The initial solution
Consider the initial system of equations, compared to their rewrites as shown below. Where
equation 4.9 and 4.10 include two more variables than equations. Therefore, two of the
variables, also called the non-basic variables, can be arbitrarily assigned a value of zero in
order to obtain an initial solution, the basic solution for the other two variables, the basic
variables. This solution will be feasible if the value of each basic variable is nonnegative. The
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best of the BFS is known to be an optimal solution, so the simplex method finds a sequence of
better and better BFS until it finds the best one.
Z − 20x1 − 10x2 + 0x3 + 0x4 = 0 (4.8)
1x1 − 1x2 + 1x3 + 0x4 = 1 (4.9)
3x1 + 1x2 + 0x3 + 1x4 = 7 (4.10)
To begin the simplex method, the slack variables are chosen to be the basic variables, hereby
x1 and x2 are the non-basic variables and set to be equal to zero. By solving the system of
equation the basic feasible solution is found to be (x1, x2, x3, x4) = (0, 0, 1, 7).
Optimality test #1
To see if the new BFS is the optimal solution rewrite equation 4.8 to:
Z = 20x1 + 10x2 (4.11)
The optimal solution, when maximizing, is found when no positive coefficient are present in
object function, where Z is isolated. Since both x1 and x2 have positive coefficients, increasing
either one of these variables can increase Z. Therefore, the current BFS is not optimal, so there
need to be preformed an iteration of the simplex method to obtain a better BFS.
Choosing the entering basic variable
x1 has the largest positive coefficient in equation 4.8, hence increasing x1 will increase Z at
the fastest rate (most positive variable in equation 4.8). Therefore, x1 is the entering basic
variable. This selection rule tends to minimize the number of iterations needed to reach an
optimal solution.
Choosing leaving basic variable
x2 is not changed because x1 is chosen as the entering basic variable. The leaving basic
variable is found by applying the minimum ratio test:
x1 − x2(= 0) + x3 = 1 ⇒ x1 ≤ 1 ← min (4.12)
3x1 + x2(= 0) + x4 = 7 ⇒ x4 ≤ 7 (4.13)
x3 is the leaving basic variable because it is the first variable to reach zero when increasing the
entering basic variable (x1).
Finding new BFS
The coefficient of the entering basic variable, in the equation of the leaving basic variable
(equation 4.9) must be 1. The current value of the coefficient is 1, therefore nothing needs to
be done to this equation.
Next a coefficient of zero for the entering basic variable needs to be obtained in every other
equation (equation 4.8 and 4.10). The coefficient of x1 in equation 4.8 is -20, so to obtain a
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coefficient of 0 it needs to be added 20 times equation 4.9 to equation 4.8. The coefficient of x1
in equation 4.10 is 3, so to obtain a coefficient of 0 it needs to be subtract 3 times equation 4.9
from equation 4.10. New system of equation is then:
Z − 0x1 − 30x2 + 20x3 + 0x4 = 20 (4.14)
1x1 − 1x2 + 1x3 + 0x4 = 1 (4.15)
0x1 + 4x2 − 3x3 + 1x4 = 4 (4.16)
New BFS is (x1, x2, x3, x4) = (1, 0, 0, 4) and Z = 20. This is a better solution than the
previous one - the intial BFS with Z equal to zero.
Optimality test #2
Isolate Z in equation 4.14 as: Z = 20+30x2 −20x3 The optimal BFS is found when no variable
in the rewrite object function has a positive variable. Because x2 have a positive coefficient
the current BFS is not optimal. Then repeat the steps: Find entering basic variable → Find
leaving basic variable → Rewrite system of equation → Find new BFS → Optimality test until
the optimal solution is found.
4.4 Pipe dimensioning
This chapter will concern itself with the governing principles behind the choice of pipe
diameter. It will present the concerns associated with this problem, as well as the formulas used
to quantify it. It will not present the optimal choice of diameter, but merely present the tools
with which the optimization will determine said diameter. Nor will it concern itself with the
theory behind said tools, but merely provide references to the academic literature from which the
tools were garnered.
When warm water flows through a pipe, two immediate concerns present themselves:
• The water needs to actually be moved from point a to point b, which represents a loss in
pressure – if the flow is to be maintained, to satisfy the Bernoulli equation, so must the
pressure, and as such pumps needs to be installed in key locations along the pipeline -
pumps that require electricity in order to run;
• In transit, the water will transfer some of its thermal energy to its surroundings. As
it happens, these two problems are directly tied to the correct choice in diameter. A
larger diameter means a slower flow, resulting a reduced pressure differential – meaning
lower pump costs. On the other hand, a larger diameter is followed by a larger heat loss.
Consequently the optimal diameter for a given flow would be a cost minimizing diameter
– the diameter in which the sum of the cost of the heat loss, and pressure differential is
as low as possible.
It should also be noted that at this point in the process, the pipe has been determined to lay
in the main road from Assens to Ejby. As such the positioning of the pipe is already known,
and will not be altered, even if this might have reduced pressure and heat loss. Furthermore,
the pipes considered in this porject are made by Isoplus, single steel pipes, series 3, as
described by product catalogue. The reason Isoplus is chosen is due to a preference stated by
Assens FJV, and series 3 is chosen due to their thicker insulation.
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Velocity
As production will be optimized, the energy consumption in Ejby will be known. And since
the temperatures of the feed and return of the water is also known, it is possible to determine
exactly how much water is needed in order to satisfy the demand:
˙m =
Φ
cp,1 · T1 − cp,2 · T2
[
kg
s
] (4.17)
• ˙m: Mass flow [kg
s ]
• Φ: Heat consumption [kJ
s ]
• cp,1: Specific heat capacity, supply line [ KJ
kg·K ]
• cp,2: Specific heat capacity, return line [ KJ
kg·K ]
• T1: Water temperature, supply line [°C]
• T2: Water temperature, return line [°C]
Once the mass flow has been determined, it is a simple matter of dividing the result with the
water’s density in order to find the volume flow:
Q =
˙m
ρ
[
m3
s
] (4.18)
• Q: Volume flow [m3
s ]
• ρ: Density of the water [Kg
m3 ]
Once the volume flow has been determined, velocity can be easily calculated as a function
based on diameter – the larger the diameter, the slower the velocity, and vice versa:
V =
Q
D2 · π
4
[
m
s
] (4.19)
• V: Velocity of the pipe [m
s ]
• D: Diameter of the pipe [m]
Pressure differential
When it comes to pressure differential in regards to pipe flow, three causes are relevant:
• The major head loss, stemming from the friction and turbulence
• The minor head loss, stemming from the twists and turns
• The pressure loss associated with the height difference
These three can be summarized into the total pressure differential, and then converted into
the total pump costs. The major head loss can be determined through a relatively simple
equation:
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HL,Major = f ·
L
D
·
V 2
2 · g
[mWc] (4.20)
• HL,Major: Pressure loss due to friction and turbulence [mWc]
• f: Friction factor [-]
• L: Length of the pipe [m]
• g: Gravitational acceleration [m
s2 ]
Of these factors, only f is at this point unknown – normally f can be easily read from a Moody
Chart, a diagram that visualises the relationship between the friction factor, the roughness of
a pipe and the Reynolds number of a pipe flow. However, since optimization is the governing
principle behind this paper, it is useful to utilize a formula, which estimates the friction factor,
thereby speeding up the process. For the purpose of this report, the following formula is used:
f =
1, 6364
ln(6,9
Re + 0, 27 · (D )1,11)2
[−] (4.21)
• is rougness, a factor dependant on the pipes used
• Re: Reynolds number [-]
Defined as:
Re =
ρ · V · L
µ
[−] (4.22)
• µ: Dynamic viscosity [ N
s·m2 ]
The minor head loss is determined through the following equation:
HL,minor = KL ·
V 2
2 · g
[mWc] (4.23)
• HL,minor: Pressure loss due to bends [mWc]
• KL: factor, based on the degree of turns and bends of the pipe [-]
This has been estimated to reach a value of approximately 3,1, based on the Termis model.
The pressure difference from the height difference is easy to determine, as the heights of the
pipe is already known.
The full pressure difference is therefore a combination of the three equations:
∆p = (f ·
L
D
+ KL) ·
V 2
2 · g
+ ∆h[mWc] (4.24)
• ∆p: Pressure difference [mWc]
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• ∆h: Change in height [m]
The result is the sum of the pressure differential, in terms of meters of water column.
For the result in pascal:
∆p[
Pa
m
] = ∆p[mWc] · ρ · g (4.25)
It can be ascertained by the above equation, a smaller diameter and higher velocity results in
a drastically higher pressure differential. It should also be noted, that since the water needs
to be transported both back and forth via the pipe, the total effect of the height differential
on the pressure is equal to zero - if the water goes up, it will eventually come back down
again. The height does have a relevance as to the placements of pumps however, which will
be discussed later.
The work done by the pump is calculated by the following equation:
Wpump = Q · ∆p · ρ · g[W] (4.26)
• Wpump: Work done by pump [W]
Heat loss [21]
The heat loss calculations are a more complicated matter, as they consists of several elements
and are based on several assumptions.
Naturally the total heat loss from the pipes is equal to the sum of the heat loss from the two
pipes:
qtotal = q1 + q2[
W
m
] (4.27)
• qtotal: Total heat loss [W
m ]
• q1: Heat loss, supply pipe [W
m ]
• q2: Heat loss, return pipe [W
m ]
The heat loss from the warmer pipe (the supply) will be equal to the heat loss to the ground
plus the heat transfer to the colder (return) pipe:
q1 = qs + qa[
W
m
] (4.28)
• qa: Anti symmetrical heat loss [W
m ]
• qs: Symmetrical Heat loss, supply pipe [W
m ]
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Figure 4.1. The anti-symmetrical & symmetrical heat loss
These two heat losses will henceforth be referred to as the “symmetrical problem” for the heat
loss to the ground and the “anti-symmetrical problem” for the heat transfer inbetween the
pipes. The heat loss from the colder pipe, on the other hand is equal to the heat loss to the
ground minus the heat transfer between the pipes (as it gains some heat from the warmer feed
pipe):
q2 = qs − qa[
W
m
] (4.29)
The heat transfer to the ground, that is to say, the symmetrical problem, is calculated using
the following equation:
qs = (Ts − T0) · 2 · π · λg · hs[
W
m
] (4.30)
• Ts: Temperature used in symmetrical problem [°C]
• T0: surface temperature [°C]
• λg: is the thermal conductivity of the ground [ W
mK ]
• hs: is the heat loss factor of the symmetrical problem [-]
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Figure 4.2. Visualization of the pipes and surrounding
The heat transfer between the pipes (the anti-symmetrical problem) is:
qa = Ta · 2 · π · λg · ha[
W
m
] (4.31)
• Ta: Temperature used in anti symmetrical problem [°C]
• ha: The heat loss factor of the anti symmetrical problem [-]
Ts and Ta are calculated as follows:
Ts =
T1 + T2
2
[℃] (4.32)
Ta =
T1 − T2
2
[℃] (4.33)
• T1: Temperature of the water in pipe – supply pipe [°C]
• T2: Temperature of the water in pipe – return pipe [°C]
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Whereas the heat loss factors, hs and ha are calculated likes this:
h−1
s = ln(
2 · H
ro
) + β + ln( 1 + (
H
D1
)2)[−] (4.34)
h−1
a = ln(
2 · H
ro
) + β − ln( 1 + (
H
D1
)2)[−] (4.35)
• H: the depth between the surface and the center of the pipe [m]
• D1: Half the distance between the centers of the two pipes [m]
• β: A parameter describing the insulation of the pipe [-]
• ro: The outer radius of the pipe [m]
β is determined by:
β =
λg
λi
· ln(
ro
ri
)[−] (4.36)
• λi: The thermal conductivity of the insulation [ W
mK ]
• ri: The inner radius of the pipe [m]
The ﬁnal result of the heat loss, qtotal is in Watts per meter.
On top of these formulas, several assumptions have to be made. Everything from thermal
conductivities to the surface temperature is subject to change. The list of assumptions
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Symbol Description Value
T1 Supply temperature 72℃
T2 Return temperature 38℃
T0 Ambient temperature 9℃
λi Thermal conductivity of isolation 0, 027 W
m·K
λg Thermal conductivity of ground 1, 6 W
m·K
H
ri → H
Distance under ground <80mm → 400mm +ro
>80mm → 600mm +ro
D
ri → D
Distance between pipes <80mm → 150mm
2 + ro
>80mm → 250mm
2 + ro
Table 4.3. Assumptions used to calculate the pressure drop and heat loss
The temperatures used are based on information given to us by Assens FJV. The rest of the
assumptions are based on information provided by course lecturers Chan Nguyen and Lasse
Elmelund Pedersen. The pipes used for these calculations will made by the company Isoplus,
the single steel pipes series 3, as described at their website.[22]
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4.5 Degree-day correction
Degree-days is a tool used in this project to manipulate arbitrary heat demand data, obtained
from Fjernvarme Fyn, in order to estimate the heat demand for Assens and Ejby during a
standard day for every month. These standard day heat demands are subsequently used to
calculate a gradient, which results in a unique heat profile for every hour of the year.
In general [23]
Degree-days are a way to measure heat demand and can be used to compare the energy
consumption for space heating with degree days for a reference year. In other words, degree-
days can be used to monitor energy consumption and as a tool for district heating production
planning. Degree-days are defined as:
Degree days = (Ti,Base − To,Mean) · Time (4.37)
Where:
• Ti,Base: Indoor base temperature. Standard indoor temperature is usually 20°C, but
when calculating degree-days, 17 °C is used due to the assumption that the temperature
difference is being obtained as casual heat gains from e.g. people, lighting and solar
gains.[°C]
• To,Mean: Mean temperature for the chosen time period. Usually the mean temperature
for a month.[°C]
• Time: The time period chosen for calculation of degree-days.[hr]
Degree-days are freely available online and can be found e.g. at the homepage of The Danish
Meteorological Institute (DMI).
Example: Projected heat demand day profile for February
For this project, approximate annual total heat demands were obtained from Assens FJV
(80.000 MWh). By using these heat demands and data of degree-days for 2015, the total heat
demand for one standard day each month can be calculated.
Standard day - day profile
To be able to calculate the heat demand day profile for a standard day, hourly heat demand
data has been obtained from Fjernvarme Fyn A/S. These hourly heat demands are used for
calculating a factor, which is used to calculate the heat demand day profile.
The heat demand day profile for one standard day in February can be calculated as:
Day profile HDStandardday =
hr=24
hr=1
FactorFV F,February,hr · HDFebruary,Standardday,Degree−daycorrected (4.38)
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Where:
• FactorFV F,February,hr: Calculated as
HDF ebruary,hr
HDF ebruary,total
• HDFebruary,hr: The sum of heat demands for the same hour every day of February, e.g.
the sum of heat demands for hour 1. [MWh]
• HDFebruary,total: Total heat demand for February. [MWh]
• HDFebruary,Standardday,Degree−daycorrected: Calculated as
Degree−dayF ebruary
Dec
Jan Degree−daymonth
· HDAssens,Annual,Total · Days−1
February
• Degree − dayFebruary The sum of degree days in February
• Dec
Jan Degree − daymonth The sum of degree days in 2015
• HDAssens,Annual,Total The total heat demand in Assens in 2015
• DaysFebruary Days in February
Figure 4.3. The day profile for a standard day in February.
Heat demand gradient The heat demand gradient used to give every day a different heat
demand is calculated as:
Gradient =
HDMarch − HDFebruary
Days in February
(4.39)
Where:
• HDMarch: Total heat demand in March [MWh]
• HDFebruary: Total heat demand in February [MWh]
The standard day is defined as the 15th in each month. The gradient is then added once every
day between the standard days in two months e.g. added every day between the standard day
in February and March and the demand will gradually increase og decrease depending on the
15th in the following month, in this scenario March. See figure 4.4
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Figure 4.4. Heat demand gradient in Februar
4.6 Termis
Termis is a modelling software which simulates district heating infrastructures. It can give
the user detailed information concerning velocities, temperatures, pressure and heat losses.
Its primary purpose in this report will be to help in terms of planning the design of the
transmission line, as it can help create digital a version of this. The reasoning behind this
prioritization is that the primary concern of this report is optimizing production, using
linear optimization – a task which Termis can not solve for us. Instead it will be necessary
to calculate things e.g. pump costs and heat loss, in a way that can be integrated in the
optimization – this is explained in section 4.4.
4.6.1 Designing the transmission line
Based on conversations with Assens FJV, it was brought to light that they wished for the
transmission line to follow the existing road structure from Assens to Ejby. The “design”
process, is therefore not a question of choice between various paths, but instead merely a
question of knowing the coordinates of the roads in question. When designing a system in
Termis, it is possible to import GPS coordinates, which can then be used as a base for the
system. This will include vectors and X/Y/Z coordinates on structures like roads and building.
As such, with the right data set, designing the Transmission line is a simple prospect. This
will give the user information like heights, pipe length and bends – all of which are relevant
when calculating the pressure diﬀerential.
With gps data provided by kortforsyningen.dk a map is generated detailing the pipe and
surrounding cities:
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Figure 4.5. Map generated by Termis [24]
It should be reiterated that this process will not include designing a distribution infrastruc-
ture, as one is already in place. Instead a pipe will be put in place from the production facili-
ties of Assens FJV to the proximity of Ejby. When all of this is done, Termis reports that the
pipe line will be 19.304 meters long. The second aspect of the design process will be the choice
of pump stations, and where to place them. Over the course of the transmission line, pressure
changes will occur, due to friction, turbulence bends and height differences. In order to main-
tain a flow in the right direction, pumps need to compensate for this. Furthermore the place-
ment of said pumps needs to take into account the full pressure differential – if, e.g. a pump
is placed on top of a hill, but the flow does not have enough pressure to actually get up the
hill, the pump is obviously not placed in the optimal spot. Therefore, in order to conclude on
pump placement, it is necessary to know what the height is at any point in the pipe line, in
order to make sure that the flow is maintained. When the pipeline has been designed in ter-
mis using the GPS coordinates, reading the heights is a simple prospect. All of this however,
is reliant on the result of the optimization - pumps can not be put in place, if the pressure has
not been calculated - and as such, this topic will be returned to, upon the presentation of the
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results.
4.7 Optimization of electricity and heat production
4.7.1 Delimitation
The delimitation of this model is shown in figure 4.6. The distribution network in Assens and
Ejby are not considered as a part of the optimization, but heat loss is assumed to be constant
20% of the heat demand.[12] Heat loss and pump loss cost in the pipe between Assens and
Ejby are not a part of the Matlab model, but will afterwards be included as described in
table 4.4. The Matlab model produces the extra amount of heat to cope with the loss in the
distribution network. The extra amount of electricity are being sold. Notice the storage unit is
still inside the delimitation of the model because they are used in the model, but is not a part
of the optimization.
Figure 4.6. Delimitation of this model
Assens FJV provides all relevant data to the simulation. The data is listed in Appendix 8.4.
4.7.2 - Without optimization of storage (Matlab model 1 year)
The first part of this section describes the overall ideas of how to find the optimal heat and
electricity production and optimal pipe dimension between Assens and Ejby. The optimal
production is used to investigate whether an investment in a pipeline to Ejby from Assens give
a positive return. Afterward the optimization equations, which can be used to solve the problem
by using Simplex in Matlab, will be described.
According to Assens FJV, they do currently not produce in accordance with an optimal
production plan. Instead, Assens FJV uses the two storage units in Assens to cope with peak
load exceeding the capacity of the woodchip boiler, hence they operate as safety buffers. To
simulate the current operation, a Matlab-model will be created without optimizing the storage.
This model will use linear optimization for the reason described in section 4.1. The objective
will be to minimize the production cost for Assens and Ejby. The Matlab simulation will run
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for 24 hour intervals at a time, in order to simulate the real world scenario, where the spot
price is known in 24 hour intervals. This choice will give some more realistic results, compared
to if a simulation for an entire year were carried out at once. The simulation will be based on
the spot price for 2015 extracted from Energinet.dk Markedsdata because the simulated year is
chosen to be year 2015. To get a more accurate result for the cost, a better solution will be
to simulate the future, but thereby the model will take much longer to simulate, introduce
uncertainty in data input and a lot more data output. Because of this reason the model is
chosen only to simulate the year 2015.
The pipe dimension between Assens and Ejby is a balance between lower pump cost,
production cost and heat loss cost. This relationship is described further in section 4.4. To
find the balance which minimizes the total cost, an iterative method is used. This consists of 6
steps, which are described in the table below.
Steps Description
1
At first, the optimal production without any restriction capacity,
between Assens and Ejby, will be found. The result is cost in DKK.
Notice negative cost means profit.
2
Based on the heat production and heat demand, the transferred
capacity between Assens and Ejby will be calculated on an hourly
basis.
3
Based on step 2, a qualified guess of the optimal maximum
transmission capacity will be chosen. This hopefully reduces the
number of calculations, when finding the real optimal pipe dimension.
4
The maximum transmission capacity between Assens and Ejby is
inserted as a restriction in the Matlab model. The result is that
expensive production may occur if the maximum capacity is met. The
result is cost in DKK.
5
Based on the maximum transmission capacity a pipe dimension which
minimize the pump cost and heat loss cost and also complies to the
restriction on fluid velocity and pressure gradient will be chosen. The
calculation of the costs is described in section 4.4. The pump cost and
heat loss cost is added to the result from step 4.
6
Step 4-5 is repeated with changing transmission capacity between
Assens and Ejby until the lowest possible cost is found. The result is
cost in DKK, optimal pipe dimension and optimal heat- and electricity
production.
Table 4.4. Iterative method to find the optimal pipe dimension and heat and electricity production
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Assumptions
• Start-up cost is not included because this model is chosen to be linear4.2.2.
• Heat loss from the storage units are not included, based upon an interview with Assens
FJV.
• Heat loss in the distribution network is 20% of the heat demand.
• The start-up time for a production unit is assumed to be zero hour.
Definition of decision variables and constraints
General definition:
t
Time dependent variables. t = 1 for the first hour and t = 8760
for the last simulated hour.
Production optimization:
cn
Heat production cost [DKK / MWh heat production]. Notice
that the heat production cost is fixed because the heat
production cost is assumed constant for all hours of the year and
thereby it is not a time dependent variable. The heat production
cost is calculated as:
cn = (Fuel price + V ar O&M Heat + Energy tax) · (1 + (1 −
Total Efficiency))
pk,t
Electricity income for a given hour. The spot price used is
from Nord Pool Spot [DKK / MWh electricity production].
Calculated as: pk,t = SPt − O&M Electricity
HDAssens,t Heat demand Assens for a given hour [MWh]
HDEjby,t Heat demand Ejby for a given hour [MWh]
CMn Slope counter pressure line for heat unit n [-]
xmin,n
Minimum heat production capacity for a given heat production
unit n. Notice xmin,n is not included in the optimization because
of the reason is described in the constraint-section [MWh]
xmax,n
Maximum heat production capacity for a given heat production
unit n. [MWh]
Capacity between Assens and Ejby:
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∆Xmax
Heat transport max capacity between Assens and Ejby. Heat can
go in both directions. Defined as positive value. [MWh]
XAssens,t
Sum of heat production (xn,t) located in Assens for a given hour
[MWh]
XEjby,t
Sum of heat production (xn,t) located in Ejby for a given hour
[MWh]
Decision variables
This optimization model finds the hourly optimal heat and electricity production defined as:
xn,t Heat production for the given unit at a given hour [MWh]
zk,t Electricity production for the given unit at a given hour [MWh]
Object function
Z Production cost for 24 hour production [DKK]
The object function minimizes the production cost for 24 hours and is repeated for the whole
year. The cost function includes heat production cost (cn) and electricity revenue (zk,t), where
cost is defined as a positive value. Notice the object function expand in a row for every hour
(t=[1;24]) added. This is only needed in Matlab, but not necessarily in Excel described later in
section 4.7.3.
minZ =
t=24
t=1 n=heat production unit
cn · xn,t −
k=electricity production unit
pk,t · zk,t
Notice, a negative result means negative profit for the 24 hour production and positive result
means positive profit for the 24 hour production.
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Constraints
Heat demand = heat delivered.
Notice, because storage not is
included, this is the same as heat
demand = heat production.
HDAssens,t + HDEjby,t =
n
xn,t
Fixed production between electricity
and heat.[14, p. 269] It applies to
all CHP in Assens and Ejby that
produces on the counter pressure line.
Notice that this restriction is only
viable if the unit produces both heat
and electricity.
xn,t · CMn = pk,t
Maximum heat production for
production unit n (xn,t) in a given
hour need to be below the maximum
heat production capacity for the given
heat production unit (xmax,n). Notice
that maximum electricity production
is not included as a constraint
because it is indirectly included in
the equation xn,t · CMn = pk,t that
described the correlation between
electricity and heat production for a
counter pressure plant.
xn,t ≤ xmax,n
Non negative constraints: No negative
heat and electricity production.
xn,t ≥ 0
pk,t ≥ 0
Notice that xn,t ≥ xmin,n is not included, because then zero could not be possible. This
can result in a small error in the end result because in reality, a cheaper device lower the
production and a more expensive unit start (above xn ≥ xmin,n) to balance the heat
production and heat demand.
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Figure 4.7. The used production relationship between heat and electricity will be CMn for the CHP-
units. In the real world, it is not possible to produce at the red dot line, but in this
model it will be possible in order to make the model linear.
Constraint: Heat transfer capacity between Assens and Ejby
The model starts the cheapest production unit (based on production cost and possibly
electricity income) first subject to the constraints set in the model. The model does not
prioritize between production units based on their location, but rather production cost
included possible electricity income to minimize production cost.
The transferred capacity from Assens to Ejby must be the production in Assens subtracted
from the heat demand in Assens. This value is positive when heat is transferred to Ejby and
should be smaller or equal to the maximum transmission capacity (∆Xmax). Resulting in the
restriction:
−HDAssens,t + XAssens,t ≤ ∆Xmax (4.40)
On the other hand, the LHS value of the above inequality is negative when heat is transferred
from Ejby to Assens. This negative value must be larger than or equal the maximum negative
transmission capacity:
−HDAssens,t + XAssens,t ≥ −∆Xmax (4.41)
Optimal pipe dimension (and heat and electricity production)
The steps described in Tabel 4.4 to find the overall minimum cost will now be described
further included the relevant variable defined above.
First run
1. The Matlab-model run with no restriction on the heat transport capacity between
Assens and Ejby, thereby set ∆Xmax → ∞. In practice ∆Xmax is just set to a very
large number, which is above the total possible heat demand in Assens and Ejby taken
together.
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2. Calculate heat transferred between Assens and Ejby (|−HDAssens,t +XAssens,t|) for every
hour, set ∆Xmax to the calculated value. Since the heat transferred can be negative if
heat from Ejby to Assens the absolute value is taken to the result.
3. Calculate the necessary pipe dimension for every given hour based on ∆Xmax and
minimizes the total cost and complies to restriction on fluid velocity and pressure
gradient.
4. Analyze the data – find one pipe dimension that will be the most optimal pipe dimension
by use of Figure 4.8.
Figure 4.8. Optimal pipe dimension for every hour of the year from MATLAB simulation. The figure
is used to choosen the diameters there will be used in the second run. The real figure
maybe have a different form than show.
Second run
1. Run optimization for every hour with a restriction on heat transfer capacity equal
to ∆Xmax based on the chosen pipe dimension from the first run. Result is yearly
production cost (Z).
2. Based on the given ∆Xmax calculate associated heat loss cost and pump loss cost. The
calculation method for the loss can be found in the section 4.4. Find possible pipes in
pipe catalogue data sheet from the extracted ∆Xmax and constraints.
• Velocity ≤ 3, 5 m/s
• Max pressure loss ≤ 8 bar.
3. Add heat loss cost and pump loss to cost the production cost (Z).
4. Change ∆Xmax and run the optimization with new ∆Xmax and repeat step 2-3.
Continue until the lowest total cost is found, which is assumed to be the result right
before the cost begins rising. A possible result is shown in the table below. In reality
this could be a local minimum - a better choice would be to calculate the total cost for
∆Xmax=[0 MW; maximum possible heat demand in Ejby], but it would be very time
consuming, in comparison to this method.
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Chosen ∆Xmax
[Constraint in
MATLAB model]
The diameter of the
pipe between Assens
and Ejby for the given
∆Xmax [not included
in MATLAB model –
can be calculated based
on ∆Xmax independent
of production mix]
Sum of Z for the whole
year added heat loss
cost and pump loss
cost.
Ex. 10 MW Ex. 50 cm Ex. 66 · 106 2DKK
Ex. 20 MW Ex. 60 cm Ex. 45 · 106 DKK
Ex. 50 MW Ex. 70 cm
Ex. 41 · 106 DKK
[lowest]
Ex. 70 MW Ex. 80 cm Ex. 51 · 106 DKK
Table 4.5.
Hence, the optimal pipe dimension will be the one with the lowest Z. In this example 70
cm 4.5. Based on the ∆Xmax the optimal production mix changes. When the optimal pipe
dimension is found by this method the best production mix is also found.
4.7.3 - With storage (Excel model 24 hours)
This section describes how to include storage as a part of the optimization. It has not been
possible to implement storage in Matlab by using the linprog command. Instead an Excel-model,
see appendix 8.6 is created for 24-hour optimization simulations. Due to storage optimization
not being possible to implement in Matlab, storage optimization will be done seperatly in the
Excel model for further use in the What-If Analysis in section 6.2
Assens FJV provides all relevant data to the simulation. The data is listed in Appendix 8.4.
Assumptions
The assumptions for the model with optimization of storage is the same as the optimization
without optimizing the storage. Notice that the optimization of the storage is assumed to be
linear.
Deﬁnition of decision variables and constraints
Capacity between Assens and Ejby:
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SAssens,t
Sum of discharge or charge of storage (Sm,t) for all units in
Assens for a given hour. Positive if heat out of the storage
and negative if heat into the storage for the given hour.
Table 4.6.
Storage:
m
m represents a given heat storage unit. Notice properties
for the two storages in Assens are the same.
m=[1;3] for the tree storage units – two in Assens and one
in Ejby. The two storage units in Assens can be seen as
one unit and hence m=[1;2]. In the Excel-model the two
storage units in Assens is seen as one.
Scurrent,m
Capacity in storage m before the optimization start / start
current capacity in storage m [MW]
Scurrentcap,m
Capacity in storage when starting the optimization for a
given storage unit m [MWh]
Scapmax,m
Maximum storage capacity for a given storage unit m
[MWh]
Scapmin,m
Minimum storage capacity for a given storage unit m
[MWh]
Sflow,m
Maximum ﬂow out / into storage per hour for a given
storage unit m [MWh]
Table 4.7.
Decision variables
In order to include storage an extra decision variable is created. The storage decisions variable
is not a part of the object function because it does not have any cost associated.
Sm,t
Discharge or charge of storage for a given hour, for a given
storage unit. Positive if heat out of storage and negative if
heat into storage. [MWh]
Table 4.8.
Object function
The object function does not change, see section 4.7.2 for more detailed
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Constraints
Heat demand is not equal to heat
production when adding a storage unit.
Instead heat demand = heat delivered,
where heat delivered is the sum of the
heat production and the heat delivered
from the storage, which can be negative,
if some of the produced heat is going into
the storage.
HDAssens,t + HDEjby,t =
n
xn,t +
m=storage unit
Sm,t
Constraint for the maximum
allowed heat transport
between Assens and Ejby.
The storage variable
(SAssens,t) is positive when
the heat ﬂows out of the
storage and represents extra
production, hence the storage
variable have the same sign
as the heat production in
Assens (XAssens,t).
−HDAssens,t + SAssens,t + XAssens,t ≤ ∆Xmax
−HDAssens,t + SAssens,t + XAssens,t ≥ −∆Xmax
The maximum charge (and
discharge) of storage is for
every hour (t=[1;24]) less
than or equal the maximum
storage capacity for all m
units.
t=[1;24]
t=1
Sm,t + Scurrent,m ≤ Smax,cap,1
Example for 3 hours for storage unit 1:
There are assumed that the storage has Scurrrent,1 MW
when starting the model. If series of days or the whole
year is simulated the current storage capacity for the hour
24 will be set as Scurrent,m for the next simulation.
Hour one: S1,t + Scurrent,1 ≤ Scap,max,1
Hour two: S1,1 + Scurrent,1 + S1,2 ≤ Scap,max,1
S1,1 + Scurrent,1 + S1,2 + S1,3 ≤ Scap,max,1
S1,1 + Scurrent,1 = Capacity from hour 1
S1,2 = Capacity from hour 2
Table 4.9.
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The maximum (charge and) discharge of
storage is for every hour (t=[1;24]) larger than
or equal the minimum storage capacity for all
m units.
t=[1;24]
t=1
Sm,t +
Scurrent,m ≥ Smin,cap,1
The maximum discharge must be equal or less
than maximum possible discharge.
St,m ≤ Sflow,m
The maximum charge must be equal or less
than the maximum possible charge. Notice
that charging the storage result in a negative
value for Sm,t.
Sm,t ≥ −Sflow,m
Optimal pipe dimension (heat and electricity production)
Results are based on the following process: First run
• The Excel-model run with no restriction on the heat transport capacity between Assens
and Ejby, thereby set ∆Xmax → ∞. In practice ∆Xmax is just set to a very large
number, there is above the total possible heat demand in Assens and Ejby taken
together.
• Calculate heat transferred between Assens and Ejby for every hour. Since the heat
transferred can be negative if heat from Ejby to Assens the absolute value is taken to
the result.
(| − HDAssens,t + XAssens,t + SAssens,t|) With storage
(| − HDAssens,t + XAssens,t|) Without storage
• Extract the average value of the district heating transfer capacity, ∆Xavg. Be advised
that a worst case scenario test run has been performed to validate the value of Z does
not change.
• Find possible pipes in pipe catalogue data sheet from the extracted ∆Xavg and
constraints.
– Velocity ≤ 3, 5m/s
– Max pressure loss ≤ 8 bar.
• Calculate total expenses due to loss, LossHeat,Pump, (heat loss, pump loss) for possible
pipes and choose optimal pipe with lowest expenses due to loss.
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Results
5
At first the validation of the model will be illuminated, then the production and transmission
results will be presented and the assumptions will be analyzed, finally a socioeconomical
analysis will be provided, in terms of an NPV analysis, to determine if the project is viable.
5.1 Validation of model
In this section the model will be validated and its strengths and weakness’ will be illuminated.
Deviation
To validate the model the results are compared with the actual production by Assens FJV, as
per the information given to us by the company.
MWh Power production Heat production
Actual 26.686 88.472
Calculated 27.357 80.229
Difference -671 8.243
Table 5.1. Total heat and electricity production in 2015. The data is based on the Matlab model
described in section 4.7.2 with data from Appendix 8.4 ,[25]
The result does not correlate exactly with the real figures, but under the given circumstances
it is within the limitations of what could be expected - as the deviation is less than 3 percent
in electricity and 9 percent in heat production, of the actual production, the model is deemed
to be reasonable accurate. The reasons as to why the result deviates is a summation of
different factors, ranging from inaccuracies in our assumptions, to slight variation in the
effectiveness of the various components. These inaccuracies are deemed slight enough as to
have a negligible effect on the end result. In conclusion, based on the above, the model is
considered to have a serviceable degree of accuracy, and as such, useful to the extent of this
report.
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Loadplan
Figure 5.1. Assens heat demand, created based on degree days
This approach is very useful, because it can be used in correlation with weather forecasts to
predict the amount of heat needed a given day.
Storage
Storage is difficult to implement into a model without dictating a plan as to how it should run.
In this model the storage is primarily being used as a safety buffer. In times of great demand
it will provide heat to the system to prevent a second boiler from starting. This plan has been
incorporated because of its simultaneities to the actual way they are being used.
Figure 5.2. Assens storage current capacity as a
function of time
Figure 5.3. Assens storage current capacity as a
function of time - Zoom
As figure 5.1 show, the storage units are only being used in the winter months, which has the
highest demand. This fits with the current production plan.
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If the model should optimize further it would need a prognosis, for the heat demand and
electricity price. With these informations, the model could be equipped with a sliding window
to optimize over an entire day.
Analysis of constants
The downsides of having a fixed CM value are rather small, it is only when the heat
production is at the lowest the power production will be lower than in reality.
The downsides of neglecting the start-up cost can on the other hand be quite large, in this
project the repercussions are somewhat small, due to the fact that one boiler handles pretty
much all the load.
Another misfortune of neglecting the start-up cost is that the three wood pellet boilers all
start up at once and not one at a time. To get rid of this inconsistency, just summarize the
three boilers and if the vector at any point exceeds 6 MWh, a second boiler is needed.
With predetermined O&M the production plan is being ignored, which may tip the balance
so boilers with actual high O&M will be used in the model, when in reality another boiler has
lower O&M and will be cheaper to use.
All in all, the model takes most factors into account when calculating the optimal production
plan, but the largest margin of error is the storage units. Programming a sliding window into
the model would eliminate this and make the model much more reliable
5.2 Validation of optimization with storage
In this section the Excel model will be verified. This verification is based on very high variation
in a randomly chosen spot price in order to highlight the effects more strongly. The data used
in the validation is placed in Appendix 8.4.
Value of object function (Z) as function of total storage capacity
The graph below shows the change in the value of the object function as a function of the
total storage capacity (two storages in Assens and one storage in Ejby). Assuming a storage
capacity for both storage units in Assens on 10 MWh and 2 MWh for the storage in Ejby
equal to the minimum storage capacity. Afterwards the storage capacity is increased to the
maximum size equal to 95 MWh for the storage units in Assens unit and 55 MWh for the
storage unit in Ejby.
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Figure 5.4. Value of object function (Z) as function of total storage capacity
Figure 5.4 show that the production cost, including electricity income, is falling when the total
storage capacity is increasing, indicating that the model work as it should. On the other hand,
if there is any increase in production cost as the storage capacity is increasing, then there is an
error in the model - as the model is free to choose not to use the storage.
Storage balance in Assens and Ejby
The graph below shows the heat in the storage, marked with green dots for Ejby and blue
dots for Assens, as a function of hours for a random day. The two Storage units in Assens,
including the minimum and maximum storage capacity, is summed together, as they have the
same properties. “-“ marked minimum capacity and “x” marked maximum capacity, where blue
color is for Assens and green color is for Ejby. The spot price is marked as orange dot where
the spot price is divided by 10 to make the graph more easy to read.
Figure 5.5. Storage balance in Assens and Ejby
Figure 5.5 shows generally that when there is a high spot price, the stored amount of heat is
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increasing, as more electricity can be produced, thereby lowering the cost i.e. in hour 2 to 5
and hour 16 to 22. On the other hand, when the spot price is low, then the storage units are
discharging, in order to reduce the otherwise expensive heat production, thereby lowering the
cost i.e. in hour 11 to 13.
Value of object function as function of maximum heat transport capacity between
Assens and Ejby
The graph below shows the change in the value of the object function as a function of the
maximum heat transport capacity between Assens and Ejby (∆Xmax).
Figure 5.6. Value of object function as function of maximum heat transport capacity between Assens
and Ejby
It is not possible to move heat between Assens and Ejby when ∆Xmax = 0, whereby i.e.
natural gas in Ejby start up despite the possible free capacity of the woodchips boiler in
Assens. It is therefore expected that the total production cost is highest for ∆Xmax = 0
and the total production cost decreases as ∆Xmax = 0 increases. This is also the case for the
model as shown in ﬁgure 5.6.
5.3 Result presentation
In this section, the Matlab models results will be displayed.
5.3.1 Production
The production plan is some what trivial. The woodchip boiler is producing approximately
enough to supply both cities throughout a full year. Looking at table 5.2 it is clear to see just
how small the production of the other boilers are. The only other boiler who has production
are the wood pellets, but as mentioned in section 5.1 it is easy to see only one of the wood
pellet boilers are required.
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Unit woodchip Pellet 1 Pellet 2 Pellet 3
x1 x2 x3 x4
Production 96236 0,16 0,16 0,16
Unit Gas engine Gas boiler 1 Gas boiler 2 Heat cartridge
x6 x7 x8 x5
Production 0 0 0 0
Table 5.2. Total production [MWh], [25]
To obtain a more detailed view of the production plan, look at ﬁgure 5.7. From the graph it
can be concluded that the woodchip boiler is highly strained in the winter months, this tells
that expanding to more city’s than to Ejby could result in more expenses, because the increase
in demand will trigger a second boiler to start-up and there raise the total cost.
On the other hand, in 3/4 of the year the production is lower, which prevents Assens FJV
from producing electricity and thereby lowering their income. To further information see
chapter 6.1.
Figure 5.7. Power production as function of the year
5.3.2 Optimal pipe
The pipe chosen for this transmission line is "Isoplus DN 250". It is the pipe with the lowest
expenses caused by pumps and heat loss, meanwhile still satisﬁes the velocity and pressure
requirements set by the law see chapter. 3.3
5.3.3 Pressure and heat loss
According to the optimization, the optimal diameter of the transmission line is 0,273 m – the
Isoplus DN 250 – resulting in 0,477 MW of heat loss for calculation please see equation 4.27
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and a pressure drop for calculation please see equation 4.24 of 38,24 mWc, at a transfer rate of
3,4 MW – resulting a volume flow of 0,024 m3
s - over the full length of the supply and returns
pipes. The heat loss is equivalent to:
Cheat,loss = 0, 477MW · c1(= 179, 4
DKK
MW
) = 85, 57DKK (5.1)
Due to the fact that the woodchip boiler produces 99,9% of the heat production, it can be
assumed that the woodchip boilers fuel cost is the total cost 8.4. This result in a pressure
gradient in Pa · m−1 it is converted by equation 4.25 and divided by the length:
38, 24mWc · 9, 82m
s2 · 980 kg
m3
19.304m · 2
= 9, 53
Pa
m
(5.2)
To visualize the effect of this pressure gradients effect on the transmission line, the following
graph has been constructed.
Figure 5.8. Pressure head diagram - Isoplus DN 250
In which the green line represents the terrain heights in comparison to the “baseline”
established at the production plant. This means that for this pipes diameter, no pumps are
needed in order to maintain a positive flow, under the assumptions that the line is fed at a
pressure of 6,5 bar, and that the distribution line is self-sufficient in terms of pressure.
In order to calculate the pumps requirements, the following equation is used, for more
information see equation 4.26:
Wpump = 0, 024
m3
s
· 38, 24mWc · 980
kg
m3
· 9, 82
m
s2
= 8.832W (5.3)
This results in absolutely marginal pump cost, in comparison to the heat loss.
Cpump = Wpump · 170, 73
DKK
MWh
(5.4)
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• 170,73 DKK
MWh : Average spot price i DK1 [26]
The reason for this choice in diameter is due to the fact that the pressure drop, and the
consequent pump costs, rises dramatically the smaller the diameter is – so much so that
the pressure would increase to 61,37 mWc and 252 mWc at the next two smaller choices of
diameter, the Isoplus DN 200 and Isoplus DN 150, respectively.
If the Isoplus DN 150 was used instead, it would result in a need for pumps equal to:
Wpump = 0, 024
m3
s
· 252mWc · 980
kg
m3
· 9, 82
m
s2
= 58.203W (5.5)
For comparisons sake, here is the same diagram as before, merely with Isoplus DN 150 pipes
and its 128 pa/m pressure gradient instead:
Figure 5.9. Pressure head diagram - Isolus DN 150
Here, several 3,45 bar pumps have been introduced in order to maintain a positive ﬂow –
note that this is not an attempt to dimension the pumps, or even to introduce the pump
placements – this is merely a visualization of the need for extra pumps in order to maintain
a positive ﬂow, in comparison to the Isoplus DN 250.
The heat loss on the other hand would merely change from 0,477 MW at Isoplus DN 250 to
0,458 MW at Isoplus DN 150. What little that might be gained in terms of heat loss – not
even 20 kW – would in no way economically justify the thirteen fold increase in pressure drop
and pumps costs.
5.4 Net Present Value
The purpose of this chapter is to determine whether the project of establishing the transmission
line is socio economically sound
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Optimization result
AssensTotal,cost = Z + CHeat,loss + Cpump (5.6)
AssensEjbyTotal,cost = Z + CHeat,loss + Cpump (5.7)
Citys Total Cost [Year/DKK]
Assens -2.031.730
Assens + Ejby -1.692.874
Table 5.3. Result from Matlab simulation with fixed price (600 DKK
MW h ) for electricity production by
biomass
In order to estimate the validity of an investment, a Net Present Value (NPV) analysis is often
the used tool. This section will calculate the socioeconomically gain. If the NPV is positive
there is an increase in the socioeconomically for Ejby, whereby Assens FJV is allowed to
expand to Ejby according to the legislation described in section 3.3.2.
The present value can be calculated as:
If single payment:
PV = FV · (1 + i)−N
(5.8)
If several payments (based on annuity):
PV = A ·
(1 + i)N − 1
i · (1 + i)N − 1
(5.9)
In which:
• N is investment write-off period [Yr]
• FV the future value [DKK]
• PV is the present value [DKK]
• i is the interest [%]
The establishment cost for the pipeline is assumed to constitute 2000 DKK · m−1 by Assens
Fjernvarme. Thermis (see section 4.6) have calculated the pipe length to be 19.304 m long
thereby the total cost of the pipeline is considered to constitute 38.608.000 DKK. However,
this is an exuberant cost, not likely to be paid at once. Instead, this figure is to be borrowed
at an interest. Therefore, first order of business is to determine the net present value of this
loan. The loan is considered to be paid off in yearly increments (annuity), over 50 years
(typical life time of transmission pipe), at a 4% interest rate.
This analysis will not consider any potential compensation to be paid to any competitors, to
compensate for lost profit.
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The yearly payment before write-back for the pipe line is equal to:
A =
−38.608.000DKK
((1 + 4%)50 − 1)
4% · (1 + 4%)50 − 1
= −1.797.210DKK (5.10)
However, the payment changes according to the inflation - meaning the total present value of
the 50 yearly payments sum up to:
−
50
N=1
1.797.210 DKK · (1 + 2%)−N
= −56.474.823 DKK (5.11)
Where the discount rate (average inflation) is assumed 2 %. The net benefit of the investment
is whatever savings occur due to the cheaper district heating cost, plus the profit from the
electricity produced due to the increased production capacity accrued from the expansion, see
table 5.3. The total savings due to the district heating cost is equal to 1.882.000 DKK and
the electricity profit equals 405.740 DKK. Therefore, the benefit is equal to:
50
N=1
(2.352.000 DKK + 405.740 DKK) · (1 + 2%)−N
= 86.658.135 DKK (5.12)
Meaning the total net present value of the investment is equal to:
NPV = 86.658.135 DKK − 56.474.823 DKK = 30.183.312 DKK (5.13)
Income Expenses
Increase in Z when expanding to Ejby
(electricity income – heat production cost).
(∗) Result from Matlab model with 2015 data
used, see table 5.3
Additional pump cost. (∗) Based
on calculation described in
section 8.1
Additional heat sales calculated as: (Current
Ejby price – Current Assens Price) * Head
demand Ejby
Additional heat loss. (∗) Based
on calculation described in
section 4.4
Investment in pipeline between
Assens and Ejby (based on
annuity with 4% investment
interest)
Table 5.4. Income and expenses used in calculation of NPV.
(∗)
The transmission capacity is designed to minimize the total expenses.
The positive gain in the preformed NPV-calculation meaning that the investment result in a
gain in the socioeconomically for Ejby, whereby Assens FJV is allowed to expand to Ejby. This
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result is based on the current price structure at Assens FJV with ﬁxed electricity price equal to
600 DKK/MWh for the wood chips. This NPV-calculation is recalculated in section 6.1 with
the possible change in electricity price at the end of year 2018.
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What-if analysis
6
6.1 What-if analysis - 1 year (Matlab model)
Assens FJV have currently a fixed price on electricity produced on biomass, which will end
in 2018 as described in the section 3.3. When this is applied to the expansion to Ejby, it
results in a loss of 1,8 million DKK, when heat loss, pump costs, electricity income and heat
production cost is summarized calculated based on tabel 6.1.
This analysis will not consider any potential compensation to be paid to any competitors, to
compensate for lost profit.
Assuming that the heat price in Assens is constant, despite the change of the electricity price,
a new NPV-calculation is made. The yearly savings without the investment cost before write-
back is calculated as: [27]
Savings = Current price in Ejby − Current price in Assens − Increased production cost
including heat loss cost and pump cost due to the expansion to Ejby
The yearly savings are calculated to be around 527.803 DKK. When the investment cost
is added, this result in a loss of 39,9 million DKK based on the NPV-method described in
more detail in section 5.4. Thereby, there is no socioeconomic gain for Ejby by making the
expansion to Ejby from Assens. This result is in reality even worse than calculated because
the heat price in Assens will rise due to the lower electricity income, which the calculation
does not reflect.
If the change from fixed electricity price to spotprice plus 150 DKK/MWh is expected to
become a reality, then the investment should not be executed. On the other hand, there is
a socioeconomic gain when the electricity price is fixed as the current price. If a deal can be
reached, which is between the values of the two mentioned beforehand, then the analysis ought
to be reapplied in order to evaluate whether the project is viable.
Citys Total Cost [Year/DKK]
Assens 5.354.660
Assens + Ejby 7.178.857
Table 6.1. Result from Matlab simulation with spot price plus 150 DKK/MWh for electricity
production by biomass
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