EXAMENSARBETE INOM MASKINTEKNIK, Maskinteknik, högskoleingenjör 15 hp SÖDERTÄLJE, SVERIGE 2019
Energy flow model for a
scrapper conveyor unit.
André Meza Kahn
SKOLAN FÖR INDUSTRIELL TEKNIK OCH MANAGEMENT
1
Energy flow model for a
scrapper conveyor unit.
André Meza Kahn
Examensarbete TRITA-ITM-EX 2019:599 KTH Industriell teknik och management
Tillämpad maskinteknik
2 Examensarbete
TRITA-ITM-EX 2019:599 Energiflödesmodell till en
skraptransportörsmodul. André Meza Kahn
Godkänt 2019 September 9 Examinator KTH Mark W. Lange Handledare KTH Jafar Mahmondi Uppdragsgivare
Värmevärden AB Företagskontakt/handledare Roger Silver
Sammanfattning
Värmevärden AB är ett energibolag som distribuerar fjärrvärme. En av deras utmaningar idag är att minska energiförlusterna och slitaget som uppstår i
anläggningens skrapstransportörs modul. Transportbandet är i konstant rörelse vare sig det transporterar bränsle eller går olastad i tomgång. Förståelse för modulens energiflöde behöver kartläggas
En beteendemodell för skraptransportörens modul presenteras. Genom analys av energiflödet i systemet simuleras startförloppet när bandet körs i tomgång och vid full last. Simuleringarna visualiserar bandets hastighet och acceleration som funktion av tid och även motorns strömförlopp under uppstartsfasen. Modellen tillåter ändringar på diverse parametrar såsom motorns varvtal, strömbegränsning, verkningsgrad och möjliggör även framräkningen av nödvändigt material för att uppnå en viss leverans uteffekt.
Med en strömbegränsad motor på 140 [A] fås en lägre acceleration när bandet transporterar bränslemassorna. Energiförlusterna minskar också eftersom bandet sätts igång av massornas egenvikt när transporteringen behövs. Detta medför att slitaget på anläggningens komponenter minskar i och med att transporten sker i mera kontrollerade former.
Nyckelord
3 Bachelor of Science Thesis
TRITA-ITM-EX 2019:599 Energy flow model for a scrapper conveyor unit.
André Meza Kahn
Approved
2019 September 9 Examiner KTH Mark W. Lange Supervisor KTH Jafar Mahmondi
Commissioner
Värmevärden AB
Contact person at company
Roger Silver
Abstract
Värmevärden AB is an energy company that distributes district heating. One of their challenges today is to reduce the energy losses and wear that occur in the plant's scrapper conveyor unit. The conveyor belt is in constant motion whether it is
transporting fuel or going unloaded at idle. An understanding of the module's energy flow needs to be mapped.
An energy flow behavior model for the scrapper conveyor unit is presented. The energy flow in the startup sequence when the belt runs at idle and at full load is simulated. The simulations visualize the velocity and acceleration of the belt as a function of time and the current flow in the engine when getting started. The model allows changes on various parameters such as engine velocity, current limitations, work efficiency and enables the calculation of necessary material to achieve a desired power delivery.
With a current-limited engine of 140 [A], a lower acceleration is obtained as the belt transports the fuel masses. Energy losses are also reduced as the belt is started by their own mass when transportation is needed. This means that the wear on the components of the plant decreases as transportation takes a more controlled form. Keywords
Scrapper conveyor unit, behavior model, simulations, energy flow, empty load, full load.
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Preface
By introducing this thesis, I would like to give my deepest appreciations to my tutor Jafar Mahmondi at KTH Södertälje for his guidance expertise and patience. I would also like to thank the company tutor, Roger Silver at Värmevärden AB for giving me the opportunity to do my thesis at the power plant. As well the staff at Värmevärden AB in Nynäshamn for making me feel welcomed and a part of their facility.
Lastly, I would thank my family members for supporting me during my time at the university. This couldn’t be possible without you.
5 Table 1 : Variables used in Equations
Designation Denotation Unit Data*
𝑊𝑊𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Stored mechanical energy. J [Nm]
𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠 Effect input. W
𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠 Effect output. W
𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Effect losses. W
𝑇𝑇 Rotor torque. Nm
𝑀𝑀1 Worm gear torque. Nm
𝜔𝜔1 Angular velocity (rotor to worm gear). 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠
𝐽𝐽𝑅𝑅 Rotor inertia. kgm2
𝑀𝑀2 Worm gear torque (outgoing). Nm
𝜔𝜔2 Angular velocity (worm gear to transmission chain). 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠
𝐽𝐽𝑊𝑊𝑊𝑊1 Worm screw inertia. kgm2
𝐽𝐽𝑊𝑊𝑊𝑊2 Worm wheel inertia. kgm2
𝜂𝜂𝑊𝑊𝑊𝑊 Worm gear work efficiency. Dimensionless 0,85*
𝑖𝑖𝑊𝑊𝑊𝑊 Angular velocity ratio for worm gear. Dimensionless 35*
𝑀𝑀3 Transmission chain torque. Nm
𝜔𝜔3
Angular velocity (transmission chain to
transportation line). 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠 𝐽𝐽𝑇𝑇𝑇𝑇1 Transmission chain inertia (small gear). kgm2
𝐽𝐽𝑇𝑇𝑇𝑇2 Transmission chain inertia (big gear). kgm2
𝑚𝑚𝑇𝑇𝑇𝑇 Mass transmission chain. kg
𝜂𝜂𝑇𝑇𝑇𝑇 Transmission chain work efficiency. Dimensionless 0,98*
𝑣𝑣𝑇𝑇𝑇𝑇 Transmission chain velocity. 𝑚𝑚/𝑠𝑠
𝑟𝑟𝑇𝑇𝑇𝑇1 Ingoing transmission chain’s gear radius. m
𝑟𝑟𝑇𝑇𝑇𝑇2 Outgoing transmission chain’s gear radius. m
𝑖𝑖𝑇𝑇𝑇𝑇 Angular velocity ratio for transmission chain. Dimensionless 3*
𝐽𝐽𝑇𝑇𝑇𝑇 Transportation line drive wheel inertia. kgm2
𝑚𝑚𝐹𝐹𝐹𝐹 Mass for fuel scrappers and transportation band. kg 3500*
𝑚𝑚𝑅𝑅𝑇𝑇 Total fuel mass carried by all fuel scrappers. kg
𝑣𝑣𝑇𝑇𝑇𝑇 Transportation line velocity. 𝑚𝑚/𝑠𝑠
𝑟𝑟𝑇𝑇𝑇𝑇 Gear radius for transportation line. m 0,38*
6 𝑣𝑣𝑇𝑇𝑇𝑇𝑇𝑇 Y-axis velocity component. 𝑚𝑚/𝑠𝑠
𝛼𝛼 Pitch angle. Dimensionless 30*
𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇1 Horizontal friction force in the transportation line. N
𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇2 Elevated friction force in the transportation line. N
𝑔𝑔 Gravitational force. 𝑁𝑁/𝑘𝑘𝑔𝑔 9,81
𝜇𝜇 Friction coefficient for fuel scrapper and the base of
transportation line. Dimensionless 0,12*
𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙 Transportation line’s total friction force. N
𝑚𝑚𝐹𝐹𝐹𝐹−𝑖𝑖𝑖𝑖𝑠𝑠𝑠𝑠𝑖𝑖 Index for fuel scrapper masses in various positions. kg 15*
𝑚𝑚𝑅𝑅𝑇𝑇−𝑖𝑖𝑖𝑖𝑠𝑠𝑠𝑠𝑖𝑖 Index for fuel masses in various positions. kg 20*
𝐼𝐼𝑚𝑚𝑡𝑡𝑖𝑖 Maximum limited current for the whole system. A 140*
𝐼𝐼 Ingoing current for the whole system without a
current limitation. A 330*
𝐶𝐶1 Variable for determine the limited current supply. Dimensionless 300*
𝐶𝐶2 Variable for determine the
limited current supply. Dimensionless 2*
𝐶𝐶3 Variable for determine the
limited current supply. Dimensionless 0,04* 𝑛𝑛𝑠𝑠 Synchronized number of
revolutions. rpm 1500*
𝑛𝑛𝑚𝑚 Asynchrony number of
revolutions. rpm
𝜌𝜌 Rotor material density
(steel). 𝑘𝑘𝑔𝑔/𝑚𝑚3 7870
𝑃𝑃𝑅𝑅𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 Engine rated power. kW
𝐹𝐹𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙 Fuel force. N
𝑃𝑃𝐵𝐵𝑖𝑖 Outgoing boiler effect. MW i=12, 18, 24
𝑚𝑚𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙 Fuel mass needed to reach 𝑃𝑃
𝐵𝐵𝑖𝑖. kg
𝑟𝑟𝑅𝑅 Rotor radius. m 0,20*
ℎ𝑅𝑅 Rotor height. m 0,04*
𝑉𝑉𝑅𝑅 Rotor volume. 𝑚𝑚3
ℎ𝑇𝑇𝑇𝑇 Height of the transmission lines drive wheel. m 0,05*
𝑉𝑉𝑇𝑇𝑇𝑇 Transmission line drive wheel volume. 𝑚𝑚3
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Table of Contents
1 Introduction ... 8 1.1 Background ... 8 1.2 Purpose ... 9 1.3 Boundaries ... 9 1.4 Methodology ... 10 2 Pre-study ... 11 2.1 District heating ... 11 2.2 Recycled woodchips ... 122.3 Scrapper conveyor unit... 13
3. Scrapper conveyor unit – Energy losses ... 14
3.1 Energy balance in a limited system ... 15
3.2 Energy balance for an asynchronous engine ... 15
3.3 Energy balance for the gearbox ... 16
3.4 Energy balance for the transmission chain ... 18
3.5 Energy balance for the transportation line ... 20
3.6 Energy balance for the scrapper conveyor unit ... 22
4. Engine for the transportation line ... 23
4.1 Choice of electrical engine ... 23
4.2 Electrical supply ... 24
5. Inertia calculation ... 25
5.1 Determination of inertia for the rotor ... 25
5.2 Determination of inertia for the drive wheel ... 26
6. Modelling procedure ... 27
7. Results ... 29
7.1 System simulation with current limited engine and fully loaded transportation line... 29
7.2 System simulation with current limited engine and unloaded transportation line ... 30
7.3 System simulation with current unlimited engine and fully loaded transportation line ... 31
7.4 System simulation with current unlimited engine and unloaded transportation line ... 32
7.5 Mass calculation for the desired boiler effect ... 33
7.6 Result summary ... 34 8. Discussion ... 35 9. Conclusion ... 38 9.1 Recommendations ... 38 10. References ... 39 Appendix ... 40
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1 Introduction
1.1 Background
Värmevärden AB is an energy company that delivers district heating to homes, industries and other premises(Värmevärden AB, 2019). One of their energy production plants at Nynäshamn operates with recycled woodchips as its fuel source and can deliver a power energy up to 24 MW.
Today the plant is facing energy loss and wear issues in the used equipment since some of the modules in the process are not synchronized and are still unable to regulate its operational reliability to the required production pace.
Figure 1 shows a schematic picture of the modules and the operation flow. The fuel flow (---) along the process is also shown.
9 A fueling sequence starts when a fuel cartridge is filled with recycled woodchips by an overhead crane. The fuel is driven through a controlled doses screw that regulates the fuel intake. The amount of fuel passes through a magnet chamber where metal residuals are removed. The processed fuel is later tossed in various quantities to the scrapper conveyor unit and transported to a fuel boiler for further energy transformation. The plant can deliver a power window between 12-24 MW.
One of the major issues in the plant is the energy loss and wear that appears in the scrapper conveyor unit when its feed with an unlimited current flow. The engine, gearbox, the transmission chain and transportation line connected to the scrapper conveyor unit runs with constant velocity and are not synchronized with the production pace. This leads to energy loss since the scrapper conveyor unit is working continuously.
1.2 Purpose
1. The purpose is to develop a behavior model of the energy flow through the scrapper conveyor unit and simulate the startup sequence.
2. Calculate the amount of required fuel according to the expected effect window.
1.3 Boundaries
• The thesis applies an investigation of the energy flow for the scrapper conveyor unit only.
• Temperature calculations of the scrapper conveyor unit will not be calculated either taken into consideration in this project.
• Offers for eventually recommended parts will not be taken into consideration in this project.
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1.4 Methodology
This chapter clarifies and shows the methodical approach to the energy-related study, see Figure 2.
Figure 2: A process flow chart.
1. Identify and understand the deviations in the plant.
2. The purpose is to develop a behavior model of the energy flow through the scrapper conveyor unit and simulate the startup sequence. And to calculate the amount of required fuel according to the expected effect window.
3. Application of energy conservation laws for a limited system.
4. Evaluation of the energy flow in the system. Used parameters are available in Table 1. The study puts focus on velocity, acceleration and current flow as a
function of time.
5. The required total fuel mass is calculated by using the velocity parameter held from the Matlab simulations.
6. Proposals to optimize the system of the power plant.
7. A behavior model of the energy flow through the scrapper conveyor unit is available.
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2 Pre-study
The following chapter provides information on the plant and its operating modules.
2.1 District heating
District heating is a method for distributing energy through an underground pipe system to households and other premises.
Fundamentally the heat transportation comes from a power plant, instead of a local pump or other heating equipment. In practice, it’s an essential solution for having the aid of a plant warming a premise for a reduced electrical expense and its low environmental emissions. But not everyone can afford or have the privilege to have district heating due for not having a plant near the desired living area. Mainly for reducing the length of the underground pipe system which risks for further energy losses if the distance is too great. Advantages and disadvantages with a district heating system for a consumer. (Energiradgivningen, 2017)
+Utilizes little to no space when used. -Costly to link up. +Often lower prices than oil/electrical
heat.
-Not available in your region. +Minimal wear and maintenance. -No change of supplier.
Below is a brief explanation of how the procedure of the district heating system works. The energy distribution starts when an energy plant processes its fuel option (fossil fuels, biofuel, etc.) and transports hot water through an underground insulated pipe system to a household’s heat receiver. The heat energy from the heat receiver is provided by underfloor heating or radiators and can be manually regulated to one’s temperature requirement (Energimyndigheterna, 2015). Lastly, the excess cold water from the premises is transported back through the insulated pipe system to the energy plant so it can be reheated. And another loop of hot water can be distributed.
See Figure 3 for the water distribution.
Figure 3: Red arrows indicate the hot water delivered to the premises and the blue arrows is the cold water
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2.2 Recycled woodchips
Recycled woodchips is the fuel source that the plant in Nynäshamn uses. It originates from recycled wood items, such as pallets and tree waste. The unused wood is transported to a recycling station for cleansing and later being broken-down, producing woodchips (see Figure 4) with a specific width.
Its high attributes for being a reliable fuel source come with the low hazardous emissions as well that it doesn’t consume the earth’s resources as any other type of fuel.
The disadvantages are when contaminated wood (i.e. pressure impregnated wood), are mixed with the processed woodchips and is used for burning.
It releases toxic waste to the atmosphere and can destroy valuable equipment in the plant, as well as having a decreased burning efficiency.
Therefore, the wood needs to be well recycled and processed for a maximum burn efficiency (Waste360, 1996).
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2.3 Scrapper conveyor unit
A scrapper conveyor unit’s main task is to transport the fuel to an elevator unit. It consists of an asynchronous engine, a worm gear (gearbox), a transmission chain and a transportation line. These four modules will be studied further in order to obtain the behavior model for the energy flow in the system. See Figure 5 for an illustration of the mechanical assembly.
Figure 5: The scrapper conveyor unit with its modules. (1)Asynchronous engine, (2)Worm gear,
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3. Scrapper conveyor unit – Energy losses
The plant delivers district heating in a power window between 12-24 MW. An undefined amount of mass is feed into the system and energy losses are unavoidable (See Figure 6). This will be presented in the next coming chapters.
Figure 6: Schematic picture of the energy flow with losses in the scrapper conveyor unit.
12-24 MW
𝐏𝐏
𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐏𝐏
𝐨𝐨𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐏𝐏𝐥𝐥𝐨𝐨𝐥𝐥𝐥𝐥𝐥𝐥
15
3.1 Energy balance in a limited system
Energy conservation laws connected to the energy balance in a limited system are used to calculate the energy flow for the concerned modules in the scrapper conveyor unit. The mechanical energy flow is according to equation [1] (Sören Andersson & Claes Tisell & Anders Folkeson, 2002).
𝑟𝑟𝑊𝑊𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑟𝑟𝑑𝑑 = 𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠− 𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠+ 𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 [1] 𝑊𝑊𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 The stored mechanical energy (i.e. potential and kinetic energy)
𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠 Input energy flow to the system (energy that drives the system)
𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠 Output energy flow from the system (energy that drives the connected system)
𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 The energy per time unit, converted inside the system (basis or descend)
To facilitate the energy balance, the system is divided into four subsystems. The subsystems are the engine, gearbox, transmission chain and transportation line. Equation [1] is used to construct the models and will later be added to a singular expression that illustrates the behavior for the whole system.
3.2 Energy balance for an asynchronous engine
As current enters the engine an electric magnetic field is created. This generates torque T for the rotor and an angular velocity 𝜔𝜔1. Torque T generates torque 𝑀𝑀1 which enters the
gearbox with the same angular velocity 𝜔𝜔1 (Sören Andersson & Claes Tisell & Anders
Folkeson, 2002). See Figure 7 for an illustration. The assumed energy losses emitted in the form of heat are relatively small and therefore are neglected in this model, although the mechanical energy stored in the engine relies on the rotor’s inertia 𝐽𝐽𝑅𝑅.
Figure 7: A free body diagram of the engine. With the generated rotor torque T and outgoing effect 𝑀𝑀1 with
the angular velocity 𝜔𝜔1.
16 According to equation [1], the engines energy balance leads to:
𝑟𝑟 𝑟𝑟𝑑𝑑 �
𝐽𝐽𝑅𝑅 ∗ 𝜔𝜔12
2 � = 𝑇𝑇 ∗ 𝜔𝜔1− 𝑀𝑀1 ∗ 𝜔𝜔1− 0 [2] With derivation on the right side of equation [2] and breakout of the independent variable 𝜔𝜔1, equation [3] is the concluded energy balance for the asynchronous engine.
�𝐽𝐽𝑅𝑅 ∗𝑟𝑟𝜔𝜔𝑟𝑟𝑑𝑑 − 𝑇𝑇 + 𝑀𝑀1 1� 𝜔𝜔1 = 0 [3]
3.3 Energy balance for the gearbox
The gearbox is illustrated in Figure 8. It is driven by torque 𝑀𝑀1 with the angular velocity
𝜔𝜔1 produced from the engine. An outgoing torque 𝑀𝑀2 with the angular velocity 𝜔𝜔2 is
created in return. The stored energy in the gearbox is produced by the rotation energy of the worm screw and worm gear. The inertia for the worm screw and worm gear is designated as 𝐽𝐽𝑊𝑊𝑊𝑊1 respectively 𝐽𝐽𝑊𝑊𝑊𝑊2.
Figure 8: A free body diagram of the gearbox. It illustrates the torque 𝑀𝑀1from the engine and the outgoing
torque 𝑀𝑀2 which leads the mechanical energy to the third subsystem, the transmission chain.
Energy losses that occur in continuous operation are referred to its work efficiency 𝜂𝜂𝑊𝑊𝑊𝑊.
Shown in equation [4] (Sören Andersson & Jan Hölcke, 2001).
𝜂𝜂𝑊𝑊𝑊𝑊 =𝑃𝑃𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠 [4]
For a gearbox running in constant operation the 𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 related to equation [1] leads to
equation [5]. 0 = 𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠− 𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠 − 𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ⟹ 𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠− 𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠 [5]
Gearbox
Worm gear, 𝐽𝐽𝑊𝑊𝑊𝑊1 Worm screw, 𝐽𝐽𝑊𝑊𝑊𝑊2 𝑖𝑖𝑊𝑊𝑊𝑊, 𝜂𝜂𝑊𝑊𝑊𝑊17 With an index insertion for 𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠 from equation [4] to equation [5], gives the complete
effect losses to the gearbox. Visualized in equation [6] 𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠= (1 − 𝜂𝜂𝑊𝑊𝑊𝑊)𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠
[6] The energy balance to the gearbox is set up by equation [1] and [6].
𝑟𝑟 𝑟𝑟𝑑𝑑 � 𝐽𝐽𝑊𝑊𝑊𝑊1∗ 𝜔𝜔12 2 + 𝐽𝐽𝑊𝑊𝑊𝑊2∗ 𝜔𝜔22 2 � = 𝑀𝑀1∗ 𝜔𝜔1− 𝑀𝑀2∗ 𝜔𝜔2− (𝑀𝑀1∗ 𝜔𝜔1− 𝑀𝑀1∗ 𝜔𝜔1∗ 𝜂𝜂𝑊𝑊𝑊𝑊) [7]
Equation [7] is simplified and leads to the concluded energy balance for the gearbox, equation [8]. 𝑟𝑟 𝑟𝑟𝑑𝑑 � 𝐽𝐽𝑊𝑊𝑊𝑊1∗ 𝜔𝜔12 2 + 𝐽𝐽𝑊𝑊𝑊𝑊2∗ 𝜔𝜔22 2 � = 𝑀𝑀1∗ 𝜔𝜔1∗ 𝜂𝜂𝑊𝑊𝑊𝑊− 𝑀𝑀2∗ 𝜔𝜔2 [8]
The gearbox two angular velocities 𝜔𝜔1 and 𝜔𝜔2 are related to the total angular velocity
ratio, 𝑖𝑖𝑊𝑊𝑊𝑊 to the worm gear, according to equation [9] (Sören Andersson & Jan Hölcke,
2001).
𝑖𝑖𝑊𝑊𝑊𝑊 = 𝜔𝜔𝜔𝜔1
2 ⟹ 𝜔𝜔2 =
𝜔𝜔1
𝑖𝑖𝑊𝑊𝑊𝑊 [9]
By deriving equation [8] and indexing 𝜔𝜔2 as the expression 𝑖𝑖𝜔𝜔𝑊𝑊𝑊𝑊1 from equation [9], obtains
18
3.4 Energy balance for the transmission chain
The outgoing torque 𝑀𝑀2 from the gearbox drives a small gear in the transmission chain
with the angular velocity 𝜔𝜔2. Meanwhile, the transmission chain bigger gear has the
outgoing torque 𝑀𝑀3 and rotates with the angular velocity 𝜔𝜔3. The stored energy in this
subsystem is created by the two gears connected to the chain, where the gears inertia are designated as 𝐽𝐽𝑇𝑇𝑇𝑇1 respectively 𝐽𝐽𝑇𝑇𝑇𝑇2 and the chains kinetic energy with its mass 𝑚𝑚𝑇𝑇𝑇𝑇. See
Figure 9.
Figure 9: The free body diagram of the transmission chain. Where 𝑀𝑀2 is the combining ingoing torque from
the engine and gearbox. And the outgoing torque 𝑀𝑀3 which drives the last subsystem, the transportation
line.
Equation [10] shows the energy balance of the transmission chain, according equation [1]. 𝑟𝑟 𝑟𝑟𝑑𝑑 � 𝑚𝑚𝑇𝑇𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇2 2 + 𝐽𝐽𝑇𝑇𝑇𝑇1∗ 𝜔𝜔22 2 + 𝐽𝐽𝑇𝑇𝑇𝑇2∗ 𝜔𝜔32 2 � = 𝑀𝑀2∗ 𝜔𝜔2− 𝑀𝑀3∗ 𝜔𝜔3+ 𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 [10]
The transmission chain mechanical energy is defined when it is on stationary positioning with its work efficiency, leading to equation [11].
𝜂𝜂𝑇𝑇𝑇𝑇 = 𝑃𝑃𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠 ⟹ 𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠 = 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠
[11]
By indexing 𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠 from equation [11] to equation [1] leads to an expression for 𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
shown in equation [12]:
0 = 𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠− 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠+ 𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ⟹ 𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = (𝜂𝜂𝑇𝑇𝑇𝑇 − 1) ∗ 𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠 [12]
When inserting equation [10] and [12] to equation [1], leads to equation [13]:
𝑟𝑟 𝑟𝑟𝑑𝑑 � 𝑚𝑚𝑇𝑇𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇2 2 + 𝐽𝐽𝑇𝑇𝑇𝑇1∗ 𝜔𝜔22 2 + 𝐽𝐽𝑇𝑇𝑇𝑇2∗ 𝜔𝜔32 2 � = 𝑀𝑀2∗ 𝜔𝜔2− 𝑀𝑀3∗ 𝜔𝜔3+ (𝜂𝜂𝑇𝑇𝑇𝑇− 1) ∗ 𝑃𝑃𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑠𝑠 [13] 𝐽𝐽𝑇𝑇𝑇𝑇1
𝐽𝐽
𝑇𝑇𝑇𝑇219 Simplification of equation [13] leads to equation [14]:
𝑟𝑟 𝑟𝑟𝑑𝑑� 𝑚𝑚𝑇𝑇𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇2 2 + 𝐽𝐽𝑇𝑇𝑇𝑇1∗ 𝜔𝜔22 2 + 𝐽𝐽𝑇𝑇𝑇𝑇2∗ 𝜔𝜔32 2 � = 𝑀𝑀2∗ 𝜔𝜔2− 𝑀𝑀3∗ 𝜔𝜔3+ (𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑀𝑀2∗ 𝜔𝜔2− 𝑀𝑀2∗ 𝜔𝜔2) [14]
Further simplification of equation [14] leads to: 𝑟𝑟 𝑟𝑟𝑑𝑑 � 𝑚𝑚𝑇𝑇𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇2 2 + 𝐽𝐽𝑇𝑇𝑇𝑇1∗ 𝜔𝜔22 2 + 𝐽𝐽𝑇𝑇𝑇𝑇2∗ 𝜔𝜔32 2 � = 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑀𝑀2 ∗ 𝜔𝜔2− 𝑀𝑀3∗ 𝜔𝜔3 [15]
Below shows the relation between the transmission chain´s velocity and both angular
velocities where 𝑟𝑟𝑇𝑇𝑇𝑇1 and 𝑟𝑟𝑇𝑇𝑇𝑇2 are the ingoing and outgoing radius for the gears.
(R.C Hibbeler & Kai Beng Yap, 2013).
𝑣𝑣𝑇𝑇𝑇𝑇 = 𝜔𝜔2∗ 𝑟𝑟𝑇𝑇𝑇𝑇1 = 𝜔𝜔3∗ 𝑟𝑟𝑇𝑇𝑇𝑇2 [16]
By rewriting equation [15] with the combination of [16] leads to equation [17]. To build an expression for the energy balance to the transmission chain:
𝑟𝑟 𝑟𝑟𝑑𝑑 � 𝑚𝑚𝑇𝑇𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇2 2 + 𝐽𝐽𝑇𝑇𝑇𝑇1∗ 𝑣𝑣𝑇𝑇𝑇𝑇2 2 ∗ 𝑟𝑟𝑇𝑇𝑇𝑇12 + 𝐽𝐽𝑇𝑇𝑇𝑇2∗ 𝑣𝑣𝑇𝑇𝑇𝑇2 2 ∗ 𝑟𝑟𝑇𝑇𝑇𝑇22 � = 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑀𝑀2∗ 𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇1− 𝑀𝑀3∗ 𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇2 [17]
With derivation on the left side of equation [17] and breakout of the variable 𝑣𝑣𝑇𝑇𝑇𝑇 gives the
concluded energy balance for the transmission chain. Shown in equation [18]. �𝑚𝑚𝑇𝑇𝑇𝑇∗𝑟𝑟𝑣𝑣𝑟𝑟𝑑𝑑 +𝑇𝑇𝑇𝑇 𝑟𝑟𝐽𝐽𝑇𝑇𝑇𝑇1 𝑇𝑇𝑇𝑇12∗ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 + 𝐽𝐽𝑇𝑇𝑇𝑇2 𝑟𝑟𝑇𝑇𝑇𝑇22∗ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 + 𝑀𝑀3 𝑟𝑟𝑇𝑇𝑇𝑇2− 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑀𝑀2 𝑟𝑟𝑇𝑇𝑇𝑇1 � 𝑣𝑣𝑇𝑇𝑇𝑇 = 0 [18]
The transmission chain two angular velocities are related to each other with the angular velocity ratio, 𝑖𝑖𝑇𝑇𝑇𝑇.
𝑖𝑖𝑇𝑇𝑇𝑇 =𝜔𝜔𝜔𝜔2 3
20
3.5 Energy balance for the transportation line
The stored energy for this subsystem according to equation [29] is determined by the inertia of the drive wheels 𝐽𝐽𝑇𝑇𝑇𝑇. As well for the fuel scrappers mass 𝑚𝑚𝐹𝐹𝐹𝐹 and the material
mass 𝑚𝑚𝑅𝑅𝑇𝑇 that travels upwards with the velocity 𝑣𝑣𝑇𝑇𝑇𝑇. See Figure 10.
Figure 10: A free body diagram of the transportation line. Where the transportation line is driven by the
outgoing transmission chain torque. With 𝑚𝑚𝐹𝐹𝐹𝐹−𝑖𝑖𝑖𝑖𝑠𝑠𝑠𝑠𝑖𝑖 as the fuel scrappers mass and 𝑚𝑚𝑅𝑅𝑇𝑇−𝑖𝑖𝑖𝑖𝑠𝑠𝑠𝑠𝑖𝑖 as the fuel
mass in various positions.
The torque 𝑀𝑀3 and angular velocity 𝜔𝜔3 from the transmission chain drive the upper drive
wheel of the transportation line by its radius 𝑟𝑟𝑇𝑇𝑇𝑇. Equation [20] calculates the velocity for
the transportation band (R.C Hibbeler & Kai Beng Yap, 2013). 𝑣𝑣𝑇𝑇𝑇𝑇 = 𝜔𝜔3∗ 𝑟𝑟𝑇𝑇𝑇𝑇 ⟹ 𝜔𝜔3 = 𝑣𝑣𝑟𝑟𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇
[20]
The outgoing energy flow is the fuel scrapper force 𝐹𝐹𝑅𝑅𝑇𝑇 multiplied with the velocity
component in vertical direction 𝑣𝑣𝑇𝑇𝑇𝑇𝑇𝑇 according to equation [21] (Sören Andersson & Jan
Hölcke, 2001).
𝑃𝑃𝑠𝑠𝑖𝑖𝑠𝑠𝑖𝑖𝑖𝑖𝑠𝑠 =∗ 𝑣𝑣𝑇𝑇𝑇𝑇𝑇𝑇 [21]
As 𝑣𝑣𝑇𝑇𝑇𝑇𝑇𝑇 is determined by multiplying the transmission line’s velocity with the pitch angle
𝛼𝛼, according to equation [22] (J.L. Meriam & L.G. Kraige, 1998).
21 𝑃𝑃𝑙𝑙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 to the subsystem is determined by the total friction force 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙, in equation [23]
(J.L. Meriam & L.G. Kraige, 1998).
𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙 = 𝑚𝑚 ∗ 𝑔𝑔 ∗ 𝜇𝜇 [23] The total frictional force is calculated by adding the friction force occurring along the
horizontally in/out passages of the transportation line 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇1; and by determine the
horizontal friction component 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇2 at the leaning part of the band where the pitch angle
𝛼𝛼 is used. Shown in equation [24] (J.L. Meriam & L.G. Kraige, 1998).
𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙 = 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇1+ 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇2 [24]
The moving masses in the subsystem are the fuel scrapper mass 𝑚𝑚𝐹𝐹𝐹𝐹−𝑖𝑖𝑖𝑖𝑠𝑠𝑠𝑠𝑖𝑖 and the fuel
mass 𝑚𝑚𝑅𝑅𝑇𝑇−𝑖𝑖𝑖𝑖𝑠𝑠𝑠𝑠𝑖𝑖. The indexing factor shows the position of masses. No friction is assumed
at the turning points of the band, point 3 and 7, observe Figure 10.
There are 70 fuel scrappers in total. The study estimates that 30 of them are working horizontally in positions 1 and 6. This gives a factor of 15 to be multiplied with the mass affecting the friction for a fuel scrapper. 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇1 is calculated in equation [25].
𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇1 = �(𝑚𝑚𝐹𝐹𝐹𝐹1+ 𝑚𝑚𝑅𝑅𝑇𝑇1) ∗ 15 + (𝑚𝑚𝐹𝐹𝐹𝐹6∗ 15)� ∗ 𝑔𝑔 ∗ 𝜇𝜇
The masses involved along the leaned part of the transportation band are related to the amount of working scrappers according the calculated factor in equation [26].
𝐴𝐴𝑚𝑚𝐴𝐴𝐴𝐴𝑛𝑛𝑑𝑑 𝐴𝐴𝑜𝑜 𝑠𝑠𝑠𝑠𝑟𝑟𝑟𝑟𝑠𝑠𝑠𝑠𝑠𝑠𝑟𝑟𝑠𝑠 =𝐻𝐻𝑠𝑠𝑖𝑖𝑔𝑔ℎ𝑑𝑑 𝐴𝐴𝑜𝑜 𝑟𝑟 𝑜𝑜𝐴𝐴𝑠𝑠𝑙𝑙 𝑠𝑠𝑠𝑠𝑟𝑟𝑟𝑟𝑠𝑠𝑠𝑠𝑠𝑠𝑟𝑟 =𝐻𝐻𝑠𝑠𝑖𝑖𝑔𝑔ℎ𝑑𝑑 𝐴𝐴𝑜𝑜 𝑑𝑑ℎ𝑠𝑠 𝑙𝑙𝑠𝑠𝑟𝑟𝑛𝑛𝑠𝑠𝑟𝑟 𝑠𝑠𝑟𝑟𝑟𝑟𝑑𝑑 5,5 𝑚𝑚 0,2 𝑚𝑚 = 27,5
The horizontal friction force component 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇2 on the leaned part of the
transportation band is calculated in equation [27].
[25]
[26]
𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇2 = �(𝑚𝑚𝐹𝐹𝐹𝐹2+ 𝑚𝑚𝑅𝑅𝑇𝑇2) ∗ cos(30) ∗ 27,5 + (𝑚𝑚𝐹𝐹𝐹𝐹5∗ cos(30) ∗ 27,5)� ∗ 𝑔𝑔 ∗ 𝜇𝜇
[27]
Using equation [24] with the insertion of equation [25] and [27] gives the total friction force for the whole transportation line:
𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙 = 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇1+ 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇2= 883 + 1401 = 2284 ≈ 2,3 [𝑘𝑘𝑁𝑁] [28]
Since the ingoing factors are determined for the energy balance, equation [29] is derived from equation [1]. 𝑟𝑟 𝑟𝑟𝑑𝑑 � 2 ∗ 𝐽𝐽𝑇𝑇𝑇𝑇∗ 𝜔𝜔32 2 + 𝑚𝑚𝐹𝐹𝐹𝐹∗ 𝑣𝑣𝑇𝑇𝑇𝑇2 2 + 𝑚𝑚𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇2 2 � = 𝑀𝑀3∗ 𝜔𝜔3− 𝐹𝐹𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇𝑇𝑇− 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙∗ 𝑣𝑣𝑇𝑇𝑇𝑇 [29]
22 Derivation of equation [29] and substitutions by equations [20] and [21] holds:
𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 � 2 ∗ 𝐽𝐽𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇2 + 𝑚𝑚𝐹𝐹𝐹𝐹+ 𝑚𝑚𝑅𝑅𝑇𝑇� 𝑣𝑣𝑇𝑇𝑇𝑇 = 𝑀𝑀3∗ 𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇 − 𝐹𝐹𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗ sin(𝛼𝛼) − 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙∗ 𝑣𝑣𝑇𝑇𝑇𝑇 [30]
𝑣𝑣𝑇𝑇𝑇𝑇 is broken out of equation [30] leading to the transportation line’s total energy balance.
Shown in equation [31]: �𝑟𝑟𝑣𝑣𝑟𝑟𝑑𝑑 �𝑇𝑇𝑇𝑇 2 ∗ 𝐽𝐽𝑟𝑟 𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇2 + 𝑚𝑚𝐹𝐹𝐹𝐹+ 𝑚𝑚𝑅𝑅𝑇𝑇� − 𝑀𝑀3 𝑟𝑟𝑇𝑇𝑇𝑇+ 𝐹𝐹𝑅𝑅𝑇𝑇 ∗ sin(𝛼𝛼) + 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙� ∗ 𝑣𝑣𝑇𝑇𝑇𝑇 = 0 [31]
3.6 Energy balance for the scrapper conveyor unit
By adding all four subsystems balance equations [3], [8], [18] and [31], a general expression for the energy balance for the whole system can be made.
An important step when adding all subsystems energy balances is to eliminate the dependent variables (such as 𝑣𝑣𝑇𝑇𝑇𝑇 or 𝜔𝜔1, 𝜔𝜔2, 𝜔𝜔3). In this case, the angular velocities are
chosen to be eliminated. �𝑟𝑟𝑣𝑣𝑟𝑟𝑑𝑑 � =𝑇𝑇𝑇𝑇 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇 ∗ 𝑇𝑇 ∗ 𝑖𝑖 𝑊𝑊𝑊𝑊∗ 𝑖𝑖𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇 − 𝐹𝐹𝑅𝑅𝑇𝑇 ∗ sin(𝛼𝛼) − 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙 2 ∗ 𝐽𝐽𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇2 + 𝑚𝑚𝐹𝐹𝐹𝐹+ 𝑚𝑚𝑅𝑅𝑇𝑇+ 𝐽𝐽𝑅𝑅 ∗ �𝑖𝑖 𝑊𝑊𝑊𝑊∗ 𝑖𝑖𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇 � 2 ∗ 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇 [32]
An extended and more detailed calculation of equation 32 is shown in Annex 1. A behavior model of the energy flow through the scrapper conveyor is achieved. The simulations are presented in chapter 6.
23
4. Engine for the transportation line
4.1 Choice of electrical engine
To determine a suitable engine for the system, it is necessary to determine the rated power 𝑃𝑃𝑅𝑅𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 (Sv: märk effekt). This is calculated from equation [33] (Sören Andersson &
Claes Tisell & Anders Folkeson, 2002).
𝑃𝑃𝑅𝑅𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑇𝑇 ∗ 𝜔𝜔1 [33]
During the engine’s stationary operation that runs in a constant velocity, the acceleration is equal to zero according to equation [34] (J.L. Meriam & L.G. Kraige, 1998).
𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇
𝑟𝑟𝑑𝑑 = 0 [34]
Combining equations [32] and [34] gives an expression for solving the rotors torque T in equation [35]:
𝑇𝑇 =�𝐹𝐹𝑅𝑅𝑇𝑇𝜂𝜂∗ sin(𝛼𝛼) + 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙�𝑟𝑟𝑇𝑇𝑇𝑇
𝑊𝑊𝑊𝑊 ∗ 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑖𝑖𝑊𝑊𝑊𝑊∗ 𝑖𝑖𝑇𝑇𝑇𝑇
[35]
The force that the material creates on the fuel scrapper, 𝐹𝐹𝑅𝑅𝑇𝑇 is determined in equation [36]
(Jan Schnittger & Anders Folkseon, 2004):
𝐹𝐹𝑅𝑅𝑇𝑇 = 𝑚𝑚𝑅𝑅𝑇𝑇 ∗ 𝑔𝑔 [36]
The mass of the fuel 𝑚𝑚𝑅𝑅𝑇𝑇 is multiplied with the number of loaded steps while being
transported along the slope part according to equation [26].
The total mass lifted by the fuel scrappers is 𝑚𝑚𝑅𝑅𝑇𝑇∗ 27,5. The force 𝐹𝐹𝑅𝑅𝑇𝑇 is calculated
according equation [36]:
𝐹𝐹𝑅𝑅𝑇𝑇 = 𝑚𝑚𝑅𝑅𝑇𝑇 ∗ 27,5 ∗ 9,81 ⟹ 𝐹𝐹𝑅𝑅𝑇𝑇 = 5395 ≈ 5,4 [𝑘𝑘𝑁𝑁] [37]
Solving T from equation [35] with a combination of equation [28] and [37] gives:
𝑇𝑇 =�𝐹𝐹𝑅𝑅𝑇𝑇𝜂𝜂∗ sin(𝛼𝛼) + 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙�𝑟𝑟𝑇𝑇𝑇𝑇
𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑖𝑖𝑊𝑊𝑊𝑊∗ 𝑖𝑖𝑇𝑇𝑇𝑇 = 21,64 ≈ 22 [𝑁𝑁𝑚𝑚]
24 Equation [39] is used to calculate the angular velocity 𝜔𝜔1 for the searched engine. As 𝑛𝑛𝑠𝑠 is
the engines synchronized number of revolutions (Sören Andersson & Claes Tisell & Anders Folkeson, 2002).
𝜔𝜔1 = 2 ∗ 𝜋𝜋 ∗ 𝑛𝑛60 𝑠𝑠 [39]
Using equation [33] with the combination of [38] and [39] gives 𝑃𝑃𝑅𝑅𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠:
𝑃𝑃𝑅𝑅𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑇𝑇 ∗ 𝜔𝜔1 = 22 ∗2 ∗ 𝜋𝜋 ∗ 150060 = 3456 ≈ 3,5 [𝑘𝑘𝑊𝑊] [40]
Based on the calculations an optimal engine choice for these values would be a 3,5 [kW] power rated engine.
4.2 Electrical supply
As the engine is supplied with an unlimited electrical supply, it generates a torque 𝑇𝑇1for
the rotor according to equation [41] (Sören Andersson & Claes Tisell & Anders Folkeson, 2002).
𝑇𝑇1 = 𝑠𝑠300 ∗ 𝑠𝑠(1 + 2 ∗ 𝑠𝑠)2+ 0.04 ∗ (𝑠𝑠 + 1) [41]
The slip “s” is between the engine’s asynchrony number of revolutions 𝑛𝑛𝑚𝑚 and the engines
synchronized number of revolutions 𝑛𝑛𝑠𝑠 (Jan Schnittger & Anders Folkseon, 2004).
𝑠𝑠 =𝑛𝑛𝑠𝑠 𝑛𝑛− 𝑛𝑛𝑚𝑚
𝑠𝑠
[42]
When the same engine is provided with a limited current supply, the torque 𝑇𝑇2 is
determined by equation [43](H. Johansson, 2002).
𝑇𝑇2 = 8,60 ∗ 𝐼𝐼 2,42 + 𝑠𝑠 𝑚𝑚𝑡𝑡𝑖𝑖 [43]
As the maximum current 𝐼𝐼𝑚𝑚𝑡𝑡𝑖𝑖 is determined in equation [44]:
25
5. Inertia calculation
The rotors inertia, 𝐽𝐽𝑅𝑅 and the transportation lines drive wheel inertia, 𝐽𝐽𝑇𝑇𝑇𝑇 are undefined
parameters in equation [32]. General equations as [45], [46] and [47] are used to calculate the inertia 𝐽𝐽𝑅𝑅 and for 𝐽𝐽𝑇𝑇𝑇𝑇 as the body is spun in a centered axis (KTH, 2003). The choice of
material is iron with its density being; 7870 𝑘𝑘𝑔𝑔/𝑚𝑚3 (KTH, 1991).
𝐽𝐽 =𝑚𝑚 ∗ 𝑟𝑟2 2 [45]
The mass is calculated by equation [46], (KTH, 1991).
𝑚𝑚 = 𝜌𝜌 ∗ 𝑉𝑉 [46]
The drive wheel and rotor have a cylindrical body shape and is calculated in equation [47] (KTH, 1991).
𝑉𝑉 = 2 ∗ 𝜋𝜋 ∗ 𝑟𝑟2∗ ℎ [47]
5.1 Determination of inertia for the rotor
Using equation [47] gives the volume for the rotor 𝑉𝑉𝑅𝑅:
𝑉𝑉𝑅𝑅 = 2 ∗ 𝜋𝜋 ∗ 𝑟𝑟𝑅𝑅2∗ ℎ𝑅𝑅 = 2 ∗ 𝜋𝜋 ∗ (0,22) ∗ 0,04 = 0,01 [𝑚𝑚3] [48]
Mass calculation using equation [46] and [48]:
𝑚𝑚𝑅𝑅 = 7870 ∗ 0,01 = 79,11 ≈ 79 [𝑘𝑘𝑔𝑔] [49]
The inertia is calculated from equation [45] and equation [49]:
𝐽𝐽𝑅𝑅 = 79 ∗ (0,2) 2
2 = 1,58 ≈ 2 [𝑘𝑘𝑔𝑔𝑚𝑚3]
26
5.2 Determination of inertia for the drive wheel
Using equation [47] gives the volume 𝑉𝑉𝑇𝑇𝑇𝑇:
𝑉𝑉𝑇𝑇𝑇𝑇 = 2 ∗ 𝜋𝜋 ∗ 𝑟𝑟𝑇𝑇𝑇𝑇2∗ ℎ𝑇𝑇𝑇𝑇 = 2 ∗ 𝜋𝜋 ∗ (0,382) ∗ 0,05 = 0,045 ≈ 0,05 [𝑚𝑚3] [51]
Calculate the mass with equation [46] and [51]:
𝑚𝑚𝑇𝑇𝑇𝑇 = 7870 ∗ 0,05 = 393,5 ≈ 394 [𝑘𝑘𝑔𝑔]
[52] The inertia is calculated by equation [45] and [51]:
𝐽𝐽𝑇𝑇𝑇𝑇 =394 ∗ (0,383) 2
2 = 28,4 ≈ 28 [𝑘𝑘𝑔𝑔𝑚𝑚3]
27
6. Modelling procedure
As the energy flow in the scrapper conveyor unit is determined through the sum of the four subsystems (engine, gearbox, transmission chain and transportation line) energy flow, a singular expression describing its behavior is derived and presented in the differential equation [32]. In order to visualize the energy flow and the relation of the ingoing parameters such as stored energy, the out-in going effects and losses, a numerical mathematical program Matlab is used.
A function file (kop_ffil) was created with a main program (see Annex 3) to simulate the startup sequence. The function file is used to express the sum of the four subsystems (equation [32]), see Figure 11.
Figure 11: Equation [32] expressed in Matlab environment.
In order to visualize the behavior of the system (velocity and acceleration) and the dependency of the model on the chosen engine and current supply, the if-condition below (Figure 12) illustrates this performance.
Figure 12: This segment in the function file shows the coding and application of the current limitation to
the system (optimized function). When simulating without a current limitation (present procedure) the if-function needs to be disabled.
28 The required parameters (e.g. inertias, applied forces, calculated masses, work efficiencies, etc.) are defined in the main program (kop_hfil) and solved with the help of the ode23 command. This function solves the differential equation [32] by using the function file. See figure 13.
Figure 13: Code that solves the differential equation [32] with help of the Matlab function ode23.
The simulation for the transportation line displays three cases. The behavior of the system regarding the velocity, acceleration and current flow is plotted when the engine operates with limited and unlimited current supply.
29
7. Results
The obtained graphs from the Matlab simulations illustrates four different scenarios, as the engine is current limited or unlimited as well when the band runs in idle or with a full load.
7.1 System simulation with current limited engine and fully loaded
transportation line
An engine with current limitation of 140 [A] runs the transportation line in a velocity of 0,57 m/s with an acceleration of 0,81 m/s2 in 0,91 seconds when the band is fully loaded.
See Figure 14.
Figure 14: The graphs illustrate the velocity, acceleration and current as a function of time. The used engine
30
7.2 System simulation with current limited engine and unloaded transportation
line
With the same supplied current of 140 [A], the velocity is unchanged while running the
band unloaded. However, the acceleration is slightly increased reaching 0,84 m/s2 in
0,79 seconds. See Figure 15.
Figure 15: The graphs illustrate the velocity, acceleration and current as a function of time. The used engine
31
7.3 System simulation with current unlimited engine and fully loaded
transportation line
As the system runs with no current limitation, the engine keeps on feeding the system
with an unnecessary amount of current, 330 [A]. The acceleration reaches up to 1,74 m/s2 in 0,45 seconds with the unchanged velocity 0,57m/s. See Figure 16.
Figure 16: The graphs illustrate the velocity, acceleration and current as a function of time. The used engine
32
7.4 System simulation with current unlimited engine and unloaded
transportation line
With the same current, 330 [A] the transportation runs with the unchangeable velocity 0,57 m/s but with the changed acceleration 1,8 m/s2 in 0,40 seconds . See Figure 17.
Figure 17: The graphs illustrate the velocity, acceleration and current as a function of time. The used engine
has a non-limited current supply and the band is unloaded from woodchips.
The Matlab simulations according to the scrapper conveyor unit’s behavior model in chapter 3 is presented. First thesis purpose is fulfilled.
33
7.5 Mass calculation for the desired boiler effect
The given velocity (0,57m/s) from the simulation with the chosen engine
(see 4.1) is used in equation [54] to calculate the total fuel mass 𝑚𝑚𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙 needed to reach the
index boiler effect that the plant distributes (J.L. Meriam & L.G. Kraige, 1998).
𝑃𝑃𝐵𝐵𝑖𝑖 = 𝐹𝐹𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙∗ 𝑣𝑣𝑇𝑇𝑇𝑇 ⟹ 𝑃𝑃𝐵𝐵𝑖𝑖 = 𝑚𝑚𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙∗ 𝑔𝑔 ∗ 𝑣𝑣𝑇𝑇𝑇𝑇 [54]
𝑃𝑃𝐵𝐵𝑖𝑖 is the effect windows 12, 18 and 24 MW for the plant. Equation [54] is rewritten to
solve 𝑚𝑚𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙. 𝑃𝑃𝐵𝐵𝑖𝑖 = 𝑚𝑚𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙∗ 𝑔𝑔 ∗ 𝑣𝑣𝑇𝑇𝑇𝑇 ⟹ 𝑚𝑚𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙 = 𝑔𝑔 ∗ 𝑣𝑣𝑃𝑃𝐵𝐵𝑖𝑖 𝑇𝑇𝑇𝑇 [55] 𝑚𝑚𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙 is established below: 𝑃𝑃𝐵𝐵𝑖𝑖 = 24 𝑀𝑀𝑊𝑊 ⟹ 𝑚𝑚𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙= 24 ∗ 10 6 9,81 ∗ 0,57 = 4300 𝑑𝑑𝐴𝐴𝑛𝑛𝑠𝑠 [56] 𝑃𝑃𝐵𝐵𝑖𝑖 = 18 𝑀𝑀𝑊𝑊 ⟹ 𝑚𝑚𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙= 18 ∗ 10 6 9,81 ∗ 0,57 = 3220 𝑑𝑑𝐴𝐴𝑛𝑛𝑠𝑠 [57] 𝑃𝑃𝐵𝐵𝑖𝑖 = 12 𝑀𝑀𝑊𝑊 ⟹ 𝑚𝑚𝑓𝑓𝑖𝑖𝑠𝑠𝑙𝑙= 12 ∗ 10 6 9,81 ∗ 0,57 = 2150 𝑑𝑑𝐴𝐴𝑛𝑛𝑠𝑠 [58]
34
7.6 Result summary
The four subsystems balance equations put together summarize a general expression for the energy flow in the scrapper conveyor unit. See equation [32].
�𝑟𝑟𝑣𝑣𝑟𝑟𝑑𝑑 � =𝑇𝑇𝑇𝑇 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇 ∗ 𝑇𝑇 ∗ 𝑖𝑖 𝑊𝑊𝑊𝑊∗ 𝑖𝑖𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇 − 𝐹𝐹𝑅𝑅𝑇𝑇 ∗ sin(𝛼𝛼) − 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙 2 ∗ 𝐽𝐽𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇2 + 𝑚𝑚𝐹𝐹𝐹𝐹+ 𝑚𝑚𝑅𝑅𝑇𝑇+ 𝐽𝐽𝑅𝑅 ∗ �𝑖𝑖 𝑊𝑊𝑊𝑊∗ 𝑖𝑖𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇 � 2 ∗ 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇 [32]
The simulation results (from Figure 14-17) of the startup sequence for the scrapper conveyor unit are summarized and presented in Table 2.
Simulations Load Imax
[A] Acceleration [m/s2] Startup sequence [s]
With current limitation (Figure 14) Full 140 0,81 0,91 With current limitation (Figure 15) Empty 140 0,85 0,79 Without current limitation (Figure 16) Full 330 1,74 0,45 Without current limitation (Figure 17) Empty 330 1,8 0,40
Table 2: Simulation results.
Since the velocity of the transportation band is defined, the required material of fuel to achieve a certain boiler effect is calculated. These are shown in Table 3.
Boiler effect Fuel mass needed
24 MW 4300 𝑑𝑑𝐴𝐴𝑛𝑛𝑠𝑠
18 MW 3220 𝑑𝑑𝐴𝐴𝑛𝑛𝑠𝑠
12 MW 2150 𝑑𝑑𝐴𝐴𝑛𝑛𝑠𝑠
35
8. Discussion
Each scenario showcases three graphs, the velocity, acceleration and startup sequence. The velocity parameter is an important parameter to visualize the movement of the band as it transports woodchips for further processing. Including that the velocity parameter is used to calculate the amount of used woodchips for the desired energy distribution. The acceleration parameter illustrates the movement of the band when it is on idle. The importance of determining the acceleration is to observe how the scrapper conveyor unit’s machinery behaves to see what components wear out and how the abrupt and fast movement treats the carried fuel.
The startup sequence is also shown to visualize how long time it takes for the band to run from idle till its maximum velocity depending on the current that enters the engine. As it is limited or unlimited the startup sequence is an important parameter for further improvements to the scrapper conveyor units components.
The evaluation shows that the acceleration without a current limitation is high regardless of the load of the transportation line. This behavior could be experienced as abrupt and to fast due to its short startup sequence. See Table 2.
A module that provides a current limitation to the system could be a potential choice since it provides a smooth and calm startup sequence with an extended time span when transporting the material. The time for the startup sequence might be longer, but it could be preferable than having too fast or too sudden motions that could damage the plant’s equipment.
The risk of upcoming wear on the parts run by an engine without current limitation is explained due to the high acceleration that appears. Another risk is the decreased material transport efficiency as material falls to the bottom of the band.
It is recommended to apply a limited current engine in the system in order to obtain more adequate transportation for the fuel.
The results of the startup sequence and acceleration vary depending on the current supply fed to the system. Due to the chosen engine and the fixed number of revolution 𝑛𝑛𝑠𝑠 the velocity
of the band (velocity of 0,57 m/s) remains unchanged regardless of the load. In case the engine is changed to another number of revolutions, the simulation results will be different.
36 This could be appreciated in the example below as the synchronized number of revolutions is changed to 1800 rpm (in comparison for being assumed to 1500 rpm) with no other changes in parameters and boundaries in the Matlab codes (e.g. inertias, applied forces, calculated masses, work efficiencies, etc.). Observe the results in Figure 18.
Figure 18: The graphs illustrate the velocity, acceleration and current as a function of time. With the change of
37 The velocity reached 0,76 m/s with an acceleration of 0,81 m/s2 with a time span of 1,26
seconds. This scenario shows that the number of revolutions is a key parameter to visualize the system behavior, shown in Table 4.
Fully loaded and limited current engine, 𝑛𝑛𝑠𝑠 = 1500
Fully loaded and limited current engine, 𝑛𝑛𝑠𝑠 = 1800
Velocity: 0,57 m/s Velocity: 0,76 m/s
Acceleration: 0,81 m/s2 Acceleration: 0,81 m/s2
Startup sequence: 0,91 seconds Startup sequence: 1,26 seconds
Table 4: Result variation depending the change of the synchronized number of revolutions 𝑛𝑛𝑠𝑠.
It is mentioned that a longer startup sequence gives the best result for not damaging the system modules by the fast and abrupt start. Since the band has a new velocity the amount of fuel mass to distribute will also vary depending on the boiler effect, suggesting for less consumption of a yearly intake of woodchips.
As the fuel mass is calculated to reach a certain boiler effect (shown in Table 3) Värmevärden AB clarified that with the assumed parameters that it only reached a 10% part of a yearly consumption of woodchips in Nynäshamn. Meaning that with correct measurements and accurate parameters it is possible to achieve an actual method to optimize the system modules.
As the Matlab programs are good references to visualize the system behavior it can be updated with more equations that can change the outcoming simulation results. Such to determine the energy losses in the asynchronous engine as it was neglected in this study (see chapter 3.2) or to insert other disturbances in the transportation line as it must be regularly checked for clearing out debris which leads to longer lead times.
38
9. Conclusion
The purpose of this thesis is to present a behavior model regarding the energy flow in the scrapper conveyor unit and to determine the amount of fuel needed to distribute the desired boiler effect. The starting point was to analyze the four subsystems separately to evaluate the energy-relation between each other. Simulations of the systems behavior were required and differential equations were resolved in Matlab. This made possible to observe the velocity and acceleration of the band by the generated current flow from the engine.
Although during the simulations, it was necessary to define parameters in order to illustrate the systems behavior.
The simulation shows that if the system is used with correct current flow and correct engine parameters, it could be possible to archive more reliable results in the future to calculate the losses and the energy flow in each subsystem.
9.1 Recommendations
• Parameters were assumed during the Matlab simulations. It would be interesting to see how the plant’s energy flow is with more accurate parameters.
• Since the energy balance is presented for the scrapper conveyor unit, it could be interesting to create more balance equations for other systems (i.e. the elevator unit or the magnetic chamber). In order to prevent further energy losses.
• Due to the importance of the weight carried by the scrappers, it is important to optimize its design in order to increase the amount of transport material. Some proposals are shown in Annex 4.
39
10. References
Energradgivningen, Fördelar och Nackdelar, 2017–09
https://energiradgivningen.se/system/tdf/faktablad_fjarrvarme_2017.pdf?file=1 (2019-06-03). Energimyndigheterna, District heating substation, 2015-03-27
http://www.energimyndigheten.se/en/sustainability/households/heating-your-home/district-heating/ (2019-05-22).
H Johansson, (2002) Elektronik Del 1
Stockholm Kungliga Tekniska Högskolan, Institutionen för Maskinkonstruktion Elektroteknik (2019-06-07)
Jan Schnittger & Anders Folkseon, (2004) Funtionsanalys och optimering i maskintekniken Stockholm: Institutionen för maskinkonstruktion KTH (2019-06-11)
J.L. Mermiam & L.G. Kraige, (1998) Engineering mechanics Statics 4th edition SI-version. (ISBN:0-471-24164-4) John Wiley and Sons, Inc, (2019-06-15)
KTH. (1991). Handbok och formelsamling i Hållfasthetslära. Institutionen för Hållfasthetslära KTH (2019-06-19)
KTH. (2003). Maskinelement Handbok. Stockholm: Institutionen för maskinkonstruktion KTH (2019-06-19)
R.C Hibbeler & Kai Beng Yap, (2013) Mechanics for engineers, Dynamics. (ISBN:978-981-06-9261-2) Nanyang Technological University, (2019-06-14)
Sören Andersson & Jan Hölckle, (2001) Modeller och functioner. Stockholm: Institutionen för maskinkonstruktion KTH (2019-06-12)
Sören Andersson & Claes Tisell & Anders Folkeson, (2002) Maskinelement Maskindynamik. Stockholm: Institutionen för maskinkonstruktion KTH (2019-06-10)
Värmevärden AB, Kraftvärmen i Nynäshamn, 2017-17-02
https://www.varmevarden.se/om-varmevarden/orter/nynashamn/ (2019-05-23). Waste360, Recycling, Wood Wastes Offer Select Markets A Solid Cash Crop, 1996-05-01
40
Appendix
1. Calculations of the subsystems balance equation. 2. Drawing of the scrapper conveyor unit.
3. Matlab program files. 4. Fuel scrapper concepts.
41
Annex 1
Create the balance equation [32] with the help of the subsystems balance equations [3][8][18][31] and with the linking formulas [A1] [A2] [A3] [A4] [A5] [A6] [A7] : Asynchrony engine: �𝐽𝐽𝑅𝑅 ∗𝑟𝑟𝜔𝜔𝑟𝑟𝑑𝑑 − 𝑇𝑇 + 𝑀𝑀1 1� 𝜔𝜔1 = 0 [3] Worm gear: 𝑟𝑟 𝑟𝑟𝑑𝑑 � 𝐽𝐽𝑊𝑊𝑊𝑊1∗ 𝜔𝜔12 2 + 𝐽𝐽𝑊𝑊𝑊𝑊2∗ 𝜔𝜔22 2 � = 𝑀𝑀1∗ 𝜔𝜔1∗ 𝜂𝜂𝑊𝑊𝑊𝑊− 𝑀𝑀2∗ 𝜔𝜔2 [8] Transmission chain: �𝑚𝑚𝑇𝑇𝑇𝑇∗𝑟𝑟𝑣𝑣𝑟𝑟𝑑𝑑 +𝑇𝑇𝑇𝑇 𝑟𝑟𝐽𝐽𝑇𝑇𝑇𝑇1 𝑇𝑇𝑇𝑇12∗ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 + 𝐽𝐽𝑇𝑇𝑇𝑇2 𝑟𝑟𝑇𝑇𝑇𝑇22∗ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 + 𝑀𝑀3 𝑟𝑟𝑇𝑇𝑇𝑇2− 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑀𝑀2 𝑟𝑟𝑇𝑇𝑇𝑇1 � 𝑣𝑣𝑇𝑇𝑇𝑇 = 0 [18] Transportation line: �𝑟𝑟𝑣𝑣𝑟𝑟𝑑𝑑 �𝑇𝑇𝑇𝑇 2 ∗ 𝐽𝐽𝑟𝑟 𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇2 + 𝑚𝑚𝐹𝐹𝐹𝐹+ 𝑚𝑚𝑅𝑅𝑇𝑇� − 𝑀𝑀3 𝑟𝑟𝑇𝑇𝑇𝑇+ 𝐹𝐹𝑅𝑅𝑇𝑇 ∗ sin(𝛼𝛼) + 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙� ∗ 𝑣𝑣𝑇𝑇𝑇𝑇 = 0 [31]
Linking formulas from the balance equations:
𝑀𝑀1 = 𝑇𝑇 − 𝐽𝐽𝑅𝑅∗𝑟𝑟𝜔𝜔𝑟𝑟𝑑𝑑 1 [A1] 𝑀𝑀2 = 𝑀𝑀1∗ 𝜔𝜔𝜔𝜔1∗ 𝜂𝜂𝑊𝑊𝑊𝑊 2 [A2] 𝑀𝑀3 = 𝑀𝑀2∗ 𝜔𝜔𝜔𝜔2∗ 𝜂𝜂𝑇𝑇𝑇𝑇 3 [A3] 𝜔𝜔1 =𝑖𝑖𝑇𝑇𝑇𝑇∗ 𝑖𝑖𝑟𝑟𝑊𝑊𝑊𝑊∗ 𝑣𝑣𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 [A4] 𝜔𝜔2 =𝑖𝑖𝑇𝑇𝑇𝑇𝑟𝑟∗ 𝑣𝑣𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 [A5] 𝜔𝜔3 = 𝑣𝑣𝑟𝑟𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 [A6]
42 𝑣𝑣𝑇𝑇𝑇𝑇 =𝑟𝑟𝑇𝑇𝑇𝑇∗ 𝑖𝑖𝑟𝑟𝑇𝑇𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇
𝑇𝑇𝑇𝑇
[A7]
Energy balance for the whole system, to achieve equation [32]:
2 ∗ 𝐽𝐽𝑇𝑇𝑇𝑇∗ 𝜔𝜔3∗𝑟𝑟𝜔𝜔𝑟𝑟𝑑𝑑 + 𝑚𝑚3 𝐹𝐹𝐹𝐹∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗𝑟𝑟𝑣𝑣𝑟𝑟𝑑𝑑 + 𝑚𝑚𝑇𝑇𝑇𝑇 𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗𝑟𝑟𝑣𝑣𝑟𝑟𝑑𝑑 =𝑇𝑇𝑇𝑇 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑇𝑇 ∗ 𝜔𝜔1− 𝐽𝐽𝑅𝑅∗𝑟𝑟𝜔𝜔1𝑟𝑟𝑑𝑑 ∗ 𝜔𝜔1∗ 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇− 𝐹𝐹𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇𝑇𝑇− 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙∗ 𝑣𝑣𝑇𝑇𝑇𝑇 ⟹ ⟹2 ∗ 𝐽𝐽𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇2 ∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 + 𝑚𝑚𝐹𝐹𝐹𝐹∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 + 𝑚𝑚𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 = 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑇𝑇 ∗𝑖𝑖𝑇𝑇𝑇𝑇∗ 𝑖𝑖𝑟𝑟𝑊𝑊𝑊𝑊∗ 𝑣𝑣𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 − 𝐽𝐽𝑅𝑅∗ � 𝑖𝑖𝑇𝑇𝑇𝑇∗ 𝑖𝑖𝑊𝑊𝑊𝑊 𝑟𝑟𝑇𝑇𝑇𝑇 � 2 ∗𝑟𝑟𝑣𝑣𝑟𝑟𝑑𝑑 ∗ 𝑣𝑣𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇∗ 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇− 𝐹𝐹𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇𝑇𝑇− 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙∗ 𝑣𝑣𝑇𝑇𝑇𝑇 ⟹ ⟹2 ∗ 𝐽𝐽𝑟𝑟 𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇2 ∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 + 𝑚𝑚𝐹𝐹𝐹𝐹∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 + 𝑚𝑚𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 + 𝐽𝐽𝑅𝑅∗ � 𝑖𝑖𝑇𝑇𝑇𝑇∗ 𝑖𝑖𝑊𝑊𝑊𝑊 𝑟𝑟𝑇𝑇𝑇𝑇 � 2 ∗𝑟𝑟𝑣𝑣𝑟𝑟𝑑𝑑 ∗ 𝑣𝑣𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇∗ 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇 = 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑇𝑇 ∗𝑖𝑖𝑇𝑇𝑇𝑇∗ 𝑖𝑖𝑟𝑟𝑊𝑊𝑊𝑊∗ 𝑣𝑣𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 − 𝐹𝐹𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗ sin(𝛼𝛼) − 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙∗ 𝑣𝑣𝑇𝑇𝑇𝑇 ⟹ ⟹𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇𝑟𝑟𝑑𝑑 �2 ∗ 𝐽𝐽𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇2 ∗ 𝑣𝑣𝑇𝑇𝑇𝑇+ 𝑚𝑚𝐹𝐹𝐹𝐹∗ 𝑣𝑣𝑇𝑇𝑇𝑇+ 𝑚𝑚𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇+ 𝐽𝐽𝑅𝑅∗�𝑖𝑖𝑇𝑇𝐶𝐶𝑟𝑟∗ 𝑖𝑖𝑊𝑊𝑊𝑊𝑇𝑇𝑇𝑇 � 2 ∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗ 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝐶𝐶�= 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑇𝑇 ∗𝑖𝑖𝑇𝑇𝑇𝑇∗ 𝑖𝑖𝑟𝑟𝑊𝑊𝑊𝑊∗ 𝑣𝑣𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 − 𝐹𝐹𝑅𝑅𝑇𝑇∗ 𝑣𝑣𝑇𝑇𝑇𝑇∗ sin(𝛼𝛼) − 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙∗ 𝑣𝑣𝑇𝑇𝑇𝑇 ⟹ 𝑟𝑟𝑣𝑣𝑇𝑇𝑇𝑇 𝑟𝑟𝑑𝑑 = 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇∗ 𝑇𝑇 ∗ 𝑖𝑖𝑇𝑇𝑇𝑇𝑟𝑟∗ 𝑖𝑖𝑇𝑇𝑇𝑇𝑊𝑊𝑊𝑊− 𝐹𝐹𝑅𝑅𝑇𝑇∗ sin(𝛼𝛼) − 𝐹𝐹𝑓𝑓𝑇𝑇𝑇𝑇−𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑙𝑙 2 ∗ 𝐽𝐽𝑇𝑇𝑇𝑇 𝑟𝑟𝑇𝑇𝑇𝑇2 + 𝑚𝑚𝐹𝐹𝐹𝐹+ 𝑚𝑚𝑅𝑅𝑇𝑇+ 𝐽𝐽𝑅𝑅∗ �𝑖𝑖𝑇𝑇𝑇𝑇𝑟𝑟∗ 𝑖𝑖𝑇𝑇𝑇𝑇𝑊𝑊𝑊𝑊� 2 ∗ 𝜂𝜂𝑊𝑊𝑊𝑊∗ 𝜂𝜂𝑇𝑇𝑇𝑇
43
Annex 2
44
Annex 3
3.1 Function File
function vprim = kop_ffil(t,v) %Function file for simulating the results in
the main program (equation 32)
global Jr Jtl ns iwg itc etawg etatc rtl Ff Frt Imax alfa mfs
mrt;%Variables for the function file and main program
%Designation of variables: %Jr=Rotor inertia
%Jtl=Transportation line inertia
%ns=Engines synchronized number of revolutions %iwg=Angular velocity ratio for worm gear
%itc=Angular velocity ratio for transmission chain %etawg=Worm gear work efficiency
%etatc=Transmission chain work efficiency %rtl=Gear radius for transportation line %Ff=Transportation line total friction force
%Frt=Performed force from the fuel towards the fuel scrapper %Imax=maximum current flow
%alfa= 30 degree angle
%mfs=Mass for fuel scrappers and chain scrapper %mrt=Fuel mass/scrapper
w1=iwg*itc*v/rtl; %Expression for w1, when w2 and w3 are eliminated (equation A4)
nm=w1*60/2/pi; %Expression to calculate the asynchrony number of revolutions
s=(ns-nm)/ns; %Slip formula used to calculate the torque and max current (equation 42)
T=300*s*(1+2*s)/(s^2+0.04*(s+1));%The engines torque (equation 41) I=T*(2.42+s)/8.60;%Maximum current to the system (equation 44)
%The if-function get erased when simulating the system without current limitation to
%the main program.
if Imax<I %The boundary for the max current to work I=Imax;%Defines I as the maximum current
T=(8.60*I)/(2.42+s);%The other torque when the current limitation is on (equation 44)
end %Closes the function
vprim=(((etawg*etatc*T*itc*iwg)/rtl)-Frt*sin(alfa)-Ff)/((2*Jtl/rtl^2+mfs+mrt+((Jr*iwg^2*itc^2*etawg*etatc)/rtl^2)));
%vprim is the sum of the energy balance to the whole system (equation 32). %As the integration of vprim happens in the main program with the ode23 function,
%it will show the systems velocity and acceleration as functions through time for only one loop when transporting fuel.
45 3.2 Main program
global Jr Jtl ns iwg itc etawg etatc rtl Ff Frt Imax alfa mfs mrt; %Defined
variables used in the function file
tstart = 0;%Time interval start when time=0 tfinal = 3;%Time interval closing when time=10 tspan = [tstart tfinal];%Interval of the time v0=0; %The velocity start and is set to 0
%Values for the transmission chain alfa=pi/6; %angle
g=9.81; %gravitation force
itc=3.0; %Angular velocity ratio for transmission chain* rtl=0.766/2; %Upper drive wheel diameter in the
transportation line*
Ff=2300; %Transportation line total friction force with full load (2300 N) % And empty load (1400 N)
Imax=140; %Max current to the system with current limitation*
%Transmission values
etatc=0.98; %Transmission chain work efficiency* etawg=0.85; %Worm gear work efficiency*
Jtl=29; %Inertia for the drive wheel (equation 53) %Engine values
ns=1500; %Engines synchronized number of revolutions* iwg=35; %Angular velocity ratio for worm gear*
Jr=2; %Inertia for the rotor (equation 50)
mfs=3500; %Mass for fuel scrappers and chain (3500/70=50)* mrt=1000; %Total mass carried from fuel scrappers. Full load (1000 kg) "50*20=1000".
%Puts to 15 when running empty.
Frt=5400; %Performed force from the fuel towards the fuel scrappers (5400N = 15*9,81*27,5)
%Frt=0 when running empty, as Frt= (0*9.81*cos(pi/6)*0.12)
%%% Change Ff, mrt, Frt When plotting!
[t,v] = ode23('kop_ffil',tspan,v0,odeset('RelTol',1e-6)); %Integration function ode23 is called
%RelTol 1e-6 is the tolerance
v1=v(:,1); %A vector for v1 w1=iwg*itc*v/rtl; %Function file code nm=60*w1/2/pi; %Function file code s=(ns-nm)/ns; %Function file code T=300*s.*(1+2*s)./(s.^2+0.04*(s+1)); %Vector multiplication I=T.*(2.42+s)/8.60; %Vector multiplication
%The for-function get erased when simulating the system without current limitation
