HAN Saving Energy in a Vehicle READER Saving Energy in a Vehicle Reader Date: 17 August 2021 G. Angelino R. Beem J. de Vries HAN Saving Energy in a Vehicle READER Content/ Inhoud Chapter 1 Energy in General .................................................................................................................. 4 Learning Objectives............................................................................................................................. 4 Introduction ........................................................................................................................................ 5 First Law of Thermodynamics ............................................................................................................. 6 Joule, the unit of energy ..................................................................................................................... 7 Is there a relation between Power, Work and Force? ........................................................................ 9 What is the difference between Work and Energy?......................................................................... 10 Power in combustion engines. .......................................................................................................... 11 What is the relation between these parameters? ........................................................................ 11 Efficiency ........................................................................................................................................... 12 WTW, WTT, TTW ........................................................................................................................... 13 Durability........................................................................................................................................... 13 Other sources of Energy.................................................................................................................... 14 Kinetic Energy ............................................................................................................................... 14 Spring energy ................................................................................................................................ 15 Potential Energy ............................................................................................................................ 16 Chemical Energy............................................................................................................................ 16 Heat ............................................................................................................................................... 16 Electrical Energy ............................................................................................................................ 17 References .................................................................................................................................... 18 Chapter 2 Energy losses while driving ................................................................................................. 19 Learning objectives of this lesson: .................................................................................................... 19 Introduction ...................................................................................................................................... 20 The Driving Resistances. ................................................................................................................... 20 Rolling resistance: ......................................................................................................................... 21 Slope resistance ............................................................................................................................ 23 Air resistance................................................................................................................................. 25 Acceleration resistance ................................................................................................................. 26 References ........................................................................................................................................ 27 Chapter 3: Power Conversions in Vehicles .......................................................................................... 28 Study Material (in advance) .............................................................................................................. 28 Learning Objectives........................................................................................................................... 28 What is Power ................................................................................................................................... 29 Drivetrain .......................................................................................................................................... 32 Saving Energy in a Vehicle Content page 1 HAN Saving Energy in a Vehicle READER Hybrid vehicles .................................................................................................................................. 34 Series Hybrid ................................................................................................................................. 35 Parallel Hybrid ............................................................................................................................... 35 Series – Parallel Hybrid ................................................................................................................. 35 Power output for ICE and MG ........................................................................................................... 36 Chapter 4: Motor performance and efficiency.................................................................................... 43 Study Material (in advance) .............................................................................................................. 43 Learning Objectives........................................................................................................................... 43 Fuel Economy Characteristics of Internal Combustion Engines (page. 49, 50) ............................ 44 Specific Fuel Consumption and Efficiency (page. 67, 68) ............................................................. 45 Energy Consumption electric drive (page. 114, 115) .................................................................... 46 Hydraulic power converter ........................................................................................................... 47 Fuel Cell’s ...................................................................................................................................... 49 Overview total efficiency .................................................................................................................. 50 Chapter 5 Aerodynamics ...................................................................................................................... 51 Learning Objectives........................................................................................................................... 51 Introduction ...................................................................................................................................... 51 Bernoulli’s Law .................................................................................................................................. 52 Gravitational pressure – hydrostatic pressure.............................................................................. 52 Dynamic pressure ......................................................................................................................... 52 Static pressure............................................................................................................................... 52 Bernoulli’s Law in practice (example 1) ............................................................................................ 53 Summary. .......................................................................................................................................... 56 Aerodynamics in practice.................................................................................................................. 57 The border layer (very important) ................................................................................................ 57 Pressure around the object........................................................................................................... 57 Turbulence and Laminar ............................................................................................................... 58 Drag Coefficient (cd) De cw waarde – (weerstands coëfficient [NL])................................................. 59 Example drag forces ...................................................................................................................... 60 Law of conservation of mass............................................................................................................. 61 CHAPTER 6 Thermodynamics (Basics and First Law) ......................................................................... 63 Introduction ...................................................................................................................................... 63 Thermodynamics definitions ............................................................................................................ 63 Ideal Gas Law .................................................................................................................................... 63 Gas constant ................................................................................................................................. 64 Specific heat .................................................................................................................................. 64 Saving Energy in a Vehicle Content page 2 HAN Saving Energy in a Vehicle READER Conservation of Energy – First Law of Thermodynamics.................................................................. 65 Special processes in a pV diagram .................................................................................................... 65 Heat energy and work....................................................................................................................... 66 Normal volume ................................................................................................................................. 66 Adiabatic process .............................................................................................................................. 67 CHAPTER 7 Thermodynamics (Cyclical Processes and Polytropes) ................................................. 68 Cyclical processes.............................................................................................................................. 68 Polytropic process ............................................................................................................................. 69 Saving Energy in a Vehicle Content page 3 HAN Saving Energy in a Vehicle READER Chapter 1 Energy in General Learning Objectives At the end of this lecture the student is able to apply the knowledge and skills related to: • • • • • • • • • • • First Law of Thermodynamics. Name the most important forms of energy including calculation of the amount of energy in this state. This counts for: o Kinetic Energy o Potential Energy o Spring Energy o Chemical Energy o Electrical Energy o Heat Calculations for Work Calculations for Energy transfer Calculation of Energy density Explaining the difference between Energy and Power Calculation of Power density Efficiency calculations for simple processes Awareness of energy consumption related to the energy crisis Link between driving strategies and energy consumption Explaining (automotive) concepts: WTW, WTT and TTW Chapter 1 Energy in General page 4 HAN Saving Energy in a Vehicle READER Introduction Figure 1 Cambridge Dictionary meaning of Energy (Cambridge, 2017) Energy is defined as "The ability to work". This is still difficult to understand because “the ability” is used here in the context of “possessing” and not in the context of the ability of an internal combustion engine or electric motor to operate. We need energy to survive. If you don’t eat or drink you will die. Your food contains energy to make your heart, brain and muscles function, and to keep yourself warm. A vehicle need energy to drive – but also to switch on the lighting. Therefore, we need to stop regularly at the gas-station for refueling, where we take on new energy. We can save energy and use it later. When we use it, we often convert energy from one form to another. The dynamo from a bicycle is mounted to the sidewall of the tire. Movement energy is converted into light. We also produce heat, another form of energy. This doesn’t only occur in the lamp but also at the contact with the tire and the dynamo. Even in the wires to the lamp we produce a little bit of heat. So, we also produce heat where we only want to produce light (to see better in the dark). For the biker, the heat is not useful. We call this an “energy loss”, but this is wrong. We don’t lose this energy, but it’s useless for us. With a dynamo on your bike, you can generate the energy by yourself, you’re the one who’s makes the wheels go round. Nowadays most bicycles travel with LED lighting using batteries. These LEDs are “low-energy” lights so we can use the batteries for a long time. But how do we determine the energy efficiency and when can we classify it as economical? That’s what you are going to find out in this lesson. Figure 2 Energy from the dynamo to the lamp by wires. (Haas, 2017) Chapter 1 Energy in General page 5 HAN Saving Energy in a Vehicle READER First Law of Thermodynamics The first law states: “In all cases in which work is produced by the agency of heat, a quantity of heat is consumed which is proportional to the work done; and conversely, by the expenditure of an equal quantity of work an equal quantity of heat is produced” [Rudolf Clausius – 1850] or “Energy is never lost” & “Energy can’t originate out of nothing” If this is true (and it is…), that energy is never lost, then: (1) we cannot have something like an energy-crisis because (2) the total amount of energy is always constant. That brings up an interesting question because that means the total amount of energy has always existed for eternity (we can’t make energy out of nothing). The second statement is true and counts for the whole universe as we know it. The first statement is not true, because an energy crisis exists when we do have a shortage of energy in a certain form. The most important form from this and the previous century is fossil energy and this form is, as we all know, limited. That’s why mankind is searching for alternatives like solar, wind and tide-energy or even in nuclear form. Also, in automotive we can see a movement towards hybrid (vehicle’s using two energy sources) and towards fully electrical propulsion. We also see middle ground solutions like dual fuel (two fuel sources in one engine). These can sometimes be more efficient. The example of the dynamo from the introduction shows clearly what the first law of thermodynamics means: Wheel Kinetic Dynamo Electric heat Lamp light heat heat Figure 3 Energy is never lost but can change from one form into another (Beem, 2017) The wheel has a certain amount of kinetic energy and makes the dynamo rotate. This is also movement but because of the friction between dynamo and tire we already loose some energy in the form of heat. In the dynamo this kinetic energy is transformed into electrical energy, plus an amount of heat in the lamp a major part of the electrical energy is transformed into heat (thick arrow) and only a small part of it in light. Chapter 1 Energy in General page 6 HAN Saving Energy in a Vehicle READER As you can see, we can only transfer the energy that is not transferred into heat, but the total amount of energy is still the same. There are also machines, like engines and other thermal circuit processes that can produce work out of heat. All internal combustion engines work is based on this principle. Joule, the unit of energy Every form of energy is expressed in the same unit, the Joule. If you eat a ‘Mars’ bar the commercial slogan said: ‘a mars a day helps you work, rest and play’ and in Dutch it is even better ‘Mars geeft je nieuwe energie’. So, a ‘Mars’ bar contains ‘Energy” … Figure 4 How much energy is in a ‘Mars’ chocolate bar (The next Corner, 2017) The energy is expressed in an old-fashioned unit, the ‘Calorie’. One calorie is the amount of energy needed to heat up 1 gram of water 1 degree Celsius and is equivalent to 4,18 Joules. But how is the Joule defined: 1 Joule is the amount of energy needed to move an object with a force of 1 newton over a distance of 1 meter. So, if we push a mower with a force of F = 80 [N] over a distance d = 5 [m] the work needed is: π = πΉ β π = 80 β 5 = 400 [π½] And of course, when the force or the distance increases the work will increase too. You may also use a prefix for the unit for larger or smaller values. 1 [kJ] = 1000 [J] and 1 [mJ] = 0.001 [J] Figure 5 Work (Pinsdaddy, 2017) In this case the man in the picture above is pushing the mower in the same direction. What if it wasn’t in the same direction? Assume he was pushing in the direction of the rod…. In that case, the force would be higher, but the work needed to move the mower would be the same. That is because the force isn’t just a value, the direction is also important. We say: Chapter 1 Energy in General page 7 HAN Saving Energy in a Vehicle READER Force is a vector (it has a value and a direction), the direction is also a vector1 So actually, we must not take the ordinary multiplication of the two values, but we must take the scalar-product/ dot product/ inner product (Inwendig product [NL]) from the two vectors. Figure 6 Work delivered by Force F is zero. (Pinsdaddy, 2017) So, the man in the picture above needs to apply a force F to lift the suitcase, but the work, delivered by him is zero if he moves perpendicular at this force. So, we need to adapt the formula a bit. π = πΉβ β πβ = |πΉ| β |π| β cos(ο±) Then, there are some other important laws of physics (discovered by Isaac Newton) according to the forces. 1. An object will remain at rest, or moving at a constant velocity, unless it is acted on by an unbalanced force. 2. Force equals mass times acceleration πΉβ = π β πβ 3. For every action there is an equal opposite reaction. If we apply these laws on a vehicle, we can see what this means: If the thrust forces are higher than the drag forces we will accelerate until the forces are in equilibrium. Then the vehicle will move on with the same speed. Figure 7 Forces on a vehicle (Schoolphysiscs, 2017) Drag forces on a car exist because of friction, wind and gravity when the car is going uphill. And friction occurs between the tires and the road but also in the bearing in the driveline 1 A vector has a value/ magnitude and a direction. When a quantity only has a magnitude, like mass in [kg] this is named a scalar. Chapter 1 Energy in General page 8 HAN Saving Energy in a Vehicle READER Example: All the drag forces are 1000 [N], the thrust forces are 1300 [N], the weight of the car is 1500 [kg], calculate the acceleration of this vehicle. If we assume that all the forces are in the same direction of the movement, we don’t need to take in account the force is a vector… πΉ = π β π => π = πΉ 1300 − 1000 π = = 0.2 [ 2 ] π 1500 π When the speed increases the drag forces will increase too. This means the acceleration will decrease while the net-forces will increase. So, the vehicle will accelerate until we have a new equilibrium. Thrust Force Drag Force Speed Acceleration Figure 8 Relation between forces, acceleration and vehicle speed. (Beem, 2017) In the graph above you can see the relation between the forces and the vehicle speed and its acceleration. When the speed increases the drag, forces increase too (normally not linear but more likely parabolic) until the forces are equal. At that moment the acceleration is zero and the speed has reached a new value. If the vehicle has reached its new value, we see at the odometer the speed is 45 [km/h]. So, we know: To drive 45 [km/h] we need a force of 1500 [N] to overcome all the drag forces. What power do we need to overcome all these forces? Is there a relation between Power, Work and Force? Yes there is: First we need to convert the unit into standard units: 45 [km/h] = 45/3.6 = 12.5 [m/s] So, we know, if we see for 1 second, the car has travelled 12.5 meter and this will cost: π = πΉ β π = 1500 β 12.5 = 18 750 [π½] and after 10 seconds this will be 187 500 [J] and in one hour this will be 67.5 [MJ] (Mega-Joule) The Power is the amount of work delivered per second. So, we can easily see the power needed to overcome all the drag forces will be 18 750 [J/s] of 18.75 [kW] πππ€ππ (π) = Chapter 1 ππππ (π) πΉ β π π = = πΉ β = πΉ β π£ [πππ‘π‘] π‘πππ (π‘) π‘ π‘ Energy in General page 9 HAN Saving Energy in a Vehicle READER It’s clear that if we are going uphill the drag forces will increase. For instance, if we will not increase the thrust force (the motor delivers the same amount of power) and the angle of the hill we encounter is 5ο° the additional drag forces because of this hill will be: πΉβπππ = π β π β sin(ο‘) = 1500 β 9.81 β sin(5) = 1282.5 [π] So, the total drag forces will be 1500 + 1282.5 ο» 2782.5 [N] and the speed will decrease until: π 18 750 π ππ π£= = = 6.7 [ ] ππ 24.3 [ ] πΉ 2782.5 π β That is exactly what we experience during driving. If the drag forces increase the speed will decrease unless we adapt to in by pressing the throttle pedal more downwards to increase the Power. What is the difference between Work and Energy? Work and Energy have the same unit, the Joule [J]. Where ‘Work’ is normally abbreviated by ‘W’ and Energy by ‘E’ – but this is not always written so. For instance, when we talk about heat (what is a form of energy) this is commonly written by a capital Q (for the Quantity of heat, rather than meaning the intensity of heat because of its temperature). As soon as a force is necessary to overcome a certain resistance (physically spoken…) we talk about Work. We can use work however to increase the amount of energy. Example A go-kart + driver (m = 100 [kg]) is on a hill and rolls down by gravity. When we assume a constant slope there is a constant force on the go-kart. At the beginning the thrust force is larger than the drag force so the go-kart accelerates. But on a certain moment it has a constant speed and we need a constant force to maintain this speed. Figure 9 Gravity Racer (youtube, 2019) Let’s assume F = 50 [N] and vmax = 20 [m/s]. So, at that moment we add every second 50 β 20 = 1000 [J] of Work into the system to maintain this speed. 1 But the kinetic energy (πΈπππ = 2 π π£ 2 = 20 000 [π½] ) doesn’t change because the speed is constant. π =πΉβπ Here the work is used to overcome the physical resistances and during acceleration the part of thrust force larger than the drag force is used to increase the kinetic energy. We can also derive another formula. Because the gravity adds 1000 [J] of energy into the system every second, so this is Joules per second and that is the unit of Power: π = πΉ β π£ [π] Chapter 1 Energy in General page 10 HAN Saving Energy in a Vehicle READER Power in combustion engines. Work done by a combustion engine follows the same rules as normal work, but here we know that the force on the piston and the piston speed constantly change. In lesson 6 and 7 of this module you will learn how to calculate how much work is done by one cycle of a combustion engine and if we know the rotation speed we can also calculate the Power. What are the normal parameters of an engine? • • • Power [kW] Torque [Nm] Engine speed [rpm] The engine speed is given in [rpm], but the standard unit for this is the frequency in [Hz] – and for most engines counts that one cycle takes two rotations (four-stroke engines) Figure 10 Internal combustion engine. (Kaiserscience, 2019) What is the relation between these parameters? The combustion engine has a heavy flywheel that acts like an energy buffer. In a combustion stroke it gains energy and in between it gives energy. The mass of the flywheel is chosen that this results in a more or less constant torque output The torque is Force x Radius: π = πΉ β π of πΉ = Radius π π The Force is always perpendicular to the radius, and for one rotation the travelled distance (d) is π = 2 π π The work delivered in one rotation will be: π1−πππ‘ππ‘πππ = πΉ β π = Force π β 2 β π β π = 2 β π β π [π½] π When the engine has a speed of 3600 [rpm] this motor delivers this work 60 times per second. The frequency is 3600/60 = 60 [Hz] The Power of this engine is de delivered work per second, here: π = π1 β π = 2 β π β π β π [π] We will rewrite this a little bit and we get: π =2 βπβ πβπ =πβ π Were π =2ππ Chapter 1 Energy in General page 11 HAN Saving Energy in a Vehicle READER Efficiency We have seen that for each energy conversion we lose some energy. The second law of thermodynamics says that it is impossible to transfer heat completely into work. That means that all cyclical processes, like the 4-stroke process for internal combustion engines, have a certain efficiency. Further you will experience that for every step between the engine and the tires we will lose some energy: the power into the gearbox is higher than the power we can take from the gearbox. How far do we want to go back into the energy chain if we talk about efficiency? Is it just the relation between fuel consumption and distance travelled, or do we need to take in account the process from the oil fields to the gas station as well? We do have common sense about fuel consumption. Nowadays, modern vehicles can travel between 8 and 30 kilometers on one liter of petrol (expressed as 12.5 [l]/100 [km] up to 3.3 [l]/100 [km]). Example: Let take an average fuel consumption of 10 [l]/100 [km] for our car from the previous example when it is driving steadily over the road at 45 [km/h]…(The drag forces are still 1500 [N]) If we travel for 1 hour, the distance will be 45 [km] and this will cost 45/100 * 10 = 4.5 liters of petrol Now we need to compare the energy needed to travel this distance and the energy consumed. Therefore, we need to know the energy density of petrol. Energy density of fuels is the amount of energy / kilogram or energy/liter that we get if we let this fuel react completely with oxygen. For fuels this is named the calorific value (Verbrandingswarmte [NL]) and for fluids, like petrol, you will find an upper and a lower limit/net calorific value (Bovenste Verbrandingswarmte BVW resp. Onderste Verbrandingswarmte OVW [NL]). In combustion engines we must take the lower limit. For petrol, the lower Limit is around 42 [MJ/kg] while the density is 800 [kg/m3]. πΈπππππ¦ππππ π’πππ = 4.5 β 0.800 β 42 β 106 = 151 [ππ½] πΈπππππ¦ππππππ = πΉ β π = 1500 β 45 000 = 67.5 [ππ½] 67.5 Now the efficiency is: 151 β 100 [%] = 45 [%] (This is an example, in reality it is more likely to be lower) Efficiency: ο¨= πΈππππππ πππππππ β 100 [%] = β 100 [%] πΈπ’π ππ ππ’π ππ In the formula above you can see that we can use the ratio of the energy (work) and/ or the ratio of the power because when we multiply the Power with the duration (time) we get the work and vice versa. The power we can take from the gearbox is less that we put into the gearbox. To illustrate this, we can make a 2-pole drawing. We will show this for our example with the dynamo on the bike. The biker can deliver 100W, 10 [%] of it is used for the lamp. PBiker=10 [W] Chapter 1 Drive line Pwheel=8 [W] Dynamo Energy in General Pdynamo=7.5 [W] Lamp PLamp=6 [W] page 12 HAN Saving Energy in a Vehicle READER WTW, WTT, TTW The efficiency we have calculated is known as the Tank-To-Wheel efficiency (or TTW). But to get the fuel in the tank we need more energy than we can take from it. Imagine the machinery to take the oil out of the soil, the refinery, the factory and all the transportation of the fuel through pipes and by trucks. So, you can even consider the energy for the production from the pipelines divided by the total amount of fuel that will pass through it and for these pipelines we need raw materials, transport, production etc. All this is known as the Well to Tank (WTT) efficiency Interesting article (Dutch): http://www.chemischefeitelijkheden.nl/Uploads/Magazines/h087Benzine.pdf (Werkt niet) In total we can make a chain from Well to Wheel (WTW). In India below you can find a calculation from “PluginIndia” where the polution of vehicles is a major problem in big cities. The QR-code on the right forwards to a video about transporting fuel. (https://youtu.be/K9m9WDxmSN8) Figure 11 The Well-to-Wheel calculated by Plugin India. (Pluginindia, 2017) Durability Global warming, greenhouse gases and pollution are problems which this and future generations will encounter. We must focus on durability to keep the environment safe and to keep the weather changes limited. That means we must focus on new energy sources that will use energy from renewable sources such as the sun, wind and water. At the same time, when we use fossil sources, we must try to keep the efficiency as high as possible. To give you an idea: you need 4-liter crude oil for 1 liter gasoline, the rest is converted into other products such as kerosene (jet fuel), lubricants etc. Chapter 1 Energy in General page 13 HAN Saving Energy in a Vehicle READER Other sources of Energy Here you can find the most common energy forms. Energy can be converted from one form into another form (losing some of its original energy). Kinetic Energy When a vehicle with a mass of 1500 [kg] has a speed of 45 [km/h] (see example) hits another vehicle you can see real damage. The vehicle has energy because of its speed and this energy is absorbed by the chassis during the collision. You can imagine how much energy you must deliver to cause the same damage to the cars yourself. Figure 12 Kinetic Energy? (Snyder, 2017) During your previous education the following formula must look familiar to you: 1 πΈπππ = 2 π π£ 2 We can derive this from the definition for Work. For those who understand integration, the full derivation can be found below. π = πΉ β π ππππ = πΉππππ π₯ πππ π‘ππππ πΉ = π β π (πππ€π‘ππ′ π πΏππ€) π = π£ . π‘ (πππ π‘ππππ = π ππππ π₯ π‘πππ) Now we use the more mathematical method for distance, we introduce ‘x’ variable. We express a small amount of Work as dW, so: ππ = πΉ β ππ₯ Here “d” does not mean distance but it is the derivate operator. If we want to know the total amount of energy, we must add all the small dW’s, this is known as taking the integral of the function. π = ∫ ππ = ∫ πΉ ππ₯ We can’t take the integral from F to dx, but now we use Newton’s Law and the formula for motion: ∫ πΉ ππ₯ = ∫ π β π ππ₯ And the acceleration (a) is the derivate of the speed (v); a=dv/dt ∫ πΉ ππ₯ = ∫ π β π ππ₯ = ∫ π β ππ£ ππ₯ ππ‘ And from π = π£ . π‘ (πππ π‘ππππ = π ππππ π₯ π‘πππ) we can see that ππ₯ = π£ β ππ‘ Chapter 1 Energy in General page 14 HAN Saving Energy in a Vehicle ∫ πΉ ππ₯ = ∫ π β π ππ₯ = ∫ π β READER ππ£ ππ£ ππ₯ = ∫ π β π£ ππ‘ = ∫ π β π£ ππ£ ππ‘ ππ‘ Therefore, we must find the integral from v with respect to dv which is ½ v2 and the mass is a constant. This means: 1 ∫ πΉππ₯ = π π£ 2 2 If you couldn’t follow the mathematical explanation, just remember the outcome. During the lessons in Toolbox and in the next few months you will understand how this process works. Spring energy All vehicles have springs for comfort. It can take the energy from mounds and bumps in the road. If you compress a spring, you must apply a force to it. As long as the force is present the spring is compressed, and as soon as you release the force the spring will expand and you will get energy in return. In vehicles this can lead to harmonic movement of the body, which will decrease the handling of the car. To prevent loss of control the energy is absorbed by shock-absorbers. Figure 13 Spring Energy. (CNX, 2017) In the picture above you can see that the force to expand the spring is a constant ‘k’ times the distance ‘x’. Instead of ‘k’ you can also find other characters (like ‘c’ and/ or ‘u’ instead of ‘x’). Now we can derive easier the spring energy because this is indicated by the area under the graph where F is expressed against the deformation ‘x’: 1 πΈπ πππππ = π π₯ 2 2 Here we can also use mathematics to find the energy. The energy is the integral of the force to the distance and we must express the force F as a function of distance (x). In figure 13 we can see that πΉ = ππ₯ se we get: 1 πΈπ πππππ = ∫ πΉ ππ₯ = ∫ π β π₯ ππ₯ = π ∫ π₯ ππ₯ = π β π₯ 2 2 This is exactly the same as the area under the graph if we take the spring force (F) as a function of ‘x’ Chapter 1 Energy in General page 15 HAN Saving Energy in a Vehicle READER Potential Energy Potential Energy (Height Energy) is the energy an object has because of its position. In ancient Greece men knew that solid rock thrown from a wall could really damage the enemy. That is because we must deliver work to lift the rock up and this energy is released as soon as we drop in down. This energy is proportional to the height and the mass. πΈπππ‘πππ‘πππ ~ π β β “Proportional to” means we can use an equal sign if we add a constant to it. In this case the constant is the gravitational acceleration ‘g’ because π β π is the gravity force: πΈπππ‘πππ‘πππ = π β π β β Chemical Energy This is already partly explained at the efficiency paragraph in this chapter. Fuels release heat when they react with oxygen. Normally we say that we burn fuels to get energy. This can be used for warming our homes, for cooking or to get mechanical energy in an engine. A part of the heat is used to vaporize the water that is used during oxidation and this heat is released as soon as the water condenses. In combustion engines, this normally happens in the open air, after the tailpipe. That means that we can’t use this extra energy. Therefore, we most use the lower limit of the calorific value (see efficiency section). Also, other chemical reactions can release energy, such as redox reactions in batteries. In this case the chemical energy transformed into electrical energy. Heat The general symbol for heat energy is ‘Q’ and its unit is the Joule. If you want to heat up material, you must put heat energy into it. The amount of heat energy depends on the rise of temperature, the mass of the material and a constant, the specific heat ‘c’. π = π β π β βπ οT is the ‘delta’ Temperature and means the change (rise) in temperature. Chapter 1 Energy in General page 16 HAN Saving Energy in a Vehicle READER Electrical Energy To calculate the electrical energy, we must know the voltage over the component and the current through it. The quantity voltage (U), expressed in the Unit Volt [V], is defined as the energy per unit of charge. Charge is expressed in Coulombs. One Coulomb is equal to 6.25ο1018 electrons (6 250 000 000 000 000 000 electrons…). So, the electrons can transfer the electrical energy. The more electrons we have, the more energy we get. The movement of charge per second through a wire is known as the current (I) expressed in Amperes (Amps) [A]. 1 ππππ‘ = 1 π½ππ’ππ π½ ππ 1 [π] = 1 [ ] 1 πΆππ’ππππ πΆ 1 π΄ππèππ = 1 πΆππ’ππππ πΆ ππ 1 [π΄] = 1 [ ] 1 π πππππ π If we multiply these: π½ πΆ π½ π β πΌ = [ ] β [ ] = [ ] = 1 [π] (πππ‘π‘) πΆ π π To know the total amount of electrical energy we only need to multiply the Power with time: ππππππ‘πππππ = ππππππ‘πππππ β π‘πππ = π β πΌ β π‘ If we switch on the wipers that take a voltage of 12 [V], and a current of 8 [A] for 2 minutes this will cost: π = π β πΌ β π‘ = 12 β 8 β (2 β 60) = 11 520 [π½] ππ 11.52 [ππ½] Electrical energy is often expressed in the unit [Wh] (Power * hours), the power is here: π = π β πΌ = 12 β 8 = 96 [π] π = π β π‘ππ βππ’ππ = 96 β ( 2 ) = 3.2 [πβ] 60 In homes the energy consumption is expressed in [kWh] = 1000 [Wh] = 3.6 [MJ] Chapter 1 Energy in General page 17 HAN Saving Energy in a Vehicle READER References Beem, R. (2017). Energie algemeen. HAN Automotive, Arnhem. Cambridge. (2017, 06 12). Cambridge Dictionary. Opgehaald van http://dictionary.cambridge.org/dictionary/english/energy CNX. (2017, 06 13). Potential Energy of a Spring. Opgehaald van http://cnx.org/contents/Xm00PHQL@5/Conservative-Forces-and-Potent Haas, P. d. (2017, 05 18). Youtube. Opgehaald van https://www.youtube.com/watch?v=jrF7NzGlZ-M Kaiserscience. (2019, 08 26). Opgehaald van https://kaiserscience.wordpress.com/physics/heat/internal-combustion-engines/ Pinsdaddy. (2017, 06 13). Bb Nsci 100 72 Work The Scientific Definition Openstax Cnx. Opgehaald van Pinsdaddy: http://www.pinsdaddy.com/bb-nsci-100-72-work-the-scientific-definitionopenstaxcnx_XfysRfZY0KE*XjDhUONNX1anyEoYRvQfmQFkbKZK9slR1rUtbkpvtj0sxboYgG%7CYJVjuvFR lKz7iLyWV9oe2*g/GWfsKwprYkGIxgSsF2*JKX*nx6I%7C8AHzr8Q6denVq2S3uwvE8bA%7CbRg hcczaA30huZl8RacG*QEe8y Pluginindia. (2017, 06 13). Pluginindia. Opgehaald van Calculating Well to Wheel efficiency: http://www.pluginindia.com/blogs/calculating-well-to-wheel-efficiency-of-electric-vehicles Schoolphysiscs. (2017, 06 13). Friction and Drag in Cars. Opgehaald van Schoolphysics: http://www.schoolphysics.co.uk/age1114/Mechanics/Forces%20in%20motion/text/Friction_and_drag/index.html Snyder, E. (2017, 06 13). Car Accident. Opgehaald van https://www.edgarsnyder.com/images/large550/car-accident/car-van-accident.jpg The next Corner. (2017, 05 17). Opgehaald van http://thenextcorner.com/i/mars-gda.png vandale. (2017, 05 18). Opgehaald van http://www.vandale.nl/opzoeken?pattern=energie&lang=nn youtube. (2019, 08 26). Opgehaald van https://www.youtube.com/watch?v=qmZtOm3Qb4A Chapter 1 Energy in General page 18 HAN Saving Energy in a Vehicle READER Chapter 2 Energy losses while driving Learning objectives of this lesson: Ability to specify the formulas for calculation of the various driving resistances and can apply these especially: • Rolling resistance • Hill resistance • Air resistance • Acceleration Resistance Understanding the meaning of acceleration resistance and how this relates to the additional mass because of the rotation of several components in the driveline Chapter 2 Energy in Losses in a Vehicle Page 19 HAN Saving Energy in a Vehicle READER Introduction Driving resistances are a part of vehicle dynamics. Vehicle dynamics deals with the forces acting on the vehicle and the consequences caused by these forces. These forces are: • driving resistance forces (driving resistances) • propulsive forces (drive forces) • brake forces • side forces • forces on the vehicle as a result of road irregularities (vertical) This module mainly focusses on the driving resistances and their impact on vehicle performance. The Driving Resistances. Anyone with experience in driving various vehicles knows that: • A car accelerates if you ‘step on it’ • A car with a "heavy" engine accelerates faster and has a higher top speed than the same model with a "lighter" engine • The acceleration of the vehicle is less if we pull a trailer or drive on an upward slope While driving a vehicle must overcome several resistances. These resistances are called driving resistances. The following driving resistances on a vehicle occur: • rolling resistance • slope resistance • air resistance (drag) • acceleration resistance • internal resistance of the powertrain These resistances will be discussed separately as well as their influence on the vehicle performance. The driving resistances are actually forces that must be delivered by the engine. While driving, the resistances exert a certain force on the vehicle. The engine must be able to deliver enough power to overcome this force. If the engine is able to deliver surplus power, then this can be used for acceleration. Once the engine delivers its maximum power to be used for overcoming only the driving resistances, then the top speed of the vehicle is reached. Chapter 2 Energy in Losses in a Vehicle Page 20 HAN Saving Energy in a Vehicle READER For a vehicle with trailer or caravan the resistances of the trailer are simply added, although a trailer might have a positive influence on the air resistance of the towing vehicle. Rolling resistance: Rolling resistance is caused mainly by deformation of the tyre on the road surface and partly by the deformation of the road itself. The deformation of the surface depends, of course, on the type of road surface that is being driven on. While driving through loose sand, the deformation is a rather large factor. To clarify the concept of rolling resistance we use the following approach: Figure 14 Rolling resistance (Kamerling, 2010) Let's look at a non-deformable wheel on a ramp. Arch A-B is the contact surface of the tyre with the road. The force that the road exerts on the wheel is assumed to be concentrated into a force (F*) at point C which points towards the wheel axle. In terms of mechanics, the wheel is in equilibrium, therefore: ο horizontal forces = 0: FH = Froll (= Fr) ο vertical forces = 0: FV = G ο moments about C = 0: πΊ β π = πΉππππ β π π From the last formula follows the rolling resistance equation πΉππππ = (π) β πΊ = ππ β π β π. The factor (e/r) is called the rolling resistance coefficient ππ , π is the mass [kg] and π is the gravitation constant [m/s²] the rolling resistance becomes: πΉππππ = ππ β π β π [N] ATTENTION: When the car is on a hill the normal force decreases and the rolling resistance needs to be compensated for this! Chapter 2 Energy in Losses in a Vehicle Page 21 HAN Saving Energy in a Vehicle READER The rolling resistance coefficient depends on the following factors: βͺ surface: type, material, condition βͺ tyre: type (diagonal or radial), material, condition βͺ tyre pressure βͺ tyre diameter βͺ vehicle speed The rolling resistance coefficient is not simply a tyre characteristic. It is a value associated with a combination of a certain tyre and road surface under certain conditions. Given the number of factors that influence the rolling resistance it is clear that it won’t be constant during driving. In practice however, one usually takes a constant value. The value of the rolling resistance coefficient f is determined by measurements. For calculations the values from the table below may be used Type of road surface rolling resistance coefficient froll concrete or asphalt paving 0.010 - 0.020 grass court or country road 0.05 loose soil 0.10 - 0.35 loose, dry sand > 1.9 Table 1 Rolling resistance coefficients on several road surfaces (Kamerling, 2010) The power required to overcome the rolling resistance is: πππππ = πΉππππ β π£ with: (3-7) Froll = Force rolling resistance [N] v = vehicle speed [m/s] Chapter 2 Energy in Losses in a Vehicle Page 22 HAN Saving Energy in a Vehicle READER Slope resistance Figure 15 slope resistance (Kamerling, 2010) If a vehicle drives on a slope, the force caused by the weight can be resolved in two components: • • with: a component perpendicular to the slope: πΊ β cos(πΌ) = π β π β cos(πΌ) a component parallel to the slope: πΊ β sin(πΌ) = π β π β sin(πΌ) (ο‘) = angle of inclination [o] G = weight [N] m = mass [kg] g = Gravitation constant [m/s2] The force parallel to the slope, the slope resistance Fh equals : πΉβπππ = π β π β sin(πΌ) Conversion from percentage to degrees Attention! The slope is normally given as a percentage instead of an angle. Then you need to convert this. Slope = 5 [%] then the angle ο‘ is tan−1 0.05 = 2.86 πππππππ [°] The power to overcome the hill resistance is πβπππ = πΉβ β π£ Chapter 2 Energy in Losses in a Vehicle Page 23 HAN Saving Energy in a Vehicle READER Compensation rolling resistance according to the hill Rolling resistance is caused by the force (weight) perpendicular to the road surface, as stated in the previous section. A vehicle on a slope therefore experiences a rolling resistance equal to: πΉππππ = ππ β π β π β πππ (πΌ) The total resistance of the vehicle due to the slope is the sum of the slope resistance and the rolling resistance: πΉππππ+βπππ = π β π β π ππ(πΌ) + ππ β π β π β cos(πΌ) The needed torque To be able to drive of on a slope, a force is needed at the driven wheels. This force comes as a torque that is ultimately delivered by the engine and transferred through the driveline. To get a drive force from the wheel torque we need the dynamic rolling radius: π = πΉ β πππ¦π The torque needed at the wheels can decrease/ increase for the torque needed at the final drive and or clutch depending on the gear ratios. When we have a gear ratio of 4:1 for the final drive the torque at the input is 4 times less. Example: A car with a weight of 1200 [kg] is on a slope of 15 [%], two wheels (rdyn=0.28 [m]) are driven and can deliver equal torque. The ratio of the final drive is 4:1, the first gear is engaged (3:1). Calculate • • • the hill resistance the torque at the wheels The engine torque First find the according slope in degrees: ο‘ = tan−1 0.15 = 8.5 [°] πΉβπππ = π β π β sin(ο‘) = 1746 [π] Needed torque is π = πΉ β πππ¦π = 1746 β 0.28 = 489 [ππ] Each wheel will need half of it because of the power split in the differential But for the total torque from the engine we count with 489 [Nm] The input torque from the gearbox is 489/4 = 122 [Nm] The input torque from the engine to the gearbox (via the clutch) is 122 / 3 = 40 [Nm] Chapter 2 Energy in Losses in a Vehicle Page 24 HAN Saving Energy in a Vehicle READER Air resistance The air resistance is a collective name for all aerodynamic resistances. A vehicle experiences three resistances of this kind: 1. Pressure resistance: while driving the shape of the vehicle creates air pressure differences on the vehicles’ surface. There is usually an overpressure on the front of a car, creating a pushing force, and an under-pressure at the rear, that creating a pulling back force. 2. Friction resistance: the speed of the vehicle driving through the surrounding air (a gas), causes the air to rub against the vehicle. This causes turbulences that produce frictional resistance. 3. Flow resistance: there is important air flow through a car that provides the cooling of the engine and the air supply of the interior. This air stream also causes resistance. All three resistances together form the air resistance. The magnitude of this resistance is experimentally determined in large wind tunnels. The total air resistance Fl follows from: 1 πΉπππ = 2 β ππππ β πΆπ β π΄ β π£π2 In which: πΉπππ = Force air resistance [N] ππππ = air density [kg/m3] πΆπ ππ πΆπ€ = drag coefficient [-] π΄ = frontal area [m2] π£π2 = relative air speed [m/s] The coefficient cd is measured in a wind tunnel. The frontal area of the vehicle is the effective crosssectional area: the projection of the vehicle in a plane perpendicular to the direction of motion. The relative air speed vr is the speed of the air relative to the vehicle. Headwind or tailwind must therefore be taken into account. In case the wind hits the vehicle at an angle the component of the speed in the longitudinal direction of the car must also be taken. If one wants to compare the air resistance of different cars, it's not enough to just compare the cd values. It is the product of the air drag coefficient and the frontal area (πΆπ β π΄) that determines the difference. The formula for the air resistance remains the same for a combination of a vehicle with trailer. However, the combination changes the cd value and it might change the frontal area. The power to overcome air drag is equal to: ππππ = πΉπππ ∗ π£ (Note that v οΉ vr) If πΉπππ is entered into this formula, it becomes clear that the power required to overcome the aerodynamic drag relates to the 3rd power of the speed. Therefore, the air drag is by far the most important resistance at higher speeds. Chapter 2 Energy in Losses in a Vehicle Page 25 HAN Saving Energy in a Vehicle READER In the table below the πΆπ = πΆπ€ value for several cars is given, as well as the frontal area and the product of both. Brand and type Cw value Alfa Romeo GT Alfa Romeo 3.2 V6 BMW 120 d Mercedes-Benz CLS 350 Mercedes-Benz B Chrysler 300 c 5.7 HEMI Peugeot 1007 Porsche 911 Porsche Carrera S Citroen C4 2.0 Audi A2 1.2 TDI Audi A4 2.0 TFSI Audi A6 3.0 TDI Audi Q7 0.34 0.31 0.30 0.31 0.34 0.31 0.29 0.32 0.25 0.33 0.30 0.28 0.37 0.37 0.26 Frontal area (A) 1.99 2.09 2.22 2.43 2.36 2.31 2.00 2.17 2.18 2.18 2.26 2.31 2.99 1.97 2.20 Cw* A 0.677 0.648 0.666 0.753 0.802 0.716 0.580 0.694 0.544 0.719 0.678 0.647 1.106 0.729 0.572 Table 2 Air drag coefficients of several passenger cars (Kamerling, 2010) Acceleration resistance When a vehicle accelerates there are two things that should be taken into account. The first is to accelerate a vehicle’s mass. The second is to accelerate the rotating parts such as engine crank, flywheel, parts of the transmission, drive shafts, differential and the wheels. In other words; the resistance due to acceleration is the effect of translational acceleration resistance as well as rotational acceleration resistance. For translational accelerations, Newton’s equation applies: πΉπππ = π β π where: m = mass of the vehicle [kg] a = longitudinal acceleration of the vehicle [m/s2] For rotational Newton’s equation applies (note this is a Torque, not a force): π = π½ β πΌπ where: π½ = mass moment of inertia [kg m2] πΌπ = rotational or angular acceleration [rad/s2] Chapter 2 Energy in Losses in a Vehicle Page 26 HAN Saving Energy in a Vehicle READER When we have a flywheel of 15 [kg] with a diameter of 40 [cm] the mass inertia moment is bigger than when we have a flywheel with the same mass with a diameter of 30 [cm] and when most mass is on the outer diameter the mass inertia moment is bigger than when most mass is centred. This means we need an additional torque to increase the speed of a flywheel and this torque is not transferred to the wheels. The vehicle behaves as if it is heavier than it is and we need to compensate this by adding additional mass to the vehicle mass. This additional mass is expressed in a rotation factor (οͺ), when οͺ = 1.2 the vehicles behaves as if it is 20 [%] more heavy than it’s curb weight. The rotation factor increases for lower gears because in lower gears we speed up rotational parts faster than in higher gears. The rotation factor is defined as follows: οͺ= m + m red m For the acceleration resistance then follows: πΉπππ = φ β π β π The table below shows guideline values for the rotation factor. Gear οͺ 1 2 3 4 5 1.25 - 1.50 1.12 - 1.20 1.07 - 1.12 1.05 - 1.07 1.05 Table 3 rotation factor οͺ, guidelines for passenger cars (Kamerling, 2010) References Kamerling. (2010). Rijweerstanden. Dictaat HTS-Autotechniek. Reimpel, J., & Sponagel, P. (1988). Fahrwerktechnik: Reifen und Räder. Würzburg: Vogel Verlag. Chapter 2 Energy in Losses in a Vehicle Page 27 HAN Saving Energy in a Vehicle READER Chapter 3: Power Conversions in Vehicles Study Material (in advance) Study this reader and read from Bosch Automotive Handbook (10th edition) the following pages/ paragraphs: 1. P35 Quantities and units 2. P39 Basic principles of mechanics 3. P812 -837 Drivetrain with hybrid drives Learning Objectives At the end of this lecture the student is able to apply the knowledge and skills related to: • • • • • • • • • • Name the most important forms of Power including calculation of the amount of Power. This counts for: o Chemical Power o Electrical Power o Mechanical Power o Heat Power Power losses Power transfers and converts Block diagrams for Power transfers in the driveline The importance of Energy and Power density for propulsion Understanding the relation between the Power en Torque characteristics. Both for combustion engines and Electric Motors Explanation why we need gears for vehicles with combustion engines Explanation when we need gears for vehicles with electric motors Difference between series, parallel and series/parallel hybrid vehicles Explanation why we need a minimum number of wheels and when we need multiple driven axles on it. Chapter 3 Power Conversions in Vehicles page 28 HAN Saving Energy in a Vehicle READER What is Power Power is the work done in one second. So, it’s not just the work that has been done, but also the time is relevant. π= π π½ [π] ππ [ ] π‘ π For many customers it is important to have a certain amount of Power in their car, because they think that more power means higher speed, better acceleration performances and they can pull a bigger trailer…and that’s true (in most cases…) First, we’re still used to the unit [hp], “horsepower”, it counts: 1 [βπ] = 0.735 [ππ] ππ 1 [ππ] = 1.36 [βπ] This counts for the metric / DIN units while an UK-hp is slightly more (745,7 W) and for electric power motors the conversion is 746W. But the standard unit for power is the Watt. Let’s look at an ordinary vehicle, how does the power flow from the tank to the wheels? FuelTank Clutch Engine Gearbox Final Drive Drive shafts Wheels Let’s start with the fuel tank, does it contain “Power”? Not really, it contains “energy”! But when we take an amount of energy in a certain time, we can say we take power from it. Let’s make a rough calculation for the Power efficiency. My car has a fuel consumption of 1 liter for 18 [km] of petrol when I drive 108 [km/h] at a constant speed. The drag forces are 450 [N] (air and rolling resistance). What can we say about the Power efficiency of this car? Answer: π£ = 108 [ ππ π ] = 30 [ ] β π π = πΉ β π£ = 450 β 30 = 13.5 [ππ] π 18000 πβπ£ 13500 β 30 πΈππ’π‘ πβπ‘ 8.1 β 106 π= = = ≈ ≈ ≈ 26.3% Μ πΈππ πΈππ’ππ πππ’ππ β πππ’ππβπΏπΆπ 1[π] β 0.75 [ππ] β 41 β 106 [ π½ ] 30.75 β 106 π ππ So, from tank to wheel we lose ≈ 73% of the energy/ power – this is converted into heat. Chapter 3 Power Conversions in Vehicles page 29 HAN Saving Energy in a Vehicle READER To find the efficiency of conversions we can compare energy-in versus energy-out of power-in versus power-out. Here we will use the Power more than the energy For the propulsion of a vehicle, we need mechanical power to spin the wheels. In order to drive a certain track or drive cycle, it is necessary to provide the mechanical power at the wheels during a certain time span. Therefor we need to have an energy storage system. To increase the driving range, we like to use an energy storage system with a high energy density. For that reason, chemical energy, stored in fuel tanks, is perfect because it has a very high energy density and we can easily fuel up. When we have electrical vehicles or hybrid vehicles, we also need electrical energy and this is usually stored in battery packs. The energy density is the amount of energy divided by the total weight for the storage. For fuel we need to take in account the appendages as well (tank, pump filter,…) as for batteries the housing and frame also contribute for the total weight πΈπππππ¦ π·πππ ππ‘π¦: π ′ = πΈ π [ π½ ] ππ Note the ‘accent added W = work, E = energy but they mean the same here… Besides the total amount of energy that we can take on the road it is also necessary that we can use the energy in a limited time to reach the desired power. When we need 150 [kW] to reach a top speed of 250 [km/h] we need a bigger engine then when we use a city car that has a top speed of 100 [km/h]. Simply by injecting more fuel we get more energy in one combustion stroke – but then we also need more air and as a result we need a bigger engine and/ or we need to use turbo’s to get more air in the cylinders. For batteries the manufacturer will say what the maximum current is the battery can deliver and this states the maximum power it can deliver. When we look at the power, we can take to the weight we need we talk about the power density: πππ€ππ ππππ ππ‘π¦: π′ = π π [ π ] ππ Depending on the choice of the energy storage system, we need to have a certain type of power converter to convert the power available from the energy storage into the required mechanical power. For example, the following converters are used for the vehicle propulsion: - ICE (Internal Combustion Engine) converts chemical energy into heat and mechanical energy Electric machines/ Motor-Generator (MG) converts electrical energy into heat and mechanical energy Hydraulic pumps [Hyd P] converts mechanical energy into heat and liquid flow energy Fuel Cell converts chemical power into electrical power πΜ πΏπΆπ π ICE βπππ‘ π πΌ π π MG π π π βπππ‘ πΜ πΏπΆπ pump βπππ‘ π βπππ‘ π Hyd. P πΌ H2 FC πΜ π π πΉ Wheel / hub βπππ‘ Figure 16 1Different types of power converters (G. Angelino - 2017) Chapter 3 Power Conversions in Vehicles page 30 π£ HAN Saving Energy in a Vehicle READER Where: ππ ππ • πΜ : is the fuel consumption [ β ] ππ [ π ] • LCV : is the lower calorific value [ππ] also known as π»π ππ π»π’ • T : is the torque [ππ] • π : is the angular velocity [ • • πΌ π : is the current [π΄] : is the voltage [π] • p : is the pressure [π2 ] • πΜ : is the volume flow [ • F : is the force [π] • π£ : is the velocity [ π ] π½ πππ ] π π π3 ] π π You can see that a dot above the quantity means it is related to the time, so normally ‘m’ means mass [kg] while πΜ means mass/hour or mass/second Further notice that every power conversion has a certain efficiency, and the power losses generate heat. Now that we know what the difference is between power and energy and know how the energy can be stored in a mobile application, we are now investigating the path of the energy flow from storage to movement. The main objective of the drivetrain is to spin the wheels in such a way that the desired performance can be accomplished. For now, we didn’t bother about the number of wheels we need for the propulsion. But The number of wheels which are driven by a powertrain depend on several factors, like: Single-, double- or triple track vehicles, on-road or off-road applications, maximum traction force for each wheel and is expressed like: • • • • • 4 x 2 4 wheels and 2 wheels are driven 4 x 4 4 wheels and all 4 are driven (off road vehicles, SUV’s etc.…) 6 x 4 6 wheels and 4 wheels are driven (Like weight trucks) 8 x 6 Indeed heavy transport 20 x 20 Very very very heavy transport… Normally we use more wheels when we need to carry more weight and we need more driven wheels when we need more traction or the traction for each wheel is to low (off road conditions). So, if we need a total force of 10 000 [N] to get up to a hill and each wheel can only deliver 3000 [N] of traction because the road is slippery a 4x4 will manage the job while ordinary cars won’t reach the top. Chapter 3 Power Conversions in Vehicles page 31 HAN Saving Energy in a Vehicle READER Drivetrain Most drivetrains in modern vehicles are front wheel drive with the engine and gearbox trans versed positioned in the vehicle and the final drive is combined with the gearbox. Figure 17 Drivetrain (Bosch Automotive Handbook 9th ed.) Block diagram front wheel drivetrain with the major efficiencies η(ICE)≈30% B Fuel tank T ICE LCV η(clutch)≈95% T Flywheel ω η(trans)≈98% T Clutch ω Transmission ω Axle T ω Differential Axle η(diff)≈95% Hub F Hub v Wheel F v Wheel Figure 18 Block diagram of the front wheel driven car (G. Angelino - 2017) The efficiency of the engine is very poor. From thermodynamics the efficiency is described as: ο¨π‘β = π πππ’π‘ =1− πππ πππ Where W is the work generated in the combustion chamber, Qin is the heat that is released when fuel is burned and Qout is the heat released to the exhaust. Then there is also friction before the power is transferred to the flywheel… we lose 70 [%] of the energy in this process. How can we lose 5 [%] in the clutch? Now it is more a combination of flywheel and clutch, springs need to fade out all the vibrations from the engine so most of the energy is absorbed by the springs in the clutch plate or the double mass flywheel. When we use torque converters the efficiency is even worse until the lock-up is activated The losses in the gearbox are mainly because of friction, as also for the final drive/ differential. But because the teeth in a differential are extremely hypoid, we have more friction in this section. Further we know that because of mass moments of inertia every rotating part contributes to energy losses when we want to accelerate and when we use bearing for these parts every bearing has a certain friction to overcome. But the total efficiency for this drivetrain will be: ο¨π‘ππ‘ = ο¨πΌπΆπΈ β ο¨πππ’π‘πβ β ο¨π‘ππππ πππ π πππ β ο¨ππππ = 0.30 β 0.95 β 0.98 β 0.95 = 0.265 Chapter 3 Power Conversions in Vehicles page 32 HAN Saving Energy in a Vehicle READER Some other drivetrains (made by G. Angelino, lecturer HAN Automotive): Drivetrain A: η(ICE)≈30% B T Fuel tank η(clutch)≈95% T ICE T Flywheel LCV ω η(trans)≈98% Clutch Transmission ω ω T ω Axle Axle Differential Axle η(diff)≈95% Hub F Hub F v Wheel v Wheel Drivetrain B: η(ICE)≈30% B Fuel tank η(clutch)≈95% T ICE T T Flywheel LCV η(trans)≈98% Clutch ω ω Transmission ω T ω powersplit T ω η(Psplit)≈95% Front Differential Axle Axle Rear Differential Axle η(diff)≈95% η(diff)≈95% Hub F Axle v Wheel Hub Hub F F v Wheel Hub F v Wheel v Wheel Drivetrain C: η(ICE)≈30% B T Fuel tank ICE LCV ω η(pump)≈95% Hydraulic pump η(hyd.motor)≈98% p Hoses Vο¦ Vο¦ F T p Hydraulic motor Hub ω v Wheel η(hoses)≈98% Chapter 3 Power Conversions in Vehicles page 33 HAN Saving Energy in a Vehicle READER Drivetrain D: η(ch)≈95% η(inv)≈95% I η(MG)≈90% I I charger Battery Inverter U η(red)≈98% T MG η(bat)≈92% T differential reduction ω ω U U η(diff)≈98% F v wheels AC DC DC Drivetrain E: η(inv)≈95% I Inverter η(ch)≈95% I charger I η(MG)≈90% T wheels MG U ω I η(MG)≈90% T F v Battery U η(inv)≈95% U η(bat)≈92% Inverter wheels MG ω U F v Drivetrain F: AC ο¦ m H2 tank I FC LCV I DC /DC U T I DC inv MG U U Wheel ω F v Battery Hybrid vehicles A hybrid vehicle is driven by two different energy sources but mostly we mean a fossil source and an electric source. When two different fossil fuels can be used it’s is commonly called bi-fuel. Hybrid vehicles are invented to lower the fuel consumption and to increase the efficiency. Many customers feel better when they drive in a hybrid and governments encourage them with fiscally benefits for the owner. Then these kinds of vehicles become more popular, and manufacturers will put more effort in the investigation towards hybrid. From hybrid to electrical vehicles looks like a small step but it is not that easy. The range for vehicles is important, even when most people drive less than 100 [km] a day, they want a range of 1000 [km] and a fast refueling time And an acceptable range and fast charging time result in heavy expensive battery packs and as a result expensive car for only a few lucky ones. Chapter 3 Power Conversions in Vehicles page 34 HAN Saving Energy in a Vehicle READER Two major differences for hybrid drivetrains exist: Series and Parallel. Series Hybrid Figure 19 series hybrid drive AH9th pag.730 A series hybrid system uses both an ICE (1) and Electromotor (3), but the ICE gets petrol from the tank (2) and drives the generator (3) that powers the Electro/Generator-motor (3), and the inverter (5) makes this possible. The electromotor relates to the final drive (4). It can be that a battery pack (6) is used to store recovered (brake) energy and/ or the take energy from the grid when it is parked. The advantage of a series hybrid with battery pack buffer is that we can drive with zero emission for e certain time and when we need to charge the batteries, we can use the optimum settings for the ICE. Parallel Hybrid Figure 20 parallel hybrid drive AH9th pag.728 Parallel hybrid cars can drive the final drive between 0%-ICE and 100% electric and 100% ICE and 0% electric. Often, they use a small or medium battery pack and the electric motor assist when the circumstances are poor for the ICE (low speeds, acceleration, etc…). The speed of both motors is the same, but the torque of both engines is added together. Series – Parallel Hybrid Finally, it is possible to combine these two options with a clutch (4) between the ICE driveline and the ICE-EM driveline ánd when they are connected a Power-Split is used (often with planetary gears – like Toyota Prius does – this will be explained later in semester three) Chapter 3 Power Conversions in Vehicles page 35 HAN Saving Energy in a Vehicle READER Figure 21 series-parallel hybrid drive AH9th pag.731 Power output for ICE and MG Now that we have a clear view about the powertrain of a vehicle and its corresponding functions, let’s have a better look at the power output of the Internal Combustion Engine (ICE) and the Motor Generator (MG). The mechanical power can be calculated as follow π =πβπ =πβ2βπβ π 60 Where π is the torque in [Nm], ο·= rotational speed [rad/sec] and π is the speed in [rpm]. With the use of a transmission, we can transform the elements of the power – here the torque and the speed. Although we always have power losses during a power conversion and we can reduce the power if required, we are not able increase power in the driveline. In case of mechanical power, we can use a transmission (also known as a gear) to transform the power elements. The gear ratio is obtained by: π= πππ πππ’π‘ ππ πππ’π‘ πππ In the literature, the gear ratio π can also be indicated as the inverse of the gear ratio mention above. The definition, which will be chosen for the gear ratio, can be different for some countries. But nevertheless, the principle is similar for both cases of course. Example: when the first gear ratio is expressed like 1 : 4.333 it is not meant that when the prise axle (connected to the flywheel) rotates one time the output shaft rotates 4.333 times – so actually they mean 4.333 : 1 or 1 : 0.231. Common sense will make you understand what they mean while first gear have normally a lower output speed than the input speed and fifth / sixth gear can have a lower speed and are named “overdrive”. When ο·ππ = ο·ππ’π‘ this is named “prise-direct”. During this course we look at the ICE as a black box, where we will mainly focus on what is going inand out the ICE. In the next term, we will investigate more what is happening inside the ICE. Page 458 of the AH9th depicts a torque and power curve as function of speed for the diesel and gasoline engine. Chapter 3 Power Conversions in Vehicles page 36 HAN Saving Energy in a Vehicle READER For the explanation of the torque, speed and power curves related to the vehicle propulsion, a few pages from the book “Modern Electric, Hybrid Electric and Fuel Cell Vehicles” (M. Ehsani, 2005) are copied and depicted below. Engine Performance Parameters (page. 70) The practical engine performance parameters of interest are power, torque, specific fuel consumption, and specific emissions. The maximum values for torque and speed are either flow limited (in naturally aspirated engines) or stress limited (in turbocharged engines). These parameters (torque and power) have both brake and indicated values. The difference between these two quantities is the engine’s friction (and pumping) requirements and their ratio is the mechanical efficiency. The relative importance of these parameters varies over an engine’s operation speed and load range. The maximum rated or normal rated brake power and the quantities such as brake mean effective pressure (bmep) define an engine’s full potential. The maximum brake torque (and bmep derived from it) over the full speed range indicates the engine’s ability to obtain a high airflow through itself over the full speed range and to use that air effectively. Over the whole operating range, particularly in those parts of the range where the engine operates for a long period of time, engine fuel consumption, efficiency, and engine emissions are important. Indicated and Brake Power and Torque (page. 71) The wide-open throttle operating characteristics of an SI engine are shown in Figure 22. The indicated power is the average rate of work transfer from the gases in the cylinders to the piston during the compression and expansion strokes. The brake torque is obtained by subtracting the friction power from the indicated power. The brake power shows a maximum value at about a speed slightly less than the maximum speed of the engine. The indicated torque shows a maximum value in the mid-speed range, which approximately corresponds to the speed at which the volumetric efficiency has the maximum value. The brake torque decreases more than the indicated torque at high speed because of more friction. Figure 22 Indicated and brake power, torques and specific fuel consumptions varying with the engine speed. Chapter 3 Power Conversions in Vehicles page 37 HAN Saving Energy in a Vehicle READER At partial load and fixed throttle position, these parameters behave similarly; however, at high speeds, torque decreases more rapidly than at full load as shown in Figure 23. The partially opened throttle causes more resistance to air flowing at a higher speed, and the volumetric efficiency decreases. The pumping components of total friction also increase as the engine is throttled. Figure 23 Torque characteristics with engine throttle opening and engine speed. Vehicle Power Plant and Transmission Characteristics (page. 33) There are two limiting factors to the maximum tractive effort of a vehicle. One is the maximum tractive effort that the tire-ground contact can support and the other is the tractive effort that the power plant torque with given driveline gear ratios can provide. The smaller of these two factors will determine the performance potential of the vehicle. For on-road vehicles, the performance is usually limited by the second factor. In order to predict the overall performance of a vehicle, its power plant and transmission characteristics must be taken into consideration. Power plant characteristics (page. 34) For vehicular applications, the ideal performance characteristic of a power plant is the constant power output over the full speed range. Consequently, the torque varies with speed hyperbolically as shown in Figure 2.10. At low speeds, the torque is constrained to be constant so as not to be over the maxima limited by the adhesion between the tire–ground contact areas. This constant power characteristic will provide the vehicle with a high tractive effort at low speeds, where demands for acceleration, drawbar pull, or grade climbing capability are high. Since the internal combustion engine and electric motor are the most commonly used power plants for automotive vehicles to date, it is appropriate to review the basic features of the characteristics that are essential to predicating vehicle performance and driveline design. Representative characteristics of a gasoline engine in full throttle and an electric motor at full load are shown in Figure 2.11 and Figure 2.12, respectively. The internal combustion engine usually has torque–speed characteristics far from the ideal performance characteristic required by traction. It starts operating smoothly at idle speed. Good combustion quality and maximum engine torque are reached at an intermediate engine speed. As the speed increases further, the mean effective Chapter 3 Power Conversions in Vehicles page 38 HAN Saving Energy in a Vehicle READER pressure decreases because of the growing losses in the air-induction manifold and a decline in engine torque. Power output, however, increases to its maximum at a certain high speed. Beyond this point, the engine torque decreases more rapidly with increasing speed. This results in the decline of engine power output. Figure 24 Figures for the ideal Power and Torque characteristic compared to the characteristics of an ICE and Electromotor. Page copied from “Modern Electric, Hybrid Electric and Fuel Cell Vehicles” (M. Ehsani, 2005) In vehicular applications, the maximum permissible speed of the engine is usually set just a little above the speed of the maximum power output. The internal combustion engine has a relatively flat torque–speed profile (compared with an ideal one), as shown in Figure 2.11. Consequently, a multigear transmission is usually employed to modify it, as shown in Figure 2.13 (next page indicated as figure 25 in this reader). Electric motors, however, usually have a speed–torque characteristic that is much closer to the ideal, as shown in Figure 2.12. Generally, the electric motor starts from zero speed. As it increases to its base speed, the voltage increases to its rated value while the flux remains constant. Beyond the base speed, the voltage remains constant and the flux is weakened. This results in constant output power while the torque declines hyperbolically with speed. Chapter 3 Power Conversions in Vehicles page 39 HAN Saving Energy in a Vehicle READER Since the speed–torque profile of an electric motor is close to the ideal, a single-gear or double-gear transmission is usually employed, as shown in Figure 2.14. Figure 25 Gears used with ICE driveline and Electromotors. Pages copied from “Modern Electric, Hybrid Electric and Fuel Cell Vehicles” (M. Ehsani, 2005) π₯ is the “speed-ratio” this is the ration between the maximum motor speed and the base motor speed. In figure 4.5 here above the maximum motor speed is set at 5000 [rpm] and when the base speed (the speed the torque in the motor is constant) is 2500 [rpm] the speed ratio is: π₯= max π ππππ πππ π π ππππ Chapter 3 = 5000 2500 =2 Power Conversions in Vehicles page 40 HAN Saving Energy in a Vehicle READER Variable-speed electric motor drives usually have the characteristics as shown in Figure 4.5 at the previous page. At the low-speed region (less than the base speed as marked in Figure 4.5), the motor has a constant torque. In the high-speed region (higher than the base speed), the motor has a constant power. This characteristic is usually represented by a speed ratio x, defined as the ratio of its maximum speed to its base speed. In low-speed operations, voltage supply to the motor increases with the increase of the speed through the electronic converter while the flux is kept constant. At the point of base speed, the voltage of the motor reaches the source voltage. After the base speed, the motor voltage is kept constant and the flux is weakened, dropping hyperbolically with increasing speed. Hence, its torque also drops hyperbolically with increasing speed. Figure 4.5 shows the torque–speed profiles of a 60 [kW] motor with different speed ratios x (x = 2, 4, and 6). It is clear that with a long constant power region, the maximum torque of the motor can be significantly increased, and hence vehicle acceleration and grade ability performance can be improved and the transmission can be simplified. However, each type of motor inherently has its limited maximum speed ratio. For example, a permanent magnet motor has a small x (> 2) because of the difficulty of field weakening due to the presence of the permanent magnet. Switched reluctance motors may achieve x > 6 and induction motors about x = 4. The use of a multigear or single-gear transmission depends mostly on the motor speed–torque characteristics. That is, at a given rated motor power, if the motor has a long constant power region, a single-gear transmission would be sufficient for a high tractive effort at low speeds. Otherwise, a multigear (more than two gears) transmission must be used. Figure 27 shows the tractive effort of an EV, along with the vehicle speed with a traction motor of x=2 and a three-gear transmission. The first gear covers the speed region of a–b–c, the second gear covers d–e–f, and the third gear covers g–f–h. Figure 26 shows the tractive effort with a traction motor of x=4 and a two-gear transmission. The first gear covers the speed region of a–b–c and the second gear d–e–f. Figure 28 shows the tractive effort with a traction motor of x=6 and a single-gear transmission. These three designs have the same tractive effort vs. vehicle speed profiles. Therefore, the vehicles will have the same acceleration and gradeability performance. Figure 27 Tractive effort vs. vehicle speed with a traction motor of x=2 and three-gear transmission Chapter 3 Figure 26 Tractive effort vs. vehicle speed with a traction motor of x=4 and two-gear transmission Power Conversions in Vehicles page 41 HAN Saving Energy in a Vehicle READER 6 Figure 28 Tractive effort vs. vehicle speed with a traction motor of x=6 and single-gear transmission Chapter 3 Power Conversions in Vehicles page 42 HAN Saving Energy in a Vehicle READER Chapter 4: Motor performance and efficiency Study Material (in advance) Study this reader and read from Bosch Automotive Handbook (10th edition) the following pages/ paragraphs: 4. 5. 6. 7. P75 Flow mechanics P460 Typical max. efficiency Thermal engines P461 Efficiency P952 – 957 Fuel consumption and power socket consumption Learning Objectives • • • • • Efficiency calculations of ICE, MG, Hydraulic machine and a fuel cell Know what influences the efficiency and making an estimated guess for common power transfers in automotive. Understand the relation between specific fuel consumption and the efficiency for combustion engines Know how to use an efficiency map for combustion-, electro- and hydraulic motors and fuel cells. Understand how the efficiency can be calculated for any power converter. Chapter 4 Motor performances and efficiency page 43 HAN Saving Energy in a Vehicle READER Please read the pages of the Bosch AHB 10th edition as mentioned above and also the following pages from the book “Modern Electric, Hybrid Electric and Fuel Cell Vehicles” (M. Ehsani, 2005), depicted below. Fuel Economy Characteristics of Internal Combustion Engines (page. 49, 50) The fuel economy characteristic of an internal combustion engine is usually evaluated by the amount of fuel per kWh energy output, which is referred to as the specific fuel consumption [g/kWh]. The typical fuel economy characteristic of a gasoline engine is shown in Figure 29 and in Figure 30. The fuel consumption is quite different from one operating point to another. The optimum operating points are close to the points of full load (wide-opened throttle). The speed of the engine also has a significant influence on fuel economy. With a given power output, the fuel consumption is usually lower at low speed than at high speed. For instance, when the engine shown in Figure 29 has a power output of 40 [kW], its minimum specific fuel consumption would be 270 [g/kWh] at a speed of 2080 [rpm]. Figure 29 brake-specific-fuel consumption (bsfc) for Power output of a typical gasoline engine For a given power output at a given vehicle speed, the engine operating point is determined by the gear ratio of the transmission. Ideally, a continuous variable transmission can choose the gear ratio, in each driving condition, to operate the engine at its optimum operating point. This advantage has stimulated the development of a variety of continuous variable transmissions, including frictional drive, hydrodynamic drives, hydrostatic drives, and hydro mechanical variable drive. Chapter 4 Motor performances and efficiency page 44 HAN Saving Energy in a Vehicle READER Figure 30 (brake) specific fuel consumption for Torque output of a typical gasoline engine Specific Fuel Consumption and Efficiency (page. 67, 68) In engine tests, fuel consumption is measured as a flow rate mass flow per unit time, πΜπ . A more useful parameter is the specific fuel consumption (sfc), which is the fuel flow rate per useful power output. It measures how efficiently an engine is using the fuel supplied to produce work. ππ ππ = πΜπ π ππ π = [πβπ ] ππ πππππ‘πππ → [ππβ] Where πΜπ is the fuel flow rate and P is the engine power. If the engine power P is measured as the net power from the crankshaft, the specific fuel consumption is called brake-specific fuel consumption (bsfc). The sfc or bsfc is usually measured in SI units by the gram numbers of fuel consumed per kW power output per hour [g/kWh]. Low values of sfc (bsfc) are obviously desirable. For SI engines, typical best values of bsfc are about 250 to 270 [g/kWh]. Normally, a dimensionless parameter that related the desirable engine output (work per cycle or power) to the necessary input (fuel flow) would have more fundamental value. The ratio of the work produced per cycle to the amount of fuel energy supplied per cycle that can be released in the combustion process is commonly used for this purpose. It is a measure of the engine efficiency (fuel conversion efficiency) as shown below: ππ‘ππ‘ = ππ ππ β πΏπΆπ = π πΜπ β πΏπΆπ = πΜπ ππ ππ πΜπ β πΏπΆπ 1 = ππ ππ β πΏπΆπ Where Wc is the work done in one cycle, mf is the fuel mass consumed per cycle, and LCV is the lower calorific value of the fuel (also known as NCV, Q HV ππ π»π’ , π»π , ππ ), which is defined as the heat released from unit fuel with complete combusting at standard conditions and the combustion products cooling down to their original temperature. The typical heating values for the commercial hydrocarbon fuels used in engines are in the range of 42 to 44 [MJ/kg] (11.7 to 12.2 [kWh/kg]). Chapter 4 Motor performances and efficiency page 45 HAN Saving Energy in a Vehicle READER Using the SI units, the efficiency can be expressed dimensionless as, ππ‘ππ‘ = π ππ 1 πΏπΆπ Energy Consumption electric drive (page. 114, 115) In transportation, the unit of energy is usually kilowatt-hour (kWh) rather than joule or kilojoule (J or kJ). The energy consumption per unit distance in kWh/km is generally used to evaluate the vehicle energy consumption. However, for ICE vehicles the commonly used unit is a physical unit of fuel volume per unit distance, such as liters per 100 [km] ([l]/100 [km]). In the U.S., the distance per unit volume of fuel is usually used; this is expressed as miles per gallon [mpg]. On the other hand, for battery-powered EVs, the original energy consumption unit in [kWh], measured at the battery terminals, is more suitable. The battery energy capacity is usually measured in [kWh] and the driving range per battery charge can be easily calculated. Similar to ICE vehicles, [l/100 km] (for liquid fuels) or [kg/100 km] (for gas fuels, such as hydrogen) or [mpg], or miles per kilogram is a more suitable unit of measurement for vehicles that use gaseous fuels. Energy consumption is an integration of the power output at the battery terminals. For propelling, the battery power output is equal to resistance power and any power losses in the transmission and the motor drive, including power losses in electronics. The power losses in transmission and motor drive are represented by their efficiencies π π and ππ , respectively. The efficiency of the electric machine can be obtained as follows. πππΊ = ππππβ πππππ = πβπ πβπΌ Depending on the type of electric machine (DC or AC) different formulas are used to compute the electric power. In practice, the battery output power (voltage and current) is mainly used as representative electric power. I should be noted that in this case, the efficiency of the inverter is also taken into account. Figure 31 Typical electric motor efficiency characteristics Chapter 4 Motor performances and efficiency page 46 HAN Saving Energy in a Vehicle READER Hydraulic power converter As you already know, it is also possible to use a hydraulic pump or motor to convert hydraulic power into mechanical power or the other way around. Please watch the following YouTube video about hydraulic transmission. This video gives a practical illustration of the working principle of a hydraulic transmission. You can also watch other alternative videos about hydraulic drive systems as well of course. https://www.youtube.com/watch?v=MTeJWE_Ou0g Please have a look at page 75 of the Bosch AHB 10th and read also the following text. On the next page, several graphs are plotted with respect to the efficiency of a hydraulic motor. Source: Bent Axis variable displ. Motors, Series 51 and 51-1 Figure 32 provides the volumetric and overall efficiencies for a typical Series 51 and 51-1 motor operating at maximum displacement, system pressures of 210 and 420 [bar] (3050 and 6090 [psi]), and a fluid viscosity of 8.2 [mm²/s] (53 [SUS]). These efficiencies can be used for all frame sizes. Figure 33 shows typical overall efficiencies for Series 51 and 51-1 motors operating at maximum displacement and system pressures up to 420 [bar] (6090 [psi]), and a fluid viscosity of 8.2 [mm²/s] (53 [SUS]). These efficiencies can also be used for all frame sizes. Figure 32 Overall and volumetric efficiency at maximum displacement The overall (total) efficiency is caused by hydraulic-mechanical friction losses ππβ and volumetric losses ππβ . ππ‘ = ππ£ππ β ππβ = Chapter 4 πππ’π‘ πππ Motor performances and efficiency page 47 HAN Saving Energy in a Vehicle READER The volumetric efficiency is the ratio between the theoretical volume flow and the real volume flow, the higher the system pressure the more leakage we can expect. The efficiency increases with the pump speed while there is less time for leakage anyway. ππ£ππ = πβ ππ πππ‘ππ β π πΜ Figure 33 Overall efficiency at maximum displacement The input flow of a hydraulic motor can be obtained as. πβ π β π πΜππ,πππ‘ππ = π π π£ππ Where: π = ππππ’π‘ ππππ€ [ π3 ] π this can also be written as πΜ ππ = π π‘ππππ π£πππ’ππ [π3 ] 1 π = π ππππ [π ] ππ πππππ‘πππ [πππ] ππ£ = π£πππ’πππ‘πππ ππππππππππ¦ [– ] π = ππ’ππππ ππ πππ’πππππ With the help of this input flow, we can calculate the input power of a hydraulic motor. πππ = βπβπ ππ π ππ£ππ π Where βπ = πβππβ − ππππ€ [π2 ] is the pressure difference between the in- and output flow of the machine. Chapter 4 Motor performances and efficiency page 48 HAN Saving Energy in a Vehicle READER Note that this formula is about the input flow of a hydraulic motor. For a hydraulic pump, the volumetric efficiency should be multiplied instead of divided. In that case, it will be output flow instead of input flow. The output power can be obtained as follows. πππ’π‘ = βπ β π β ππ β π β ππ£ππ Fuel Cell’s The basic principle of a fuel cell is the direct conversion from hydrogen and oxygen into electricity. The fuel cell works on the principle of an electrochemical process and has in basic no moving parts, only membranes. Only the peripheral equipment such as pumps and cooling fans have moving parts. The “fuel” for the fuel cell is hydrogen and oxygen and as long as these 2 gases are provided the fuel cell can provide an external device with energy. The emission of a fuel cell is clear water. One of the main problems with a fuel cell is the fact that hydrogen cannot be liquefied (unless cooled down to 252 [°C]) so the “fuel tank“ must be filled up to a very high pressure (up to 1000 [Bar]) and this sets high (safety) demands on the construction of the fuel tank. (e.g., collision resistant) Hydrogen is very flammable too. The Voltage of a single fuel cell is 0.5 to 0.8 Volt, so if we need a higher Voltage, we must stack the cells. The efficiency of a fuel cell is higher than a conventional combustion engine, the “fuel” is directly converted into energy, in contrast to a combustion engine where the fuel is converted into heat first. Hydrogen can be made in different ways: out of natural gas or by electrolysis of water. Figure 34 Fuel Cell systems The output power of a fuel cell is electric power, Voltage times Current. The input power of a fuel cell is the hydrogen fuel consumption (πΜπ»2 ) times the calorific value of hydrogen (πΏπΆππ»2 ). So, the efficiency can be calculated as follows. ππΉπΆ = πΜ πβπΌ π»2 βπΏπΆππ»2 Chapter 4 Motor performances and efficiency page 49 HAN Saving Energy in a Vehicle READER Overview total efficiency The total efficiency can also be calculated by looking at the ratio between the output and input power or the ration between the energy IN and the energy OUT. This counts in general for all types of engines whether it is an ICE, MG or hydraulic motor. For ICE motors the efficiency can also be calculated with the heat we put into the system (fuel) (Qin) and take out of the system (Quota) ο¨= πππ’π‘ πππ πΈ π = πΈ ππ = π = ππ’π‘ ππ πππ −πππ’π‘ πππ Engine ICE motor (πΌπππ ) Electromotor (πΌπππ ) Hydraulic motor (πΌπππ = πΌππ β πΌπππ ) Hydraulic pump = 1− πππ’π‘ πππ Pin Pout πΜ β π³πͺπ½ ο·π» πΌβπ° ο·π» βπ β π β π½π β π πΌπππ ο·π» ο·π» βπ β π π½π π πΌπππ πΜ β π³πͺπ½ πΌβπ° (πΌπππ = πΌππ β πΌπππ ) Fuel Cell (πΌπππ ) Chapter 4 Motor performances and efficiency page 50 HAN Saving Energy in a Vehicle READER Chapter 5 Aerodynamics Learning Objectives At the end of this lecture the student can apply the knowledge and skills related to: • • • • • • • • • static and dynamic pressure The relation between pressure and force Bernoulli’s Law Laminar and Turbulent Flows The meaning of the number of Reynolds The influence of the border layer around objects Methods to lower the air resistance The drag force coefficient Calculate the power needed to overcome air resistance Introduction Historically, aerodynamics has played a big role in transport. When the Vikings sailed around the world, they understood how the shape of the boat influenced its agility and speed and how to take as much wind as possible with their sails. The first cars didn’t bother about any aerodynamic rule at all, that was because the speeds of these cars were the same as walking. Later when scientists tried to best the speed record time after time the shape of cars changed from rectangular to drop-shaped. With airplanes it was more important to understand how the shape could influence the drag force or how it could create lift. Figure 35 a brief overview of shapes for transport ( (Carini, 2017) From all the drag forces the force to propel yourself through a medium like air doubles with the speed. That means the forces will be four times higher if the speed doubles. This is theoretical… in practice it might differ. In this lesson you will learn more about flow theory, Bernoulli’s law and a brief introduction into the principles of aerodynamics which is a very complex topic. We can easily define the factors which influence the size of the drag force through air. First of all is speed, we already mentioned that. Then of course the air density, or perhaps the air pressure… we’ll see. We then consider the shape… and the frontal area and whether or not we have holes in our car, does it have bumps and dents etc. etc. So, a lot to take in account. Chapter 5 Aerodynamics page 51 HAN Saving Energy in a Vehicle READER Bernoulli’s Law In simplified form, in fluid dynamics, Bernoulli’s principle can be described as a sum of pressures. Where the sum of all pressures is a constant – similar to the principle of conservation of energy we can’t lose pressure ergo it must have changed into another form of pressure… What kind of pressures exist? Gravitational pressure – hydrostatic pressure First of all, pressure because of height (or depth) – is called gravitational pressure. You can experience it when you are in an airplane or when you dive deep into the sea. In the first case the pressure is lower than you experienced before, in the other case the pressure increases. In water this phenomenon is also known as hydrostatic pressure The deeper you get the greater this hydrostatic pressure will be. If we assume just at the water level, we have 1 bar than the additional hydrostatic pressure at 140 cm deep is: πππππ£ = πβπ¦ππ = ο² β π β β πππππ£ = πβπ¦ππ = 1000 β 9.81 β 1.40 = 13.734 [ππ] = 0.137 [πππ] Where ο² (rho) is the specific weight (density), π is the gravity acceleration and β the height in meters! Figure 36 Hydrostatic pressure (Swimex, 2017) The unit of pressure is the Pascal [Pa]. One Pascal is 1 Newton per square meter [N/m2] which is 0.00001 [bar] Dynamic pressure When a medium, like a gas or fluid, moves the pressure changes. It doesn't matter if the medium moves or that we move inside the medium. Dynamic pressure depends on the relative speed. For instance, if we travel over the road we need to plough through the air; if we have tail wind, we can travel easier because the relative speed has decreased. If we go through water the forces are much bigger. So, it depends on the speed and the density of the medium. 1 πππ¦π = ο² π£ 2 [ππ] 2 Static pressure. The static pressure is present in a certain place at a certain moment. We are familiar with the absolute pressure at sea level which is around 1 [bar]. And if we press the brake pedal the pressure in the calipers can rise up to 200 [bar]. And the pressure in common rail diesel systems can reach over 2000 [bar]. Chapter 5 Aerodynamics page 52 HAN Saving Energy in a Vehicle READER There are multiple units to describe pressure, but in formulae you must always use the standard unit, that is the Pascal. Also, when it’s more suitable N/m2 is often used. • • • • • 1 Pa 1 bar 1at 1psi 1 torr 1 [N/m2] 0.00001 [bar] 1000 [mbar] 100 000 [Pa] 1013.25 [mbar] 101 325 [Pa] 68.95 [mbar] 6894.76 [Pa] 1.33 [mbar] 133 322 [Pa] (this is the Standard Unit) (atmosphere) (Pound per Square Inch) (1 [bar] ο»14.5 [psi]) (Torricelli = 1 [mm Hg]) Bernoulli’s Law in practice (example 1) Assume somebody carries a glass plate of 2 [m2] in the open air. If there is a side wind, what will be the force on the glass? Now we must know some values: • • • • Windspeed: 5 Beaufort (10 [m/s]) air density: 1.3 [kg/m3] Glass Area 2 [m2] Air Pressure: 1 [bar] Figure 37 (left) Carrying glass in the open air (Kerulit, 2017) First, we make a simple sketch: Figure 38 Simplified situation for the glass carrier (below) (Beem, 2017) 1 2 Wind Glass We can see three situations, 1, 2 and 3. Position 1 is far away from the glass plate, here the static pressure is 1 [bar] and the windspeed is 10 [m/s]. At position 2 the windspeed is zero and at position three the wind speed is also zero but the static pressure will there be 1 [bar]. The height of all positions is the same, so we don’t need to take the gravitational pressure in account, but for completeness we also include it in the formula. 3 You may choose your own zero line for the height (Where h = 0 [m]), in this case it’s practical to take the dashed line as h = 0 [m]. 1 1 π1 + ο² π β1 + ο² π£12 = π2 + ο² π β2 + ο² π£22 2 2 1 1 5 2 10 + 1.3 β 9.81 β 0 + β 1.3 β 10 = π2 + 1.3 β 9.81 β 0 + β 1.3 β 02 2 2 π2 = 100 065 [ππ] Chapter 5 Aerodynamics page 53 HAN Saving Energy in a Vehicle READER This doesn’t seem to be a huge difference… the pressure is only 65 [Pa] bigger than the static pressure. When we assume that the force all over is the same, we can calculate the force on the glass plate. If the pressure on both sides is the same the net-force is zero. So, we must look at the pressure difference across the plate. πΉ => πΉ = π β π΄ = 65 β 2 = 130 [π] π΄ Bernoulli’s law can be derived from the first law of thermodynamics, conservation of energy. π= Out of thermodynamics Work can be described as: π1−2 = π1 β π1 − π2 β π2 But when an object falls down to earth potential energy is converted into kinetic energy. Let’s combine these two in an equation for conservation of energy… 1 1 π1−2 = π1 β π1 − π2 β π2 = (π β π β β2 − π β π β β1 ) + ( π β π£22 − π β π£12 ) 2 2 Now write the start situation (1) on the left and the final situation (2) on the right… 1 1 π1 β π1 + π β π β β1 + π β π£12 = π2 β π2 + π β π β β2 + π β π£22 2 2 And we divide both sections by the mass (m) 1 1 2 2 π1 β π1 π β π β β1 2 π β π£1 π2 β π2 π β π β β2 2 π β π£2 + + = + + π π π π π π The π π (specific volume) is the reciprocal of the density; so π π = ο² that means π π = 1 ο² π1 1 π2 1 + π β β1 + π£12 = + π β β2 + π£22 ο² 2 ο² 2 Now multiply both sides with the density and we get: 1 1 π1 + ο² β π β β1 + ο² π£12 = π2 + ο² β π β β2 + ο² β π£22 2 2 Chapter 5 Aerodynamics page 54 HAN Saving Energy in a Vehicle READER Example 2 (Gravitational Pressure) Perhaps you’ve heard you can better pump water upwards instead of using a suction pump. That is because the hydrostatic pressure can’t be ignored here. If we lower the pressure above the water level the water will rise in the pipe, similarly to when you drink through a straw. But the water in the pipe wants to stream back because of gravitational forces. Let’s ignore the speed of the water, we just want to get the water out of the source. 1 1 π1 + ο² β π β β1 + ο² β π£12 = π2 + ο² β π β β2 + ο² β π£22 2 2 Position 1 is at the water level and position 2 is at the top of the pipe. The minimum pressure here will be 0 [bar] (theoretically, because you need a very good pump to reach this value…). Figure 39 How high can we get the water up? (Wikipedia, 2017) At position 1 the static pressure is 1 [bar], because that is the air pressure and the air pressure doesn’t change much in the ground. If we dig a hole 5 meters deep the air pressure will change as much as if we rose 5 meters up and the air pressure at the first or second floor on a building is essentially the same. (In reality the pressure changes logarithmic because air is compressible but roughly up to 3500 [m] the pressure declines with 1 [mbar] per 9 meter). The density of water is 1000 [kg/m3]. 105 + 0 + 0 = 0 + 1000 β 9.81 β β2 + 0 β2 = 10.19 [π] So, it’s impossible to take water out of the source when the water level is below 10 meters, in practice it gets difficult when the water is below 7 meters. In that case you cannot use a suction pump, but you must use a pressure pump or rise the water in stages. Chapter 5 Aerodynamics page 55 HAN Saving Energy in a Vehicle READER Example 3 (dynamic pressure) Before fuel injection was widely used in combustion engines many vehicles relied on carburetors. In a carburetor the fuel level in the inlet just didn’t flow out of the nozzle unless the pressure above the nozzle decreased. To achieve this, we need a narrowing in the inlet manifold to lower the pressure. This is called the Venturi. Figure 40 The carburetor principle. (Madhvan, 1017) To understand this, we need to use…Bernoulli’s principle! On the right we focus on the venturi. Because the mass airflow must be a constant everywhere (otherwise we accumulate air somewhere). This means that de speed in the wide section is lower than the speed in the narrow section. Figure 41 The venturi. (Instrument, 2017) For air we can have a deviation in specific weight, for fluids the speed is related to the cross sections. Let’s take two points, 1 in the wide section and 2 in the narrow section. The height of both points is the same so we can ignore the gravity pressure 1 1 π1 + ο² β π β β1 + ο² β π£12 = π2 + ο² β π β β2 + ο² β π£22 2 2 Now we can see, that if we have atmospheric pressure at the beginning, where the speed is low we must have a lower pressure at point 2 where the speed is higher! Because: 1 1 π1 + ο² β π£12 = π2 + ο² β π£22 2 2 π1 + π ππππ = π2 + πππ π2 < π1 Summary. So, we have two different kind of pressures. Static pressure which is partly gravitational and partly because of circumstances, mostly because π = πΉ π΄ or because of the air pressure that surrounds us. Dynamic pressure because of the flow from the medium (gas / fluid). Within a flow the sum of the pressures is a constant. That means that if the static pressure rises, the dynamic pressure lessens and vice versa. Chapter 5 Aerodynamics page 56 HAN Saving Energy in a Vehicle READER Aerodynamics in practice We can imagine that if we make cars out of square boxes the aerodynamics of the vehicles would be horrible. Suffice it to say that small children have no knowledge of aerodynamics at all. In this case the drag forces because of the air would be very high. All the air would slow down completely so the speed of air would be zero and create a high thrust pressure (Stuwdruk [NL]). A better arrangement is to guide the air around the obstacle and make it more aerodynamic. Figure 42 a car in the eyes of a child. (Merriam, 2017) The border layer (very important) Actually, the speed of the medium at the border layer (grenslaag [NL]) of the object is always zero… This insight is sometimes hard to understand, because…how can we blow away a fluffy if the airspeed near the fluffy is zero? That has to do with pressure. The pressure on the fluffy, how small this fluffy might be, acts a force to it and so you can blow it away. But the influence of the border layer on the total friction force that the object experiences is a new science. Therefore, the idea to polish the surfaces of objects to lessen the drag forces is not always true. Figure 43 Smooth surfaces are not always better... (rideapart, 2017) Outside the border layer, the medium behaves like a frictionless surface. Within the border layer the friction causes a drag force. The dimples on the surface of a golf ball makes it easier for the air to leave the surface, the border layer becomes thinner and the friction is lower. Lower friction means that the air is guided better around the surface and the pressure on the other side of the ball is less low than without dimples, so the drag forces are lower… This phenomenon was discovered in 1910 by Prandtl. Pressure around the object We saw by the example with the glass plate there was a pressure because of the wind. We assumed the pressure on the other side of the plate was the same as the static pressure. If air goes around an object this can result in a lower pressure at the end of the object. Mostly that is the point where the air leaves the border area which is at the point of the lowest pressure. That Chapter 5 Aerodynamics page 57 HAN Saving Energy in a Vehicle READER means, if we look at the golf ball above the pressure on the right, near the ball is lower than the pressure further away from the ball. We create a low pressure point and that creates an air stream to fill up this low pressure. And to create an air flow… this takes energy, so the drag forces also depend on the pressure after the object. If you drive close behind a truck you can profit from this air-stream. If you are further away from the truck turbulences can make the drag forces higher. For the total air resistance, it’s not only important to look at the front side, but also at the tail side of the object. So, make the border layer very thin, guide the air well around the object and try to make the air stream as small as possible. Figure 44 Air stream at the tail. (Lammertse, 2008) Turbulence and Laminar In the drawing before you have seen the term laminar and turbulent. In a laminar stream each particle will follow the same path as its predecessor. When turbulence occurs, this will not be the case. If turbulence occurs depends on the diameter of the tube, the viscosity and the flow speed of the medium. If you want to know more about this, please search on the world wide web for the Reynolds number. . Chapter 5 Aerodynamics page 58 HAN Saving Energy in a Vehicle READER Drag Coefficient (cd) De cw waarde – (weerstands coëfficient [NL]) In the 1970s one would think that the wipers on the windshield had to have had a dramatic influence on the air resistance. Pictured here you can see the Volkswagen Golf1. And we can also see that ahead of the wipers we have the ‘paravane’ that looks like a dimpled surface where the border layer lifts off from the bonnet. In practice the wipers didn’t have much influence on the air resistance at all. Figure 45 The knick between bonnet and windshield from a VW-Golf-1. (Ralfons, 2017) The drag force coefficient is the relation between the drag force and the dynamic pressure and can derived in experimental way if we can measure the resistance force for a certain airspeed. Figure 46 cd values for different types of transport. (Wikipedia, 2017) Formula 1 cars do have a cd value near or more than one. That is because the want to produce downforce that makes it possible to corner faster. The power in the engine is not only used for speed but also for handling. But the air resistance does not only depend on the drag coefficient and the (relative) speed but also from the frontal surface / area of the vehicle. And now we have all the ingredients to calculate the air resistance: 1 πΉπππ = ππ β π΄ β πππ¦π = ππ β π΄ β ο² β π£ 2 2 A few tips: • • • • • Avoid sharp edges at the front when you need to use obstacles (like wipers…) make a dimple or edge in your construction to release the border layer. Let the frontal area increase slowly at the front, but also at the rear! Look at good practices on the internet. The drag force coefficient has the minimum value when you can reach the speed of sound. If not, then review the tips above Chapter 5 Aerodynamics page 59 HAN Saving Energy in a Vehicle READER Example drag forces A Mc Laren F1 has a power of 465 [kW] (around 630 [hp]) and has a top speed of 380 [km/h]. How much power is needed to overcome only the air resistance? We need some numbers of course: ππ ] π3 ππ = 0.32 [−] π΄ = 1.8 [π2 ] ο²πππ = 1.3 [ First, we calculate the speed in [m/s]: π£= 380 = 105.5 [π/π ] 3.6 Now we fill in the formula for the Air resistance: 1 1 πΉπππ = ππ β π΄ β ο² β π£ 2 = 0.32 β 1.8 β β 1.3 β 105.52 = 4.17 [ππ] 2 2 The power is Force times velocity: π = πΉ β π£ = 4.17 β 105.5 = 440 [ππ] This means that only the air resistance takes 95 [%] of all the power and the other frictional resistances take 5 [%]. This shows that air resistance is one of the most important drag forces for high speeds. Figure 47 Mc Laren F1. (NickD, 2017) Chapter 5 Aerodynamics page 60 HAN Saving Energy in a Vehicle READER Law of conservation of mass Aerodynamics is a part of the general subject of Fluid Mechanics. In automotive technology the motion of fluids also plays an important role. If we are dealing with a steady flow, then the amount of mass that flows per unit time in a flow field remains constant. This is a very important concept and is the so-called law of conservation of mass in stationary flow. If the flow is a liquid, the same applies to the volume flow because a liquid is effectively incompressible (not usually the case for gases!). The law of conservation of mass is also known as the law of continuity. Definitions: ο© kg οΉ οͺ m mass flow or mass flux οͺ οΊ ο«s ο» ο© m3 οΉ οͺV volume flow or volume flux οͺ οΊ ο« s ο» ο ο A flow cross-sectional area m 2 ο©mοΉ v average speed οͺ οΊ ο«sο» ο© kg οΉ ο² specific mass (density) οͺ 3 οΊ ο«m ο» For flow through a cylindrical pipe it applies that: οͺ m = ο² οͺ v = ο² A v in which the flow cross-sectional area is A = 14 ο° d 2 If multiple pipes come together and further on divide again into other pipes, the total incoming mass flow is equal to the total outgoing mass flow (compare to Kirchhoff’s Laws in Electrical Theory). (ο₯ οͺ m )in = (ο₯ οͺ m )out so, for example for three incoming and two outgoing pipes: οͺ m1 + οͺ m2 + οͺ m3 = οͺ m4 + οͺ m5 ο 1 4 ο° d12 v1 + 14 ο° d 2 2 v2 + 1 4 ο° d 3 2 v3 = 1 4 ο° d 4 2 v4 + 1 4 ο° d 5 2 v5 If it is one liquid, one can divide by the density because of the incompressibility (so in the second equation the density has already gone). Chapter 5 Aerodynamics page 61 HAN Saving Energy in a Vehicle READER Such a combination will probably hardly occur in practice, but a pipe that splits in two pipes or the joining of pipes into one pipe are constructions that we regularly see in practice. See the figure below. Figure 48 Conservation of mass means you need to add the same mass as you take from it. In the left part of figure 48 we see a pipe that narrows. Based on the continuity rule, we can deduce the following: ππ1 = ππ2 ⇒ π1 β π΄1 β π£1 = π2 β π΄2 β π£2 Because liquids are incompressible there follows π1 = π2 So π΄1 β π£1 = π΄2 β π£2 2 π π£2 = (π1 ) π£2 2 1 πβπ12 4 βΉ βΉ β π£1 = π1 > π2 1 πβπ22 4 β π£2 βΉ βΉ π12 β π£1 = π22 β π£2 π£2 > π£1 In the right part of figure 48 we see a pipe that splits and therefore: (ππ )1 = (ππ )2 + (ππ )3 ⇒ (ππ )1 = (π β 14π β π2 β π£) 2 Chapter 5 (ππ )1 = (π β π΄ β π£)2 + (π β π΄ β π£)3 + (π β 14π β π2 β π£) 3 Aerodynamics page 62 HAN Saving Energy in a Vehicle READER Chapter 6 Thermodynamics I (Basics and First Law) Introduction In this part of the course, we will look at the topic of Thermodynamics which plays an important role in combustion engines. Although there are all kinds of options for alternative engines, it seems, for the time being, that combustion engines in vehicles will still be on the street. But we also need Thermodynamics in material science and electrical engineering. Everybody knows that when batteries are charged the temperature increases in this chemo-electric process. And when dealing with heat transfer (inside materials) thermodynamical processes take place. In this chapter we will only look at the Thermodynamics of processes involving gasses (not that of liquids or solids) Thermodynamics definitions The word Thermodynamics comes from twee Greek words: thermos = temperature and dynamos = force (think of the word dynamite) Three important thermodynamical quantities are pressure p, temperature T and volume V Often processes are draw in a pV-chart where V is de x-coordinate and p the y-coordinate In thermodynamics, units that do not originate from SI are still often used 1 [πππ] = 1000 [ππππ] = 105 [π/π2] 1 [ππ‘π] = 76 [ππ] ππππ π π’ππ ππ πππππ’ππ¦ = 1013 [ππππ] T (in [K]) = 273.15 + t (in [°πΆ]); usually the rounded-off value 273 is used, so that: 0[°πΆ] = 273 [πΎ] Ideal Gas Law The simplest way to describe a gas is to interpret it as a large set of very small particles. Particles that are point masses without any interaction between each other (no potential energy) only collisions between the particles (exchange of kinetic energy) Kinetic energy of the particles is the microscopic quantity which has to do with the macroscopic quantity temperature. As the particles go faster, the temperature increases (increase of kinetic energy) and as the particles move slower, the temperature decreases. It is shown that the following equation applies: ππ = ππ π Whit pressure p, volume V, mass m, gas constant R and temperature T This law is called the ideal gas law. Ideal gases do not exist because each gas (when cold enough) finally condenses at a certain pressure and temperature. But if a gas, at some temperature, is far from its condensation point, this comparison is good. Chapter 6 Thermodynamics (part 1) page 63 HAN Saving Energy in a Vehicle READER Air at 1 atm and room temperature behaves well as an ideal gas because the main constituents of oxygen and nitrogen condense at very low temperatures (-183 [°C] and -196 [°C], respectively) In many calculations, this law can only be used. If one considers a closed system, the mass remains constant and, of course, the gas constant. If you go from a starting situation 1 to a second thermodynamic situation 2 and then 3 etc. then one can write: π1 π1 π2 π2 π3 π3 = = = ......... π1 π2 π3 Figure 49 ideal gas: no internal friction nor cohesion Gas constant The quantity R is the so-called specific gas constant which is different for each gas. The unit is [J/kgK] This can be seen as follows because the product pV has the unit of energy [π’πππ‘ π] ⋅ [ π’πππ‘ π] π = ππ ππ ⇒ π ] ⋅ [π3 ] = = [ π ⋅ π] = [ π½ππ’ππ ] = [ π½] π2 π½ [ π’πππ‘ π ] = [ ] =[ πππΎ Specific heat c is the specific heat of a substance in [J/kgK]. It applies in most cases that the specific heat is (near enough) independent of the temperature. In many calculations, use is made of cp and cv, the specific heats at constant pressure and volume. (For solids and liquids, it applies that c depends on the temperature.) Frequently this is a near-linear relationship e.g. π(π) = π0 + π1 π (with a0 and a1 coefficients to be defined), but it may also be approximated by a quadratic equation. This depends on the accuracy that is aimed for.) Note the different notations in the formulae that define c. π12 = π π ( π2 − π1 ) (if c is constant) π = Chapter 6 π₯π π π₯π π = Thermodynamics (part 1) ππ π ππ page 64 HAN Saving Energy in a Vehicle READER Conservation of Energy – First Law of Thermodynamics Just as in other physical subjects, the Law of Conservation of Energy plays an important role in Thermodynamics. This law is therefore titled the First Law of Thermodynamics. (The Second and Third Laws also exist, but these are not covered in this series of lessons.) Many kinds of energy may be defined, such as nuclear, chemical, mechanical (work), heat and electrical energy. All types of energy may be traced finally to the kinetic and potential energy of the microscopic structure of the material and the interaction among elementary particles (but we will not go that far in this course). We will restrict ourselves in this part to three types of energy, namely Q = heat, W = work (mechanical energy) and U = internal energy. Internal energy is the total amount of kinetic and potential energy of the molecules. For an ideal gas, this is only the amount of kinetic energy, and it may be derived that U is a function of the temperature: π12 = π ππ ( π2 − π1 ) or expressed as π₯π = π ππ π₯π . Application of the Law of Conservation of Energy to Q, W and U yields the equation: π12 = π12 + π12 or expressed as π₯π = π₯π + π₯π Special processes in a pV diagram In thermodynamical technology there are often processes where one of the quantities of pressure, volume or temperature remains the same. These processes have the following names. (the prefix iso is derived from the Greek word isos which means equal) Isobar process at equal pressure Isotherm process at equal temperature T = constant Isochore process at equal volume Chapter 6 p = constant V = constant Thermodynamics (part 1) page 65 HAN Saving Energy in a Vehicle READER Heat energy and work Heat energy is represented by the letter Q and mechanical energy (= work) by the letter W. If there is a process of a thermodynamic situation 1 to another situation 2 (for example, because the pressure or temperature changes) then the heat energy supplied or removed is noted as Q12 and the work done by the system as W12 (the same for situation 2 to 3, 3 to 4 Q23, Q34, Q45 ....... etc.) The following is defined for positive or negative energy Q12 > 0 positive heat-energy at heat supply Q12 < 0 negative heat-energy at heat dissipation (emission) W12 > 0 positive work on expansion (volume increase) Work is being done by the system W12 < 0 negative work on compression (volume decrease) Work is being done on the system U12 > 0 increase in internal energy Temperature rises (particles move faster) U12 < 0 decrease in internal energy Temperature drops (particles move slower) Normal volume An important thermodynamic reference point (p0 , V0, T0 ) of a gas is the situation under so-called normal conditions, namely p0 = 1 [atm] and t0 = 0 [°C]. The volume that the gas occupies in this situation is called the normal volume V0. These normal conditions are easy to achieve in a laboratory (why?). Different thermodynamic situations can be easily compared together with each other with regard to this reference point. Chapter 6 Thermodynamics (part 1) page 66 HAN Saving Energy in a Vehicle READER Adiabatic process In nature and technology, there are often thermodynamic processes that take place so quickly that there is no time to exchange heat with the environment. These processes are called adiabatic processes. π₯π = 0 The thermodynamic environment in which this happens is considered a “closed system”. (there is no matter and no energy exchange) Filled in the First Law of Thermodynamics, this provides: π₯π = π₯π + π₯π ⇒ 0 = π₯π + π₯π ⇒ π₯π = − π₯π Work and internal energy are opposite to each other. (the minus sign is important!!) This means the following: Adiabatic expansion Adiabatic compression ο ο provides work W12 > 0 ο temperature drops U12 < 0 negative work W12 < 0 ο temperature increases U12 > 0 If an adiabatic process is thermodynamically reversible between two situations, that process is called isentropic. In this lesson series we will not go into the concept of reversibility, but it is important to know the two concepts. The following statements can be made: All isentropes are adiabatic processes but not all adiabatic processes are isentropes. (compare that to the statement: all squares are rectangles but not all rectangles are squares) (An isentrope is a process of constant entropy. The term 'entropy' is not discussed in this first-year course but is a subject in the second year) Chapter 6 Thermodynamics (part 1) page 67 HAN Saving Energy in a Vehicle READER CHAPTER 7 Thermodynamics II (Cyclical Processes and Polytropes) Cyclical processes Cyclical processes occur in combustion engines. If from a thermodynamic starting situation, after a number of sub-processes, the same situation is returned to then a cyclical process has taken place. In calculations a so-called energy balance is often drawn up. In this energy balance the work, heat added or emitted and the internal energy is calculated for each sub-process (usually expressed as amount of energy per unit mass, [J/kg]). Important here is the Law of Conservation of Energy, the First Law of Thermodynamics. Because internal energy is a situation-dependent quantity, this energy is unaltered after a cyclical process, and so the total work done is equal to the algebraic sum of the supplied and emitted heat (follows from the Law of Conservation of Energy). So ∑ π = ∑ π = (∑ π)π π’ππππππ − (∑ π)ππππ‘π‘ππ If the total work is positive then this is called a positive cyclical process, and for negative work, a negative cyclical process. The ratio between the total W and the Q supplied finally yields the thermal efficiency. ππ‘β = ∑π ∑π (∑ π)π π’π − (∑ π)ππππ‘π‘ππ (∑ π)ππππ‘π‘ππ = = =1− (∑ π)π π’π (∑ π)π π’π (∑ π)π π’π (∑ π)π π’π As already mentioned, thermodynamical processes are draw in a pV-chart (or pV-diagram) where V is de x-coordinate and p the y-coordinate. If the pressure and volume are measured at every moment in a cyclic process like the four-stroke cycle of a combustion engine the pV- diagram becomes the so-called indicator diagram. How to construct this indicator diagram will be shown in the lesson. This diagram can be simplified using the First Law of Thermodynamics ( Conservation of Energy ) An amount of positive work can be conveniently crossed out with the same amount of negative work. The remaining processes can be approximated by so-called polytropes.(see paragraph below ) How this can be done will be shown in the lesson. Figure 50 Four stroke cycle of an Internal Combustion Engine. Chapter 7 Thermodynamics (part 2) page 68 HAN Saving Energy in a Vehicle READER Polytropic process A polytropic process is a thermodynamical process of an ideal gas that obeys the relation: π ππ = πΆ Where n is the polytropic index and C is a constant. By making use of the ideal gas law, this equation can be transformed to an equation with T, V or p, T π ππ = πΆ ππ π π π−1 = πΆ ππ−1 = πΆ Another definition can be given as follows: A polytrope is a process where the specific heat capacity remains constant. Every polytropic process can be characterized by its index and specific heat capacity. The following processes are of special interest in Thermodynamics: isobar n=0 c = cp isotherm n=1 c=∞ isentrope (adiabatic) n = k = cp/cv c=0 isochore n=∞ c = cv Figure 51 Basic 4 polytropic processes in the pV diagram (Source is from the book: Warmteleer voorTechnici page 60, Author: Ir. AJM van Kimmenaede, Chapter 7 Thermodynamics (part 2) ISBN 90 401 04492) page 69
