Basic Thermodynamic Concepts-Thermodynamics-Handout, Exercises of Thermodynamics

Download Basic Thermodynamic Concepts-Thermodynamics-Handout and more Exercises Thermodynamics in PDF only on Docsity! Chapter 3: Basic Thermodynamic Concepts Introduction Thermodynamics is often applied in a rather pragmatic way with more concern for useful answers than a rigorous treatment of the basic science. This is because “real” thermodynamic processes defy full analysis, so we restrict ourselves to a useful combination of basic science and experience gleaned from years of application. We treat thermodynamics as part of the art of applying science profitably - the essence of engineering. As engineers, we deal with many types of machine. “Thermodynamic machines” are those in which heat flows or temperature changes are a significant part of their operation (almost all machines have heat flows, if only because of friction). This obviously covers “engines”, but includes other common machines like compressors and turbines. Thermal processes are similar in most thermodynamic machines, with gases or vapours as the usual working fluids. Underlying our treatment of these processes are basic concepts you need to understand. These are outlined below to ensure you have a good grasp of the concepts, many of which you may previously have taken for granted. Some will be examined more rigorously later. At this stage, you need only familiarize yourself with their simple form. Energy People often use terms like “energy”, “inertia”, “time”, etc. without any clear understanding of what they are. Often, a full definition is difficult. Energy is a convenient generic term for “something” which is transferred in the doing of work. For our present purposes, we need to note that energy can be converted between different forms. Those of most interest are: potential energy (via gravity, or “spring” storage, including pressurized substances like gases), kinetic energy, chemical energy, and internal energy (which is really kinetic energy of molecules, but we treat it separately from “bulk” kinetic energy). We are not generally concerned about other forms such as electrical, magnetic, etc. It is useful to note that all the forms above can “store” what we call energy. “Work” can not, so it is only a transient manifestation of energy, it is a process of transferring energy rather than a form. For similar reasons, so is “heat”. The thermodynamic term “heat” refers to “heat flow”. Heating and Cooling Heating and cooling both involve transfer of energy across a boundary between two “bodies” by means of a temperature difference, e.g. a cold drink and the atmosphere. For energy to “flow” as heat we need three things: 1. a flow path with finite cross sectional area (A) to convey heat, 2. a non-zero temperature difference (ΔT), and 3. non-zero time. The amount of heat transferred will be proportional to the product of these three things. In practical situations, if any of these is zero or very small, we may treat the heat flow as negligible even though it may not be zero. In thermodynamics, “Heating” and “Cooling” refer to the process of transferring heat across a system boundary. A change in temperature is not essential. A system’s temperature can also be changed by other kinds of process (e.g. compressing or expanding a gas). It is even possible to heat a gas and have its temperature decrease! Heat Transfer within Solids Some basic understanding of heat transfer is needed to understand thermodynamic machines, and what affects their performance. The three main mechanisms for heat transfer are Conduction (mostly in solids), Convection (in fluids) and Radiation (electromagnetic). In many engineering applications, the combined effect of these is often simplified into a single “Overall Heat Transfer Coefficient” (often denoted as “U” and given in W/m2.K) which is then assumed to hold true over a reasonable (but limited) range of operating conditions. The rate of heat flow is estimated by multiplying the heat transfer coefficient by the Temperature Difference “driving” the heat flow, and the Area through which the heat is flowing, thus heat flow rate is given by: ˙ Q = U.ΔT.A Conduction is probably the simplest heat transfer mechanism to understand. All substances conduct heat to some extent. The measure is Conductivity (often denoted “k” and given in W/m.K). As you would expect, heat flow rate increases with Conductivity, Flow Area, and Temperature Difference, and decreases with a longer flow path docsity.com length. Thus heat flow rate is given by: ˙ Q = k.ΔT .A L Where heat flows through a composite wall with layers of different conductivity (e.g. an insulated wall), we can analyse it like an electrical circuit where we sum up the “resistances”. In such a case, the “resistance to thermal flow” is L/k.A. Thus total flow “resistance” is: R = L1 k1.A1 + L2 k2 .A2 Heat flow rate through a composite wall is therefore: ˙ Q = ΔT R = ΔT L1 k1.A1 + L2 k2 .A2       While the resulting equation looks tedious, the concept is simple. Importantly: if even one part of the heat flow path has high thermal resistance, the heat flow rate is enormously reduced. This is very significant in many situations. Heat Exchange with Fluids Heat transfer flow rates from a solid to a fluid is heavily dependent on the fluid velocity. This is because liquids and (especially) gases are poor conductors. We therefore rely heavily on the fluid flow and turbulence within that flow to disperse the heated or cooled fluid, and bring further fluid in contact with the solid surface. The mathematics of this will not be handled in this unit, but it is important to understand that higher fluid velocities give greater heat transfer between solids and fluids. Specific heat We know that, when heat is transferred into a mass while no other major process is happening, its temperature rises. The rise is proportional to the amount of heat entering. We know intuitively that energy entering is somehow stored in the mass, and can later be removed. It is, in fact, stored as kinetic energy of individual molecules, and we call this “internal energy”. In physics, you will have encountered the concept of “specific heat” which is the “thermal capacity” of a material, i.e. the degree to which the temperature will rise when a mass receives a given amount of energy as heat. Specific heat is thus related to the “internal energy increase” required to produce a given rise in temperature. We now need to look more closely at what we mean by “specific heat”. If we supply a quantity of heat to some gas in a “closed” system (e.g. a gas-tight cylinder fitted with a piston), we can obtain different temperature rises depending upon how we constrain the system volume. If we fix the piston to keep the volume constant, the gas pressure rises with temperature, i.e. p α T. The temperature rise from the same heat input is different if we allow piston movement. Overall temperature rise will vary, depending upon how pressure varies during the volume change. We therefore find that what is loosely described as “specific heat” depends upon conditions during the heat transfer. The normal “limiting” extremes would be constant volume and constant pressure. It is therefore common for specific heats of substances to be quoted for both constant pressure and constant volume cp and cv. Heat balance Most people are familiar with the notion of “conservation of energy”. Although not universally true when considering nuclear processes, it is of very practical value in thermodynamics. We frequently use a “heat balance” to identify where energy has gone, and in what form. Essentially, we total up the amounts of energy entering and leaving a system in the form of heat and work. If the state of the system is not changing (i.e. it is in equilibrium), then the energy entering must equal that leaving. We often use this to help calculate heat losses which would otherwise be impossible to measure. Work done by a piston If a piston of area A is pushed inwards against a net constant pressure P, the force F required is: F = p.A P would in fact be the pressure difference between the two sides of the piston. Now, if the piston moves distance L, the work done is: W = p.L.A Note also that, because L.A = ΔV (change in volume), the work done can also be defined as: W = - p.ΔV = - p. V2 - V1( ) (only for constant pressure process, +ve when energy enters system) If this work is done N times per second (e.g. in a machine), the power required to move the piston is: Power = p.L.A.N docsity.com which will be greater than the compression work because the pressure forces are larger at all points of the expansion process (i.e. area under curve 34 is greater than area under 12). To return the cylinder pressure from P4 to the original pressure P1, we need to remove heat to lower the temperature back to T 1 . We can choose to do this without allowing the volume to change (i.e. W 41 = 0 ). Our system has now returned to its original condition through a series of processes, and we have obtained net work output from the system. We find the amount of heat put in (Q 23 ) is greater than that removed at the end (Q 41 ), the difference being exactly equal to the net work extracted because we have returned to the original condition and the “law” of conservation of energy must be satisfied (i.e. “what goes in must come out”). In putting our system through the series of processes and returning it to its original condition, we have converted some of our input heat into work (and rejected the rest). We have in effect put an engine through a cycle. The processes employed can vary, but all heat engine cycles are made up in a similar manner. The cycle described above (provided the expansion and compression processes have zero heat transfer) is called the “Otto cycle” after its originator, and is commonly used as an approximate representation of spark ignition (petrol) engines. Gauge pressure Most devices for pressure measurement have a sensing element with atmospheric pressure acting on its other side (see figure). A gauge pressure is therefore the comparison between the actual pressure and atmospheric pressure (and is called “gauge” pressure). We must add the atmospheric pressure to the reading to obtain the “absolute” pressure. i.e. pabs = pgauge + patmos Even piezo-electric devices need correction for atmospheric pressure because the “zero” setting still has atmospheric pressure acting on it, though many commercial devices now have this feature built-in. Indicator diagrams The pressure in an engine (or compressor) varies throughout the cycle, and produces a net output (or input) of work. A device called an “indicator” was originally used to observe this variation. It used a piston and spring mechanism to make a plot of pressure against volume over the cycle (a pV diagram, but often called an “indicator diagram”). These devices have mostly been replaced by piezo-electric pressure sensing devices with electronic recording equipment, but the general principle remains the same. Steam engines (venting to atmosphere) are simple devices where high pressure steam is passed into a cylinder, pushing the piston outwards. The steam supply is cut-off part way through the process, and the pressure starts to drop as piston movement continues. At the end of the stroke, an exhaust valve opens and the steam is pushed out to atmosphere as the piston returns. The upper figure shows a typical indicator diagram for a steam engine. Arrows go clockwise, so there is a net output of work. In most practical applications, steam is exhausted below atmospheric pressure into a low pressure condenser in order to extract the maximum amount of work from the steam. The lower figure is typical of a reciprocating (piston) air compressor. Arrows go anti- clockwise, so there is a net input of work. Note that the “suction” stroke to admit gas into the cylinder requires a pressure below atmospheric. The bumps in the shape are due to the behaviour of the spring loaded valves which need an extra pressure difference to push them open. Indicators show “gauge pressure” and need correction for atmospheric pressure. We generally put an “atmospheric line” on each diagram (the “natural” zero level) as a reference. Units You should already be familiar with SI units (kg, N, m , s). In thermodynamics, pressures and heat flows are generally large, so we use kiloPascals (kPa=kN/m2) and kiloJoules (kJ=kN.m) instead of Pa and J. This slightly unconventional approach still results in a consistent set of units except when calculating kinetic energy for which using kg,m,s gives J, not kJ, so we need to divide kinetic energy values obtained by 1000 to give kJ. Absolute Temperature Commonly used temperature scales (eg Celcius) have an arbitrary “zero point” and “interval size”. There is an “absolute” zero temperature, at which a “perfect gas” would have zero volume. When using numerical values for temperature in thermodynamics, we use “absolute temperatures” (all other scales have arbitrarily numbers). The absolute scale corresponding to Celcius is the Kelvin scale, where 0°K is –273.15°C docsity.com 3.1. Two large, identical, warm metal blocks are placed in a store room. Block A is in a quiet corner while block B is located near a busy doorway. Assuming the room temperature to be constant throughout, which block will cool to room temperature quicker, and why? Theoretically, will either of them ever reach room temperature? 3.2. Two large, identical, warm metal blocks are placed in the same part of a store room. Block A is 20°C hotter than block B. Which block will cool to room temperature quicker, and why? 3.3. A large, warm metal block is placed in a disused freezer cabinet, the door closed, and left until the temperatures stabilise. Which has lost more heat, the block or the freezer, and why? 3.4. Heat flows along a rod of 75mm diameter, 0.85 m long, and having conductivity of 120 W/mK. If the temperature difference across its length is 50°C, at what rate is heat flowing along the rod? What assumption was necessary? 3.5. An 8m2 wall comprises bricks 90mm thick and a 15mm thick fibre-board facing (in good thermal contact). Thermal conductivities of the bricks and fibre-board are 1.04 W/mK, and 0.048 W/mK respectively. If the outer wall surface is at 50°C and the other at 30 °C, at what rate will heat flow through the wall? 3.6. A 95mm diameter piston in a room at atmospheric pressure (101 kPa absolute) pushes fluid at 800 kPa absolute slowly out of a horizontal cylinder by moving 150mm.† a. Assuming negligible friction, what is the force required to move the piston? b. If the cylinder is steel and the fluid density is 1800 kg/m3, how much work is required to move the piston? c. If the fluid then pushes the piston back to its original position at 1800 kPa, what is the net work done? d. If sliding friction required a force of 150 N, what would be the revised net work done? † This question contains some redundant information. docsity.com