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##### Energy in thermal processes

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**Energy in thermal processes**PHY231**Where are we going?**• We learnt about ideal gases • And we learnt that temperature is proportional to the average kinetic energy of particles in a gas • (monoatomic gas) • We learnt about transfer of thermal energy (heat) • Changing the thermal energy of a system can • change its temperature • be linked to a phase change • Be linked to mechanical work: Engine, heat pump… !! Thermodynamics PHY231**Work in thermodynamic processes**Work >0 • During a process where the Pressure P and Volume V of a system are changed, the work done on the system is the area under the PV curve • An arrow indicates the direction of the process • Arrow points right (V increases): the work done on the system is negative • Arrow points left (V decreases): The work done on the system is positive Work <0 i f PHY231 Vf Vi**More (on) work**• W>0, the system is gaining energy, • W<0, the system is losing energy and provides work to the outside,. (engine mode) • Limiting cases in the PV space • Isobaric process (P constant) • Isovolumetric process (V constant) Pf Pi PHY231 Vi,f**A piston is pushed down and pulled up on a volume of gas.**The following four diagrams show the pressure P and volume V during the processes. • Which one of the following has the largest increase in energy ? A) B) C) D) PHY231**Work is the area under the PV curve**• Arrow to the right (V increases) W<0 • Arrow to the left (V decreases) W>0 A) B) C) D) PHY231 6**First Law of Thermodynamics**• If Q is the energy transferred to a system by heat and W the work done on it, the change in the system’s internal energy DU is given by the sum • The change in internal energy is related to the position of the atoms/molecules and to their kinetic energy • W and Q >0 means the system gains energy • W and Q <0 means the system loses energy PHY231**Compressed gas**• A gas is compressed at a constant pressure of 0.800 atm from a volume of 9.00 L down to 2.00 L. In the process 400 J of energy leaves the gas by heat. • What is the work done on the gas and the change of its internal energy? PHY231**Thermal processes**• The change of internal energy for an ideal gas is given by • Cv is the molar specific heat at constant volume • For monoatomic gases Cv= 3R/2, diatomic Cv= 5R/2**Partial summary**• First law of thermodynamics • W given by area under PV diagram • P=constant then • Internal energy change is • W and Q > 0 mean the internal energy increases • W and Q <0 mean the internal energy decreases • Also remember ideal gas law**Moving forward**• Consider the two important equations • DU=Q+W • PV=nRT • These equations tell us that: • The change in a gas’ internal energy can be due to heat and work simultaneously • The three main quantities P, V and T are not independent and can vary simultaneously • So, in general, the changes for an ideal gas can be quite complicated… • To simplify things, processes where one of the variable is kept constant are usually considered • There are four important basic processes**Four basic processes**• Isobaric process (P=constant) • Adiabatic process (Q=0) • Isovolumetric process (V=constant) • Isothermal process (T=constant) • These are simplification of a general process for which none of the quantities is kept constant PHY231**Isobaric process**• In this process, the pressure P is constant • The heat transferred to an ideal gas is given by • Cp is the molar specific heat at constant pressure • For most gases R=8.31 J/(K.mol) PHY231**Adiabatic process -**• No heat enters or leaves the system • The first law simplifies to • As a result • With the definition of the adiabatic index g PHY231**Specific heat table**• For most gases Cp-Cv=R PHY231**Isovolumetric process**• In isovolumetric process the volume is constant • It follows that the work on the gas in this case is always zero • The first law simplifies to • Thus in this case • Isovolumetric process is sometimes also called isochoric process PHY231**Isothermal process**• In this process the temperature is kept constant. Since DU is linked to the change in temperature it follows that • And then • T=constant and • The work on the gas can be shown to be given by**Summary**Advice: copy that table on your equation sheet PHY231**Visualizing the four processes in PV space**Isobaric Isothermal Isovolumetric Adiabatic**PV diagram with T=constant lines plotted**Isobaric Isovolumetric Isothermal Adiabatic**Isothermal process:(additional materialnot needed**forhomework nor exam)**HW problem**• Adiabatic process, find V2 knowing P1, V1 and P2 PHY231**PV diagram**• A quantity of monatomic ideal gas goes through the following process. Calculate the work, the change in internal energy, and the heat associated with this process PHY231**Work done during the process**PHY231**DU during the process**PHY231**Q during the process**PHY231**Let’s build a heat engine…**• What’s a heat engine? • Basically it’s an energy converter • Takes energy from a heat source • Convert part of it into mechanical energy (= work) • Cyclic process (closed path in PV diagram) • Second law of thermodynamic says it’s impossible to convert all the heat into mechanical energy… • This means some (wasted) energy will be released in the exhaust through heat • Second Law of Thermodynamics • No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely for performance of an equal amount of work**Heat engine concept**• Schematically a heat engine looks like this • Conservation of energy (what comes in the engine leaves the engine): Qhot=|Qcold|+|Weng| 1) Heat (Qhot) is introduced into the engine from a hot reservoir Qh 2) Work is done by the engine (the engine loses that energy ) 3) Some heat (Qcold) is expelled by the engine to a cold reservoir**Heat engine cycle**• An example of heat engine cycle in PV diagram • Enclosed area is the work provided to the engine • W<0 since the engine loses energy through work Path B to C Isothermal T=constant Path A to B Isovolumetric V=constant Path C to A Isobaric P=constant**Basics properties of heat engines**• Cyclic process • Starting point and end points are the same • DU for a complete cycle is 0 PHY231**Heat engine - thermal efficiency**• For a heat engine, the thermal efficiency e gives an idea of how much work is done with respect to the heat supplied • For an ideal engine (= Carnot engine explained later) Tc = Temperature of the cold reservoir Th = Temperature of the hot reservoir T in K !!!!**Examples**• What is the efficiency of a heat engine working with 20 kJ heat supply and 10 kJ heat exhaust? • What is the maximum efficiency of a heat engine operating between 20 ºC and 100 ºC**Heat engine – full example**• Let’s go back to our heat engine example using a monatomic ideal gas Path A to B Isovolumetric V=constant Path B to C Isothermal T=constant Reminder: 1 L = 10-3 m3 1 atm = 101,325 Pa R=8.31 J/(mol.K) Monatomic ideal gas Cv=3R/2 Cp=5R/2 Path C to A Isobaric P=constant PHY231**Before starting**• This is a fully worked example illustrating all the concepts of the chapter • You are not supposed or expected to know the following formula • Especially, the end results are specific to this example and NOT general formulas applicable to all heat engines • Nonetheless, you are encouraged to look at the following to gain some insight • Much simpler problems are asked from you in homework-set 11 • You are expected to know the Carnot-engine (also called ideal engine) results PHY231**Numerically (non SI units)**PHY231**Remark about the work W**• You must remember that the work is the area under the PV curve. Let’s look back at our results and picture them graphically PHY231**Remark about the work W done on the gas**• When we sum the last two we get the initial area as we should (note that the WBC <0 because V increases whereas WCA>0 because V decreases) + = (If we were reversing the directions of all arrows then W would become 6 L.atm) PHY231**Carnot engine**• The Carnot engine is an ideal engine and is the most efficient engine possible • Carnot Theorem • No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs • The Carnot cycle uses 2 isothermals at Tc and Th and 2 adiabatic processes • Thermal efficiency of a Carnot engine is fixed by the temperature of the two reservoirs (Tc<Th) T in K !!!!**Refrigerators and Heat pumps**• These are engine working in reverse • (same as engine but with arrows reverse directions in PV diagram) Engine (W<0) Refrig./heat pump (W>0)**Coefficient of performance (COP)**• Refrigerator • Refrigerator uses work energy to take some heat away from the cold reservoir • Heat pump • Heat pump uses work energy to add-up heat to the hot reservoir • Both options need work because heat never flows naturally from the cold to the hot reservoir…**Engine or fridge/heat pump ?**• A device work along the cycle ABC. Is it an engine or a refrigerator/heat pump ? • A) Engine • B) Refrigerator/heat pump • C) Both • D) Neither PHY231**A) Engine**• B) Refrigerator/heat pump • C) Both • D) Neither WBA>0 WAC<0 WCB=0 W=WBA+WAC+WCB>0**PV diagram**• On a PV diagram, 2 curves are plotted. • Both start at (P1, V1) and both end at the same increased volume (V2). One curve is for an isothermal process; the other is for an adiabatic process. Except for the common starting point, which curve is the upper one? • A) Isothermal • B) Adiabatic • C) Adiabatic and isothermal are the same curve • D) The curves cross in the middle PHY231**A) Isothermal**• B) Adiabatic • C) Adiabatic and isothermal are the same curve • D) The curves cross in the middle PHY231