Lesson 9: Second Principle of Thermodynamics PDF

Second Principle Degree Industrial Technologies Engineering Lesson 9: Second Principle of Thermodynamics 9.1 Introduction 9.2 Thermal Engines 9.3 Second principle: Kelvin-Planck statement 9.4 Refrigerator engines. Heat Pump 9.5 Second principle: Clausius s...

Second Principle Degree Industrial Technologies Engineering Lesson 9: Second Principle of Thermodynamics 9.1 Introduction 9.2 Thermal Engines 9.3 Second principle: Kelvin-Planck statement 9.4 Refrigerator engines. Heat Pump 9.5 Second principle: Clausius statement 9.6 Carnot Cycle. Carnot theorem 9.7 Cycles with ideal gases First Principle Degree Industrial Technologies Engineering 9.1 Introduction The first thermodynamics principle establishes that the net energy must remain constant in a thermodynamic process. So the first principle is about the energy conservation The first principle does not tell us about the “direction” in which the process will happen or the probability that the process takes place In nature, the processes seem to happen in a certain direction Examples Let two bars be, one of the them is at temperature T1 and the other at temperature T2. Both bars are in thermal contact and isolated from the surroundings. If T1 < T2, nature tells us Initially In the equilibrium A T1 T2 B A T3 T3 B 𝑇1 < 𝑇2 𝑇1 < 𝑇3 < 𝑇2  Heat flows from B to A  The heat given by B, QB, is taken by A, QA. Besides 𝑄𝐴 = 𝑄𝐵  After some time, in the equilibrium both bars have the same temperature First Principle Degree Industrial Technologies Engineering The following process does not take place, even if, in this case the energy remains constant Initially In the equilibrium A T1 T2 B A T3 T4 B 𝑇1 < 𝑇2 𝑇3 < 𝑇1; 𝑇4 > 𝑇2  Heat flows from A to B  The heat given by A, QA, is taken by B, QB. Besides 𝑄𝐴 = 𝑄𝐵  After some time, in the equilibrium both bars have different temperatures The second principle of thermodynamics is concerned about the direction of the natural processes First Principle Degree Industrial Technologies Engineering 9.2 Thermal engines  The energy can appear in different forms. Work and heat are two different forms of energy  A system can increases its internal energy by receiving work or by absorbing heat It seems  All the work can be transformed into heat (Joule´s experiment)  All the heat cannot be transformed into work… It seems that some part of the heat is lost. This fact will be seen on considering the thermal engines How does a thermal engine work?  It removes heat energy from a Hot reservoir  It does a work  It gives off heat energy to a cold reservoir  The process is cyclic, so it is continuously repeated Important: The thermal engine uses a certain substance: water vapor, air, oil The hot and cold reservoirs keep the temperature constant. The have a very high heat capacity First Principle Degree Industrial Technologies Engineering Hot reservoir TH The engine removes heat energy from the hot reservoir 𝑄𝐻 > 0 QH A part of QH is used by the engine to do work 𝑊 0 and QC < 0 𝑊 = 𝑄𝐶 − 𝑄𝐻 The efficiency of the thermal engine is defined as: 𝐺𝑜𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑐ℎ𝑖𝑛𝑒: 𝑤ℎ𝑎𝑡 𝑤𝑒 𝑤𝑎𝑛𝑡 𝑡𝑜 𝑜𝑏𝑡𝑎𝑖𝑛 𝜀= 𝐻𝑜𝑤 𝑚𝑢𝑐ℎ 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑝𝑎𝑖𝑑 𝑓𝑜𝑟 𝑖𝑡 Our goal is Work: 𝑊 = 𝑄𝐻 − 𝑄𝐶 The price: 𝑄𝐻 𝑊 𝑄𝐻 − 𝑄𝐶 𝑄𝐶 𝜀= = =1− 𝜀 0 QH reservoir A work is done on the refrigerator 𝑊>0 Refrigerator W The engine gives off heat energy to the hot reservoir 𝑄𝐻 < 0 QC Let us apply the first thermodynamics principle: ∆𝑈 = 𝑄 + 𝑊 Cold reservoir The process is cyclic: DU = 0 𝑄 = −𝑊 ⇒ 𝑊 = −𝑄𝐻 − 𝑄𝐶 TC As QH < 0 and QC > 0 𝑊 = 𝑄𝐻 − 𝑄𝐶 First Principle Degree Industrial Technologies Engineering The efficiency of the refrigerator is defined as: Our goal is take heat out of the cold reservoir: 𝑄𝐶 The price: 𝑊 𝑄𝐶 𝑄𝐶 1 𝜂>1 𝜂= = = 𝑊 𝑄𝐻 − 𝑄𝐶 𝑄𝐻 −1 𝑄𝐶  First Principle Degree Industrial Technologies Engineering Example: refrigerator engine Expansion valve QC (3) Condenser Evaporator reservoir reservoir Hot Cold (2) Win (4) (1) QH Compressor W  (1) The gas is compressed by the compressor. The gas is initially at room temperature. The temperature and pressure of the gas increases. The compressor does work  (2) The gas enters into the condenser (hot reservoir). The gas gives heat QH, and the gas transforms into liquid  (3) The liquid passes through the expansion valve (abrupt decreases of the pressure) and the gas decreases it temperature. A part of the energy received by the compressor is used to keep vacuum in the expansion valve.  (4) The liquid enters into the evaporator with a lower temperature than it. It takes heat, QC, and transform into gas. Later, the gas enters into the compressor and the cycle starts again First Principle Degree Industrial Technologies Engineering Heat pump A heat pump operates like a refrigerator engine: it removes heat energy from a cold reservoir and gives off heat energy to a hot reservoir, However the goals of both devise are quite different: For a refrigerator engine: The goal is to keep cold the cold reservoir For a heat pump: The goal is to keep hot the hot reservoir On applying the first thermodynamics principle: ∆𝑈 = 𝑄 + 𝑊 Hot reservoir TH The process is cyclic: DU = 0 𝑄 = −𝑊 ⇒ 𝑊 = −𝑄𝐻 − 𝑄𝐶 QH As QH < 0 and QC > 0 𝑊 = 𝑄𝐻 − 𝑄𝐶 Refrigerator W The performance of the heat pump is: QC Our goal is to give heat to the hot reservoir: 𝑄𝐻 The price: 𝑊 Cold reservoir 𝑄𝐻 𝑄𝐻 1 TC 𝜂𝐻𝑃 = = = 𝑊 𝑄𝐻 − 𝑄𝐶 𝑄 1− 𝐻 𝑄𝐶 The efficiencies of a refrigerator engine and a heat pump are related 𝜂𝐻𝑃 = 1 + 𝜂 First Principle Degree Industrial Technologies Engineering 9.5 Second principle: Clausius statement It is impossible for any device that operates on a cycle to transfer heat from a cold reservoir to a hot reservoir without any other effect Hot reservoir On applying the first thermodynamics principle: TH ∆𝑈 = 𝑄 + 𝑊 QH The process is cyclic: DU = 0; as W = 0 𝑄 = 0 ⇒ 𝑄𝐻 + 𝑄𝐶 = 0 Refrigerator 𝑄𝐻 = 𝑄𝐶 QC Cold reservoir TC So, an ideal refrigerator is impossible!!!!!!! First Principle Degree Industrial Technologies Engineering Let us point out that the Kelvin-Planck and the Clausius statements are equivalent Hot reservoir TH Hot reservoir TH QH Q Refrigerator Thermal W engine QC Cold reservoir TC The second law of thermodynamics says that it is impossible to construct a thermal engine or a refrigerator of the second kind First Principle Degree Industrial Technologies Engineering 9.6 Carnot cycle. Carnot theorem Carnot cycle establishes the maximum efficiency that a thermal engine can achieve when it is operating on a cycle between a hot reservoir and a cold reservoir. The different thermodynamic processes of the Carnot cycle are reversible Processes of the Carnot cycle Step 1: Heat is absorbed from a hot reservoir at temperature Th during an isothermal expansion from state 1 to state 2. Step 2: The gas expands adiabatically from state 2 to state 3 and its temperature is reduced to Tc. Step 3: The gas gives off heat to the cold reservoir as it is compressed isothermally at Tc from state 3 to state 4. Step 4: The gas is compressed adiabatically until its temperature is again Th. First Principle Degree Industrial Technologies Engineering Let us assume that an ideal gas undergoes all the thermodynamic processes of the Carnot cycle 1 2 Isothermal expansion. T = TH 𝑉2 𝑇1 = 𝑇2 = 𝑇𝐻 Δ𝑈12 = 0 𝑄𝐻 = −𝑊 𝑊12 = −𝑛𝑅𝑇𝐻 𝐿𝑛 < 0 𝑉1 𝑃1𝑉1 = 𝑃2𝑉2 Work done by the gas 𝑄 >0 𝐻 2  3 Adiabatic expansion. Q23 = 0 𝛾 𝛾 𝑃2𝑉2 = 𝑃3𝑉3 𝛾−1 𝛾−1 Δ𝑈23 = 𝑊23 𝑊23 = 𝐶𝑉 (𝑇𝐶 − 𝑇𝐻 ) 𝑇2𝑉2 = 𝑇3𝑉3 𝑇2 = 𝑇𝐻 ; 𝑇3 = 𝑇C 3  4 Isothermal compression. T = TC 𝑉4 𝑇3 = 𝑇4 = 𝑇𝐶 Δ𝑈34 = 0 𝑄𝐶 = −𝑊 𝑊34 = −𝑛𝑅𝑇𝐶 𝐿𝑛 >0 𝑉3 𝑃3𝑉3 = 𝑃4𝑉4 Work done on the gas 𝑄𝐶 < 0 4  1 Adiabatic compression. Q41=0 𝛾 𝛾 𝑃4𝑉4 = 𝑃1𝑉1 Δ𝑈41 = 𝑊41 𝛾−1 𝛾−1 𝑇4𝑉4 = 𝑇1𝑉1 𝑊41 = 𝐶𝑉 (𝑇𝐻 − 𝑇𝐶 ) 𝑇4 = 𝑇𝐶 ; 𝑇1 = 𝑇H First Principle Degree Industrial Technologies Engineering The net work is: 𝑊 = 𝑊12 + 𝑊23 + 𝑊34 + 𝑊41 𝑉2 𝑉4 𝑊 = −𝑛𝑅𝑇𝐻 𝐿𝑛 + 𝐶𝑉 (𝑇𝐶 − 𝑇𝐻 ) − 𝑛𝑅𝑇𝐶𝐿𝑛 + 𝐶𝑉 (𝑇𝐻 − 𝑇𝐶 ) 𝑉1 𝑉3 𝑉2 𝑉4 𝑊 = −𝑛𝑅𝑇𝐻 𝐿𝑛 − 𝑛𝑅𝑇𝐶𝐿𝑛 𝑉1 𝑉3 The efficiency of the cycle is: 𝑊 𝑄𝐻 − 𝑄𝐶 𝑄𝐶 𝜀= = =1− 𝑄𝐻 𝑄𝐻 𝑄𝐻 𝑉2 𝑉4 𝑉3 𝑄𝐻 = 𝑛𝑅𝑇𝐻𝐿𝑛 𝑄𝐶 = 𝑛𝑅𝑇𝐶𝐿𝑛 ⇒ 𝑄𝐶 = 𝑛𝑅𝑇𝐶𝐿𝑛 𝑉1 𝑉3 𝑉4 𝑉 𝑛𝑅𝑇𝐶𝐿𝑛 𝑉3 𝜀 =1− 4 𝑉 𝑛𝑅𝑇𝐻𝐿𝑛 𝑉2 1 𝛾−1 𝛾−1 𝑇𝐻𝑉2 = 𝑇𝐶 𝑉3 𝑉2 𝑉3 = 𝛾−1 𝛾−1 𝑉 1 𝑉4 𝑇𝐶𝑉4 = 𝑇𝐻 𝑉1 𝑇𝐶 The efficiency only depends on the temperatures of 𝜀 = 1− 𝑇𝐻 the hot and cold reservoirs (in Kelvin) First Principle Degree Industrial Technologies Engineering Carnot cycle for a refrigerator 1 The Carnot cycle for a refrigerator is similar to that of a P thermal engine, but it is travelled in the inverse direction Qout In this case: Isothermal compression 𝑉1 𝑉4 at TH 𝑄𝐻 = 𝑛𝑅𝑇𝐻𝐿𝑛 𝑄𝐻 = 𝑛𝑅𝑇𝐻𝐿𝑛 Adiabatic 𝑉4 𝑉1 expansion 4 𝑉3 𝑉3 Adiabatic 𝑄𝐶 = 𝑛𝑅𝑇𝐶𝐿𝑛 𝑄𝐶 = 𝑛𝑅𝑇𝐶𝐿𝑛 compression 𝑉2 𝑉2 2 Isothermal 3 𝑄𝐶 𝑄𝐶 expansion at TC Qin The efficiency is: 𝜂= = 𝑊 𝑄𝐻 − 𝑄𝐶 V 𝑉3 𝛾−1 𝛾−1 𝑛𝑅𝑇𝐶𝐿𝑛 𝑉 𝑇𝐶𝑉3 = 𝑇𝐻 𝑉4 𝑉3 𝑉4 𝜂= 2 As = 𝑉 𝑉 𝑇𝐻𝑉1 𝛾−1 𝛾−1 = 𝑇𝐶 𝑉2 𝑉2 𝑉1 𝑛𝑅𝑇𝐻𝐿𝑛 𝑉4 − 𝑛𝑅𝑇𝐶𝐿𝑛 𝑉3 1 2 𝑄𝐶 𝑇𝐶 Therefore: 𝜂= = 𝑊 𝑇𝐻 − 𝑇𝐶 First Principle Degree Industrial Technologies Engineering For a heat pump 𝑄𝐻 𝑄𝐻 𝑇𝐻 𝜂𝐻𝑃 = = = 𝑊 𝑄𝐻 − 𝑄𝐶 𝑇𝐻 − 𝑇𝐶 First Principle Degree Industrial Technologies Engineering Carnot theorem The second principle of thermodynamics tells us that it is impossible to construct an engine whose efficiency is 100 %. Then, what is the maximum efficiency that a thermal engine can achieve? This is established by the Carnot theorem. “No engine working between two given hot and cold reservoirs can be more efficient than a reversible engine working between those two reservoirs” “All reversible engine working between the same two hot and cold reservoirs have the same efficiency” Any irreversible engine working between two hot and cold reservoirs have a less efficiency than a carnot machine working between the same reservoirs First Principle Degree Industrial Technologies Engineering Conditions for a reversible cycle  There are not dissipative forces, for instance no friction  There are only heat transferences between systems which are at the same temperature or the temperature difference is very small dT.  The processes are quasi static, It means that the processes take a long time so that all the states are equilibrium states When the cycle is reversible, everything, the work substance and the rest of the universe, come back to the initial point All real life processes are irreversible First Principle Degree Industrial Technologies Engineering 9.7 Cycles with ideal gases Otto cycle This cycle is the closer one to a internal-combustion engine a  b Adiabatic compression, Q = 0 The temperature increases 2 b  c Isochoric process Both the temperature and the pressure increase c  d Adiabatic expansión, Q = 0 3 The temperature decreases d  a Isochoric process 1 4 The temperature and the pressure decrease Internal-combustion engine 1. The gasoline- air mixture enters at a and is adiabatically compressed to b. 2. It is then heated (by ignition from the spark plug) at constant volume to c. 3. The power stroke is represented by the adiabatic expansion from c to d. 4. The cooling at constant volume from d to a represents the exhausting of the burned gases and the intake of a fresh gasoline--air mixture. First Principle Degree Industrial Technologies Engineering Diesel cycle This cycle is the closer one to a diesel engine P QH A  B Adiabatic compression, Q = 0 B C The temperature increases B  C Isobaric expansion D QC The temperature increases C  D Adiabatic expansión, Q = 0 A The temperature decreases D  A Isochoric Compression The temperature and the pressure decrease V First Principle Degree Industrial Technologies Engineering Thermodynamic scale of temperature For a reversible carnot cycle, It is verified 𝑄𝐶 𝑇𝐶 = 𝑄𝐻 𝑇𝐻 QH is the heat removed from the hot reservoir at TH and QC is the heat given off to the cold reservoir at TC The expression is INDEPENDENT from the substance used in the cycle. This relationship is used to the define the thermodynamic absolute scale of temperature. For it, it is necessary to consider a reference point. The reference point is the water triple point, 273,16 K. For instance, the cold reservoir is water at a triple point and the hot reservoir is the system whose temperature must be measured. This temperature scale coincides with the scale obtained with an ideal gas thermometer