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Chapter 7. Fundamentals of thermodynamics. Study what rules must follow when the macroscopic parameters of thermodynamic system change. §7-1 Internal Energy Heat &Work.
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Chapter 7 Fundamentals of thermodynamics Study what rules must follow when the macroscopic parameters of thermodynamic system change.
§7-1 Internal Energy Heat &Work is a system that can interact with its surroundings in at least two ways, one of which must be heat transfer. Thermodynamic system: : everything outside the system. Surroundings(exterior) Thermodynamic system its surroundings 1. Two basic concepts:
Thermodynamic system Thermodynamic systems in engineering: Gas: such as air. Vapor: such as steam. Liquid: such as freon(氟利昂). Solid: such as semiconductor…….. They are also called as working substance working substance
Ideal gas state quantity 2. Internal energy Internal energy Depends on T of the system only.
Work is process quantity 3.The work done by system during equilibrium process The value of the work equals the total area under the process curve of p~Vdiagram.
Process quantity ---- macroscopic displaying of thermal motions 4. Heat Transfer heat:one form of exchanging energy between system and its environment as the temperature difference. the molecules of the system and the molecules of the environment exchange their thermal kinetic energies.
stirring 5. The relationship between work and heat Heating Work and heat have same effect for the system. Joule proved that 1 cal of heat causes the same temperature increment as that of 4.18 joule of work does. i.e. 1cal= 4.186 J
§7-2 The First Law of Thermodynamics 1. The first law … Stipulation: Q:“+” the system absorbs heat from exterior “-” the system gives out heat to exterior E:“+”the internal energy of system increases “-” the internal energy of system decreases W:“+” thesystem does work on exterior. “-” exterior does work on the system.
Note The FLT …can be used for any thermodynamic process of any thermodynamic system. ---Conservation law of energy including the heat phenomena and thermal motion. differential form: for the equilibrium process of ideal gas:
Other description of the FLT: There is not any system(machine) that can do work forever without the supply of heat. Or It is impossible to make out the first kind of perpetual motion machine.
第一类永动机: E2 - E1= 0(循环) Q = 0(外界不供给能量) W > 0 (对外界作功)
system Heat Work “Heat” and “Work” can not be transformed without thermodynamic system.
§7-3 Application of the First Law to isochoric, isobaric & isothermal Processes 1. Isochoric (Constant volume) process(V=Const.) If Q>0, then E>0: Absorb heat = increment of internal energy
Discussion When Q>0,we have V2>V1 or p1 >p2.The isothermal expansion process. Absorb heat from exterior = work for exterior. When Q<0,we have V2<V1. The isothermal compression process . Exterior does work to the system = the system withdraws heat to the exterior. The isothermal expansion process
Discussion WhenQ>0,T2>T1(V2>V1),E>0 , W>0. Isobaric expansion. The system absorbs heat. Part of heat is used for increasing the internal energy. Part of heat is used for doing work to exterior. When Q<0,T2<T1,E< ,W<0:Isobaric expression. The work done by exterior for the system+the decrease of the internal energy = the heat withdrawning from the system.
Q, AandEalways have the same “ positive-negative sign” in isobaric process. Furthermore Isobaric expansion
4.Differential forms of three processes above Constant volume Constant temperature Constant pressure
[Example] A monatomic ideal gas is contained in a cylinder closed with a movable piston. The initial pressure is 3atm and the initial volume 1l. The gas is heated first at constant pressure until the volume become to 2l. Then expands with constant temperature until the volume become to 3l. Finally cooled down at constant volume until the pressure drops to 1atm. Calculate: The increment of internal energy, the work done by the gas and the heat supplied to the gas in the three process.
Solution Draw every process in p-V diagram Internal energy is the state quantity. It has nothing to do with process.
dQ is a process quantity 1. Molar heat capacity §7-4 Heat Capacities of an Ideal Gas ---The amount of heat added to cause unit rise of temperature (or withdrawn to cause unit fall of temperature) for 1mol of a material. i.e. C depends on process.
2. The molar heat capacity at constant volumeCv Under constant volume process, Absorb heat for increasing internal energy only
Note Cv depends on i only. Physical meaning:The average kinetic energy of each freedom degree is (1/2)kT. The more numbers of freedom degree are, the more heat is needed. For a random process:
3. The molar heat capacity at constant pressureCp Under constant pressure process,
Note Cp>Cv. Physical meaning : Absorb heat not only for increasing internal energy but for working for exterior. For Mkg gas The ratio of molar capacity:
[Example] Suppose the molar heat capacity of an ideal gas is proportional with its temperature C=aT (a is a constant). Calculate the thermodynamic equation for 1mol ofthis gas. Solution: According to the first law of thermodynamics molar heat capacity of 1mol ideal gas
Separate variables Integration Integration constant We get
then As And §7-5 Application of the first Law of thermodynamics to Adiabatic Process No heat transfer between a system and exterior. dQ = 0 i.e. 1.character T, p when adiabatic expanding T ,p when adiabatic compressing.
For ideal gas, 2. Equation of adiabatic process for ideal gas Take differential , ··· And ··· Eliminating dT from and :
i.e. Or Integration --The equation of adiabatic process
Using state equation of ideal gas, other forms of process equation can be gotten. Three constants are different. 3. Adiabatic curve and isothermal curve (1)Mathematical method:compare the two slopes of the two curves.
isotherm i.e. Iso-curve adiabat i.e. Adia-curve
Find the reasons of pchanging (2)Physical method: Isotherm:--Vincreasing causes n decreasing p decreasing Adiabat:--Vincreasing causes n and T decreasing p decreasing --the adiabat is steeper than the isotherm
[Ex.]I is adiabatic process for a gas as shown in diagram. Judge: does J and K process withdraw heat form outside or reject heat to outside respectively? Solution:For I process For J process --Withdraw heat For Kprocess --Reject heat
[Ex.] A gas goes through three processes respectively as shown in figure. AB is an isobaric process. AC is an isothermal process. AD is an adiabatic process. Judge: (1) which process does the maximum work? (2) which process withdraw the maximum heat from outside?(3)which process has the maximum change of internal energy? Solution: (1) Work = the area under the process curve ABprocess does the maximum work
T T Compare (2) AB process: V AC process: V AD process: (3) AB process:
AD process: I.e. AB process has the maximum change of internal energy.
Non-equilibrium process A B 4. Adiabatic free expansion Removing baffle Molecules AB quickly E= 0 W = 0 Q = 0 When the system becomes equilibrium state, T1= T2 Use state equation
Note (1) It is not adiabatic process though Q = 0 It is not isothermal process though T1= T2 ---Non-equilibrium process (2) For real gas, T changes as the system gets equilibrium state after adiabatic process Reasons: There are interactions between molecules potential energies of molecules
If there are repulsive forces between molecules The repulsive forces do positive work during the gas expands T Kinetic energies Potential energies If there are attractive forces between molecules The attractive forces do negative work during the gas expands T Kinetic energies Potential energies
p2 p1 F 多孔塞 piston piston Application —throttling (节流) process Keep p1> p2 when moving the pistons The gas through 多孔塞 adiabatic free expansion T --Joule-Thomson effect
§7-6 Cyclical Processes 1. Characters When a system is carried through a cyclical process, p、V、T return their initial values Classify: positive cycle(clockwise)—heat engine Inverse cycle(counterclockwise)--refrigerator
= the area enclosed by the cyclical curve 2. positive cycle Working substance:for cycle Assume working substance withdraws heat Q1 from a high-temperature source, and rejects heat Q2 into lower temperature source, and does work W to the exterior meanwhile. the net heat absorbed during a complete cycle Q1, Q2 are all positive Net work
热机工作原理图 蒸汽 热水 进气阀 锅炉 排气阀 泵 冷凝器 冷水
Heat engine A simplified heat engine: Hot reservoir at T1 Cold reservoir at T2
i.e. what you get divided by what you pay for The efficiency of heat engine:the ratio of the useful work to the heat supplied. We hope that the work output from heat engine is as large as possible and the heat thrown away is as small as possible.
In 1794~1840, was about 3%~4%. So low!!! Now, is about 30%~40%.
[Ex.]1mol oxygen is carried out a cycle as shown if Fig. AB is isothermal process.BC is isobaric process. CAis isochoric process. Find =? Solution Find “withdrawing heat” process --“AB,CA “process Then