第4章理想气体的热力过程 Chapter 4 Thermodynamic processes of Ideal Gas 4.1基本热力过程 Basic thermodynamic processes p.134p313 44.2多变过程 Polytropic process ---P 135 44.3活塞式压气机的理论压缩过程 Theoretical compression process---P 336 44.4多级压缩中间冷却 Multistage Compression with inter-cooling P337~P343
第4章 理想气体的热力过程 Chapter 4 Thermodynamic processes of Ideal Gas 4.1基本热力过程 Basic thermodynamic processes--- p.134,p.313 4.2多变过程 Polytropic process ---P.135 4.3活塞式压气机的理论压缩过程 Theoretical compression process ---P.336 4.4多级压缩中间冷却 Multistage Compression with inter-cooling- -- P.337~P343
1.本章的研究任务 The task of this chapter A. It is to determine the change in the properties of working fluid during a process; (在于确定在一个过程中工质状态参数的变化量) B. It is to determine the amount of heat, work interaction between the system and its surroundings during a process. (确定在一个过程中系统与外界所交换的功量和热量的多少) 2.本章研究的对象 The object of this chapter Reversible processes of Ideal gases in closed systems (闭口系统中理想气体的可逆过程)
A. It is to determine the change in the properties of working fluid during a process; (在于确定在一个过程中工质状态参数的变化量) B. It is to determine the amount of heat, work interaction between the system and its surroundings during a process. (确定在一个过程中系统与外界所交换的功量和热量的多少) 2. 本章研究的对象 The object of this chapter Reversible processes of Ideal Gases in closed systems (闭口系统中理想气体的可逆过程) 1.本章的研究任务 The task of this chapter
3.所采取的步骤 The procedures adopted PVERT pi p2l T △=c.△T △h=c.△T &g=c, dT+ pdv q=cpdr-vdp Tas=C,dT+ pd Tds=c. dr-vdp R R ds===di ds=PdT-dp As=c In 2+rhn 2 As=c, hn2-Rhn △s=c.h2+Rh T pi Sw= pdv 6=-1 'dp dq= tas
u c T h = c p T = v q c dT pdv = v + Tds c dT pdv = v + dv v R dT T c ds v = + 1 2 1 2 ln ln v v R T T s c = v + Tds = cp dT − vdp q = c p dT − vdpdp p R dT T c ds p = − 1 2 1 2 ln ln p p R T T s c = p − 1 2 1 2 ln ln p p R v v s c = p + w = pdv wt = −vdp q = Tds 3. 所采取的步骤 The procedures adopted 2 2 2 1 1 1 T p v T p v Pv=RT =
4. 1 Basic Thermodynamic Process (4.1基本热力过程) 1等容过程 Isochoric-----Constant Volume process V=0.1979m3 P=101420Pa 合合合合
4.1 Basic Thermodynamic Process (4.1 基本热力过程) 1.等容过程 Isochoric-----Constant Volume process
(1)过程方程( Process equation For a constant volume process the addition or removal of heat will lead to a change in the temperature and pressure of the gas, as shown on the two graphs above. Substitute v=C into p,v, p2v2
For a constant volume process, the addition or removal of heat will lead to a change in the temperature and pressure of the gas, as shown on the two graphs above. Substitute into 2 2 2 1 1 1 T p v T p v v = C = 2 2 1 1 T p T p = (1) 过程方程 (Process equation)
(2)内能、焓及熵的变化量 The change in Internal Energy, Enthalpy and Entropy dg=c, dr+ pdy du d =c, dr dh=c dt The amount of heat added to a closed system during a constant volume process equals to the increase in internal energy. 定容过程中加入闭口系统的热量等于系统的内能的增 加量
q c dT pdv = v + du = q = cv dT dh = c p dT (2)内能、焓及熵的变化量 The change in Internal Energy, Enthalpy and Entropy The amount of heat added to a closed system during a constant volume process equals to the increase in internal energy. 定容过程中加入闭口系统的热量等于系统的内能的增 加量
Entropy Change To find the Entropy change, start with the expression derived from the first law, replacing du using the definition of specific heat at constant volume and using the definition of entropy △ Ax三C. In 3) Work done and Heat transferred(功量和热量) Applying the first law of thermodynamics to the process dU=-6 Replacing ow with the reversible work du=5o-pdy since the volume is constant dv= o W=0∞=l=ln
Entropy Change To find the Entropy change, start with the expression derived from the first law, replacing dU using the definition of specific heat at constant volume and using the definition of entropy (3) Work done and Heat Transferred(功量和热量) Applying the first law of thermodynamics to the process Replacing δW with the reversible work since the volume is constant dV = 0 q = du Tds = cv dT 1 2 ln T T s c = v dU = Q −W dU = Q − PdV W = 0 Q = dU
using the definition of the specific heat at constant volume c du dt to replace du in the first law ∞Q=mCpd7 0=mCat The technical work done W1=|wc=v(p1-P2)
using the definition of the specific heat at constant volume to replace dU in the first law. The technical work done Q = mCV dT = 2 1 T T Q m CV dT = − = − 2 1 ( ) 1 2 p p wt vdp v p p
2等压过程 Isobaric-- Constant pressure P=250000Pa T=298K Y=0.0991m3 Heat +ye
2.等压过程 Isobaric-----Constant Pressure
(1)过程方程 Process equation For a constant pressure process the addition or removal of heat will lead to a change in the temperature and volume of the gas, as shown on the two graphs above Substitute p=c into 1v112V2 V v2 7172 112
(1) 过程方程 Process equation 2 2 2 1 1 1 T p v T p v = p = C 2 2 1 1 T v T v = For a constant pressure process, the addition or removal of heat will lead to a change in the temperature and volume of the gas, as shown on the two graphs above. Substitute into
