工程热力学(英文版)第9单元课件

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(精品)工程热力学(全套467页PPT课件)

(精品)工程热力学(全套467页PPT课件)
从能源结构来看,2004年一次能源消费中,煤炭占 67.7%,石油占22.7%,天然气占2.6%,水电等占 7.0%;一次能源生产总量中,煤炭占75.6%,石油 占13.5%,天然气占3.0%,水电等占7.9%。
我国能源现状
据预测,目前中国主要能源煤炭、石油和天然气的储 采比分别为约80、15和50,大致为全球平均水平的 50%、40%和70%左右,均早于全球化石能源枯竭 速度。
工程热力学
Engineering Thermodynamics
绪论
工程热力学属于应用科学(工程科学) 的范畴,是工程科学的重要领域之一。
工程热力学 是一门研究热能有效利用及 热能和其 它形式能量转换规律的科学
工程热力学所属学科

工程热力学

传热学 Heat Transfer

流体力学 Hydrodynamics
工程热力学是节能的理论基础
能量转化的一般模式

次 能
热能

电能 机械能
问题:下面哪些是热机,哪些不是?
燃气轮机、蒸气机、汽车发动机、燃料电池、制冷机、 发电机、电动机
能量转化的一般模式
风 能

水 能
化 学 能
料 电 池
风 车
水 轮 机
水 车
燃 烧
核 能
聚裂 变变

生物质
地太 热阳 能能
利 光转 用 热换
大气压(at),毫米汞柱(mmHg),毫米水柱(mmH2O)
1 kPa = 103 Pa
1bar = 105 Pa
换 1 MPa = 106 Pa
算 关
1 atm = 760 mmHg = 1.013105 Pa

工程热力学与传热学9)_气体动力循环PPT课件

工程热力学与传热学9)_气体动力循环PPT课件

1压缩 2-3:定容吸热 3-4:定压吸热
4-5:绝热膨胀
5-1:定容放热
三、柴油机理想循环及其热效率
分析循环吸热量,放热量,热效率和功量
p
3
4
T
4 3
2
2
5
5
1 1
v
s
定义几个柴油机特性参数
p
3
2
压缩比 v1
反映 气缸
4
v2 容积
5
定容升压比
p3 p2
1 定压预胀比 v4
工程热力学研究方法,先对实际动力循 环进行抽象和理想化,形成各种理想循 环进行分析,最后进行修正。
§9-1 柴油机实际循环和理想循环
一、四冲程柴油机实际工作循环
进气
压缩 燃烧和膨胀
排气
温度370~400 K, 压力
0.07~0.09MPa
进气行程
排气门关闭
下止点 上止点
活塞
P
进气门开启
大气压力线 r a
下止点 上止点
活塞
Z P
c
大气压力线 r
作功终了:温度 1300~1600 K, 压 力0.3~0.5 MPa
示功图
b
a V
下止点 上止点
活塞
进气门关闭 排气行程
排气门打开
Z P
残余废气
c b
大气压力线 r
V 示功图
温度900~1200 K 压力 0.105~0.115 MPa
温度300-370K 压力0.0785~ 0.0932MPa
第九章 气体动力循环
动力循环研究目的和分类
动力循环:工质连续不断地将从高温热源取得的 热量的一部分转换成对外的净功

9 mass-action kinetics——工程热力学课件PPT

9 mass-action kinetics——工程热力学课件PPT

9.1.1 Introduction
The Gibbs free energy (J/mole) is defined as:
G=H-TS For the isothermal process(定温过程):
∆G= ∆ H-T ∆ S
For a certain set of chemical species concentrations, temperature, and pressure, a chemical process will proceed in the direction that decreases the free energy.
9. Mass-action kinetics(质量作用动力学)
Reacting streams of busting gases are among the most important and difficult flow problems studied.
1)the species continuity equation: provide source and sink terms for homogeneous reactions
If ∆G<0, the process will proceed usly in the direction of the forward reaction.
If ∆G>0, the process will proceed spontaneously in the reverse direction of the reaction.
Section 9.3: Gas-phase mass-action kinetics General expressions for species production/

工程热力学英语ppt

工程热力学英语ppt
2 c1
状态参数

( Q W ) ( Q W )
1a 2 1b 2
内能及闭口系热一律表达式
定义 dU = Q+W
闭口系热一律表达式
!!!两种特例 绝功系 Q = dU 绝热系 W = - dU
heat transfer to a system and work done by a system are positive; heat transfer from a system and work done on a system are negative. Heat is transferred by three mechanisms: conduction, convection, and radiation. Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interaction between particles. Convection is the transfer of energy between a solid surface and the adjacent fluid that is in motion, and it involves the combined effects of conduction and fluid motion. Radiation is the transfer of energy due to the emission of electromagnetic waves (or photons). An overview of the three mechanisms of heat transfer is given at the end of this chapter as a Topic of Special Interest.

工程热力学与传热学(英文) 绪论

工程热力学与传热学(英文)   绪论
2. Engineering Thermodynamics(工程热力学)
deals with the principles of energy conversion among thermal energy and mechanical energy, as well as other form energies.
Thermodynamics and Heat Transfer
Thermodynamics and Heat Transfer
--- Introduction ---
Introduction
❖ Energy and Energy Resources ❖ The Contents of Thermodynamics
Mechanical energy Thermal energy
Electrical energy Chemical energy
Nuclear energy
Radiation energy
0-1-2 Thermal Energy and Its utilization
1. Energy Resources(能源):
such as: cooking, warming, drying, smelting. etc
厨房用热
太阳能热水器
熔炼炉
➢ The powered utilization:
Converting heat into other form energy.
• Steam power equipment • Gas power equipment • Thermoelectricity power generation • Nuclear electric power generation • Solar power

工程热力学第九章图文ppt课件

工程热力学第九章图文ppt课件

活塞式内燃机各种理想循环热力比较
Tmax 和 pmax 保持不变
T
3
q2 相等
2p
t
1
q2 q1
1 T2 T1
2m 2v
4
1
q1p q1m q1v
tp tm tv
s
为 了 规 范 事 业单位 聘用关 系,建 立和完 善适应 社会主 义市场 经济体 制的事 业单位 工作人 员聘用 制度, 保障用 人单位 和职工 的合法 权益
为 了 规 范 事 业单位 聘用关 系,建 立和完 善适应 社会主 义市场 经济体 制的事 业单位 工作人 员聘用 制度, 保障用 人单位 和职工 的合法 权益
分析循环的步骤:
将简化好的理想可逆循环表示在p-v、T-s图上
对理想循环进行分析计算
计算循环中有关状态点(如最高压力 点、最高温度点)的参数,与外界交换的 热量、功量以及循环热效率或工作系数。
为 了 规 范 事 业单位 聘用关 系,建 立和完 善适应 社会主 义市场 经济体 制的事 业单位 工作人 员聘用 制度, 保障用 人单位 和职工 的合法 权益
研究目标:
分析以气体为工质的内燃机循环、 燃气轮机循环的热力性能,揭示能量利 用的完善程度与影响其性能的主要因素, 给出评价和改进这些装置热力性能的方 法与措施。
q2p q2m q2v
T
2p 2m 2v 1
3p 3m 3v 4v
4p4m
tp tm tv
s
q ??Tmax和 1相同,图示 tp ,tm ,t大v 小
为 了 规 范 事 业单位 聘用关 系,建 立和完 善适应 社会主 义市场 经济体 制的事 业单位 工作人 员聘用 制度, 保障用 人单位 和职工 的合法 权益

工程热力学英文版9

工程热力学英文版9

A + B AB.
(10.3)
Suppose we introduce NA,0 moles of A and NB,0 moles of B into a container with volume V at time t = 0. We allow the system to evolve with no interaction with the surroundings (isolated). As time proceeds, we find that the mole numbers change, due to formation of AB molecules. If we wait long enough, the mole numbers come to steady-state values, which we will denote NA∗ , NB∗ , and NA∗B. Evidently, once steady state is reached, the rate at which AB is formed by reaction (10.1) just balances the rate at which it is destroyed by reaction (10.2).
When this condition is reached, we say the system is in chemical equilibrium. Depending on the nature of the chemical reaction, and also on the temperature
10.3 Reactions in Ideal Gases
If the mixture of A, B, and AB is an ideal gas mixture, then we have already

《工程热力学》课件

《工程热力学》课件

热力学状态由压力、容积和温度 等多个参数所定义。
热力学循环和周期
热力学循环将热量转换为功,有 多种应用,如蒸汽循环、空气循 环、涡轮循环等。
热能和功
1

在物理学中,功是由力作用于物体时所
热能
2
做的功。
热能可以转化为功,例如燃料在发动机
里的燃烧可以形成热能,进而转化为引 擎的动力。
3
热力学第一定律
热力学第一定律表明能量守恒,即能量 不能被创造或破坏,只能从一种形式转 化为另一种形式。
工程热力学循环
理想气体循环
理想气体循环有多个阶段,包括 等压加热、等容冷却、等压膨胀 和等容加热。
蒸汽循环
气轮发动机循环
蒸汽循环的主要组成部分包括锅 炉、汽轮机、冷凝器和再生器等。
气轮发动机循环的主要组成部分 包括压缩机、燃烧室、高压涡轮 和低压涡轮等。应用领域1 Nhomakorabea能源领域
热力学原理和循环在能源领域和能源的
制造业
2
开发利用中有着广泛的应用,例如火电 站、核电站、风电场等。
热力学在制造过程中的应用可以提高产
品质量,减少污染和能源浪费的发生,
例如冶炼、焊接、淬火、加热等。
3
空调与制冷
热力学原理在空调和制冷领域可以提高 制冷效率,从而降低能源消耗和对环境 的影响。
工程热力学
工程热力学是研究热、功、能的转化和传递过程的一门学科。此课程将覆盖 基本概念、能量转化、热力学循环以及应用领域等内容。
为什么学习工程热力学
1 领域广泛
工程热力学应用广泛,包括能源、制造业和空调等领域。
2 提高效率
学习热力学可以帮助你理解能量转换的过程并且提高能源利用的效率。

工程热力学第三版答案【英文】第9章

工程热力学第三版答案【英文】第9章

9-13The three processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the back work ratio and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air are given as R = 0.287 kJ/kg.K, c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4.Analysis (a) The P -v and T -s diagrams of the cycle are shown in the figures. (b) The temperature at state 2 is K 2100kP a100kP a 700K) 300(1212===P P T TK 210023==T TDuring process 1-3, we havekJ/kg516.600)K 21K)(300kJ/kg 287.0()()(3131113,13=-⋅-=--=--=-=⎰-T T R P Pd w in V V vDuring process 2-3, we havekJ/kg8.1172n7K)(2100)Kl kJ/kg 287.0(7ln 7ln ln22233232,32=⋅======⎰⎰-RT RT RT d RTPd w out V VV V v Vv The back work ratio is then0.440===--kJ/kg8.1172kJ/kg6.516,32,13outin bw w w rHeat input is determined from an energybalance on the cycle during process 1-3,kJ/kg2465kJ/kg 1172.8300)K)(2100kJ/kg 718.0()(,3213,3231,3131,32,31=+-⋅=+-=+∆=-∆=--------outv outin out in w T T c w u q u w qThe net work output issvkJ/kg 2.6566.5168.1172,13,32=-=-=--in out net w w w(c) The thermal efficiency is then26.6%====266.0kJ2465kJ656.2in net th q w η9-21An air-standard cycle executed in a piston-cylinder system is composed of threespecified processes. The cycle is to be sketcehed on the P -v and T -s diagrams; the heat and work interactions and the thermal efficiency of the cycle are to bedetermined; and an expression for thermal efficiency as functions of compression ratio and specific heat ratio is to be obtained.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air are given as R = 0.3 kJ/kg·K and c v = 0.3 kJ/kg·K. Analysis (a) The P -v and T -s diagrams of the cycle are shown in the figures. (b) Noting that1.4297.00.1KkJ/kg 0.13.07.0===⋅=+=+=vv c c k R c c p pProcess 1-2: Isentropic compressionK 4.584)5)(K 293(429.01112112===⎪⎪⎭⎫ ⎝⎛=--k k r T T T vvkJ/kg 204.0=-⋅=-=-K )2934.584)(K kJ/kg 7.0()(12in 2,1T T c w v0=-21qFrom ideal gas relation,2922)5)(4.584(3212323==−→−===T r T T v v v v Process 2-3: Constant pressure heat additionkJ/kg701.3=-⋅=-=-==⎰-K )4.5842922)(K kJ/kg 3.0()()(2323232out 3,2T T R P Pd w v v vskJ/kg2338=-⋅=-=∆=∆+=----K )4.5842922)(K kJ/kg 1()(233232,32in 3,2T T c h u w q p outProcess 3-1: Constant volume heat rejectionkJ/kg 1840.3=⋅=-=∆=--K 293)-K)(2922kJ/kg 7.0()(1331out 1,3T T c u q v0=-13w(c) Net work isK kJ/kg 3.4970.2043.701in 2,1out 3,2net ⋅=-=-=--w w wThe thermal efficiency is then21.3%====213.0kJ2338kJ497.3in net th q w η9-32The two isentropic processes in an Otto cycle are replaced with polytropic processes.The heat added to and rejected from this cycle, and the cycle’s thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa·m 3/kg·K, c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The temperature at the end of the compression isK 4.537K)(8) 288(13.11112112===⎪⎪⎭⎫ ⎝⎛=---n n r T T T vvAnd the temperature at the end of the expansion isK 4.78981K) 1473(113.11314334=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=---n n r T T T vvThe integral of the work expression for the polytropic compression giveskJ/kg 6.238)18(13.1K) K)(288kJ/kg 287.0(1113.1121121=--⋅=⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫ ⎝⎛-=---n n RT w vvSimilarly, the work produced during the expansion iskJ/kg 0.65418113.1K) K)(1473kJ/kg 287.0(1113.1143343=⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎭⎫⎝⎛-⋅-=⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫ ⎝⎛--=---n n RT w vv Application of the first law to each of the four processes giveskJ/kg 53.59K )2884.537)(K kJ/kg 718.0(kJ/kg 6.238)(122121=-⋅-=--=--T T c w q v kJ/kg 8.671K )4.5371473)(K kJ/kg 718.0()(2332=-⋅=-=-T T c q vkJ/kg 2.163K )4.7891473)(K kJ/kg 718.0(kJ/kg 0.654)(434343=-⋅-=--=--T T c w q vkJ/kg 0.360K )2884.789)(K kJ/kg 718.0()(1414=-⋅=-=-T T c q vThe head added and rejected from the cycle arekJ/kg419.5kJ/kg 835.0=+=+==+=+=----0.36053.592.1638.6711421out 4332in q q q q q qThe thermal efficiency of this cycle is then0.498=-=-=0.8355.41911in out th q q η9-37An ideal Otto cycle with air as the working fluid has a compression ratio of 8. Theamount of heat transferred to the air during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.Properties The properties of air are given in Table A-17E. Analysis (a) Process 1-2: isentropic compression.32.144Btu/lbm92.04R 540111==−→−=r u T v()Btu/lbm 11.28204.1832.144811212222=−→−====u r r r r v v v v v Process 2-3: v = constant heat addition.Btu/lbm241.42=-=-===−→−=28.21170.452419.2Btu/lbm452.70R 240023333u u q u T in r vvP(b) Process 3-4: isentropic expansion.()()Btu/lbm 205.5435.19419.28434334=−→−====u r r r r v v v v v Process 4-1: v = constant heat rejection.Btu/lbm 50.11304.9254.20514out =-=-=u u q53.0%=-=-=Btu/lbm241.42Btu/lbm113.5011in out th q q η (c) The thermal efficiency of a Carnot cycle operating between the same temperature limits is 77.5%=-=-=R2400R54011C th,H L T T η9-40The expressions for the maximum gas temperature and pressure of an ideal Otto cycleare to be determined when the compression ratio is doubled.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis The temperature at the end of the compression varies with the compression ratio as1112112--=⎪⎪⎭⎫⎝⎛=k k r T T T v vsince T 1 is fixed. The temperature rise during thecombustion remains constant since the amount of heat addition is fixed. Then, the maximum cycle temperature is given by11in 2in 3//-+=+=k r T c q T c q T v vThe smallest gas specific volume during the cycle isr13v v =When this is combined with the maximum temperature, the maximum pressure is given by ()11in 1333/-+==k r T c qRrRT P v v v9-47An ideal diesel cycle has a compression ratio of 20 and a cutoff ratio of 1.3. The maximum temperature of the air and the rate of heat addition are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis()K 6.95420K) 288(14.11112112===⎪⎪⎭⎫ ⎝⎛=---k k r T T T vvK 1241===⎪⎪⎭⎫ ⎝⎛=K)(1.3) 6.954(22323c r T T T vv Combining the first law as applied to the various processes with the process equations gives6812.0)13.1(4.113.12011)1(1114.111.41th =---=---=--c k c k r k r r ηAccording to the definition of the thermal efficiency,kW 367===0.6812kW 250th net inηW Q9-59An ideal dual cycle has a compression ratio of 15 and cutoff ratio of 1.4. The net work,heat addition, and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft 3/lbm.R (Table A-1E), c p = 0.240 Btu/lbm·R, c v = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea).Analysis Working around the cycle, the germane properties at the various states are()R 158015R) 535(14.11112112===⎪⎪⎭⎫ ⎝⎛=---k k r T T T vvout()psia 2.62915psia) 2.14(4.112112===⎪⎪⎭⎫ ⎝⎛=k kr P P P vvpsia 1.692psia) 2.629)(1.1(23====P r P P p xR 1738psia 629.2psia 692.1R) 1580(22=⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=PP T T xxR 2433R)(1.4) 1738(33===⎪⎪⎭⎫⎝⎛=c x xx r T T T vvR 2.942151.4R) 2433(14.11314334=⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=---k c k rr T T T vvApplying the first law to each of the processes givesBtu/lbm 7.178R )5351580)(R Btu/lbm 171.0()(1221=-⋅=-=-T T c w v Btu/lbm 02.27R )15801738)(R Btu/lbm 171.0()(22=-⋅=-=-T T c q x x vBtu/lbm 8.166R )17382433)(R Btu/lbm 240.0()(33=-⋅=-=-x p x T T c qB t u /l b 96.47R )17382433)(R Btu/lbm 171.0(Btu/lbm 8.166)(333=-⋅-=--=--x x x T T c q w vBtu/lbm 9.254R )2.9422433)(R Btu/lbm 171.0()(4343=-⋅=-=-T T c w vThe net work of the cycle isBtu/lbm 124.2=-+=-+=---7.17896.479.25421343net w w w w x and the net heat addition isBtu/lbm 193.8=+=+=--8.16602.2732in x x q q q Hence, the thermal efficiency is0.641===Btu/lbm193.8Btu/lbm124.2in net th q w η9-61An expression for cutoff ratio of an ideal diesel cycle is to be developed.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potentialenergy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis Employing the isentropic process equations,112-=k rT Toutwhile the ideal gas law gives1123T r r r T T k c c -==When the first law and the closed system work integral is applied to the constant pressure heat addition, the result is)()(111123in T r T r r c T T c q k k c p p ---=-=When this is solved for cutoff ratio, the result is11in1T r c q r k p c -+=9-81A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.Properties The properties of air are given in Table A-17E. Analysis (a ) Noting that process 1-2 is isentropic,T h P r 11112147=−→−==520R124.27Btu /lbm .()()Btu/lbm 240.11 147.122147.110221212==−→−===h T P P P P r r R 996.5(b ) Process 3-4 is isentropic, and thus()Btu/lbm38.88283.26571.504Btu/lbm115.8427.12411.240Btu/lbm 265.834.170.1741010.174Btu/lbm 504.71R 200043out T,12inC,43433343=-=-==-=-==−→−=⎪⎭⎫⎝⎛====−→−=h h w h h w h P P P P P h T r r rThen the back-work ratio becomess200052048.5%===Btu/lbm238.88Btu/lbm115.84outT,in C,bw w w r(c ) 46.5%====-=-==-=-=Btu/lbm264.60Btu/lbm123.04Btu/lbm123.0484.11588.238Btu/lbm264.6011.24071.504inout net,th in C,out T,out net,23in q w w w w h h q η9-87A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10.The air temperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.Properties The properties of air are given in Table A-17E. Analysis (a ) Noting that process 1-2 is isentropic,T h P r 11112147=−→−==520R124.27Btu /lbm .()()Btu/lbm 240.11 147.122147.110221212==−→−===h T P P P P r r R 996.5(b ) Process 3-4 is isentropic, and thus()Btu/lbm38.88283.26571.504Btu/lbm115.8427.12411.240Btu/lbm 265.834.170.1741010.174Btu/lbm 504.71R 200043out T,12inC,43433343=-=-==-=-==−→−=⎪⎭⎫⎝⎛====−→−=h h w h h w h P P P P P h T r r rThen the back-work ratio becomes48.5%===Btu/lbm238.88Btu/lbm115.84outT,in C,bw w w rs2000520(c ) 46.5%====-=-==-=-=Btu/lbm264.60Btu/lbm123.04Btu/lbm123.0484.11588.238Btu/lbm264.6011.24071.504inout net,th in C,out T,out net,23in q w w w w h h q η(d) The expression for the cycle thermal efficiency is obtained as follows:⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛-=⎪⎭⎫⎝⎛---=⎪⎪⎭⎫ ⎝⎛---=-⎪⎪⎭⎫ ⎝⎛--=---=----=-==-----------1111111111111111111231223in in 2,1out 3,2in net th 11)1(11111)1(11)1(1)1(1)()()()()(k k p k p k p k k v p k k p k v p p v r r k k r r k c R r T T r k c R r r T c r T T r T c c R r T r rT c T r T c c RT T c T T c T T R q w w q w η since 111kc c c c c c R p v p v p p -=-=-=。

《工程热力学》第九章 气体动力循环 (2)

《工程热力学》第九章  气体动力循环 (2)

27
一、概述--燃气轮机工作过程
燃料 燃烧室 2 压气机 发电机
3
空气
1
4
废气
开式装置示意图
28
闭式燃气轮机装置示意图
空气 2 加热器 压气机 燃料
3
Ws
1
冷却器
4
29
二、定压加热理想循环(Bryden Cycle)
1、定压加热理想循环组成 2、定压加热参数计算 3、循环热效率计算及其提高途径 4、轴功计算及其最大值与增温比关系
1、汽油机的实际工作循环 2、对汽油机的工作循环的简化及抽象--定容加 热循环 Otto Cycle 3、定容加热循环参数计算 定容加热循环参数关系 热效率计算及分析 循环净功计算 提高循环热效率的途径
15
定容加热循环 Otto Cycle
P 3
T 1-2:绝热压缩过程 2-3:定容燃烧过程 3-4:绝热膨胀过程 4-1:定容放热过程 4
1、三种循环压缩比ε 相同,放热量q2相 同 (书中以吸热量q1相同分析) η t,V> η t,C> η t,P
2、循环最高压力Pmax> η , η t,P> η t,C及最高温度Tmax相同 t V
以P-V图、T-S图分析!
23
9-5 活塞式热气发动机及循环
自学
24
作业: 9-5,9-11
C P 0 (T3 T4 ) C P 0 (T2 T1 )
1 ( k 1) / k w0 C P 0 T3 (1 ( k 1) / k ) T1 ( 1)
当W0达最大,可有增压比选择
max, w 0
T3 ( ) T1
k 2 ( k 1)

工程热力学(英语)

工程热力学(英语)

1-11 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building. The height of the building is to be determined. Assumptions The variation of air density with altitude is negligible.Properties The density of air is given to be ρ = 1.18 kg/m 3. The density of mercury is 13,600 kg/m 3.Analysis Atmospheric pressures at the top and at the bottom of the building arekPa100.70N/m 1000kPa1m/s kg 1N1m) )(0.755m/s )(9.807kg/m (13,600)(kPa97.36N/m 1000kPa1m/s kg 1N1m) )(0.730m/s )(9.807kg/m (13,600)(2223bottombottom 2223toptop =⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅===⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅==h g P h g ρP ρTaking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtainkPa 97.36)(100.70N/m 1000kPa 1m/s kg 1N1))(m/s )(9.807kg/m (1.18)(/2223topbottom air topbottom air -=⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅-=-=h P P gh P P A W ρIt yields h = 288.6 mwhich is also the height of the building.1-21 The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined.Assumptions The manometer fluid is an incompressible substance.Properties The density of the liquid is given to be ρ = 0.81 kg/L = 810 kg/m 3. Analysis The gage pressure in the duct is determined fromPa636=⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅==-=2223atm abs gage N/m 1Pa1m/s kg 1N1m) )(0.08m/s )(9.81kg/m (810ghP P P ρ The length of the differential fluid column iscm 13.9=︒==35sin /)cm 8(sin /θh L730 mmHgDiscussion Note that the length of the differential fluid column is extended considerably by inclinin g the manometer arm for better readability.2-4 No. This is the case for adiabatic systems only.2-6 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined.Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined fromQ Q Q Q cooling lights people heat gain =++whereQ Q Q lights people heat gain 10100W 1kW40360kJ /h 4kW 15,000kJ /h 4.17kW=⨯==⨯=== Substituting,.Q cooling 9.17kW =++=14417Thus the number of air-conditioning units required is units 2−→−=1.83kW/unit5kW9.172-18 The flow of air through a flow channel is considered. The diameter of the wind channel downstream from the rotor and the power produced by the windmill are to be determined. Analysis The specific volume of the air is/kg m 8409.0kP a 100K) K)(293/kg m kP a 287.0(33=⋅⋅==P RT v The diameter of the wind channel downstream from the rotor ism 7.38===−→−=−→−=m/s9m/s10m) 7()4/()4/(21122221212211V V D D V D V D V A V A ππ The mass flow rate through the wind mill iskg/s 7.457/kg)m 4(0.8409m/s)(10m) 7(3211===πvV A mThe power produced is thenkW 4.35=⎪⎭⎫ ⎝⎛-=-=22222221/s m 1000kJ/kg 12)m/s 9()m/s 10(kg/s) 7.457(2V V m Wcool·2-19 The available head, flow rate, and efficiency of a hydroelectric turbine are given. The electric power output is to be determined.Assumptions 1 The flow is steady. 2 Water levels at the reservoir and the discharge site remain constant. 3 Frictional losses in piping are negligible. Properties We take the density of water to beρ = 1000 kg/m 3 = 1 kg/L.Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy,which is gz per unit mass, and gz m for a given mass flow rate.kJ/kg 177.1/s m 1000kJ/kg 1m ) 120)(m /s (9.81222mech =⎪⎭⎫⎝⎛===gz pe e The mass flow rate iskg/s ,000100/s)m 0)(10kg/m 1000(33===V ρmThen the maximum and actual electric power generation becomeMW 7.117kJ/s 1000MW 1kJ/kg) 7kg/s)(1.17 000,100(mech mech max =⎪⎭⎫ ⎝⎛===e mE WMW 94.2===MW) 7.117(80.0max overall electric W W η Discussion Note that the power generation would increase by more than 1 MW for each percentage point improvement in the efficiency of the turbine –generator unit.3-9 A rigid container that is filled with R-134a is heated. The temperature and total enthalpy are to be determined at the initial and final states.Analysis This is a constant volume process. The specific volume is/kg m 0014.0kg10m 014.03321====m Vv vThe initial state is determined to be a mixture, and thus the temperature is the saturation temperature at the given pressure. From Table A-12 by interpolation C 0.61︒==kPa 300 @sat 1T TandR-134a 300 kPa 10 kgvkJ/kg52.54)13.198)(009321.0(67.52009321.0/kgm )0007736.0067978.0(/kg m )0007736.00014.0(113311=+=+==--=-=fg f fgf h x h h x v v vThe total enthalpy is thenkJ 545.2===)kJ/kg 52.54)(kg 10(11mh HThe final state is also saturated mixture. Repeating the calculations at this state,C 21.55︒==kPa 600 @sat 2T TkJ/kg64.84)90.180)(01733.0(51.8101733.0/kgm )0008199.0034295.0(/kg m )0008199.00014.0(223322=+=+==--=-=fg f fgf h x h h x v v vkJ 846.4===)kJ/kg 64.84)(kg 10(22mh H3-22 rigid tank contains an ideal gas at a specified state. The final temperature is to be determined for two different processes.Analysis (a ) The first case is a constant volume process. When half of the gas is withdrawn from the tank, the final temperature may be determined from the ideal gas relation as()K 400=⎪⎭⎫ ⎝⎛==K) 600(kP a 300kP a 1002112212T P P m m T (b ) The second case is a constant volume and constant mass process. The ideal gas relation for this case yieldskPa 200=⎪⎭⎫ ⎝⎛==kP a) 300(K 600K 4001122P T T P3-32 Complete the following table for H 2 O : P , kPa T , ︒C v , m 3 / kgu , kJ/kg Phase description 200 30 0.001004 125.71 Compressed liquid 270.3130--Insufficient information200 400 1.5493 2967.2 Superheated steam 300133.520.5002196.4Saturated mixture, x=0.825500 473.1 0.6858 3084 Superheated steam4-14 Oxygen is heated to experience a specified temperature change. The heat transfer is to be determined for two cases.Assumptions 1 Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 154.8 K and 5.08 MPa. 2 The kinetic and potential energy changes arenegligible, 0pe ke ≅∆≅∆. 3 Constant specific heats can be used for oxygen.Properties The specific heats of oxygen at the average temperature of (25+300)/2=162.5︒C=436 K are c p = 0.952 kJ/kg ⋅K and c v = 0.692 kJ/kg ⋅K (Table A-2b ).Analysis We take the oxygen as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for a constant-volume process can be expressed as)(12in energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in T T mc U Q E E E -=∆=∆=-vThe energy balance during a constant-pressure process (such as in a piston-cylinder device) can be expressed as)(12in out ,in out ,in energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsferenergy tra Net out in T T mc H Q U W Q UW Q E E E p b b -=∆=∆+=∆=-∆=-since ∆U + W b = ∆H during a constant pressure quasi-equilibrium process. Substituting for both cases, kJ 190.3=-⋅=-==K )25K)(300kJ/kg 692.0(kg) 1()(12const in,T T mc Q v VkJ 261.8=-⋅=-==K )25K)(300kJ/kg 952.0(kg) 1()(12const in,T T mc Q p P4-25 A rigid tank filled with air is connected to a cylinder with zero clearance. The valve is opened, and air is allowed to flow into the cylinder. The temperature is maintained at 30︒C at all times. The amount of heat transfer with the surroundings is to be determined.Assumptions 1 Air is an ideal gas. 2 The kinetic and potential energy changes are negligible,∆∆ke pe ≅≅0. 3 There are no work interactions involved other than the boundary work. Properties The gas constant of air is R = 0.287 kPa.m 3/kg.K (Table A-1). Analysis We take the entire air in the tank and the cylinder to be the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this closed system can be expressed asoutb,in 12out b,in energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in 0)(W Q u u m U W Q E E E ==-=∆=-∆=-since u = u (T ) for ideal gases, and thus u 2 = u 1 when T 1 = T 2 . The initial volume of air is33112212222111m 0.80)m (0.41kP a200kP a 400=⨯⨯==−→−=V V V V T T P P T P T PThe pressure at the piston face always remains constant at 200 kPa. Thus the boundary work done during this process iskJ 80m kPa 1kJ 10.4)m kPa)(0.8 (200)( 3312221out b,=⎪⎪⎭⎫⎝⎛⋅-=-==⎰V V V P d P W Therefore, the heat transfer is determined from the energy balance to bekJ 80==in out b,Q W4-27 An insulated cylinder is divided into two parts. One side of the cylinder contains N 2 gas and the other side contains He gas at different states. The final equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and moving freely.Assumptions 1 Both N 2 and He are ideal gases with constant specific heats. 2 The energy stored in the container itself is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible. Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m 3/kg.K is c v = 0.743 kJ/kg·°C for N 2, and R = 2.0769 kPa.m 3/kg.K is c v = 3.1156 kJ/kg·°C for He (Tables A-1 and A-2)Analysis The mass of each gas in the cylinder is()()()()()()()()kg0.808K 298K /kg m kPa 2.0769m 1kPa 500kg 4.77K 353K /kg m kPa 0.2968m 1kPa 50033He111He33N 111N 22=⋅⋅=⎪⎪⎭⎫ ⎝⎛==⋅⋅=⎪⎪⎭⎫ ⎝⎛=RT P m RT P m V VTaking the entire contents of the cylinder as our system, the 1st law relation can be written as()()He12N 12HeN energiesetc. potential, kinetic,internal,in Change system massand work,heat,by nsferenergy tra Net out in )]([)]([0022T T mc T T mc U U U E E E -+-=∆+∆=∆=∆=-v vSubstituting,()()()()()()0C 25C kJ/kg 3.1156kg 0.808C 80C kJ/kg 0.743kg 4.77=︒-︒⋅+︒-⋅f fT TIt givesT f = 57.2︒Cwhere T f is the final equilibrium temperature in the cylinder.The answer would be the same if the piston were not free to move since it would effect only pressure, and not the specific heats.Discussion Using the relation P V = NR u T , it can be shown that the total number of moles in the cylinder is 0.170 + 0.202 = 0.372 kmol, and the final pressure is 510.6 kPa.6-9 An inventor claims to have developed a heat engine. The inventor reports temperature, heat transfer, and work output measurements. The claim is to be evaluated.Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from42%or 42.0K500K29011C th,max th,=-=-==H L T T ηη The actual thermal efficiency of the heat engine in ques tion is42.9%or 0.429kJ700kJ300net th ===H Q W η which is greater than the maximum possible thermal efficiency. Therefore, this heat engine is a PMM2 and the claim is false .6-11 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the power consumption of the heat pump are given. It is to be determined if this heat pump can do the job.Assumptions The heat pump operates steadily.Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined from()()()14.75K 27322/K 273211/11COP rev HP,=++-=-=H L T TThe required power input to this reversible heat pump is determined from the definition of the coefficient of performance to bekW 2.07=⎪⎪⎭⎫ ⎝⎛==s 3600h 114.75kJ/h 110,000COP HP minin,net,H Q W This heat pump is powerful enough since 5 kW > 2.07 kW.7-8A reversible heat pump with specified reservoir temperatures is considered. The entropy change of two reservoirs is to be calculated and it is to be determined if this heat pump satisfies the increase in entropy principle.Assumptions The heat pump operates steadily. Analysis Since the heat pump is completely reversible, the combination of the coefficient of performance expression, first Law, and thermodynamic temperature scale gives73.26)K 294/()K 283(11/11COP rev HP,=-=-=HL T T The power required to drive this heat pump, according to the coefficient of performance, is thenkW 741.326.73kW 100COP rev HP,in net,===HQ WAccording to the first law, the rate at which heat is removed from the low-temperature energy reservoir is5 kWnetkW 26.96kW 741.3kW 100in net,=-=-=W Q Q H L The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, iskW/K 0.340===∆K294kW 100H H H T Q Sand that of the low-temperature reservoir iskW/K 0.340-=-==∆K283kW 26.96L L L T Q S The net rate of entropy change of everything in this system iskW/K 0=-=∆+∆=∆340.0340.0total L H S S Sas it must be since the heat pump is completely reversible.7-11 Steam is expanded in an isentropic turbine. The work produced is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The process is isentropic (i.e., reversible-adiabatic).Analysis There is one inlet and two exits. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form asoutin energiesetc. potential, kinetic,internal,in change of Rate (steady) 0systemmassand work,heat,by nsferenergy tra net of Rate out in 0E E E E E==∆=-332211out out332211h m h m h m W W h m h m h m --=++=From a mass balance,kg/s 75.4)kg/s 5)(95.0(95.0kg/s 25.0)kg/s 5)(05.0(05.01312======m mm mNoting that the expansion process is isentropic, the enthalpies at three states are determined as follows:6)-A (Table KkJ/kg 6953.7kJ/kg4.2682 C 100 kPa 503333 ⋅==⎭⎬⎫︒==s h T P 6)-A (Table kJ/kg 3.3979K kJ/kg 6953.7 MPa 41311=⎭⎬⎫⋅===h s s Ps6)-A (Table kJ/kg 1.3309K kJ/kg 6953.7 kPa 7002322=⎭⎬⎫⋅===h s s P Substituting,kW6328=--=--=kJ/kg) .4kg/s)(2682 75.4(kJ/kg) .1kg/s)(3309 25.0(kJ/kg) .3kg/s)(3979 5(332211outh m h m h m W 7-13 The entropy change relations of an ideal gas simplify to∆s = c p ln(T 2/T 1) for a constant pressure processand ∆s = c v ln(T 2/T 1) for a constant volume process.Noting that c p > c v , the entropy change will be larger for a constant pressure process.7-22 Air is compressed in a piston-cylinder device. It is to be determined if this process is possible. Assumptions 1 Changes in the kinetic and potential energies are negligible. 4 Air is an ideal gas with constant specific heats. 3 The compression process is reversible.Properties The properties of air at room temperature are R = 0.287 kPa ⋅m 3/kg ⋅K, c p = 1.005 kJ/kg ⋅K (Table A-2a).Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed asin,out 12out in ,12out in ,12out in ,energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in )(since 0)( )(b b p b b W Q T T Q W T T mc Q W u u m U Q W E E E ===--=--=∆=-∆=-The work input for this isothermal, reversible process iskJ/kg 8.897kP a100kP a250K)ln K)(300kJ/kg 287.0(ln12in =⋅==P P RT w That is,kJ/kg 8.897in out ==w qThe entropy change of air during this isothermal process isK kJ/kg 0.2630kP a100kP a250K)ln kJ/kg 287.0(ln ln ln121212air ⋅-=⋅-=-=-=∆P P R P P R T T c s p The entropy change of the reservoir isK kJ/kg 0.2630K300kJ/kg 89.78R R ⋅===∆R T q s Note that the sign of heat transfer is taken with respect to the reservoir. The total entropy change (i.e., entropy generation) is the sum of the entropy changes of air and the reservoir:K kJ/kg 0⋅=+-=∆+∆=∆2630.02630.0R air total s s sNot only this process is possible but also completely reversible.Heat8-1 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.Properties The properties of air at room temperature are c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).Analysis (b ) From the ideal gas isentropic relations and energy balance,()()K 579.2kPa 100kPa 1000K 3000.4/1.4/11212=⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛=-kk P P T T()()()K3360==−→−-⋅=-=-=3max 32323in 579.2K kJ/kg 1.005kJ/kg 2800T T T T T c h h q p(c )()K 336K 3360kP a1000kP a1003344444333===−→−=T P P T T P T P v v ()()()()()()()()21.0%=-=-==-⋅+-⋅=-+-=-+-=+=kJ/kg2800kJ/kg 221211kJ/kg2212K 300336K kJ/kg 1.005K 3363360K kJ/kg 0.718in out th14431443out 41,out 34,out q q T T c T T c h h u u q q q p ηvDiscussion The assumption of constant specific heats at room temperature is not realistic in this case the temperature changes involved are too large.8-5 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a ) Process 1-2: isentropic compression.()()()()kP a 2338kP a 100K 308K 757.99.5K757.99.5K 3081122121112220.412112=⎪⎪⎭⎫⎝⎛==−→−===⎪⎪⎭⎫ ⎝⎛=-P T T P T P T P T T k v v v v vvProcess 3-4: isentropic expansion.()()K 1969==⎪⎪⎭⎫ ⎝⎛=-0.4134439.5K 800k T T vvvvsProcess 2-3: v = constant heat addition.()kPa 6072=⎪⎪⎭⎫⎝⎛==−→−=kPa 2338K 757.9K 19692233222333P T T P T P T P v v (b ) ()()()()kg 10788.6K 308K /kg m kPa 0.287m 0.0006kPa 100433111-⨯=⋅⋅==RT P m V()()()()()kJ0.590=-⋅⨯=-=-=-K 757.91969K kJ/kg 0.718kg 106.78842323in T T mc u u m Q v(c) Process 4-1: v = constant heat rejection.()()()()kJ0.2 40K 308800K kJ/kg 0.718kg 106.788)(41414out =-⋅⨯-=-=-=-T T mc u u m Q v kJ 0.350240.0590.0out in net =-=-=Q Q W59.4%===kJ0.590kJ0.350inout net,th Q W η(d ) ()()kPa 652=⎪⎪⎭⎫⎝⎛⋅-=-=-===kJm kPa 1/9.51m 0.0006kJ0.350)/11(MEP 331outnet,21outnet,max 2min r W W rV V V V V V8-7 An ideal diesel cycle has a a cutoff ratio of 1.2. The power produced is to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The specific volume of the air at the start of the compression is/kg m 8701.0kPa95K)288)(K /kg m kPa 287.0(33111=⋅⋅==P RT vThe total air mass taken by all 8 cylinders when they are charged is kg 008665.0/kgm 8701.0m)/4 12.0(m) 10.0()8(4/3212cyl 1cyl===∆=ππv v VS B N N m The rate at which air is processed by the engine is determined fromkg/s 1155.0rev/cycle2rev/s) 1600/60kg/cycle)( (0.008665rev ===N n m msince there are two revolutions per cycle in a four-stroke engine. The compression ratio is2005.01==routAt the end of the compression, the air temperature is()K 6.95420K) 288(14.1112===--k r T TApplication of the first law and work integral to the constant pressure heat addition giveskJ/kg 1325K )6.9542273)(K kJ/kg 005.1()(23in =-⋅=-=T T c q pwhile the thermal efficiency is6867.0)12.1(4.112.12011)1(1114.111.41th =---=---=--c k c k r k r r ηThe power produced by this engine is thenkW105.1====kJ/kg) 67)(1325kg/s)(0.68 (0.1155in th net netq m w m W η8-12 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).Analysis For the isentropic compression process,K .1527K)(10) 273(0.4/1.4/)1(12===-kk p r T TThe heat addition iskJ/kg 500kg/s1kW500in in===m Q qApplying the first law to the heat addition process,K 1025KkJ/kg 1.005kJ/kg500K 1.527)(in 2323in =⋅+=+=-=p p c q T T T T c q The temperature at the exit of the turbine isK 9.530101K) 1025(10.4/1.4/)1(34=⎪⎭⎫⎝⎛=⎪⎪⎭⎫⎝⎛=-kk p r T TApplying the first law to the adiabatic turbine and the compressor producekJ/kg 6.496K )9.5301025)(K kJ/kg 1.005()(43T =-⋅=-=T T c w pskJ/kg 4.255K )2731.527)(K kJ/kg 1.005()(12C =-⋅=-=T T c w pThe net power produced by the engine is thenkW 241.2=-=-=kJ/kg )4.2556kg/s)(496. 1()(C T net w w m W 0.482===kW 500kW 241.2in net th Q W ηFinally the thermal efficiency is。

工程热力学(高教社第四版)课件 第9章2

工程热力学(高教社第四版)课件 第9章2

9-6 燃气轮机装置循环用途:航空发动机尖峰电站移动电站大型轮船燃气轮机装置燃气轮机的利用燃气轮机装置简介燃气轮机示意图和理想化(布雷顿循环)23燃烧室工质:数量不变,定比热理想气体2)闭口⇒3)布雷顿循环(Brayton Cycle )图示s12341234布雷顿循环的计算Ts1234吸热量:()1p 32q c T T =−放热量:()2p 41q c T T =−热效率:12241t 1113211w q q q T T q q q T T η−−===−=−−布雷顿循环热效率的计算s1234热效率:t 12111k kp p η−=−⎛⎞⎜⎟⎝⎠循环增压比21p p π=111k kπ−=−πtηktη布雷顿循环净功的计算s1234循环增温比31T T τ=()()324134211111p p p w c T T c T T T T T c T T T T =−−−⎛⎞=−−+⎜⎟⎝⎠净1111k k k kp c T ττππ−−⎛⎞=−−+⎜⎟⎝⎠对净功的影响s123431T T τ=1111k k kkp w c T ττππ−−⎛⎞=−−+⎜⎟⎝⎠净3’4’当不变π不变τw 净但T 3 受材料耐热限制111t k kηπ−=−τ对净功的影响s31T T τ=1111k k kkp w c T ττππ−−⎛⎞=−−+⎜⎟⎝⎠净当不变τ太大πw 净π3T 太小πt ηt ηw 净存在最佳,使最大πw 净111t k kηπ−=−1T最佳增压比(w 净)的求解s1111k k kkp w c T ττππ−−⎛⎞=−−+⎜⎟⎝⎠净令opt π3T 2(1)opt ()k k w πτ−=净1T 0w π∂=∂净最大循环净功()211opt p w c T τ=−9-7 燃气轮机装置的定压加热实际循环s1234压气机:绝热压缩燃气轮机:绝热膨胀2’4’'21c 12h h h h η−=−定义:'34oi 34h h h h η−=−燃气轮机的实际循环的净功Ts12342’4’()()'''314221oi 34cw h h h h h h h h ηη=−−−−=−−净净功吸热量''2113312ch h q h h h h η−=−=−−'21c 12h h h h η−=−'34oi 34h h h h η−=−'21c 12h h h h η−=−燃气轮机的实际循环的热效率s12342’4’1'''111111oik ckk c kw q τηηπητηπ−−−==−−−净t 热效率影响燃气机实际循环热效率的因素1'''111111oik ckk c kw q τηηπητηπ−−−==−−−净t·oi ηc η'tη·π一定,τ't η·τ一定,有最佳()'opt t πη·τ()'opt t πη右移和的关系()'opt tπη()'optw π净()'optw π净()'opt tπη()'opt wπ净tητπ受材料耐热限制取最佳()'opttπη有无其它途径2T 4T 4 500o C 1344p T 4>T 2回热一、回热9-8 提高燃气轮机装置循环热效率的措施布雷顿循环回热示意图234压气机燃气轮机燃烧室回热器4R2A回热在Ts 图上的表示21344R2R2A回热度2222A R h h h h σ−=−0.6~0.9t t 1w q ηη=>净回简2R 4R 2A压气机间冷的图示23燃气轮机燃烧室间冷器5压气机62’压气机间冷在Ts 图上的表示21342’65AB t 1w q η=净间1234162’256联合工作?压气机间冷热效率的推导A B tA 1A tB 1B t 1A 1B 1A 1B 1A 1B tA tB 1A 1B 1A 1Bw w q q q q q q q q q q q q ηηηηη++==++=+++净净间tA tBηη>tA tBt ηηη>>间tA tB ηη<tA tB t ηηη<<间tA tBηη=tA tBt ηηη==间间冷+回热示意图3燃气轮机燃烧室间冷器5压气机62’回热器4R 2R间冷+回热在Ts 图上的表示21342’65t t 1w q ηη=>净间+回简4R2R再热示意图23压气机燃气轮机燃烧室1燃烧室23’5再热在Ts 图上的表示2133’4’4t t ηη<再简w w >再简5结论:再热+回热示意图123压气机燃气轮机燃烧室2回热器燃烧室14R2R53’再热+回热在Ts图上的表示2 133’4’454R2R2t+t11qqηη=−>再回回w w>再+回回再热+间冷+回热示意图1234压气机燃气轮机燃烧室2回热器间冷器燃烧室12R4R结论:再热+间冷+回热在Ts 图上的表示3T s 214t t +1w q ηη=>净再+间+回再回t t t t ηηηη>>>再+间+回再+回回简w w w w >>=再+间+回再+回回简+w w >再+间+回再回2R4R无穷多级的极限情况2 13 4两个等温过程两个等压过程+回热概括性卡诺循环2~3第9章小结活塞式内燃机循环:燃气轮机循环:提高热效率的手段:t ηη=简124w 净1’2’0 w=净动力循环的一般规律:热能代价以作功为目的升压是前提加热是手段作功是目的放热是必须顺序不可变步骤不可缺。

工程热力学课件完整版

工程热力学课件完整版
的热消失时,必产生相应量的功;消耗一定量的功时 ,必出现与之对应的一定量的热。
第三章 理想气体的性质
基本要求: 1、熟练掌握并正确应用理想气体状态方程式; 2、正确理解理想气体比热容的概念,熟练应用比热容计算理想 气体热力学能、焓、熵及过程热量; 3、掌握有关理想气体的术语及其意义; 4、掌握理想气体发生过程; 5、了解理想气体热力性质图表的结构,并能熟练应用它们获得 理想气体的相关状态参数。
T
不可逆过程的熵增(过程角度)
q
T
0
克劳休斯积分不等式(循环角度)
dsiso 0
孤立系统角度
ds sf sg 非孤立系统角度
熵、热力学第二定律的数学表达式
1. 熵的定义
ds qre
T
2. 循环过程的熵
3. 可逆过程的熵变
qre Tds
ds 0,则 q 0 可逆过程中ds 0,则 q 0
dv
q cndT Tds
T s
n
T cn
T ,定容过程 cV
T ,定压过程 cp
4个基本过程中的热量和功的计算
2
2
1、定容过程
w pdv 0 1
wt 1 vdp v( p2 p1)
2、定压过程
qv u cv (T2 T1)
2
w 1 pdv p(v2 v1)
热力学上统一规定:外界向系统传热为正,系统向外界传热为负。
可逆过程的热量
T
1
B
qre = Tds
T
A
2
q
ds qrev
T
S1
S dS S2
q “+”
q “-”
热力循环
功:工质从某一初态出发,经历一系列热力状态后,又回到原来 初态的热力过程称为热力循环,即封闭的热力过程,简称循环。

工程热力学与传热学(英文) 绪论

工程热力学与传热学(英文) 绪论

3. Basic Principles
(1)The first law of thermodynamics (2)The second law of thermodynamics
4. Main Contents
(1)The thermal properties of substances (2)Processes and cycles (3)Ways and technical measures to enhance the energy conversion efficiency
0-1 Energy
0-1-1 Energy
1. Energy(能量):
The measurement of the substance movement.
2. Various forms of energy
(1)Mechanical energy (2)Thermal energy
(3)Electrical energy
3. Research Approaches
Three basic approaches and their combination • Theory analysis • Numerical simulation • Experiment research
思考题
什么是能量?目前被人们认识和利用的能量形式有哪些? 什么是能源?能源是如何分类的? 热能利用的两种形式是什么? 热工基础的研究是什么? “热工基础”课与节能有怎样的关系? 以任意一种热能动力装置为例,分析其在热功转换过程 中所经历的过程。 7. 你能够从各种热能动力装置的工作过程中,初步概括出 它们在实现热功转换时的某些共同特性吗? 1. 2. 3. 4. 5. 6.
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CI engine-Diesel Cycle
1-2 isentropic compression 2-3 constant P heat addition 3-4 isentropic expansion 4-1 constant V heat rejection (等熵压缩) ; (定压吸热); (等熵膨胀); (定容放热)
SHANDONG UNIVERSITY
Chap7 Summary
Clausius inequality: the cyclic integral of is always ≦ zero
Entrop T
Entropy change of a closed system: Reversible process
– internal combustion(内燃机) : like automobile engines. It is done by burning the fuel within the system boundaries. – and external combustion engines(外燃机) : like steam power plants. Heat is supplied to the working fluid from an external source such as a furnace, a geothermal well, a nuclear, or even the sun.
Four strokes:compression stroke, expansion or power stroke, exhaust stroke,intake stroke. (压缩冲程、做功(燃烧、膨胀)冲程 、排气冲程和吸气冲程)
SI engines-Otto cycle The ideal Otto cycle:
1, Entropy change of pure substances:
sf at 0.01℃ =0 kJ/kg.k
2, Isentropic process
0,
adibatic
0,
Reversible
3, T-S, h-s diagrams
=
Some remarks
4, The 3rd law of thermodaynamics: The entropy of a pure crystalline substance at absolute zero temperature is zero
Increase of Entropy Principle(熵增原理)
Irreversible process
>
≥0
Increase of entropy principle (孤立系统熵增原理,简称熵增原理):the entropy of an isolated system during a process always increase or , in the limiting case of a reversible process remains constant. (孤立系统的熵可以增大,或保持不变,但不可能减少)
refrigeration cycles chapter 11
SHANDONG UNIVERSITY
Application of thermodynamics
• Thermodynamics cycles, according to different phase of
work fluid
– Gas cycles (working fluid remains in gaseous phase during the cycle) – Vapor cycles (working fluid exists in vapor phase and liquid phase during different part of the cycle)
1-2 isentropic compression 2-3 constant V heat addition 3-4 isentropic expansion 4-1 constant V heat rejection (等熵压缩) ; (定容吸热); (等熵膨胀); (定容放热)
The ideal Diesel cycle:
2
SHANDONG UNIVERSITY
Application of thermodynamics
• Two important areas:
– Power generation (engines-produce work) power cycles
Chapter 9 and chapter 10 – Refrigeration (refrigerators, air conditioners, heat pumps)
TDC BDC Stroke Bore Intake valve Exhause valve MEP
Reciprocating engines
Reciprocating engines are classified as spark-ignition (SI) engines (点
燃式内燃机) compression-ignition (CI) engines (压燃式内燃机)
air standard assumptions(空气标准假设): 1) air=ideal gas, Cv = const; 2) internal reversible process; 3) combustionheat addition process; 4) exhaust heat rejection process
• Thermodynamics cycles can also be categorized as closed cycles and open cycles.
SHANDONG UNIVERSITY
Application of thermodynamics
• Heat engines are categorized as
5, T ds relations: 6, reversible work output
1
SHANDONG UNIVERSITY
Chap9 Summary
Assumptions of gas power cycles
Basic considerations : actual cycle, ideal cycle, carnot cycle, P-V, T-S
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