化工热力学授课教案设计 ch2

Chapter 2 The First Law and Other Basic Concepts

2.1 The First Law of Thermodynamics 2.2 Energy Balance for Closed System 2.3 Mass and Energy Balance for Open System

2.1 The First Law of Thermodynamics

(1) Statement of the first law of thermodynamics:

Although energy assumes many forms, the total quantity of energy is constant, and when energy disappears in one form it appears simultaneously in other forms.

(2) The basic form of the first law of thermodynamics:

△( Energy of the system) + △( Energy of surroundings) =0 Characteristics of energies (1) Heat and work

(2) Potential, kinetic, and internal energy 能量分类:

(1)系统蓄积的能量，如动能、势能和热力学能，是系统状态的函数。 (2)过程中系统和环境传递的能量，如功、热量，不是状态函数，与过程有关。

2.2 Energy Balance for Closed System

Definition of closed system

if the boundary of a system does not permit the transfer of matter between the system and its surroundings, the system is said to be closed system Energy Balance for Closed System

△( Energy of the system) =Q+W That is Energies of a system include internal, potential, kinetic energy, etc..

Closed systems often undergo processes that cause no change in the

W

Q E +=?

system other than in its internal energy.

Thinking(2-1)

Decreasing temperature of room shutting door and window by refrigerator Chose room as the system Thinking(2-2)

Decreasing temperature of room shutting door and window by air conditioner Chose room as the system

2.3Mass and Energy Balance for Open System

2.3.1 Measures of Flow Four common measures of flow

A: is the cross-sectional area of a conduit

ρ: is specific or molar density

Mass flowrate

Molar flowrate V olumetric flowrate q Velocity u

i

m i

n m uA n uA ρρ

==

Q and W , transferring into the system from the surroundings, are positive

and

U Q W

?=+dU dQ dW

=+W

Q E +=?

Q

T ?

Refrigerator

u: average speed of a stream 2.3.2 Mass Balance for Open System Definition of control volume (CV)

The region of space identified for analysis of open systems For control volume

The flow is positive when directed into the CV and negative when directed out

The continuity equation （连续性方程） For steady-state flow process

When only a single entrance and a single exit

一些常见的属于稳流体系的装置

Energy analysis for open system

()0CV fs dm m dt +?=

Control volume

Control surface

3

m 2

m 1

m /CV dm dt

()0CV

fs dm uA dt

ρ+?= ()0

fs uA ρ?=2221110

u A u A ρρ-=2211

21

0u A u A V V -=换热装置

压缩机

喷嘴 扩压管 透平机

混合装置

节流阀

The net rate of flow of mass into CV

Each unit mass of a stream carries with it a total energy

Each stream transports energy at the rate

The net energy transported into the system by the flowing stream (in-out)

Energy balance for open system

The rate of energy accumulation within CV

Work rate

work rate =the flowing work rate + the other work rate （功传递速率=流动功速率+其他功）

the flowing work rate ：moving the flowing streams through entrances and

exits through the entrance The work rate

through the exit The work rate the flowing work rate the other work rate:

2()()2CV fs

d mU u U gz m Q work rat

e dt ??

=-?++++???? 2

2

i i u U 2p i

Total energy entering

the system

Total energy leaving

the system

Change in the total energy of the system

=

-

2

2

u U gz

++2()2

u U gz m ++ 2()2fs u U gz m ??-?++??

?? ()CV

d mU dt

()i i i

pV m i i p V ()j j j

p V m j j p V [

]()fs pV m -?

=the shaft work rate + the expansion work rate + contraction work + stirring work rate + …

Energy balance for open system

2.3.4 Energy Balance for Steady-State Flow Process Review

稳态流动是指流体流动途径中所有各点的状况都不随时间而变化，系统中没有物料和能量的积累。

features ：The changes of total properties such as ?H 、?V 、?E 、?S 、?G all are zero

General energy balance for open system

H =U +pV

2()()2CV fs

d mU u H gz m Q W dt ??

=-?++++???? Omit potential, kinetic energy

[]()CV fs d mU Hm Q W dt

=-?++

For closed system

()CV

d mU Q W dt

=+ W

[]()fs work rate pV m W

=-?+

2()()[()]2CV fs fs

d mU u U gz m Q pV m W dt ??=-?+++-?+????

The first law of a steady-state flow process Attention!

使用时注意单位必须一致。按照SI 单位制，每一项的单位为 J ·kg-1。 Omit the potential, kinetic energy

Attention! for closed system

例题6-1 将90℃的热水，以12m 3·h-1速率从储罐1输送到高度为15m 的储罐2，热水泵的电动机功率为1.5kW ，并且热水经过一个冷却器，放出热量的速率为2.5×106 kJ ·h-1，试问：储罐2的水温度是多少？

思路 A steady-state flow process

Omit kinetic energy

For one entrance and one exit

For a steady-state flow process

()0CV d mU dt =2()2s fs u H gz m Q W ???++=+???? 2()2s

u H gz m Q W ?++=+

Divided by m 2()2s

s

W u Q H gz Q W m m

?++=+=+ 2()()2CV fs

d mU u H gz m Q W dt ??

=-?++++???? 2

2

s

u H g z Q W

??++?=+or s

H Q W ?=+

Attention!

for closed system

s

U Q W ?=+2

2

s

u H g z Q W ??++?=+s H Q W g Z

?=+-?21H H H

=+?近似为饱和液体的焓

Solution

从水性质表中可查得

90℃时水的密度为 965.3kg ·m-3

水的质量流率为 965.3×12 = 11583.6 kg ×h-1

得到放出的热量

从附录C-1饱和水性质表中可查得90℃时饱和液体的焓

再从饱和水性质表中可插查到此时的温度约为38.5℃。

5.1.3 Application for energy balance equation (1) For closed system

d m 1(U + pv +u 2/2 + gz )1- d m 2(U + pv+u 2/2 + gz )2 + d Q + d W s = d(mE )system

δm1=δm2=dm=0 δQ+δWs=mdE

E=U+E P +E K mdE=mdU δW ＝δWs

(2) For Steady-State Flow System

16

kg kJ 8.2156

.11583105.2-?-=?-

=Q 1kg kJ 466.06

.1158336005.1-?=?=s W 轴功

1

3kg kJ 147.010158.9--?=??=?z g 势能

1

kg kJ 48.215147.0466.08.215-?-=-+-=?-+=?z g W Q H S 1

1kg kJ 92.376-?=H 1

12kg kJ 44.16192.37648.215-?=+-=+?=H H H W

Q U +=?dU Q W

δδ=+12()0

sys d mE m m m

δδδ===22121

1

()()0

22S Q W H gZ u m H gZ u m δδδδ++++-++=s

W Q z g u H +=++???22For differential process 2

2s du dH gdz Q W δδ++=+ Attention!

使用时要注意单位必须一

稳态流动是指流体流动途径中所有各点的状况都不随时间而变化，系统中没

有物料和能量的积累。

features ：The changes of total properties such as ?H 、?V 、?E 、?S 、?G all are zero

①For eqipment such as turbine and compresser, etc. g ?Z ≈0 ?u2/2 ≈0

②For eqipment such as valve, pipe, heat- exchange chamber, etc. g ?Z ≈0

?u2/2 ≈0 Ws=0 ③For adiabatic reaction, adiabatic mixing, etc.

Ws=0 g ?Z= ≈0 Q=0 ?u2/2 ≈0

④ Balance equation of mechanical energy dH=TdS+Vdp

For reversible process δQ=TdS

For real fluid, Consider energy lost owing to friction

For a non-viscous, non-compressible fluid, and Ws=0

Bernoulli equation

For example

Turbine and compresser,etc.

g ?Z=? ≈0 ?u2/2 =? ≈0

s

dH udu gdz Q W δδ++=+s

W Q z g u H +=++???22

s H W Q

?=+s

W Q z g u H +=++???2

2

H Q

?=s W Q z g u H +=++???2

20H ?=s

dH udu gdz Q W δδ++=+s VdP udu gdz W δ++=s

VdP udu gdz F W δδ+++=0

VdP udu gdz ++=2

02

P

u g z ρ??++?=

透平机是借助流体的减压和降温过程来产出功 压缩机可以提高流体的压力，但是要消耗功 Valve g ?Z=? ≈0 ?u2/2 =? ≈

Ws=? =0

Thinking!

理想气体通过节流阀温度?

不变 Mixing chamber g ?Z=? =0 ?u2/2 =? =0

Ws=? =0

For adiabatic reaction, adiabatic mixing, etc.

Ws=? =0 g ?Z=? =0 Q=? ≈0 ?u2/2 =? =0

Adiabatic mixing chamber

Q=? ≈0

s

W Q z g u H +=++

???2

2

s H W Q

?=+s W Q z g u H +=++

???2

2

H Q

?=混合器

s

W Q z g u H +=++

???2

2

H Q

?=s

W Q z g u H +=++

???2

2

H ?=

g ?Z=? =0 ?u2/2 =? =0

Ws=? =0

Nozzle and Diffuser

Ws=? =0 Q=? ≈0

g ?Z=? =0

流体通过焓值的改变来换取动能的调整

喷嘴与扩压管的结构特点是进出口截面积变化很大。流体通过时，使压力沿着流动方向降低，而使流速加快的部件称为喷嘴。反之，使流体流速减缓，压力升高的部件称为扩压管。

例 5-1 1.5MPa 的湿蒸汽在量热计中被节流到0.1MPa 和403.15K ，求湿蒸汽的干度

解：节流过程无功的传递，忽略散热、动能变化和位能变化

T ℃ H kJ/kg 120 2716.6 130 H2 160

2796.2

1.5MPa 饱和液体焓值 Hl=844.9 饱和蒸汽焓值 Hg=279

2.2

s

W Q z g u H +=++

???2

2

H ?=s

W Q z g u H +=++

???2

2

2

2

u H ??+=22

12

212

u u H H --=

喷嘴

扩压管

s W Q z g u H +=+???2

2

21

H H =6

2716227966

27161201601201302...H --=

--kg

/kJ .H 527362=212736.5

H H ==()x

H x H H g l +-=11

例 5-2

将空气当作理想气体，并忽略压降时

换热器的动能变化和位能变化可以忽略不计

9709

09

844227929

844527361.....H H H H x l g l =--=--=

s

W Q z g u H +=++

???2

2

()()kJ .T T C m H m P 603030342300515012=-??=-=?1

1

22V T V T =A

u T

A u T 11

22=s /m .T T u u 986303

4235121

2=?==kJ .J .u m 59305932

598650212

22==-?=?kJ

.J .z mg 47211472381950==??=?kJ

..Q 6032472159306030=++=