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《热质交换原理与设备》习题答案(第3版)

《热质交换原理与设备》习题答案(第3版)
《热质交换原理与设备》习题答案(第3版)

第一章绪论

1、答:分为三类。动量传递:流场中的速度分布不均匀(或速度梯度的存在);

热量传递:温度梯度的存在(或温度分布不均匀);

质量传递:物体的浓度分布不均匀(或浓度梯度的存在)。

2、解:热质交换设备按照工作原理分为:间壁式,直接接触式,蓄热式和热管式等类型。

●间壁式又称表面式,在此类换热器中,热、冷介质在各自的流道中连续流动完成热量传

递任务,彼此不接触,不掺混。

●直接接触式又称混合式,在此类换热器中,两种流体直接接触并且相互掺混,传递热量

和质量后,在理论上变成同温同压的混合介质流出,传热传质效率高。

●蓄热式又称回热式或再生式换热器,它借助由固体构件(填充物)组成的蓄热体传递热

量,此类换热器,热、冷流体依时间先后交替流过蓄热体组成的流道,热流体先对其加热,使蓄热体壁温升高,把热量储存于固体蓄热体中,随即冷流体流过,吸收蓄热体通道壁放出的热量。

●热管换热器是以热管为换热元件的换热器,由若干热管组成的换热管束通过中隔板置于

壳体中,中隔板与热管加热段,冷却段及相应的壳体内穷腔分别形成热、冷流体通道,热、冷流体在通道内横掠管束连续流动实现传热。

3、解:顺流式又称并流式,其内冷、热两种流体平行地向着同方向流动,即冷、热两种流体由同一端进入换热器。

●逆流式,两种流体也是平行流体,但它们的流动方向相反,即冷、热两种流体逆向流

动,由相对得到两端进入换热器,向着相反的方向流动,并由相对的两端离开换热器。

●叉流式又称错流式,两种流体的流动方向互相垂直交叉。

●混流式又称错流式,两种流体的流体过程中既有顺流部分,又有逆流部分。

●顺流和逆流分析比较:

在进出口温度相同的条件下,逆流的平均温差最大,顺流的平均温差最小,顺流时,冷流体的出口温度总是低于热流体的出口温度,而逆流时冷流体的出口温度却可能超过热流体的出口温度,以此来看,热质交换器应当尽量布置成逆流,而尽可能避免布置成顺流,但逆流也

有一定的缺点,即冷流体和热流体的最高温度发生在换热器的同一端,使得此处的壁温较高,为了降低这里的壁温,有时有意改为顺流。

第二章 传质的理论基础

1、答:单位时间通过垂直与传质方向上单位面积的物质的量称为传质通量。传质通量等于传质速度与浓度的乘积。

以绝对速度表示的质量通量:,,A A A B B B A A B B m u m u m e u e u ρρ===+ 以扩散速度表示的质量通量:(),(),A A A B B B B A B j u u j u u u j j j ρρ=-=-=+

以主流速度表示的质量通量:1()()

A A A A

B B A A B e u e e u e u a m m e ??

=+=+????

()B B A B e u a m m =+

2、答:碳粒在燃烧过程中的反应式为22C O CO +=,即为1摩尔的C 与1摩尔的2O 反应,

生成1摩尔的2CO ,所以2O 与2CO 通过碳粒表面边界界层的质扩散为等摩尔互扩散。

3、从分子运动论的观点可知:D ∽31

2

p T -

两种气体A 与B 之间的分子扩散系数可用吉利兰提出的半经验公式估算:

4

10D -=

若在压强5

001.01310,273P

Pa T K =?=时各种气体在空气中的扩散系数0D ,在其他P 、T 状态下的扩散系数可用该式计算

32

00

0P T D D P T ??= ???

(1)氧气和氮气:

2233025.610/()32o V m kg kmol μ-=??=

223331.110/()28N N V m kg kmol μ-=??=

52

11

5233 1.5410/1.013210(25.6)D m s -==???+

(2)氨气和空气:

51.013210P Pa =? 25273298T K =+=

5

0 1.013210P Pa =? 0273T K =

3

221.0132980.2()0.228/1.0132273D cm s

=??=

2-4、解:气体等摩尔互扩散问题

12423

0.610(160005300)

()0.0259/()8.3142981010A A A D N P P kmol m s RT z --??-=-==?????

m 2s R 0通用气体常数单位:J/kmol ﹒K

5、解:250

C 时空气的物性:3

51.185/,

1.83510,kg m Pa s ρμ-==??

6242015.5310/,0.2210/m s D m s υ--=?=?

32

42

00066

4

0.2510/40.08

Re 2060515.531015.53100.620.2510

o c P T D D m s P T u d v v S D ----??==? ???

?=

==??===?

用式子(2-153)进行计算

0.830.440.830.444

0.0230.023206050.6270.95

70.950.25100.0222/0.08

m e c m m sh R S sh D h m s

d -==??=??===

设传质速率为A G ,则

211

2

2000

0()()()

4

4ln

4A A A m A s A A l

A m A s A

A s A m A s A dG d dx h d u d du d dx h du l h ρρπ

πρρρρρρρρρρ????=-=

=

--=

-?

?

2-6、解:20℃时的空气的物性:(注:状态不同,D 需修正)

353

35

2

2

442

005

05

5

4

1.205/, 1.8110,

1.013102930.22100.2410/1.0132102730.053 1.205

Re 99901.81101.81100.626

1.2050.2410o c kg m Pa s P T D D m s P T u d

v S D ρμρμρ------==???????==???=? ? ????????=

=

=??===??

(1)用式0.830.44

0.023m e c sh R S =计算m h

0.830.444

0.02399900.6260.24100.01875

0.05m m sh D h d -????===

(2)用式13

3

4

0.0395e c sh R S =计算m h

13

4

3

40.0395(9990)(0.626)0.24100.01621/0.05m sh D h m s

d -??===

2-7、错解:氨在水中的扩散系数92

1.2410/D m s -=?,空气在标准状态下的物性为;

3535

9

1.293/, 1.7210,

Pr 0.708, 1.00510/()

1.721010727.741.293 1.2410p c kg m Pa s c J kg k S D ρμμρ----==??==???===?? 由热质交换类比律可得

23

1Pr m p c h h c S ρ??= ???

2

23

3

5

1Pr 560.7087.0410/1.293100110727.74m p c h m s h c S ρ-????==?=? ? ??????

? 1)(第3版P25)用水吸收氨的过程,气相中的NH3(组分A )通过不扩散的空气

(组分B ),扩散至气液相界面,然后溶于水中,所以D 为NH3在空气中的扩散。 ? 2)刘易斯关系式只对空气——水系统成立,本题为氨——空气系统,计算时类比关

系不能简化。

? 3)定压比热的单位是J/kgK

正解:组分A 为NH3,组分B 为空气,空气在0℃时物性参数查附录

3-1

)2-2P36(/102.0708

.0Pr 664

.0102.01028.13244

6

表查s m D D S c ---?===??==υ

23

1Pr m p c h h c S ρ??= ???

h m s m Sc c h h p

m /10161/98.44708.0664.0005.1293.156Pr 33

/23

/2?==??

? ????=??

? ???=

--ρ

8、解:325

100.04036/8314(27325)i CO P C kmol m

RT ===+

22N CO C C = 2

2

2220.5

N N CO N CO C x x C C ===+ 3225

4410 1.776/8314298CO i

CO M P kg m

RT ρ??===?32

252810 1.13/8314298N i N M P kg m

RT ρ??===?

2

22

2

0.611

CO

CO CO N

a ρρρ=

=+ 20.389

N a =

9、解:(a )已知A M ,B M ,A x ,B x

A A A A A

A A

B A A B B A A B B M n M x M a M M n M n M x M x M =

==

+++ B B B B B

B A B A A B B A A B B M n M x M a M M n M n M x M x M ===

+++ 已知B a ,A a ,A M ,B M

A

A A

A A

A A

B A B A B A B A B m a n M M x m m a a n n M M M M =

==+++

B B

B B B

B A B

A B

A B A B A B m a n M M x m m a a n n M M M M ===

+++

(b )22222222232

0.3077

322844O O O O O N N CO CO x M a x M x M x M ===++++ 20.2692N a = 20.4231CO a =

若质量分数相等,则2

222222

2

2

1320.3484111322844

O O O O N CO O N CO a M x a a a M M M =

==+++

+

20.3982N x = 20.2534CO x =

10、解;(a )2O ,2N 的浓度梯度沿垂直方向空气由上部向下部运动: (b )2O ,2N 的浓度梯度沿垂直方向空气由下部向上部运动,有传质过程。

2-11、解;

12()aV

A A aV A A DA G N A C C z

==

-?212

1

2()

ln aV L r r A r r π-=

? 1)柱形:L d V r r r r L A av 21

21241

,ln )(2ππ=-=

球形:3

213

4,4d V r r A av ππ==

? 2)d=100mm 为内径,所以r 1=50,r 2=52

若为球形A av =0.033,质量损失速率为1.46×10-12

kg/s ;压力损失速率3.48×10-2Pa/s

2-12、解:98

123

10(0.020.005)() 1.510/()110A A A D N C C kmol m s z ---?-=-==????

? 1)j A 为A 的质量扩散通量,kg/m 2s ;J A 为A 的摩尔扩散通量kmol/m 2s ;

? 2)题中氢氦分子量不同

2-13、解: 氨---空气

425

0000.210/, 1.01310,273,350,O a D m s P P T K T K P P -=?=?===

3

32

2

442

003500.2100.2910/273O D P T D m s

P T --????=

=??=? ? ?????

氢—空气

420.51110/O D m s -=?

3

32

2

442

003500.511100.74210/273O D P T D m s

P T --????=

=??=? ? ?????

2-14溶解度s 需先转化成摩尔浓度:

3

4

3

11/105.103.0105m kmol sP C A A --?=??==

()()

s kmol C C Z DA A N G A A av av A A /1025.20105.101

.05

310104921---?=-???=-?=?= s kg M G M A A A

/1005.4181025.2910--*

?=??=?=τ

2-15、解、3

2

212

1

2()20.5100.12420

ln ln 19.5aV L r r A m r r ππ--??=== 3111602320/A A C sP kmol m ==?=

3221600.116/A A C sP koml m ==?=

11

6123

1.8100.124()(32016) 1.35710

0.510aV

A A A DA G C C z ---??'=-=?-=???/koml s 质量损失66

1.357102

2.71410/A G kg s --=??=?

16、解:02225CO N C 和在时,扩散系数420.16710/D m s -=?

33121013.6109.86664A A a

P P P --=????=(100-50)51112 1.67106664()48.810/83142981

A A A A D P P G N A koml s

RT z

π

--???

-==

=

=????

18、解、该扩散为组分通过停滞组分的扩散过程

(),0A

A A A

B B A

A A A

dG N D

x N N N dr dG N D x N dr =-++==-+ ,A A

A A P P C x RT P == A A

A A

dP P D N N RT dr P =-+

整理得

()A A A dP DP

N RT P P dr =-

- 24A A r G N A r π==

24()A A A G dP DP

r RT P P dr π=--

分离变量,并积分得0024AS A A

r P A

G RT dP dr DP r P P π∞=--??得

4ln

AS

A P P DPr G RT P π-=-

第3章传热传质问题的分析和计算

1、答:当物系中存在速度、温度和浓度的梯度时,则分别发生动量、热量和质量的传递现象。动量、热量和质量的传递,(既可以是由分子的微观运动引起的分子扩散,也可以是由旋涡混合造成的流体微团的宏观运动引起的湍流传递)

动量传递、能量传递和质量传递三种分子传递和湍流质量传递的三个数学关系式都是类似的。

2、答:将雷诺类比律和柯尔本类比律推广应用于对流质交换可知,传递因子等于传质因子

①22

33

r P 2m H D t t c G

J J S S S ===?=?

② 且可以把对流传热中有关的计算式用于对流传质,只要将对流传热计算式中的有关物理

参数及准则数用对流传质中相对应的代换即可,如:r ,,,P ,,m c u h t t t c a D D S N S S S λ??????

③当流体通过一物体表面,并与表面之间既有质量又有热量交换时,同样可用类比关系由传

热系数h 计算传质系数m h 2

3

m h

h Le e φ-=?

3:答:斯密特准则

c i v S D =

表示物性对对流传质的影响,速度边界层和浓度边界层的相对关系

刘伊斯准则r P c v

S D a Le v D a ===

表示热量传递与质量传递能力相对大小 热边界层于浓度边界层厚度关系

4、解:定性温度为

02520

22.5,

2

g t C +=

=此时空气的 物性

ρυ?2

3-6

=1.195kg/m ,=15.29510m /s 查表得:?-42o D =0.2210m /s,0C 25饱和水蒸汽的浓度30.02383/v kg m ρ=

3

32

2

44

00 1.0132980.22100.2510/1.0132273O D P T D m s

P T --????==???=? ? ?????

022

20

20

9.48/3.140.025

3600 1.195360044u m s d πρ=

=

=?????

0e 9.480.025R 15488

u d υ?===?-615.29510

40.25100.61

c D S υ-??===-6

15.29510

用式(2--153)计算

0.830.440.830.440.0230.023154880.6155.66,m e c sh R S ==??=

4255.660.2410 5.56610/0.025m m sh D h m s

d --??===?

设传质速率为A G ,则 20()()()

4

A m A s A A dG d dx h d u d π

πρρρ?=-=

2100

4A A l

A

m A s A du d dx h ρρρρρ?=

-?

?

1

2

04exp(

)A s A

A A s m

h du ρρρρ??-=-

020C 时,饱和水蒸汽的浓度3

0.0179/A s kg m ρ?=

1

1A

A

d

ρρρ=-

1

33

0.003 1.195

3.5710/110.003

A d kg m d ρρ-??∴===?++

∴ 代入上面的式子得:23

0.01193/A kg m ρ=

1

1

2.23/A A

d g kg

ρρρ==-

5、解:

040,C 时空气的物性ρυ?23-6=1.128kg/m ,=16.9610m /s 6

0e 210R 1.1810

u l

υ?=

=

=??-616.9610

转折点出现在56

e 51010

1.1810e R , 4.24R c x l m μ

ν???=== 因此,对此层流---湍流混合问题,应用式(2-157)3

0.8

(0.037870)e

c

L

R S Sh γ=-

查表2—4得,定性温度为350

C 时,

3

2

4000.26410O D P T D P T -??=

=? ???2

m /s

4

0.264100.64c D

S υ-??=

==-6

16.9610

3

60.8[0.037(1.1810)870]0.641548.9

L

Sh γ=??-?=4

30.288101548.9 4.4610/10mL L D h Sh m s

L --???

==?=? ???

每2

m 池水的蒸发速率为()m A A S A n h ρρ?∞=-

300

C 时,3030.03037/;40,0.05116/A S A S kg m C kg m ρρ??'==时

()

35

4.4610(0.030370.50.05116) 2.1410m A A S A S n h ρ?ρ--??'=-=??-?=?

6、解:在稳定状态下,湿球表面上水蒸发所需的热量来自于空气对湿球表面的对流换热,即可得以下能量守衡方程式2()s fg H O h T T h n ∞-=其中fg h 为水的蒸发潜热

22

2

()

H O H O H O m S n h ρρ??∞=-

22

()H O H O m

s fg

S h T T h h ρρ∞??∞=+-

2

3

r P 1m p c h h c S ρ??= ????

查附录2—1,当s T =0

35C 时,水蒸汽的饱和蒸汽压力5808S P

=于是 32

580818

0.0408/

8314308H O

S S s

P M kg m

RT ρ?=

=

=?

0ρ∞=

3-7、三种方法

? 1)含湿量是什么?d 与相对湿度的区别 ? 2)主体空气为湿空气,其C f 不等于0。 ? 2-14分析

方法3解:2()()s H O m S h T T r n r h ρρ∞∞-=?=?-

其中0

26,20S t C t C ∞== 查表2—1,当0

20S t C =时水蒸汽的饱和蒸汽压力

2330S a P P = 于是

2

233818

0.017278314293H O

S S s

P M kg

RT ρ?=

=

=?

2454.3/r kJ kg =

1V d d ρρρ∞?==

+

当0

26t C ∞=,时定性温度为

023,2s

t t t C ∞+=

=31.193/ 1.005/()

p kg m c kJ kg k ρ=?=?

由奇科比拟知22

3

3

4

r P 110.749.59101.197 1.0050.6m p c h h c S ρ-????===? ? ???????

()1S s m h d T T d rh ρρ∞?=--+ 41.19326200.0172712454700905910d d

-?-=-

+?? d=12.5g/kg

3-8、解:()A m A S A n h ρρ??∞=-

查表得当温度为270

C 时,3

0.026446/A S kg m ρ?=

A A S ρ?ρ?∞?=

1

54.02/(13090)0.026446A m A S A n h m h

ρρ??∞===--?

3-9、解:(a )当温度为23

C 时,A S ρ?=0.0212143/kg m 0A ρ?∞=

1

21.510170.27/()0.0230.021214

4A m A S A n h m s

A πρρ-??∞?===-??

(b) 2()170.270.0230.021214(10.5)0.075/4A m A S A S n h A kg s

π

ρ?ρ??=-=?????-=

(c) 当温度为470

C ,A S ρ?=0.0734623/kg m

2170.270.0230.0734620.519/4

A m A S n h A kg s

π

ρ?==?

???=

求h m 时需除以面积A

3-10、解:

320.1

110 2.7810/3600A n kg s -=

??=?

当温度为305K 时,

A S ρ?=0.034533/kg m

2

2.7810 1.34/(1)10.3453(10.4)A m A S n h m s

A ρ?-??===-??-

11、解:2446

12()10(610510)10A m A A m h ρρ----=-=??-?=

第四章 空气的热湿处理

1、(1)大气是由干空气和一定量的水蒸汽混合而成的。我们称其为湿空气,其主要成分是:

氮、氧有、氩、二氧化碳、水蒸气等。

(2)在湿空气中水蒸气的含量虽少,但其变化确对空气环境的干燥和潮湿程度产生重要的影响。且使湿空气的物理性质随之改变。因此研究湿空气中水蒸气的含量在空气调节行业中占重要地位.

2、(1)湿空气的密度等于干空气密度与蒸汽密度之和。

3

0.001315/287s P B

kg m T T φρ=

-

在大气压力B 和T 相同情况下,湿度增大时,湿空气的密度将变小。

天气由晴转阴时,空气中水蒸汽的含量增加,由此降低了空气的密度,于是大气压要下降。 (2) 在冬季。天气干燥。水蒸汽在空气中含量减少,而且温度T 也减少了,所以密度增加了,于是冬季大气压高于夏季的。

3、(1)在大气压强。温度一定的条件下,湿空气的水蒸汽分压力是指,在与湿空气同体积的条件下,将干空气抽走,水蒸汽单独存在时的压力。湿空气的水蒸汽饱和分压力是指,在与饱和湿空气同体积的条件下,将干空气抽走,水蒸汽单独存在时的压力。 湿空气的水蒸汽饱和分压力是湿空气的水蒸汽分压力的上限。 (2)它们的大小是受大气压力影响的。

4、(1)会有凝结水产生。

(2)由附录4—1可知:当房中漏点温度为9.5 ℃ 而冷水管表面温度为8℃ 所以会有凝结水产生。

(3)若想管道表面不产生凝结水,则可以对房间内空气进行除湿。 5、

由附录4—1可知:湿空气20℃ 1φ=50%时,i=39kJ/kg(干空气) ;

湿空气15℃,

2φ=90% 时,i=39kJ/kg(干空气); 所以空气的焓值没有发生变化。

6、由已知得,ε=Q/W =14000/2=7000 (kJ/kg ) 由初始状态B=0.1MPa, 1t

=18℃, 1φ=50%

终状态

2t =25℃,查 附录4—1 得2φ=40%,2i =45.5 kJ/kg(干空气) 2d =7.9g/kg(干空气)

4-7、由已知得,ε=5000 (kJ/kg ) 由初始状态 1t

=20℃, 终状态2t =30℃, 2φ=50% 查 附录4—1 得1φ=62%,1i =43 kJ/kg(干空气) 1d =9g/kg(干空

气)

8、解:(a,b,c )

由室内空气状态:温度20℃,压力101325Pa 水蒸汽分压力为1400Pa ,查附录4—1 得d=8.8g/kg(干空气) φ=6 0%,i=42 kJ/kg(干空气)

(d ) 已知干空气的气体常数为287J/(kg*k )干空气分压力B-Pq=101325-1400=99925(Pa)

干空气密度:

3

99925

1.188/287287293g g p kg m T

ρ=

=

=?

室内干空气质量;

1.1885 3.3358.8g g M V kg

ρ==???=

(e):室内水蒸汽质量:Mq=8.8*58.8=517.5g

(f):如果使室内空气沿等温线家湿至饱和状态,则角系数ε=2500 kJ/kg 当空气的状态是温度为20℃,φ=100%时,则d=14.6g/kg(干空气) 水蒸汽分压力2350 Pa 此时室内的干空

气的密度为

3

1013252350

1.177/287293g kg m ρ-=

=?

室内干空气质量为Mg=1.177?5?3.3?3=58.26kg 室内水蒸汽质量为14.6?58.26=850.6g 加入的水蒸汽量;850.6-517.5=333.1g 4-9、解:

由题意得,可知,假设室内空气的露点温度为7℃,则在标准大气压下,初始温度为20℃,露点温度为7℃的空气参数。可由附录4—1 得d=6.2g/kg(干空气) φ=42.5%,所以允许最大相对湿度为42.5%,最大允许含湿量是6.2g/kg(干空气) 10、解:a ,由附录4—1 得1t

=25℃,φ1=70%时,

1d =14g/kg(干空气)

2t =15℃, 2φ=100%时,2d =10.5g/kg(干空气)

失去的水蒸汽△d=1d -2d =14-10.5=3.5g

(b,c,d )

(2500)

1000p g p q d i c t c t ??=?++?

()

1(2515)10/p g Q c t kJ kg ??=??=?-=干空气

114

(2500 1.8425)35.644/1000Q kJ kg =+?= 210.5

(2500 1.8415)

26.54/1000Q kJ kg =+?=

29.1/Q kJ kg ?=

空气状态变化时失去的总热量是19.1 kJ/kg

11、 当大气压发生变化时,空气所有的状态参数都会发生变化。

4-12、 A B C D

设过一段时间后A 、B 、C 、D 温度分别为

A B C D t t t t 、、、环境温度为f , 则有

A f

B f

t t t t ->-

C f

D f

t t t t ->-

A 、C 与环境进行热交换主要是通过外表面热辐射和外表面与环境进行热交换。

B 、D 除拥有A 、

C 的换热特点外,还有液体表面与环境直接进行的热质交换,因此它们的热量传递速率较A 、C 的快,更能在短时间内接近

f

t

足够长的时间,A 、B 、C 、D 与环境平衡,而且A 、C 的温度应等于环境干球温度B 、D 应等于环境湿球温度。

13、解:a 由初始状态湿球温度为25℃,室内空气温度为24℃,相对湿度为50%3i

查附录4—1 则新风的焓为76 kJ/kg(干空气) 回气的焓为48 kJ/kg(干空气) 由能量守衡,

()M i M i M M i +=+混

回回回新新新

2?76+3?48=5?

i 混 i 混=59.2 kJ/kg(干空气)

(b) 由已知查附录4—1得

1d =15.8g/kg(干空气) 2d =9.3g/kg(干空气) 则由质量守衡

M1?d1+M2?d2=(M1+M2)d3 2?15.8+3?9.3=5?d3 d3=11.9 g/kg(干空气)

(c) / 1.005(2501 1.86)11.9

59.2 1.005(2507 1.86)

1000

kJ kg h t d t t t =++=+

+29t =℃ (d)

1122cm t cm t ?=? 2?(35-t)=3?(t-24) t=28.4℃

14、解:由题意的空气温度为15℃,相对湿度为100%时,查附录4—1得当加热到22℃时,含湿量为d3=10.5 g/kg(干空气) 当1t

=30℃, 1φ=75%时,1i =82 kJ/kg(干空气) 1d =20.2g/kg(干空气)

2t =15℃, 2φ=100%时,2i =42kJ/kg(干空气) 2d =10.5当1t =30℃, 1φ=75%g/kg(干空气) 当3t

=22℃,

3d =10.5g/kg(干空气) 时i 3=49 kJ/kg(干空气)

则在冷却器中放出的热量为500 kg/min ?(82 kJ/kg-42 kJ/kg )=20000 kJ/min 凝结水量500 kg/min ?(20.2g/kg(干空气)- 10.5g/kg(干空气))=4850g/min

加热器加入的热量500 kg/min ?49 kJ/kg (干空气)- 42 kJ/kg (干空气))=3500 kJ/min 4-15、查焓湿图i-d 图

错解:查附录4—1得 初态为50℃时,1i

=62 kJ/kg(干空气) 1d =4.3g/kg(干空气)

末状态为35℃时2i

=129 kJ/kg(干空气)

2d =36.5g/kg(干空气)

△d=36.5-4.3=32.2 g/kg(干空气) 所以从被干燥的物体中吸收1 kg 水分时所需的干空气量G=1000/32.2=31 kg 加热量Q=G ?△i=31?(129-62)=2077 kJ 正解:热量是由于加热过程是1到2加入的。干燥过程是2到3过程完成的。

2状态为50℃时,1i

=62 kJ/kg(干空气) 1d =4.3g/kg(干空气)

3状态为35℃时2i

=129 kJ/kg(干空气)

2d =36.5g/kg(干空气)

△d=36.5-4.3=32.2 g/kg(干空气) 所以从被干燥的物体中吸收1 kg 水分时所需的干空气量

G=1000/32.2=31 kg

加热量Q=G ?△i=31?

30.5=945.5 kJ

4-16、由附录4—1得

空气:初态:t=15℃,φ=50% 得1i

=28.5 kJ/kg(干空气)

1d =5.3g/kg(干空气)

末态:t=30℃,φ=100% 得

2i =100 kJ/kg(干空气)

2d =27.3g/kg(干空气)

所以△i=71.5 kJ /kg(干空气) △d=22 g/kg(干空气) 由能量守衡的

c m t G i

?=?水水气 4.2?100?3

10?15=G 气?71.5 G

气=88?3

10kg/h

M 水蒸汽=

G 气

△d=88?3

10?22=1936 kg/h

查附得从塔府进入的空气的温度为15℃,相对湿度为50%时其湿球温度为为9.7℃则冷却塔水从38℃冷却至9.7℃

G 水?m ?△t=G 气△I 4.2?100?310?20.3=G

气?71.5

G 气

=166?3

10kg/h

令解:

17、解:总热交换量以空气初状态的湿球温度Ts 为界,

显热交换量以空气初状态的干球温度T1 为界,

潜热交换量以空气初状态的露点温度T2 为界,由T1=30℃ ,水蒸汽的分压力为2000Pa 得Ts=21.4 ℃ T2=17.5℃=18

水温t 50 ℃ 30 ℃ 18 ℃ 10 ℃ 传热方向 气 ←水 气=水 气→水 气→水 传质方向 气←水 气←水 气=水 气→水 18、解:(a )常压下气温为30℃ ,湿球温度为28℃ ,由附录4—1得1d =23g/kg(干空

气)

被冷却到10℃的饱和空气由附录4—1得知2d =7.5g/kg(干空气)

所以每千克干空气中的水分减少了15.5g

(b )若将气体加热到30℃,由附录4—1得湿球温度为17.8℃。 19、解:因为不计喷入水的焓值,则可以认为是等焓变化。查附录得 末状态:含湿量为26g/kg 干空气 水蒸汽分压力:4100Pa 相对湿度为:42% 湿球温度为:32.4 焓值为:113kJ/kg 干空气

第5章吸附和吸收处理空气的原理与方法

1.解:物理吸附是被吸附的流体分子与固体表面分子间的作用力为分子间吸引力,它是一种可逆过程,物理吸附是无选择的,只要条件适宜,任何气体都可以吸附在任何固体上。吸附热与冷凝热相似。适应的温度为低温。吸附过程进行的急快参与吸附的各相间的平衡瞬时即可达到。

化学吸附是固体表面与吸附物间的化学键力起作用的结果。吸附力较物理吸附大,并且放出的热也比较大,化学吸附一般是不可逆的,反应速率较慢,升高温度可以大大增加速率,对于这类吸附的脱附也不易进行,有选择性吸附层在高温下稳定。人们还发现,同一种物质,在低温时,它在吸附剂上进行物理吸附,随着温度升到一定程度,就开始发生化学变化转为化学吸附,有时两种吸附会同时发生。

2、硅胶是传统的吸附除湿剂,比表面积大,表面性质优异,在较宽的相对湿度范围内对水蒸汽有较好的吸附特性,硅胶对水蒸汽的吸附热接近水蒸汽的汽化潜热,较低的吸附热使吸附剂和水蒸汽分子的结合较弱。

缺点是如果暴露在水滴中会很快裂解成粉末。失去除湿性能。 与硅胶相比,活性铝吸湿能力稍差,但更耐用且成本降低一半。

沸石具有非常一致的微孔尺寸,因而可以根据分子大小有选择的吸收或排除分子,故而称作

“分子筛沸石”。

3、目前比较常用的吸附剂主要是活性炭,人造沸石,分子筛等。 活性炭的制备比较容易,主要用来处理常见有机物。

目前吸附能力强的有活性炭纤维,其吸附容量大吸附或脱附速度快,再生容易,而且不易粉化,不会造成粉尘二次污染,对于无机气体如

2SO 2X 、H S 、NO 等有也很强的吸附能力,吸

附完全,特别适用`于吸附去除693

1010/g m --、 量级的有机物,所以在室内空气净

化方面有着广阔的应用前景。

4、有效导热系数通常只与多孔介质的一个特性尺度----孔隙率有关。

第6章 间壁式热质交换设备的热工计算

1、解:间壁式 换热器从构造上可分为:管壳式、胶片管式、板式、板翘式、螺旋板式等。 提高其换热系数措施:⑴在空气侧加装各种形式的肋片,即增加空气与换热面的接触面积。⑵增加气流的扰动性。⑶采用小管径。

6-2、解:空气的湿球温度越高所具有的焓值也愈大,在表冷器减湿冷却中,推动总热质交换的动力是焓差,焓差越大,则换热能力就愈大。 6-3、

表冷器的传热系数定义为

1

11s m n y K AV Bw ρξ-??=+?????? Ks 随迎风面积Vy 的增加而增加:随水流速w 的增加而增加。

析水系数ξ与被处理的空气的初状态和 管内水温有关,所以二者改变也会引起传热系数Ks 的变化。

6-4、解:总热交换量与由温差引起的热交换量的比值为析湿系数,用 表示,定义 为

()

t b

p b dQ i i dQ c t t ξ-=

=

- 表示由于存在湿交换而增大了换热量,其值大小直接反映了表冷器上

凝结水析出的多少。

5、解:逆流流动时,t '?=100-90=100C ,t ''?=120-50=700

C

90

270t t '?=<''? ∴m t ?=(90+70)/2=800C

管束未加肋光管,管壁很薄,所以

f

R 、

w R 可不记,则

150

11

580050

k =

=+

∴传热量为Q=FK m t ?=10?50?80=40000W

顺流流动时:t '?=120-10=1100C t ''?=100-50=500

C

11050

76.1110ln 50

m t -?=

=0

C Q=10?50?76.1=38050W 6-6、解:设冷水的温度为

2t '

Q Q =吸

11p G C ''''''2112p 22(t -t )=G C (t -t )

20.63 2.09(19365) 1.05 1.67(149)

t '??-=??-

解得

2t '

=52.90

C

(193149)(6552.9)

24.6

193149ln 6552.9

m t ---?=

=--0

C Q=KA

m t ?

2

3

0.63 2.09(19365)

9.80.71024.6m Q A m K t ??-=

==???

即保持这样的负荷需要换热面积为2

9.8m 7、解:设机油出口温度为

1t ''

11p G C ''''''2112p 22(t -t )=G C (t -t )

12.6 1.9(100) 1.0 4.18(8040)167.2t ''??-=??-=

0166.2t C

''=

66.24026.2t '?=-=

1008020t ''?=-= 26.2220t t '?=<''?

221180400.67

10040t t P t t '''--===-'''-

112210066.20.8458040t t R t t '''--=

==-'''

-

()026.220

23.12t f

m C

t C +?=

=

Q=KA

m t ?由P-R 值图5—27得φ=0.78

现代大学英语精读3_第二版_unit1、2课文翻译

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define the word resent the treatment frustrate the students declare war evaluate the result perform one’s duty narrow the gap expand business present the facts decorate a room be my haunting memory take a deep breath give a hint stretch one's neck drift on the river let it go at that prick up its ears ripple in the breeze hover over the trees a bond between brothers a carpet for you to walk on snap the door to be serious out of proportion to the occasion go for a picnic put one's arms round her scratch each other’s backs Publish or perish. make RMB convertible seek the truth discard conventions satirize people’s vanity and extravagance The island is uninhabited, and being without drinking water, is also uninhabitable. admire their courage/guts/bravery neglect one’s duty escape/avoid the consequence erase/delete sth from one’s memory take command block the way / stand in the way / be in one’s way ruin one’s reputation/name

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Unit1中国书法 中国书法(calligraphy)是一门独特的艺术、是世界上独一无二的艺术瑰宝。中国书法艺术的形成,发展与汉文字的产生与演进存在着密不可分的关系。汉字在漫长的演变发展过程中,一方面起着交流思想、继承文化的重要作用,另一方面它本身又形成了一种独特的艺术。书法能够通过作品把书法家个人的生活感受、学识、修养、个性等折射出来,所以,通常有“字如其人”的说法。中国书法不仅是中华民族的文化瑰宝,而且在世界文化艺术宝库中独放异彩。 Chinese calligraphy is a unique art and the unique art treasure in the world. The formation and development of the Chinese calligraphy is closely related to the emergence and evolution of Chinese characters. In this long evolutionary process, Chinese characters have not only played an important role in exchanging ideas and transmitting culture but also developed into a unique art form. Calligraphic works well reflect calligraphers' personal feelings, knowledge, self-cultivation, personality, and so forth, thus there is an expression that "seeing the calligrapher's handwriting is like seeing the person". As one of the treasures of Chinese culture, Chinese calligraphy shines splendidly in the world's treasure house of culture and art. Unit2互联网 近年来,随着互联网技术的发展,我国的数字化教育资源建设取得了巨大的成就。很多高校建立了自己的数字化学习平台,数字化教学在教育中发挥着越来越大的作用。和传统教学方式相比,数字化教学方式有很大的优势。一方面,数字化教学使教学资源得以全球共享;另一方面,它拓展了学习者的学习时间和空间,人们可以随时随地通过互联网进入数字化的虚拟学校学习。这使得人类从接受一次性教育走向终身学习成为可能。 In recent years, with the development of Internet technology, the construction of digital education resources of our country has made great achievements. Many universities have set up their own digital learning platforms, and digital teaching is playing an increasingly important role in education. Compared with the traditional way of teaching, the digital way has a lot of advantages. On one hand, digital teaching makes global sharing of teaching resources possible on the other hand, it expands the learner's study time and space to learn, allowing people to get access to the digital virtual schools through the Internet anytime and anywhere. These advantages make it possible for people to shift from one-time learning to lifelong learning. Unit3 孝道(filial piety)是中国古代社会的基本道德规范(code of ethics)。中国人把孝视为人格之本、家庭和睦之本、国家安康之本。由于孝道是儒家伦理思想的核心,它成了中国社会千百年来维系家庭关系的道德准则。它毫无疑问是中华民族的一种传统美德。孝道文化是一个复合概念,内容丰富,涉及面广。它既有文化理念,又有制度礼仪(institutional etiquette)。一般来说,它指社会要求子女对父母应尽的义务,包括尊敬、关爱、赡养老人等等。孝道是古老的"东方文明"之根本。 Filial piety is the basic code of ethics in ancient Chinese society. Chinese people consider filial piety as the essence of a person's integrity, family harmony, and the nation's well-being. With filial piety being the core of Confucian ethics, it has been the moral standard for the Chinese society to maintain the family relationship for thousands of years. It's undoubtedly a traditional Chinese virtue. The culture of filial piety is a complex concept, rich in content and wide in range. It includes not

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