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第八章可压缩流体的流动

8－1 假定声音在完全气体中的传播过程为等温过程，试证其音速计算式为T R a =T 。

已已知知：：等温过程，T ＝常数。

根据连续性方程和动量方程可得 ρ

d p d a =

将完全气体状态方程T R p ρ=代入上式，即得 T R a =

T 。

8－2 重量为2.5kN 的氧气，温度从30℃增加至80℃，求其焓的增加值。

已已知知：：W ＝2.5kN ，T 1＝30℃＝303K ，T 2＝80℃＝253K ，k ＝1.395，R ＝259.82J/kg·K 。单位质量的氧气，温度从30℃增加至80℃的焓增为 J /k g 6.45879)303353(1

395.182.259395.1)(1

12p =-?-?=

--=

=T T k kR T C i ??

总焓增为 kJ 11692J 101692.16.4587981

.925007

=?=?=

=i m I ??

8－3 炮弹在15℃的大气中以950m/s 的速度射出，求它的马赫数和马赫角。已已知知：：t ＝15℃，T ＝288K ，u ＝950m/s ，k ＝1.4，R ＝287J/kg·K 。炮弹飞行的马赫数为 793.2288

2874.1950=??=

k R T

u M

马赫角为

21793

.21sin

1sin

===--M

8－4 在海拔高度小于11km 的范围内，大气温度随高度的变化规律为aH T T -=0。其中T 0

＝288K ，a ＝0.0065K/m 。现有一飞机在10000m 高空飞行，速度为250m/s ，求它的飞行马赫数。若飞机在8000m 高空飞行，飞行马赫数为1.5，求飞机相对于地面的飞行速度及所形成的马赫角。

已已知知：：aH T T -=0，T 0＝288K ，a ＝0.0065K/m ，H 1＝10000m ，u 1＝250m/s ，H 2＝8000m ，M 2＝1.5，k ＝1.4，R ＝287J/kg·K 。

(1) K 22310105.628843101=??-=-=-aH T T

835.0223

2874.12501

11=??=

k R T u M

(2) K 236108105.62883

02=???-=-=-aH

T T

m /s 9.4612362874.15.12

22=???==k R T M u

那么，飞机飞行的马赫角为

8.415

.11sin

1sin

2===--M

8－5 作绝热流动的二氧化碳气体，在温度为65℃的某点处的流速为18m/s ，求同一流线上温度为30℃的另一点处的流速值。

已已知知：：T 1＝65℃＝338K ，u 1＝18m/s ，T 2＝30℃＝303K ，k ＝1.288，R ＝188.92J/kg·K 。根据能量方程

222

11u k kRT u k kRT +

+-，得

m /s 24418

)303338(1

288.192

.188288.12)(1

1212=+--??=

+--=

u T T k kR u

8－6 等熵空气流的马赫数为M ＝0.8，已知其滞止压力为p 0＝4.9×105N/m 2，滞止温度为t 0

＝20℃，试求其滞止音速a 0、当地音速a 、气流速度u 及压力p 。

已已知知：：M ＝0.8，p 0＝4.9×105N/m 2，t 0＝20℃，T 0＝293K ，k ＝1.4，R ＝287J/kg·K 。滞止音速为 m /s 3432932874.100=??=

kRT a

由212

0)2

11(M k a

a -+=，得当地音速为

m /s 323)

8.02

14.11(343)

11(2

0=?-+

?=-+

M k a a

气流速度为 m/s 4.2583238.0=?==a M u

由1

k k

11(--+=M k p

p ，得气流压力为

4.14.12

k k 2

0N /m 10

21.3)

8.02

14.11(109.4)

11(?=?-+

??=-+

=--

M k p p

8－7 氦气作绝热流动，已知1截面的参量为t 1＝60℃，u 1＝10m/s ，2截面处u 2＝180m/s ，求t 2、M 1和M 2及p 2/p 1。

已已知知：：t 1＝60℃，T 1＝333K ，u 1＝10m/s ，u 2＝180m/s ，k ＝1.659，R ＝2078.2J/kg·K 。 (1) 由能量方程式

222

11u k kRT u k kRT +

-，可得

K 330)10180(2

.2078659.121659.1333)(212

212=-??--

=---=u u kR

k T T

即T 2＝330 K ，t 2＝57℃。 (2) 马赫数 009.0333

2.2078659.110

11=??=

kRT u M

169.0330

2.2078659.1180

=??=

kRT u M

977.0)

333

330(

)

(

659.1659

.11

2===--k k

T T p p

8－8 空气流经一收缩形管嘴作等熵流动，进口截面流动参量为p 1＝140kN/m 2

，T 1＝293K ，u 1＝80m/s ，出口截面p 2＝100kN/m 2，求出口温度T 2和流速u 2。

已已知知：：p 1＝140kN/m 2，T 1＝293K ，u 1＝80m/s ，p 2＝100kN/m 2，k ＝1.4，R ＝287J/kg·K 。 (1) 由等熵过程方程式

1k k

2)(-=T T p p ，得

K 266)

140

100(

293)

(

.11

4.1k

k 1

212=?==--p p T T

(2) 由能量方程式

222

11u T R k k u T R k k +

-，得

m /s 3.24680

)266293(1

4.12874.12)(1

1212=+--??=

+--=

u T T k kR u

8－9 有一充满压缩空气的储气罐，其内绝对压力p 0＝9.8MPa ，温度t 0＝27℃，打开气门后，空气经渐缩喷管流入大气中，出口处直径d e ＝5cm ，试求空气在出口处的流速和质量流量。

已已知知：：p 0＝9.8MPa ，t 0＝27℃，T 0＝300K ，d e ＝5cm ，取p a ＝0.1Mpa ，k ＝1.4，R ＝287J/kg·K 。 (1) 由流速计算式得 m /s 4.663])

.91.0(

1[1

4.1300

2874.12])

(

1[1

.11

4.1k

k 0

0=--???=

--=

--p p k T R k u a

(2) 由流量方程式得

k g /s

6.5])

.91.0()8

.91.0[(

300

28710

8.91

4.14.1205

.04

)

(

)[(

124

1])

(

)[(

.11

4.14.12

k 0

a k 2

a 02

k 0

a k 2

a 00=-???

-??=--=

--=+++ππρp p p p RT p k k

d p p p p p k k A

8－10 空气经一收缩形喷管作等熵流动，已知进口截面流动参量为u 1＝128m/s ，p 1＝400kN/m 2

，T 1＝393K ，出口截面温度T 2＝362K ，喷管进、出口直径分别为d 1＝200mm ，d 2＝150mm ，求通过喷管的质量流量G 和出口流速u 2及压力p 2。

已已知知：：u 1＝128m/s ，p 1＝400kN/m 2，T 1＝393K ，T 2＝362K ，d 1＝200mm ，d 2＝150mm ，k ＝

1.4，R ＝287J/kg·K 。

(1) 由流量方程式得

k g /s 25.142.04

1128393

28710

111

1111=??

???=

=πρA u RT p A u G

(2) 由能量方程式可得 m /s 280128)362393(1

4.12874.12)(1

1212=+--??=

+--=

u T T k kR u

(3)由等熵过程方程式1k k

2)(-=T T p p ，得

4.14

.11

k k

212k N /m 300)

393

362(

400)

(

=?==--T T p p

8－11 试计算流过进口直径d 1＝100mm ，绝对压力p 1＝420kN/m 2，温度t 1＝20℃；喉部直径d 2＝50mm ，绝对压力p 2＝350kN/m 2的文丘里管的空气质量流量。设为等熵过程。

已已知知：：p 1＝420kN/m 2，t 1＝20℃，T 1＝293K ，p 2＝350kN/m 2，d 1＝100mm ，d 2＝50mm ，k ＝1.4，R ＝287J/kg·K 。

(1) 由等熵过程方程式

1k k

2)(-=T T p p ，可得

K 278)

420

350

(293)

(

.114.1k

k 1

212=?==--p p T T

由气体状态方程式T R p ρ=，可得

111k g /m 995.429328710

420=??=

RT p ρ

22k g /m 387.4278

28710

350=??=

RT p ρ

(2) 由连续性方程式222111A u A u ρρ=，可得 12

12))(

(

u d d u ρρ=

代入能量方程式

222

11u T R k k u T R k k +

-，得

m /s 07.39]

1)50

100(

)387

.4995.4)[(

14.1()278293(2874.12]

1)(

))[(

1()(24

1211=---???=

---=

d d k T T kR u ρρ

m /s 94.17707.39)50

100)(

387

.4995.4(

))(

(

12=?==u d d u ρρ

(3) 由流量方程式得 kg/s 53.11.04

107.39995.42

111=??

?==πρA u G

8－12 氨气由大容器中经喷管流出，外界环境压力为100kN/m 2，容器内气体的温度为200℃，压力为180kN/m 2

，如通过的重量流量为20N/s ，求喷管直径。设流动为等熵。氨的气体常数为R ＝482J/kg·K ，绝热指数k ＝1.32。

已已知知：：p 0＝180kN/m 2，t 0＝200℃，T 0＝473K ，p a ＝100kN/m 2，W ＝20N/s ，，k ＝1.32，R ＝

482J/kg·K 。

(1) 由气体状态方程式T R p ρ=，可得

00k g /m 79.0473

48210

180=??=

RT p ρ

(2) 由质量流量计算式])

(

)[(

002

k a k a p p p p p k k d

G +--=

ρπ，可得喷管直径为

m 10.0]

)

180

100(

)180

100[(

79.0101801

32.132.1214

.381.9204]

)

(

)[(

244

1.32

32.11.322

k 0

a k

a 00=-????-???=

--=

++p p p p p k k G

d ρπ

8－13 空气流等熵地通过一文丘里管。文丘里管的进口直径d 1＝75mm ，压力p 1＝138kN/m 2，温度t 1＝15℃，当流量G ＝335kg/h 时，喉部压力p 2不得低于127.5kN/m 2，问喉部直径为多少？

已已知知：：d 1＝75mm ，p 1＝138kN/m 2，t 1＝15℃，T 1＝288K ，G ＝335kg/h ，p 2≥127.5kN/m 2，k ＝1.4，R ＝287J/kg·K 。

(1) 由气体状态方程式T R p ρ=，可得

11k g /m 670.1288

28710

138=??=

RT p ρ

由等熵过程方程式

k 1

2)(

p p =ρρ，可得

4.11

k 1

212k g /m 578.1)138

5.127(

670.1)(

=?==p p ρρ

(2) 由流量方程式111A u G ρ=，可得

m /s 6.12075.014.3670.13600335

442

11=????=

d G

u πρ

(3) 由能量方程式

u p k k

+-ρρ，可得

m /s 1.1146.1210)578

.15.127670

.1138

(

14.14

.12)(

122

2=+?-

-?=

u p p k k

u ρρ

(4) 由流量方程式222A u G ρ=，得 mm 7.25m 0257.01

.114578.114.33600335

442

22==????=

u G

d ρπ

8－14 空气在直径为10.16cm 的管道中等熵流动，其质量流量为1kg/s ，滞止温度为38℃。在管道某截面处的静压为41360N/m 2，试求该截面处的马赫数M 、流速u 及滞止压力p 0。

已已知知：：d ＝10.16cm ，G ＝1kg/s ，t 0＝38℃，T 0＝311K ，p ＝41360N/m 2，k ＝1.4，R ＝287J/kg·K 。 (1) 由连续性方程A u G ρ=和能量方程

p k k

k T R k +-=

-ρ，可得

21202

=--

k T R k u G

p k k u

代入数据解之，得 062479904.23462=-+u u 求解上述方程，得 m /s 5.241=u

(2) 由连续性方程 A u G ρ=，得

k g /m 511.01016.014.35.2411

4=???=

=A u G ρ

则马赫数为 717.0511.0413604.15.241=?===ρ

k u a

u M

(3) 由压力比计算式

1k k 2

0)2

11(--+=M k p

p ，得

14.14

.121k k

0N /m 58256)717.02

14.11(41360)2

11(=?-+

?=-+

=--M k p p

8－15 用毕托管测得空气流的静压为35850N/m 2(表压)，全压与静压之差为49.5cmHg ，大气压力为75.5cmHg ，气流滞止温度为27℃。假定(1)空气不可压缩；(2)空气等熵流动。试计算空气

的流速。

已已知知：：p m ＝35850 N/m 2，Δp ＝p 0－p ＝49.5 cmHg ＝66018.4 N/m 2

，p a ＝75.5cmHg ＝100694.7

N/m 2，t 0＝27℃，k ＝1.4，R ＝287J/kg·K 。

根据已知条件可得 p ＝p m ＋p a ＝136544.7 N/m 2，p 0＝p ＋Δp ＝202563.1 N/m 2，T 0＝300 K ，

00k g /m 353.2300

2871.202563=?=

T R p ρ (1) 根据伯努利方程2

1u p p ρ+=，得

m /s 9.236353

.2)

7.1365441.202563(2)

(20=-?=

p p u

(2) 根据可压缩流的流速计算式得 m /s 4.253])

.2025637.136544(

1[1

4.1300

2874.12])

(

1[1

.11

4.1k

k 0

0=--???=

--=

--p p k T R k u

8－16 已知正激波后气流参量为p 2＝360kN/m 2

，t 2＝50℃，u 2＝210m/s ，试求波前气流的马赫数M 1及气流参量p 1、t 1和u 1。

已已知知：：p 2＝360kN/m 2，t 2＝50℃，T 2＝323K ，u 2＝210m/s ，k ＝1.4，R ＝287J/kg·K 。

(1) 激波后的马赫数为 583.0323

2874.12102

=??=

kRT u a u M

由正激波前后马赫数的关系，得

872.3)

14.1(583

.04.12583.0)14.1(2)

1(2)1(22

112

=--???-+=

---+=

=k M

k M k k M

k M

则 968.11=M

(2) 由正激波前后气流参量比与波前M 1数的关系式：

2+--

k k M k k p p ；

]1)1(2)[

2+---+-=M

k M k k k k T T ；

1)1(22

2+-+

k k M k u u

可得到 2

121k N /m 7.82)

4.114.1968

.11

4.14.12(360)

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

+=--k k M k k p p

194]

1968

.1)14.1(2

[

)1968.11

4.14.12()

4.114.1(

323]

1)1(2[)112(

)

212

21=+?--?-?+-?=+---+-=------M k M k k k k T T

m /s 550]

4.114.1968

.1)14.1(2

[

210]

1)1(2

[

21=+-+

?+?=+-+

+=--k k M

k u u

8－17 空气流在管道中产生正激波，已知激波前的参量M 1＝2.5，p 1＝30kN/m 2，t 1＝25℃。试求激波后的参量M 2、p 2、t 2、u 2及p 02。

已已知知：：M 1＝2.5，p 1＝30kN/m 2

，t 1＝25℃，T 1＝298K ，k ＝1.4，R ＝287J/kg·K 。

(1) 由正激波前后马赫数的关系式，得

263.0)

14.1(5.24.125

.2)14.1(2)

1(2)1(22

112

=--???-+=

---+=

k M k M k k M k M k M

则 513.02=M

(2) 由正激波前后气流参量比与波前M 1数的关系式：

2+--

k k M k k p p ；

]1)1(2)[

2+---+-=M k M k k k k T T ；

1)1(22

2+-+

k k M k u u ；

2101

02)1

])1(2)1([--+--

+-++=k k k

k k M k k

k M k p p

可得到 2

112k N /m 75.213)1

4.114.1

5.21

4.14.12(

30)1

=+--

?+??=+--

+=k k M k k p p

K 637]15

.2)14.1(2

)[

15.214.14.12(

)14.114.1(

298]

1)1(2)[112(

12=+?--?-?+-?=+---+-=M

k M k k k k T T

m /s

5.259]1

4.114.15

.2)14.1(2

[

2982874.15.2]

1)1(2[]1

1)1(2

[

121

12=+-+

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

+=k k M

k k R T M k k M

k u u

4.11

4.14.12

k 1

k k

2121

k 1

k k

0102kN/m

78.255)1

4.114.1

5.21

4.14.12(

.2)14.1(25.2)14.1()

5.2214.11[(30)1

2(])1(2)1()211[()1

])1(2)1([=+--

?+??-+?+?-+

+--

+-++-+=

+--

+-++=

------k k M

k k

M k M k M k p k k M

k k M k M k p p

8－18 已知正激波上游空气的参量为p 1＝80N/m 2，T 1＝283K ，u 1＝500m/s ，求激波下游的参量p 2，T 2，ρ2和u 2。

已已知知：：p 1＝80N/m 2，T 1＝283K ，u 1＝500m/s ，k ＝1.4，R ＝287J/kg·K 。波前马赫数及气流密度分别为 483.1283

2874.15001

11=??=

kRT u M

11k g /m 10

85.9283

28780-?=?=

RT p ρ

由式(8－125)可得 m /s 273]1

4.114.1483

.1)14.1(2

[

500]1

1)1(2[

12=+-+

?+?=+-+

+=k k M

k u u

由式(8－123)可得 2

112N /m 192)1

4.114.1483

.11

4.14.12(

80)1

=+--

?+??=+--

+=k k M k k p p

由式(8－122)可得 3

12k g /m 10

81.1483

.1)14.1(2483

.1)14.1(10

85.9)1(2)1(--?=?-+?+??=

-++=

k M k ρρ

由式(8－124)可得

4.370]1483

.1)14.1(2

)[

1483.11

4.14.12(

4.114.1(

283]

1)1(2)[112(

12=+?--?-?+-?=+---+-=M

k M k k k k T T

8－19 已知管路中正激波前的气流参量为u 1＝920m/s ，p 1＝70kN/m 2

，t 1＝395℃，求激波后的气流参量u 2，p 2和t 2值。

已已知知：：u 1＝920m/s ，p 1＝70kN/m 2，t 1＝395℃，T 1＝668K ，k ＝1.4，R ＝287J/kg·K 。正激波前的马赫数为 776.1668

2874.19201

11=??=

kRT u M

由式(8－125)可得 m /s 396]1

4.114.1776

.1)14.1(2

[

920]1

1)1(2[

12=+-+

?+?=+-+

+=k k M

k u u

由式(8－123)可得 2

112k N /m 246)1

4.114.1776

.11

4.14.12(

70)1

=+--

?+??=+--

+=k k M k k p p

由式(8－124)可得

1011]1776

.1)14.1(2

)[

1776.11

4.14.12(

4.114.1(

668]

1)1(2)[112(

12=+?--?-?+-?=+---+-=M

k M k k k k T T

或者 ℃73827310112=-=t

8－20 已知一超音速气流以u 1流过2θ＝20°的尖楔，在楔形物的顶点处产生斜激波，测得激波角β＝50°，激波前的滞止温度T 0＝288K 。求激波前的马赫数M 1和速度u 1。

已已知知：：θ＝10°，β＝50°，T 01＝288K ，k ＝1.4，R ＝287J/kg·K 。由式(8－133)，即

ββ

θβ2

212

1sin

)1(2

sin

)1(tg )tg(M k M k ++-=

-，可得

βββθββtg 2tg sin )1()tg(sin )1(212212=---+M k M k 则激波前的马赫数为

.150

tg 50sin )14.1()1050tg(50sin )14.1(0

5tg 2tg sin )1()tg(sin

)1(tg 22

1=---+?=

---+=

βθβββ

k k M 由式(8－53)，即

11M

k T

T -+

=，可得

K 188)

63.12

14.11(288)

11(1

1011=?-+?=-+

=--M k T T

所以 m /s 4481882874.163.111111=???===kRT M a M u

8－21 一超音速空气流的马赫数M 1＝2.5，压力p 1＝0.85×105N/m 2，温度t 1＝25℃，流过θ＝10°的楔形物体产生斜激波。求激波角β和激波后的参量M 2、p 2、t 2和u 2。

已已知知：：M 1＝2.5，p 1＝0.85×105N/m 2，T 1＝298K ，θ＝10°，k ＝1.4，R ＝287J/kg·K 。

斜激波前的气流速度为 m/s 8652982874.15.2111=???==kRT M u (1) 先由式(8－134)，即)

1sin

(2)1()

1sin

(ctg 2tg 2

21--+-=

βββθM M k M ，解出激波角β(略)；

(2) 由式(8－132)，即)

1(sin

2sin )1(2)(sin 2

212

---+=

-k kM

M k M ββ

θβ，可得到激波后的马赫数为

)(s i n )]1(sin

2[sin

)1(22

12θβββ

----+=

k kM M k M

(3) 由式(8－129)，即

1sin 1

2+--

k k M k k p p β，可得激波后的压力为

1s i n 1

112+--

+=k k M k k p p β

(4) 由式(8－130)，即]1sin )1(2

)[

1sin 1

212

2+---+-=β

βM k M k k k k T T ，可得到激

波后的温度为

]1s i n )1(2

)[

1sin

212

12+---+-=β

βM k M k k k k T T

(5) 由式(8－131)，即

)

sin(sin

)1(sin ]sin )1(2[2

2θβββ

β-+-+=

M k M k u u ，可得到激波后的气流速度为

)

s i n (s i n )1(s i n ]s i n )1(2[2

112θββββ-+-+=

k M k u u

8－22 空气流以u 1＝650m/s 的超音速绕过θ＝18°的楔形物体流动，在折角处产生斜激波，已知激波角β＝51°，求激波后的气流速度及通过激波后的焓值增量。

已已知知：：u 1＝650m/s ，θ＝18°，β＝51°。 (1) 由图8－27所示的斜激波图形可知 βs i n 11n u u =； ββ

ββ

c o s tg sin tg 111n 1τ2τu u u u u ===

)t g (c o s )t g (

12τ2n θββθβ-=-=u u u

m /s

488)

1851cos(51

cos 650)

cos(cos )sin()

tg(cos )

sin(112τ2=-?=

--=

θββθβθββθβu u u u

(2) 由能量方程2

222

11u i u i +

=+，得焓值增量为

J /k g 92178)488650(2

1)(2

112=-?=

-u u i i

8－23 已知一超音速气流u 1＝500m/s ，T 1＝300K ，p 1＝105Pa ，现绕外钝角折转15°。试求折转后气流的速度、压力和温度。

已已知知：：u 1＝500m/s ，T 1＝300K ，p 1＝105Pa ，θ＝15°，k ＝1.4，R ＝287J/kg·K 。超音速气流绕外钝角折转后将产生膨胀波，波前的马赫数为 44.1300

2874.15001

11=??=

kRT u M

由于本题所给的初始马赫数M 1＝1.44＞1，所以气体膨胀波函数表不能够直接应用。可以设想初始来流马赫数M 1＝1.44是由M ＝1绕某一个θ*角膨胀而得来的，然后在此基础上再绕一个θ＝15°的折转角继续膨胀加速而得到M 2，这样所求得的M 2和p 2完全相当于M 1＝1.44直接转折θ＝15°角的结果。这是因为M 2只与总折转角有关，而与折转过程无关，所以可按以下步骤解题：

(1) 先求出由M ＝1膨胀到M 1＝1.44的折转角θ* 由M 1＝1.44查气体膨胀波函数表，得θ*＝10° (2) 再求出M ＝1膨胀到M 2的总折转角Σθ Σθ＝θ*＋θ＝10°＋15°＝25° (3) 最后求出M 2、p 2、T 2和u 2

依总折转角Σθ＝25°查膨胀波函数表，得M 2＝1.951，p 2/p 02＝0.138，T 2/T 02＝0.568；根据M 1＝1.44查得p 1/p 01＝0.299，T 1/T 01＝0.708。注意到等熵流p 01＝p 02，T 01＝T 02，则

Pa 10

6.410299.01138.04

010222?=??

=??=

p p p p p p

K 241300708

.01

568.011

0102

22=??

=??=

T T T T T T

m /s 6072412874.1951.12

2222=???===k R T M a M u 8－24 已知一超音速气流M 1＝2.32，p 1＝1.2×105

Pa ，t 1＝30℃，现绕150°的凸钝角流动。试求下游气流的压力、温度、马赫数和折转角。

已已知知：：M 1＝2.32，p 1＝1.2×105Pa ，T 1＝303K ，θ＝180°－150°＝30°，k ＝1.4，R ＝287J/kg·K 。因为M 2只与总折转角有关，而与折转过程无关，所以可按上题相似的步骤解题： (1) 先求出由M ＝1膨胀到M 1＝2.32的折转角θ* 由M 1＝2.32查气体膨胀波函数表，得θ*＝35° (2) 再求出M ＝1膨胀到M 2的总折转角Σθ Σθ＝θ*＋θ＝35°＋30°＝65° (3) 最后求出M 2、p 2、T 2和气流折转角φ

依总折转角Σθ＝65°查膨胀波函数表，得M 2＝3.949，p 2/p 02＝7.12×10－3

，T 2/T 02＝0.244，φ2

＝140°20′；根据M 1＝2.32查得p 1/p 01＝0.0764，T 1/T 01＝0.480，φ1＝99°33′。注意到等熵流p 01＝p 02，T 01＝T 02，则

Pa 10

12.1102.10764

.0110

12.74

0102

22?=???

?=??=

-p p p p p p

K 154303480

.01244.011

0102

22=??

=??

T T T T T T

744033990214012'='-'=-= ???

8－25 氧气在直径d ＝200mm ，阻力系数λ＝0.02的管道中作绝热流动，已知入口截面的流动参量p 1＝450kN/m 2，t 1＝33℃，u 1＝600m/s 。求极限管长及相应的出口压力、温度和速度值。

已已知知：：d ＝200mm ，λ＝0.02，p 1＝450kN/m 2，t 1＝33℃，T 1＝306K ，u 1＝600m/s ，k ＝1.4，R ＝260J/kg·K 。

(1) 进出口马赫数分别为 80.1306

2604.16001

11=??=

kRT u M ； 1*2==M M

由 21

1max )1(2)1(ln

211M

k M k k

k M

k M D

L -++++

，得极限管长为

42.2]8

.1)14.1(28

.1)14.1(ln

.1214.18

.14.18

.11[

.02.0])1(2)1(ln 21

1[2

1212

1max

=?-+?+?++

?-?=-++++-=M k M k k k M k M D L λ

(2) 由 2

12112

])1(2)1(2[M k M k M M p p -+-+=，得出口压力为 2

222

1211*2k N /m 949]1

)14.1(28.1)14.1(2[18.1450])1(2)1(2[=?-+?-+??=-+-+==M k M k M M p p p (3) 由

2)1(2)1(2M

k M k T T -+-+=

，得出口温度为

K 4201

)14.1(28.1)14.1(2306)1(2)1(22

*2=?-+?-+?

=-+-+==M

k M k T T T

(4) 出口速度为 m /s 3914202604.1**2=??=

==kRT u u

8－26 氮气在直径d ＝300mm 的管道中作绝热流动，已知通过的质量流量为42kg/s ，入口截面的流动参量为p 1＝980kN/m 2，t 1＝60℃，出口截面的密度为ρ2＝0.8ρ1，沿程阻力系数λ＝0.015，求管道长度。

已已知知：：d ＝300mm ，G ＝42kg/s ，p 1＝980kN/m 2，t 1＝60℃，T 1＝333K ，ρ2＝0.8ρ1，λ＝0.015，

k ＝1.4，R ＝296.8J/kg·K 。

(1) 由气体状态方程得

11k g /m 916.9333

8.29610

980=??=

RT p ρ

12k g /m 933.7916.98.08.0=?==ρρ

(2) 由连续性方程得 m /s 95.593.014.3916.94242

11=???=

A G

u ρ

m /s 94.748

.095.598

.012

112====

u u u ρρ

(3) 马赫数 161.0333

8.2964.195.591

11=??=

kRT u M

由

221211

2])1(2)1(2[M k M k M M u u -+-+==

ρρ，得

.0161

.0)14.1(]161.0)14.1(2[8.0161

.02)1(])1(2[)(

2121

=?--?-+??=

---+=

M k M k M M

ρρ

(4) 由])1(2)1(ln 211[2

max

k M k k k M k M D L -++++-=λ，得对应于M 1和M 2的极限管长分别为

477]161

.0)14.1(2161

.0)14.1(ln

.1214.1161

.04.1161

.01[

015

.03.0])1(2)1(ln 211[2

121211

max =?-+?+?++

?-?=

-++++-=M k M k k k M k M D L λ

291]2

.0)14.1(22

.0)14.1(ln

.1214.12

.04.12

.01[

015

.03.0]

)1(2)1(ln 211[2

max =?-+?+?++

?-?=

-++++-=M k M k k k M k M D L λ

则管道长度为 m 1862914772max 1max =-=-=L L L 。

8－27 空气在d ＝300mm 的管道中绝热流动，已知入口截面参量p 1＝1.0MPa ，T 1＝330K ，出口截面的压力和密度分别为p 2＝0.78MPa ，ρ2＝0.8ρ1，沿程阻力系数λ＝0.02，求质量流量和管长。

已已知知：：d ＝300mm ，p 1＝1.0MPa ，T 1＝330K ，p 2＝0.78MPa ，ρ2＝0.8ρ1，λ＝0.02，k ＝1.4，R

＝287J/kg·K 。

(1) 由气体状态方程得 3

11k g /m 559.10330

28710

0.1=??==

RT p ρ

12k g /m 447.8559.108.08.0=?==ρρ

(2) 由连续性方程2211u u ρρ=和能量方程

u p k k

-ρρ，得

)(

12])(1[2

2ρρρρp p k k

u -

代入数据得

m /s 5.214])559

.10447.8(1[)]447.81078.0559.10100.1(14.14.12[]

)(

1[)](

12[

166

221

2=-?-?-?=--

-=--

ρρρρp p k k

m /s 6.1715.2148.021

21=?==

u u ρρ

(3) 质量流量 kg/s 0.1283.04

16.171559.102

11=???==πρA u G

(4) 马赫数 471.0330

2874.16.1711

11=??=

kRT u M

597.0447

.810

78.04.15.2146

=??=

=ρp k u M

由])1(2)1(ln 211[2

max

k M k k k M k M D L -++++-=λ，求得对应于M 1和M 2的极限管长分别为

20]471

.0)14.1(2471

.0)14.1(ln

.1214.1471

.04.1471

.01[

.03.0])1(2)1(ln 211[2

1212

max =?-+?+?++

?-?=-++++-=M k M k k k M k M D L λ

54.7]597

.0)14.1(2597

.0)14.1(ln

.1214.1597

.04.1597

.01[

.03.0]

)1(2)1(ln 211[2

max =?-+?+?++

?-?=-++++-=M k M k k k M k M D L λ

则管道长度为 m 46.1254.7202max 1max =-=-=L L L

8－28 空气在d ＝150mm 的管道中绝热流动，流量为2.7kg/s ，沿程阻力系数λ＝0.018，起始截面的压力为180kPa ，温度为50℃，求不发生壅塞的最大管长及相应的出口截面温度和压力。

已已知知：：d ＝150mm ，G ＝2.7kg/s ，λ＝0.018，p 1＝180kPa ，t 1＝50℃，T 1＝323K ，k ＝1.4，R ＝287J/kg·K 。

(1) 由气体状态方程得 3

11k g /m 942.1323

28710

180=??=

RT p ρ

由连续性方程得 m/s 7.7815

.04

1942.17.22

11=??

πρA

马赫数为 218.0323

2874.17

.781

11=??=

kRT u M

(2) 由])1(2)1(ln 211[2

max

k M

k k k M k M D L -++++-=λ，可得不发生壅塞的最大管长为

8.98]218

.0)14.1(2218

.0)14.1(ln

.1214.1218

.04.1218

.01[

018

.015.0])1(2)1(ln 211[2

1212

1max

=?-+?+?++

?-?=-++++-=M k M k k k M k M D L λ

(3) 由式

)1(21M

k k T T -++=

及式

212

])1(21[

k k M

p p -++=

，得到

K 2721

4.1218

.0)14.1(23231

)1(22

*=+?-+?

=+-+=k M k T T

k P a 36]

218

.0)14.1(21

4.1[

218.0180]

)1(21[

11*=?-++??=-++=-

k k M p p

8－29 16℃的空气在d ＝20cm 的管道中等温流动，沿管道3600m 的压降为98kPa ，假设入口压力为490kPa ，沿程阻力系数λ＝0.03，求质量流量。

已已知知：：t ＝16℃，T ＝289K ，d ＝20cm ，L ＝3600m ，Δp ＝98kPa ，p 1＝490kPa ，λ＝0.03，R ＝287J/kg·K 。

因为 p 2＝ p 1－Δp ＝490－98＝392kPa ，所以质量流量为

k g /s

38.110

)392490

(289

287360003.0162

.014.3)

(166

2215

2=?-?????=

p p T

R l D

G λπ

8－30 空气在水平等直径长管中等温流动，断面1的参数p 1＝700kN/m 2，u 1＝40m/s ，断面2的参数p 2＝280kN/m 2

，温度为303K ，求在断面1和2之间每kg 气体所增加的动能值。

已已知知：：p 1＝700kN/m 2，p 2＝280kN/m 2，u 1＝40m/s ，T ＝303K 。由式(8－185)，可得 m /s 10040280

70012

12=?=

u p p u

则在断面1和2之间每kg 气体所增加的动能值为

k J /k g

4.2J /k g 4200)40100

1)(2

2==-?=

-u u 8－31 已知煤气管路的直径为200mm ，长度为3000m ，进口截面绝对压力p 1＝980kPa ，温度T 1＝300K ，出口截面的压力为p 2＝490kPa ，管道摩擦阻力系数λ＝0.012，煤气的气体常数R ＝490J/kg·K ，绝热指数k ＝1.3，煤气管路按等温流动考虑，求通过的质量流量和出口马赫数。另计算极限管长及相应的出口压力和流速。

已已知知：：d ＝200mm ，L ＝3000m ，p 1＝980kPa ，T 1＝300K ，p 2＝490kPa ，λ＝0.012，k ＝1.3，R ＝490J/kg·K 。

(1) 由质量流量计算式，得

k g /s

18.510

)490980

(300

4903000012.0162

.014.3)

(166

2215

2=?-?????=

p p T

R l D

G λπ

(2) 由式2

1121M k D

l p p λ

-=，得

0566.0])980

490(

1[3

.13000012.02

.0])(

1[2

21=-??=

p p k

l D

M λ 由2

2M M u u p p =

ρρ，得出口马赫数为

113.00566.0490

980

12=?=

M p p M (3) 由极限管长计算式)ln(12

max M k M

k M k D

L +-=λ

，得极限管长为

3894)]0566.03.1ln(0566

.03.10566.03.11[

012

.02.0)]ln(1[2

212

1max

=?+??-?=

+-=M k M k M k D L λ

由出口压力及流速计算式得

k P a 24.630566.09803.111=??=

=M p k p ?

m /s 4.383300490=?==

T R u ?

8－32 甲烷气体通过直径d ＝600mm ，长度L ＝40km 的钢管，从p 1＝420kN/m 2的上游压缩机站输送至p 2＝140kN/m 2

的下游出口处，设流动是等温的，t ＝20℃，沿程阻力系数λ＝0.02，求其质量流量。

已已知知：：d ＝600mm ，L ＝4×104m ，p 1＝420kN/m 2，p 2＝140kN/m 2

，T 1＝293K ，λ＝0.02，R ＝

518.3J/kg·K 。

由质量流量计算式(8－180)，得

k g /s

86.710

)140420

(293

3.51810402.0166

.014.3)

(166

2215

2=?-??????=

p p T

R l D

G λπ

8－33 一等温管路长800 m ，入口压力和温度分别为800kPa 和15℃，出口压力为600kPa ，沿程阻力系数为0.01，要求输送的空气量为0.3kg/s ，试计算管路应有的直径。

已已知知：：L ＝800m ，p 1＝800kPa ，p 2＝600kPa ，T 1＝288K ，λ＝0.01，R ＝287J/kg·K ，G ＝0.3kg/s 。由质量流量计算式(8－180)，得 mm 51m 051.010

)600800

(3.140.32882878000.0116)

(165

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

p p TG

R l d πλ

8－34 空气流过一直径d ＝300mm 的管道排入大气中，大气压力为98.1kN/m 2，管长l ＝100m ，空气温度t ＝32℃，通过的质量流量为13.2kg/s ，沿程阻力系数λ＝0.015，如保持等温流动，进口压力为多少(按短管考虑)？

已已知知：：d ＝300mm ，p 2＝p a ＝98.1kN/m 2，l ＝100m ，t ＝32℃，T ＝305K ，G ＝13.2kg/s ，λ＝0.015，k ＝1.4，R ＝287J/kg·K 。

(1)由气体状态方程得 3

22k g /m 12.1305

28710

1.98=??=

RT p ρ

由连续性方程得 m/s 8.1663

.014.312.12.1342

22=???=

u ρ

(2)由短管压降计算式)ln

2(1

212

11222

l u u p u p p λ

ρ+=-及

2u u p p =

ρρ，可得

)ln

2)()(()ln 2)()((2

122221

211112

221D

l p p u p u D

l u u u p u p p λ

ρλ

ρ+=+=-

代入已知数据，并整理后可得

084.1410

27.3ln 2

110

1=-?--p p

采用试算法解上述方程，得 2

1kN/m

91.167=p 。

8－35 空气在光滑水平管道中输送，管长200m ，管径50mm ，摩阻系数λ＝0.016，进口处绝对压力为106

N/m 2

，温度为20℃，流速为30m/s ，求沿此管道的压力降。(1)气体为不可压缩流体；(2)气体为可压缩绝热流动；(3)气体为可压缩等温流动。

已已知知：：L ＝200m ，d ＝50mm ，λ＝0.016，p 1＝106N/m 2

，t 1＝20℃，T 1＝293K ，u 1＝30m/s ，k

＝1.4，R ＝287J/kg·K 。

(1) 当气体为不可压缩流体时， 2

621

1k N /m 5.342293

28730

102

105

.0200016.0212

1=???

===u RT

p D l u D l p λ

ρλ

(2) 当气体为可压缩绝热流动时，

14.14

.12

k k

112k N /m

9.596)293

28705.0230

200016.04

.114.11(10)

211(=??????

?=+-

=++DRT u l k k p p λ

则 2

21k N /m 1.4039.5961000=-=-=p p p ?

(3) 当气体为可压缩等温流动时， 2

12k N /m 3.561293

28730

.0200016.0110001=??

-?=-=T

R u D l p p λ

则 221k N /m 7.4383.5611000=-=-=p p p ?

8－36 空气无摩擦地流过等截面管道，入口气流参量为T ＝280K ，p ＝2.4×105N/m 2，M ＝0.24，试求：(1)管道出口达到M ＝1时，每千克气体需要加入多少热量？(2)出口截面上的临界压力和临界滞止压力为多少？(3)熵值增加多少？已已知知：：T ＝280K ，p ＝2.4×105N/m 2，M ＝0.24，k ＝1.4，R ＝287J/kg·K 。

(1) 根据式(8－53)和式(8－168)，得 K 283)24.02

14.11(280)2

11(2

1101=?-+

?=-+

=M k T T

K 1182]

4.124

.0)14.1(2[

)24

.04.114.11(

24.0283

]

1)1(2[

)11(

*0=+?-+?++?=

+-+++=

k M k M

k k M T T

由式(8－156)可算得管出口M ＝1时，单位质量气体所获得的热量为

J /k g 10

03.9)2831182(1

4.12874.1)(1

)(5

01*001*0p ?=-?-?=

--=-=T T k kR T T C q

(2) 根据式(8－162)和式(8－160)，可以得到出口截面上的临界温度和临界压力为

K 986)

4.11(24.0)24.04.11(280)

1()1(2

222

2211

*=+??+?

=++=k M M k T T

k P a 1008.14

.1124

.04.11104.2115

*?=+?+?

?=++=k

M k p p

由式(8－55)和式(8－170)可得到出口截面上的临界滞止压力 k P a 105.2)24.02

14.11(104.2)2

11(5

14.14

.125

1k k

21101?=?-+

??=-+

=--M k p p

k P a

1005.2]24

.0)14.1(21

4.1)[

.1124

.04.11(

105.2])1(21)[

11(

14.14.12

k k

01*0?=?-+++?+??=-++++=--M

k k k

M k p p

由式(8－172)可得到熵值增加量为 K J /k g 1494)]08

.14.2(

)

280

986ln[(

287)](

)

ln[(

4.14

.1*

k k

*max ?=?==---p p T

T R s s

8－37 设空气无摩擦地流入等截面管道，入口处马赫数M ＝0.2，温度T ＝290K ，滞止压力p 0＝2.75×105N/m 2，沿管道有热量加入，使气流出口温度T 2＝1200K ，试计算气流出口马赫数M 2。

已已知知：：M 1＝0.2，T 1＝290K ，p 01＝2.75×105N/m 2

，T 2＝1200K ，k ＝1.4，R ＝287J/kg·K 。

由式(8－163)，即

121

2212)11(

k M k M

M T T ++=

，可得到

1()1(2

2221=-++M T T kM kM M

代入已知数据，并整理后得 00359.0141.00703.02

242=+-M M 解上述方程，得 547.02=M

8－38 空气从一个滞止压力为p 0＝4.8×105N/m 2和滞止温度为T 0＝320K 的容器中流入内径为D ＝75mm 的管道，然后排入大气。沿管道流动过程中，外界输入335kJ/kg 的热量。试求：(1)管道出口的马赫数M 、温度T 和压力p ；(2)空气的质量流量G 。

已已知知：：D ＝75mm ，p 01＝4.8×105N/m 2，T 01＝320K ，p a ＝105

Pa ，q ＝335kJ /kg ，k ＝1.4，R ＝287J /kg·K 。

依题意，比较p 01和p a 的数值可知，空气流在管道出口处将达到临界状态，即M 2＝M *＝1。

(1) 由式(8－156)知，)(1

0102T T k kR q

--= ，则管道出口处空气的滞止温度为

Chapter Eight Time as communication

Chapter Eight Time as communication Learning objective: .By the end of the chapter, you should be able to understand time is a personal phenomenon, and how we perceive and treat it that expresses our character. .How members of different cultures value and respond to time. .Understand the three perspectives (informal time, perceptions of past, present, and future and Hall’s monochronic and polychronic classifications) and their reflection in different cultures and learn to determine a culture’s conception of time from three different perspectives mentioned above. Warm-up activities In a small group, read the following paragraph and explain what went wrong, have a discussion on the questions. Jan was in Brazil on business. Ciro a Brazilian associate, invited her to a dinner party he and his wife were giving. The invitation was for “around 8, this Friday night.”Jan arrived at Ciro’s house at exactly 8:00. Ciro and his wife were still dressing and had not even begun to prepare the food. Why did these problems take place? What are some common problems when treating appointment? What do you think theCiro and his wife’s time concept? Reading Reading1 Time Orientation When Shakespeare wrote “The inaudible and noiseless foot of Time,’’ we cannot hold or see time, we respond to it as if it had command over our lives. Because time is such a personal phenomenon, all of us perceive and treat it in a manner that expresses our character. If we arrive thirty minutes late for an important appointment and offer no apology, we send a certain message about ourselves. Telling someone how guilty we feel about our belated arrival also sen ds a message. A culture’s use of time can also provide valuable clues to how members of that culture value and respond to time. In America, we hear people saying, “Time is money” and “He who hesitates is lost.” All Chinese know the Confucian proverb, “Think three times before you act.” Reflect for a moment on how differently each of these cultures perceives time. A culture’s conception of time can be examined from three different perspectives:(1) informal time; (2) perceptions of past, present, and future; and (3) Hall’s monochronic and polychronic classifications. Informal Time: Most of the rules for informal time, such as pace and tardiness, are not explicitly taught. Like most of culture, these rules usually function below the level of consciousness. Argyle makes much the same point when he compares cultural differences in punctuality standards: How late is “late”? This varies greatly. In Britain and America only maybe 5 minutes late for a business appointment, but not 15 and certainly not 30 minutes late, which is perfectly normal in Arab countries. On the other hand in Britain it is correct to be 5-15 minutes late for an invitation to dinner. An Italian might arrive 2 hours late, an Ethiopian after, and a Javanese not at all—he had accepted only to prevent his host from losing face. Our reaction to punctuality is rooted our cultural experiences. In the United States, we have

语言学教程第八章知识点

Chapter Eight Pragmatics ?Definition ?Pragmatics is generally the study of natural language understanding, and specifically the study of how context influences the interpretation of meanings. In another word it is the study of the relationship between symbols and their interpreters. ?In 1937,the American philosopher Charles William Morris introduced the word “Pragmatics” into literature. ?莫里斯（C.Morris）和卡耐基（R.carnap）在1938年《符号基础理论》中提出符号三分说： ?句法学（符号关系学）Syntactics 是研究符号与符号之间的关系；语义学semantics是研究符号与符号所指对象的关系；语用学pragmatics则是研究符号与符号解释者的关系。 ?Teaching Focus ? 1. Some basic notions ? 2. Speech act theory ? 3. The theory of conversational implicature ? 4. Post-Gricean Developments ? 1. Some basic notions ? 1.1 The definition of pragmatics ? 1.2 Pragmatics and semantics ? 1.3 Context ? 1.4 Sentence and utterance ? 1.1 The definition of pragmatics ?Various definitions: ?The study of how speakers of a language use sentences to effect successful communication. ?The study of language in use. ?The study of meaning in context. ?The study o f speakers’ meaning, utterance meaning, & contextual meaning. ? 1.2 Pragmatics and semantics ?Both semantics and pragmatics study the meaning of language.

ChapterEightSemantics---WordMeani

Chapter Eight Semantics --- Word Meaning and Sentence Meaning 0. Introduction 0.1 Definition: Semantics is the study of linguistic meaning: the meaning of words, phrases, and sentences. 0.2 Sub-branches: i. Lexical semantics: Lexical semantics deals with word meaning. ii. Sentence semantics: Sentence semantics deals with sentence meaning. ii. Two types of meaning: In general, we may say that a linguistic form has two types of meaning: denotation（外延意义）and connotation（内涵意义）. A. The denotative meaning of a linguistic form is the person, object, abstract notion, event, or state which the word or sentence denotes. B. The connotation of a linguistic form has to do with its overtones of meaning, that is, what the linguistic form suggests. Such overtones may be good or bad, and thus we can speak of a positive connotation褒义内涵and a negative connotation 贬义内涵.

写作篇 Chapter Eight

---------------------------------------------------------------最新资料推荐------------------------------------------------------ 写作篇Chapter Eight Progressive WritingPart Eight Sentence Errors 1/ 31

遣词造句中的常见错误大学英语教学实践及历年CET-4作文考试表明，我们在作文方面普遍存在“词不达意”、“语法结构混乱”、‘‘思想表达不清”、“主题不突出”等问题。以下是用词及句子结构方面的典型错误。

---------------------------------------------------------------最新资料推荐------------------------------------------------------ 1. 结构不完整[例1]原句：I look for ward to holidays．For example，theMid-Autumn Festival and the Spring Festival．分析：斜体部分可以是句子的一个成分，但不能独立成句。试改：I look for ward to holidays like the MidAutumn Festival and the Spring Festival．[例2] 原句：I refused to go to the show. Because I had been up late last night and needed sleep. 分析：斜体部分是原因状语从句，不能独立存在。试改：I refused to go to the show because I had been up late last night and needed sleep 3/ 31

Chapter Eight Packing and Shipment

Chapter Eight Packing & Shipment Exercises I.Put in the missing words: Dear Sirs， We thank you ____ your Letter of Credit No.F-120 amounting ____ US $ 1,050,000 issued in our favor through The Hong Kong & ShangHai Banking Corporation. ____ regard to shipment, we regret very much to inform you that, despite strenuous efforts having been ____by us, we are still unable to book space of a vessel sailing ____Jakarta direct. The shipping companies boat told us that, ____ the time being, there is no regular boat sailing between ports in possible, for us to ship these 10,000 metric tons of sugar to Jakarta direct. In view ____ the difficult situation faced by us, you are requested to amend the L/C to allow transshipment of the ____ in Hong Kong where arrangements can easily be made ____ transshipment. Since this is something beyond ____control, your agreement ____our request and your understanding of our position will be highly appreciated. We are anxiously awaiting the amendment ____ the L/C. Yours faithfully, II.Fill in the blanks: 1)Owing ___ the delay ___ the part of the suppliers, we must ask you to

语言学第八章

Chapter Eight Exercise I. Choose the best answer. 1. What essentially distinguishes semantics and pragmatics is whether in the study of meaning ______ is considered. A. reference B. speech act C. practical usage D. context 2. “Logic and Conversation” was written by ______. A. H.P. Grice B. William James C. Stephen Levinson D. John Austin 3. According to the conversational maxim of _______ suggested by Grice, one should speak truthfully. A. quantity B. quality C. relevance D. manner 4. Speech act theory did not come into being until _____. A. in the late 50’s of the 20the century B. in the early 1950’s C. in the late 1960’s D. in the early 21st century 5. ____ is the act performed by or resulting from saying something; it is the consequence of, or the change brought about by the utterance. A. A locutionary act B. An illocutionary act C. A perlocutionary act D. A performative act 6. All of the following are characteristics of implicature EXCEPT_______. A. conventionality B. cancellability C. non-detachability D. calculability 7. The theory of meaning which relates the meaning of a word to the things it refers to or stands for is known as the ________. A. An integrated Theory B. Speech Act Theory C. The Conceptual Theory D. The Referential Theory 8. Which of the following can best describe the relationship between “They have six cows” and “They have some animals”? _______ A. presupposition B. synonym C. antonym D. entailment 9.In which of the following aspects conversational implicature theory and speech act theory are different ? A. how contextual meaning is generated. B. how much the implied meaning is dependent on the context. C. how much role the conventional meaning of words plays. D.how indeterminate the implied meaning is. 10.‘I hereby declare the starting of the war’ reflects the ______ function of language. A. informative B. interpersonal C. performative D. emotive 11.Which of the following statements is true about John Langshaw Austin ? A. He classified the following five types of speech acts performed by sentences: representative, directive, interrogative, expressive and declarative acts. B. He accounted for why people always spoke indirectly in daily conversation. C. He finally realized that there was no clear boundary between performatives and constatives. D. Austin proposed the felicity conditions to stress performatives can be true or false.

语言学Chapter Eight Pragmatics

Chapter Eight Pragmatics Aims: ? To know what is pragmatics and its main concern; ? To have a general idea about context and the notions of deixis, reference and anaphora; ? To understand the Speech Act theory, the Cooperative Principles in Conversation and Politeness Principles in Conversation. ⅠIntroduction Pragmatics is concerned with the interpretation of linguistic meaning in context. Two kinds of contexts are relevant. When we read or hear pieces of language, we normally try to understand not only what the words mean, but what the writer or speaker of those words intended to convey. Take a look at the following dialogue: 1 A: I have a fourteen-year-old son. B: Well that?s all right A: I also have a dog. B: Oh I?m sorry. In making sense of the quote above, it may help to know that A is trying to rent an apartment from B. the real meaning is beyond the literal interpretation. It seems that pragmatics is the study of …invisible? meaning, or how we recognize what is meant even when it isn?t actually said/written. In order for that to happen, speakers/writers must be able to depend on a lot of shared assumptions and expectations. The investigation of those assumptions and expectations provides us with some insights into how more gets communicated than is said. ⅡContext It can be seen from the above discussion that context plays a very import role in making sense of utterances we hear. There are different kinds of context to be considered. One kind is best described as linguistic context, also known as co-text. The context of a word is the set of other words used in the same phrase or sentence. For example, the word bank is a homonym, a form with more than one meaning. How do we usually know which meaning is intended in a particular