工程流体力学答案(周云龙第三版)

工程流体力学答案(周云龙第三版)

第一章

1-1

90610500453.06

=⨯=

=

-V m ρkg/m 3

906.01000

906

==

d

1-2

544

.01400

273273

34.11013252732730

=⨯+⨯=+=p t ρρkg/m 3

1-3 1

1

21211V V V t t V dV dt V

--==

α

98

.616060)2080(10550)(611122=+⨯-⨯⨯=+-=-V V t t V V αm 3

/h

1-4

9

3

36661121210

51011011099510102111----⨯=⨯⨯-⨯-⨯-=---=-=V V V p p V dV dp κ1/

Pa

1-5 4

7

109.26781028.4--⨯=⨯⨯==νρμ Pa·s

1-6 6

3

103.14

.99910

3.1--⨯=⨯==ρμνm 2/s 1-7 (1) 17

.2660

5000

1.014.360

=⨯⨯=

=

dn

u π m/s 5

21023.510005.017.260⨯=⨯=-=-δu dy du 1/s

(2)

2

22d

dy du dL d dy du A d F

M μπμ===

3

5221033.510

23.5108.01.014.35.322-⨯=⨯⨯⨯⨯⨯==

du dy L d M πμ Pa·s

(3)

3531079.21023.51033.5⨯=⨯⨯⨯==-dy

du

μ

τPa

1-8 (1)y dy

du μμτ2==

(2)μμμμτ2122=⨯===y dy

du 1-9 (1)

h

u bL dy du

A

F 022μμ==

(2) 当2

h y =时，h u dy du

μμτ== (3)当h y 2

3=时，0

u u = 所以0==dy du

μτ 1-10

2903

.03

.0133)(112121=⨯⨯==+=+=μμμμdy du A dy du A

F F F N

967

.01=μ Pa·s

933

.1212==μμ Pa·s 1-11 dr r

r dr r r r dA dy du r dF dM α

δπωμαπδωμμsin 2sin 203

=-=⋅=⋅= α

δαπωμααδπωμαδπωμαδπωμαα

α

cos 24)(sin 2sin 2sin 234403030tg H Htg dr r dr r dM M Htg Htg Htg ==

===⎰⎰⎰1-12

62

.260

200

25.014.360

=⨯⨯=

=

dn

u πm/s

3925

.050.025.014.3=⨯⨯==dL A πm 2

3

3

1022.410

2.00

62.23925.082.0⨯=⨯-⨯⨯==-dy du A

F μN 05

.1162.21022.43=⨯⨯==Fu P kW 1-13

0841

.0100092.0109144.04=⨯⨯⨯==-νρμ Pa·s 1459

.03048.01524.014.3=⨯⨯==dL A πm 2

2

.736102

4.1526.152061459.00841.03

=⨯--⨯⨯==-dy

du A

F μN

42

.462.736=⨯==Fv P kW

1-14

dr r r r rdr r dy du dA

r dF dM 3

202δ

μπωδωπμμ=-⋅==⋅=

δ

μπωδ

μπω

322420

3

d dr r dM M d A

=

==⎰

⎰ 1-15 785

.0125.014.3=⨯⨯==dL A πm 2

3610258.4001

.003.0785.01008.18--⨯=-⨯⨯⨯==dy du A

F μN 1-16

1884

.03.02.014.3=⨯⨯==Db A πm 2

δ

μδμμ2

0u A

u u A u dy du A Fu N =-=== 9374.01884.0245.01008.07.502=⨯⨯⨯==

-A N u μδm/s

9056.892

.014.39374.06060≈=⨯⨯==

D u n πr/min

1-17 082.091810893.04

=⨯⨯==-νρμ Pa·s

75.1410

3.00

3.01.08.1082.03

=⨯-⨯⨯⨯==-dy du A F μN 1-18 由1-14的结果得

2

.791023.096046.09014.31044003032323

4

24424=⨯⨯⨯⨯⨯⨯=⨯==--δμπδμπωnd d M N·m

1-19

dy

du A

F 00μ=

dy

du

A

F 120120μ=

%7.86015

.0002

.0015.00120001200=-=-=-μμμF F F

1-20

3.29105.032

4.010

5.08.910000728.098.1324.098.133

2=⨯⨯-⨯⨯⨯⨯=-=

--r gr h O H ρσmm