太子爷小说网 > 文学电子书 > 黄万里文集 >

第16节

黄万里文集-第16节

小说: 黄万里文集 字数: 每页4000字

按键盘上方向键 ← 或 → 可快速上下翻页,按键盘上的 Enter 键可回到本书目录页,按键盘上方向键 ↑ 可回到本页顶部!
————未阅读完?加入书签已便下次继续阅读!




slope  J  at  the  given  section  of  flow  are  all  given;  it  is  required  to  find  the  velocity

profile  u~y  along  the  vertical  of  the  section。  Since
u*  =
ghJ
and
um   = Q / Bh


are  known;  from
um   = u*C /
g ;  the  Chezy  constant  reflecting  roughness  of  the

boundary of flow is also given。

Evidently;  the  u~y  relation  must  satisfy  the  requirement  of  total  discharge  Q

after integration; i。e。;

h
? udy = um h = Q / B
(13)
o

Among  an  infinite  number  of  such  u~y  relations;  there  exists  only  one  relation  true
&
that  obeys  the  law  of  minimum  rate  of  energy  reserved
Er     in  the  section;  which  is
equivalent to the law of maximum rate of energy dissipation:

E
r
h
& = ?
? u 2
( )
?Budy ?
p ?
+ + y ?

o ? 2 g ?
87


=
?B   h
     ?

u 3 dy ? ?hQ = min 。

(14)


since
2 g

p
    + y = h ; and

o

h
?o  Budy = Q 。

Thus the problem reduces to one of finding the function of velocity distribution

along the vertical such that

h
? u 3 dy = min 。
(15)

whith
o
u = u( y ) = u(y;? ) ; where
? = ?(y ) 。
Substituting Eq。 (12) into (15); we have

u 3
h h ?
3
y  dy ?
? u 3 dy = ?
*  ??
    + 1?
dy = min 。
(16)
o o ? o  ?y ?



The   equation   shows   that   the   required   condition   of   minimum   holds   for   any

multiple of u; i。e。; u=kua ; as
u*   =
ghJ
is given; This offers a method to determine

the  absolute  values  of  u  so  as  satisfy  Eq。  (13)。  Let  us  change  u  to  ua  in  Eq。  (12)  and

(16) for the time being:

h h
? u 3 dy = k ?
u 3 dy = min 。
(17)
o o a

According   to   the   variational   principle;   the   Euler ’s   Formula   provides;   the

extremum occurs at
u 2   ?ua    = 0
a    ??

since
ua   ? 0 ; the minimum condition is
? y  dy
   ?    = 0



(18)
??  o  ?y

The integral equation is solved through the following processes:
y  d ln y 1   dy
?   +   = 0
?o ? 2
?y d?



88


d ln y
d (d ln y ) d ln y
d (d? 2  / 2)
( ) ( ) 
  =   ?   ?
? 2 d ? 2  / 2 d ? 2  / 2
d (? 2  / 2)
1 d (? 2  / 2)
d (d ln y )
d (d? 2  / 2)
  =   ?
2   ? 2  / 2 d ln y
d (? 2  / 2)

?   ? 2  ?
1
1
   ln? 2  / 2 = ln(d ln y ) ? ln? d   ? + ln C
2 ? 2  ?


1
? ? 2  ? 2
?     ?
?  2  ?

? ? 2  ?
1
d ?     ? = C d ln y
?  2  ?

3
2 ? ? ? 2
  ?    ?
3 ? 2 ?

2
= ln y C1    + C

? 3
2
  = ln y C1    + C

(20)

The
3
? ~  y
2
curve  is  fixed  at  the  two  ends  by  the  following  relations:  At  the

water surface;



y = h;



? = 0;



u = umax



(21)

1 y
This  is  seen  from  the  Nikuratze  experiments  of        ~     
ro ro
curves;  although   ?   is  a
little over zero when the Reynold’s number is high。 M。S。Yalin (page 31; (2)) explains

that  when  approaching  the  free  surface;  the  turbulent  fluctuations  will  be  dampened
du
so that 1 tends to zero while    
dy
is not necessary to be zero。

At the top of sand bed; we already have

y = y0   = v / u*   ? 0;
? = 1;
u = u*
(11)

Thus; by substituting Eq。 (21) into (20;
C1
ln h
? C2   = 0
(22)
Also Eq。 (11) into (22);



89


1 1
y ? ?
? = ? ln
? ln  1  ? 3

 
? ?
? h ?

u h
? ?
R
? *  ?
(23)
where
R*


 ua  
=     *    
v


y  dy



1
? 1  ? 3


d ln y
y   h  
(24)
*
u   = ?o  ?y
+ 1 = ? ln

?  ?
o
R*  ? ?

1
1
y ? 3
? ln     ?
? h ?

2
3
1 ? y ?
*
= (1 + 1。5 ln R
) ? 1。5(ln R
) ? ln     ? 3
(25)
*
? h ?

To find the mean of ua ;

d
? y ?
?    ?
 uam    = 1 ?h  ua   1 ? y    ? h ?
u
   dy = (1 + 1。5 ln R  ) ? 1。5(ln R  )3
u* h  o *
* * o
2
? ln y ? 3
? ?
*
1
? h ?


*
= (1 + 1。5 ln R
) ? 1。5(ln R
)3 ?(5 3)

(26)

To transform ua  back to u in order to satisfy Eq。 (13); we can let
m
*
 ua    ? 1
u ? u
= (u
? u  )  u ? u*     =   u*  (u
? u  )
(27)
* m *
um  ? u*
 uam    ? 1
u*


ln R


? (ln R

2
)1 ? ln y ? 3
?  Q ? *
*   3 ? ?
? h ?

 
? u = ?      ? u* ? 1 + u*
Bh
(28)
*
? ?  ln R*
? (ln R
)3 ?(5 3)

u* h

h 3 2 gJ

? 5 ?
in which
u*  =
ghJ  ;
R*   =   =   ;
??   ? = 0。9028
v v ? 3 ?



90


2
?  Q ? (ln R
)=3 (ln y
h)3
u = u*
+ ?      ? u*
  * 
2
2
(29)

? Bh
? (ln R

*
C
)=3   ? ?(5 3)

hJ



  ?   (30)
ghJ
When y=h;
u = umax   = 2 2
1 ? ?(5 3)  (ln
ghJ )=3 (ln
ghJ )3
?(5 3) ? 1



When y=ym;

u = um ; (ln y

2
h)3
±1
= ?(5 3) = 0。9028; ln   y
? ?
?    ?
? h ?

= 0。8578 ;

ym  / h = 0。424
(31)
which  is  the  exact  location  of  the  mean  velocity  um   along  the  vertical  of  profile;
while   the   conventional   stipulation   of   measuring   um     is

downward from the water surface。
1 ? ym  / h = 0。6 ? 0。576

It  is  seen  that  Formulae  (23)  and  (29)  without  any  empirical  coefficient  are

derived from purely theoretical analys


Example
Data from “Summary of Alluvial Channel Data From Flume Experiments”; 1956…61。

U。S。Geological Survey Professional Paper 462…1; 1966。 By H。P。Guy; D。B。Simmons; and E。V。Richardson。
Run 1。 P。 162; 163。

Slope J=0。00034

Depth 0。58 ft。 = 0。177m。 from bottom of flume。

h = 0。535 ft。 = 0。163m。 from sand bed。 Width B = 8。00 ft。 = 2。439m。
Discharge Q=3。42 c。f。s。=。0969m3 /s。  ≈。984m3 /s (from pitot tube measurement)
q=Q/B=0。428ft2 /s;=0。0394m2 /s。≈。04036m2 /s
Temprature T=13。6o C

Suspended load = 0; Bedload=0。000032 lb。/s。ft。=0。0476 t/s。m。
Bed particle size = 0。643×10…3 ft。; D50 =0。000196m。
Mean Velocity; apparent;
uam   = 3。42 / 0。58 × 8 = 0。737 ft / s = 0。225m / s
Mean Velocity; actual;
um   = Q / Bh = 3。42 / 0。535 × 8 = 0。799 ft。 / s。。
= 0。0394 / 。163 = 0。242m / s
91


。04036
Mean Velocity from pitot measurement= um   = ?u?y / h =   = 0。248m / s
1
。163

Shear Velocity=0。080 ft。/s=0。0244m/s。 for h=0。177m。


*
Shear Velocity actual;
u   = (9。81× 。163 × 。00034)= 2   = 0。0233m / s

Kinematic Viscocity

v = 1。28 ×10 ?5  ft 2  / s = 1。19 ×10 ?6 m 2  / s。
Shear stress at flume bed=0。012 lb。/sq。ft。=58。6×10…6 t/m2 。

    *
u h
Reynolds Number R=328; Shear Reynolds No。= = R


。0233 × 。163
=
  = 3。192

Froude Number F=0。17。
v * 1。19 × 10 ?6

1 1
Chezy   C
g  = 9。3 ft 2  / s = 5。14m 2  / s。

Manning n=0。026

Bed configulation; ripple。

The curve fitted    u=0。47 lg y+1。15 ft。/s。

=0。143 1g y+0。425 m。/s

=0。312+0。0621 ln y/h m。/s。


Original Data from the U。S。G。S Experiments


SEDIMENT TRANSPORT IN ALLUVIAL CHANNELS

TABLE 12 – Vclocity…profle data for 0。19…mm sand in 8…foot…wile flume

(The lateral section used for measuring was 95…115 ft。 from the headbox。
w。s。…water surface)


 


Run 2。0 ft from left wall of flume 4。0 ft from left wall of flume 6。0 ft from left wall of flume    
  Distance above sand
bed (ft)
Velocity
(fps) Distance above sand
bed (ft)
Velocity
(fps) Distance above sand
bed (ft)
Velocity
(fps)    
1 。564 w。8。 。535 w。3。 。523 w。3。    
2 。547 。88 。506 1。15 。492 1。09    
3 。447 。76 。406 1。07 。392 1。09    
4 。347 。76 。306 。91 。292 。91  


92


 
5 。247 。68 。206 。91 。192 。84    
6 。197 。68 。156 。77 。142 。76    
7 。147 。58 。106 。77 。092 。80    
8 。097 。52 。056 。52 。042 。76    
9 。017 。52 。006 。38 …… ……  


The averaged velocity profile of the above data
Run 1。 G。S。Prof。 Paper 462…1; 1966

By Guy; Simmons and Richardson


 

y
(ft)
h
(m)
y/h Velocity;(ft/s)
Δy
Interval
u
(m/s)
Δy?u     
      2ft left
Middle 6ft left
Average          
。535 。163 1。000 w。s。 w。s。 w。s。 1。03 。004 。314 。00126    
。506 。154 。9450 。82 1。15 1。10 1。02 。020 。312 。00624    
。406 。124 。7610 。76 1。07 1。09 。97 。030 。296 。00888    
。306 。0933 。5720 。72 。91 。92 。85 。031 。259 。00803    
。206 。0628 。3850 。68 。91 。85 。81 。031 。247 。00568    
。156 。0476 。2920 。60 。77 。78 。72 。023 。219 。00328    
。106 。0323 。1980 。53 。77 。77 。69 。015 。210 。00315    
。056 。0717 。1050 。52 。52 。77 。60 。015 。183 。00275    
。006 。00183 。0112 。52 。38 …。 。40 。009 。121 。00109  


Q
   = 。0394 ? ?u?y = 0。04036
B

=
Q 。0394
u o      =   = 。242(m / s)
?u?y 。04036
=
? u   =   = 。248 (m/s)
m Bh
。163
m h 。163


putation Sheet
The Fundamental Formula

2
3
=2 ? y ?
*

(ln R
)  ? ? ln     ? 3
m
u = u*
+ (u
? u*
= ?   h ?  
2
*
(ln R
)=3   ? ?(5 / 3)


93




um…u* =0。248…0。0233=0。225
putations


ln R*

= 8。068;

*
(ln R

2
)= 3   = 4。023;

?(5 / 3) = 0。9028


2
4。023 ? (ln y / h)3
u = 0。0233 + 0。225 ×  
4。023 ? 0。9028


2
? y ? 3
u = 0。313 ? 0。0720? ln     ?
 The proposed Formula (1)



The Karman Formula:
h ?

u  = 2。5 ln
u*


 u* 
v


y
y + 5。5 = 5。5 + 2。5 ln R*  + 2。5 ln
h
u = 0。598 + 。0583 ln y / h
u = 0。312 + 。0621ln y / h
 The Karman formula。 (2)

 The fitted formula。 (3)



Results of putation and Measurement


 

y
(m) y
h ln  y
h (ln  y )2/3 
h
u(1)
proposed
u(2)
Karman’s u(3)
data fitted
u
measured    
0。163 1。000 0 0 0。313 0。598 0。312 0。314    
。154 。945 …。0566 。147 。302 。594 。308 。312    
。124 。761 …。2731 。421 。283 。582 。295 。296    
。0933 。572 …。5586 。678 。264 。565 。277 。259    
。0628 。

返回目录 上一页 下一页 回到顶部 2 3

你可能喜欢的