Hydraulics Tutorial Open Channel Flow Solution 001

(())  == 0.75    ( )    =     6.525 = 6.867   ( )    =     4.1238 = 5.15

 S bT3RZR0RRRd. S Y0TYd.  SR0dXZbT3RZR0RRRd.S X03X.

W'[&+9

. S R0bXG W'[&+9

. S R0TG

!'[: *G'((:% <'(": $; .0 . S R0TG Finding b using Manning’s equatio+ equatio + D8$+*&7:%&+9 #%'?:c$&7'( 85'++:(E

)   == (+++2√  2√ )1+= (+30. 8 0. 8 = 0. 8 +1. 9 2    =  +2 + 20. 0. 8 1 + 3   =  +5. + 5. 06 0 6 √        .  .    =  = .  =   // .. / / 6.(.5.= .))/(0.8+1.92) .  0.0007 (.))/   = 5.159 4$(<&+9

- S Y0XG !5: 6&7#5 $; 85'++:( &* Y0XG0 _%$<&7: ;%::-$'%7 $; R0>X. S R0>G0 !5: 85'++:( 7:?#5 &* R0T`R0> S 30RG0 Q

Y. Design a channel by Lacey’s theory to carry 50 m K* $; 6'#:%0 !5: 85'++:( &* #$ -: 8"# &+ '+ '(("<&'( *$&( $; G:7&'+ *&c: R0bGG0 4$("#&$+@ Q

/&*85'%9: D^E S XR G K* W:7&'+ *&c: $; ?'%#&8(: D7GGE S R0bGG 4&(# ;'8#

%$( ) = 1.76 6  . = 1./76√  0.0..9./ () = / = /

N$+9&#"7&+'( *($?:

 S 30Yd

S R0RRRQYd

L.7%'"(&8 %'7&"*

() = 0.4848 / = 0.4848 . / () = 4.75 5  = 4.75√ 5√ 5050

\:##:7 ?:%&G:#:%

 S 30UbG

 S QQ0XTG >

P%$** *:8#&$+'( '%:' D=E S _a S QQ0XTZ30Ub S XR0RQ G 4&7: *($?:* DH@3E S R0X@3 ;$% ;&+'( %:9&G: 85'++:( - S 6&7#5 $; 85'++:(V . S 7:?#5 $; ;($6

)  50. =0 3(= =++ (+0.+)0.=5)5 ()+0. 5 )    33.=5858 =+ 2√  2+2.+√ 2.1 2+36 =  +2 + 2√ 1+0. 1+0.5 =  +2.+ 2.236 236 ) 50. −0319.19= =. 3(33.4+ 5 8−2. 2 36+0. 5 )   4 + 28.28.819 = 0  

DfE

 

DffE

- S QQ0XT,>0>QY.

4"-*#&#"#&+9 - &+ f

4$(<&+9 ;$% .

. S 3d0d3GV 30YQG 2:9(:8#&+9 5&95:% <'(": $; . 65&85 &* &G?%'8#&8'-(:V /:?#5 D.E S 30YQG \&7#5 D-E S QQ0XT,>0>QYZ30YQ S >b0bG Q

d0 /:*&9+ #5: 9:$G:#%&8 *&c: $; ' #%'?:c$&7'( 85'++:( #$ 8'%%. ' 7&*85'%9: :C"'( #$ XX G K* &+ G$-&(: -$"+7'%. $; 8$'%*: &+ '(("<&"G 6 ($+9&#"7&+'( *($?: 3@3RRR '+7 *&7: *($?:* $; 30dX@30 !'[: 8%&#&8'( 7&G:+*&$+(:** #%'8#&<: *5:'% *#%:** '* R0RXV 7&'G:#:% $; ?'%#&8(: X0RT8GV '+9(: $; &+#:%+'( ;%&8#&$+ QdRV Manning’s n 0.023 and sp gr of sediment particle as 2.65. 4$("#&$+@ Q

/&*85'%9: D^E SXX G K* A:7 *($?: D4RE S3K3RRR H@3 S 30dX@3 Angle of repose (φ) = 37

R

_'%#&8(: *&c: D7E S X0RT8G + S R0R>Q P5'++:( 7:?#5 D.E S ] P5'++:( 6&7#5 D-E S ] Tanθ = 1/1.75

 = 1.75/√  5/√ 1+1. 1 +1.75/  =  1 −   = 0.86868 1 − (/.)) / S R0TYT

P$+*&7:% *#%'&95# 85'++:(V 1> S R0b

 SR0XYY

 () = 0.05 (. =) ). = 0.05  (()) = = (())  == 0.75    ( )    =     36.99 = 9.81 >

U303 2KG

!%'8#&<: *5:'% $+ #5: -:7

 S R0bZU303 S QY0bb 2KG

!%'8#&<: *5:'% $+ #5: *&7:

> >

 S R0XYYZQY0bb S >R0bU 2KG

 S bT3RZD3K3RRRE. S b0T3.  SR0dXZbT3RZD3K3RRRE.S d0XQdX.

W'[&+9

. S Q0ddG

  ( )    =     20.94 = 7.5375 W'[&+9

. S >0TXG

!'[: *G'((:% <'(": $; .0 . S >0TXG Finding b using Manning’s equation (considering tr apezoidal channel)

)   == (+++2√  2√ )1+= (+1. = 7+2 52. 8 5) 5 2. 85 8 5 = 2. 8 5+ 5 + 14. 14 . 2 14  + 22. 2. 85√  8 5 1+1. 1 +1. 7 5   =  +11 + 11. . 4 88 √        .  .    =  = .  =  // .. / / 55.. = .(2./85+14.214) .  (1/1000) (.))/ = 40 4$(<&+9 ;$% - S U0TG P5'++:( 6&7#5 S U0TGV P5'++:( 7:?#5 S >0TXG _%$<&7: ;%::-$'%7 $; R0>X. T0 = 85'++:( >0R G 7::?V 3X G -:7 6&7#5V 6 >@3 *&7: *($?:* &* :Z8'<'#:7 &+ 9%'<:( $; 7S XRGG0 \5'#

 = 37

&* #5: G'Z&G"G ?:%G&**&-(: 85'++:( *($?:V '+7 65'# 7&*85'%9: 8'+ #5: 85'++:( 8'%%. 6$"# 7&*#"%-&+9 &#* *#'-&(&#.] !'[:

 '+7 1> S R0b0

4$("#&$+@ /:?#5 D.E S >G A:7 6&7#5 D-E S 3XG H@3 S >@3 Angle of repose (φ) = 37

R

_'%#&8(: *&c: D7E S XRGG W'Z&G"G ?:%G&**&-(: -:7 *($?: D4RE S] /&*85'%9: D^E S] Tanθ = 1/2

 = 2/√  2/√ 5   /  / /  =  1 −   = √  1 −  () = 0.056 (. =) ) = 0.056  ./ =  (..)/ (()) = = (())  == 0.75   = ()

 SR0dQ

F$% 7gYGGV

UX0Q> 2KG

>

Manning’s n =

S R0R>T

!%'8#&<: *5:'% $+ #5: -:7

!%'8#&<: *5:'% $+ #5: *&7:

 S R0bZUX0Q> S UR0dTT 2KG

>

 S R0dQZUR0dTT S >b0ddX 2KG

 S bT3RZ4RZ> S 3TQYR 4R

 SR0dXZbT3RZ 4RZ>S 3QddR 4R

W'[&+9

UR0dTTS3TQYR 4R 4R S R0RR>> W'[&+9

>

 = ()

>b0ddXS3QddR 4R 4R S R0RR>3

!'[: 9%:'#:% $; #6$ <'(":* 4R S R0RR>>

)   == (+++2√  2√ )1+= (15+22) 15+22 2 = 38    = 1 5 + 2  2 1 + 2   = 23. 9 4 √   =  = 1.587

 = // / / = . 38(1.587) 0.0022

Q

 S TY0Y G K*

Q

b0 = 85'++:( 65&85 &* #$ 8'%%. 3R G K* #5%$"95 G$7:%'#:(. %$((&+9 #$?$9%'?5. $+ ' *($?: $; R0RR3Y &* #$ -: :Z8'<'#:7 &+ 8$'%*: '(("<&"G 6 XRi $; ?'%#&8(:* -:&+9 Q8G $% G$%: &+ 7&'G:#:%0 =**"G: #5'#



85'++:( &* #$ -: "+(&+:7 '+7 $; #%'?:c$&7'( *:8#&$+0 F&+7 *"&#'-(: <'(": $; -'*: 6&7#5 '+7 *&7: *($?:0 !'[:

R

 S QU  '+7 1> D%'#&$ -:#6::+ -:7 *5:'% *#%:** '+7 8%&#&8'( *5:'% *#%:**E S R0dX0 h*: #%'8#&<: ;$%8:

G:#5$70 4$("#&$+@ Q

/&*85'%9: D^E S 3R G K* A:7 *($?: D4RE SR0RR3Y Angle of repose (φ) = 34

R

_'%#&8(: *&c: D7E S Q8G S QRGG Q RGG 1> S R0dX side slope (θ) = ? P5'++:( 6&7#5 D-E S ]

/ /           =  1 −   =   1 −   = ( (1−2.2)/

F$% 7gYGGV F$% 7gYGGV

() =0. 056 (.) ) = 0.056  /  (.)/ . = . ( )    =      () = = ()  = 20.3925925  S >d03b 2KG

>

Manning’s n =

!%'8#&<: *5:'% $+ #5: -:7

S R0R>Y

 S R0dXZ>d03X S >R0Qb>X 2KG

>

W'[&+9

bT3RZR0RR3Y. S >R0Qb>X

. S 30QG

)/ ( ) (    =     = 20. 3 925( 925 1−2. 2     () = =0.7 (5) 20.3925( 925(1−2.2)/ = 0.75 !%'8#&<: *5:'% $+ #5: *&7:

W'[&+9

20.3925( 925 −2.(2−2. =20.563)/ = 0.7598100.00161.3  −2.2 = 0.563 1−  = 0.369  = 1/21.64 = 2.5 θ = 21.65

R

H@3 S >0X@3

Finding b using Manning’s equation

)   == (+++2√  2√ )1+= (+2. 5 1. 3 1. 3 = 1. 3  +4. + 4. 2 25    =  +2 + 21. 1. 3 1+2. 1 +2. 5   =  +7 √   =  = ..  =  // .. / / 10(..= .))(1./ 3+4.225)   0.0016 () )/  = 6.5 4$(<&+9

- S >0YG 3R0 = *#%:'G 5'* ' *:7&G:+# -:7 $; G:7&'+ *&c: R0QXGG0 !5: *($?: $; #5: 85'++:( &* 30XZ3R ,U0 4#%:'G &* 8$+*&7:%:7 '* #%'?:c$&7'( 6 -'*: 6&7#5 QG '+7 *&7: *($?: 30XL@3J0 'E f; #5: 7:?#5 $; ;($6 &+ #5: 85'++:( &* R0>XGV :Z'G&+: 65:#5:% #5: -:7 ?'%#&8(:* 6&(( -: &+ G$#&$+ $% +$#0 -E P'(8"('#: G&+&G"G *&c: $; 9%'<:( #5'# 6&(( +$# G$<: &+ #5: -:7 $; 85'++:(0 h*: :G?&%&8'( :C"'#&$+ $; 8%&#&8'( *5:'% *#%:** '*@

(/) = 0.155155 + .  . /

4$("#&$+@ W:7&'+ *&c: $; *:7&G:+# D7E S R0QXGG 4($: $; 85'++:( D4E S 30XZ3R

,U

-'*: 6&7#5 D-E S QG H@3 S 30X@3 /:?#5 $; ;($6 D.E S R0>XG

)250.√ 21+1.   == (+++2√  2√ )1+= (3+1. = 3+35+0.20. 220.5)525√  51+1.5 () =  'E

>

 S R0TUQdX G S Q0b

a S =K_ S R0>3YQ A:7 *5:'%

P%&#&8'( *5:'% *#%:**

,U

 S bT3RZR0>3YQZ 30XZ3R S R0Q> 2KG

>

 = 0.155+ .  . / = 0.155+ (.  ...)/  >    == 0.32 0.155+ .  ./ = 0.32    / = 0.4034 .  =*

V #5: -:7 ?'%#&8(:* 6&(( -: &+ G$#&$+0

-E F$% #5: G&+&G"G *&c: $; #5: 9 %'<:( D78E #5'# 6&(( +$# G$<: &+ #5: -:7V

78 S R0YXGG

>

S R0>RUX 2KG