Hydrodynamics

1. A nozzle discharges 2cfs with a velocity of 50ft/sec. The nozzle is i s inclined downward so that as the jet strikes a fixed curved vane it is directed 30 degrees from the horizontal. The jet is deflected upward 90 degrees, making an angle of 60 degrees with the horizontal as it leaves the vane. Determine X and Y components of the force exerted.

Sol’n. 0

V1 = 50ft/s

V1x = 50cos30

Q = 2cfs

V2x =50cos60

FX = =

0

0

V1y = - 50sin30 0

V2y =50sin60

 (V – V )  1x 2x

$&{$ (50cos300 - 50cos600) %$$

FX = 71lb

 (V – V )  1y 2y

FY = =

to the right

$&{$ (-50sin300 – 50sin00) %$$

FY = 265lb downwards

2. A jet having a diameter of 2 inches and a velocity of 50ft/sec is deflected by a vane which is curved trough an angle of 60 degrees and which is moving with a velocity of 20ft/sec in the same direction as the jet. Determine X and Y components of the force exerted and direction and velocity of water leaving the vane.

 (V – V )  1x 2x

FX = Sol’n. u = V1 – V’

=

u = 50 – 20 = 30ft/s 3 0ft/s

FX = 19lb

Q’ = Au

FY =

$  $ = Ә ә (30) & #$

=

Q’ = 0.6545cfs

 (V – V )  1y 2y

$&{"'&' (0 – 25.98) %$$

FY = 33lb

V1x = V1 = 50ft/s

{ ˢ$ $ + ˢ$$) V = { ŷŹ$ + ŶŹ$ ) V2 =

V1y = 0 0

2

0

V2y = u sin60 = 30sin60

V2 = 43.59 ft/s

= 25.98 ft/s 0

0

V2x =V’ + u cos60 = 20 + 30cos60 = 35ft/s

$&{"'&' (50 - 35) %$$

Tanβ = V2y / V2x β= 25.98 / 35 β = 36.58

0

4. A submarine when running submerged in sea water has a drag coefficient of 0.15 and a projected area of 65 sq. ft. Determine the drag force for a velocity of 15miles per hour and the horsepower  required to overcome this force.

Sol’n. Specific gravity of Sea Water = 1.03 V = 15mi/hr. = 22ft/s FD = CD γ A

FD

 $

 $$ = 0.15(62.2 x 1.03) (65) [ ${%$$]

FD = 4860lb

3. A jet having a diameter of 1 1/2 inch and a velocity of 60ft/sec strikes a series of vanes so arranged in the periphery of a wheel that the entire jet strikes the vanes. The deflection angle of the vane is 170 degrees. Determine the maximum amount of work that can be done and the direction and absolute velocity of water leaving the vanes

1. An earth canal carries a depth of water of 6ft. The canal is 20ft wide on the bottom and has side’s slopes of 1.5 horizontal to 1 vertical. S = 0.0002. Using a value of n of 0.025, compute the discharge by the Manning formula and with this discharge determine: a. the value of  n in the Kutter formula. b. the value on m in the Basin formula. Sol’n:

Θ = 56.30

a.) By Kutter Formula

y = 6sec56.30 0

 A = 20(6) + 2( (6) (9)

&#'   C=  # {&#'   &#'   75.33 = # {&#'  

 A = 174ft2

n = 0.0253

y = 10.81ft P = b + 2y P = 20 +2(10.81) P = 41.62ft

# $

 ˓9˜ = ŵŻŸ9ŸŵźŶ

R=

b.) by Bazin Formula

R = 4.18

#&) R S  #&) (4.18) Q = 174 ( ""$' 2/3

Q=A(

1/2

2/3

(0.0002) 1/2

Q = 379cfs. Q = AC C=

{˞˟

 = %  #&{&#{""""$

C = 75.33

#' # #' 75.33 = # C=

m = 2.23

2. What should be the slope of a planed timber flume rectangular in cross-section, to carry 30cfs if the width is 4ft and the depth of flow is to be 1.5ft? Is the flow at upper or lower stage? Determine the critical depth and critical slope for the given discharge and width.

FN =

ˢ9 ; Q = VA ˧ˤ

30 = V (6), V = 4.997 FN =

ŸŻ9 ŷŶŶ{ŵŹ

FN = 0.719 < 1 (Upper stage)  A = bd = (1.5x6)  A = 6ft2

 P = + 2(d)   + 2 (1.5) P= #' P=7

 ˓9˜ = ź9Ż

R=

#&) R S  #&) (ź9 ) 30 = 6 ( ""#$ Ż 2/3

Q=A(

1/2

; n = 0.012

2/3

S1/2

S = 0.002

Ŷ ˝ J d =  ˧ ; q = 9I = 7.5 Ŷ ŻŹ d =  ŷŶŶ c

c

dc = 1.2ft

#&) R 

VC = (

c

2/3

Sc1/2; n = 0.012

J9 = ŻŹ9ŵŶ = 6.25 ft/s VU Ÿ{ŵŶ: = ˓9U = Ÿ-Ŷ{ŵŶ = ŷ9Ÿ

VC = RC

6.25 = (

#&) (ŷ9  Ÿ 

Sc = 0.0037

2/3

Sc1/2

3. Find the most efficient cross-section and the required grade of a trapezoidal canal in clean earth with good alignment to carry 470cfs at a velocity of 3ft per sec. assuming side slopes of 2 horizontal to 1 vertical Sol’n.

 Ź – 4)] d + #$ (d) (2d) (2)

Q = 470cfs

 A = [d (2

V = 3ft/s y=

 A = .472d2 + 2d2

 Ź d

Θ = 63.43

 A = 2.472 d 2; A = 156.67ft 2

0

d = 7.96 ft

Q = AV 470 = A (3)

x = 2y b + 2 (2d) = 2 (

 Ź d)

 Ź d b = d (2 Ź – 4) # (d) (2d) (2)  A = bd + $ b + 4d = 2

ˤ9Ŷ R = Żź9Ŷ = 3.98 b = 7.96 (2  Ź – 4) R=

 A = 156.67 ft 2

b = 3.76ft

#&) R S ; n = 0.02  #&) (3.98) S 470 = 156.67 ( ""$ Q=A(

2/3

1/2

2/3

S = 0.00026

1/2

; n = 0.02

4. What should be the bottom width and the depth of flow for a concrete-lined canal of most efficient cross-section with side slopes 3/4 horizontal to 1 vertical to carry 1200cfs on a grade of 5ft per mile? What is the velocity of flow?

Sol’n:  A = d2 + 0.75d2

Q = 1200cfs n=

' = 0.00095 '$'

Q=A(

#&) R 

2/3

S1/2; n = 0.013

#&) () ""#% $

x = 2y

1200 = (d 2 + 0.75d2) (

b + 2(0.75d) = 2 (1.25 d)

b = d = 8.58ft

b + 1.5d = 2.5d

Q = VA

b=d  A = bd +

# (d) (0.75d) (2) $

V=

2/3

(0.00095) 1/2

#$"" {'"'{'

V = 9.31 ft/s

5. Determine the slope in feet per mile that a circular sewer 5ft in diameter must have when flowing at its maximum capacity if the mean velocity is 8ft per sec.

cos

ß Ŷŵ =

, θ = 57.67

$ $'

0

d = 0.938(5) = 4.69 2

A = πr -

^Ŷõ ŵ

%" + $ r sin θ Ŷ{ŹŻźŻ ŵ {   ŶŹ = π (2.5) %" + $ (2.5) sin (57.67) 2

2

A = 19.13ft

Q = AV = 19.13(8) =153.04

 ˓9˜ =

R=

#%

b{{!! = 0.4949 

#&) R 

2/3

Q=A(

1/2

S

; n = 0.000087

#& ) (0.4949) """""

153.04 = 19.13 ( -7

S = 5.639x10 (5280) S = 0.00296

2/3

1/2

S

6. A smooth wooden flume with vertical sides is 8ft wide and has a grade of 0.002. The discharge is 190cfs, the depth of water being 3ft. It is proposed to construct a weir in the flume which will raise the water surface 1ft at a section 1000ft upstream. Assuming the water to have a plane surface, what will be the depth of water 500ft upstream from the weir?

7. Water upon leaving the spillway of a dam passes over a level concrete apron 200ft wide. Conditions are such that a hydraulic jump will form on the apron. When the discharge is 50cfs per ft. width of channel and the velocity where the water leaves the spillway is 45 ft/sec and the depth after  the jump is 10ft., Determine the distance downstream from the dam to the place where the jump occurs.

Sol’n: 3

q = 50ft /s

A2 = bd2 = 200(1.37)

V1 – 45ft/s

A2 = 274

b = 200ft

P2 = 200 + 2 (1.37)

d3 = 10ft

P2 = 202.74

q = v1d1

R2 =

50 = 45d1 d1 = 1.11ft

 = {  $ '" = {#"{#" %$$ $ d2 = 1.37ft q = v2d2 50 = v2 (1.37) v2 = 36.5 ft/s A1 = bd1 = 200(1.1) A1 = 222 P1 = 200 + 2 (1.11) P1 = 202.22

 ˓9˜ = ŶŶŶ9ŶŴŶŶŶ = 1.098

R1 =

 ˓9˜ = ŶŻŸ9ŶŴŶŻŸ = 1.35

#"#%' = 1.224 $ %'&' = 40.75 V = $    S= #&{È n = 0.0113 {&"' {""##% = #&{#$$&È Rm =

m

= 0.1089

   {&'  H = $ + d1 = ${%$$ + 1.11 = 32.55   {%' + 1.37 = 22.06  + d2 = H = $ ${%$$  = $$"%$'' L=  ""#" 1

2

L = 96ft

5. A 24-inch cast-iron pipe, 3/4 thick and 2,000ft long discharges water from a reservoir under a head of 80ft. If a valve at the discharge end is closed in 6 seconds, determine the magnitude of the resulting water hammer.