Ansewr exam 2 hydraulics ii all groups

Beirut Arab University Faculty of Engineering

Exam -2 Sec-5 Eleventh Week

Civil Engineering Dept. CVLE 342 Hydraulics II

Time allowed: 40 Minutes (20%)

Name ID No

QUESTION ONE (5 %)

a- What is the meaning of best hydraulic section.

Best Hydraulic Section (BHS)

Q = .

1

n. 2/1

3/2

3/5

.SP

A

For giver n , S , A, Q = 3/2P

C

For Q = max -------------------- P = mini

So the best hydraulic section can be defined as the channel section which

conveys maximum discharge for a given value of cross-section area.

b- Prove that the triangular section is best hydraulic when Z = 1.0.

Best hydraulic Section of Triangle :

P = 2y 12 Z (1)

A = zy2 y =

Z

A (2)

P = 2 )1

(212

ZZAZ

Z

A

Z

P

= 0 2 *

2

1[ A (1 -

2

1

Z) ] = 0

zy2 (1 -

2

1

Z ) = 0 1 -

2

1

Z = 0

z = 1 (means 45o side slope)

Exam -2 Sec-5 QUESTION TWO (15 %)

The extensive velocity measurements at a station on a certain river have shown that the velocity distribution equation is given by :

u = 2.5 y – y2 + C where u is the general velocity, m/s & y is the height, m, measured from the

bottom and C is a constant. The bottom velocity is measured to be 0.1 m/s and the total water depth is 2.0 m.

If the channel is wide, and S = 8 cm/km, find the following: a. The constant, C. b. The surface velocity, c. The mean velocity , d. The maximum velocity. e. Manning coefficient. f. The unit discharge.

Solution a- Constant, C.

0.1 = 2.5 * 0.0 – 0.02 + C C = 0.1

u = 2.5 y – y2 + 0.1

b- Surface velocity:, Substitute y = 2.0 in the equation:

us = 2.5 *2.0 – 2.02 + 0.1 = 1.1 m/s

c- Mean velocity:

Umean =

= (1/y) [1.25 y2 – y3/3 + 0.1 y]

= [1.25 y – y2/3 + 0.1]

Substitute by y = 2.0

Umean = 1.25*2 – 4/3 + 0.1 = 1.267 m/s

d- Max. velocity: , Umax means = 0’

= 2.5 – 2 y = 0 , y = 1.25 m,

Substitute the value of y in the equation:

Umax = 2.5 *1.25 – 1.252 + 0.1 = 1.66 m/s

e- Umean = .

1

n. 3/2R 2/1S --- 1.267=

.

1

n. 3/22 2/15 )10*8( & n = 0.0112

f- q = y Umean = 2 * 1.267 = 2.534 m3/s/m





Name ID No

QUESTION ONE (5 %)

Prove that the circular open channel section is best when the water level is at half of the diameter.

Best Hydraulic Section of Circular

P = RD

2

A = 8

2D ( sinR ) , D =

sin

8R

A

P = 2

R

sin

8R

A =

2

8A R ( sinR )

-1/2

P = 0 00.1sincos1sin

2

1 2/15.1

RRR

- 2

1 R ( sinR )-1.5

( 1 – cos Θ ) + ( sinR )-1/2

(1.0) = 0

Multiply by ( sinR )1.5

2

R ( 1 – cos Θ ) + ( sinR )

1.0 = 0

Θr ( 1 – cos Θ ) = 2 ( sinR ) by trial Θ = or R =

4

D

Exam -2 Sec-6

QUESTION TWO (15%)

A new area is to be irrigated using Q = 500 *103 m3/hr. The canal is of best hydraulic section (BHS) and has side slopes 2 : 1, longitudinal bed slope =12 cm/km, Manning coefficient, n = 0.0182, free board = 0.5 m, and lining thickness = 10 cm, lining extension = 0.75 m. The unit cost of excavation is 20 $/m3 and the unit cost of lining is 200 $/m3. You are asked to:

1- mention the advantages and disadvantages of BHS in this case. 2- estimate the cost of constructing 1.0 m of canal.

Solution 1- The advantages are that the section convey maximum Q with

minimum p or minimum lining cost. The disadvantages are that the section has b < y. In this case the excavation height may be deep than usual.

2- Q = 500 *103 m

3/hr = 138.9 m

3/s

S = 12 cm/km , Side slope 2 : 1, n = 0.0182 (lined thickness 10 cm)

Excavation cost = 20 $/m3 , Lining cost = 200 $/m

3

y =

8/1

5

223

2/1

12

ZC

ZC

S

Qn

Best hyd. Section (BHS), R = y/2 and for Z = 2, C = y

b = 0.472

y = 6.517 m and b = 3.076 m

Take b = 3.0 y = 6.5 m F = 0.5

Aex = (6.5+0.5)(3+2*(6.5+0.5)) = 119 m2

Plin = 3 + 2 (6.5+0.5)( 22 + 1)

0.5) + 0.5*2 = 34.3 m

exc cost = 119 *1.0 * 20 = 2380 $/m

lin cost = 34.3 *0.10 *1.0 * 200 = 686 $/m

Total cost = 2389 + 686 = 3066 $/m

best hyd. Section, C = 0.472





Name ID No

QUESTION ONE (5%)

a- What is the meaning of isovel?

Isovels are lines of equal velocities in open channel.

Fig. shows Isovels

b- Define four hydraulic quantities can be estimated by using isovels.

Isovels' are used to estimate Q = passing discharge, V = mean velocity,

= correction of energy , correction of momentum and o = actual

boundary shear stress. c- Draw two lines of isovels for the shown open channel.

Exam -2 Sec-3

QUESTION (2) (15 %)

For a given open channel, your are given the following data:

Q = 15.0 m3/s, z = 1.5 , 1/n = 40 , S = 12*10-5 , b/y = 4 , angle of repose of soil = 360 , ds = 0.4 mm. The following table is given for constant a & b, where

c = a (ds)b where c = kg / m2, ds = mm

ds, mm a b

0.15 < dS < 0.28 0.0280 0.33025

0.28 < dS < 0.56 0.04913 0.83371

0.56 < dS < 1.14 0.05791 1.12241

You are asked to: 1- Find values of y & b and 2- Check stability of section against scour. 3- If there is a scour, how can you prevent it by two methods.

Solution

y =

8/1

5

223

2/1

12

ZC

ZC

S

Qn

y = 2.15 m & b = 8.6 m

For y

b ≥ 4 , C1 = 0.7 , C2 = 0.97

(os)max = 0.75 y S --(ob )max =0.75(1000) (2.15)(12*10-5

) 0.1935 kg/ m2

(ob)max = 0.97 y S --(ob )max =0.97(1000) (2.15)(12*10-5

) 0.25 kg/ m2

c = a (ds)b

0.28 < dS < 0.56 a = 0.04913 b = 0.83371

c = 0.04913 (0.4)0.83371 = 0.0229 kg/m

2

tan = (1/1.5) ------------------------------------------ = 33.7o

K =

2

2

sin

sin1

= [1- (sin 33.7)

2/(sin 36.0)

2]0.5

= 0.33

(s)cr = K (b)cr = 0.33*0.0229 kg/m

2 = 0.0076 kg/m

2

(b)max > (b)cr scour occurs on bed

(s)max > (s)cr scour occurs on bed

General solution:

1- Decrease passing discharge.

2- Make lining.





Name ID No

QUESTION (1) (5%)

a- Derive the relationship between (n) & (C) and (f). The Relationship Between:

Manning Coefficient (n) & Chezy Coefficient (C) and Fiction Factor (f)

Manning Eq. V = n

1 R

2/3 . S

1/2 (1)

Chezy Eq. V = RSC (2)

Darcy Eq. hf = f . D

L .

g

V

2

2

(3)

where D is used for circular pipe and for open channel, use D = 4R hf = f

R

L

4 .

g

V

2

2 (4)

L

h f = gR

f

8 V

2 ------------------- S =

gR

f

8 V

2

V = f

gRS8 (5)

Eq. 1 = Eq. 2 n

1 R

2/3 S

1/2 = C RS

C = n

1 R

1/6

Eq. 2 = Eq. 5 --------------------- C RS = f

gRS8

C = f

g8

So C = n

1 R

1/6 =

f

g8

Exam -2 Sec-4

QUESTION (2) (15 %)

a- Draw a sketch to show the relation between the water depth and the velocity inside a pipe of certain D. Show location of max V.

b- Draw a sketch to show the relation between the water depth and the discharge inside a pipe of certain D. Show location of max Q.

c- A sewer pipeline is laid on a slope 0.00062, is designed to convey a maximum

discharge of 120 liters/sec. Considering n = 0.013. Find: 1. The diameter of the sewer pipeline.

2. The maximum velocity. 3. the maximum boundary shear stress

Q = Qmax , y = 0.94 d , θ = 302.4o, θ = 5.278 rad

Ap = (θr – sin θ), Pp = (d/2) θr ,

Ap = 0.765 d2, Pp = 2.639 d,

By applying Manning’s equation: Q = n

A R2/3

. S1/2

0.12 = (1/0.013)(0.00062)0.5 (0.765d2)(5/3)/(2.639d)(2/3)

0.187 = d(8/3)

d = 0.533 m V = Vmax , y = 0.84 d , θ = 257.5o, θ = 4.494 rad

Ap = (θr – sin θ), Pp = (d/2) θr ,

Ap = 0.684 d2, Pp = 2.494 d,

By applying Manning’s equation:

Vmax = n

1 Rmax

2/3 . S

1/2

Vmax = (1/0.013)(0.00062)0.5 [(0.684*0.5332/(2.494*0.533)](2/3)

Vmax = 0.53 m/s

max = Rmax . S

max = 1000 [(0.684*0.5332/(2.494*0.533)] (0.00062)

max = 0.09 kg/m2





Name ID No

QUESTION (1) (5 %) 1- Draw a sketch to show the distribution of average

boundary shear stress distribution acting on the perimeter of an open channel.

Fig. average boundary shear stress for different cross-sections

2- Draw a sketch to show the distribution of maximum

boundary shear stress distribution acting on the perimeter of an open channel.

( os)max = maximum boundary stress on sides

( ob)max = maximum boundary stress on bed

Fig. Location of maximum boundary shear stress on bed and side

(trapezoidal)

( os)max = C1 y S

( ob)max = C2 y S

C2 & C1 an be estimated using the shown figures.

3- Show by using the following velocity equation, how can

you estimate the actual shear stress at a vertical section.

U = 2.5 U* Ln Ks

y30

Solution Actual Boundary Shear Stress Distribution

By applying the velocity distribution Equation

U = 2.5 U* Ln Ks

y30 (turbulent rough flow)

U1 = 2.5 U* Ln Ks

y130 (1)

U2 = 2.5 U* Ln Ks

y230 (2)

eq. (2) – eq. (1) Leads to

U2 – U1 = 2.5 U* Ln Ks

y230 .

130y

Ks

U2 – U1 = 2.5 U* Ln 1

2

Y

Y

U* =

1

2

12

5.2Y

YLn

UU

U* = 0t

0 =

1

2

12

5.2Y

YLn

UU

o = (

1

2

12

5.2Y

YLn

UU )2

Values of o1 , o2, and o3 are shown in figure.

Exam -2 Sec-1

QUESTION (2) (15%)

Three identical channels running side by side with the same slope, S, and

the same roughness coefficient, n. Each of them is rectangular best

hydraulic section. What would be the relative saving in the cost of

excavation if the total discharge of three channels is to be carried by one

channel of rectangular best hydraulic section?

For three channel of BHS, the relative saving in excavation,

Q =

For one rectangular channel carry Q,

Q*n/ 3= = = = 2 Y8

Y = (1/2)(1/8) Q*n/ (3/8)

A = BY = 2 Y2 = 2 (1/2)(2/8) Q*n/

(6/8)= 1.68 K

Where K = Q*n/ (6/8)

For three rectangular channels, each carry Q/3,

Q*n/ 3

= = = = 2 y8

y = (1/2)(1/8) Q*n/ (3/8)

a = by = 2 y2 = 2(1/2)(28) (6/8)

Q*n/ (6/8)

= 0.738 K

the relative saving in excavation isgiven by:

(3a - A)/(3a) = [3*0.738 K -1.68 K]/ (3*0.738 K)= 24 %





Name ID No

QUESTION ONE (5 %)

For a laminar uniform flow, prove that = RS0 Solution

Momentum Equation

ρ ( Q V ) out - ρ ( Q V ) in = forces

( /g) ( Q V )out - ( /g) (Q V ) in = forces

= momentum correction factor

ρ 2 Q2 V2 - ρ 1 Q1 V1 = Forces

( /g) QV2 - ( /g) QV1 = FP1 + WS0 - FP2 - F

( /g) QV2 - ( /g) QV1 = FP1 + WS0 - FP2 - PL

g

QV2 -

g

QV1 = FP1 + WS0 – FP2 - PL

FP1 = hC.G A & W = ( 2

21 AA ) (L) & F = P L

Uniform flow : water depths are the same along distance.

y1 = y2 = y & V1 = V2 = V & FP1 = FP2 & W = AL

= PL

WS0 = ( AL) /(PL)

= RS0 = is called average boundary shear stress .

Exam -2 Sec-2

QUESTION (2) (15 %)

In a wide steady uniform channel, the velocities at 0.2 yo & 0.6yo & 0.8yo from the surface at a vertical section were measured by a current meter and were found to be 0.952 m/s & 0.8 m/s and 0.738 m/s, respectively. The measured total water depth was 12.0 m. It is required to: 1- Estimate the mean velocity, V, by two methods.

Which of them is more accurate?. 2- Estimate the unit discharge. 3- Given following two velocity equations, what is the difference

between them and which one can represent the flow: u = 2.5 ln (30 y/ks) u = 2.5 u* Ln (9 u* y /)

4- Estimate the actual average boundary shear stress, , 5- Estimate the longitudinal bed slope, S, and n. 6- Check stability of canal bed assuming the maximum bed shear stress

equal the actual average boundary shear stress where the mean diameter of soil particle is ds = 0.7 mm. Given that:

c = a (ds)b where c = kg / m2, ds = mm

ds, mm A b

0.15 < dS < 0.28 0.0280 0.33025

0.28 < dS < 0.56 0.04913 0.83371

0.56 < dS < 1.14 0.05791 1.12241

Solution R = yo = yo =1 2.0 m

738 m/sec

V0.6y = 0.80 m/sec

952 m/sec

= 0.845 m/sec

V = V0.6y = 0.80 m/sec

1- The more accurate is v = 0.845 m/s

2- The unit discharge q = V yo = 0.845 * 12 = 10.14 m3/s/m

3- Velocity distribution

Turbulent flow on rough surface

u = 2.5 ln (30 y/ks)

Turbulent flow on smooth surface

u = 2.5 u* Ln (9 u* y /)

The first one can represent the flow.

4- actual shear stress, t, is estimated by two velocity values close to bed.

U 0.4 y = 0.8 = 2.5 u*ln ( 0.4yo/ks)

U 0.2 y = 0.738 = 2.5 u*ln ( 0.2yo/ks)

Subtracting the two equations

0.8 - 0. 738 = 2.5 u*ln ( 0.4yo/ks) - 2.5 u*ln ( 0.2yo/ks)

= 2.5 u*ln ( 0.4yo/0.2yo)

= 0.0357 m/sec = (/)0.5

= 0.00013 t/ m2 = 0.13 kg/m2 = max

5- u* = (gRS)0.5

S = 1.083 * 10-5 cm/km V = (1/n)*(R)2/3 (S)0.5 = (1/n)*(yo)2/3 (S)0.5

0.845 = (1/n)*(12)2/3 (1.083 * 10-5)0.5

n = 0.02

6- c = 0.05791 (0.7) 1.12241 = 0.0388kg/

Comment:

In this case, the discharge must be decreased or the channel be must be

protected by using rocks or PC lining.