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Hydraulics for civil engineering Book · February 2017

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Hydraulics By prof. Dr. Saleh Issa Khassaf

.......................................................................

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Q  2.18 m 3 / sec

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

LOCAL AND MINOR PIPE LOSSES A local is any energy loss, in addition to that of pipe friction alone, caused by some localized disruption of the flow by some appurtenances , such as valves, bends and other fittings . Local losses are usually computed from the equation : v2 hL  kL 2g In which : V=Q/A is normally the down stream mean velocity for enlargements the following alternate formula applies : v12  v 22 hL  k L 2g V1= u.s. velocity V2= d.s. velocity KL= 1.0 for all a sudden enlargement and takes on values between 0.2 and 1.2 for assorted gradual conical enlargements. Local losses coefficient (KL) for some common valve and pipe fitting are listed in table (4)

 Ke=0.5 Ke=1.0 ke  0.5  0.3 cos  0.2 cos2 

r

Ke= 0.04

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

KL

Fitting Globe valve, fully open Angle valve, fully open Butterfly valve, fully open Gate valve, fully open 3/4 open 1/2 open 1/4 open Check valve, swing type, fully open Check valve, lift type, fully open Check valve, ball type, fully open Foot valve, fully open Elbow , 45 Long radius , 90 Medium radius, 90 Short radius (standard) elbow 90 Close return bend, 180 Pipe entrance, rounded, r/D < 0.16 Pipe entrance, square-edged Pipe entrance, re-entrant

D2/D1

KL

1.0 0.8 0.6 0.4 0.2

0 0.15 0.35 0.43 0.48

10.0 5.0 0.4 0.2 1.0 5.6 17.6 2.3 12.0 70.0 15.0 0.4 0.6 0.8 0.9 2.2 0.1 0.5 0.8

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Example : A water is flow from pipe (D=300 mm) to a pipe (D=600 mm) if

the discharge = 900 L/sec and the pressure in the small pipe = 100 kpa , Find the pressure in the other pipe ? Neglect the losses due to friction . Solution : A1 

 * D12

 0.0215 m 2

4 V1  12.47 m. / sec A2 

  D22

 0.09 m 2

4 V2  3.18 m / sec

losses due to exp ansion :  k L  1.0 for sudden exp ansion v12  v 22 12.74 2  3.18 2  hL  1.0 *  7.74 m 19.6 2g from Bernoulli Eq. between (1) and (2)

hL  k L

z1 

p1





p v2 v12  z 2  2  2  hL 2g  2g

center of pipe as datum   9800 N m 3 p 100  10 3 12.74 2 3.18 2   0 2   7.74 9800 19.6 9800 19.6 p 2  100.2456 kpa

0

(2) (1) Q=900 L/sec D1=300 mm

D2= 600 mm

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

THE PUMP WITH PIPE LINE The pump increase the pressure into the flowing fluid in the pipe. The energy addition is called the net head (hp) of the pump. The water power (Pw) that is delivered to the fluid stream is the product of the net head, the discharge and the unit weight of the fluid or : Pw   * Q * h p The mechanical power to operate the power must be larger, it is called the brake horse power bh p  T * 

In which : T= the torque  = angular velocity of the pump drive shift The ratio 

:

Pw bh p

Is the pump of efficiency, which may be larger than (0.8) usually are between 70% and 85% . The net positive suction head (NPSH) for a pump is used to determine the head (Zi) that is needed at the pump inlet so that cavitation is avoided in the pump. Cavitation is the conversion of liquid into vapor by locally low absolute pressure. A useful form of (NPSH) relation is : NPSH 

Patm.





Pvap.



 hL  Zi

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Example : Calculate the horse power of the pump to lift the discharge of the pipe line from 140 L/sec to 180 L/sec , if the coefficient of friction decrease 10% from the value before using the pump and the minor losses in the inlet is 0.5 ? Draw the H.G.L and E.L when using the pump? Show the fig. Solution: A150  0.01767 m 2  v150  A75  0.00442 m 2  v75 

Q  7.9 m / sec A150

Q  31.6 m / sec A75

Applying Bernoulli Eq. between tank level and end of pipe hL  90  30  0.5 f

2 v150 L v2 v2  f * 150  75 2g D 2g 2g

L  2.256 D

When using the pump and increase the discharge 2 150

2 150

:

2 75

L v v v  0. 9 * f *  D 2g 2g 2g hp  38 m  E P

60  h p  0.5

The horse power in KW is

:

hp   * Q * E p hp  9.8 * 180 * 103 * 38 hp  67 KW

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

SERIES PIPE FLOW The basic tools for analysis here are equations of continuity , work-energy and Darcy-Weisbach equation or : Q=A1V1=A2V2

z1 

p1



hf  f



continuity equation

p v2 v12  z2  2  2   hlosses  2g 2g

L v2 * D 2g

work-energy Eq.

Darcy Weisbach Eq.

All series pipe flow problem fit one of three computational categories, depending on which factors are known or given and which is sought, as listed in the table :

Category 1 2 3

Problem Types Known quantities To find Q, pipe line properties

hL

hL, pipe line properties

Q

Q, hL

Smallest size D

The problem in categories (1) and (2) are analysis problem , analysis of type (1) problems is direct, without iteration, but iteration may be required for the second group. Category (3) is a design problem, which normally requires more assumptions and more iterative computations to solve. Pipeline properties include the length, diameter and material type so that the relative roughness is known.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Example : pump is used to deliver a discharge of 0.05 m3 / sec from a river to a storage reservoir. The system consists of 500 m of 0.15 m cast iron pipe with a 90 bend with a radius of 0.75m of which then transitions to a 0.1 m cast iron pipe 100 m long, as shown in fig. , If the difference between the water levels in the river and the reservoir is 20 m and the water discharges into the atmosphere at the reservoir through a fully open gate valve, what horse power pump would be required? Given  water  1.003 *10 6 m 2 / sec e=0.00026 for cast iron K cont.  0.054 , Kbend  0.2 , K valve  0.15

solution

A1 

D12



 * 0.75 2

 0.0177 m 2

4 4 2  * 0.1 A2   0.00785 m 2 4 Q v1   2.82 m / sec A1 v2 

Q  6.36 m / sec A2 v1 * D1

2.82 * 0.15  4.22 *10 5 6  1.003 *10 6.36 * 0.1 Re 2   6.61*10 5 6 1.003 *10 from table or Moody diagram find ( f ) f1  0.022, f 2  0.025

Re1 



friction loss  h f1 

f1 L1v12  29.7 m d1 * 2 g

h f 2  51.5 m

Minor losses v2 contraction  k  0.11 m, valve  0.31 m, bend  0.08 m 2g  total pump head Hp  Z  h f1  h f 2  helbow  hbend  hvalve  20  29.7  57.5  0.11  0.08  0.31  101.7 m  power  Q *  * h p  9800 * 0.05 *101.7  49833 W

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Example: Two reservoir are connected by a pipe line consisting of two pipes, on of 15 cm diameter and length 6 m and the other of diameter 22.5 cm and 16 length. If the difference of water levels in two reservoir is 6 m. Calculate the discharge and draw the energy gradient line. Take f=0.04, K inlet  0.5 , K outlet  1 Solution:

From continuity 

*15 2 * v1 

4 v1  2.25v 2

 4

Eq.:

Q1=Q2

* 22.5 2 * v 2

loss head at entrance : hL1  0.5(

v12 v2 )  2.53 2 2g 2g

loss of head due to friction : h f1  0.04

6 v2 v2 * 1  8.1 2 0.15 2 g 2g

loss of head due to sudden enl arg ement : hL2  h f2  f

v12  v22 v2  1.56 2 2g 2g

L2 v22 v2  2.84 2 2g D2 2 g

loss of head at exit  k  hL1  h f1  hL2  h f 2 

v22 2g

v 22 6 2g

 v2  2.7 m / sec Q  A2V2  0.108 m 3 / sec

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

PARALLEL PIPE FLOW (EQUIVALENT PIPES) To solve this problem, replace the set of parallel pipes by single "equivalent pipe". This equivalent pipe, which is devised so it has the same head loss as the original set of parallel pipes and conveys the same total discharge. The equivalent pipe formula can be constructed so it can be used with any pipe combination having head loss characteristics that can be described by the exponential of formula :

h f  KQ n For Hazen Williams : n  1.852 C *L k  1.852k C HW * D 5.33 C k  10.67 (meter system) C k  4.73 ( ft  system )

D, L in meter

D, L in ft

For Manning Equation n  2.0 Ck * n 2 * L k D 5.33 Ck  10.29 (meter system) Ck  4.66 ( ft  system)

D, L in meter

D, L in ft

If two pipe are parallel

:

h f  KQ n  K 1Q1n  K 2 Q2n Q  Q1  Q2 1

solving the two Eq. 1

1

1 1 1 ( )n  ( )n  ( )n K K1 K2 For the reminder of the problem the equivalent variables K and Q are then used in place of original parallel pipes. Once Q has been found, n then a back-substitution into ( h f  KQ ) determines Q1 and Q2

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Example: Two reservoir have difference in water surface elevation of 40 ft. Water flows from the higher reservoir through 4000 ft of 12 inch diameter pipe, which then joins a pair of parallel 2000 ft-long pipes which end at the lower reservoir. One parallel pipe has a 10 inch in diameter, the diameter of other pipe is 8 inch , f = 0.02 for all pipe. Find the discharge for each pipe ? Solution : For each pipe n=2 L 1 1 D 2 g A2 with the given data : K12  2.01 K f

K10  2.51 K 8  7.65 1 2

1

1

1

1

 1 2  1 2  1 2  1 2  1               K K 2 . 51  Ke     7.65   10   8 

Ke=1.014 omitting local losses H  40  ( K12  Ke)Q 2 Q  3.64 ft 3 / sec  h f  KQ n  K1Q1n  K 2Q2n  Q10  2.31 ft 3 / sec  Q8  1.33 ft 3 / sec

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

H.W : 1- Two reservoirs are connected by 2 pipes of the same length laid in parallel. The diameter of the pipe are 10 cm and 30 cm respectively. If the discharge through 10 cm diameter pipe is 0.01 m 3 / sec . What will be the discharge through 30 cm pipe? Assume that "f" is the same for all pipes. Ans. (Q2 = 0.156 m 3 / sec )

2- Two pipes with diameters 2D and D are first connected in parallel and when a discharge Q passes the loss of head is H1. When the same pipes are connected in series for the same discharge, the loss of head is H2. Find the relationship between H1 and H2. Neglect minor losses. Both the pipes are equal in length and have the same friction factor. Ans. (H2 = 45.86 H1)

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Q : What is the difference between pump and turbine in pipe line flow ? Answer :

Pipe line with Pumps : The pump are installed in the pipe line to increase the head so that the fluid can be lifted to higher levels. The pumps not only lifts the water , but also overcomes the friction loss in the suction and delivery pipes (hf) Pipe line with Turbine : A Turbine converts hydraulic energy into mechanical work. The gross head (H1) is equal to difference of levels in the reservoir and tail race. The net head available (H) is less than the gross head and is given by : H  H1  h f

Q : What do you understand by " water hammer " in pipes ? If a fluid flowing in a pipe is suddenly brought to rest by closing the valve, there is an abrupt rise in pressure, because the momentum of fluid is destroyed. A pressure wave transmitted along the pipe. The pressure wave and the accompanying pulsations in a pressure create noise as knocking. The sudden rise in pressure has the effect of hammering actions on the walls of the pipes. The phenomena is known as water hummer. This is a case of unsteady flow and the inertia of fluid is to be taken into account.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Three reservoir problems Problems involving pipe flow between more than reservoirs will always require some form of iterative solution. Here we examine briefly an economical solution strategy for these problems. Example : The figure below is a diagram of the three reservoirs problems, the reservoir are connected by three pipes with an external demand at the common junction of the pipes. The highest reservoir has a water surface elevation of 100 m , the middle reservoir water surface elevation is 85 m , and the lowest reservoir has a water surface elevation of 60 m . Determine the discharge in each pipe ? e  0.0005 m,   1.31 * 10 6 m 2 sec

Pipe

k

n

1 2 3

1469 2432 5646

1.974 1.927 1.971

Solution : It is clear that flow is out of the upper reservoir and into the lowest reservoir. What is unclear is the direction of flow in the pipe that connects the middle reservoir to the system. The key step is to determine the direction is only one trial. Let ( H J ) be the head at the junction , the discharge in pipe (1) and (2) can the be found from these two head loss equation : 100  K1Q1n1  H J H J  K 3Q3n3  60

Now select H J = 85 , the water surface elevation of the middle reservoir ,so that there is no flow in pipe (2) for the first trial solution. Inserting values of (K) and (n) from the table We find Q1  0.098 m3 / sec and Q3  0.0639 m3 / sec . These values, combined with the external demand QJ , do not satisfy continuity at the junction J . To satisfy junction continuity we need more inflow to the junction, so H J must be less that (85)m , thus, we find that the flow in pipe (2) will be toward the junction and will be governed by : 1

85  K 2 Q2n2  H J

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

The junction continuity error for each trial will be : Qe  Q1  Q2  Q3  QJ 1

Now we select trial values for H J until approach Q  0 as shown in the table : H J (m)

85.0 80.0 83.0 83.5 83.7

Q1 m 3 / sec 0.0980 0.1134 0.1045 0.1029 0.1023

Q2 m 3 / sec 0.0 0.0403 0.0251 0.0216 0.0200

Q3 m 3 / sec 0.0693 0.0571 0.0613 0.0620 0.0622

Qe m 3 / sec -0.0259 0.0366 0.0083 0.0083 0.0001

Ans. Q1  0.1023 m 3 / sec Q2  0.02 m 3 / sec Q3  0.0622 m 3 / sec

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

PIPE NETWORK ANALYSIS The water supply system consists of several loops and branches of pipes. The system is known as pipe network. Design criteria are that specified minimum flow rates and pressure head must be attained at the out flow points of network. The flow and pressure distributions across a network are effected by the arrangement and sizes of the pipe and distributions of the outflow. Since a change of diameter in one pipe length will affect the flow and pressure distribution A Hardy Cross Method is one of important method to design of pipe network. A pipe network analysis involves the determination of the pipe flow rates and pressure heads and must satisfy the continuity and energy conservation equations The following three basic conditions: 1- At any junction the total inflow must be equal to the total outflow. 2- The algebraic sum of the head losses around any closes circuit is zero. 3- The head loss equation must be satisfied for each pipe ( h f  kQ n )

A

D

G

B

C

E

H

F

J

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Steps Of Hardy Cross Method The method consists six steps : 1- Assume a reasonable distribution of flow in various pipes satisfy condition (1) . 2- Compute the head loss (hf) in each pipe using ( h f  kQ n ). 3- Divide the network into a number of closed circuit so that each pipe is included in at least one circuit . 4- Compute the algebraic sum of the head losses in each circuit (  h f ) . Take suitable sign convention , unless the assume distribution of flow happens to be correct ,  h f is not zero and the assumed discharge needs correction. 5- Revise the flow by applying the correction Q obtained as under : Let Q be the assumed discharge in a pipe . Let the correct discharge be Q . thus : Q  Q  Q

Where

Q

is the correction in discharge , Now

h f  kQ n  k Q  Q 

n

The head loss along a single pipe is h f If the flow is estimated with an error



2

hL  K Q  Q   K Q2  2Q Q  Q 2

Neglecting



Q 2

hL  K Q2  2Q Q

, assuming

Q



to be small "



Now round a closed loop  hL  0 and maintain continuity

h

L

 kQ n

Q

is same for each pipe to

  KQ2  2Q  KQ  0

i.e.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

2

Q  

College Of Engineering

2

 KQ    KQ KQ 2 KQ 2

2

Q

Which may be written : Q  

h 2 h Q

Which h is the head loss in a pipe based on the estimated

Q

In general :

Q  

h n KQ f

n 1

It must be noted that in the numerator the algebraic sum is taken. In the denominator, the arithmetic summation is done, without considering the sign. While applying the correction, the sign of the correction obtained the general Eq. must be considered. Since some pipes are common to more than one circuit , more than are correction will be applied to such pipes. After the corrections have been applied, new values of assumed discharge are obtained. 6- Assume the discharge as a found in step (5) . repeat the procedure till the corrections become negligibly small . Example : Find the discharge in each pipe of the network shown in figure (a). the values of k corresponding to that in the head loss equation h f  kQ 2 are also shown in the figure.

A

B ١٥

C 1.25

10

5

٥

١.٢٥

١٥

٣.٧٥ ٤.٩٧

5

٤.٠٢

١.٥٧

١٢.٥ ٢.٥

٢.٥

٢.٧٧

٦.٠٧ ٣.٠٩

٢.٥

٢.٥

١٠.٠ ٣

١٢.٥ ١.٨٨ ١.٢٥

١.٨٣ ١.٢

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept. ٢.٥

٠

١.٨٨

٠.٠٥

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Solution:

1-

2-

3-

4-

pipe

circuit

Assume the discharge as shown in fig. (b) . Let us take the head loss in clockwise direction as positive. The correction have been worked out in the correction table . Correction in circuit 1 = +0.03 Correction in circuit 2 = - 0.98 Correction in circuit 3 = + 0.62 Correction in circuit 4 = - 0.05 The discharge corrected after these corrections are shown in fig. (c) . It is to be noted that the correction will have the same sign as found in table if the flow in the pipe under consideration is in the clockwise direction. The correction will be of the opposite sign if the flow is in counter-clockwise direction. The procedure can be repeated by assuming the discharge as given in the table and the fig. (c) till correction become negligible .

AB BE AD DE BC CF FE BE DE DG GH HE FJ HE FE HJ

hL  kQ 2

 hL

nKQn 1

n 1

 nKQ

Q  

1*10*10=100 1.5*5*5=37.5 -5*5*5=-125

-3.1

-2.5*2.5*2.5=-15.6

1*5*5=25 5*3.75*3.75=70.3 2.5*2.5*2.5=15.6 -1.5*2.5*2.5=-9.4 2.5*2.5*2.5=15.6 -5*2.5*2.5=-31.2 -1*2.5*2.5=-6.25 -1.5*2.5*2.5=-9.4 5*1.25*1.25=7.82 1.5*2.5*2.5=9.4 -2.5*2.5*2.5=15.6 0

+73.4

-31.25

+1.62

2*1*10=20 2*1.5*5=15 2*5*5=50 2*2.5*2.5=12.5 2*1*5=10 2*5*3.75=37.5 2*5.5*2.5=12.5 2*1.5*5=15 2*2.5*2.5=12.5 2*5*2.5=25 2*1*2.5=5 2*1.5*2.5=7.5 2*5*1.25=12.5 2*1.5*2.5=7.5 2*2.5*2.5=12.5 0

h  nKQ

97.5

+ 0.03

75

- 0.98

50

+ 0.62

32.5

- 0.05

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

L

n 1

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Multiple Pump System The typical performance curve for centrifugal pump is shown in the fig. below

Discharge L/sec

Pumping station frequently contain several pumps in a 'parallel arrangements' see fig. below , the objective being to deliver a range of discharges. This is a common feature of sewage pumping stations , whole the inflow rate varies during the day.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

The series pumps operating , the same discharge passes through each pump. All pumps system must be operating simultaneously. The configuration is the basis of multistage and borehole pumps ; the discharge from the first pump ( or stage is delivered to the inlet of the second pump and so on .

Pumps operating in series

Note : Efficiency (  ) = output power / input power Also : 

H  hf H

 100

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

OPEN CHANNEL HYDRAULIC When flow takes place in a channel or pipe such that the water has a free surface exposed to the atmospheric, we speak of open channel flow. Rivers, sewer, irrigation ditches and drainage channels, culvert, spillways and similar human made structures are designed and analyzed by method of open channel hydraulics.

Comparison Between Pipe Flow and Open Channel Flow :

Two kinds of flow are compared in figure below :

Pipe flow 1- The water must fill the whole conduit. 2- It a confined in a closed conduit, exerts no direct atmospheric pressure but hydraulic pressure only. 3- The water pressure in the tubes are maintained by pressure in the pipe elevation called hydraulic grade line. 4- The total energy with reference of a datum line is the sum of elevation (z), the piezometric height (y) and velocity head ( v 2 2 g ) and is called energy grade line . 5- The loss of energy that result when water flow between two section is represented by hf . 6- It easily to solve problems. 7- Cross section of flow is fixed defined by geometry of the conduit, generally round. 8- Section of roughness coefficient is easily.

Open channel flow 1- Open channel flow must have a free surface. 2- A free surface is subject to atmospheric pressure. 3- The water surface is the hydraulic grade line and the depth of flow corresponding to piezometric height. 4- The total energy with reference to datum v2 = z y 2g

5- The loss of energy between two section is represented by hf 6- It is much more difficult to solve the problem because the free surface is change with respect to line and space 7- May be any shape from circular to irregular from of natural. 8- Selection of roughness coefficient not easily.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Types of flow in open channel 1- steady and unsteady flow : time as criterion. 2- uniform flow and varied flow : space as criterion . a- Steady uniform flow : the depth of flow does not change during the time interval under consideration. b- Unsteady uniform flow : the water surface fluctuate from time to time while remaining parallel to the channel bottom this is a practically impossible condition . c- Unsteady varied flow. Rapidly varied flow : the depth changes abruptly over a comparatively short distant ; otherwise , it is gradually varied . - Gradually varied flow . Q : Give the summary classification of open channel flow : 1- steady flow a- steady flow b- uniform flow * Rapidly varied flow * Gradually varied flow 2- unsteady flow a- Unsteady uniform flow (rare) b- Unsteady varied flow Q : Give an example for the following type of flow : 1- Uniform flow : flow in a laboratory channel . 2- Unsteady uniform flow : Rare . 3- Rapidly varied flow : Hydraulic jump, Flood wave and Hydraulic drop. 4- Gradually varied flow : The flow on the flow or basin after gate or weir .

State Of Flow : 1- Laminar or Turbulent : Depending on the effect of viscosity relative to inertia and defined by Reynolds Number :

Re 

V *L



The range of Re between the critical value 500 from laminar to high value 12500 to turbulent when the channel flow is uniform , the friction factor when d   4 R and S 

hf L

in Darcy-Weisbach Eq.

f 

8 gRS V2

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

R= Hydraulic radius . 2- Subcritical or Supercritical : depending on the effect on the gravity ( by ratio of inertial forces gravity forces ) this ratio is given by the Froude Number (Fr) defined as :

Fr 

v g * yh

V  mean velosity m

s

y h  Hydraulic depth m

If : Fr < 1 subcritical flow Fr = 1 critical flow Fr > 1 supercritical flow

Chezy's Formula Chezy's formula equation is : V  C RS

Where : C= Chezy's formula . R= Hydraulic radius . S= slope of the channel. Several investigator gave their own expression for the Chezy's coefficient (C). Two commonly used formulas are given below : (a) – Ganguillet Kutter formula (1869) 0.00155 1  S N C N 0.00155  1   23   R S  23 

The coefficient N is known Kutter's coefficient. Its value depends on the nature of the surface of the channel (N= 0.0225 for unlined canal in earth ) (b) – Bazin's formula (1897) C

157.6 m 1.81  R

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

m = Bazin's coefficient m= 1.54 for earth channel in perfect condition m= 2.36 for earth channel in average condition m= 3.14 for earth channel in rough condition m= 0.11 very smooth cement .

Manning formula In 1889, Manning proposed the following formula : 2

1

1 V  R3S 2 N Where : V=mean velocity R= Hydraulic radius . S= slope of the channel. N= Manning's roughness coefficient . For lined cement canal (n=0.015) For unlined earthen canal (n=0.0225) Comparing Manning's formula with Chezy's formula : 1

1 C  R6 N Example : A canal has a bottom width 0f (3 m) and a side slope 2:1 . If the slope of the bed 1/5000 and the depth of flow is (1.5 m) , Calculate the discharge using : (a) Kutter formula (b) Manning formula Take N=0.025 Solution : 0.00155 1  S N C N 0.00155  1   23   R S  23 

(a) :

A= (b+zy) y = 9 m 2 ,

P=9.72 m, R=

A = 0.925 m P

C= 39.3 V  C RS = 39.3 0.925 *

1  0.534 m s 5000

Q  AV  4.81 cumecs

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

/

College Of Engineering

(b) 2

1

1

2 1 3 2 1  1 2 V  R S  * 0.925 3 *    0.537 m s N 0.025  5000 

Q  AV  4.84 cumecs

H.W: The cross-section of an open channel is a trapezium with a bottom width of (4 m) and a side slope 2:1 , Calculate the discharge if the depth of flow (1.5 m) and S=1/1600 use : (a) Chezy's formula C=50 Ans. (13.02 cumecs) (b) Bazin's formula m=2.3 Ans. (9.9 cumecs) .

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Channel of Composite Roughness When the channels have different (n) values for the bed and sides , the equivalent (n) must be used. The water area is divided into N parts having wetted perimeters P1,P2,P3…. Pn with the roughness N1,N2…..Nn and each section has a mean velocity (same mean velocity) Horton and Einstein find equivalent (n) as follow:   n Pi ni 3 / 2   n   i 1 P    

2/3

  n ( P i ni 2 )   n   i 1 P    

Horton and Einstein

1/ 2

Pavlovskij

5 3

n

PR 5  n  Pi Ri3   i 1  ni 

Lotter     

P= total wetted perimeter The equation of Horton and Einstein gives the least error and its good to application

Channel of compound section A compound cross section may be defined as section in which various sub-areas have different flow properties. E.g. surface roughness, etc. . A natural stream having overbank flow during a flood is a typical example of compound cross the roughness of overbanks is usually higher than of a main channel and therefore; the flow velocity in the main channel is higher than that in overbank flow. If assumed that the bed slope is same for the three sub-areas and the total discharge from Manning is :

=

⁄

+

⁄

+

⁄

°

The above assumption leads to large discrepancies between computed and measured discharge when flood flow conditions. By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

H.W1 : A circular pipe of (1 m) radius is laid at an inclination of 5° with the horizontal . Calculate the discharge through the pipe if the depth of water in the pipe is (0.75 m) using Chezy's equation = 65? Ans.

= 13.24

⁄

H.W2 : A channel with a cross section shown in Fig. has a flow of 150 m3/s. The slope of the channel bottom is two per thousand, and the Manning n for the flow surfaces is 0.03. Compute the normal and critical depths in the channel.

Choose the correct answer: The Chezy-Manning equation is valid if: a- The acceleration of the flow is zero. b- The discharge, slope and depth are constant. c- The flow in the channel is normal. d- All the above. e- None of the above.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Channel Of Most Economical Cross Section A channel which gives maximum discharge for a given cross-section area and bed slope is called a channel of most economical cross section or in other words, it is a channel involves lesser excavation for a designed amount of discharge . Also defined as channel which has a minimum wetted perimeter. For a channel of most economical cross section the cross sectional area is assumed to be constant and a relation between depth and width of the section is found out, to gives maximum discharge . Q: find the best hydraulic section of a rectangular channel? A=by ……(1) P= b+zy ……(2) From (1) = Substituting in (2)

For the best hydraulic section; the wetted perimeter is minimum

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Example: A trapezoidal channel carries a discharge of 2.5 cumecs. Design the section if the slope is 1 in 1200 and the side slopes are 1 in 1. Use Chezy's formula c=55. Solution : For best hydraulic section

=

=

+

− =

= .

= . √

∗

∗

= .

= . = =

∗ .

+

− √ +−

B= 0.9 m Width of the base = 0.9 m Depth of the channel = 1.085+2*1.085 Side slope = 1:1

H.W : 1- Prove that the best hydraulic section of trapezoidal channel when the b=base width , y=depth of flow, side slope 1:z is = 2- Prove that the best side slope of the trapezoidal channel is 1:√3 . 3- Prove that the most efficient section of triangular channel is when side make an angle of 45 with the vertical. 4- Prove that the maximum discharge will take place when the depth of water is (0.95) times the diameter of the circular channel.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Total Energy In Open Channels The total energy of liquid at any point is equal to the sum of the elevation head, =

+ .

=

+ .

∝= =

+

+

2

2

ℎ ℎ ℎ

ℎℎ ∝= 1 ℎ

.

ℎ

For the channel slope is small cos ≈ 0 ∴

=

+

+

=T.E.L

In 1912, Boris A-Bakhmeteff introduced the concept of specific energy and defined as with respect to the channel bed as datum : Speci ic energy (E) = y + ∴

=

=

+

2

2

The last equation indicates that for a channel with constant discharge, specific energy is a function of depth of flow (y). For constant discharge , a curve can be plotted between (E) and (y) curve is known specific energy diagram. The curve has two limbs AC and BC, the limb BC approaches the horizontal axis asymptotically and the limb AC approaches the line OD asymptotically . the line OD is indicate at 45 to the horizontal. It should be noted that the curve is applicable when the discharge is Q. If the discharge is changed to ( > ) ( < ) different curve are obtained. If the channel shape or size is changed, a new diagram should be drawn.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

The following points should be noted from the specific energy curve : 1- There are two depths of flow for a given specific energy, except at one point (C) at which there is only on depth. The two depth for a given specific energy are known as alternative depths and are represented by [ < ]. the depth at point (C) is known at the critical depth . 2- When < ; the flow is subcritical and specific energy increases with increase in depth of flow. 3- When > ; the flow is supercritical and specific energy decreases with increase in depth of flow. 4- At the critical depth, the specific energy is minimum. A uniform flow in which the depth if flow is equal to the critical depth is known as critical flow.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Critical Depth In Rectangular Channels Let B be the base width of rectangular channel. Discharge per unit width of the channel is = ∴

=

+

=

2

+

= ∴

2

= =

+

2 The critical depth is at minimum specific energy =0

∴ =1−

2 2

=0

= in a specific energy eq.

Substituting this value of = = =

+

+

2

=

+

2

= It is give also maximum discharge

=

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Critical Depth In Non-rectangular Channels E  y

v2 2g

E  y

Q2 2 gA2

dE Q2 d  1  .    1 2 g dy  A2  dy 2Q 2  dA  dE    1 2 gA3  dy  dy 

dE  0 at critical depth dy

1

Q 2  dA   0 gA3  dy 

 dA   represents the rate of increase of area with respect to (y). a little  dy 

The term 

reflection will show that it is equal to the top width T . therefore; Q 2T 1 0 gA 3 Q2 A3  g T

 A    hydraulic depth T 

Using trial and error to find the critical depth Yc in A and T will occur. H.W1 : A rectangular channel 3 m wide, carries a uniform flow of water of 6 cumecs. Compute and plot the specific energy curve, and determine (a) the minimum specific energy and critical depth. (b) the alternate depth for a specific energy of 2.5 kg.m/kg (c) the depth which is the alternate depth for 1.5m depth. (d) the bed slope required to maintain a uniform depth of 0.6 m. H.W2 : A trapezoidal channel of bed width 3m and side slope 1.5:1 (V:H) carries a discharge 7

m3 . Calculate the critical depth. sec

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Velocity Distribution In Open Channel The velocity in open channel is not uniform over the section. It is retarded near the boundary because of the resistance. Had there been no resistance other than the boundary resistance, the maximum velocity would have occurred at free surface. But the surface tension and wind produce resistance to flow at free surface therefore; the maximum velocity occurs at some depth below the free surface.

The figure represent the curves of equal velocities in the channel. It will be noticed that the velocity decreased as the sides and bed are approached. The mean velocity in any vertical section occurs at the depth of approximately (0.6 D) below the free surface. A more accurate value of the mean velocity is obtained by measuring the velocities at depth (0.2 D) and (0.8 D) from the surface and then taking the average of these velocities as Shown in fig. below.

The surface velocity is generally (1.1) times the mean velocity. However , this ratio is greatly affected by the wind and other factors and as such is not reliable. For shallow channels, the maximum velocity occurs near the free surface ; whereas for deep channels, it occur at

from the free surface.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Measuring Of Velocity The velocity of flow in an open channel can be determined by various method. Some of the more common method are discussed below : (a)- Pitot tube : A Pitot tube is held with its nose facing upstream at the point at which the velocity is required. The rise of liquid above the free surface is measured. The velocity is calculated from the formula : Where: C= is the coefficient to be determined experimentally. h= is the rise of liquid above the free surface. C<1.0

Example: A Pitot tube having a coefficient of 0.98 is used to measure the velocity of water at a center of the pipe . calculate the velocity when the stagnation pressure and static pressure are 6m and 5m respectively. Solution: = 2 ℎ = 0.98 2 ∗ 9.81 ∗ (6 − 5) = 4.33

/

(b)- Current meter : A current meter consists of a horizontal wheel on which buckets(or cups) are fixed (as shown in figure). When the current meter is held in a flowing liquid , the liquid strikes with the buckets and the wheel starts rotating. The current meter is calibrated before use. The calibration is done in a channel in which the velocity is either known or can be calculated. By varying the velocity , a calibration chart between the number of revolution per minute and the velocity is obtained. The velocity (v) can be expressed as : V=CN By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Where (c) is the coefficient of meter obtained from the calibration chart and (N) is the speed in revolution per minute. Using the calibration chart or the equation above, the velocity at any point can be obtained from the number revaluations per minute. (c)- Floats :A small float is permitted to travel a known distance and the time taken by the float is noted. The surface velocity is calculated from the measured distance and time. The accuracy is greatly affected by the resistance produced by winds. From the surface velocity , the mean velocity can be estimated by experience. The mean velocity is (0.8) to (0.95) time the surface velocity.

Measuring Of Discharge The discharge in a small channel can be determined by constructing a notch across the channel and measuring the head over the crest. The discharge is calculated by using the weir formula: (V − notch) Q = Cd. L. H ⁄ ⁄ (rectangle or board crested weir) Q = Cd. L. H Where: Cd= coefficient of discharge. L= length of weir. H= head over the crest. The discharge can be also be determined be Venturi flume. =

−

2 ( − ℎ)

Measuring Of Discharge In Irregular Channels If the channel is very large and irregular in cross section the width of the river or channel divided into a number of vertical segments, the mean velocity in each segment can be determined by measuring the velocity at (0.6D) from the free surface, the discharge in the channel is obtained by product the mean velocity and the area of the segment . By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

For more accuracy , the velocity should be measured at several depths in the segment and the mean velocity is taken as the average of these velocities. Total discharge in the channel is obtained by adding up the discharges in the individual segments.

River Bends When a river flows around the bend, scouring takes place on the outer bank and silting takes place on the inner bank. The scouting is mainly due to impact of water striking the outer bank. The pressure of water on the outer bank is more than the inner bank, due to curvature, the pressure at the surface is more than that near the bottom. This causes the water at the outer bank to flow down ward and form a cross-current.

Hydraulic Jump A hydraulic jump (H.J.) occurs when flow changes from supercritical state to a subcritical state. There is a sudden rise in water level at the point where the hydraulic jump occurs. Rollers of turbulent water from at this point. These rollers causes dissipation of energy. A hydraulic jump occurs in practice at the toe of spillway or below a sluice gate where the velocity is very high. A hydraulic jump can be used for one or more of the following purposes: 1- To dissipate excessive energy. 2- To increase the water level on the downstream of a hydraulic structure. 3- To reduce the net uplift force by increasing the downward weight due to increase depth. 4- To increase the discharge from a sluice gate by increasing the effective head causing flow. 5- To provide a control section. 6- For thorough mixing of chemicals in water. 7- For aeration of drinking water. 8- For removing air pockets in a pipe line. Despite the complex appearance of a hydraulic jump due to excessive turbulence , it has been successfully analyzed by application of the impulsemomentum equation. The impulse-momentum principle gives fairly good results. By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Equations For The Hydraulic Jump  The energy dissipated in H.J. is giving by: ∆ = 1− 2 ∆ =

+

+

-

Or: − + 2 2  The depth before the jump is called (the initial depth y1) and the depth after the jump is called ( the sequence or conjugate depth y2).  The relation between y1 and y2 is : ∆ =



+

= 0.5

1+8

−1

= 0.5

1+8

−1

In rectangular channel, it can be calculated the energy dissipated using the following equation:

∆ =

(

)



Efficiency of hydraulic jump The ratio of specific energy after and before the jump is known as the height of jump:

= 

=

∆

Height of hydraulic jump (hj)(H.J.) The difference of the depths before and after the jump is known as the height of jump :

ℎ = 

−

Length of the hydraulic jump(Lj) The distance between the front face of the jump to a point on the downstream where the rollers terminate and the flow becomes uniform is known as the length of the H.J. By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

The length of the jump varies from (5 to 7) times it height. An average value of (6 hj) is usually taken. Thus

=

= (

−

)

Example: A rectangular channel which is 7m wide carries a discharge of 12 cumecs. If the mean velocity of flow before the hydraulic jump is 7 m/sec, Calculate the height of the jump and the energy dissipated. Solution : = 0.5

=

=

√

1+8

−1

√

12 = 1.71 7 . 1.71 = = = 0.245 7 7 ∵ = = 4.515 √9.81 ∗ 0.245 ∴ = 1.4375 Hj = y − y = 1.4375 − 0.245 = 1.926 m ∆ = 1− 2 =

∆ =

+

=

+

-

= ∆ =

.

+

∗

.

−

.

+ ∗

. .

∗ .

= .

.

Or ∆ =

(

)

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Homework 1- A horizontal rectangle channel of constant

width is fitted with a sluice gate. When the sluice gate is opened, water issues with a velocity of 6 m/sec and depth of 0.5 m at the venacoutreata. Determine whether a hydraulic jump will form or not. If so, calculate the energy dissipated. Ans. (yes, 0.492 kg.m/kg ) 2- A trapezoidal channel of a bottom width 3m and side slope 1.5:1 (V:H) carries a discharge of 7.02 cumecs. If the depth before the jump is 33 cm, calculate the conjugate depth after the jump. Ans. (1.5 m) 3- Determine the flow rate in a horizontal rectangular channel 1.5 m wide in which the depth before and after the hydraulic jump are respectively 0.25 and 1.0 m. Ans. (1.85 cumecs)

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Non-Uniform Flow In Open Channel Most of artificial channel are designed for uniform flow, but there are certain reaches in these channels where the flow is non-uniform. Some of these locations are : 1- The reach upstream of a weir or dam. 2- The reach downstream of a sluice gate 3- The reach is sudden change of slope. In gradually varied flow problems, we are mainly concerned with the method of predicting water surface curve and computing their length ( such as upstream of weir or dam). In rapidly varied flow problems, we are generally interested in the location of2 the rapidly varied phenomena and its effect on the flow ( such as hydraulic jump).

Dynamic Equation of Gradually Varied Flow Dynamic Equation of Gradually Varied Flow is an expression giving the relationship between the water surface slope and other characteristics of flow , this equation is : dy  dx

S  S f 1

d  v2    dy  2 g 

S = bed slope

S f = the slope of the energy line

The equation gives the slope of the water surface

dy with respect to the bottom dx

of the channel. In other word, it gives the variation of depth (y) with respect to the distance along the bottom of the channel. The value of

dy may be zero, positive or negative depending upon the type of dx

flow as described below : 1.

dy = 0. This indicates that the slope of the water surface is equal to the dx

bottom slope. In other words, the water surface is parallel to the channel bed. The flow is uniform. 2.

dy = positive. This indicates that the slope of the water surface is less than dx the bottom slope ( S ) in this case, the water surface rises in the direction of

flow. The profile so obtained is called the backwater curve. 3.

dy = negative. This indicates that the slope of the water surface is greater dx than the bottom slope ( S  ) in this case, the water surface falls in the direction

of flow. The profile so obtained is called the drawdown curve. By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

It may be noted that the slope of water surface with respect to the horizontal ( S f ) is different from the slope of water surface with respect to the bottom of the channel (

dy ). dx

A relationship between the two slope, can be used the following equation: S w  S 

dy dx

This equation can be used to calculate the water surface slope with respect to horizontal.

Classification of channel slopes 1234-

Critical slope S   S c or y n  y c Mild slope S   S c or y n  y c S   S c or y n  y c Steep slope Horizontal slope S  = 0 the normal depth is infinite. 5- Adverse slope : the bottom of the channel rises in the direction of flow. In this case ( S  = negative) the normal depth is imaginary.

Characteristics of Flow Profiles The surface curves of water are also called flow profiles. The flow profile is a backwater curve if the depth of flow increase in the direction of water (i.e. dy = positive ). The flow profile is drawdown curve when the depth of flow dx dy decrease in the direction of flow ( i.e. = negative ). dx

Classification of Flow Profiles For a given discharge, the normal depth ( y n ) and the critical depth ( y c ) can be calculated. On the longitudinal section of the channel, a line parallel to the channel bottom and at a height of ( y n ) is designed as a normal depth line By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

(N.D.L.). A line parallel to the channel bottom and the height of ( y c ) is designed as critical depth line (C.D.L.). Depending upon the slope and zone, the flow profiles are classified into (13) types as follows: 1- Mild slope curves M1, M2, M3 2- Steep slope curves S1, S2, S3 3- Critical slope curves C1, C2, C3 4- Horizontal slope curves H2, H3 5- Adverse slope curves A2, A3 For example S2 curves occurs in the zone (2) of steep slope.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Notes : 1234-

All profile in zone (1) and (3) are backwater curves. All profile in zone (2) are drawdown curves. The profile in zone (3) , theoretically commence from the channel bed. C1 and C3 profiles are practically horizontal.

Summary of flow profile gives the shapes of the profiles - If mild slope and y  y n  y c  M 1 backwater curve - If steep slope and y c  y n  y  S 3 backwater curve

- If horizontal slope y n  y c  y  H 2 drawdown curve

 All flow in zone 1 is subcritical flow and = +ve  All flow in zone 3 is supercritical flow and = +ve  All flow in zone 2 is

=

−ve except c ; = 0 But M , H , A are subcritical and S is supercritical C is uniform critical Example: Draw the water surface at profile Solution: = Supercritical →

= > =+

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Example: A rectangular channel 6m wide consists of three reaches of different bed slopes which are in continuations of another. The first reach has a slope of 1:100, the second 1:2500 and the third 0.0036. the three reaches are, respectively, 200 m, 600 m and 400 m long. Sketch the surface profile when a discharge of 12 cumecs flows in the channel. Take Manning n=0.015. Solution: / = =2 = Let

=

4 = 0.742 9.81

ℎ

ℎ

=

1

=

1 6 0.015 6 + 2

, ℎ

ℎ

ℎ

ℎ

ℎ ∗ (0.01) ℎ, = 0.72 m

ℎ,

.

∗ (6 ) = 12 = 0.52 = 1.50

As the critical depth is 0.742, the first and third reaches indicate steep slope. A hydraulic jump will be formed when the water surface crosses the critical depth line when passing from steep slope to mild slope. There are two possibilities : 1- Hydraulic jump followed by S1 curve. 2- Hydraulic jump preceded by M3 curve, depending upon whether the (H.J) is formed on steep slope or mild slope. Let us whether case (2) is possible : 2/1.5 = = = 0.347 √9.81 ∗ 1.5 2 = 0.5

1+8

−1 ⇒

= 0.3

Thus case (2) is not possible, as the depth if 0.3 m is not available. Let us now try the case(1) . the depth before the (H.J) would be 0.52 m 2/0.52 = = 1.7 √9.81 ∗ 0.52 = 0.5

1+8

−1 ⟹

= 1.06

This case is possible , as the depth 1.06 , is available. When the flow passes from the second to the third reach, M2 and S2 profile are formed. By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Computation Of Gradually Varied Flow An engineer often requires to know the distance up to which a surface profile of a gradually varied flow will extend. For instance, he may require the distance on the upstream of the weir up to which the effect of backwater curve exists to know the extent of submergence so that the land may be acquired accordingly . Three common method of computation : 1- Direct step method. 2- Graphical integration method. 3- Direct integration method. We deals the Direct step method in our studies.

Direct Step Method In this method, the channel is divided into short reaches. The computations of surface profile is carried out step by step from one section to another one. The direct step method for the prismatic channel is explained below. Let us consider a short reach of channel of length (dx) see fig. below Equating the total energy at two section (1-1) and (2-2) v12 v2  y 2  2  S f dx 2g 2g S  dx  E1  E2  S f dx S  dx  y1 

dx 

E 2  E1 S  S f

The method of computation consists of the following steps : 1- Calculate the specific energy at the section, say section (1-1) where the depth is known. This section is usually a control section. 2- Assume an appropriate value of the depth (y2) at the other end of the small reach. The assumed at the section (2-2) will be greater than that the section (1-1) if the profile is rising curve and less if the profile is a falling curve. 3- Calculate the specific energy at section (2-2) for the assumed depth y2. 4- Calculate the slope of the energy line ( S f ) at section (1-1) and (2-2) by 1 1 2 / 3 1/ 2 R 22 / 3 S 1f 2/ 2 the using Manning formula : V 2  R1 S f , V  N N 1

average slope in the reach : S

fm



S

f1

 S

f2

2

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

5- Compute the length of the curve between section (1-1) and (2-2) using the equation : L1 2 

E 2  E1  S f  S f2 S    1 2 

  

6- Now the depth at section 2-2 being known, assume the depth at another section , say 3-3, repeat the procedure to get the length L23 . 7- Repeating the procedure , the total length of the curve may be obtained. Thus : L  L1,2  L2,3  ......Ln1,n where (n-1) is the number of reaches into which the channel is divided.  It may be noted that the subcritical flow the computation is carried from downstream to upstream, whereas in supercritical flow from upstream to downstream.

Example: A rectangular channel of a base width 7m is laid on a slope of 1 in 1000 and it carries a discharge of 35 cumecs. The channel terminates in a free overfall. Compute the length of the flow profile from the free overfall to a point where the depth is 0.95 times the normal depth. (N=0.025). Solution: From Manning formula : Q 

A 2 / 3 1/ 2 R S N

7 yn  7 yn  If y n is the normal depth : 35  0.025  7  2 y n

  

2/3

 1     1000 

1/ 2

By trial and error: y n =2.9 m  0.95 times the normal depth = 0.95*2.9 = 2,75 m The depth at the free overfall is equal to critical depth q 2 3 35 / 7   1.364m g 9.81 2

yc  3

Now make a table to continue the solution

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

y

A

P

R

1.364

9.55

9.73

0.982

R

4

3

0.976

V Q A 3.66

College Of Engineering Sf 

E

V2

v 2n2 4 R3

y

13.4

0.0086

v2 2g

2.046

S fm

E

S  S fm

x

0.00755

0.024

0.00655

3.6

L

3.6 1.50

10.50

10.0

1.05

1.07

3.33

11.10

0.0065

2.070

1.75

12.25

10.5

1.17

1.22

2.85

8.10

0.0042

2.163

2.00

14.00

11.0

1.27

1.36

2.50

6.25

0.0029

2.318

2.25

15.70

11.5

1.37

1.51

2.22

4.94

0.0020

2.501

2.50 2.75

17.50 19.20

12.0 12.5

1.46 1.54

1.66 1.76

2.00 1.82

4.00 3.30

0.0015 0.0012

0.00535

0.093

0.00435

21.4

25

0.00355

0.155

0.00255

61.7

86.7

0.00245

0.183

0.00145

126

212.7

0.00175

0.202

0.00075

285

470.7

0.00135

0.216

0.00035

635

1105.7

2.703 2.919

H.W : 1- A 13.5 cumecs flow in a rectangular channel 7.5 m wide, lined with rubble masonry and laid on a slope of 1 in 1000. The channel ends in a free outfall and at a point in the channel, the depth is 1.86 m. How far upstream from this point the depth be 1.95 m ? (N = 0.017). Ans.(997 m) 2- For a rectangular channel 6 m wide and having Manning N = 0.015, Determine the following : a- The normal slope at the a normal depth of 0.41 m and a discharge 0f 6 cumecs. b- The critical slope and corresponding normal depth for a discharge of 6 cumecs. c- The critical slope at a normal depth of 0.41 m and corresponding discharge. Ans. a- 0.0052 b- 0.00345, 0.467m c- 0.00353, 4.93 cumecs

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Dimensional Analysis and Hydraulic Similitude Dimensional analysis is a mathematical technique which deals with the dimensions of the physical quantities involved in the phenomena. The physical quantities are measurable entities such as length, time, electric current, temperature, density and specific weight. Dimensional analysis reduces the number of variables in a fluid phenomena by combining some variables to form non-dimensional parameter. Instead of observing the effect of individual variables, Dimensional analysis is widely used in research work for developing design criteria and also for conducting model test.

Dimensions The fundamental dimension are mass (M), length (L) and time (T) in M.L.T system, and force (F), length (L) and time (T) in F.L.T system. The two system are interrelated by Newton's second law of motion:  L  F  MX  2  T   FT 2 M    L

  

Temperature,  is taken as the fundamental dimensions.

Dimensional Homogeneity A physical equation is said to be dimensionally homogenous if quantities in both sides of the equation have identical dimensions. Let us consider the following two equations : P  h v

...(1)

1 2 / 3 1/ 2 R S N

 ( 2)

Equation (1) is dimensionally homogenous: F F F  2    3  L    2  L  L  L 

Equation (2) is dimensionally non-homogenous: Left-hand side = (L/T) Right-hand side = L2 / 3 (1)

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Notes: 1- The quantities which are dimensionless are represented by (1) such as slope. 2- A dimensionally homogenous equation is applicable to all system of units. On the hard a dimensionally non-homogenous equation is applicable on to the system of units for which it had been derived. 3- Two dimensionally homogenous equations can be multiplied or divided without effecting. But the equation can't be added or subtracted (as the resulting equation may not be dimensionally homogenous. 4- The principle of dimensional analysis can be used to convert units from one system to another system. Example : convert the pressure from kg / cm 2 to lb / in 2 Solution : 1 lb = 453.6 gm 1 kg = 2.204 lb 1 inch = 2.54 cm conversion factor 

kg / cm 2 kg inch 2  *  2.204 * 2.54 * 2.54  14.22 lb / inch 2 lb cm 2

Therefore; the pressure in kg / cm 2 can be convert into lb /in 2 by multiplying it with (14.22). in other words Pressure ( lb /in 2 )= 14.22* pressure in kg / cm 2

−

Buckingham's  -Theorem states that if there are (n) variables in a dimensionally homogenous equation and if these variables contain (m) fundamental dimension (such as M.L.T), they may be grouped into (n-m) non-dimensional parameters as  -term. The Buckingham method can be summarized as follows: 1- List all the (n) variables which affect the phenomena. 2- Choose (m) repeating variables, when selecting repeating variables, the point mentioned above must be kept in view. 3- Write the general equation giving the functional relationship in the form of equation : F ( 1 ,  2 ,  3 ,)

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

4- Write separate expression for each a1 1

b1 2

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 -term

in the form of equation:

c1 3

1  X X X   2  X 1a X 2b X 3c  2

2

2

 n m  X 1an m X 2bn  m X 3cn  m 

Each  -term contain the repeating variables and one of the remaining variables. The repeating variables are written in exponential form. 5- The exponents of the repeating variables on the basis that each  -term is non-dimensional. 6- After the  -term have been determined, the functional relationship ( [ F ( 1 ,  2 ,  3 ,  m n )  const.] and the required expression can be obtained. The following artifice, if necessary, can be used to get the required expression: a- Any  -term may be replaced by any power of it because the power of a non-dimensional term is also non-dimensional. For example may be . replaced by , , b- Any  -term may be replaced by the product of that term with an absolute numerical constant. For example can be replaced by (2 , 3 , ). c- Any  -term can be replaced by another term which is the product or quotient of two  -term. For example can be replaced by ( ∗ ). d- Any  -term can be replaced by another  -term which is the sum or difference of the  -term with an absolute numerical constant for can be replaced ( - 1) or ( + 1). example

Note: The repeating variables selected are ( , ) i.e., the first representing the fluid property, the second representing the fluid characteristics and third representing the geometrical characteristics of the body. This in general. Example: Assume that the friction factor (f) depends upon the diameter of pipe (D), density ( ), viscosity( ), the height of roughness(k) and the velocity (v). Drive  -term of the non-dimensional variables of the friction factor of the flow through the pipe? Solution: = ∅( , , , , ) { , , , , , } Where F stand for 'a function of f ' The number of primary dimension involved is (3), i.e. m=3 The total number of variables is (6) i.e. n=6 By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

Therefore; the number of

 -term is 6-3=3, Thus (

Now taking , ,

College Of Engineering

,

,

)=

as repeating variables

= = = The exponents can be determined as under (Re) = Writing the dimensions =

[ ]

Equating exponents of M,L and T For : M: 0= + 1 = −1 T: 0=- − 1 = −1 L: 0=−3 + + − 1 = −1 Therefore; = As reciprocal of a non-dimensional parameter is also non-dimensional, the expression for can be written as: = Likewise, writing the dimensions in expression for = =

[ ] [ ]

Equating exponents of M,L and T M: 0= T: 0=− L: 0 = −3 + + + 1 → Therefore;

= −1

=

Likewise, =

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

[ ] [

=

]

Equating exponents M: 0 = =0 T: 0 = − + 0 =0 L: 0 = −3 + + + 0 =0 It may be noted that the non-dimensional variables, such as f, itself becomes the  -term . Thus the functional relationship becomes , =∅

,

= ,

→

=∅

,

The friction factor depends upon Re and

(roughness ratio)

H.W : 1- the discharge Q of a centrifugal pump depends upon the mass density of fluid ( ),the speed of the pump (N), the diameter of the impeller (D), the monometric head (Hm) and viscosity of fluid (u), show that : = ∅ ,

2- show that the viscosity through a circular orifice is given by =

2 ℎ

,

H= head causing flow D= diameter of orifice

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Example : If the drag force on the ship (Fd) depends on length of ship (L), viscosity of the liquid ( ), velocity of ship (v) and ( ),(g). Explain the terms of dimensionless parameter. Solution : By Rayleigh's method = ( , , , , ) = = :1 = + : −3 − + + : −2=− − −2 Solve the equation in term of c and e : =2+ − , =1− , = , ∴ =

=2− −2 ,

=

= ∴

=

∴

(

,

)

=

(

,

)

By Buckingham's method ( , , , , , )=0 6-3=3 repeating variables= , , ( , , )=0 = In term MLT =

∴

=

→

=

=

→

=

,

,

H.W : A capillary rise (h) of a liquid in tube varies with tube diameter (d), gravity(g), fluid density ( ), surface tension ( ) and contact angle ( ). Find a dimensionless statement of this relation.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Comparison of Rayleigh and Buckingham method In Rayleigh method, a series of terms is used. The variables are expressed in exponential forms. The values of the exponents are determined from the principle of dimensional homogeneity. The method become tedious when a large number of variables affect the phenomenon. In Buckingham method, there is no series of terms, the expression for  -term are simple and straight forward, irrespective of the number of variables involved. In flow phenomenon in which the number of variables is large, this method is more convenient than the Rayleigh method . It may, however , be noted that both the methods are intrinsically the same . both methods are based on the principle of dimensional homogeneity. The algebraic steps in the two methods are essentially the same.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

Hydraulic Similitude Model analysis is frequently used to study the flow phenomenon which are complex and are not amenable to mathematical analysis. In model analysis, investigations are made on model which is similar to the full-size structure known as prototype. Model study of proposed hydraulic structures and mechanies are generally under taken to product the behavior of the prototypes. The civil engineering model such as dams, spillways and canals to know the working of full size structures. It is noted essential that the model should always be smaller than its prototype. Sometimes a full size model or even a model larger than the prototype is used. In fact, a model is a mechanical analog of the prototype. The advantages of model testing of for most economical and for safe design in case of its failure.

Types of hydraulic similarity To complete working and behavior of the prototype , from its model, there should be a complex similarity between the prototype and its scale model. This similarity is known as hydraulic similitude or hydraulic similarity. Following three types of similarity : a- Geometric similarity The model and its prototype are geometrically similar when they are identical in shape but differ only in size. Thus the scale ratio (Lr) is: =

=

=

Similarly, area ratio and volume ratio : = =

= =

b- Kinematic similarity Kinematic similarity if the prototype and model have identical motions or velocities. On other words, the kinematic similarity is said to exist between the model and prototype, if the ratio of the corresponding velocities at corresponding points are equal. If , the velocity of liquid at point 1 and 2 The kinematic similarity exist, then the velocity ratio of prototype to the model (Vr): By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

( ( ( = (

=

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( ) ) = = ) ( ) ( ) ) = = ) ( )

c- Dynamic similarity Dynamic similarity means the similarity of forces. The corresponding points of prototype and model are equal. ( ) ( ) = = ( ) ( ) The force mean : gravity force, pressure force, viscosity force, elasticity force surface and the tension force : ( ) ( ) ( ) = = = = ( ) ( ) ( ) The technique of hydraulic model involves the selection of suitable scale, operation of hydraulic model and correct prediction. Example : A spillway design to be studied by means of a geometrical similar model constructed to a scale of 1:20. Neglecting viscous and surface tension effects, calculate : a- The discharge in model corresponding to a discharge of 1500 cumecs in the prototype. b- The velocity in the prototype corresponding to a velocity of 2 m/sec in the model. c- The height of the hydraulic jump corresponding to a jump of 5 cm in the model. d- The energy dissipated in the prototype corresponding to 0.25 H.P in the model. Solution : a)( ) = ( ) ∴( ) =( ) → ∴

=

→

=

= ∴

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

B)-

= C)-

Height of hydraulic jump, from geometric similarity, = = 20 ∗

=

= 20 ∗ 0.05 = 1

D) Therefore; ( . ) = ( . )

H.W: The mean velocities in the river and its model are respectively 3 m/sec and 1 m/sec. if the slopes in the river and model are 1:2500 and 1:200 respectively. Calculate the length ratio.

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

1 2

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

CLASSIFICATION OF MODEL All hydraulic models may be broadly classified into following two types : 1- Undistorted model 2- distorted model Undistorted model: which is geometrically similar to the prototype (i.e. having geometric similarly in length, breadth, height and head of water etc.) the protection of undistorted model is comparatively easy and some of the result, obtained from the models, can be easily transferred to the prototypes as the basic condition (of geometric similarity) is satisfied. Example : the discharge over a model, which is reduced to 1:100 in all its dimensions is 1.5 l/sec. what is the corresponding discharge in the prototype ? Solution: = 1.5 / Q =over prototype q=a*v Q=A*V

Distorted models A model is said to be perfect or undistorted when it is perfectly similar to its prototype. In undistorted models because the conditions of similitude are completely satisfied. The results of model investigation can be directly used for the prototype . A model is said to be distorted when one or more characteristics of the model are not identical with their corresponding characteristics in the prototype. In distorted model, the result obtained are primarily qualitative. However, in certain cases, by applying the laws of distortion, the results obtained from model investigation can be transferred to prototype quantitatively as well. Distortion of the model can be due to any one of the following models: 1-Deometric distortion : this type of distortion occurs when the scale ratio not constant usually ,the vertical model scale is exaggerated . 2- distortion of configuration : in this type of distortion, the configuration of the model is different from the configuration of prototype, such as the slope of model is different from the slope of prototype , although the geometric similarly exist. By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

College Of Engineering

3- material distortion : in this case , the material used in the model is not of the required grade. 4- Distortion of hydraulic quantities : in such a distortion, there is a distortion of some hydraulic quantity such as discharge, velocity and time. It means that the required scale ratio of that quantity is not maintained. Example of the distorted model is the pressure distribution, velocity distribution and wave patterns are not correctly responded and it is difficult to prepare distorted models of some structure such as river bends, earth cuts and dikes. The following table represent a scale ratios for distorted models for some quantity: Quantity Symbol Scale ratio Reynolds's law Froude's law Length, Breadth, Height Lr, Br, Hr Lr,Dr Lr,Dr Horizontal Area Vertical Area Lr Dr Lr Dr Slope Sr / / . . Velocity Vr / . . discharge Qr / Example : A distorted model of a rigid bed river has a horizontal scale ratio of 1:1000 and a vertical scale ratio of 1 in 100. What is the flow in the model corresponding to a discharge of 5000 cumecs in the river ? also calculate the value of Manning's roughness (N) for model if that for the river is 0.03. If the flood peak requires 1 hour to travel through 100m in the model, how much time would the flood peak take to travel the corresponding distance in the river? Solution : According to the Froude law : 1 1 = 100 10 1 1 1 = = ∗ = 100 ∗ 1000 10 10 5000 1 = = = 10 200 10

=

=

= ≈

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

Prof. Dr. Saleh Issa Khassaf

= =

.

College Of Engineering

= 1.47

∗ 1.47 = 0.03 ∗ 1.47 = 0.0441 1 1 : . = 1000 = 100 1 100 = ∗ 100 = 1 ∗ 100 = 100 ℎ

By Abbas Al-DuJailY. Uni. Of Kufa Col. Of Eng. Structures and Water Resources Dept.

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