Water Flow in Pipes - Islamic University of...

Chapter (3)

Water Flow in

Pipes

Page (2)

Water Flow in Pipes Hydraulics

Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Bernoulli Equation Recall fluid mechanics course, the Bernoulli equation is:

P1ρg

+v1

2

2g+ z1 =

P2

ρg+

v22

2g+ z2 − hP + hT + ∑hL

Here, we want to study how to calculate the total losses(∑hL):

∑hL = Major Losses + Minor Losses

Major Losses Occurs mainly due to the pipe friction and viscous dissipation in the flowing

water.

The major head loss is termed by(hf).

There are several formulas have been developed to calculate

major losses:

1. Darcy-Weisbach Formula:

Is the most popular formula used to calculate major losses and it has the

following form:

hf = f (L

D)(

V2

2g) , V =

Q

A=

Qπ4

D2→ hf =

8 f L Q2

π2g D5

L = length of pipe (m)

D = diameter of pipe (m)

V2

2g= velocity head (m)

Q = flow rate (m3/s)

f = friction factor

Friction Factor(𝐟): For laminar flow (Re<2000) the friction factor depends only on Re :

f =64

Re (smooth pipe)

If the pipe is smooth (e= 0) and the flow is turbulent with (4000 <Re < 105)

→ The friction factor depends only on Re :

f =0.316

Re0.25

Page (3)



For turbulent flow (Re > 4000) the friction factor can be founded by Moody

diagram.

To use Moody diagram you need the followings:

The Reynolds number: Re

The relative roughness: e

D

Reynolds Number: Re

Re =ρ V D

μ but ν =

μ

ρ→ Re =

V D

ν

μ = dynamic viscosity (Pa. s) or kg/m. s

ν = kinemaic viscosty (m2/𝑠)

V = mean velocity in the pipe

D = pipe diameter

Note:

Kinematic viscosity depends on the fluid temperature and can be calculated

from the following formula:

ν =497 × 10−6

(T + 42.5)1.5 T: is fluid temperature in Celsius

Relative Roughness: 𝐞

𝐃

e(mm) = Roughness height (internal roughness of the pipe)

and it depends mainly on the pipe material.

Table (3.1) in slides of Dr.Khalil shows the value of e for different pipe

materials.

The following figure exhibits Moody diagram:

Page (4)



Note:

Moody diagram is a graphical representation of Colebrook-White formula:

1

√f= −2 log (

e/D

3.7+

2.51

Re√f)

Empirical Formulas for Friction Head Losses

These formulas gives exact value for friction head losses and each formula

extensively used in a specific field, for example, Hazen-Williams formula is

used extensively in water supply systems and most software’s (like

WaterCAD) used it in analyzing and design of water networks. However,

Manning equation us extensively used in open channel, and in designing of

waste water networks.

See these formulas from the slides of Dr.Khalil.

Page (5)



Minor Losses Occurs due to the change of the velocity of the flowing fluid in the

magnitude or in the direction.

So, the minor losses at:

Valves.

Tees.

Bends.

Contraction and Expansion.

The minor losses are termed by(hm) and have a common form:

hm = KL

V2

2g

KL = minor losses coefficient and it depends on the type of fitting

Minor Losses Formulas

Minor Losses

Entrance of a pipe: hent = KentrV2

2g Exit of a pipe: hexit = Kexit

V2

2g

Sudden Contraction: hsc = KscV2

2

2g

Sudden Expansion: hse = KseV1

2

2g

or hse =(V1 − V2)

2

2g

Gradual Enlargement: hge = Kge(V1

2−V22)

2g Gradual Contraction: 𝐡𝐠𝐜 = 𝐊𝐠𝐜

(𝐕𝟐𝟐−𝐕𝟏

𝟐)

𝟐𝐠

Bends in pipes: hbend = KbendV2

2g Pipe Fittings: hv = Kv

V2

2g

The value of (K) for each fitting can be estimated from tables exist in slides

of Dr.Khalil.

You must save the following values of K:

For sharp edge entrance: K = 0.5

For all types of exists: K = 1.

Page (6)



Problems

1. In the shown figure below, the smaller tank is 50m in diameter. Find the

flow rate, Q. Assume laminar flow and neglect minor losses.

Take μ = 1.2 × 10−3 kg/m. s ρ = 788 kg/m3

Solution

For laminar f =64

Re

Re =ρ V D

μ=

788 × 2 × 10−3 × V

1.2 × 10−3

→ Re = 1313.33 V (substitute in f)

f =64

Re=

64

1313.33 V=

0.0487

V≫ (1)

Now by applying Bernoulli’s

equation from the free surface of

upper to lower reservoir

(Points 1 and 2).

P1ρg

+v1

2

2g+ z1 =

P2

ρg+

v22

2g+ z2 + ∑hL

P1 = P2 = V1 = V2 = 0.0

Assume the datum at point (2) →

0 + 0 + (0.4 + 0.6) = 0 + 0 + 0 + ∑hL → ∑hL = 1m

∑hL = 1m = hf in the pipe that transport fluid from reservoir 1 to 2

hf = f (L

D)(

V2

2g) L = (0.4 + 0.8) = 1.2m , D = 0.002m

1 = f (1.2

0.002)(

V2

19.62) ≫ 2 → Substitute from (1)in (2) →

1 =0.0487

V× (

1.2

0.002)(

V2

19.62) → V = 0.671 m/s

Q = A × V =π

4× 0.0022 × 0.671 = 2.1 × 10−6 m3/s✓.

1

2

Page (7)



2. A uniform pipeline, 5000m long, 200mm in diameter and roughness size of

0.03mm, conveys water at 20℃ (ν = 1.003 × 10−6 m2/s) between two

reservoirs as shown in the figure. The difference in water level between the

reservoirs is 50m. Include all minor losses in your calculations, determine

the discharge.

Note: the valve produces a head loss of (10 V2/2𝑔) and the entrance to and

exit from the pipe are sharp.

Solution

Applying Bernoulli’s equation between the two reservoirs

P1ρg

+v1

2

2g+ z1 =

P2

ρg+

v22

2g+ z2 + ∑hL

P1 = P2 = V1 = V2 = 0.0

Assume the datum at lower reservoir (point 2) →

0 + 0 + 50 = 0 + 0 + 0 + ∑hL → ∑hL = 50m

∑hL = hf + hm = 50

Major Losses:

hf = f (L

D)(

V2

2g) = hf = f × (

5000

0.2) (

V2

19.62) = 1274.21 f V2

Minor losses:

For Entrance:

hent = Kentr

V2

2g (For sharp edge entrance, the value ofKentr = 0.5) →

2

1

Page (8)



hent = 0.5 ×V2

19.62= 0.0255 V2

For Exit:

hexit = Kexit

V2

2g (For exit: Kexit = 1) →

hexit = 1 ×V2

19.62= 0.051 V2

For Valve:

hvalve = 10 ×V2

2g= 0.51V2

hm = (0.0255 + 0.051 + 0.51)V2 = 0.5865 V2

∑hL = 1274.21 f V2 + 0.5865 V2 = 50 → V2 =50

1274.21 f + 0.5865> (1)

e

D=

0.03

200= 0.00015

In all problems like this (flow rate or velocity is unknown), the best initial

value for f can be found as following:

Draw a horizontal line (from left to right) starts from the value of e

D till

intercept with the vertical axis of the Moody chart and the initial f value is

the intercept as shown in the following figure:

Page (9)



So as shown in the above figure, the initial value of f = 0.0128

Substitute in Eq. (1) → V2 =50

1274.21 × 0.0128 + 0.5865= 2.96

→ V = 1.72 m/s

→ Re = V D

ν=

1.72 × 0.2

1.003 × 10−6 = 3.4 × 105 and

e

D= 0.00015 → Moody

f = 0.016 → Substitute in Eq. (1) → V = 1.54 → Re = 3.1 × 105

→ Moody → f ≅ 0.016

So, the velocity is V = 1.54 m/s

Q = A × V =π

4× 0.22 × 1.54 = 0.0483 m3/s ✓.

3. The pipe shown in the figure below contains water flowing at a flow rate of

0.0065 m3/s. The difference in elevation between points A and B is 11m.

For the pressure measurement shown,

a) What is the direction of the flow?

b) What is the total head loss between points A and B?

c) What is the diameter of the pipe?

Given data:

ν = 10−6 m2/s , e = 0.015mm , Pipe length = 50m , 1atm = 105 Pa.

Page (10)



Solution

a) Direction of flow??

To know the direction of flow, we calculate the total head at each point, and

then the fluid will moves from the higher head to lower head.

Assume the datum is at point A:

Total head at point A:

PA

ρg+

vA2

2g+ zA =

2.5 × 105

9810+

vA2

2g+ 0 = 25.484 +

vA2

2g

Total head at point BA:

PB

ρg+

vB2

2g+ zB =

2 × 105

9810+

vB2

2g+ 11 = 31.39 +

vB2

2g

But, VA = VB (Since there is the same diameter and flow at A and B)

So, the total head at B is larger than total head at A>> the water is flowing

from B (upper) to A (lower) ✓.

b) Head loss between A and B??

Apply Bernoulli’s equation between B and A (from B to A)

PB

ρg+

vB2

2g+ zB =

PA

ρg+

vA2

2g+ zA + ∑ hL

31.39 +vB2

2g= 25.484 +

vA2

2g+ ∑hL , But VA = VB → ∑hL = 5.9 m✓.

c) Diameter of the pipe??

∑hL = hf + hm = 5.9m (no minor losses) → ∑hL = hf = 5.9m

hf = f (L

D)(

V2

2g)

5.9 = f (50

D) × (

V2

19.62)

V =Q

A=

0.0065π4

× D2=

0.00827

D2→ V2 =

6.85 × 10−5

D4→

5.9 = f (50

D) × (

6.85 × 10−5

19.62 D4) → 5.9 = f ×

0.000174

D5

Page (11)



→ D5 = 2.96 × 10−5 f → D = (2.96 × 10−5)15 × f

15 → D = 0.124 f

15

Re = 0.00827

D2 × D

10−6=

0.00827

10−6 D

Now, assume the initial value for f is 0.02 (random guess)

→ D = 0.124 × 0.0215 = 0.0567 m → Re =

0.00827

10−6 × 0.0567= 1.46 × 105

→e

D=

0.015

56.7= 0.00026 → Moody → f ≅ 0.019

→ D = 0.124 × 0.01915 = 0.056 m → Re =

0.00827

10−6 × 0.056= 1.47 × 105

→e

D=

0.015

56= 0.00027 → Moody → f ≅ 0.019

So, the diameter of the pipe is 0.056 m = 56 mm ✓.

4. In the shown figure , the connecting pipe is commercial steel 6 cm in

diameter having a roughness height of 0.045mm. Determine the direction of

flow, and then calculate the flow rate. The fluid is water (μ = 1 × 10-3

kg/m.s).Neglect Minor losses.

Solution

Page (12)



Assume the datum is at point 1:

Total head at point A:

P1ρg

+v1

2

2g+ z1 =

200 × 103

9810+ 0 + 0 = 20.387 m

Total head at point 2:

P2

ρg+

v22

2g+ z2 = 0 + 0 + 15 = 15 m

So, the total head at 1 is larger than total head at 2>> the water is flowing

from 1 to 2 ✓.

Applying Bernoulli’s equation between 1 and 2:

P1ρg

+v1

2

2g+ z1 =

P2

ρg+

v22

2g+ z2 + ∑hL

20.387 = 15 + ∑hL → ∑hL = 5.38 m

∑hL = hf + hm = 5.38 m

hm = 0 (given) → hf = 5.38 m

5.38 = f (L

D)(

V2

2g) = hf = f × (

50

0.06)(

V2

19.62) = 42.47 f V2

→ V2 =0.126

f→ Eq. (1)

Re = ρ V D

μ=

1000 × V × 0.06

1 × 10−3 = 60,000V → Eq. (2)

e

D=

0.045

60= 0.00075

Assume fully turbulent flow → f = 0.018 → Sub. in (1) →

V = 2.64 m/s → Sub. in (2) → Re = 1.58 × 105 → Moody⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑

→ f = 0.02 → V = 2.51 → Re = 1.5 × 105 → Moody⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑

→ f ≅ 0.02 → V = 2.51 m/s

Q = A × V =π

4× 0.062 × 2.51 = 0.0071 m3/s ✓.

Page (13)



5. Two reservoirs having a constant difference in water level of 66 m are

connected by a pipe having a diameter of 225 mm and a length of 4km. The

pipe is tapped at point C which is located 1.6km from the upper reservoir,

and water drawn off at the rate of 0.0425 m3/s. Determine the flow rate at

which water enters the lower reservoir. Use a friction coefficient of f = 0.036

for all pipes. Use the following K values for the minor losses:

Kent = 0.5, Kexit = 1, Kv = 5

Solution

Applying Bernoulli’s equation between points A and E:

PA

ρg+

vA2

2g+ zA =

PE

ρg+

vE2

2g+ zE + ∑hL

PA = PE = VA = VE = 0.0

Assume the datum at point E →

0 + 0 + 66 = 0 + 0 + 0 + ∑hL → ∑hL = 66m

∑hL = hf + hm = 66

Major Losses:

Since the pipe is tapped, we will divide the pipe into two parts with the same

diameter ; (1) 1.6 km length from B to C, and (2) 2.4 km length from C to D.

Page (14)



hf = f (L

D)(

V2

2g)

For part (1)

hf,1 = 0.036 × (1600

0.225) (

V12

19.62) = 13.05 V1

2

For part (2)

hf,2 = 0.036 × (2400

0.225) (

V22

19.62) = 19.6 V2

2

hf = 13.05 V12 + 19.6 V2

2

Minor losses:

For Entrance: (due to part 1)

hent = Kentr

V12

2g (Kentr = 0.5) →

hent = 0.5 ×V1

2

19.62= 0.0255 V1

2

For Exit: (due to part 2)

hexit = Kexit

V22

2g ( Kexit = 1) →

hexit = 1 ×V2

2

19.62= 0.051 V2

2

For Valve: (valve exists in part 2)

hvalve = 5 ×V2

2

2g= 0.255V2

2

hm = 0.0255 V12 + 0.051 V2

2+0.255V22 = 0.0255 V1

2 + 0.306V22

∑hL = (13.05 V12 + 19.6 V2

2) + (0.0255 V12 + 0.306V2

2) = 66

→ 66 = 13.0755 V12 + 19.906 V2

2 ≫ Eq. (1)

Continuity Equation: π

4× 0.2252 × V1 = 0.0425 +

π

4× 0.2252 × V2

V1 = 1.068 + V2 (substitute in Eq. (1)) →

Page (15)



66 = 13.0755 (1.068 + V2)2 + 19.906 V2

2 → V2 = 0.89 m/s

Q2 =π

4× 0.2252 × V2 =

π

4× 0.2252 × 0.89 = 0.0354 m3/s✓.

Chapter (4)

Pipelines and Pipe

Networks

Page (17) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks

Hydraulics

Flow through series pipes

Is the same in case of single pipe (Ch.3), but here the total losses occur by

more than 1 pipe in series. (See examples 3.9 and 4.1 in text book).

Flow through Parallel pipes

Here, the main pipe divides into two or more branches and again join

together downstream to form single pipe.

The discharge will be divided on the pipes:

Q = Q1 + Q2 + Q3 + ⋯

But, the head loss in each branch is the same, because the pressure at the

beginning and the end of each branch is the same (all pipes branching from

the same point and then collecting to another one point).

hL = hf,1 = hf,2 = hf,3 = ⋯

Pipelines with Negative Pressure (Siphon Phenomena)

When the pipe line is raised above the hydraulic grade line, the pressure

(gauge pressure) at the highest point of the siphon will be negative.

The highest point of the siphon is called Summit (S).

If the negative gauge pressure at the summit exceeds a specified value, the

water will starts liberated and the flow of water will be obstructed.

The allowed negative pressure on summit is -10.3m (theoretically), but in

practice this value is -7.6 m.

If the pressure at S is less than or equal (at max.) -7.6, we can say the water

will flow through the pipe, otherwise (Ps >-7.6) the water will not flow and

the pump is needed to provide an additional head.

Note

Pabs = Patm + Pgauge

The value of Patm = 101.3 kPa =101.3×103

9810= 10.3m

Pabs = 10.3 + Pgauge

So always we want to keep Pabs positive (Pgauge = −10.3m as max, theo. )

To maintain the flow without pump.



Hydraulics

Problems:

1. Three pipes A,B and C are interconnected as shown. The pipe are as follows:

Find:

1) The rate at which water will flows in

each pipe.

2) The pressure at point M.

Hint: Neglect minor losses.

Solution

Applying Bernoulli’s equation between the points 1 and 2.

P1ρg

+V1

2

2g+ z1 =

P2

ρg+

V22

2g+ z2 + ∑hL

P1 = V1 = P2 = 0.0 , V1 = VC =? ? , ∑hL =? ?

0 + 0 + 200 = 0 +VC

2

19.62+ 50 + ∑hL

0.0509 VC2 + ∑hL = 150m → Eq. (1)

hf = f (L

D)(

V2

2g)

hf,A = hf,B (Parallel Pipes) (take pipe A)

∑hL = hf,A + hf,C

hf,A = 0.02 (600

0.15)(

VA2

2g) = 4.07VA

2

Pipe D (m) L (m) f

A 0.15 600 0.02

B 0.1 480 0.032

C 0.2 1200 0.024



Hydraulics

hf,C = 0.024 (1200

0.2)(

VC2

2g) = 7.34VC

2

∑hL = 4.07VA2 + 7.34VC

2 (substitute in Eq. (1)) →

0.0509 VC2 + 4.07VA

2 + 7.34VC2 = 150m

→ 7.39VC2 + 4.07VA

2 = 150m → Eq. (2)

How we can find other relation between VA and VC? ?

Continuity Equation

QA + QB = QC π

4× 0.152VA +

π

4× 0.12VB =

π

4× 0.22VC

→ 0.025VA + 0.01VB = 0.04VC → Eq. (3)

But, hf,A = hf,B →

hf,A = 4.07VA2 (calculated above)

hf,B = 0.032 (480

0.1)(

VB2

2g) = 7.82VB

2

4.07VA2 = 7.82VB

2 → VB2 = 0.52 VA

2

→ VB = 0.72 VA (substitute in Eq. (3))

→ 0.025VA + 0.01(0.72 VA) = 0.04VC → VC = 0.805 VA (Subs. in Eq. 2)

→ 7.39(0.805 VA)2 + 4.07VA2 = 150m → VA = 4.11 m/s

→ VB = 0.72 × 4.11 = 2.13 m/s

→ VC = 0.805 × 4.11 = 3.3 m/s

QA =π

4× 0.152 × 4.11 = 0.0726 m3/s✓.

QB =π

4× 0.12 × 2.13 = 0.0167 m3/s✓.

QC =π

4× 0.22 × 3.3 = 0.103 m3/s✓.

Pressure at M: → Bernoulli’s equation between the points M and 2.

PM

ρg+

VM2

2g+ zM =

P2

ρg+

V22

2g+ z2 + ∑hL(M→2) VM = V2

∑hL(M→2) = hf,C = 7.34 × 3.32 = 79.93m

→PM

9810+ 120 = 0 + 50 + 79.93 → PM = 97413.3 Pa.✓.



Hydraulics

2. Three pipes A,B and C are interconnected as shown. The pipe are as follows:

Find:

1) The rate at which water will flows in

each pipe.

2) The pressure at point M.

Solve the problem in the following two cases:

a) The valve (V) is closed.

b) The valve (V) is opened with K= 5

Solution

Case (a): Valve is closed

When the valve is closed, no water will flowing in pipe B because the valve

prevents water to pass through the pipe. Thus the water will flowing

throughout pipe A and pipe C (in series) and the system will be as follows:

Pipe D (m) L (m) f

A 0.15 600 0.02

B 0.1 480 0.032

C 0.2 1200 0.024



Hydraulics

Now, you can solve the problem as any problem (Pipes in series).

And if you are given the losses of the enlargement take it, if not, neglect it.

Case (b): Valve is Open

Here the solution procedures will be exactly the same as problem (1) >>

water will flow through pipe A and B and C, and pipes A and B are parallel

to each other .

But the only difference with problem (1) is:

Total head loss in pipe A = Total head loss in pipe B

Total head loss in pipe A = hf,A

Total head loss in pipe B = hf,B + hm,valve

hm,valve = KValve

VB2

2g= 5

VB2

2g

Now, we can complete the problem, as problem 1.

3.

The flow rate between tank A and tank B shown in the figure below is to be

increased by 30% (i.e., from Q to 1.30Q) by the addition of a second pipe in

parallel (indicated by the dotted lines) running from node C to tank B. If the

elevation of the free surface in tank A is 7.5m above that in tank B,

determine the diameter, D, of this new pipe. Neglect minor losses and

assume that the friction factor for each pipe is 0.02

Solution

1 2

3



Hydraulics

Before addition of pipe:

Apply Bernoulli’s equation between reservoirs A and B.

PA

ρg+

VA2

2g+ zA =

PB

ρg+

VB2

2g+ zB + ∑hL(A→B)

0 + 0 + 7.5 = 0 + 0 + 0 + ∑hL(A→B) → ∑hL(A→B) = 7.5 m

∑hL(A→B) = 7.5 m = 0.02 ×180 + 150

0.15×

V2

2g→ V = 1.828 m/s

→ Qbefore = A × V =π

4× 0.152 × 1.828 = 0.0323 m3/s

After addition of pipe:

Qafter,1 = 1.3 Qbefore = 1.3 × 0.0323 = 0.0412 m3/s

V1 =0.0412

π4

× 0.152= 2.33 m/s

Apply Bernoulli’s equation between reservoirs A and B.

0 + 0 + 7.5 = 0 + 0 + 0 + ∑hL(A→B) → ∑hL(A→B) = 7.5 m

∑hL(A→B) = hf1 + hf2

7.5 = 0.02 ×180

0.15×

2.332

2g+ 0.02 ×

150

0.15×

V22

2g→ V2 = 0.918 m/s

→ Q2 = A2 × V2 =π

4× 0.152 × 0.918 = 0.0162 m3/s

→ Q3 = Q1 − Q2 = 0.0412 − 0.0162 = 0.025 m3/s

hf2 = hf3

0.02 ×150

0.15×

0.9182

2g= 0.02 ×

150

D3×

V32

2g→ D3 = 0.178V3

2

Q3 = A3 × V3 → 0.025 =π

4× D3

2 × V3 → 0.025 =π

4× (0.178V3

2)2× V3

→ V3 = 0.99 m/s → D3 = 0.178 × 0.992 = 0.174 m = 174 mm✓.



Hydraulics

4. A 500 mm diameter siphon pipeline discharges water from a large reservoir.

Determine:

a. The maximum possible elevation of its summit, B, for a discharge of 2.15

m3/s without the pressure becoming less than 20 kN/m

2 absolute.

b. The corresponding elevation of its discharge end (ZC).

Neglect all losses.

Solution


Pabs,B =20 × 103

9810= 2.038 m

Patm = 10.3m

Pgauge,B = 2.038 − 10.3 = −8.26m

Q = AV → V =Q

A=

2.15π4

× 0.52= 10.95 m/s

Applying Bernoulli’s equation between the

points A and B.

PA

ρg+

VA2

2g+ zA =

PB

ρg+

VB2

2g+ zB + ∑hL

(Datum at A) , hL = 0 (given)

0 + 0 + 0 = −8.26 +10.952

19.62+ zB + 0 → zB = 2.15 m ✓.

Calculation of ZC:

Applying Bernoulli’s equation between the points A and C.

PA

ρg+

VA2

2g+ zA =

PC

ρg+

VC2

2g+ zC + ∑hL

(Datum at A) , hL = 0 (given)

0 + 0 + 0 = 0 +10.952

19.62+ (−zC) → zC = 6.11 m ✓.



Hydraulics

5.

In the shown figure, if the length of the pipe between reservoir A and point

B is 180 m and the absolute pressure at point B is 7.3 mwc. Calculate the

minor loss coefficient (K) of the valve at point C.

Given data:

The entrance is sharp edged.

Total length of pipe (L= 730 m), Pipe diameter (d = 150mm), friction factor

(f = 0.02).

Solution

Pabs,S = 7.3m , Patm = 10.3m → Pgauge,S = 7.3 − 10.3 = −3m

Apply Bernoulli’s equation between the points A and D.

Datum at (D):

PA

ρg+

VA2

2g+ zA =

PD

ρg+

VD2

2g+ zD + ∑hL(1→D)

0 + 0 + 16 = 0 +V2

2g+ 0 + ∑hL(1→D)

16 =V2

2g+ ∑hL(1→D) → Eq. (1)

∑hL(1→D) = hf + hm

hf = 0.02 (730

0.15)(

V2

2g)

hm = 0.5V2

2g+ Kv

V2

2g

∑hL(1→D) = 4.86V2 + 0.0509KvV2 → Sub. in (1) →

16 =V2

2g+ 4.86V2 + 0.0509KvV

2 But V =? ?

Apply Bernoulli’s equation between the points A and B.

Datum at (B):



Hydraulics

PA

ρg+

VA2

2g+ zA =

PB

ρg+

VB2

2g+ zB + ∑hL(1→B)

0 + 0 + 0.8 = −3 +V2

2g+ 0 + 0.02 ×

180

0.15×

V2

2g→ V = 1.727 m/s

→ 16 =1.7272

2g+ 4.86 × 1.7272 + 0.0509Kv × 1.7272 → Kv = 8.9✓.

6. The difference in surface levels in two reservoirs connected by a siphon is

7.5m. The diameter of the siphon is 300 mm and its length 750 m. The

friction coefficient is 0.025. If air is liberated from solution when the

absolute pressure is less than 1.2 m of water, what will be the maximum

length of the inlet leg (the portion of the pipe from the upper reservoir to the

highest point of the siphon) in order the siphon is still run if the highest point

is 5.4 m above the surface level of the upper reservoir? What will be the

discharge.

Solution

The graph is not given so you should understand the problem, and then

drawing the system as follows:


Pabs,S = 1.2m , Patm = 10.3m → Pgauge,S = 1.2 − 10.3 = −9.1m

Applying Bernoulli’s equation between the points 1 and S.

Datum at (1):



Hydraulics

P1ρg

+V1

2

2g+ z1 =

PS

ρg+

VS2

2g+ zS + ∑hL(1→S)

0 + 0 + 0 = −9.1 +VS

2

2g+ 5.4 + ∑hL(1→S) → Eq. (1)

∑hL(1→S) = hf(1→S) = 0.025 (X

0.3)(

V2

19.62) But V =? ?

Applying Bernoulli’s equation between the points 1 and 2.

Datum at (2):

P1ρg

+V1

2

2g+ z1 =

P2

ρg+

V22

2g+ z2 + ∑hL(1→2)

0 + 0 + 7.5 = 0 + 0 + 0 + ∑hL(1→2) → ∑hL(1→2) = 7.5m

∑hL(1→2) = 7.5m = hf(1→2) = 0.025 (750

0.3)(

V2

19.62) → V = 1.534m

hf(1→S) = 0.025 (X

0.3)(

1.5342

19.62) = 0.01X (Substitute in Eq. (1))

0 + 0 + 0 = −9.1 +VS

2

2g+ 5.4 + 0.01X (VS = V = 1.534)

0 + 0 + 0 = −9.1 +1.5342

2g+ 5.4 + 0.01X → X = 358m✓.



Hydraulics

7. For the three-reservoir system shown. If the flow in pipe 1(from reservoir A

to junction J)is 0.4m3/s and the friction factor for all pipes is 0.02, calculate:

The flow in the other pipes (pipe 2, 3 and 4).

The elevation of reservoir B (ZB).

Neglect minor losses.

Solution

If we put a piezometer at point J, the will rise at height of Zp, which can be

calculated by applying Bernoulli’s equation between A and J:

PA

ρg+

VA2

2g+ zA =

PJ

ρg+

VJ2

2g+ z + ∑hL(A→J)

0 + 0 + 100 = ZP + ∑hL(A→J)

∑hL(A→J) =8fLQ2

π2gD5=

8 × 0.02 × 400 × 0.42

π2g × 0.45= 10.32m

→ 100 − 10.32 = ZP = 89.68 m

Note that ZP = 89.68 > ZC = 80, So the flow direction is from J to C, and

to calculate this flow we apply Bernoulli’s equation between J and C:

ZP − ZC = ∑hL(J→C)

89.68 − 80 =8 × 0.02 × 300 × Q4

2

π2g × 0.35→ Q4 = 0.2177 m3/s✓.

Now, by applying continuity equation at Junction J:

∑Q@J = 0.0 → 0.4 = 0.2177 + (Q2 + Q3) → Q2 + Q3 = 0.1823 m3/s



Hydraulics

Since pipe 2 and 3 are in parallel, the head loss in these two pipes is the

same:

hL,2 = hL,3 →8fL2Q2

2

π2gD25 =

8fL3Q32

π2gD35

→8 × 0.02 × 200 × Q2

2

π2g × 0.25=

8 × 0.02 × 180 × Q32

π2g × 0.155→ Q3 = 0.513Q2

But, Q2 + Q3 = 0.1823 → Q2 + 0.513Q2 = 0.1823 → Q2 = 0.12 m3/s✓.

Q3 = 0.513 × 0.12 = 0.0618 m3/s✓.

Now we want to calculate the elevation ZB:

Apply Bernoulli’s equation between J and B:

ZP − ZB = ∑hL(J→B) (take pipe 2)

89.68 − ZB =8 × 0.02 × 200 × 0.122

π2g × 0.25→ ZB = 74.8 m✓.

8.

In the shown figure, determine the total length of pipe 3 (L3) and the head

delivered by the pump (hp). Neglect minor losses and take f = 0.032 for all

pipes.



Hydraulics

Solution

hL,1 =8fL1Q1

2

π2gD15 → 40 =

8 × 0.032 × 950 × Q12

π2 × 9.81 × 0.455→ Q1 = 0.542 m3/s

hL,2 =8fL2Q2

2

π2gD25 → 40 =

8 × 0.032 × 450 × Q22

π2 × 9.81 × 0.455→ Q2 = 0.352 m3/s

∑Q@J = 0.0 → Q3 = 0.542 − 0.352 = 0.19 m3/s

→ (direction shown in the figure)

To find the total head at point J (ZP): apply Bernoulli equation between J and

A:

ZP = 30 + ∑hL(J→A) → ZP = 30 + 8 = 38m

To find the length of pipe 3 (L3): apply Bernoulli between J and B:

ZP = 25.5 + ∑hL(J→B) → 38 − 25.5 =8 × 0.032 × L3 × 0.192

π2 × 9.81 × 0.35

→ L3 = 318.22 m✓.

To find the pump head (hP): apply Bernoulli between C and J:

18 = ZP + hL,1 − hP → hP = 38 + 40 − 18 = 60m ✓.



Hydraulics

9. The network ABCD is supplied by water from reservoir E as shown in the

figure. All pipes have a friction factor f = 0.02.

Calculate:

1. The flow rate in each pipe of the system using the Hardy Cross method.

Consider the following:

Assume for the first iteration that:

QCB = 0.15 m3/s from C to B and QBA = 0.05 m

3/s from B to A.

Do only one iteration (stop after you correct Q)

2. The pressure head at node A.

Given Also:

Nodes

elevation

Node/Point A B C D Reservoir E

Elevation(m) 20 25 20 30 50

Pipe AB BC CD DA BD CE

Length (m) 400 400 400 400 600 200

Diameter (m) 0.30 0.30 0.30 0.30 0.30 0.40

Pipes

dimension



Hydraulics

Solution

Firstly, from the given flows, we calculate the initial flow in each pipe and

the direction of flow through these pipes using continuity equation at each

note.

We calculate the corrected flow in each pipe, from the following tables:

Loop Pipe L D Qinitial hf

hf

Qinitial

∆ Qnew

I

AB 400 0.3 -0.05 -0.68 13.6 -0.025 -0.075

BD 600 0.3 +0.05 +1.02 20.4

DA 400 0.3 +0.1 +2.72 27.2 -0.025 +0.075

∑ +3.06 61.2

hf calculated for each pipe from the following relation:8fLQ2

π2gD5

∆=−∑hf

2∑hf

Qinitial

=−3.06

2 × 61.2= −0.025 m3/s

Qnew = Qinitial + ∆

Note that, we don’t correct pipe BD because is associated with the two

loops, so we calculate the flow in the associated pipe after calculating the

correction in each loop.

Always we assume

the direction of the

loop in clock wise

direction



Hydraulics

∆=−∑hf

2∑hf

Qinitial

=−(−1.02)

2 × 102= +0.005 m3/s

Now, to calculate Qnew for pipe BD, we calculate the associated value for

correction:

Qnew,BD = Qinitial,loop(1) + (∆loop(1) − ∆loop(2))

Qnew,BD = +0.05 + (−0.025 − 0.005) = +0.02m3/s

Or:


Qnew,BD = −0.05 + (0.005 − (−0.025)) = −0.02 m3/s

Now, we put the corrected flow rate on each pipe of the network:

✓✓✓✓✓✓✓✓

Loop Pipe L D Qinitial hf

hf

Qinitial

∆ Qnew

II

BC 400 0.3 -0.15 -6.12 40.8 +0.005 -0.145

BD 600 0.3 -0.05 -1.02 20.4

CD 400 0.3 +0.15 +6.12 40.8 +0.005 +0.155

∑ -1.02 102

Note: If the sign of the

corrected flow rate is the same

sign of initial flow rate, the

direction of flow will remain

unchanged, however if the

sign is changed, the flow

direction must reversed.



Hydraulics

To calculate the pressure head at point A, we must starts from point which

have a known head, so we start from the reservoir.

Important Note:

The total head at any node in the network is calculated as following:

htotal = hpressure + helevation → ht =P

γ+ Z

By considering a piezometer at each node, such that the water rise on it

distance: P

γ+ Z and the velocity is zero.

Starts from reservoir at E:

Apply Bernoulli between E and C:

50 + 0 + 0 =PC

γ+ 0 + 20 +

8fLQ2

π2gD5

→PC

γ= 50 − 20 −

8 × 0.02 × 200 × 0.32

π2 × 9.81 × 0.45= 27.1m

Apply Bernoulli between C and B:

20 + 27.1 + 0 =PB

γ+ 0 + 25 +

8fLQ2

π2gD5

→PB

γ= 20 + 27.1 − 25 −

8 × 0.02 × 400 × 0.1452

π2 × 9.81 × 0.35= 16.38m

Apply Bernoulli between B and A:

25 + 16.38 + 0 =PA

γ+ 0 + 20 +

8fLQ2

π2gD5

→PA

γ= 25 + 16.38 − 20 −

8 × 0.02 × 400 × 0.0752

π2 × 9.81 × 0.35= 19.84m ✓.



Hydraulics

10.

For all pipe segments, the initial estimate of the flow in each pipe (m3/s) is

shown in the figure below.

a) Show on the sketch all sources and sinks (in and out flows) of water from

the pipe network. Use an arrow to indicate direction (in or out of system)

and give the magnitude of the flow.

b) Use Hardy-Cross method to correct the flow rate through each pipe. Do

only one iteration (stop after you correct Q), assume = 0.02 for all pipes

and neglect minor losses.

c) If the pressure head at A = 90m, determine the pressure head at E.

Solution



Hydraulics

a) By applying continuity equation at each junction, the in and out flow from

the system are shown in the following figure:

✓✓✓✓✓✓✓✓

b) We calculate the corrected flow in each pipe, from the following tables:

Loop Pipe L D Qinitial hf hf

Qinitial

∆ Qnew

I

AB 250 0.5 +0.5 +3.305 6.61 +0.1028 +0.6028

BD 180 0.2 -0.2 -37.18 185.9

DC 200 0.6 -0.8 -2.72 3.4 +0.1028 -0.6972

CA 350 0.6 -0.8 -4.76 5.95 +0.1028 -0.6972

∑ -41.35 201

∆=−∑hf

2∑hf

Qinitial

=−(−41.35)

2 × 201= +0.1028 m3/s



Hydraulics

∆=−∑hf

2∑hf

Qinitial

=−(+110)

2 × 623.525= −0.088 m3/s


Qnew,BD = −0.2 + (0.1028 − (−0.088)) = −0.0092 m3/s

Or:


Qnew,BD = +0.2 + (−0.088 − 0.1028) = +0.0092 m3/s.

The corrected flows are shown in the following figure:

✓✓✓✓✓✓✓✓

Loop Pipe L D Qinitial hf hf

Qinitial ∆ Qnew

II

BE 400 0.2 +0.2 +82.62 413.1 -0.088 +0.112

BD 180 0.2 +0.2 +37.18 186

DE 380 0.4 -0.4 -9.81 24.52 -0.088 -0.488

∑ +110 623.525



Hydraulics

c)

@ point A: P

γ= 90m and Z = 20m → ht,A = 90 + 20 = 110m

Apply Bernoulli between A and B:

110 =PB

γ+ 25 +

8fLQ2

π2gD5

→PB

γ= 110 − 25 −

8 × 0.02 × 250 × 0.6032

π2 × 9.81 × 0.55= 80.2 m

Apply Bernoulli between B and E:

80.2 + 25 =PE

γ+ 30 +

8fLQ2

π2gD5

→PE

γ= 105.2 − 30 −

8 × 0.02 × 400 × 0.1122

π2 × 9.81 × 0.25= 49.29m ✓.

Chapter (5)

Pumps


Pumps

Hydraulics

1.

The figure below shows a portion of a pump and pipe system. The 30-m long pipeline

connecting the reservoir to the pump is 1m in diameter with friction factor f = 0.02. For

the pump, the required net positive suction head (NPSHR ) is 2 m.

If the flow velocity V is 2.5 m/s check the system for cavitation. Take the atmospheric

pressure = 100 kPa and the vapor pressure = 2.4 kPa.

Solution

(NPSH)A = ±hs − hfs − ∑hms +Patm

γ−

Pvapor

γ

hs = −(10 − 5) = −5m

(−ve sign because the supply tank is below the pump).

hfs = f (L

D)(

V2

2g) = 0.02 × (

30

1)(

2.52

2 × 9.81) = 0.191 m

∑hms = (0.5 + 2 × 2 + 3) ×V2

2g= 7.5 ×

2.52

2 × 9.81= 2.39 m

→ (NPSH)A = −5 − 0.191 − 2.39 +100

9.81−

2.4

9.81= 2.36 m

(NPSH)A = 2.36 > (NPSH)R = 2 → System is adequate for cavitation✓.


Pumps

Hydraulics

2.

A pump is required to supply water to an elevated tank through a 0.2m

diameter pipe which is 300m long and has a friction factor f equal to 0.025.

Minor losses causes an additional head loss of 4V2/2g where V is the

velocity in the pipe in m/s. The static head between the pump wet well and

the elevated tank is 40 m. The relationship between head, H, and flow, Q,

for the pump is given by the following equation:

H= 50 - 600Q2

Where Q in m3/s and H in meters.

a) under these conditions determine the flow in the pipeline

b) if the maximum efficiency of the pump is achieved when the flow is 70

ℓ/s then how could the system be redesigned so that the pump would operate

at this efficiency.

Solution

Pump Curve Equation is: H= 50 - 600Q2 and since Q in m

3/s we calculate

the head at the following values of Q as shown in the following table:

Q (m3/s) 0 0.05 0.1 0.15 0.2 0.25

H (m) 50 48.5 44 36.5 26 12.5

Now, we want to find the system head equation:

To find the relation between the required head and flow for the system, the

best way is to apply Bernoulli equation on the given system and then

find the relation between pump head and flow rate (general case), or by

applying the following equation (special case):

H = HStatic + ∑hL

When the pump is used to transfer water from one reservoir to the other.

H = HStatic + ∑hL

HStatic = 40 m (given)

∑hL = hf + hm


Pumps

Hydraulics

∑hL =8fLQ2

π2gD5+ hm

hm = 4v2

2g= 4 ×

Q2

2gA2 , but A =

π

4× D2 =

π

4× 0.22 = 0.0314

→ hm =206. 7Q2

→ ∑hL =8 × 0.025 × 300Q2

π2 × 9.81 × 0.25+ 206.7Q2 = 2143.26Q2

So, the following equation is for system curve:

H = 40 + 2143.26Q2

Now, we calculate the head at the same values of Q (in pump curve) as

shown in the following table:

Q (m3/s) 0 0.05 0.1 0.15 0.2 0.25

H (m) 40.0 45.4 61.4 88.2 125.7 174.0

Now, we draw the pump curve and the system curve, and the point of

intersection between the two curves is the operating point:

As shown in figure above, the pump flow rate is 0.06 m3/s (60 ℓ /s) and the

pump head is 49m.


Pumps

Hydraulics

The required B, means if we get maximum efficiency if the operating point

gives flow rate of (70 ℓ/s) = 0.07 m3/s, what should you modify in the

system to meet this efficiency??

From the above graph, the corresponding head at Q = 0.07 is 47.5m.

Now, we know the system curve equation:

H = 40 + ∑hL

At maximum efficiency, H = 47.5 m and Q = 0.07 →→

47.5 = 40 + ∑hL → ∑hL = 7.5m

Now the only, factor that we can change is the diameter of the pipe, or

changing material of the pipe (change f value), but assume the same

material, so change the diameter of the pipe as following:

∑hL = 7.5 =8 × 0.025 × 300 × 0.072

π2 × 9.81 × D5+ 4 ×

0.072

2 × 9.81 × (π4

× D2)2

→→ D = 0.214m (to get maximum efficiency)✓.

3.

For the shown figure below, (a) what is the operation point for the system

that shown in the figure below, the pump characteristics curve is given

below (Assume f = 0.014), (b) If the efficiency of the system is 80%, what is

the power required?, (c) What is the operation point in case of two identical

pumps in parallel?

Pipe is 20cm in diameter


Pumps

Hydraulics

Solution

a)

The pump curve is given as shown above, now we want to find the system

curve. So firstly we find the system curve equation:

By applying Bernoulli equation between the two reservoir, the equation will

be:

H = HStatic + ∑hL

HStatic = 3.5 + 15 − 10 = 8.5 m

∑hL = hf + hm

∑hL =8fLQ2

π2gD5+ hm

hm = (0.4 + 0.9 + 1)v2

2g= 2.3 ×

Q2

2gA2 ,

but A =π

4× D2 =

π

4× 0.22 = 0.0314 → hm =118.9Q2

→ ∑hL =8 × 0.014 × (200+200 + 15)Q2

π2 × 9.81 × 0.25+ 118.9Q2 = 1619Q2

So, the following equation is for system curve:

H = 8.5 + 1619Q2


Pumps

Hydraulics

But, in this equation Q is in m

3/s and the horizontal axis in the given graph is

in m3/min., so each value on the graph must be divided by 60 (to transform

to m3/s) and then we draw the system curve on the same graph as following:

The value of pump flow (at the operation point) is 3 m3/min = 0.05 m

3/s✓.

The value of pump head (at the operation point) is 12.55 m ✓.

b)

η =γ × Q × H

Pin→ Pin =

9810 × 0.05 × 12.55

0.8= 7694.7 watt✓.

c)

When we use two pumps in parallel, the system curve will never change, but

the pump curve will change (The values of head remains constant but the

values of Q multiplied by 2), so we draw the new pump curve by

multiplying each value of Q by 2 at each head as following:


Pumps

Hydraulics

So the new value of Q is 4.1 m3/min = 0.0683 m

3/s and the new value of H is

16m ✓.

Chapter (6)

Open Channels


Open Channels

Hydraulics

Formulas used to describe the flow in open channels The most common formula is Manning equation:

V =1

n× Rh

2/3× S0.5

n = manning coefficient

S = The slope of channel bed.

Rh = Hydraulic Radius =Wetted Area

Wetted Perimeter=

A

P

Most Economical Section of Channels The most efficient section is satisfied when the flow in the channel is

maximum and thereby the minimum wetted perimeter.

I.e. for most economical section:

dP

dy= 0.0 𝐎𝐑

dP

dB= 0.0 such that,

y is the depth of water in channel also known by (normal depth)

and B is the width of the channel.

Energy Principals in Open Channels

Especific = y +v2

2g→ Es = y +

Q2

2A2g (for any cross section)

Special Case for rectangular channel:

Since the width of the channel B is constant, we can calculate the discharge

per unit width; q =Q

B (m2/s)

→→ Es = y +q2

2y2g (For rectangular channel)

Sub-critical, critical and supercritical flow:

FR(Fround Number) =V

√gDh

Dh = Hydraulic depth of channel =Area of flow

Water surface depth (T)


Open Channels

Hydraulics

FR < 1 → Sub − critical

FR = 1 → Critical

FR > 1 → Sper critical

Critical Depth (yc):

Is the depth of flow of liquid at which at which the specific energy is

minimum (Emin) and the flow corresponds to this point is called critical

flow (FR = 1).

So, we can classify the flow also as following:

If y > yc → Sub − critical flow(FR < 1)

If y = yc → Critical flow(FR = 1)

If y < yc → Super critical flow(FR > 1)

For rectangular channel:

FR =V

√g y

yc3 =

q2

g→ yc = (

q2

g)

1/3

The critical velocity is:

Vc = √g × yc

Emin = Ec = 1.5 yc

Non-Rectangular Shapes (All Shapes):

Q2

g=

A3

T (To calculate yc) ,

vc2

2g=

Dh

2

Emin = Ec = yc + 0.5 (A

T)


Open Channels

Hydraulics

Hydraulic Jump

y1 = upstrem depth and y2 = downstream depth

y1 and y2 are called 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 𝐝𝐞𝐩𝐭𝐡𝐬.

𝐲𝟏 < 𝐲𝟐


y2

y1= 0.5 [−1 + √1 + 8(

yc

y1)3

] (𝐓𝐨 𝐜𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐞 𝐲𝟐)

y1

y2= 0.5 [−1 + √1 + 8(

yc

y2)3

] (𝐓𝐨 𝐜𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐞 𝐲𝟏)

Or in terms of Fround number:

y2

y1= 0.5 [−1 + √1 + 8 F1

2] → F1 =V1

√gy1

(𝐓𝐨 𝐜𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐞 𝐲𝟐)

y1

y2= 0.5 [−1 + √1 + 8 F2

2] → F2 =V2

√gy2

(𝐓𝐨 𝐜𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐞 𝐲𝟏)

𝐇𝐞𝐚𝐝 𝐥𝐨𝐬𝐬 (𝐇𝐋) = ∆E =(y2 − y1)

3

4y1y2

Height of Hydraulic jump (𝐡𝐉) = y2 − y1

Length of Hydraulic jump (𝐋𝐉) = 6hJ


Open Channels

Hydraulics

1.

A trapezoidal channel is to be design to carry a discharge of 75m3/s at

maximum hydraulic efficiency. The side slopes of the channel are 1V:2H

(1 vertical and 2 horizontal) and the Manning’s roughness n is 0.03.

• If the maximum allowable velocity in the channel is 1.75m/s m what

should be the dimensions of the channel (bottom width and height)

• What should be the longitudinal slope of the channel if the flow is

uniform?

Solution

a)

We draw the following graph (from the given data in the problem):

A =Q

V=

75

1.75= 42.86 m2

At maximum hydraulic efficiency (most economical section) →→

dP

dy= 0.0 Or

dP

dB= 0.0

P = B + 2 × √5 y → P = B + 4.47y →→ Eq. (1)

A = 42.86 = By + 2 ×1

2× 2y × y → 42.86 = By + 2y2

→ B =42.86

y− 2y (Substitute in Eq. (1)) →→→

P =42.86

y− 2y + 4.47y → P =

42.86

y+ 2.47y

dP

dy= −

42.86

y2+ 2.47 = 0.0 (for most economical)

√y2 + (2y)2 = √5 y


Open Channels

Hydraulics

→42.86

y2= 2.47 → y = 4.16 m✓.

→ B =42.86

4.16− 2 × 4.16 = 1.98 m✓.

b)

According Manning equation:

V =1

n× Rh

2/3× S0.5

V = 1.75 m/s ,

n = 0.03 (given)

Rh =A

P=

42.86

1.98 + 4.47 × 4.16= 2.083 m

→ 1.75 =1

0.03× 2.0832/3 × S0.5 → S = 0.00104 m/m✓.

2.

In the figure shown, water flows uniformly at a steady rate of 0.4 m3/s in a very long

triangular flume (open channel) that has side slopes 1:1. The flume is on a slope of 0.006,

and n = 0.012;

(a) Find normal depth (yn).

(b) Determine wether the flow is subcritical or supercritical.

(c) Find the specific energy (Es) and the critical specific energy (Ec).

√y2 + y2 = √2 y


Open Channels

Hydraulics

Solution

a)

To find the normal depth, we use Manning equation:

V =1

n× Rh

2/3× S0.5 → Q = AV =

A

n× Rh

2/3× S0.5

Q = 0.4 m3/s , n = 0.012 , S = 0.006 (Givens)

A = area of triangle = 1

2× (2y) × y = y2

Rh =A

P=

y2

2 × √2 y= 0.353y

now, substitute by all above data in Manning’s equation:

0.4 =y2

0.012× (0.353y)2/3 × 0.0060.5 → y = 0.457 m ✓.

b)

To determine the type of flow, there are two methods:

FR =V

√gDh

V =Q

A=

0.4

0.4572= 1.915 m/s

Dh =A

T=

y2

2y=

0.4572

2 × 0.457= 0.2285 m

→ FR =1.915

√9.81 × 0.2285= 1.28 > 1 → 𝐒𝐮𝐩𝐞𝐫𝐜𝐫𝐢𝐭𝐢𝐜𝐚𝐥 ✓.

Or, we calculate the critical depth:

Q2

g=

A3

T→

0.42

9.81=

(yc2)3

2yc→ yc = 0.504

Since yc = 0.504 > y = 0.457 → 𝐒𝐮𝐩𝐞𝐫𝐜𝐫𝐢𝐭𝐢𝐜𝐚𝐥 ✓.

c)

ES = y +v2

2g= 0.457 +

1.9152

2 × 9.81= 0.644 m✓.

Ec = y𝑐 +1

2(A

T) = 0.504 +

1

2(

0.5042

2 × 0.504) = 0.63 m✓.

Ec = Eminmust be less than ES✓.


Open Channels

Hydraulics

3.

You are asked to design a rectangular channel that has the minimum wetted

perimeter and that conveys flow in critical condition. Find the relationship

between the critical depth and the channel width if the flow discharge is

constant. Your answer should look something like yc = mB , where m is

constant and B is the channel width.

Solution

P = B + 2yc →→ Eq. (1)

We want other relation, between B and yc


yc3 =

q2

g, (q =

Q

B) → yc

3 =Q2

B2g

→ B2 =Q2

yc3g

→ B =Q

yc3/2√g

→ B =Q

√g yc

−3/2 (Substitute in Eq. (1)) →

P =Q

√g yc

−3/2 + 2yc

dP

dyc= 0.0 (for most economical)

dP

dyc= −1.5

Q

√g yc

−52 + 2 = 0.0 → 2 = 1.5

Q

√g yc

−52

But, Q = AVc = Byc × √gyc = B√g yc

3

2

→ 2 = 1.5B√g yc

32

√g yc

−52 → yc

52 = 1.5

B yc

32

2→ yc × yc

32 = 1.5

B yc

32

2

→ yc = 0.75 B✓.

We can solve this problem alternatively as following:

P = B + 2yc →dP

dyc=

dB

dyc+ 2 →

dB

dyc= −2 →→ Eq. (1)

Now, we want to find other relation for dB

dyc

Remember: Critical velocity in

rectangular open channel is: √gyc


Open Channels

Hydraulics

yc3 =

q2

g=

Q2

B2g→ B2yc

3 =Q2

g (Now,Derive according to yc)

(B2)(3yc2) + (2B ×

dB

dyc) × (yc

3) = 0.0

→ 3B2yc2 + 2Byc

3dB

dyc= 0.0 (Substitute from Eq. (1)) (

dB

dyc= −2) →

→ 3B2yc2 + 2Byc

3 × −2 = 0.0

→ 3B2yc2 = 4Byc

3 → yc = 0.75B✓.

4.

A rectangular channel carrying a supercritical stream is to be provided with

a hydraulic jump type of energy dissipater. It is required to have an energy

loss of 5m in the jump when the inlet Fround number is 8.5.(F1 = 8.5)

Determine the conjugate depths. (y1, y2).

Solution

y2

y1= 0.5 [−1 + √1 + 8 F1

2] →y2

y1= 0.5 [−1 + √1 + 8 × 8.52]

→ y2 = 11.53y1 →→ Eq. (1)

∆E = 5 =(y2 − y1)

3

4y1y2 (Substitute from Eq. (1)) →→

5 =(11.53y1 − y1)

3

4y1 × 11.53y1→ y1 = 0.1975 m✓.

→ y2 = 11.53 × 0.1975 = 2.27 m✓.

تذكر: مشتقة حاصل ضرب اقترانين:

األول * مشتقة الثاني + مشتقة األول * الثاني

Water Flow in Pipes - Islamic University of...

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