Water Flow in Pipes -...

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:

P1

ρg+

v12

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).

P1

ρg+

v12

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

P1

ρg+

v12

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 +vB

2

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. 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

Page (12)



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.

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) →

Page (13)



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)) →

66 = 13.0755 (1.068 + V2)2 + 19.906 V22 → 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 (15) 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.

P1

ρg+

V12

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. 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)



Hydraulics

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 ✓.

4. 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



Hydraulics

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

Datum at (1):

P1

ρg+

V12

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):

P1

ρg+

V12

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

5. 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+ ZP + ∑ hL(A→J)

0 + 0 + 100 = 0 + 0 + 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✓.



Hydraulics

6. 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 all

loops 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 = Qinitial,loop(2) + (∆loop(2) − ∆loop(1))

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 ✓.

Water Flow in Pipes -...

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