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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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
→ 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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Water Flow in Pipes Hydraulics
Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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.
Page (18) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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
Page (19) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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.✓.
Page (20) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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
Page (21) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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
Page (22) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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✓.
Page (23) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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 = Patm + Pgauge
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 ✓.
Page (24) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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):
Page (25) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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 = Patm + Pgauge
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):
Page (26) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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✓.
Page (27) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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
Page (28) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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.
Page (29) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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 ✓.
Page (30) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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
Page (31) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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
Page (32) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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.
Page (33) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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 ✓.
Page (34) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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
Page (35) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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
Page (36) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
∆=−∑hf
2∑hf
Qinitial
=−(+110)
2 × 623.525= −0.088 m3/s
Qnew,BD = Qinitial,loop(1) + (∆loop(1) − ∆loop(2))
Qnew,BD = −0.2 + (0.1028 − (−0.088)) = −0.0092 m3/s
Or:
Qnew,BD = Qinitial,loop(2) + (∆loop(2) − ∆loop(1))
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
Page (37) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pipelines and Pipe Networks
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
Page (39) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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✓.
Page (40) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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
Page (41) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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.
Page (42) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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
Page (43) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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
Page (44) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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:
Page (45) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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
Page (47) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Page (48) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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)
Page (49) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Open Channels
Hydraulics
Hydraulic Jump
y1 = upstrem depth and y2 = downstream depth
y1 and y2 are called 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 𝐝𝐞𝐩𝐭𝐡𝐬.
𝐲𝟏 < 𝐲𝟐
For rectangular channel:
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
Page (50) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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
Page (51) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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
Page (52) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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✓.
Page (53) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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
For rectangular channel:
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
Page (54) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
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✓.
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