Sep 27, 2017Dr./ Ahmed Nagib Elmekawy

REE 307

Fluid Dynamics II

Branched Pipe System

2

• Pipe in Series

• Pipe in Parallel

Branched Pipe System

3

• Pipe in Series

• Pipe in Parallel

Branched Pipe System

4

• Electrical circuits

𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 × 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒

• Fluid Flow

∆𝑝 =𝑓𝑙𝑉2

2𝑔𝑑=0.8 𝑓𝑙𝑄2

𝑔𝑑5= 𝑅𝑄2

𝑅: 𝑝𝑖𝑝𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =0.8 𝑓𝑙

𝑔𝑑5

Branched Pipe System

5

• Pipe in Series

𝑄1 = 𝑄2 = 𝑄3

ℎ𝑙𝐴−𝐵 = ℎ𝑙1 + ℎ𝑙2 + ℎ𝑙3

Branched Pipe System

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• Pipe in Parallel

𝑄𝑡𝑜𝑡𝑎𝑙 = 𝑄1 + 𝑄2 + 𝑄3

ℎ𝑙𝐴−𝐵 = ℎ𝑙1 = ℎ𝑙2 = ℎ𝑙3

𝑓1𝑙1𝑉12

2𝑔𝑑1=𝑓2𝑙2𝑉2

2

2𝑔𝑑2

𝑉1𝑉2

=𝑓2𝑙2𝑑1𝑓1𝑙1𝑑2

0.5

Branched Pipe System

7

• Loop

𝑄1 = 𝑄2 + 𝑄3𝑃𝐴𝛾+𝑉𝐴2

2𝑔+ 𝑧𝐴 =

𝑃𝐵𝛾+𝑉𝐵2

2𝑔+ 𝑧𝐵 + ℎ𝑙1 + ℎ𝑙2

𝑃𝐴𝛾+𝑉𝐴2

2𝑔+ 𝑧𝐴 =

𝑃𝐵𝛾+𝑉𝐵2

2𝑔+ 𝑧𝐵 + ℎ𝑙1 + ℎ𝑙3

ℎ𝑙2 = ℎ𝑙3

Branched Pipe System

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Supply at several points

Branched Pipe System

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• Three Tank Problem

Branched Pipe System

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• Three Tank Problem

Branched Pipe System

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• Three Tank Problem

Branched Pipe System

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• Three Tank Problem

Solution of 3 Tanks Problem

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1. Assume EJ less than Za, and greater than Zc

𝐸𝑎 > 𝐸𝐽 > 𝐸𝑐2. Applying Bernoulli's equation between A & J

𝐸𝑎= 𝐸𝐽+ ℎ𝑙

Δ𝐸𝑎−𝐽 = 𝐸𝐽 − 𝑍𝑎 = ℎ𝑙= 0.8 𝑓𝑙𝑄𝑎

2

𝑔𝑑5

∴ 𝑄𝑎 =ℎ𝑙𝑔𝑑

5

0.8 𝑓𝑙

𝑃𝑎𝜔+ 𝑍𝑎 +

𝑉𝑎2

2𝑔= 𝐸𝑗 +

0.8 𝑓𝑙𝑄𝑎2

𝑔𝑑5

Solution of 3 Tanks Problem

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Similarly between B & J : Get 𝑄𝑏between C & J : Get 𝑄𝑐

3. Checking the assumption

If 𝑄𝑎+ 𝑄𝑏+ 𝑄𝑐 = 0.0 Right Assumption

If 𝑄𝑎+ 𝑄𝑏+ 𝑄𝑐 ≠ 0.0 𝑄𝑎> (𝑄𝑏+ 𝑄𝑐)

Wrong Assumption

𝑄𝑎< (𝑄𝑏+ 𝑄𝑐)

Increase the Energy of

junction (EJ), and decrease

the losses

Increase the Energy of

junction (EJ), and decrease

the losses

Solution of 3 Tanks Problem

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4. We repeat the last step, by increasing or decreasing the value

∆𝑄 = 𝑄𝑎+ 𝑄𝑏+ 𝑄𝑐 ≅ 0.0 < 𝑇𝑜𝑙𝑒𝑟𝑎𝑛𝑐𝑒

As it won't equal to zero

Branched Systems

Supply at Several Points

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• Pipe Flows

- For known nodal demands, the rates can be partially determined.

- Flow rates & directions in the pipe routes connecting the sources

depend on the piezometrie heads at the sources and the

distribution of nodal demands.

• Velocities

- Also partially known.

• Pressures

- Conditions are the same as in case of the single source, once the

flows and velocities have been determined.

• Hydraulic calculation

- Single pipe calculation can only partially solve the system.

- Additional condition ìs necessary.

Solution of Branched Pipe

Systems

17

• Note

In case of rectangular duct or not circular cross section.

𝑑 =4𝐴

𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑒𝑡𝑒𝑟 (𝑝)

Example

18

• Given :

• Required :

Qa, Qb, Qc, Flow Direction

L1 = 3000 m D1 = 1 m f1 = 0.014 Za = 30 m

L2 = 6000 m D2 = 0.45 m f2 = 0.024 Zb = 18 m

L3 = 1000 m D3 = 0.6 m f3 = 0.02 Zc = 9 m

Solution

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1. Assume EJ = 25 m

2. Applying Bernoulli's equation between A & J

𝐸𝑎= 𝐸𝐽+ ℎ𝑙

𝑃𝑎𝜔+ 𝑍𝑎 +

𝑉𝑎2

2𝑔= 𝐸𝑗 +

0.8 𝑓𝑙𝑄𝑎2

𝑔𝑑5

Δ𝐸𝑎−𝐽 = 𝐸𝐽 − 𝑍𝑎 = ℎ𝑙= 0.8 𝑓𝑙𝑄𝑎

2

𝑔𝑑5

∴ 𝑄𝑎 =ℎ𝑙𝑔𝑑

5

0.8 𝑓𝑙=

30−25 ×9.8×15

0.8 ×0.014×3000

∴ 𝑄𝑎 = 1.2 m3/s

Solution

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3. Similarly between B & J and C & J

∴ 𝑄𝑏 =25−18 ×9.8×0.455

0.8 ×0.024×6000= 0.105 m3/s

∴ 𝑄𝑐 =25−9 ×9.8×0.65

0.8 ×0.02×1000= 0.873 m3/s

∆𝑄 = 𝑄𝑎+ 𝑄𝑏+ 𝑄𝑐 = 1.2 − 0.105 − 0.873 = 0.222 > 0.0

𝑄𝑎 = 1.2 m3/s 𝑄𝑏 = 0.105 m3/s 𝑄𝑐 = 0.873 m3/s

Solution

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4. Increase energy of junction (EJ = 26.6 m)

∴ 𝑄𝑎 =ℎ𝑙𝑔𝑑

5

0.8 𝑓𝑙=

30−26.6 ×9.8×15

0.8 ×0.014×3000= 0.996 m3/s

∴ 𝑄𝑏 =26.6−18 ×9.8×0.455

0.8 ×0.024×6000= 0.116 m3/s

∴ 𝑄𝑐 =26.6−9 ×9.8×0.65

0.8 ×0.02×1000= 0.916 m3/s

∆𝑄 = 𝑄𝑎+ 𝑄𝑏+ 𝑄𝑐 = 0.996 − 0.116 − 0.916 = −0.036 ≅ 0.0

𝑄𝑎 = 0.996 m3/s 𝑄𝑏 = 0.116 m3/s 𝑄𝑐 = 0.916 m3/s

Solution

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Solution

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• For more accuracy, use more assumptions, but the result can

be accepted, as ∆Q is very small w.r.t the smallest value, which

is Qb

• For more accuracy and saving time, use programming

languages to solve the problem.

Network of pipes

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Function of piping system:

1. Transmission ---- One single pipeline

2. Collection ---- Waste water system

3. Distribution ---- Distribution of drinking water

---- Distribution of natural gas

---- Distribution of cooling water

---- Air conditioning systems

---- Distribution of blood in veins and

arteries

Types of Network

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1. Branched network

Properties

1. Lower reliability

2. High down time

3. Less expensive

4. Used in rural area

Types of Network

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2. Looped network

Properties

1. Higher reliability

2. lower down time

3. More expensive

4. Used in urban area

Looped Networks

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Looped Networks

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• Pipe Flows

➢ Flow rates and directions are unknown.

• Velocities

➢ The velocities and their directions are known only after the

flows have been calculated.

• Pressures

➢ Conditions are the same as in case of branched networks

once the flows and hydraulic losses have been calculated

for each pipe.

• Hydraulic calculation

➢ The equations used for single pipe calculation are not

sufficient.

➢ Additional conditions have to be introduced.

➢ Iterative calculation process is needed.

Network Components

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1.Pipes

2.Pipes fitting

3.Valves

4.Pumps / Compressors

5.Reservoir and tanks

Network Analysis – Hardy Cross Method

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• Pipe/Element/Branch: 12, 23, 36, 16, 34, 65, 45

• Hydraulic node (junction): 1, 2, 3, 4, 5, 6

• Loop: 12361, 63456

Network Analysis – Hardy Cross Method

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• It is not acceptable that the outlet of consumption

is a junction

Basic Equations

32

1. Continuity equation σ𝑄 at any node = 0.0

2. Energy equation σℎ𝑙𝑜𝑠𝑠 around any closed loop = 0.0

For losses ℎ𝑙=0.8 𝑓𝑙𝑄2

𝑔𝑑5

𝑓 = fn(Re, 𝜀

𝑑) from Moody Chart

Basic Equations

33

• In case of using programming, we can't use moody chart

Hazen Williams equation Colebrook equation

ℎ𝑙𝐿=

𝑅𝑄𝑛

𝐷𝑚

𝑛 = 1.852

𝑚 = 4.8704

𝑅 =10.675

𝐶𝑛,

𝐶 = 60 ⟷ 140

𝐶 : is a constant depending on

the pipe age and roughness

(pipe condition)

1

𝑓= −2 log

Τ휀 𝐷

3.7+

2.51

𝑅𝑒 𝑓

ℎ𝑙𝐿= 𝑓

𝑉2

2𝑔𝑑

• Colebrook is a curve fitting

• 1st equation is solved by

trial & error

• Colebrook equation is the

most commonly used in the

industrial field

Rough Smooth

Hardy Cross Method

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1. Assume flow rate at each pipe 𝑄0, so that:

• The velocity range between 1 - 3 m/s

• Satisfying the continuity equation

2. If ℎ𝑙 = 0.0 (stop) Impossible

If ℎ𝑙 ≠ 0.0

2. Adjust 𝑄: 𝑄 = 𝑄0 + ∆𝑄

ℎ𝑙𝑜𝑠𝑠 = 𝑟𝑄𝑛 = 𝑟 𝑄0 + ∆𝑄 𝑛

ℎ𝑙𝑜𝑠𝑠 = 𝑟 𝑄0𝑛 + 𝑛∆𝑄𝑄0

𝑛−1 + … . .

Since σℎ𝑙𝑜𝑠𝑠 = σ𝑟𝑄𝑛 = 0.0 ∴ −σ𝑟 𝑄0|𝑄0|𝑛−1 = σ𝑟 𝑛|𝑄0|

𝑛−1∆𝑄

Hardy Cross Method

35

∴ −𝑟𝑄0|𝑄0|𝑛−1 =𝑟𝑛|𝑄0|

𝑛−1∆𝑄

∴ ∆𝑄 =−𝑟σ𝑄0|𝑄0|

𝑛−1

σ𝑟 𝑛|𝑄0|𝑛−1

ℎ𝑙𝑜𝑠𝑠 =0.8 𝑓𝑙𝑄2

𝑔𝑑5= 𝑟𝑄𝑛

where

n=2

𝑟 =0.8 𝑓𝑙

𝑔𝑑5……pipe resistance

Hardy Cross Method

36

Note:

Since ℎ𝑙𝑜𝑠𝑠 won't reach 0.0

Then Hardy Cross Method is used till

σ |∆𝑄| <Small value (Tolerance)

Network Analysis – Hardy Cross Method

37

• Pipe/Element/Branch: 12, 23, 36, 16, 34, 65, 45

• Hydraulic node (junction): 1, 2, 3, 4, 5, 6

• Loop: 12361, 63456

Kirchhoff's Laws

38

Flow continuity at junction of pipes

The sum of all ingoing and outgoing flows in each node equals

zero (σ𝑄𝑖= 0).

Head loss continuity at loop of pipes

The sum of all head-losses along pipes that compose a complete

loop equals zero (σ∆𝐻𝑖; = 0).

• Hardy Cross Method

o Method of Balancing Heads

o Method of Balancing Flows

• Linear Theory

• Newton Raphson

• Gradient Algorithm

Hardy CrossMethod of Balancing Heads

39

Step 1

Arbitrary flows are assigned to each pipe; (σ𝑄𝑖 = 0).

Step 2

Head-loss in each pipe is calculated.

Step 3

The sum of the head-losses along each loop is checked.

Step 4

lf σ∆𝐻𝑖 differs from the required accuracy, a flow

correction 𝛿𝑄 is introduced in loop ‘i’

Step 5

Correction 𝛿𝑄 is applied in each loop (clockwise or anti-

clockwise). The iteration continues with Step 2

Hardy CrossMethod of Balancing Heads

40

∴ ∆𝑄 =−σ𝑟𝑄0|𝑄0|

𝑛−1

σ𝑟 𝑛|𝑄0|𝑛−1

𝛿𝑄𝑖 =−σ𝑖=1

𝑛 ∆𝐻𝑖

2σ𝑖=1𝑛 |

∆𝐻𝑖𝑄𝑖

|

Dealing with network pressure heads

41

1. Using valves or fittings

• Valves and fittings are source of energy loss

Dealing with reservoir

42

1. Draw a pipe connecting the tanks, and it's resistance = ∞2. This pipe creates a new loop

3. The losses between the tanks = the difference between the

heads of the tanks, and it's sign depends on the direction of

the loop

Example:

ℎ𝑡𝑎𝑛𝑘(𝐴) = 100 𝑚

ℎ𝑡𝑎𝑛𝑘(𝐵) = 70 𝑚

∆ℎ = 100 − 70 = +30 𝑚(+) the flow direction in the

pipe is the same direction

of loop III

Example

43

Example

44

Example

45

L 𝑟𝑄0|𝑄0|𝑛−1 𝑟𝑛|𝑄0|

𝑛−1

3-4

4-1

1-3

5(-30)(30)

6(70)(70)

3(35)(35)

5(2)(30)

6(2)(70)

3(2)(35)

28575 1350

L 𝑟𝑄0|𝑄0|𝑛−1 𝑟𝑛|𝑄0|

𝑛−1

1-2

2-3

3-1

1(15)(15)

2(-35)(35)

3(-35)(35)

1(2)(15)

2(2)(35)

3(2)(35)

-5900 380

∆𝑄1 =−28575

1350= −21.17 𝑚3/𝑠 ∆𝑄2 =

5900

380= 15.53 𝑚3/𝑠

|∆𝑄| = 21.17 + 15.5 = 36.67 > Tolerance

Epanet Software

46

Why do we solve network problems ?

47

1. Replacement and renovation of networks

2. Network Design

Consumption of network elements

Flow rate of network

Select Velocity (1-3 m/s)

Diameter

3. Selecting pumps, valves and fittings