REE 307 Fluid Dynamics II - drahmednagib.comdrahmednagib.com/fluid_mechanics_2_2017/REE-307-Lec.2.pdf · Collection ---- Waste water system 3. ... from Moody Chart. Basic Equations

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

6

• 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

8

Supply at several points

Branched Pipe System

9

• Three Tank Problem

Branched Pipe System

10

• Three Tank Problem

Branched Pipe System

11

• Three Tank Problem

Branched Pipe System

12

• Three Tank Problem

Solution of 3 Tanks Problem

13

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

14

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

15

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

16

• 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

19

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

20

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

21

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

22

Solution

23

• 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

24

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

25

1. Branched network

Properties

1. Lower reliability

2. High down time

3. Less expensive

4. Used in rural area

Types of Network

26

2. Looped network

Properties

1. Higher reliability

2. lower down time

3. More expensive

4. Used in urban area

Looped Networks

27

Looped Networks

28

• 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

29

1.Pipes

2.Pipes fitting

3.Valves

4.Pumps / Compressors

5.Reservoir and tanks

Network Analysis – Hardy Cross Method

30

• 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

31

• 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

34

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