Download - REE 307 Fluid Dynamics II - Ahmed Nagib
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Sep 27, 2017Dr./ Ahmed Nagib Elmekawy
REE 307
Fluid Dynamics II
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Branched Pipe System
2
• Pipe in Series
• Pipe in Parallel
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Branched Pipe System
3
• Pipe in Series
• Pipe in Parallel
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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
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Branched Pipe System
5
• Pipe in Series
𝑄1 = 𝑄2 = 𝑄3
ℎ𝑙𝐴−𝐵 = ℎ𝑙1 + ℎ𝑙2 + ℎ𝑙3
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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
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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
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Branched Pipe System
8
Supply at several points
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Branched Pipe System
9
• Three Tank Problem
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Branched Pipe System
10
• Three Tank Problem
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Branched Pipe System
11
• Three Tank Problem
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Branched Pipe System
12
• Three Tank Problem
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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
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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
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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
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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.
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Solution of Branched Pipe
Systems
17
• Note
In case of rectangular duct or not circular cross section.
𝑑 =4𝐴
𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑒𝑡𝑒𝑟 (𝑝)
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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
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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
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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
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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
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Solution
22
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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.
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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
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Types of Network
25
1. Branched network
Properties
1. Lower reliability
2. High down time
3. Less expensive
4. Used in rural area
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Types of Network
26
2. Looped network
Properties
1. Higher reliability
2. lower down time
3. More expensive
4. Used in urban area
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Looped Networks
27
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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.
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Network Components
29
1.Pipes
2.Pipes fitting
3.Valves
4.Pumps / Compressors
5.Reservoir and tanks
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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
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Network Analysis – Hardy Cross Method
31
• It is not acceptable that the outlet of consumption
is a junction
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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
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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
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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
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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
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Hardy Cross Method
36
Note:
Since ℎ𝑙𝑜𝑠𝑠 won't reach 0.0
Then Hardy Cross Method is used till
σ |∆𝑄| <Small value (Tolerance)
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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
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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
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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
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Hardy CrossMethod of Balancing Heads
40
∴ ∆𝑄 =−σ𝑟𝑄0|𝑄0|
𝑛−1
σ𝑟 𝑛|𝑄0|𝑛−1
𝛿𝑄𝑖 =−σ𝑖=1
𝑛 ∆𝐻𝑖
2σ𝑖=1𝑛 |
∆𝐻𝑖𝑄𝑖
|
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Dealing with network pressure heads
41
1. Using valves or fittings
• Valves and fittings are source of energy loss
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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
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Example
43
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Example
44
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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
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Why do we solve network problems ?
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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