ree 307 fluid dynamics ii -...
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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
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
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
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
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 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
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