REE 307 Fluid Dynamics II - Ahmed Nagib
Transcript of REE 307 Fluid Dynamics II - Ahmed Nagib
Sep 27, 2017Dr./ Ahmed Nagib Elmekawy
REE 307
Fluid Dynamics II
Branched Pipe System
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β’ Pipe in Series
β’ Pipe in Parallel
Branched Pipe System
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β’ Pipe in Series
β’ Pipe in Parallel
Branched Pipe System
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β’ Electrical circuits
ππππ‘πππ = πΆπ’πππππ‘ Γ π ππ ππ π‘ππππ
β’ Fluid Flow
βπ =πππ2
2ππ=0.8 πππ2
ππ5= π π2
π : ππππ πππ ππ π‘ππππ =0.8 ππ
ππ5
Branched Pipe System
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β’ Pipe in Series
π1 = π2 = π3
βππ΄βπ΅ = βπ1 + βπ2 + βπ3
Branched Pipe System
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β’ Pipe in Parallel
ππ‘ππ‘ππ = π1 + π2 + π3
βππ΄βπ΅ = βπ1 = βπ2 = βπ3
π1π1π12
2ππ1=π2π2π2
2
2ππ2
π1π2
=π2π2π1π1π1π2
0.5
Branched Pipe System
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β’ Loop
π1 = π2 + π3ππ΄πΎ+ππ΄2
2π+ π§π΄ =
ππ΅πΎ+ππ΅2
2π+ π§π΅ + βπ1 + βπ2
ππ΄πΎ+ππ΄2
2π+ π§π΄ =
ππ΅πΎ+ππ΅2
2π+ π§π΅ + βπ1 + βπ3
βπ2 = βπ3
Branched Pipe System
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Supply at several points
Branched Pipe System
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β’ Three Tank Problem
Branched Pipe System
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β’ Three Tank Problem
Branched Pipe System
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β’ Three Tank Problem
Branched Pipe System
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β’ Three Tank Problem
Solution of 3 Tanks Problem
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1. Assume EJ less than Za, and greater than Zc
πΈπ > πΈπ½ > πΈπ2. Applying Bernoulli's equation between A & J
πΈπ= πΈπ½+ βπ
ΞπΈπβπ½ = πΈπ½ β ππ = βπ= 0.8 ππππ
2
ππ5
β΄ ππ =βπππ
5
0.8 ππ
πππ+ ππ +
ππ2
2π= πΈπ +
0.8 ππππ2
ππ5
Solution of 3 Tanks Problem
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Similarly between B & J : Get ππbetween C & J : Get ππ
3. Checking the assumption
If ππ+ ππ+ ππ = 0.0 Right Assumption
If ππ+ ππ+ ππ β 0.0 ππ> (ππ+ ππ)
Wrong Assumption
ππ< (ππ+ ππ)
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
βπ = ππ+ ππ+ ππ β 0.0 < πππππππππ
As it won't equal to zero
Branched Systems
Supply at Several Points
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β’ 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.
β’ Velocities
- Also partially known.
β’ Pressures
- Conditions are the same as in case of the single source, once the
flows and velocities have been determined.
β’ Hydraulic calculation
- Single pipe calculation can only partially solve the system.
- Additional condition Γ¬s necessary.
Solution of Branched Pipe
Systems
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β’ Note
In case of rectangular duct or not circular cross section.
π =4π΄
π€ππ‘π‘ππ ππππππ‘ππ (π)
Example
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β’ Given :
β’ 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
πΈπ= πΈπ½+ βπ
πππ+ ππ +
ππ2
2π= πΈπ +
0.8 ππππ2
ππ5
ΞπΈπβπ½ = πΈπ½ β ππ = βπ= 0.8 ππππ
2
ππ5
β΄ ππ =βπππ
5
0.8 ππ=
30β25 Γ9.8Γ15
0.8 Γ0.014Γ3000
β΄ ππ = 1.2 m3/s
Solution
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3. Similarly between B & J and C & J
β΄ ππ =25β18 Γ9.8Γ0.455
0.8 Γ0.024Γ6000= 0.105 m3/s
β΄ ππ =25β9 Γ9.8Γ0.65
0.8 Γ0.02Γ1000= 0.873 m3/s
βπ = ππ+ ππ+ ππ = 1.2 β 0.105 β 0.873 = 0.222 > 0.0
ππ = 1.2 m3/s ππ = 0.105 m3/s ππ = 0.873 m3/s
Solution
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4. Increase energy of junction (EJ = 26.6 m)
β΄ ππ =βπππ
5
0.8 ππ=
30β26.6 Γ9.8Γ15
0.8 Γ0.014Γ3000= 0.996 m3/s
β΄ ππ =26.6β18 Γ9.8Γ0.455
0.8 Γ0.024Γ6000= 0.116 m3/s
β΄ ππ =26.6β9 Γ9.8Γ0.65
0.8 Γ0.02Γ1000= 0.916 m3/s
βπ = ππ+ ππ+ ππ = 0.996 β 0.116 β 0.916 = β0.036 β 0.0
ππ = 0.996 m3/s ππ = 0.116 m3/s ππ = 0.916 m3/s
Solution
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Solution
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β’ For more accuracy, use more assumptions, but the result can
be accepted, as βQ is very small w.r.t the smallest value, which
is Qb
β’ 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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β’ Pipe Flows
β’ Flow rates and directions are unknown.
β’ Velocities
β’ The velocities and their directions are known only after the
flows have been calculated.
β’ Pressures
β’ Conditions are the same as in case of branched networks
once the flows and hydraulic losses have been calculated
for each pipe.
β’ Hydraulic calculation
β’ The equations used for single pipe calculation are not
sufficient.
β’ Additional conditions have to be introduced.
β’ 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 β Hardy Cross Method
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β’ Pipe/Element/Branch: 12, 23, 36, 16, 34, 65, 45
β’ Hydraulic node (junction): 1, 2, 3, 4, 5, 6
β’ Loop: 12361, 63456
Network Analysis β Hardy Cross Method
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β’ It is not acceptable that the outlet of consumption
is a junction
Basic Equations
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1. Continuity equation Οπ at any node = 0.0
2. Energy equation Οβπππ π around any closed loop = 0.0
For losses βπ=0.8 πππ2
ππ5
π = fn(Re, π
π) from Moody Chart
Basic Equations
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β’ In case of using programming, we can't use moody chart
Hazen Williams equation Colebrook equation
βππΏ=
π ππ
π·π
π = 1.852
π = 4.8704
π =10.675
πΆπ,
πΆ = 60 β· 140
πΆ : is a constant depending on
the pipe age and roughness
(pipe condition)
1
π= β2 log
Ξ€ν π·
3.7+
2.51
π π π
βππΏ= π
π2
2ππ
β’ Colebrook is a curve fitting
β’ 1st equation is solved by
trial & error
β’ 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 π0, so that:
β’ The velocity range between 1 - 3 m/s
β’ Satisfying the continuity equation
2. If βπ = 0.0 (stop) Impossible
If βπ β 0.0
2. Adjust π: π = π0 + βπ
βπππ π = πππ = π π0 + βπ π
βπππ π = π π0π + πβππ0
πβ1 + β¦ . .
Since Οβπππ π = Οπππ = 0.0 β΄ βΟπ π0|π0|πβ1 = Οπ π|π0|
πβ1βπ
Hardy Cross Method
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β΄ βππ0|π0|πβ1 =ππ|π0|
πβ1βπ
β΄ βπ =βπΟπ0|π0|
πβ1
Οπ π|π0|πβ1
βπππ π =0.8 πππ2
ππ5= πππ
where
n=2
π =0.8 ππ
ππ5β¦β¦pipe resistance
Hardy Cross Method
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Note:
Since βπππ π won't reach 0.0
Then Hardy Cross Method is used till
Ο |βπ| <Small value (Tolerance)
Network Analysis β Hardy Cross Method
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β’ Pipe/Element/Branch: 12, 23, 36, 16, 34, 65, 45
β’ Hydraulic node (junction): 1, 2, 3, 4, 5, 6
β’ Loop: 12361, 63456
Kirchhoff's Laws
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Flow continuity at junction of pipes
The sum of all ingoing and outgoing flows in each node equals
zero (Οππ= 0).
Head loss continuity at loop of pipes
The sum of all head-losses along pipes that compose a complete
loop equals zero (Οβπ»π; = 0).
β’ Hardy Cross Method
o Method of Balancing Heads
o Method of Balancing Flows
β’ Linear Theory
β’ Newton Raphson
β’ Gradient Algorithm
Hardy CrossMethod of Balancing Heads
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Step 1
Arbitrary flows are assigned to each pipe; (Οππ = 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 Οβπ»π differs from the required accuracy, a flow
correction πΏπ is introduced in loop βiβ
Step 5
Correction πΏπ is applied in each loop (clockwise or anti-
clockwise). The iteration continues with Step 2
Hardy CrossMethod of Balancing Heads
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β΄ βπ =βΟππ0|π0|
πβ1
Οπ π|π0|πβ1
πΏππ =βΟπ=1
π βπ»π
2Οπ=1π |
βπ»πππ
|
Dealing with network pressure heads
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1. Using valves or fittings
β’ Valves and fittings are source of energy loss
Dealing with reservoir
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1. Draw a pipe connecting the tanks, and it's resistance = β2. 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:
βπ‘πππ(π΄) = 100 π
βπ‘πππ(π΅) = 70 π
ββ = 100 β 70 = +30 π(+) the flow direction in the
pipe is the same direction
of loop III
Example
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Example
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Example
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L ππ0|π0|πβ1 ππ|π0|
πβ1
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 ππ0|π0|πβ1 ππ|π0|
πβ1
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
βπ1 =β28575
1350= β21.17 π3/π βπ2 =
5900
380= 15.53 π3/π
|βπ| = 21.17 + 15.5 = 36.67 > Tolerance
Epanet Software
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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