CWR4203 Hydraulics

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CWR4203 Hydraulics
Lecture II (d) Branching Pipe and Pipes in Series
and Parallel Instructor Dr. Ni-Bin Chang
Spring, 2006
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Learning Objectives

• Apply continuity and energy equations to pipe
  systems involving branching pipe, a series of
  pipes and/or pipes in parallel
• Design a small scale system with complexity

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Types of Engineering Problems in a Two-Reservoir
System

• Type I Two Reservoirs, calculate head loss in
  the system, given the pipe configuration.
• Type II Two reservoirs, calculate flow rate,
  given the water levels and losses through pipes.
• Type III Two reservoirs, calculate diameter of
  pipe in between.
• Hybrid Type Two Reservoirs and Open-channel
  System (complex system)

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New York Citys Water Supply System
Two Reservoirs Connected By Aqueducts
http//www.nyc.gov/html/dep/html/wsmaps.html
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Pipeline with Turbine Associated with a
Downstream Reservoir
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Types of Engineering Problems in a
Three-Reservoir System

• Type I Branching pipelines from a
  three-reservoirsystem asking for head loss
  calculations and water levels downstream
• Type II Three reservoirs connected by pipelines
  asking for discharge calculations, given the pipe
  configuration and water level
• Type III Three reservoirs connected by pipelines
  asking for sizing of pipes to meet the discharge
  goal

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Branching Pipes three-reservoirs problem
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Multiple Reservoir Connected at a Junction
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Review of Piezometer

• The energy grade line (EGL) and the hydraulic
  grade line (HGL) are defined as
• EGL shows the height of the total Bernoulli
  constant while HGL is the height to which liquid
  would rise in a piezometric tube attached to the
  pipe.

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Branching Pipes three-reservoirs problem

• These types of problems are most conveniently
  solved by trial and error method.
• Step 1 Not knowing the discharge in each
  pipe, we may first assume a piezometric surface
  elevation, P, at the junction.
• Step 2 This assumed elevation gives the head
  losses hf1, hf2, hf3 for each of the three pipes.
  
• Then, the trial computation gives a set of
  values for discharges Q1, Q2, and Q3. Check on
  the mass balance to see if Q1Q2Q3 would be
  satisfied.

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Branching Pipes three-reservoirs problem

• Step 3 If not, adjust the initial guess of
  the assumed elevation (P) and retry it in an
  iterative procedure.
• Step 4 If the assumed elevation is correct,
  the conservation of mass flow rate would hold at
  the junction eventually.

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Branching Pipes
Final solution
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Branching Pipes three-reservoirs problem
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Solution Procedure
End
Begin
Assign a P value
Calculate head loss for target pipes
Use a fast estimation formula
Calculate flow rate for all pipes
Check mass balance at junction point ?
No
Yes
Check Reynolds Number
Calculate water level at the target reservoir
Calculate flow rate at the target pipe
End
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Derivation of a Fast Formula
Darcy-Weisbach equation
Transformed equation
Equation 8.56 in textbook
Colebrook Equation
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Numerical Example I

• Given the system as above,
• Pipe 1 is 6,000 ft of 15 in diameter,
• Pipe 2 is 1,500 ft of 10 in diameter,
• Pipe 3 is 4,500 ft of 8 in diameter, all
  asphalt-dipped cast iron.
• The elevations of the water surfaces in
  reservoirs A and C are 250 ft and 160 ft,
  respectively,
• The discharge Q2 of 60oF water into reservoir B
  is 3.3 cfs.
• Find the surface elevation of reservoir B.

This is a Scenario 2 problem.
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Numerical Example I
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Numerical Example I

• From Table A.1 for water at 60 oF ?12.17x10-6
  ft2/sec

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Numerical Example I

• Find the elevation of P by trial and error.
• Elevation of P lies between 160 and 250 ft.
  Calculate V or Q to or from two reservoirs A and
  C, and the surface elevation at reservoir B

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Numerical Example I

• Find the elevation of P by trial and error.
• Elevation of P lies between 160 and 250 ft.
  Calculate V
• Interpolation (230-P)/(230-200)
  0.463/(0.4633.04) P 226.03

Q23.3 cfs
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Numerical Example I

• V2Q2/A23.3/0.545 6.055 fps
• R2D2V2/? 416,500
• f2 0.01761 in Moody Diagram
• So, h2 18.05ft
• Elevation of B Elevation P-h2 226.64- 18.05
  208.59 ft

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Numerical Example II

• With the sizes, lengths, and material of pipes
  given in last example, suppose that the surface
  elevations of reservoirs A, B, and C are 525 ft,
  500 ft, and 430 ft, respectively.
• (a) Does the water enter or leave reservoir B?
• (b) Find the flow rates of 60oF water in each
  pipe.

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Numerical Example II
(a) Assume P 500 ft
At J, S Q 5.53- 2.09 3.44 cfs. P must be
moved up.
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Numerical Example II
(a) Assume P 510 ft
At J, S Q 4.24 - 2.42 - 2.24 -0.42 cfs. P
must be moved down
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Numerical Example II
(a) Assume P 508.9 ft
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Numerical Example III

• In the same system above,
• Pipe 1 is 300 mm diameter and 900 m long,
• Pipe 2 is 200 mm diameter and 250 m long,
• Pipe 3 is 150 mm diameter and 700 m long.
• The Hazen-Williams coefficient for all pipes is
  120.
• The surface elevations of reservoirs A, B, and C
  are 160 m, 150 m, and 120 m, respectively.
• (a) Does water enter or leave reservoir B?
• (b) Find the flow rate in each pipe?

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Numerical Example III
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Numerical Example III
(a) Trial I. First, try P at the elevation of
reservoir surface B 150 m
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Numerical Example III

• At J, SQ inflow-outflow 0.1236 0.0414
  0.0822 m3/sec.
• This must be zero, so we raise P.
• Then water will flow into reservoir B.

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Numerical Example III

• Trial 2, Raise P. 150 m lt elevation P lt 160 m, so
  try P at elevation 155 m
• At J, SQ inflow -outflow 0.0850
  0.0584-0.0450 -0.0184 m3/sec.
• This must be zero, so we must lower P. By
  interpolation, P 154.09m

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Numerical Example III

• Trial 3. Try P at elevation 154 m

At J, SQ -0.0023 m3/sec. All Vs are lt 3
m/sec, so these solutions are valid.
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Multiple Reservoir Connected at a Junction
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Multiple Reservoir Connected at a Junction

• From energy conservation

• From mass conservation

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Multiple Reservoir Connected at a Junction

• The head loss may be expressed in the form of the
  Darcy-Weisbach equation

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Multiple Reservoir Connected at a Junction

• Substituting the previous relationship into Hi,
  we have

And
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Multiple Reservoir Connected at a Junction

• The above system of equations can be solved for
  the four unknowns, Q1, Q2, Q3, and Q4
• Other empirical formulas, such as the Manning
  formula and the Hazen-Williams formula, may be
  applied using corresponding expressions.

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Pipe in Series

• If a pipeline is made up of lengths of different
  diameters, conditions must satisfy the continuity
  and energy equations.

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Pipe in Series

• Q Q1 Q2 Q3 .
• hL h1 h2 h3 .
• Using rigorous equations
• Using non-rigorous equations, we have
• hf K1 Q1n K2 Q2n K3 Q3n .

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Numerical Example

• The pipes 1, 2, and 3 are 300 m of 300 mm
  diameter, 150 m of 200 mm diameter, and 250 m of
  250 mm diameter, respectively, of new cast iron
  and are conveying 15oC water.
• If ?Z 10 m, find the rate of flow from A to B.

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Numerical Example

• Table 8.1 in textbook For cast iron pipe
• e 0.25 mm 0.00025 m
• Table A.1 in textbook For water at 15oC ?
  1.139 x 10 -6 m2/sec

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Numerical Example
Energy equation
Continuity Equation
QA1V10.0836m3/sec
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Pipes in Parallel
In the case of flow through two or more parallel
pipes, conditions must satisfy the continuity and
energy equations.
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Pipe in Parallel

• Q Q1 Q2 Q3 .
• hL h1 h2 h3
• Using non-rigorous (empirical ) equations, we
  have

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Numerical Example

• A horizontal, galvanized iron pipe system
  consists of a 20-cm-diameter, 4-m-long main pipe
  between the two joints 1 and 2.
• The branch pipe is 12 cm in diameter and 6.4 m
  long. It consists of two 90o elbows (R/D2.0) and
  a glove valve.
• The system carries a total discharge of 0.26
  m3/sec water at 10oC.
• Determine the discharge in each of the pipes when
  the valves are both fully opened.

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Numerical Example
Adapted from Fundamentals of Hydraulic
Engineering Systems (Hwang and Hita, 1987)
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Numerical Example

• The cross-sectional area of pipes a and b are,
  respectively,

• Continuity condition requires that,

(a)
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Numerical Example

• The head loss between 1 and 2 along the main pipe
  is ha,

The second term in above represents the head
loss for a fully opened gate valve
kc
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Numerical Example

• The head loss between 1 and 2 along the branch
  pipe is hb

kc
Where the second term is for the elbow losses and
the third term is head loss for fully opened
globe valves.
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Numerical Example

• Since the total head losses through both pipes
  must be the same, we have

(b)
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Numerical Example

• (e/D)a0.15/2000.00075
• (e/D)b0.15/1200.00125
• We may obtain fmin from Moody diagram
• fa0.0185 and fb0.021
• as a first approximation.

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Numerical Example

• Substitute the above values into energy equation
  (equation (b)), we have

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Numerical Example

• Substitute Va into continuity equation (equation
  (a)), we have

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Numerical Example

• Corresponding Reynolds numbers are calculated to
  check the above friction factors. For pipe a

The Moody diagram gives f0.0185, which is
correct.
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Numerical Example

• For pipe b

The Moody diagram gives fb0.0225, which is not
close to 0.021
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Numerical Example

• Equations (a) and (b) are solved again for the
  new value of fb

Vb1.630 m/sec and Va 7.693
m/sec Therefore, the discharges are Qa
AaVa 0.0314x7.693 0.242 m3/sec Qb
AbVb 0.01134x1.630 0.018 m3/sec