Fluid Mechanics and Applications MECN 3110

1
Fluid Mechanics and Applications MECN 3110

• Inter American University of Puerto Rico
• Professor Dr. Omar E. Meza Castillo

2
Viscous Flow in Ducts

• Chapter 6

3
Course Objectives

• To describe the appearance of laminar flow and
  turbulent flow.
• State the relationship used to compute the
  Reynolds number.
• Identify the limiting values of the Reynolds
  number by which you can predict whether flow is
  laminar or turbulent.
• Compute the Reynolds number for the flow of
  fluids in round pipes and tubes.
• State Darcys equation for computing the energy
  loss due to friction for either laminar and
  turbulent flow.
• Define the friction factor as used in Darcys
  equation
• Determine the friction factor using Moodys
  diagram for specific values of Reynolds number
  and the relative roughness of the pipe.
• Major and Minor losses in Pipe Systems.

Thermal Systems Design Universidad del Turabo
4
Introduction

• This chapter is completely devoted to an
  important practical fluid engineering problem
  flow in ducts with various velocities, various
  fluids, and various duct shapes. Piping Systems
  are encountered in almost very engineering design
  and thus have been studied extensively.
• The basic piping problem is this Given the pipe
  geometry and its added components (such as
  fitting, valves, bends, and diffusers) plus the
  desired flow rate and fluid properties, what
  pressure drop is needed to drive the flow? Of
  course, it may be stated in alternative form
  Given the pressure drop available from a pump,
  what flow rate will ensue? The correlations
  discussed in this chapter are adequate to solve
  most such piping problems

5
Reynolds Number Regimes

• As the water flows from a faucet at a very low
  velocity, the flow appears to be smooth and
  steady. The stream has a fairly uniform diameter
  and there is little or no evidence of mixing of
  the various parts of the stream. This is called
  laminar flow.
• High-viscosity, low-Reynolds-number, laminar flow

6
Reynolds Number Regimes

• When the faucet is nearly fully open, the water
  has a rather high velocity. The elements of fluid
  appear to be mixing chaotically within the
  stream. This is a general description of
  turbulent flow.
• Low-viscosity. High-Reynolds-number, turbulent
  flow

7
Reynolds Number Regimes
8
Reynolds Number Regimes

• The changeover is called transition to turbulent.
  Transition depends on many effects, such as wall
  roughness or fluctuations in the inlet stream,
  but the primary parameter is the Reynolds number.
• Studies present the following approximate ranges
  that commonly occur
• 0 lt Re lt 1 highly viscous laminar creeping
  motion
• 1 ltRelt100 laminar, strong Reynolds number
  dependence
• 100 ltRe lt103 laminar, boundary layer theory
  useful
• 103 ltRe lt104 transition to turbulence
• 104ltRelt106 turbulent, moderate Reynolds number
  dependence
• 106ltRelt8 turbulent, slight Reynolds number
  dependence

9
Reynolds Number Regimes

• In 1883 Osborne Reynolds, British engineering
  professor was the first to demonstrate that
  laminar or turbulent flow can be predicted if the
  magnitude of a dimensionless number, now called
  the Reynolds number is known.
• The following equation shows the basic
  definition of the Reynolds number, Re
• The value of 2300 is for transition in pipes.
  Other geometries, such as plates, airfoils,
  cylinders, and spheres, have completely different
  transition Reynolds numbers.

10
Critical Reynolds Number

• For practical applications in pipe flow we find
  that if the Reynolds number for the flow is less
  than 2000, the flow will be laminar.
• Re lt 2000 Laminar flow
• If the Reynolds number is greater than 4000, the
  flow can be assumed to be turbulent.
• Regt4000 Turbulent flow
• In the range of Reynolds numbers between 2000 and
  2000, it is impossible to predict which type of
  flow exists therefore this range is called the
  critical region.

11
Application Problems
12
Problem

• Statement Determine whether the flow is laminar
  or turbulent if glycerin at 25oC flows in a pipe
  with a 150-mm inside diameter. The average
  velocity of low is 3.6 m/s.
• Solution
• Because Re708, which is less than 2000, the flow
  is laminar

13
Problem

• Statement Determine whether the flow is laminar
  or turbulent if water at 70oC flows in a 1-in
  Type K copper tube with a flow rate of 285 L/min.
• Solution For a 1-in Type K copper tube,
  D0.02527m and A5.017 x 10-4 m2. Then we have
• Because Reynolds number is greater than 4000, the
  flow is turbulent.

14
Head Loss The Friction Factor

• In the general energy equation
• Julius Weisbach in 1850 established that hf is
  proportional to (L/D), and G.H.L Hagen shown that
  for turbulent flow, hf is proportional to V2. The
  proposed correlation, still as effective today as
  in 1850, is
• This expression is called Darcys Equation. The
  dimensionless parameter f is called the Darcy
  Friction factor.

15
Friction Loss in Laminar Flow

• Because laminar flow is so regular and orderly,
  we can derive a relation between the energy loss
  and the measurable parameters of the flow system.
• This relationship is known as the
  Hagen-Pouseuille equation
• The Hagen-Pouseuille equation is valid only for
  laminar flow (Relt2000).
• If the two previous relationships for hf are set
  equal to each other, we can solve for the value
  of the friction factor

16
Friction Loss in Laminar Flow

• In summary, the energy loss due to friction in
  laminar flow can be calculated either from
  Hagen-Pouseuille equation or Darcys equation.
  The pipe friction factor decrease inversely with
  Reynolds number.

17
Application Problems
18
Problem

• Statement Determine the energy loss if glycerin
  at 25oC flows 30 m through a 150-mm-diamter pipe
  with an average velocity of 4.0 m/s.
• Solution First, we must determine whether the
  flow is laminar or turbulent by evaluating the
  Reynolds number
• Because Re768, which is less than 2000, the flow
  is laminar

19
Problem

• Using Darcys Equation
• This means that 13.2 NM of energy is lost by each
  newton of the glicerin as it flow along the 30 m
  of pipe.

20
Friction Loss in Turbulent Flow

• For turbulent flow of fluids in circular pipes it
  is most convenient to use Darcys Equation to
  calculate the energy loss due to friction.
• Turbulent flow is rather chaotic and is
  constantly varying.
• For these reasons we must rely on experimental
  data to determine the value of f.
• The following figure illustrate pipe wall
  roughness (exaggerated) as the height of the
  peaks of the surface irregularities.
• Because the roughness is somewhat irregular,
  averaging techniques are used to measure the
  overall roughness value

21
Friction Loss in Turbulent Flow

• For commercially available pipe and tubing, the
  design value of the average wall roughness has
  been determined as shown in the following table

22
Relative Roughness of Pipe Material
23
Moody Diagram

• It is the graphical representation of the
  function f(ReD, e/D)

24
Moody Diagram

• Several important observations can be made from
  these curves
• For a given Reynolds number flow, as the relative
  roughness is increased, the friction factor f
  decreases.
• For a given relative roughness, the friction
  factor f decreases with increasing Reynolds
  number until the zone of complete turbulent is
  reached.
• Within the zone of complete turbulence, the
  Reynolds number has no effect on the friction
  factor.
• As the relative roughness increases, the value of
  Reynolds number at which the zone of complete
  turbulence begins alto increases.

25
Application Problems
26
Problem

• Check your ability to read the Moody Diagram
  correctly by verifying the following values for
  friction factors for the given values of Reynolds
  number and relative roughness

27
Problem

• Statement Determine the friction factor f if
  water at 70oC is flowing at 9.14 m/s in an
  uncoated ductile iron pipe having an inside
  diameter of 25 mm.
• Solution The Reynolds number must first be
  evaluated to determine whether the flow is
  laminar or turbulent

28
Problem

• Thus, the flow is turbulent. Now the relative
  roughness must be evaluated. From previous table
  we find e2.4x10-4 m. Then , the relative
  roughness is
• The final steps in the procedure are as follows
• Locate the Reynolds number on the abscissa of the
  Moody Diagram.
• Project vertically until the curve for e/D
  0.00961538 is reached.
• Project horizontally to the left, and read f0.038

29
Problem

• Statement In chemical processing plant, benzene
  at 50oC (SG0.86) must be delivered to point B
  with a pressure of 550 kPa. A pump is located at
  point A 21m below point B, and the two points are
  connected by 240 m of plastic pipe having an
  inside diameter of 50 mm. If the volume flow rate
  is 110 L/min, calculate the required pressure at
  the outlet of the pump.

30
Problem

• Solution Using the energy equation we get the
  following relation
• Mass Balance
• Energy Balance

31
Problem

• The evaluation of the Reynolds number is the
  first step. The type of flow, laminar or
  turbulent, must be determined.
• For a 50-mm pipe, D.050 m and A1.963 x 10-3 m2.
  Then, we have

32
Problem

• For turbulent flow, Darcys equation should be
  used
• With the Reynolds number and the relative
  roughness we obtain the friction factor from the
  Moodys Diagram f0.018

33
Problem

• You should have the pressure as follows

34
Fundamental Equation of Fluid Mechanics

• In order to apply previous equation to a piping
  system, we must extend the Bernoulli equation to
  account for losses which result from pipe
  fittings, valves, and direct losses (friction)
  within the pipes themselves. The extended
  Bernoulli equation may be written as
• Additionally, at various points along the piping
  system we may need to add energy to provide an
  adequate flow. This is generally achieved through
  the use of some sort of prime mover, such as a
  pump, fan, or compressor.

35
Fundamental Equation of Fluid Mechanics

• For a system containing a pump or pumps, we must
  include an additional term to account for the
  energy supplied to the flowing stream. This
  yields the following form of the energy equation
• Finally, if somewhere in the piping system a
  component extracts energy from the fluid stream,
  such as a turbine, the energy equation takes the
  form

36
Losses in Piping System

• Friction Factor The total head loss hf in a
  piping system are typically categorized as major
  and minor losses.
• Major losses are associated with the pipe-wall
  skin friction over the length of the pipe, and
• Minor losses in piping systems are generally
  characterized as any losses which are due to pipe
  inlets and outlets, fittings and bends, valves,
  expansions and contractions, filters and screens,
  etc.
• Minor losses are not necessarily smaller than
  major losses.

37
Major Losses

• Major losses of head in a piping system are the
  direct result of fluid friction in pipes and
  ducting. The resulting head losses are usually
  computed through the use of friction factors.
  Friction factors for ducts have been compiled for
  both laminar and turbulent flows. Two widely
  adopted definitions of the friction factor are
  the Darcy and Fanning friction factors.
• The head loss due to flow of a fluid at an
  average velocity V through a length L of pipe
  with a diameter D is

(Darcy-Weisbach) or
38
Major Losses
(Fanning)
Where fD-W Darcy-Weisbach friction
factor fF Fanning friction factor L Length of
considered pipe D Pipe diameter V2/2gc Velocity
head
39
Roughness Height (e or e) for Certain Common Pipes
e
40
Friction Factor f

• The Moody diagram is sufficient to determine the
  friction factor and, hence, the head loss for
  given flow conditions. If we should desire to
  generate a computer-based solution, the
  translation of the Moody diagram into tabular
  form to use in interpolation is awkward. To say
  the least. What is needed is a simple algebraic
  expression in the form f(ReD, e/D). Historically,
  the implicit expression of Colebrook has been
  accepted as the most accurate in the turbulent
  zone.

41
Friction Factor f

• Benedict suggests the expression proposed by
  Swamee and Jain, i.e.,
• While for e/Dgt10-4 Haaland recommends

42
Friction Factor f

• For situations where e/D is very small, as in
  natural-gas pipelines, Haaland proposes
• Where n 3
• The use of the Swamee-Jain or Haaland provide an
  explicit formula of the friction factor in
  turbulent flow, and is thus the preferred
  technique.

43
Friction Factor f

• For laminar flow (Relt2000) the usual
  Darcy-Weisbach friction factor representation is
• For turbulent flow in smooth pipes (e/D0) with
  4000ltRelt105 is
• For turbulent flow (Regt4000) the friction factor
  can be founded from the Moody diagram

44
Friction Factor f

• Churchill devised a single expression that
  represents the friction factor for laminar,
  transition and turbulent flow regimes. This
  expression, which is explicit for the friction
  factor given the Reynolds number and relative
  roughness, is
• where
• and

45
Friction Factor f

• In our discussion so far we have been concerned
  only with circular pipes, but for a variety of
  reason conduit cross sections often deviate from
  circular. The appropriate characteristic length
  to use in evaluating the Reynolds number for
  noncircular cross-sectional areas is the
  hydraulic diameter. The hydraulic diameter is
  defined as
• To use the hydraulic diameter concept, the
  Reynolds number is defined as

µ ? (m2/s) T ? (oC)
46
Application Problems
47
Example 2

• Find the head loss due to friction in
  galvanized-iron pipe 30 cm diameter and 50 m long
  through which water is flowing at a velocity of 3
  m/s assume that water flowing at 20oC.

e
48
Minor Losses

• Minor losses are due to the change of the
  velocity of the flowing fluid in magnitude or
  direction. They are most often calculated using
  the concept of a loss coefficient or equivalent
  friction length method. In the loss coefficient
  method, a constant or variable factor K is
  defined as
• The associated head loss is related to the loss
  coefficient through

49
Minor Losses

• The Minor Losses occurs at
• Valves
• Tees
• Bends
• Reducers
• Valves
• And other appurtenances

50
Minor Losses
51
Minor Losses Typical Constant K-Factors
52
Minor Losses
53
Head Loss Due to a Sudden Expansion (Enlargement)

• Or

54
Head Loss Due to a Sudden Contraction
55
Head Loss Due to Gradual Enlargement (Conical
diffuser)
56
Head Loss Due to Gradual Contraction (Reducer or
nozzle)
57
Head Loss at the Entrance of a Pipe (Flow leaving
a Tank)
58
Another Typical Values for various amount of
Rounding of the Lip
59
Head Loss at the Exit of a Pipe (flow entering a
tank)

• The entire kinetic energy of the exiting fluid
  (velocity V1) is dissipated through viscous
  effects as the stream of the fluid mixes with the
  fluid in the tank and eventually comes to rest
  (V2)

60
Head Loss Due to Bends in Pipes
61
Head Loss Due to Mitre Bends
62
Head Loss Due to Piping Fittings (Valves, Elbows,
Bends, and Tees)
63
Head Loss Due to Piping Fittings (Valves, Elbows,
Bends, and Tees)
64
The loss coefficient for elbows, bends, and tees
65
General Equation

• The basis for any analysis or design in the
  energy equation written between any two points
  and incorporating multiple pumps, turbines, and
  majors and minor losses. The general
  representation that we shall use is
• The relation between head loss and pressure drop
  is given by
• And the relation between power and pressure drop
  is given by

66
Piping Network HVAC Piping System
67
Piping Network

• Most engineering systems are comprised of more
  than one section of pipe. In fact in most systems
  a complex network of piping is required to
  circulate the working fluid of a particular
  thermal system. These networks consist of series,
  parallel, and series-parallel configurations.
• Pipe flow problems fall into three categories. In
  Category I problems the solution variable is the
  head loss or pressure drop ?p. The problem is
  specified such that the volumetric flow Q, the
  length of pipe L, the size or diameter D, are all
  known along with other parameters such as the
  pipe roughness and fluid properties. These types
  of problems yield a direct solution for the
  unknown variable ?p.

68
Piping Network

• In a Category II problem, the head loss (h or ?p)
  is specified and the volumetric flow Q is sought.
  Finally in a Category III problem, both the head
  loss and volumetric flow are specified, but the
  size or diameter of the pipe D is sought.
  Category I and Category II problems are
  considered analysis problems since the system is
  specified and only the flow is calculated.
  Whereas Category III problems are considered
  design problems, as the operating characteristics
  are known, but the size of the pipe is to be
  determined. Both Category II and Category III
  problems require an iterative approach in
  solution.

69
Piping Network

• Depending upon the nature of the flow (and
  solution process), it may be required to
  recompute other parameters such as the relative
  roughness at each iterative pass, since the e/D
  ratio will change as the pipe diameter changes.
  However, with most modern computational software,
  we may solve iterative problems rather
  efficiently and need not resort to classic
  methods such as Gaussian elimination.

70
Series Piping Systems
71
Series Piping Systems

• The series flow arrangement is the simplest to
  analyze. In a series arrangement of pipes, the
  volumetric flow at any point in the system
  remains constant assuming the fluid is
  incompressible. Thus, for an arrangement of N
  pipes, the volumetric flow is given by
• Or
• The head loss in the system is the sum of the
  individual losses in each section of pipe. That
  is

72
Series Piping Systems

• The series flow arrangement is the simplest to
  analyze. In a series arrangement of pipes, the
  volumetric flow at any point in the system
  remains constant assuming the fluid is
  incompressible. Thus, for an arrangement of N
  pipes, the volumetric flow is given by
• Or
• The head loss in the system is the sum of the
  individual losses in each section of pipe. That
  is

73
Application Problems
74
Example 1-1 (Book)

• Apply the energy equation to the situation
  sketched in the following figure.

1
Water
La, Da, Va
H
Lb, Db, Vb
2
Lc, Dc, Vc
Figure 1-1
75
Example 1-1 (Book)

• The energy equation will be applied from the free
  surface at position 1 of the upper reservoir to
  the free surface at position 2 of the lower
  reservoir. The results, with each loss term
  identified, are
• From the figure, we obtain

76
Example 1-1 (Book)

• And from the continuity equation for
  incompressible steady flow, we can find that
• Which, for circular pipes, becomes
• Substitution of the preceding into the energy
  equation yields, after applying some algebra,
• Until additional information is specified, we can
  not proceed any further than this

77
Example 1-2 (Book)

• For the system illustrated in Example 1-1,
  specify the nominal size of clean commercial
  steel pipes required for a flow rate of 0.2 ft3/s
  if the following are given
• Solution This is category III problem, since the
  system is specified except for the pipe size
  required for a given flow rate. Using the values
  specified, equation of the Example 1-1 becomes
  for this system

78
Example 1-2 (Book)

• For the loss coefficients, we get the following
• The final equation then finally reduces to
• Where fT is the turbulent friction factor

79
Example 1-2 (Book)

• The solution must be by iteration, since V and f
  are functions of D. For this particular example,
  we shall use the diameter as the iteration
  variable and compute the head H necessary to
  deliver 0.2 ft3/s through the system.
• The iteration sequence is given in the following
  table

80
Generalized Series Piping System Software

• All series piping problems have similar
  characteristics 1) upstream and downstream are
  unambiguously defined, 2) major and minor losses
  are additive in the direction of the flow, and 3)
  the energy equation can be applied to any segment
  of the system.

Generalized Series Piping System Schematic
Figure 1-2
81
Generalized Series Piping System Software

• The Generalized Series Piping System Schematic
  consist of a series piping system with pipes of
  different diameter, a variety of major and minor
  losses, and a pump with an increase in head of
  Ws. Assuming that the flow is from a to b, the
  energy equation becomes

82
Generalized Series Piping System Software

• Conservation of mass for an incompressible flow
  for the generalized series piping system required
  that
• Which for a circular pipe can be expressed as
• Using the above equations for the velocity,
  combining some terms in previous equation, and
  generalizing to J pipes results in the following
  generic form of the energy equation

83
Generalized Series Piping System Software

• The example 1-2 in generic form appears as
• For a simple-pipe system, the symbolic form of
  the generic energy equation becomes

84
Parallel Piping Systems
85
Parallel Piping Systems

• Flow in parallel piping elements is also easy to
  analyze. In a parallel arrangement the total head
  loss or pressure drop across the system is
  constant. That is
• On the other hand, the volumetric flow through
  the system is the sum total of the individual
  flow in each pipe. That is

86
Parallel Piping Systems

• Parallel systems are often used to reduce the
  pumping power required for process-control
  system.
• The following figure shows a generalized parallel
  system.

87
Parallel Piping Systems

• The increase in head across the pump, Ws, is such
  that the pressures at a and b are equal. The
  change in pressures (or heads) are equal for each
  leg of the parallel arrangement hence Ws is also
  the change in head across each leg required for
  Pa Pb. If QT is the total flow rate, the
  conservation of mass yields
• The energy equation for line i, under the
  assumptions of Pa Pb and Va Vb, can be
  expressed

88
Parallel Piping Systems

• Analysis or design calculations for parallel
  piping require the use of the conservation mass
  and an energy equation, for each line. For
  example for a system composed of two parallel
  lines the requirement expressions are
• Together with the usual friction-factor and
  Reynolds-number definitions.

89
Parallel Piping Systems

• Two types of parallel-system analysis problems
  are evident in the last equations (1) given Ws,
  find QT, Q1 and Q2 and (2) given QT, find Ws, Q1
  and Q2.
• Type 1 problems are straightforward, as they can
  be solved by category II problem methodology.
• Type 2 problems are more complex, since the total
  flow rate must be apportioned to each of the
  parallel lines in such manner that the pressure
  drops across each line are equal. We can list a
  simple sequence to systematically accomplish this
  apportionment

90
Parallel Piping Systems

• Assume a discharge Qi through pipe i of the
  parallel system. Solve for the head loss hfi (or
  pressure drop ?pi) through pipe i this is a
  category I pipe-flow solution.
• Using hfi hfj(i?j), solve for the Qi (i?j).
  This is a category II pipeflow solution.
• Redistribute the total flow rate QT by the simple
  ratio process
• where K is the total number of pipes
• Check the hfk (k1, K) for the equality using the
  Qk (k1,K) obtained from the step 3. Repeat steps
  1 through 4 until convergence is obtained.

91
Application Problems
92
Example 1-5(Book)

• Consider the parallel flow network of the
  following figure with the following
  specifications

(1)
L13000 ft, PA80 psia, L23000 ft D11 ft,
ZA100 ft, D28 in e1 0.001 ft, ZB80 ft, e2
0.0001 ft e1/D10.001, e2/D20.00015 QA5.3
ft3/s Find Q1, Q2 and PB
? B
A ?
(2)
93
Example 1-5(Book)

• Solution This is a type 2 problem
• Step1 Assume Q1 3 ft3/s. Apply the energy
  equation along line 1 from A to B to obtain
• Assuming that VAVB. Finding hf1with Q1
  specified is a category I problem, and we have

94
Example 1-5(Book)

• So f10.0022, and
• Step2 The loss for pipe 2 must then become
• Finding Q2 for hf214.97 ft-lbf/lbm is category
  II problem. We shall use V2 as the iteration
  variable. The results are given in the following
  table

95
Example 1-5(Book)

• Step3 From step2, Q2V2A21.411ft3/s and
  SQl1.14134.141ft3/s. We find the corrected
  values by using

96
Example 1-5(Book)

• Step4 Using Q13.84ft3/s and Q21.46ft3/s, we
  compute V1, V2, ReD1, ReD2, f1 and f2, to find
• The heat losses in the pipes then agree to 0.54
  percent, which is sufficient.
• With the flow rates in each line and the head
  loss for the parallel segments knows, Ws is
  computed from the first equation in step1.

97
Example 1-5(Book)

• Thus, a pump with an increase in head of 3.91
  ft-lbf/lbm would be required for the system to
  pass 5.3 ft3/s while maintaining PAPB.

98
Generalized Parallel Piping System Software

• Using the example 1-5. For the two-pipe parallel
  system, conservation of mass and the energy
  equation becomes as
• The above expressions for the system under
  consideration reduce to the following

99
Series-Parallel Network
100
Series-Parallel Network

• In a series-parallel pipe network as shown in
  previous figure, we must apply rules which are
  analogous to the analysis of an electric circuit.
  In previous figure only shows the pipe network in
  2-Dimensions. In reality, a pipe network is most
  often three dimensional. Thus, the elevations of
  each nodal point need to be considered when
  writing the extended Bernoulli equation. The
  following rules apply in any network of pipes

101
Series-Parallel Network

• Application of these rules leads to a complex set
  of equations which must be solved numerically.
  These are easily dealt with in most
  mathematical/numerical analysis programs.
  However, a method of hand calculation know as the
  Hardy-Cross Method may also be applied. This
  method is the basis for most computer software
  developed for analyzing piping systems.

102
Hardy-Cross Method

• The Hardy-Cross formulation is an iterative
  method for obtaining the steady-state solution
  for any generalized series-parallel flow network.
  Its great advantage is systematicness.
• The Hardy-Cross method can be systematically
  applied to any fluid flow network, and if the
  guidelines are followed, a converged solution
  will always be obtained.
• The basis for any Hardy-Cross analysis technique
  is the same as for series-parallel flow network
  analysis (1) conservation of mass at a node and
  (2) uniqueness of the pressure at a given point
  in the loop.

103
Hardy-Cross Method

• A modified version of the head-loss of the
  Darcy-Weishbach expression called Hazen-Williams
  expression
• Where K and n are determined either by experiment
  or by curve fits using the Moody diagram.
• C is called the Hazen-Williams coefficient

104
Hardy-Cross Method

• Table 1-5 Hazen-Williams Coefficients

105
Hardy-Cross Method

• Table 1-6 values of k1 for Different Units

106
Hardy-Cross Method
107
Example 1-8

• Use the Hardy-Cross method to obtain the flow
  rates in each lines of the network as follow.
  (C130)

1 cfs
3000 6
4000 6
2000 8
2000 12
3 cfs
1000 8
2000 8
2 cfs
3000 8
108
Example 1-8

• Solution
• Step 1We begin by dividing the network into
  loops and numbering all pipes and nodes in each
  loop.
• The system has 6 nodes (s6) and 7 lines (r7)

1 cfs
F
E
3
2
4
Loop 1
A
D
1
3 cfs
Loop 2
7
5
2 cfs
B
C
6
109
Example 1-8

• Steps 2 and 3Determine the zeroth estimate for
  the flow rate and obtain the flow rates for all
  lines in such manner that conservation of mass is
  satisfied at each node.
• To start the process, we shall assume that
  Q51.0cfs. Then visiting each node in turn, we
  obtain

Node B
Q51.0 cfs
Q61.0 cfs
Node D
Q2?
Node C
Q1?
Q7-1.0 cfs
Q7-1.0 cfs
2 cfs
Q61.0 cfs
110
Example 1-8

• Here we make the second assumption, Q1-0.8cfs.

Node F
Node D
Q2-0.2cfs
Q51.0 cfs
Q3-1.2 cfs
Q1-0.8cfs
Q7-1.0 cfs
Q4-1.2 cfs
Node E
Node A
1 cfs
Q4-1.2cfs
3 cfs
Q3-1.2 cfs
Q10.8cfs
Q2-0.2 cfs
Q51.0 cfs
111
Example 1-8

• Steps 4 and 5 Now determine a correction factor
  ?Qi for each loop
• Where J is the number of line in the loop and Kj
  equals the constant for jth line.
• After obtaining ?Qi for each loop, obtain
  algebraically a new value for the flow rate in
  each line that is

112
Example 1-8

• Converged solution

1 cfs
Q3-0.2431 cfs
Q4-0.2431 cfs
Q20.7569 cfs
Q11.8596 cfs
3 cfs
Q7-1.1027 cfs
Q50.8973 cfs
Q60.8973 cfs
2 cfs
113
Generalized Hardy-Cross Analysis

• The Hardy-Cross analysis has been restricted to
  flow networks in which the pipe wall friction
  represented the only loss (i.e., minor loss has
  been neglected).
• If the line lengths are short enough so that
  minor losses are important, the equivalent-length
  approach can be used to include the losses due to
  fittings.
• The equivalent length is additional length of
  pipe needed to give the same head loss (or
  pressure drop) as a fitting at a giving flow
  rate. Hence the equivalent length Leq is obtained
  by equating the loss-coefficient expression to
  conventional head-loss expression, i.e.,

114
Generalized Hardy-Cross Analysis

• The generalized Hardy-Cross analysis can be used
  for piping network in which the lines can contain
  devices that result either in additional pressure
  drop (a heat exchanger or turbine, for example)
  or in a pressure increase ( a pump, for example).

Lj Dj Qj
Device
115
Generalized Hardy-Cross Analysis

• Consider a typical line, line j with length Lj,
  and some device in the line that causes either a
  head decrease or increase.
• In general, the change in head hfD across the
  device will depend on the flow rate
• The functional dependence can be represented by
  the polynomial expression

116
Generalized Hardy-Cross Analysis

• Where M represents the degree of polynomial. The
  Bjms can have positive, negative or zero values
  depending of the particular device being
  described.
• The head loss through a fitting is typically
  described by
• The coefficients in the polynomial expression
  then take the values

117
Generalized Hardy-Cross Analysis

• Since this represents a positive head loss for
  loop flow in the positive flow direction, Bj2gt0.
  If an ideal backward curved blade centrifugal
  pump is used, the representation is
• Note that hDj must be less than zero, because a
  pump represents a negative head loss (i.e., an
  increase in head).

118
Generalized Hardy-Cross Analysis

• Where sgn(Qj)1 When Qjgt0 and sgn(Qj)-1 When
  Qjlt0.

119
Hardy-Cross Method
120
Example 1-9

• For the network of Example 1-8, investigate the
  effects of adding
• A pump with hD -(50-0.4Q) ft-lbf/lbm to line 6
• A heat exchanger with hD 50Q2 ft-lbf/lbm to line
  6
• A very large pump with hD -1000 ft-lbf/lbm to
  line 6

121
Example 1-9

• A pump with hD -(50-0.4Q) ft-lbf/lbm to line 6

1 cfs
Q3-0.208 cfs
Q4-0.208 cfs
Q20.792 cfs
Q10.963 cfs
3 cfs
Q7-0.172 cfs
Q51.828 cfs
Q61.828 cfs
2 cfs
P
122
Example 1-9

• A heat exchanger with hD 50Q2 ft-lbf/lbm to line
  6

1 cfs
Q3-0.259 cfs
Q4-0.259 cfs
Q20.741 cfs
Q10.963 cfs
3 cfs
Q71.442 cfs
Q50.558 cfs
Q60.558 cfs
2 cfs
123
Example 1-9

• A very large pump with hD -1000 ft-lbf/lbm to
  line 6

1 cfs
Q30.230 cfs
Q40.230 cfs
Q21.230 cfs
Q14.658 cfs
3 cfs
Q75.888 cfs
Q57.888 cfs
Q67.888 cfs
2 cfs
P
124
Friction-Factor-Based Hardy-Cross Method

• If the fluid is not water, the procedure outlined
  in the previous slides can be used, but a more
  direct approach in to develop the major losses in
  a friction-factor form rather than a
  Hazen-William form

125
Friction-Factor-Based Hardy-Cross Method
126
Cost Estimates

• Economic considerations are important in most
  energy systems, and estimating the cost of a
  project in an integral part of any bidding
  procedure.
• www.RSMeans.com
• Typically values for the purchase and
  installation of schedule 40 commercial steel
  piping are as follows

127
Cost Estimates

• The values do not include the cost of purchase of
  land, trenching, backfilling or disposal of old
  (replaced) systems.
• Pump cost are more varied, but an
  order-of-magnitude estimate is given by the
  following expression
• Where hp is the power output of the pump and the
  pump cost is in dollars.
• The energy charge is based on the kW-h
  (kilowatt-hour. The cost of kW-h varies 0.02 for
  large industrial costumers to as such as 0.15.
• Typical value 8-10 per kW

128
Alternative Method to Calculate Costs
129
Cost Estimates

• The best way to select an appropriate type of
  pipeline is to compare the inside diameter of
  various pipe materials that might be part of the
  design of a project.

DIPS Ductile Iron Pipe PCCP Prestressed
Concrete Cylinder Pipe HDPE High Density
Polyethylene
130
Cost Estimates
131
Cost Estimates - Some equations
132
Cost Estimates

• For a 24-inch nominal diameter water transmission
  pipeline that is 30,000 feet long with water
  flowing at 6,000 gpm, the cost is 0.06/kW-h, 70
  percent of pumping efficiency and the pump will
  operate 24 hours per day. Calculate the pumping
  Costs of all type of pipes.

133
Homework5 ? Webpage
Due, Wednesday, March ??, 2011
Omar E. Meza Castillo Ph.D.