Fluid Mechanics and Applications MECN 3110

Fluid Mechanics and Applications MECN 3110
Inter American University of Puerto Rico Professor: Dr. Omar E. Meza Castillo

Chapter 6 Viscous Flow in Ducts

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 Darcy’s equation for computing the energy loss due to friction for either laminar and turbulent flow. Define the friction factor as used in Darcy’s equation Determine the friction factor using Moody’s 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

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

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

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

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 < Re < 1: highly viscous laminar “creeping” motion 1 <Re<100: laminar, strong Reynolds number dependence 100 <Re <103: laminar, boundary layer theory useful 103 <Re <104: transition to turbulence 104<Re<106: turbulent, moderate Reynolds number dependence 106<Re<∞: turbulent, slight Reynolds number dependence

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.

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 < 2000 : Laminar flow If the Reynolds number is greater than 4000, the flow can be assumed to be turbulent. Re>4000 : 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.

Application Problems

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 Re=708, which is less than 2000, the flow is laminar

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, D= m and A=5.017 x 10-4 m2. Then we have Because Reynolds number is greater than 4000, the flow is turbulent.

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 Darcy’s Equation. The dimensionless parameter f is called the Darcy Friction factor.

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 (Re<2000). If the two previous relationships for hf are set equal to each other, we can solve for the value of the friction factor:

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 Darcy’s equation. The pipe friction factor decrease inversely with Reynolds number.

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 Re=768, which is less than 2000, the flow is laminar

Problem Using Darcy’s 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.

Friction Loss in Turbulent Flow
For turbulent flow of fluids in circular pipes it is most convenient to use Darcy’s 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

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

Relative Roughness of Pipe Material

Moody Diagram It is the graphical representation of the function f(ReD, ε/D)

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.

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:

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:

Problem Thus, the flow is turbulent. Now the relative roughness must be evaluated. From previous table we find ε=2.4x10-4 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 ε/D = is reached. Project horizontally to the left, and read f=0.038

Problem Statement: In chemical processing plant, benzene at 50oC (SG=0.86) 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.

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

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 A=1.963 x 10-3 m2. Then, we have

Problem For turbulent flow, Darcy’s equation should be used: With the Reynolds number and the relative roughness we obtain the friction factor from the Moody’s Diagram f=0.018

Problem You should have the pressure as follows:

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.

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:

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.

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

fD-W: Darcy-Weisbach friction factor fF: Fanning friction factor
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

Roughness Height (e or ε) for Certain Common Pipes

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, ε/D). Historically, the implicit expression of Colebrook has been accepted as the most accurate in the turbulent zone.

Benedict suggests the expression proposed by Swamee and Jain, i.e.,
Friction Factor f Benedict suggests the expression proposed by Swamee and Jain, i.e., While for ε/D>10-4 Haaland recommends

Friction Factor f For situations where ε/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.

For turbulent flow in smooth pipes (ε/D=0) with 4000<Re<105 is
Friction Factor f For laminar flow (Re<2000) the usual Darcy-Weisbach friction factor representation is For turbulent flow in smooth pipes (ε/D=0) with 4000<Re<105 is For turbulent flow (Re>4000) the friction factor can be founded from the Moody diagram

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

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)

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. ε

The associated head loss is related to the loss coefficient through
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

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

Minor Losses

Minor Losses: Typical Constant K-Factors

Minor Losses

Head Loss Due to a Sudden Expansion (Enlargement)
Or:

Head Loss Due to a Sudden Contraction

Head Loss Due to Gradual Enlargement (Conical diffuser)

Head Loss Due to Gradual Contraction (Reducer or nozzle)

Head Loss at the Entrance of a Pipe (Flow leaving a Tank)

Another Typical Values for various amount of Rounding of the Lip

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)

Head Loss Due to Bends in Pipes

Head Loss Due to Mitre Bends

Head Loss Due to Piping Fittings (Valves, Elbows, Bends, and Tees)

The loss coefficient for elbows, bends, and tees

The relation between head loss and pressure drop is given by
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

Piping Network: HVAC Piping System

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 Δp. 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 Δp.

Piping Network In a Category II problem, the head loss (h or Δp) 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.

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 ε/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.

Series Piping Systems

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

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

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

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

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

For the loss coefficients, we get the following:
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 Nominal size ½” ¾” 1” 1 1/4 ” 1 1/2” 2” 2 ½, 3” Friction factor fT 0.027 0.025 0.023 0.022 0.021 0.019 0.018 Nominal size 4” 5” 6” 8-10” 12-16” 18-24” Friction factor fT 0.017 0.016 0.015 0.014 0.013 0.012

The iteration sequence is given in the following table:
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: D (ft) ε/D V (ft/s) ReD f fT (200f/D+115fT+1.78) H 0.1 25.46 1.819X105 0.0228 0.023 50.036 504.20 0.2 6.37 9.095X104 0.0213 0.018 25.126 15.82 0.15 11.32 1.213X105 0.0216 0.020 32.894 65.48 0.14 12.99 1.299X105 0.0218 35.152 92.91 0.145 12.10 1.254X105 0.0217 33.959 77.42 0.147 11.78 1.237X105 33.534 72.37 0.146 11.95 1.246X105 33.770 74.90

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

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

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:

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

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

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

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

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.

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:

Assume a discharge Q’i through pipe i of the parallel system. Solve for the head loss h’fi (or pressure drop Δp’i) through pipe i; this is a category I pipe-flow solution. Using h’fi= h’fj(i≠j), solve for the Q’i (i≠j). This is a category II pipe=flow solution. Redistribute the total flow rate QT by the simple ratio process where K is the total number of pipes Check the hfk (k=1, K) for the equality using the Qk (k=1,K) obtained from the step 3. Repeat steps 1 through 4 until convergence is obtained.

Example 1-5(Book) Consider the parallel flow network of the following figure with the following specifications: (1) L1=3000 ft, PA=80 psia, L2=3000 ft D1=1 ft, ZA=100 ft, D2=8 in ε1 =0.001 ft, ZB=80 ft, ε2 = ft ε1/D1=0.001, ε2/D2= QA=5.3 ft3/s Find Q1, Q2 and PB A ● ● B (2)

Example 1-5(Book) Solution: This is a type 2 problem Step1: Assume Q’1 = 3 ft3/s. Apply the energy equation along line 1 from A to B to obtain Assuming that VA=VB. Finding h’f1with Q’1 specified is a category I problem, and we have

Example 1-5(Book) So f’1=0.0022, and Step2: The loss for pipe 2 must then become Finding Q’2 for h’f2=14.97 ft-lbf/lbm is category II problem. We shall use V’2 as the iteration variable. The results are given in the following table

Example 1-5(Book) V’2 (ft/s) ReD f’2 h’f 2(ft-lbf/lbm) 1 22,233 0.0265 1.859 4 88,932 0.0192 21.47 3.27 72,702 0.0200 14.94 Step3: From step2, Q’2=V’2A2=1.411ft3/s and ΣQ’l= =4.141ft3/s. We find the corrected values by using

Example 1-5(Book) Step4: Using Q1=3.84ft3/s and Q2=1.46ft3/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.

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 PA=PB.

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:

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:

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.

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.

C is called the Hazen-Williams coefficient
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

Table 1-5 Hazen-Williams Coefficients
Hardy-Cross Method Table 1-5 Hazen-Williams Coefficients Types of pipe C Extremely smooth and straight pipes 140 New, smooth cast iron pipes 130 Average cast iron, new riveted steel pipes 110 Vitrified sewer pipes Cast iron pipes, some years in service 100 Cast iron, in bad condition 80

MGD (million gals/day)
Hardy-Cross Method Table 1-6 values of k1 for Different Units Units of Q k1 CFS (ft3/s) 4.727 MGD (million gals/day) 10.63 CMS (m3/s) 10.466

Hardy-Cross Method

Example 1-8 Use the Hardy-Cross method to obtain the flow rates in each lines of the network as follow. (C=130) 1 cfs 3000’ 6” 4000’ 6” 2000’ 8” 2000’ 12” 3 cfs 1000’ 8” 2000’ 8” 2 cfs 3000’ 8”

The system has 6 nodes (s=6) and 7 lines (r=7)
Example 1-8 Solution: Step 1:We begin by dividing the network into loops and numbering all pipes and nodes in each loop. The system has 6 nodes (s=6) and 7 lines (r=7) F E 1 cfs 3 4 Loop 1 2 A D 3 cfs 1 Loop 2 5 7 2 cfs B 6 C

Example 1-8 Steps 2 and 3:Determine 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 Q5=1.0cfs. Then visiting each node in turn, we obtain: Node B: Q5=1.0 cfs Q6=1.0 cfs Node D: Q2=? Node C: Q1=? Q7=-1.0 cfs Q7=-1.0 cfs 2 cfs Q6=1.0 cfs

Here we make the second assumption, Q1=-0.8cfs.
Example 1-8 Here we make the second assumption, Q1=-0.8cfs. Node D: Node F: Q2=-0.2cfs Q5=1.0 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 Q1=0.8cfs Q2=-0.2 cfs Q5=1.0 cfs

Example 1-8 Steps 4 and 5: Now determine a correction factor ΔQi for each loop Where J is the number of line in the loop and Kj equals the constant for jth line. After obtaining ΔQi for each loop, obtain algebraically a new value for the flow rate in each line; that is

Example 1-8 Converged solution 1 cfs Q3=-0.2431 cfs Q4=-0.2431 cfs

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.,

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

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

Where M represents the degree of polynomial. The Bjm’s 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

Since this represents a positive head loss for loop flow in the positive flow direction, Bj2>0. 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).

Where sgn(Qj)=1 When Qj>0 and sgn(Qj)=-1 When Qj<0.

Hardy-Cross Method

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

A pump with hD =-(50-0.4Q) ft-lbf/lbm to line 6
Example 1-9 A pump with hD =-(50-0.4Q) ft-lbf/lbm to line 6 1 cfs Q3= cfs Q4= cfs Q2=0.792 cfs Q1=0.963 cfs 3 cfs Q7= cfs Q5=1.828 cfs Q6=1.828 cfs 2 cfs P

A heat exchanger with hD =50Q2 ft-lbf/lbm to line 6
Example 1-9 A heat exchanger with hD =50Q2 ft-lbf/lbm to line 6 1 cfs Q3= cfs Q4= cfs Q2=0.741 cfs Q1=0.963 cfs 3 cfs Q7=1.442 cfs Q5=0.558 cfs Q6=0.558 cfs 2 cfs

A very large pump with hD =-1000 ft-lbf/lbm to line 6
Example 1-9 A very large pump with hD =-1000 ft-lbf/lbm to line 6 1 cfs Q3=0.230 cfs Q4=0.230 cfs Q2=1.230 cfs Q1=4.658 cfs 3 cfs Q7=5.888 cfs Q5=7.888 cfs Q6=7.888 cfs 2 cfs P

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

Friction-Factor-Based Hardy-Cross Method

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. Typically values for the purchase and installation of schedule 40 commercial steel piping are as follows: Nominal diameter (in.) Cost per foot ($) 1 11 2 20 3 30 4 40 5 55 6 70 8 90 10 130 12 160

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

Alternative Method to Calculate Costs

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

Cost Estimates

Cost Estimates - Some equations

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.

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

Fluid Mechanics and Applications MECN 3110

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Fluid Mechanics and Applications MECN 3110

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Presentation on theme: "Fluid Mechanics and Applications MECN 3110"— Presentation transcript:

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About project

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