CHAPTER 2: PIPE FLOWReferences & Suggested reading

1. Solving problems in Fluid MechanicsVol 1. = J F Douglas2. Solving problems in Fluid MechanicsVol 2. = J F Douglas3. Fluid of MechanicsJ W Ireland4. Fluid Mechanics 2 for TechniciansCF Frylinck5. Fluid Mechanics: Fundamental & applicationsCengal & Cimbala6. Principle of Fluid MechanicsCF Meyer

PIPE FLOWFlow Continuity:21Q1Q2

Fig.1:Flow ContinuityThe flow continuity principle states that:THE MASS FLOW RATE IN ANY CLOSED SYSTEM REMAINS CONSTANTie: . .

1 = v AWhere

Where:v = flow velocity [m/s]A = cross sectional area of flow [m]Q = flow rate [m/s]

Flow Energy: Bernoulli, T.T.E.L., T.E.L., H.G.

T.T.E.L.21Consider the sketch below:

12T..E.L.

Fig.2.:Flow Energy variation.BERNOULLIS Theorem states that:THE TOTAL ENERGY OF EACH PARTICLE OF A BODY OF FLUID REMAINS CONSTANT PROVIDED THAT NO ENERGY ENTERS OR LEAVES THE SYSTEM AT ANY POINT.In terms of energy per unit weight (i.e. J/N = m)H = H

2Total Energy (height) H = p / pg v / 2g + z Where: m

and p / pg=Static Pressure (energy) head [m]v / 2g=Velocity (energy) head [m] z=Potential (energy) head [m]Remember pipeline systems are used to transport fluid from one point to another.Application of Bernoulli:(a) For NO energy losses:H1 = H2 p / plg + v / 2g + z = p / pg + v / 2g + z(b) Considering energy losses () between (1) and (2):

3 P / pg + / 2g + z = p / pg + / 2g + z + / 2g + z1 = p2 / p g + v2 / 2g + z2 + hlH = H +

Where:H = Energy head lost between reference points (1) and (2)

Note:1. The straight line indicating the STATIC PRESSURE HEAD (h) above the centre line of the pipe is called the HYDRAULIC GRADIENT (H.G.)H.G. = Hydraulic Gradient2. The straight line indicating the TOTAL ENERGY HEAD () is referred to as the T.E.L.T.E.L.=Total Energy Line3. The straight line parallel to the DATUM LINE indicates the TOTAL ENERGY plus LOSSES (i.e. + H); seeing that it conforms to Bernoullis law which considers no losses, it is referred to as the total energy line without losses or the TOTAL THEORETICAL ENERGY LINE.T.T.E.L.=Total Theoretical Energy Line4. The T.E.L. is always ABOVE the H.G. by the amount equal to the velocity head at that point.

Pipe Flow Energy Losses:Pipe flow losses are due to: Shock Losses and Friction Losses.Shock Losses:This occurs as a result of a change in fluid momentum (velocity) at, for example: pipe couplings such as bends, elbows, T-pieces, Y-pieces, valves, strainers, filters, etc.

4These losses are always a function of the velocity head and is calculated by:(a) m

Where:K=Loss Coefficient, determined by experiment(b) 5K = 4f (L / d)Very often the shock loss is expressed in terms of an EQUIVALENT PIPE LENGTH (L) or EQUIVALENT PIPE LENGTH TO DIAMETER RATIO (L/d). The total pipe length (or L/d ratio) in the Darcy equation then becomes the actual pipe length plus all the equivalent lengths. In this respect:

Note:1. T.T.E.LThe K values of the following pipe components are very often approximated by calculation.

21221 2hTELhQFig:3: Sudden Enlargement1.1 Sudden Enlargement

6 m

- (a) Most commonly used.OR

At (a), the K1 - value is always LESS than 1.At (b), the K - value is often GREATER than 1.

12Q12hT.E.LT.T.E.LhFig: 4: Sudden ContractionSudden Contraction

7 m

2. In practice the K-values are determined experimentally

Loss Coefficients for Pipe FittingsPipe Roughness (mm)

BendsValvesCast Iron0.25

Regular 90 Flanged0,3Check/Non-Return Valve (ball)3,5Concrete3

Regular 90 Threaded1,5Check/Non-Return Valve (swing)2,5Copper0.12

Long Radius 90 Flanged0,2Globe Valve (Fully Open)8Drawn Tubing0.0015

Long Radius 90 Threaded0,7Gate Valve Fully Open0,2Epoxy Coated Steel0.065

Regular 45 Threaded0,4Gate Valve Fully open1,2HDPE (High Density Polyethylene)0.007

Long Radius 45 Flanged0,2Gate Valve Fully Open6

Long Radius 45 Threaded0,4Gate Valve Fully Open24PVC0.0057

VariousButterfly Valve Fully Open 900,9Glasssmooth

Pipe Union/Coupling0,08Butterfly Valve 604,6Galvanized Iron0.15

Tee straight through0,05Butterfly Valve 4510Welded Steel0.046

Tee branch flow1,8Butterfly Valve 3090Riveted Steel4

Reducing coupling/bush1Ball Valve (Fully opened)0Stainless steel (304 and 316)0.0048

Strainer0.7Angle Valve (Fully opened)3.5Uncoated seamless steel0.045

3. If no information is available for reservoir inlet pipes, or any other sharp pipe reduction the K-value is taken as K = 0,5 (i.e.. = 0,5V / 2g) because the average value of = 0,64. Where the flow velocity is low (say V < 0,6 m/s) the shock losses are considered as MINOR LOSSES and are very often neglected.In high pressure systems where flow velocities are high, shock losses are very important and must be allowed for.

Example:A 0,3 m diameter pipe through which water flows at the rate of 0,283 m / s suddenly enlarges to 0,6 m in diameter. If the axis of the pipe is horizontal and the water in the vertical tube connected to the larger pipe stands 0,36 m higher than the level in a tube connected to the smaller pipe, determine the coefficient K if the shock loss is expressed as Kv / 2g, where v is the velocity in the smaller pipe.

d = 0,6m121 20,36mQ = 0,283m/sd = 0,3mAnswer:0,496Bernoulli at 1 + 2 : Where:

And

Note P > P

Ref: JF Douglas (ref. 1)

Friction Losses:This loss can be defined, when there is no potential and kinetic energy changes (i.e. in a horizontal uniform diameter pipe), as follows:

T.T.E.L.T.E.L.lQlFig 5: Friction Losses12H.G.Bernoulli at 1 and 2 yields:

8

Hence, the pressure drop in a system where there is no changes in potential and kinetic energies, is referred to as FRICTION LOSS and is due to internal fluid friction and fluid to pipe friction (as will be discussed later in greater detail) and is calculated by any of the following equations, which was discussed in FLM 2.

The DArcy-Weissbach equation

9 m

Where:f = DArcys coefficient of friction ( = 2gq/ pg)l = pipe length (m)v = flow velocity (m/s)d = pipe diameter (m)Q = flow rate (m/s)g = gravitational acceleration (= 9.81 m/s)h = Energy per unit weight lost in friction (m)Example:

d = 150 mmH.G.ll = 360mDetermine the loss of head due to friction in a new cast-iron pipe 360 m long and 150 m in dia, which carries 42 dm / s.Take f = 0.005. Answer Where: = OR Ref: JF Douglas (ref. 1)Note:1. The answer differs slightly because is an approximation.Can you explain the significance of this difference?

2. can be found from the slope of the H.G.i.e. 3. The slope of the H.G. is equal to the slope of the T.E.L.

10 b) The CHEZY formula: m/s

Where:C = CHEZYs Constant =

11M = Hydraulic Mean Depth (H.M.D.) or Hydraulic Radius

For ROUND pipes:

12[m]

And

13 = Friction head lost per unit length= Slope of the hydraulic gradient v = velocity of flow

Example:Using the Chezy formula, find the loss of head in a circular pipe 120 m long and 75 mm dia, when the velocity of flow is 4.8 m / s.Take C = 54,6 SI units.Answer:

Ref: JF Douglas (ref. 1)

Note:1. The Darcy and Chezy constants are complex values and depend on: the flow velocity of the fluidthe density and viscosity of the fluid and pipe roughness2. Chezys equation is more commonly used for channel flow

Pipe (Mains) Systems: 1. Single Pipe Systems: Low Pressure Systems1.1 Gravity Mains: Two-Reservoir System (Revision)Consider the following system where the flow rate is constant in a uniform diameter pipe.

Fig. 6 Gravity MainNote:1. Piezo tubes installed all along the pipeline will give the STATIC PRESSURE, as at X. The line connecting the levels will produce the H.G.

2. At point X, the total energy per unit weight, according to Bernoulli is:

i.e.The T.E.L. is always above the H.G. by the amount equal to the velocity head.

3. To determine the difference in reservoir levels, take Bernoulli at A and B

14Equation 1.1.14 states that: THE DIFFERENCE IN LEVEL BETWEEN ANY TWO RESERVOIRS, OPEN TO ATMOSPHERE, AND CONNECTED BY A PIPELINE IS DUE (EQUAL) TO THE LOSSES IN THE SYSTEM.4. The slope of the hydraulic gradient is given by:Where:l =Actual pipe length (m)L= Pipe projection length (m)The reason for working with the actual length, is that the slope, in practice is small; so that tan i = sin i and it can be concluded that the H.G. is independent of the path followed by the pipe.5. If no information is given about in- and outlet losses, it can be ignored. Very often shock losses are ignored because they are small compared with the friction loss and are referred to as MINOR LOSSES in LOW pressure systems.Example:A 600 mm dia. Pipe, 1,2 km long carries 0.8 cumec water from a reservoir, whose surface level is 30 m above datum, to another reservoir whose surface level is 14 m above datum. The centre line of the pipe runs through the following points.Distance from reservoir(m)04609001200-------------------------------------------------------------------------------Height above datum(m)27 27 8 12

Calculate:1. The pressure at these points if:1.1 the water flows freely1.2 when the stop valve before the lower reservoir is closed2. The power loss as a result of frictionSolution:1. When water flows freely: (consider only friction losses)1.1 Where: Hence:

3 2Note also

2 4

6 2

2 2 1Then:i) Pressure a___ : Bernoulli between + :

3 1 3ii) Pressure a___ : Bernoulli between + :

43 iii) Pressure a___ : Bernoulli between 1 + 4 :

5 55 15 5iv) Pressure a_ __ : Bernoulli between + :

1.2 With v/v closed, no flow losses occur and the pressure at the various points is the static pressure dead. eg. at:2:3:4:5:

2. Power loss due to friction=

The Syphon:

The siphon is a tube/pipe used to convey liquid over an obstruction from a higher to a lower point.

To find the pressure at E, take Bernoulli between A and E, then with datum through A: Note:

1. If is atmospheric pressure, is given in absolute units.

2. Fluid loses energy at a uniform rate as is indicated by the H.G.

3. Hence at C and D the pressure will be atmospheric and between C and D the pressure will be less than atmospheric with a maximum at E. By definition it can be stated that: A SYPHON OCCURS WHERE THE PIPELINE RISES ABOVE ITS HYDRAULIC GRADIENT.

4. Seeing that a fluid boils when the fluid pressure is reduced to its Vapour Pressure, it is recommended that great; care should be taken to prevent this situation, because at this point the fluid separates (or cavitates) and flow stops or become intermittent.

5. A siphon therefore also exists with E lower than A

Pipeline open to Atmosphere:

Consider the system shown where water emerges from the pipe outlet with velocity, v m/s under a static pressure head, H m

Take Bernoulli between A and B, then:

15 (a)

i.e. Static Pressure Head = Outlet vel. Head + Energy head lost

Hence:

16(b)Note:1. Eq. 15 is the same as eq. 16 but for different reasons.

2. It is important to recognize the difference between systems 1 and 2. In the two-reservoir gravity system the outlet kinetic energy is converted into pressure energy and with the open pipe the k.e. at outlet is not utilized and is sometimes referred to as UNUSED KINETIC ENERGY.

Series Pipe Systems:

Fig. 9: Serie Pipe MainIn the system the in-and outflow is the same and constant, so that the difference in level, causing the flow, also remains constant.

To find out what causes the difference in level, H, take Bernoulli at A and B.

17

Hence, in this case:Note then:1. The difference in reservoir levels is equal to THE SUM of the different energy heads lost.

2. The flow rates in the different pipes remain constant i.e. = . Consider the following example and see how pipe design can be improved for better results.

Pipes in Parallel:

Fig. 10. Gravity Mains consisting of pipes in Parallel.The system above assumes a constant flow situation. The flow being carried by three different pipes in parallel.By closing the isolating valves of any two pipes at a time; or having them all open, and regulating the flow each time so that H remains the same; it can be shown, by taking Bernoulli between A and B each time, that:

Or

1 = .. Or

1 = ...

Hence, for pipes in parallel:

1. The TOTAL LOSS of the system is equal to the TOTAL LOSS OF ANY ONE PIPE in the system.

18i.e.

2. The FLOWRATE in the system is equal to the SUM OF THE FLOWRATE of each pipe in the system.

19i.e.

Example:Two reservoirs are connected by three pipes laid in parallel, their diameter are respectively: d, 2d, and 3d and they are all the same length. Assuming f to be the same for all the pipes, what will be the discharge through the larger pipes if that through the smallest is 0.03 m / s.Solution:With reference to the sketch on p.19, neglecting minor losses:

AndRef: JF DouglasPipes in Series-Parallel:

The principles of both series and parallel pipes are applicable.

Example:

A hydraulic pressure vessel contains oil under a pressure of 4 MPa. The vessel is connected to a service reservoir by means of a 20 mm diameter horizontal pipeline which is 10 m long. The relative density of the oil is 0.8. in order to meet an increase of 30% in oil demand it is decided to lay an additional pipe 25 mm in diameter from the supply vessel in parallel to the existing pipe to join them up at a suitable point.

Take f = 0.008 for all pipes and calculate the length of the additional pipe required. Ignore minor loses and take the service reservoir level at the same height as that of the pressure vessel.

ASupply12ServiceBSolution:

Bernoulli between A & B yields: And

For single (original pipeline):

1For pipes in Series-Parallel:(a) (b) For parallel pipes 1 and 2 :

12 Subject into (c) For series system: 1 + 2 in series with 3

Branching Pipelines:Consider the following situation:

Fig. 11: Branching PipesConsider unit weight of water flowing from A to C. It will lose energy in pipe 1 more than in pipe 3 and at any flow interruptions like at junction J.According to Bernoulli at A and C: Giving : Hence: 20Note then that:1. Minor losses ignored 2. 3. 4. It is a common mistake to regard the loss in the system as the sum of the losses in all the pipes.This is only true when pipes are in series i.e. when an element of fluid passes through all the pipes, losing energy en route to the discharge, but obviously wrong when pipes are in parallel or branching.

Example:A high level reservoir supplies water to a reservoir of surface level 13 m below it through a pipeline 3 657.0 m long and 610 mm in diameter. It becomes necessary to supply a second reservoir with surface level 15 m below the upper one by means of a branching pipe 1 220 m long, connected into the main line 915 m from the top reservoir. Calculate the diameter of the branching pipe so that the flow rate to both reservoirs are the same.

Take f = 0.006 and ignore minor losses.

Solution:

(a) 1(b) 2

(c) But 3 Subtract 3 into 2 : Subtract into 1 :

Pumping Mains:In pump main lines energy losses occur in both inlet and delivery lines. Pump delivery lines can be arranged in any combination discussed in section 1.2 so far.The energy supplied by the prime mover per unit weight of fluid can be summarized as follows. With reference to the sketch below:

232221 Pressure rise across the pump:Consider open suction and delivery levels and let: For Suction side:Bernoulli between 1 and 2. For Delivery side: Bernoulli between 3 and 4. ]Note the following:1. 1 2.

2 3. 3 This is the actual pressure head rise across the pump or the energy per unit weight appearing in the fluid to overcome:(a) the total static lift ()(b) all the losses (i.e. shock and friction) in both suction and delivery pipes,(c) the velocity head change across the pump, often referred to as the LINE CORRECTION. This is due to the fact that the inlet flange diameter is very often larger than the outlet flange diameter.4. Sometimes, when a line correction is applicable, some pump manufacturers refer to the pressure head rise across the pump as the EFFECTIVE HEAD, i.e. And if the inlet and outlet branches of the pump is of the same diameter i.e. , then , hence:(see Pumping Machinery P353)5. If is not atmospheric, then:

24 6. To produce equation 1.2.10 above in a more practical form, the following can be considered:6.1 6.2

Hence:6.3 Very often the following are not applicable or can be neglected:6.3.1 The line correction6.3.2 The suction pressure, when open to atmosphere6.4 Equation 1.2.10 then reduces to:

25If the shock losses are small, they can be neglected as well.Then:

26Where: = Delivery Pressure Head = Static Suction Head = Static Delivery Head = Suction Friction Head = Delivery Friction Head = Reservoir inlet loss; this is only applicable when the pump delivers fluid to a reservoir.Example:A centrifugal pump installed at a reservoir whose surface level is 6 m below datum, pumps water at the rate of 450 m / h into a reservoir whose surface level is 36 m above datum. The suction pipe diameter is 40 mm and is 10 m long. The first portion of the delivery pipe is 300 mm in diameter and 1500 m long. The second portion of the line consists of two pipes in parallel each 1200 m long; one being 150 mm in diameter, the other 200 mm in diameter.

Calculate:1. The quantity of water delivered by each branch pipe.2. The power required to drive the pump if it has an efficiency of 72%.The losses due to all fittings may be ignored. Take: for all pipes. a) 1 Water delivered by each branch? And (Parallel Pipes)

26

12Substitute into

And Power =

MINOR LOSSES IGNORED (EXIT AND ENTRY)AT THE SUCTION*: AT THE DELIVERY: = 19.29 + 52.98 = 72.278m IGNORE LINE CORRECTION: = 194.68kNPower Transmission by Pipelines: Remember that Power is calculated by

P = p x Q P = g Q H1 W Or2Where:P = pressure or pressure drop [Pa]Q = flow rate [m3 /s]H = pressure head or pressure head lost [m] = flued digtheid [kg/m3]

Variation of Power with Discharge:

Fig. 12: Variation of power with discharge.Bernoulli between A and B shows that:

3 Orm

Where:H = Total head availablehp = Outlet velocity head or Head at outlet available for generating power = Friction head lost in supply pipeIt is obvious that the friction head and power head varies with flow rate, controlled by the spear valve situated in the outlet nozzle. The curve alongside illustrates the variation of hf and hp with QNote that Qmax is obtained with the nozzle removed

Fig. 13 Variation of hf and hp with QTransmission Efficiency: Seeing that the power at pipe outlet is reduced by the losses in the pipeline, the pipeline efficiency can be defined as follows:Transmission Efficiency =

4 Eq. 1.3.2 into 1.3.3 gives:

5

Example: The input pressure of a system is 4600kPa. Determine the minimum number of pipes required to transmit 170kW to a machine 3000m from the power station.The efficiency of transmission is 90%. f (Darcy) = 0.0075Solution

Let:

1Then: When:

1 The loss in one pipe, Conditions for Maxim Power Transmission:Because the friction loss varies parabolically with Q (i.e. h Q) it is obvious that the output power and efficiency will also vary according to the law of the parabola. The curves are shown in fig. 1.3.3

Fig 14 (a) Power vs QFig 14 (b) Efficiency vs QFig 14 (a) illustrates that the maximum power is not transmitted at maximum flow but at some point in between which can be determined as follows:Power at pipe outlet: Maximum power is delivered when:

6 So that:For Maximum Power transmission:

Example: Calculate the maximum power available at the far end of a hydraulic pipeline 4.8 km long and 200 mm in diameter when water at 6900 kPa pressure is fed in at the near end. Take f = 0.007.Solution:

Output power = Where for maximum power: And Output power, 8