Mathcad - Flow in Conduits

Emilian Popa - Flow in conduits

Flow in Conduits

When one considers the convenience and necssities of everyday life, it is truly amazing to note the role played by conduits. For example , all of the water that we use in our homes is pumped through pipes so that it will be available when and where we want it. In addition, virtually all of this water leaves our homes as dilute wastes through sewers, another type of conduit. In addition to domestic use, the consumption of water by industry is enormous, from the processign of agricultural products to the manufacturing of steel and paper. All of the water used in these manufacturing processes is transported by means of piping systems. In the United States the petroleum industry alone transports approximately 20 million barrels of liquid petroleum per day in addition to the billions of cubics feet of gas transported by pipeline.

In the foregiong examples it is the transportation of the fluid that is the primary objective. However , there are numerous applications in which flow is a necessary but secondary part of the process. For example, heating and ventilating systems, as well as electric generating stations, utilize conduit flow to circulate fluids in order to transport energy from one location to another. Piping systems are also used extensively for controlling the operation of machinery.

Thus the application of flow in conduits cut across all fields of engineering. Consequently, every engineer should understand the basic fluid mechanics involved with such flow. In this chapter we shall introduce the fundamental theory of flow in conduits as well as basic design procedures.

Shear Stress Distribution Across a Pipe Section

The velocity distribution in a pipe is directly linked to the shear stress distribution ; hence it is important to understand the latter. To determine the shear-stress distribution, we start with the equation of equilibrium applied to a cylindrical element of fluid that is oriented coaxially with the pipe, as shown in Fig 10.1. For the conditions shown in Fig 10.1, it is assumed that the flow is uniform (stream lines are straight and parallel). Therefore , the pressure accross any section of the pipe will be hydrostatically distributed. Thus the pressure force acting on an end face of the fluid element will be the product of the pressure at the center of the element (also at the center of the pipe) and the area of the face of the element. With steady uniform flow, equilibrium between the pressure, gravity, and shearing forces acting on the fluid will prevail. Consequently, the equilibrium equation yields the following:

n

Fx 0

p A. p dpds

Δs. A. ΔW sinα. τ 2 π r( ) Δs 0 10.1

In Eq 10.1 ΔW γAΔs and sinα dzds

Therefore Eq 10.1 reduces to

dpds

Δs. A. γ A. Δs. dzds. τ 2 π r( ) Δs. 0 10.2

Then when we divide Eq 10.2 through by ΔsA and simplify , we obtain

τr2

dds

p γz( ).rr10.3

1


Since the gradient itself, d/ds(p+γz) is negative (see Sec 7.4) and constant across the section for uniform flow, it follows that -d/ds(p+γz) will be positive and constant across the pipe section. Thus τ in Eq (10.3) in the following section to derive the velocity distribution for laminar flow.

10.2 Laminar Flow in Pipe

We determine how the velocity varies across the pipe by substituting for τ in Eq (10.3) its equivalent μdV/dy and integrating. First, the making the substitution we have

μdVdy. r

2d

dsp γz( )..rr 10.4

Because dVdy

dVdrdVdV Eq 10.4 becomes

dVdr

r2 μ.

dds

p γ z.( )rr 10.5

When we depatate variables and integrate across the section, we obtain

V r2

4 μ.d

dsp γ z.( ). Crr 10.6

2


We can evaluate the constant of integration in Eq 10.6 by noting that when r=r0 the velocity V=0. Therefore, the constant of integration is given by

C r02

4 μ.d

dsp τz( ).r0r0 10.7

Equation (10.7) indicates that the velocity distribution for laminar flow in a pipe is parabolic across the section with the maximum velocity at the center of the pipe. Figure 10.2 shows the variation of the shear stress and velocity in the pipe.

Note

The combination p+γz is constant across the section because the streamlines are straight and parallel in uniform flow , and for this condition there will be no acceleration of the fluid normal to the streamline. Thus hydrostatic condition prevail across the flow section. For a hydrostatic condition, p/ γ+z=constant orp+τz=constant, as shown in Chapter 3.

Exemple 10.1 Velocity at center and x mm from center flow laminar

Oil S 0.90 μ 5 10 1. N s

m2. flows steadily in a 3 cm pipe. The pipe is vertical and the

3


m

pressure at an elevation of 100 m is 200 kPa If the pressure at an elevation of 85 m is

250 kPa is the flow direction up or down ? What is the velocity at the center of the pipe and at

6 mm from the center , assuming that the flow is laminar ?

Solution

First determine the rate of change of p γ z. Taking s in the z direction

dds

p γ z.( ) p100 γ z100.( ) p85 γ z85.( )z100 z85

p100p100

N

m3z100 100 m z85 85 m γ 8830

p100 200 kPa p85 250 kPa

p100 p100 103. Pa p85 250 103. Pa

p100 γ z100.( ) p85 γ z85.( )z100 z85

5.497 103.=N

m3

The quantity p γ z. is not constant with elevation it increases upward (decreases downward).

Therefore , the direction of flow is downward.. This can be seen by substituting

d p γz( )ds

5.497d p γz( )ds

d p γz( )ds

kN

m3into Eq (10.7). When this is done, V is negative for all values of r in

the flow. When r 0 center of the pipe, the velocity will be maximum. Thus

Vcenter VmaxVmaxVmax Vmax r02

4 μ.5.53 kN

m3.r0r0

Vmax 0.0152 m2.

4 5 10 1. N. s

m2..

5.53 103. N

m3.. Vmax 0.622 m s 1.=

At first it may seem strange that the velocity is in a direction oposite to the direction of decreasing pressure. However, it may not seem so peliculear if one realizes that in this example the pipe is vertical, so the gravitational force as well as pressure helps to establish the flow. What counts the flow is othe than in the horizontal direction is how the combination p+γz changes with s. If p+γz is constant, then we have the equation of hydrostatic and no flow occurs. However, if (p+γz) is not constant, flow will occur in the direction of decreasing p+γz.

Next determine the velocity at r 6 mm r r 10 3. r 6 10 3.= m

Using Eq 10.7 we find that

4


V r02 r2

4 μ.d

dsp γz( ).r0r0 10.7

r0 15 mm r0 r0 10 3. r0 0.015= m

V r02 r2

4 μ.5.497 103.. V 0.519=

ms

For many problems we wish to relate the pressure change to the rate of flow or mean velocity Vm in the conduit. Therefore, it is necessary to integrate

dQ V dA. over the corss-sectional area of flow. That is

Q AVd

Q

0

r0

rr02 r2

4 μ.d

dsp γz( ). 2. π. r. d 10.8

The factor π

d p γz( )ds

4 μ. is constant across the pipe section. Therefore, upon integration, we obtain

Q π

4 μ.dds

p γz( ). r2 r02 2

2. betweeen 0 and r0 10.9

which reduces to

Q π r02.

8 μ.d

dsp γz( ) 10.10

If we divide through the cross sectional area of the pipe, we have an expression for the mean velocity:

Vm r02

8 μ.d

dsp γz( ). 10.11

Comparing Eqs (10.11) and (10.7) reveals that Vm Vmax2

Also by substituting D2

for r0 we

have

Vm D2

32 μ.d

dsp γz( ).DD 10.12

or dds

p γz( ) 32 μ. Vm.

D2DD10.13

5


Integrating Eq 10.13 along the pipe between sections 1 and 2 we obtain

p2 p1 γ z2 z1( ). 32 μ. Vm.

D2s2 s1( ).

Here s2 s1 is the length L of pipe between the two sections. Therefore, Eq 10.14 can be rewritten

as

p1γ

z1 p2γ

z2 32 μ L. Vm.

γ D2..

10.15

It can be seen that when the general energy equation for incompressible flow in conduits. Eq 7.24 is reduced to one for uniform flow in a constant diameter pipe where

V1 V2V2V2 the result is

p1γ

z1 p2γ

z2 hfp2p2 10.16

Here hf is used instead of hL to signify head loss due to frictional resistance of the pipe. Comparison of Eqs (10.15) and (10.16) then shows that the heat loss is given by

hf 32 μ. L. V.

γ D2.DD10.17

Here the bar over the V has been omitted to conform to the standard practice of denoting the mean velocity in one dimensional flow analyses by V without the bar.

Note

γ Specific Weight

γ ρ g.ρρ N

m3γwater 9799 N

m3γair 11.97 N

m3

S specific gravity and is the ratio of the specific weight of a liquid to the specific weight of water at a standard reference temperature is defined as specific gravity S

μ dinamic viscosity μτ

dVdy

ττN s

m2.

υ kinematic vistosity υμ

ρρρ

m2

s

6


10.3 Criterion for Laminar or Turbulent Flow in a Pipe

To predict whether flow will be laminar or turbulent, it is necessary to explore the characteristics of flow in both laminar and turbulent states. Although other scientists before him and sensed the marked physical difference between laminar and turbulent flow, it was Osborne Reynolds (32) who first developed the basic laws of turbulent flow. With his analytical and experimental work he showed that the Reynolds numner was a basic parameter relating to laminar as well as turbulent flow. For example, using an experimental apparatus such as that shown in Fig 10.3 , he found that the onset of turbulence in a smooth pipe was related to the Reynolds number (VDρ/μ) in a very interesting way. If the fluid in the upstream reservoir was not completely still or if the pipe had some vibration in it, the flow reservoir was not completely still or if the pipe had some vibration in it, the flow in the pipe as it was gradually increased from a low rate to higher rates was initially laminat but when changed from laminar to turbulen flow at a Reynolds number in the neighborhood of 2100. However, Reynolds found that if the fluid was initially completely motionless and if there was no vibration in the equipment while the flow was increased, it was possible to reach a much higher Reynolds number before the flow became turbulent. He also found that, when going from high velocity turbulent flow to low velocity flow the change from turbulent flow always occurred at a Reynolds number of about 2000.

7


These experiments of Reynolds indicate that under carefully controlled conditions it is possible to have laminar flow in pipes at Reynolds numbers much higher than 2000. However,the slightest disturbances will trigger the onset of turbulences will trigger the onset of turbulence at high values of Re. Because most engineering applications involve some vibration or flow disturbance, it is reasonable to expect that pipe flow will be laminar for Reynolds numbers less than 2000 and turbulent for Reynolds numbers greater than 3000. When Re is between 2000 and 3000, the type of flow is very unpredictable and often changes back and forth between laminar and turbulent states. Fortunately, however, most engineering applications either are not in this range or are not significantly affected by the unstable flow.

Example 10.2 Head loss per 100 m of pipe oil

Oil S 0.85 with a kinematic viscosity υ 6 10 4. m2

sflows in a 15 cm pipe at a rate of

0.020 m3

sWhat is the head loss per 100 m of length of pipe ?

Solution

Q 0.020 m3

sd 0.15 m A π d2.

4A 0.018= m2

V QA

V 1.132=ms

Then

Re V d.

υRe 282.942=

Since the Reynolds number is less than 2000, the flow is laminar. The head loss per L 100 m

is obtained from Eq 10.17:

hL 32 μ. LV.

γ D2.

Here μ

γ

υ

ghence

Then hL 32 υ. L. V.

9.81 d2.hL 9.845= m

The heat loss is hL 9.845= m at L 100= m length .

8


Example 10.3 Flow rate in pipe Kerosene

Kerosene (0 degree Celsius) flows under the action of gravity in the pipe shown, which is 6 mm in diameter and 100 m long. Determine the rate of flow in the pipe.

Solution

Because the pipe diameter is small and because the head producing flow is also quite small it is expected that the velocity in the pipe will be small. Hence it will be initially assumed that the flow is laminar and V^2/2g is negligible. Then, to solve the velocity, we apply the energy equation to the problem. We write this equation between a section at the upstream liquid surface and the outlet of the pipe. Thus we have

p1γ

α1 V12.

2 g.z1 p2

γ

α2 V22.

2 g.z2 32 μ. L. V.

γ D2.

p2p2

With the assumption we have noted, the equation reduces to

0 0 1 0 0 32 μ. L. V.

γ D2.DD

or

32 μ. L. V.

γ D2.1

For 0 degree Ceslsius the viscosity (from Figs A.2 and A.3 in the Appendix) is

9


μ 3.2 10 3. N s.

m2υ 3.9 10 6. m2

s

Then

V 1 γ. D2.

32 μ. L.DD D 6 D D 10 3. D 6 10 3.= m

γ 8010 N

m3L 100 m A π D2.

4A 2.827 10 5.= m2

V 1 γ. D2.

32 μ. L.V 0.028=

ms

Now check Re to see if the flow is laminar and check V2

2 g.to see if it is indeed negligible.

Re V D.

υRe 43.323=

V2

2 9.81.4.042 10 5.= negligible

Therefore , the flow is laminar and the velocity is valid. The discharge is then calculated as follows

Q V A. Q 7.962 10 7.= m3

s

10.4 Turbulent Flow in Pipe

Turbulence and its influence in Pipe Flow

In the preceding section it was pointed out that pipe flow is turbulent when the Reynolds number is larger than approximately 3000. However, to say that the flow is turbulent is only a gross description of it. We can obtain a better "feel" for the flow by exploring the similarities between turbulent flow in a pipe and flow in a turbulent boundary layer and by relating the shear stress in the pipe to the level of turbulence. Once we understand these basic physical relationships , we will be better equipped to proceed to the development of equations for the velocity distribution and the resistance to turbulent flow in pipes.

The similarities between turbulent boundary layer flow and turbulent flow in pipes are many. In fact, it is valid to think of turbulent flow in a pipe as a turbulent boundary layer and by relating the shear stress in the pipe to the level of turbulence. Once we understand these basic physical relationshiips, we will be better equipped to proceed to the development of equations for the velocity distribution and the resistance to turbulent flow in pipes.

10


The similarities between turbulent boundary layer flow and turbulent flow in pipes are many. In fact, it is valid to think of turbulent flow in a pipe as a turbulent boundary layer that has become as thick as the radius of the pipe. With this perspective we realize that flow in a smooth pipe has a viscous sublayer just as a flat boundary layer does. In addition, the velocity gradient in the viscous sublayer will be consistent with the shear stress, as given by τ:=μ*du/dy . However, outside the viscous sublayer the viscous shear stress is ngligible compared with the resistance resulting from turbulence. We have already referred in Chapter 9 to the apparent shear stress τ=-ρn'v' which involves an exchange the momentum, but uts effect is like that of a true shear stress. It is zero at the pipe center and increases to a maximum near the wall, as shown in Fig 10.4. Here it is seen that the apparent shear stress increases linearly almost to the edge of the pipe.This linear change in τapp is in acordance with Eq 10.3 which was developed in Sec 10.1. Near the wall, in the viscous sublayer, τapp reduces to zero because all of the shear stress there is in the form of viscous shear stress.

We have shown that there are indeed many analogies between turnulent boundary layer flow and tubulent flow in pipes. The primary difference is that pipe flow is uniform and boundary layer flow is not. Of course, this difference does not apply near the inlet of the pipe, where the flow is nonuniform.

Velocity Distribution and Resistance in Smooth Pipes

Experiments have shown that, in the viscous sublayer and in turbulent zone near the wall, the velocity distribution equation are of the same form as those for the turbulent layer. This si sfor a smooth pipe.

Figure 10.4 Apparent shear stress in a pipe page 362

uux

ux y.

υ

uxux for 0 y ux.

υ< 5< 10.18

uux

5.75 log ux y.

υ. 5.5uxux for 20 y ux.

υ< 105< 10.19

Figure 10.5 is a plot of Eqs (10.18) and (10.19) as well as an indication of the range of experimental data from various surces. For flow near the center of the pipe, as for flow near the outer limit of the boundary layer, the velocity defect law is applicable, as shown in Fig 10.6. Figure 10.6 also includes the range of experimental velocity data obtained from flow in rough conduits. Again , a power law formula like that for the turbulent boundary layer is applicable everywhere except close to the wall. This formula is

uumax

yr0

myy10.20

Here y is the distance from the wall and m is an empirically determined quantity. Some references indicate that m has a value of 1/7 for turbulent flow. However, Schlichting shows that m varies from 1/6 to 1/10 depending on the Reynolds number . His value for m are given in Table 10.1

Exponents for power law equation and ratio of mean to maximum velocity

11


Re 4 103. m 16.0

VmVmax

0.791VmVmaxVm

Vmax

Re 2.3 104. m 16.6

VmVmax

0.807VmVmaxVm

Vmax

Re 1.1 105. m 17.0

VmVmax

0.817VmVmaxVm

Vmax

Re 1.1 106. m 18.8

VmVmax

0.850VmVmaxVm

Vmax

Re 3.2 106. m 110.0

VmVmax

0.865VmVmaxVm

Vmax

Figure 10.5 Velocity distribution for smooth pipe.

Figure 10.6 Velocity defect law for turbulent flow in smooth and rough pipes

In Chapter 9 the local shear stress on a flat plate was expressed as

τ0 cf ρ. V02

2.cfcf

where cf is a function of the character of flow (laminar or turbulent) and the Reynolds number. For pipe flow it is customary to express τo in a s similar manner, however, we use the mean velocity as the reference velocity, and the coefficient of proportionality is given as f/4 instead of cf. Here f is called the resistance coefficient of friction of the pipe. Thus we have

τ0 f4ρ.

V2

2.ff 10.21

Or because τ0ρ

uxuxux we have

uxV

f8ff

Nothing that τ=τ0 when r=r0 in Eq (10.3) we can eliminate τ0 between Eqs 10.3 and 10.21. Then by integrating between two sections along the pipe, we obtain

h1 h2 f LD. V2

2 g..ff

hf f LD. V2

2 g..ff 10.22

12


where hf is the head loss created by viscous effects and is equal to the change to piezometric head along the pipe. Equation 10.22 is called the Darcy _Weisbach equation.If is named after Henry Darcy, a French engineer of the nineteeth century, and Julius Weisbach, a German engineer and scientist of the same era. Weisbach first proposed the use of nondimensional resistance coeficient, and Darcy carried out numerous tests on water pipes. Brief accounts on their work are given by Rouse and Ince. For laminar flow , it can be easily shown by a simultaneous solution of Eqs (10.17) and (10,22) that the resistance coefficient is given by

f 64Re

10.23

For turbulence flow , analytical and empirical results on smooth pipes yield the following approximate relation relative for f:

1

f2 log Re f.. 0.81

f

1

ffor Re>3000 10.24

Equation 10.24 was first developed by Prandtl

Velocity Distribution and Resistance - Rough Pipes

Numerous tests on flow in rough pipes all show that a semilogarithmic velocity distribution is valid over most of the pipe section (29.34). This relationship is given in the following form:

uux

5.75 log yk

. Byy 10.25

Here y is the distance from the rough wall, k is a measure of the height of the roughness elements, and B is a function of the character of roughness. That is, B is a function of the type, concentration and size variation of the roughness. Research by Reborson and Chen (33) shows that B can be abalytically determined for artificially roughened boundaries. More recent works by Wright and Calhoun , Kumar and Eldridge show that using a numerical approach with the same theory yields solutions for B and f for natural roughness, such as found in rock-bedded streams and commercial pipes when the roughness characteristics are known. Kumar 22 in a recent paper, has developed a simplified analytical method of solution to predict the resistance coefficient of rough boundaries. His method can be applied to both artificial and natural types of roughnesses.

In 1933 Nikuradse carried out a number of tests on flow in pipes that were roughned with uniform sized sand grains. From these tests he found that the value of B with the kinf of roughness was 8.5. Thus, for his testsm Eq (10.25) becomes u/ux=5.75 log (y/ks)+8.5 In Nokurandse's tests the distance y was measured from the geometric mean of the wall surface, and ks was the size of the sand grains.

Nikuradse's tests revealed two very important characteristcs of rough pipe flow. Frist with low Reynolds numbers and with small sized sand grains, the flow resistance is virtually the same as that for a smooth pipe. Second , for high values of the Reynolds number, the resistance coeffcient is solely a function of the relative roughness ks/D. The characteristics are shown in Fig 10.7 where f is plotted as a function of Re for various values of the relative roughness ks/D. The reason the resistance is like that of a smooth pipe for low values of ks/D and Re is that for these conditions the reoughness elements are complerely within the viscous sublayer and hence have negligible influence of the main flow in the pipe.

13


However, at high values of the Reynolds number, the viscous sublayer is so thin that the roughness elements project into the main stream of flow and the flow resistance is determined by the drag of the individual roughness elements. Hence for relatively large values of ks/D and for large Reynolds numbers, the resistance to flow is proportional to V^2 thus f becomes constant for these conditions.

The unform character of the sand grains used in Nikurandse's tests produced a dip in the f versus Re curve (Fig 10.7) before the curve reached a constant value of f. However, tests on commercial pipes where the roughness is somewhat random reveal that no such dip occurs. By plotting data for commerical pipes from a number of sources, Moody developed a design chart similar to that shown in Fig 10.8 on page 368.

Figure 10,7 Resistance coefficient f versus Re for sandroughened pipe.

In Fig 10.8 the variable ks is the variable ks is the symbol used to denote the equivalent sand roughness. That is, a pipe that has the same resistance characteristics at high Re value as a sand-roughned pipe of the same size is said to have a size of roughness equivalent to that of the sand-roughned pipe. Figure 10.9 gives approximate values of ks and ks/D for various kinds of pipes. Both of these figures are used to solve certain kinds of pipe flow problems,

In Fig 10.8 the abscisa (labeled at the bottom) is the Reynolds number Re and the ordinate (labeled at the left) is the resistance coefficient f. Each blue curve is for a constant relative roughness ks/D and the values of ks/D are given on the right at the end of each curve.To find f, given Re and ks/D one goes to the right to find the correct relative roughness curve. Then one looks at the bottom of the chart to find the given value of Re and, with this value of Re, moves vertically apward until the given ks/D curve is reached. Finally, from this point one moves horizontally to the left scale to read the value of f. If the curve for the given value of ks/D is not plotted in Fig 10.8 , then one simply finds the proper posiiton on the graph by interpolation between the ks/D curves that bracket the given ks/D.

For some problems it is convenient to enter Fig 10.8 using a value of the parameter Re f12.

The parameter is useful when hf and ks/D are known but the velocity V is not. Without V the Reynolds number cannot be computed, so f cannot be read by entering the chart with Re and ks/D. But from

hf f LD

. V2

2 g.. for V 2 g. hf. L.( )

12 D

f

12

.and Re V D.

υ one can see that Re can be given

asRe D

23

f12

2 g. hf.

L

12

.

Upon multiplyig both sides of the above equation by f12 we get

Re f12. D

12

υ2 g. hf

L.

12

.Re f12.Re f12.

14


Thus a value of Re f12. can be calculated for this type of flow problem, which in turn, enables us to

determine f directly, using Fig 10.8, where curves of constant Re f12. are plotted slanting from

upper left to the lower right and the values of Re f12. for each line are given at the top of the chart.

There are basically three types of problems involved with uniform flow in a single pipe. These are

1. Determine the head loss, given the kind and size of pipe and the flow rate

2. Determine the flow rate, given the head, kind and size of pipe

3. Determine the size of pipe needed to carry the flow, given the kind of pipe, head, and flow rate.

Example 10.4 Head loss per kilometer pipe cast iron pipe

Water T 20 degree Celsius flows at a rate of Q 0.05 m3

sin a d 20 cm asphalted

cast iron pipe. What is the head loss per kilometer of pipe ?

Solution

d d 10 2. d 0.2= m

First compute the Reynolds number, V D.

υυ kinematic viscosity m2

s

υμ

gm2

sμ dinamic viscosity N s

m2.

Here V QA

Thus A πd2

4. A 0.031= m2

V QA

V 1.592=ms

υ 1.0 10 6. m2

sfrom Table A.5

Then

Re V d.

υRe 3.183 105.=

From Fig 10.9 ksD

0.0006 Then from Fig 10,8 using the values obtained for ksD

and Re

we find f 0.019 Finally the head loss is computed from the Darcy-Weisbach equation

15


L 1 km L L 103. L 1 103.= m

hf f Ld. V2

2 9.8.. hf 12.277= m

The head loss per kilometer is hf 12.277= m

In the first type of problem, the Reynolds number and ksD

are first computed and then f is read

from Fig 10.8 , after which the head loss is obtained by the ise of Eq 10.22

Re f12. D

32

υ

2 g. hf.

L

12

.Re f12.Re f12.

Figure 10.8 Resistance coefficient f versus Re. Reprinted with minor variations.

Figure 10.9 Relative roughness for various kinds of pipe

In the second type of problem ksD

and the value of D32

υ2 g. hf

L.. are computed so that the top

scale can be used to enter the chart of Fig 10.8. Then, once f is read from the chart, the velocity from Eq 10.22 is solved for and the discharge is computed from Q=VA

Example 10.5 The head loss per kilometer of D 20 cm asphalted cast iron pipe is hl 12.2

What is the discharge of water?

Solution

First compute the parameter D32

2 g. hlL.

υ. Assume T 20 degree Celsius so that

D D 10 2. D 0.2= m

L 1000 m υ 1 10 6. m2

s

D32

2 9.81 hlL..

υ. 4.376 104.=

16


From Fig 10.9 ksD

0.0006 Using Fig 10.8 we read f 0.019 We use this f in the Darcy

Weisbach equation to solve for V:

f 0.019= D 0.2= m hl 12.2= mhl f LD. V2

2 g..

L 1 103.= hl 12.2

Vp hl D. 2. 9.81.

LV Vp V 0.219=

ms

Finally we compute the discharge:

Q V A. Q 6.874 10 3.=m3

s

Example 10.4 and 10.5 are good checks on the validity of the methods of solution because their basic data are exactly the same - in one case, the head loss is unknown, in the orther, the discharge is unknown.

In examppel 10.5 the head loss in the pipe was given. Therefore, it was possible to obtain a direct solution by entering Fig 10.8 with a value of

Re f12.

However, many problems for which the discharge Q is desired cannot be solved directly. For example, a problem in which water flows ferom a reservoir through a pipe and into the atmosphere cannot be solved directly. Here part of the available head is lost to friction in the pipe, and part of the head remains as kinetic energy in the jet as it leaves the pipe. Therefore , at the outset one does not known how much head loss occurs in the pipe itself. To effect a solution, one must iterate on f. The energy equation is written and an initial value for f is guessed. Because f tends to a constant value of high values of Re, and "educated" first guess is to use this limiting value of f. Next one solves for the velocity V. With this value of V, one then computes a Reynolds number that makes it possible to determine a better value of f using Fig 10.8 and so on. This type of solution usually converges quite rapidly because f changes more showly than Re. This type of solution usually converges quite rapidly because f changes more slowly than Re. Once f and V have been determined, one calculates the discharge by using the continuity equation,

Example 10.6

Determine the discharge of water through the D 50 cm D D 10 2. D 0.5= m

steel pipe shown in the figure.

17


υ 10 6 m2

sSolution

From Fig 10.8 the value of ks 4.6 10 5. m Therefore ksD

9.2 10 5.ksDksD

m

Now write the energy equation from the rreservoir water surface to the free jet at the end of the pipe

p1γ

V12

2 g.z1 p2

γ

V22

2 g.z2 hLp2p2

z1 60 m z2 40 m

0 0 z1 0 V12

2 g.z2 f L. V22.

D 2. g.V1V1

or

V 2 9.8. z1 z2( )1 200 f.

12

First trial: Assume f 0.020 then V 8.86 ms

and Re 4.43 106.

With Re 4.43 106. and ksD

9.2 10 5.ksDksD

then f 0.012 from Fig 10.8

This f then yields V 10.7 ms

Second trial For V 10.7=ms

Re 5.35 106. and f 0.012

D 50 cm D D 10 2. D 0.5= m

A π D2.

4A 0.196= m2

18


Q V A. Q 2.101=m3

s

In the third type of problem, it is usually best first to assume a value of f and then to solve for D, after which a better value of f is computed based on the first estimate of D. This iterative procedure is continued until a valid solution is obtained . A trial and error procedure is necessary because without D one cannot compute ks/D or Re to enter Fig 10.8.

Example 10.7 Size of asphalted cast iron pipe water

What size of asphalted cast iron pipe is required to carry water at a discharge of Q 3 cfs and with

a head loss of hl 4 ft per L 1000 ft per pipe?

Solution

First assume f 0.015 Then

hf f L. V2.

d 2. g.hf f L.

DQ2

A2. 1

2 g.. hf f l. Q2.

2 g. π

4

2. D5.

ll

or D5 f L. Q2.

0.7852 2 g. hl.( ).

For this example:

f L. Q2.

0.7852 2 9.81. hl.( ).2.791= D 2.791( )

15 D 1.228= ft

Now compute a more accurate value of f

A π

4D2. A 1.184=

ksD

0.0004ksDksD

V QA

V 2.534=fts

υ 1.21 10 5. ft2

sThen Re V D.

υRe 2.571 105.=

From Fig 10.8 f 0.0175 Now recompute D by applying the ratio of f is to previous

19


calculations for D5

D5 0.01750.015

0.85 .( ).D D5

15D5D5 D 1.228= ft

Use a pipe with a 12 inch diameter

Note

If a size that is not available commercially is calculated during design, it is customary practice to choose the next larger available size. The cost will be less that for odd sized piped, and the pipe will be more than large enough to carry the flow.

Explicit Equations for hf, Q and D

In the foregoing discussion, methods were presented by which hf, Q, and D can be calculated. All of these methods involve the use of the Moody diagram (Fig 10.8). With the advent of computers, it is most desirable to be able to solve similar problems without having to resort to the Moody diagram. By utilizing the Calebrook-White formula, from which the Moody diagram was developed, Swamee and Jain developed explicit formulas relating f, hf, Q and D. It is reported that their formulas yield results that deviate no more than 3% from those obtained from the Moody diagram for the following ranges of ks/D and Re

10 5 ksD

< 2 10 2.< and 4 103. Re< 108< The formulas for f and Q are

f 0.25

log ks3.75 D.

5.74

Re0.5

210.26

Q 2.22 D52 g hf.

L.. log ks

3.7 D.1.78 υ.

D32 2 g. hf

L..

. 10.27

They also developed a formula for the explicit determination of D. A modified version of that formula, given by Streeter and Wylie is

D 0.66 ks1.25 L Q2.

g hf.

4.75. υ Q9.4. L

g hf.

5.2.

0.04

. 10.28

If you want to solve the head loss given Q, L, D, ks and υ simply solve for f by Eq 10.26 and compute hf with the Darcy Weisbach equation, Eq 10.22. Straightforward calculations for Q and D can also be made if hf is known. However, the problems involving head losses in addition to hf, an interative solution is required. For computing Q you can assuume an f and solve for Q from the energy equation after substituting Q/A can assume an f and solve for Q from the energy equation after substituting Q/A in the equation.

20


For computing Q, you can assume an f and solve for Q from the energy equation after substituting Q/A in that equation. Then compute Re and use the result in Eq (10.26) to get a better estimate of f and so on, until Q converges analogous to the procedure for determining Q using the Moody diagram. In this case, however Eq 10.26 is substituting for the Moody diagram. Similarly, you can determine D if you are given Q, υ the change in pressure or head, and the geometric configuration.

Example 10.8 Size of asphalted cast iron pipe water

Solve Example 10.5 using Eq 10.27

Solution

From Fig 10,8 ks for asphalted cast-iron pipe is given as ks 1.2 10 4. m From the given conditions

hfL

0.0122hfLhfL

Assume T 20 degree Celsius so υ 10 6 m2

s

Then , using Eq (10.27) we have

D 20 cm D D 10 2. D 0.2= m hf 12.2 m L 1000 m

D - diameter of cast iron pipe L - length of cast iron pipe

υ 1.78 10 6. m2

s

Q 2.22 D52 9.81 hf.

L.. log ks

3.7 D.1.78 υ.

D12 9.81 hf

L..

. Q 0.051=m3

s

Reduction of Head Loss with Polymers

In 1984 Toms reported on the effect on head loss of minute amouns of soluble polymer added to water. These solutions are often called drag reducing agents. Since the hundreds of experiments have confirmed his initial findings. The essential features of this phenomenon are as follows.

1. The head loss of water flow in pipes can be reduced as much as 70% with the addition of a polymer solution in concentrations as low as 5 ppm.2. Significant reductions in head loss occur only for turbulent flow. No reduction in head loss occurs in laminar flow.3.The fluid mechanics of this phenomenon is not yet well understood, however, most experiments indicate that it is related to the turbulence associated with flow very near of the pipe. Some researchers think it is associated with the thickness if the viscous sublayer , whereas at least one study indicates that it is primarily associated with the buffer zone of the boundary layer4. Polymer solutions are more effective on small pipes than on large one

21


Because of the potential for achieving significant economic advantages with reduced head loss, research in this area will undoubtelly by extensive for the foreseable future. For background information on the subject, see references

A practical application of the head loss reduction phenomenon is the use of polymer solutions in the Trans Alaska PIpeline. For more details see reference Other details about the use of drag reducing agents in oil pipelines is given by Lester.

10.5 Flow at Pipe Inlets and Losses From Fittings

In the preceding section, formula were presented that are used to determine the head loss for uniform flow in a pipe. However, pipe systems also include inlets, outlets, bends, and other appurtenance that create additional head losses. Flow separation and the generation of additonal turbulence therefrom usually cause these head losses. In this section we will consider the flow patterns and resulting head losses for some of these flow transitions.

Flow in a PIpe Inlet

If the inlet to a pipe is well rounded, as shown in Fig 10.10 the boundary layer will develop from the inlet and grow in thickness until it extends to the center of the pipe. After that point, the flow in the pipe will be uniform. The length L, of the developing region at the entrance is equal to approximately 0.05*D*Re for laminar flow and approximately 50D for turbulent flow. Velocity and pressure distribution for the inlet region of a pipe with turbulent flow are shown in Fig 10.11

22


The head loss that is produced by inlet, outlets, or fittings is expressed by the equation:

hL K V2

2 g.. 10.29

In Eq 10.29 V is the mean velocity in the pipe and K is the loss coefficient for the particular fitting that is involved. For example, K for a well rounded inlet with flow at high values of Re is approximately 0.10. Hence the head loss for such a transition is quite small compared with that for an abrupt pipe outlet for which K is 1.0.

If the pipe inlet is abrupt, of sharp edged, as in Fig 10.12 separation occurs just downstream of the entrance.

Figure 10.11 Distribution of velocity and pressure in the inlet region of a pipe

23


Hence the streamlines converge and then diverge with consequent turbulence and relatively high head loss. The loss coefiicient for the abrupt inlet is approximately 0.5

Flow through an Elbow

Although the cross sectional area of an elbow may not change from sections to section, considerable head loss is produced because separation occurs near the inside of the bend and downstream of the midsection. Thus when the flow leaves the elbow, the eddies produced by separation create considerable head loss. The approximate flow pattern for an elbow is shown in Fig 10,13.

The loss coefficient for an elbow at high Reynolds numbers depends primarily on the shape of the elbow. If it is a very short-radius elbow, the loss coefficient is quite high. For larger radius elbows, the coefficient decreases until a minimum value is found at an r/d value of about 4.(see Table 10.2).

24


However , for still larger value of r/d, an increase in loss coefficient occurs because, for larger r/d values, the elbow itself is significantly longer than elbows with small r/d values. The greater length creates an additional head loss. The loss coefficients for various types of elbows, and for a number of other fittings and flow transitions , are given in Table 10,2

Example 10.9 Smooth pipe between two reservoir

If oil υ 4 10 5. m2

sS 0.9 flows from the upper to the lower reservoir at a rate of Q 0.028

m3

sin the D 15 cm D D 10 2. D 0.15= m smooth pipe, what is the elevation of the oil

surface of the upper reservoir ?

Solution

Apply the energy equation between the surfaces of upper and lower reservoir:

p1γ

V12

2 g.z1 p2

γ

V22

2 g.z1

n

hLp2p2

p1 0 N

m2V1 0 m

sp2 0 N

m2V2 0 m

s

0 0 z1 0 0 z2 f L. V2.

D 2. g.2 Kb. V2

2 g.. Ke V2

2 g.. KE V2

2 g..

Here Kb , Ke and KE are loss coefficients for bend, entrance, and outlet, respectively. These have values of

25


Kb 0.19 Ke 0.5 KE 1.0 Table 10.2

To determine f, we get Re in order to enter Fig 10.8

Re V D.

υD 0.15= m A π

4D2. m2 A 0.018= m2

But V QA

V 1.584=ms

Then υ 4 10 5. m2

sRe V D.

υRe 5.942 103.=

Now we read f from Fig 10,8 (smooth pipe curve) f 0.035 Then

L 60 7 130 L 197= m

z1 m z2 130 m hl f L. V2.

D 2. 9.81.hl 5.882= m for length head loss

hc Kb V2

2 9.81.. hc 0.024= m bend head loss

he Ke V2

2 9.81.. he 0.064= m entrance head loss

hE KE V2

2 9.81.. hE 0.128= m outlet head loss

z1 z2 hl 2 hc. he hE z1 136.122= m

26


Table 10.2 Loss coeficients for various transitions and fittings

Pipe entrance

hL Ke V2

2 g..

Additional Data

rd

K

0.0 0.5

0.1 0.12

0.03>0.2

27


Exemple Pipe entrance

r 0.5 m

d 1 m rd

0.5= ==> Ke 0.03

V 2 ms

hL Ke V2

2 9.81.. hL 6.116 10 3.= m

Contraction

hL Kc V2

2 g..KcKc

Kc KcD2D1 θ 60 degree θ 180 degree

0.0 0.08 0.50

0.20 0.08 0.49

0.40 0.07 0.42

0.60 0.06 0.27

28


0.80 0.06 0.20

0.90 0.06 0.10

Example Contraction

D1 0.80 m D2 0.5 m θ 60 degree

D2D1

0.625= Kc 0.06

V 2 ms

hL Kc V2

2 9.81.. hL 0.012= m

Expansion

hL KE V2

2 g..

29


D1D2

KE KE

θ 20 degreeθ 180 degree

0.0 1.00

0.20 0.30 0.87

0.40 0.25 0.70

0.60 0.15 0.41

0.80 0.10 0.15

Example Expansion

D1 0.30 m D2 0.5 m D1D2

0.6= θ 20 KE 0.15

V 2 ms

hL KE V2

2 9.81.. hL 0.031= m

30


90 degree miter bend

Without vanes Kb 1.1

With Vanes Kb 0.2

hL Kb V12

2 g..

31


Example 90 degree miter bend

Kb 0.2 V1 2 ms

hL Kb V12

2 9.81.. hL 0.041= m

90 degree smooth bend

rd

Kb

1 0.35

2 0.19

4 0.16

6 0.21

8 0.28

10 0.32

32


Example 90 degree smooth bend

d 0.40 m r 0.2 m

rd

0.5= Kb 0.16 0.21( )2

Kb 0.185=

V 2 ms

hL Kb V2

2 9.81.. hL 0.038= m

Threaded pipe fittings

Globe valve - wide open Kv 10.0

Example globe valve - wide open

V 2 ms

hL Kv V2

2. 9.81.. hL 2.039= m

Angle valve - wide open Kv 5.0

Example angle valve wide open

V 2 ms

hL Kv V2

2. 9.81.. hL 1.019= m

Gate valve - wide open Kv 0.2

Example gate valve - wide open

V 2 ms

hL Kv V2

2. 9.81.. hL 0.041= m

Gate valve - half open Kv 5.6

Example gate valve half open

V 2 ms

hL Kv V2

2. 9.81.. hL 1.142= m

33


Return bend Kb 2.2

Example return bend

V 2 ms

hL Kb V2

2. 9.81.. hL 0.449= m

Tee straight through flow Kt 0.4

Example Tee straight through flow

V 2 ms

hL Kt V2

2. 9.81.. hL 0.082= m

Tee side outlet flow Kt 1.8

Example Tee side outlet flow

V 2 ms

hL Kt V2

2. 9.81.. hL 0.367= m

90 degree elbow Kb 0.9

Example 90 degree elbow

V 2 ms

hL Kb V2

2. 9.81.. hL 0.183= m

45 degree elbow Kb 0.4

Example 45 degree elbow

V 2 ms

hL Kb V2

2. 9.81.. hL 0.082= m

34


Transition Losses and Grade Lines

In Chapter 7 the effect on the energy and hydraulic grade lines (EGL and HGL) of the pipe head loss and head loss at an abrupt expansion were discussed in some detail . However, no mention was made of head loss due to entrance, bends, and other flow transitions.The primary effect, just as for abrupt expansions, is to cause the EGL to drop an amount equal to the head loss produced by that transition.

Generally , this drop occurs over a distance of several diameters downstream of the transition, as seen inFig 10.14. The HGL also drops sharply immediately downstream of the entrance because of the high velocity flow in the contracted portion of the stream. Then , as the turbulent mixing occurring in the mixing process. Thus the EGL at the entrance is steeper than it is farther downstream where the flow has become uniforme. As the additional energy loss from the transition subsides, the EGL takes on the slope created by the head loss of the pipe itself.

Even though many transitions produce grade lines that have interested local details, such as those noted for a sharp edged entrance, it is common as a gross indication to simply show abrupt changes in the EGL and to neglect local departures between the EGL and the HGL owing to local changes in

V2

2 g.Thus , Fig 10.15 is a simplified plot of the EGL and HGL for a pipe with several transitions in it.

35


Example 10.10

Refer to Fig 10.15 The difference in water surface between the reservoirs is Δz 5 m and the

horizontal distance between them is L 300 m Using the explicit formula for f. Eqa 10.26

f 0.25

log ks3.7 D.

5.74

Re0.9

2 determine the size of steel need for a discharge of Q 2 m3

s10.26

when the gate valve is wide open.

Solution

First write the energy equation:

p1γ

V12

2 g.z1 p2

γ

V22

2 g.z2

n

hl

p1 0 N

m2V1 0 m

sp2 0 N

m2

z1 z2 V2

2 g.f L.

DKe Kv KE.

Δz V2

2 g.f L.

DKe Kv KE.

V QA

ms

A π D2.

4m2 V Q

π D2.

4

V 4 Q.

π D2.

Δz 4 Q.

π D2.

2 12 g.. f L.

DKe Kv KE.

Ke 0.5 Kv 0.2 KE 1.0 Table 10,2

Δz 16 Q2.

π2 D4.

12 g.. f L.

DKe Kv KE.

Δz 5= m A 16

π2 2. 9.81.

L. A 24.788= B Ke Kv KE B 1.7=

Δz A 1

D5f.. B Δz B A f

D5.

36


The other equations for solving the problem are 10.26 f 0.25

log ks3.7 D.

5.74

Re0.9

2

V QA

and Re V D.

υ

As we did when using the Moody diagram, we make an initial assumption for D. Next we compute V and Re, after which we compute f from Eq 10.26/ Then we compute D from the energy equation (above). With this calculated value of D we go through the process again to get a better estimate of D, and so on, untilthe change of D is negligible small.

In this example υ 10 6 m2

sand ks 4.6 10 5. m from Fig 10.9

We assume an initial value for D 1 m Then

A π D2.

4A 0.785= m2 Q 2=

m3

sV Q

AV 2.546=

ms

Re V D.

υRe 2.546 106.=

f 0.25

log ks3.7 D.

5.74

Re0.9

2 f 0.012=

With this value of f, we solve the energy equation for D to obtain

Δz B A f.

D5 D A f.

Δz B

15

D 0.308=

Δz 5= m B 1.7= A 0.785=

With D 0.308= m we repeat the computational procedure again and again, until convergence.

Then D 0.308= m

A π D2.

4A 0.074= m2 Q 2=

m3

sV Q

AV 26.93=

ms

Re V D.

υRe 8.281 106.=

f 0.25

log ks3.7 D.

5.74

Re0.9

2f 0.013=

37



Δz B A f.

D5 D A f.

Δz B

15

D 0.197= m

Δz 5= m B 1.7= A 0.074=

Then D 0.197= m

A π D2.

4A 0.03= m2 Q 2=

m3

sV Q

AV 65.659=

Re V D.

υRe 1.293 107.=

f 0.25

log ks3.7 D.

5.74

Re0.9

2 f 0.014=


Δz B A f.

D5 D A f.

Δz B

15

D 0.167= m

Then

A π D2.

4A 0.022= m2 Q 2=

m3

sV Q

AV 90.777=

Re V D.

υRe 1.52 107.=

f 0.25

log ks3.7 D.

5.74

Re0.9

2 f 0.015=


Δz B A f.

D5 D A f.

Δz B

15

D 0.158= m

38


Then D 0.158= m

A π D2.

4A 0.02= m2 Q 2=

m3

sV Q

AV 102.024=

Re V D.

υRe 1.612 107.=

f 0.25

log ks3.7 D.

5.74

Re0.9

2 f 0.015=


Δz B A f.

D5 D A f.

Δz B

15

D 0.155= m

Then D 0.155= m

A π D2.

4A 0.019= m2 Q 2=

m3

sV Q

AV 106.401=

Re V D.

υRe 1.646 107.=

f 0.25

log ks3.7 D.

5.74

Re0.9

2 f 0.015=


Δz B A f.

D5 D A f.

Δz B

15

D 0.154= m

10.6 Pipe systems

Simple Pump in a Pipeline

We considered a number of pipe flow problems in which the head for producing the fow was explicitly given.

39


Now we shall consider flow in which the head is developed by a pump.However, the head produced by a centrifugla pump is a function of the discharge. Hence a direct solution is usually not immediately available. The solution (that is, the flow rate for a given system) is obtained when the system equation (or curve) of head versus discharge is solved simultaneously with the pump equatin (or curve) of head versus discharge. The solution of these two equations (or the point where the two curves intersect) yields the operating confition for the system. Consider flow of water in the system in Fig 10.16. When the energy equation is written from the reservoir water surface to the outlet stream, we obtain the following equation:

P1γ

V12

2 g.z1 hp p2

γ

V22

2 g.z2

n

KL V2

2 g..

n

f L V2.

D 2. g..

Figure 10.17 Pump and system curve

For a system with one size of pipe , this simplifies to

hp z2 z1( ) V2

2 g.1

n

KL f L.

D.

nn10.30

Hence , for any given discharge, a certain head hp, must be supplied to maintain the flow. Thus we can consider a head versus discharfe curve, as shown in Fig 10.17. Such a curve is called the system curve. Any given centrifugal pump has a head versus discharge curve that is characteristic of that pump at a given pump speed. Such curves are supplied by the pump manufacturer, a typical one for a centrifugal pump is shown in Fig 10.17.

40


Figure 10.17 reveals that , as the discharge increases in a pipe, the head required for flow also increases. However, the head that is produced by the pump decreases as the discharge increases. Consequently, the two curves intersect, and the operating point is at the point of intersection that point where the head produced by the pump is just the amount needed ro overcome the head loss in the pipe.

Example 10.11 Discharge in the system

What will be the discharge in the water system if the pump has characteristics shown in Fig 10.17.

Assume f 0.015

Solution

First we write the energy equation from water surface to water surface:

p1γ

V12

2 g.z1 hp p2

γ

V22

2 g.z2

n

hL

p1 0 N

m2V1 0 m

sp2 0 m

sV2 0 m

s z1 200 m

z2 230 m

0 0 z1 hp 0 0 z2 f L.

DKe Kb KE V2

2 g..

Here Kb, Ke and KE are coefficients for bend, entrace, and outlet, respectively. (Table 10.2)

Here Ke 0.5 Kb 0.35 and KE 1.0 Hence

hp z2 z1 Q2

A2 2. g.f L.

DKe Kb KE.

A1 z2 z1 A1 30= m L 1000 m

41


D 40 cm D D 10 2. D 0.4= m

A πD2

4. A 0.126= m2

A2 1

A2 2. 9.81.

f L.

DKe Kb KE. A2 127.007=

hp A1 A2 Q2.

Now we make a table of Q versus hp (see below) to give values to produce a

Q m3

sQ2 m6

s2127 Q2. hp A1 A2 Q2.

0 0 0 30

0.1 1 10 2 1.3 31.3

0.2 4 10 2. 5.1 35.1

0.3 9 10 2. 11.4 41.4

system curve that will be plotted with the pump curve. What the system curve is plotted on the same graph as the pump curve, it is seen (Fig 10.17) that the operating condition occurs at

Q 0.27 m3

s

42


PIpes in Parallel

Consider a pipe that branches into two parallel pipes and then rejoins, as shown in Fig 10.18. A problem involving this configuration might be to determine the division of flow in each pipe, given the total flow rate.

scale

No matter which pipe is involved, the pressure difference between the two junction points is the same.

Because hL p1γ

z1 p2γ

z2 it follows the hL between the two junction points in the same

in both of the pipes of the parallel pipe system. Thus we can write

hL1 hL2

f1 L1D1. V12

2 g.. f2 L2

D2. V22

2 g..

Then V1V2

2 f2 L2. D1.

f1 L1. D2.or V1

V2f2 L2. D1.

f1 L2. D2.

12f2f2

If f1 and f2 are known, the division of flow can be easily determined. However, some trial and error analysis may be required if f1 and f2 are in the range where they are functions of the Reynolds number.

43


Pipe Networks

The most common pipe networks are the water distribution systems for municipalities. These systems have one or more sources (discharges of water into the system) and numerous loads: one for each household and commercial establishment. For purposes of simplification, the load are usually lumped throughout the system. Figure 10.19 shows a simplified distribution system with two sources and seven loads.

The engineer is often engaged to design the original system or to recommend an economical expansion to the network. An expansion may involve additional housing or commercial developments, or it may be to handle increased loads within the existing area.

If the design of such a system, the engineer will have to estimate the future loads for the system and will need to have sources (wells or direct pumping from streams or lakes) to satisfy the loads.Also , the layout of the pipe network must be made (usually parallel to streets) and pipe sizes will have to be determined.

The object of the design is to arrive at a network of pipes that will deliver the design flow at the design pressure for minimum cost. The cost will include first costs (materials and construction) as well as maintenance and operating costs. The design process usually involves a number of iterations on pipe sizes and layouts before the optimum design (minimum cost) is achieved.

44


So far as the fluid mechanics of the problem is concerned, the engineer must determine pressures throughout the network for various conditions - that is, for various combinations of pipe sizes, sources, and loads. The solution of a problem for a given layout and a given set of sources and loads requires the two conditions be satisfied:

1. Continuity must be satisfed, That is, the flow into a junction of the network must equal the flow out of the junction. This must be satisfied for all junctions.

2. The head loss between any two junctions must be the same regardless of the path in the series of pipes taken to get from one junction point to the other. This requirement results because pressure must be continuous throughout the network (pressure cannot have two values at a given point). This condition leads to the conlcusion that the algebraic sum of the head losses around a given loop must be equal to zero. Here the sign (positive and negative) for the head loss in a given pipe is given by the sense of the flow with respect to that loop, that is, whether the flow has a clockwise or counterclockwise sense.

Only a few years ago, these solutions were made by trial and error hand computation, but modern applications using digital computers have made the older methods obsolete. Even with these advances, however, the engineer changed with the design or analysis of such a system must understand the basic fluid mechanics of the system to be able to interpret the results properly and to make good engineering decisions based on the results. Therefore, an understanding of the original method of solution by Hardy Cross may help you to gain this basic insight. The Hardy Cross method is as follows.

The engineer first distributes the flow throughout the network so that loads at various nodes are satisfied. In the process of distributing the flow through the pipes of the network, the engineer must be certain that continuity is satisfyied at all junctions (flow into a junction = flow out of the junction), thus satisfying requirement 1. The first guess at the flow distribution obviously will not satisfy requirement 2 regarding head loss therefore , corrections are applied. For each loop of the network , a discharge correcion is applied to yield a zero net head loss around the loop. For example, consider the isolated loop in Fig 10.20 . In this loop, the loss of head in the clockwise sense will be given by

i

hL hLAB hLBCLABLAB

i

hL

i

k Qcn.

ii

10.31

45


The loss of head for the loop in the counterclockwise sense is

n

hLcc

cc

k Qccn. 10.32

For a solution , the clockwise and counterclockwise head losses have to be equal or,

n

hLc

n

hLcc

n

k Qcn.

n

k Qccn.

As we noted, the first guess for flow in the network is undoubtedly in error, therefore , a correction in discharge ΔQ will have to be applied to satisfy the head loss requirement. If the clockwise head loss is greater than the counterclockwise head loss, ΔQ will have to be applied in the counterclockwise sense.

46


That is, subtract ΔQ from the clockwise flows and add it to the counterclockwise flows:

n

k Qc ΔQ( )n

n

k Qcc ΔQ( )n.

nn

10.33

Expanding the summation on either side of Eq 10.33 and including only two terms the expansion, we obtain:

n

k Qcn n Qcn 1. ΔQ..

n

k Qccn n Qccn 1. ΔQ..

nn

Then solving for ΔQ we get

10.34ΔQ i

k Qcn.

i

k Qccn.

i

n k. Qcn 1.

i

n k. Qccn 1.

ii

Thus if ΔQ as computed from Eq (10.34) is positive, the corection is applied in a counterclockwise sense (add ΔQ to countercloskwise flows and subtract it from clockwise flows)

A different ΔQ is computed for each loop of the networkand applied to the pipes. Some pipes will have two ΔQ's applied because they will be common to two loops. The first set of correctiosn usually will not yield the final desired result because the solution is approached only by successive approximations. Thus the corrections are applied successively until the corrections are negiglible. Experience has shown that for most loop configurations, applying ΔQ as computed by Eq (10.34) produces too large a correction. Fewer trials are required to solve for Q's if approximately 0.6 of the computed ΔQ is used.

For information on more modern methods of solutions of pipe network problems, see reference

10.7 Turbulent Flow in Noncircular Conduits

Basic Development

Earlier in this chapter (See 10.4) τ0 was eliminated between Eq (10.3) and (10.21) to yield the Darcy Weisback equation Eq 10.22 . It should be noted that Eq (10.3) was derived by writing the equiation of equilibrium in the longitudinal direction for an element of fluid with a circular cross section. If one derives an equation analogous to Eq 10.3 for flow in a noncircular conduit in which the shear stress acts on the conduit surface having a perimeter P (such as the perimeter of a rectangular conduit) instead of perimeter 2πr, then τ0 is given by

τ0 AP

dds

p γ z.( ).PP

10.35

47


In Eq 10.3 to which Eq 10.35 is analogous , the shear stress τ0 was everywhere constant around the perimeter of the cylindrical element. In Eq 10.35 for the noncircular conduit, the shear stress is not constant over the perimeter. However, we can still use Eq 10.21 to relate τ0 and V, where τ0 is now the average shear stress on the boundary.Eliminating τ0 between Eqs 10.21 and 10.35 and integrating along thge pipe yield

hf f4

LAP

. V2

2 g.. 10.36

In Eq 10.36 hf is the head loss between two points in the conduit, L is the length between the points, and P is the wetted perimeter of the conduit. Thus Eq 10.36 is the same as the Darcy Weisbach equation except that D is replaced by 4A/P. The ratio of the cross sectional area A to the wetted perimeter P is defined as the hydraulic radius Rh. Experiments have shown that we can solve flow problems involving noncircular conduits, such as rectangular ducts, if we apply the same methods and equations that we did for pipes but use 4Rh in place of D. Consequently , the relative roughness is

ks4 Rh.

and the Reynolds number is defined as V 4 Rh.

υ

Example 10.12 pressure drop on steel rectangular

Air T 20 degree C and p1 101 kPa absolute flows at a rate of Q 2.5 m3

sin a

commercial steel rectangular duct b 30 cm and h 60 cm .

What is the pressure drop per L 50 m of the duct ?

b b 10 2. b 0.3= m h h 10 2. h 0.6= m

Solution

First compute Re and ks4 Rh.

: Re V 4. Rh.

υ

Here V QA

A b h. A 0.18= m2

V QA

V 13.889=ms

The hydraulic radius is given by P 2 b h( ). P 1.8= m

Rh AP

Rh 0.1= m

υ 1.51 10 5. m2

sair

Re 4 V. Rh.

υRe 3.679 105.=

48


ks4 Rh.

1.15 10 4.=

Then from Fig 10.8 f 0.015 Thus L 50= m f 0.015=

hf f L. V2.

4 Rh. 2. 9.81.hf 18.435= V 13.889=

ms

Rh 0.1= m

ρ 1.2 kg

m3

Δpf γ hf. Δpf f L.

4 Rh.ρ.

V2

2. Δpf 217.014= Pa

Uniform Flow in Open Channels

You may recall from 4 that for uniform flow the streamlines must be straight and parallel. Therefore, in the case of uniform flow in open channels, the cross-sectional area and shape of the channel will not change from section to section along the channel which also means that from section to section the velocity and depth will not change. Figure 10.21 is an example of uniform flow in a channel having a rectangular cross section. Note here that the velocity varies across the section but does not change from section to section along any given streamline. Uniform flow depth is also called normal depth.

The criterion for determining whether the flow in open channels will be laminar or turbulent is similar to that for flow in pipes. Recall thta in pipe flow, if the Reynolds number (VDρ/μ=VD/υ) is less than 2000, the flow will laminar, and if it is greater than about 3000, one can expect the flow to be turbulent. The Reynolds number criterion for open channel flow is the same if we replace D in the Reynolds number by 4Rh as we did in the Darcy Weisbach equation in the preceding subsection. Thus, we can expect laminar flow to occur in open channels if V(4Rh)/υ<2000.

49


The Reynolds number for open channels is usually defined as VRh/υ. Therefore the open channels, if the Reynolds number is less than 500, we will lanimar flow, and if it is greater than about 750, we xan expect to have turbulent flow. A brief analysis of this turbulent criterion will show that water flow in channels will usually be turbulent unless the velocity and/or the depth is very small. Example 10.13 illustrates this point.

It should be noted that the wetted perimeter used for calculating the hydraulic radius is the perimeter of the channel that is actually in contact with the flowing liquid. For example, in Fig 10.21 the hydraulic radius of this channel of rectangular cross section is

Rh AP

Rh B y.

B 2 y.yy 10.37

One can see that for very wide, shallow channels the hydraulic radius approaches the depth y.

Example 10.13 Velocity for laminar flow at wide rectangular channel

Water T 60 degree F flows in a b 10 ft wide rectangular channel at a depth of

h 6 ft What is the Reynolds number if the mean velocity is V 0.1 fts

?

With this velocity, at what maximum depth can we assume of having laminar flow ?

Solution

Re V Rh.

υ

where V 0.1 ms

A b h. A 60= ft2 P b 2 h. P 22= ft

Rh AP

Rh 2.727= ft

υ 1.22 10 5. ft2

sfrom Table A.5

then Re V Rh.

υRe 2.235 104.=

The maximum Reynolds number at which we can expect to have laminar flow in open channel is

V 0.1=ftsRe 500 Thus , for this limit of Re and for a water velocity of we can solve

for the depth at which this condition will prevail:

50


Re 500

Re V Rh.

υRh Re υ.

VRh 0.061= ft

For rectangular channels Rh b y.

b 2 y.

Rh b 2 y.( ). b y. Rh b. 2 y. Rh. b y.

Rh b. b y. 2 y Rh. Rh b. y b 2 Rh.( ).

y Rh b.

b 2 Rh.y 0.062= ft

Example 10.13 shows that indeed the velocity and/or depth must be very small to yield laminar flow of water in a channel. Note also, the depth and hydraulic radius are virtually the same for this case where the depth is very small relative to the width of the channel.

To determine the head loss for uniform flow in open channels, we utilize Eq 10.36. That is, the Darcy Weisbach equation with D replaced by 4Rh is used.

Example 10.14 Discharge of water in a concrete channel

Estimate the discharge of water that a concrete channel b 10 ft wide can carry if the depth of

flow is h 6 ft and the slop of the channel is S0 0.0016

Solution

We use the Darcy-Weisback equation

hf f L. V2.

4 Rh. 2. g.or hf

Lf V2.

4 Rh. 2. g.

When we have uniform open-channel flow, the slope of the EGL Sf hfL

is the same as the channel

slope S0. Therefore hfL

S0 The foregoing equation then reduces to

V2

2 g.4 Rh S0.

f.

or V 8 g Rh. S0.

f.

Assume ks 0.005 ft Then the relative roughness is

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A b h. A 60= ft2 P b 2 h. P 22= ft

Rh AP Rh 2.727=

ks4 Rh.

4.583 10 4.=

Using ks4 Rh.

4.583 10 4.= as a guide and referring to Fig 10.8 we assume that f 0.016

Thus

V 8 32.2. Rh. S0.

fV 8.382=

fts

ft2

sυ 1.22 10 5.

Then Re V 4. Rh.

υ Re 7.495 106.=

Using the new value of Re and with ks4 Rh.

4.583 10 4.= we read f 0.016

Our initial guess was good; and now that the valocity is known, we can compute Q:

Q V A. Q 502.908= cfs

For rock bedded channels such as those in some natural streams or unlined canals, the larger rocks produce most of the resistance to flow, and essentially none of this resistance is due to the viscous effects. Thus the friction factor is independent of the Reynolds number. For this type of channel , Limerins has shown that the resistance coefficient f can be given in terms of the size of rock in the stream beam as

f 1

1.2 2.03 log Rhd84

.2

10.38

where d84 is a measure of the rock size

52


Example 10.15 Resistance coefficient f for a natural rock bedded channel

Determine the value of the resistance coefficient, f, for a natural rock bedded channel that si

b 100 ft wide has an average depth of h 4.3 ft The d84 size of the boulders in the stream

bed is d84 0.72 ft

Solution

This is a fairly wide channel relative to its depth, therefore Rh is approximated by its depth depth of

Rh 4.3 ft

f 1

1.2 2.03 log Rhd84

.2

f 0.13=

The Chezy Equation and Manning Equation

Leaders in open channel research have recommended the adoption of the methods already presented (involving the Reynolds number and relative roughness) for channel design. However, many engineers continue to use the older , more traditional methods for designing open channels. Because of this current use and for historic reasons, the Chezy equation and Manning equation are presented here.

If Eq 10.36 is rewritten is slightly different form, we have

hfL

f4 Rh.

V2

2 g.. 10.39

For uniform flow, the depth - called normal depth - is constant. Consequently, hf/L is the slope S0 of channel and, Eq 10.39 can be written as

Rh S0. f8 g.

V2. or V C Rh S0.. 10.40

where C 8 g.

f10.41

Since Q V A. we can express the discharge in a channel as

Q C A. Rh S0.. 10.47

This equation is known as the Chezy equation adter a French engineer of that name. For practical application the coefficient C must be determined , and the usual design formula for C in the SI system of units is given as

53


C Rh16

nnn10.43

When we insert this expression for C into Eq 10.42, we obtain a common from of the discharge equation for unform flow in open channels for SI units:

Q 1.0n

A. Rh23. Sh

12. 10.44

In Eq 10.44 n is a resistance coefficient called Manning's , which has different values for different types of boundary roughness. Table 10,3 gives n for various types of boundary surfaces. The major limitation of this approach is that the viscous or relative roughness effects are not present in the design formula. Hence, application outside the range of normal - sized channels carrying water in not recommended,

Manning Equation - Traditional System of Units

It can be shown that, in converting from SUI to the traditional system of units, one must apply a factor equal to 1.49 if the same value of n is used in the two systems. Thus in the traditional system the discharge equation using Manning's n is given as

Q 1.49n

A. Rh23. S0

12.

nn10.45

Table 10.3 Typical values for the roughness coefficient n

Lined Canals n

Cement plaster 0.011

Untreated gunite 0.016

Wood, planed 0.012

Wood, unplaned 0.013

Concrete , troweled 0.012

Concrete, wood forms, unifinished 0.015

Rubble in cement 0.020

Asphalt smooth 0.013

Asphalt rough 0.016

Corrugated metal 0.024

54


Unlined Canals

Earth straight and uniform 0.023

Earth winding and weedy banks 0.035

Cut in rock , straight and uniform 0.030

Cut in rock. jagged and irregular 0.045

Natural Channels

Gravel beds, straight 0.025

Gravel beds plus large boulders 0.040

Earth , straight, with some grass 0.026

Earth , winding no vegetation 0.030

Earth , winding weedy banks 0.050

Earth , very weedy and overgrown 0.080

Example 10.16 Value of Manning's for this channel

If the channel of Example 10.15 has a slope of S0 0.0030 what is the discharge in the channel

and what is the numerical value of Manning's n for this channel ?

Solution

For Example 10.15 we have an f 0.130 and the approximate value of Rh 4.3 ft

In Example 10.14 it was shown that

V 8 32.2.

fRh S0.. V 5.056=

fts

L 100 ft h 4.3 ft A L h. A 430= ft2

But Q V A. Q 2.174 103.= cfs

We solve for n using Eq 10.41 . Note: we are in the traditional system of units, therefore , we use Eq 10.45 rather than Eq 10.44

55


n 1.49Q

A. Rh23. S0

12. n 0.043=

Exanoke 10.17

Using the Chezy equation with Manning's compute the discharge in the channel described in Example 10.14.

Solution Q 1.49n

A. Rh23. S0

12.

For Example 10.14

Rh 2.73 ft A 60 ft2

S0 0.0016

Rh23 1.953= S0

12 0.04= n 0.015 Table 10.3

Then Q 1.49

nA. Rh

23. S0

12. Q 465.673= cfs

The two results (Example 10.14 and 10.17) are within expected engineering accuracy for this type of problem. For a more complete discussion of the historical development of Manning's equation and the choice of n values for use in design or analysis. refer to Yen and Chow.

Best Hydraulic Section

The quantity A R23. in Manning's equation Eqs 10.44 and 10.45 is called the section factor in which

R AP

therefore ,the section factor relating to uniform flow is given by A AP

23

.

Thus , for a channel of given resistance and slope, the discharge will increase with increasing cross sectional area but decrease with increasing wetted perimeter P. For a given area A, and a given shape of channel, for example, rectangular cross section, there will be a certain ratio of depth to width (y/B) for which the section factor will be maximum. Such a ratio establishes the best hydraulic section. That is, the best hydraulic section is the channel proportion that yields a minimum wetted perimeter for a given cross sectional area.

56


Example 10.19 Best hydraulic section for a rectangular channel

Determine the best hydraulic section for a rectanguar channel.

Solution

For the rectangular channge A B y. and P B 2 y. Let A be constant, then let

minimize P. But

P B 2 y. B Ay

or

P Ay

2 y.

Then, we see that the perimeter varies only with y for the given conditions. If we differentiate P with respect to y and set the differential equal to zero, we will have the condition for minimizing P for the given area A:

dPdy

A

y22 dP

dy0

or A

y22 But A B y. so

B y.

y22 B

y2 B 2 y. y 1

2B.

Thus , the best hydraulic section for a rectangular channel occurs when the depth in one half the width of the channel.

It can be shown that the best hydraulic section for a trapezoidal channel is half a hexagon , for the circular section, it is the half circle, and for the triangular section, it is half of the square. Of all the various shapes, the half circle has the best hydraulic section.

The best hydraulic section can be relevant to the cost of the channel. For example, if a trapezoidal channel were to be excavated and if the water surface were to be at adjancent ground level, the minimum amount of excavation (and excavation cost) would result if the channel of best hydraulic section were used.

Flood Flows

When a river is in flood stage, there is flow in the main channel of the river and also in the part of the flood plain than is flooded. The part of the flood plain over which flow is occurring is called the overbank area (see Fig 10.22).

57


The channel characteristics of the main channel and the overbank area are distinctly different. For example, the depth in the main channel is much greater than in the overbank area. Also, the overbank area is often covered with vegetation such as crops, brush, and trees, whereas the bed of the main channel may be of rock. Thus, the resistance to flow in the overbank area is usually much greater than in the main channel. See Table 10.4 for typical n values for overbank areas. Because of different resistance characteristics of the different channel parts, it is customary to consider each part separately when making discharge calculations. The procedure is explained in the next paragraph.

The total discharge for the river in flood stage will be given as

QT Ql QmQlQl 10.46

where

Ql = discharge in overbank area

Qm = discharge in the main channel

However , Q can be given by the Chezy equation Eq 10.42 therefore Eq 10.46 can be written as

QT Cl Al. Rh1 S01.. Cm Am. Rhm S0m..ClCl 10.47

The slope S01 and S0m will be equal because a line following the water surface across the channel is assumed to be horizontal and continuous over the main channle as well as the overbank area, therefore, the change in elevation of the water surface between given cross sections will be the same for both the main channel and the overbank area.

Because the change in elevation of the water surface between given cross section will be the same for the both the main channel and overbank area.

58


Because the change in elevation of the water surface for the main and overbank channels will be the same , the slopes of the water surfaces for both will be equal. Then Eq 10.47 may be given as

QT Cl Al. Rhl S0.. Cm Am. Rhm S0.. 10.48

QT Cl Al Rhl. Cm Am. Rhm. S0.

QT S0

n

C A. Rh.. 10.49

If we are using traditional units (ft-lb-s) Eq 10.48 will be given as

QT 1.49 S0.

n

1n

A. Rh23.. 10.50

The corresponding equation for the SI system is

QT S0

n

1n

A. Rh23.. 10.51

When using Eq 10.50 and 10.51 certain decisions must be made regarding the demarcation between the main channel and the overbank area. It is customary to simply assume a vertical separation plane between them, such as is shown by the vertical dashed line in Fig 10,22. The placement of this plane will be an arbitraru decision by the engineer making the calculation. Once the separation plane has been chosen, the areas and wetted perimeters associated with each channel can then be calculated. In regard to the wetted perimeters, it is customary to include only the perimeter in contact with the channel bottoms and walls - the perimeter associated with the "imaginary surface" between the two channels is not included.

Example 10.19

The cross section of a river in flood stage has the dimensions shown below.

59


Estimate the total discharge if Manning's n is 0.05 for the overbank area and 0,03 for the main channel, and the channel slope is 0.00167. Also, what is the mean velocity for each section ?

Solution

In this simplified problem, it is obvious where to locate the separation plane between the main channel and overbank area. Therefore

L1 30.5 m L2 30.5 m h1 3.04 m h2 1.52 m

Am L1 h1. Am 92.72= m2 A1 L2 h2. A1 46.36= m2

Pm L1 h1 h2 Pm 35.06= m

P1 L2 h2 P1 32.02= m

Then Rhm AmPm

Rhm 2.645= m and Rh1 A1P1

Rh1 1.448= m

Solving Eq QT S0

n

1n

A. Rh23.. 10.51

n1 0.03 S0 0.00167 n2 0.05

QT S0 1n1

Am. Rhm23. 1

n2A1. Rh1

23.. QT 290.031=

m3

s

Qm S0 1n1. Am. Rhm

23. Qm 241.538=

m3

s

Q1 S0 1n2. A1. Rh1

23. Q1 48.493=

m3

s

Thus Vm QmAm

Vm 2.605=ms

and V1 Q1A1

V1 1.046=ms

60


In the development of Eqs (10.50) and (10.51) and also in Example 10.19 only one overbank area was considered, but it is easy to envision several oiverbank areas, as shown in Fig 10.23. In any event the application of Eqs (10.50) and (10.51) to a channel with several areas simply involves extra terms in the summation process.

Uniform Flow in Culverts and Sewers

Sewers are conduits that are used to carry domestic, commercial or industrial waste (called sewage) from households, business and factories to sewage disposal sites. These conduits are often circular in cross section, but elliptical and rectangular conduits are also used. The volume rate of sawage varies throughout the day and season, but of course sewers are designed to carry the maximum design discharge flowing full or near full. At discharge less than the maximum , the sewers will operate as open channels.

Sewage usually consists of about 99% water and 1% solid waste. Because most sewage is so dilute, it is assumed that it has the same physical properties as water for purposes of discharge computations. However, if the velocity in the sewer is too small, the solid particles may settle out and cause blockage of the flow. Therefore, sewers are usually designed to have a minimum velocity of about 2 ft/s (0.60 m/s) at maximum flow condition. This condition is met by choosing a slope on the sewer line to achieve the desired velocity.

Example 10.20

A sewer line is to be constructed of concrete pipe and to be laid on a slopeof S0 0.006

61


p p p S0 0.006

If n 0.013 and if the design discharge is Q 110 cfs what size pipe (commercially available)

would you choose for a full flow conditions? What will be the mean velocity in the sewer pipe

for these conditions ? (It should be noted that concrete pipe is readily available in commercial

sizes of 8 in 10 in and 12 in diamater and then in 3 in increments up to 36 in

diameter . For 36 in diameter up to 144 in the sizes are available in 6 in increments .)

Solution

Q 1.49n

A. R23. S0

12.

where Q 110 ft3

sn 0.013 S0 0.006 assume atmospheric pressure along the

pipe)

Then A1 A R23. A1 Q n.

1.49 S012.

A1 12.39= ft83

Bur R AP R

23 A

P

23

ThenA R

23. A

53

P23

= A1 12.39= ft83

For a pipe flowing flow A π D2.

4and P π D. or

πD2

4.

53

π D.( )

23

12.39 ft83

D83 12.39 4

53. π

23.

π

53

D1 A1 453. π

23.

π

53

π

53 D

103.

.453 π

23. D

23.

12.39

D1 12.39 453. π

23.

π

53

D1 39.752= D D138 D 3.979= ft

D D 12. D 47.746= in

62


Use the next commercial size larger, what is D 48 in D D12

D 4= ft

A π D2.

4A 12.566= ft2 for pipe flowing full

Thus V QA

V 8.754=fts

A culvert is a conduit placed under a fill such as a highway embankment. It is used to convey streamflow from the uphill side of the fill to the downhill side. Figure 10.24 shows the essential features of a culvert.Culverts are designed to pass the design discharge without adverse effects on the fill. That is, the culver should be able to convey runoff from a desng storm without overtopping the fill and without erosion of the fill at either the upstream storm without overtopping the fill and without erosion of teh fill at either the upstream or downstream end of the culvert. The design storm, for example, might be the maximum storm that could be expected to occur once in 50 years at the particle site.

The flow in a culvert is a function of many variables, including cross-sectional shape (circular or rectangular), slope, length, roughness, entrance design ,and exit design. Flow in a culvert may occur as an open channel throughout its length, it may occur as a completely full pipe, or it may occur as a combination of both. The complete design and analysis of culverts are beyond the scope of this text; therefore, only a simple example is included here. For more extensive treatment of culverts, please refer to Chow (8) , Henderson (15) and American Concrete Pipe Assoc (1).

63


( ) ( ) p ( )

Example 10.21

A D 54 in diameter culvert laid uncer a highway embankment has a length of L 200 ft

and a slope of S0 0.01 This was designed to pass a 50 year flood flow of Q 225 cfs

under full flow conditions (see the figure below). For these conditions, what head H is required ?

What the discharge is only Q1 50 cfs what will be the uniform flow depth in the culvert?

Assume n 0.012

Solution

For the flood flow Q 225= cfs one must consider the entrance and exit head losses as well as

the head loss in the pipe itself. therefore , use the energy equation to solve this example.

p1γ

V12

2 g.z1 p2

γ

V22

2 g.z2

n

hL

64


Let points 1 and 2 be at the upstream and downstream water surfaces, respectively.

Thus p1 0 p2 0 gage and V1 0 ms

V2 0 ms

Also z1 z2 H

Then we have H

n

hLHH

H = pipe head loss + entrance head loss + exit head loss

H V2

2 g.Ke KE( ). pipe_head_loss H V2

2 g.Ke KE( ). pierderea_de_presiune_teva

Assume Ke 0.50 from Table 10.2

KE 1.00 from Table 10.2

For the pipe lost use Eq 10.45

Q 1.49n

A. Rh23. S0

12. 10.45

where : Q 225 ft2

s

D 54= in A π D2.

4A 2.29 103.= in2

A A

122A 15.904= ft2

P πD12. P 14.137= ft

Rh AP

Rh 1.125= ft

S0 hfL

Then Eq 10.45 is written as

Q 1.49n

A. Rh23. S0

12. 10.45

65


S0 hfL

Q 1.49n

A. Rh23. hf

L

12

.Q 225= cfs

Rh 1.125=

n Q.

1.49 A. Rh23.

hf12

L12

L12 n. Q.

1.49 A. Rh23.

hf12

hf L12 n. Q.

1.49 A. Rh23.

2

hf 2.219= ft

V QA

V 14.147=fts Ke 0.5= KE 1=

Solving for H: hf 2.219= ft

V 14.147=fts

V 14.147=ftsH1 V2

2 32.2.Ke KE( ).

H H1 hf H 6.881= ft

For Q 50 cfs we need to use Eq 10.45:

Q 1.49n

A. Rh23. S0

12.

However , this culvert will flow only partly full with a Q 50 cfs Therefore , the physical relationship

will be as shown below.

66


Thus , if the angle Q is given in degrees , the cross sectional flow area will be given as:

A π D2.

42 Q.

360. D

2

2sin Q( ). cos Q( ).

The wetted perimeter will be P π D. Q180. or

D48

1.125=Rh AP

Rh D4

1 sin Q( ) cos Q( ).

πQ

180.

.

Substituting these relationships for A and Rhh into the discharge equation and solving for Q yields:

67


Q 90 D 54 in Check Rh value:

D D12

D 4.5=D4

1.125= sin Q( ) 0.894=

We modify the Q and we check the Rh:

Rh D4

1sin Q 3.14

180. cos Q 3.14

180..

πQ

180.

. Rh 1.124=

Depth of flow: D D 12. D 54= in

y D2

D2

cos Q π

180.. y 27= in

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Mathcad - Flow in Conduits