Chapter
13 Flow in Closed Conduits Many of the theoretical relations that have been developed in the previous chapters apply to special situations such as inviscid flow, incompressible flow, and the like. Some experimental correlations were introduced in Chapter 12 for turbulent flow in or past surfaces of simple geometry. In this chapter, an application of the material that has been developed thus far will be considered with respect to a situation of considerable engineering engineering importance, importance, namely fluid flow, flow, both laminar and turbulent, turbulent, through closed conduits.
13.1 DIMENSI DIMENSIONA ONAL L ANALYSIS ANALYSIS OF CONDUIT CONDUIT FLO FLOW As an initial approach to conduit flow, we shall utilize dimensional analysis to obtain the significant parameters for the flow of an incompressible fluid in a straight, horizontal, circular pipe of constant cross section. The significant variables variables and their dimensional dimensional expressions expressions are as represented in the following table: Variable Pressure drop Velocity Pipe diameter Pipe length Pipe roughness Fluid viscosity Fluid density
Symbol
Dimension 2
DP
M/Lt L/t L L L M/Lt M/L 3
v
D L e m r
Each Each of the varia variable bless is famil familiar iar,, with with the exc except eptio ion n of the pipe pipe roug roughn hness ess,, symbo symboliz lized ed e. The roughness roughness is included to represent represent the condition condition of the pipe surface surface and may be thought thought of as the characteristic height of projections from the pipe wall, hence the dimension of length. According to the Buckingham p theorem, the number number of independent independent dimensionless dimensionless grou groups ps to be form formed ed with with these these vari variab able less is four four.. If the the core core grou group p cons consis ists ts of the the vari variab ables les v , D, and r , then the groups to be formed are as follows:
¼ p2 ¼ p3 ¼ p4 ¼ p1
168
a
v
Db rc DP
d e f
v
D r L
g
v
Dh ri e
j
v
Dk rl m
13.1
Dimensional Analysis of Conduit Flow
169
Carrying out the procedure outlined in Chapter 11 to solve for the unknown exponents in each group, we see that the dimensionless parameters become
¼ rDP2
p1
v
¼ DL e p3 ¼ D p2
and p4
¼
Dr
v
m
The first p group is the Euler number. As the pressure drop is due to fluid friction, this parameter is often written with DP / r replaced by gh L where h L is the ‘‘head loss’’; thus p 1 becomes h L 2 / g
v
The third p group, the ratio of pipe roughness to diameter, is the so-called relative roughness. The fourth p group is the Reynolds number, Re. A functional expression resulting from dimensional analysis may be written as h L 2 / g
v
¼ f1
L e ; ; Re D D
(13-1)
Experimental data have shown that the head loss in fully developed flow is directly proportional to the ratio L / D. This ratio may, then, be removed from the functional expression, giving h L
L
e f ¼ ; Re 2 2 / g D D
v
(13-2)
The function f2, which varies with the relative roughness and Reynolds number, is designated f , the friction factor. Expressing the head loss from equation (13-2) in terms of f , we have L v 2
¼ 2 f D
h L
f
g
(13-3)
With the factor 2 inserted in the right-hand side, equation (13-3) is the defining relation for f f , the Fanning friction factor . Another friction factor in common use is the Darcy friction factor , f D, defined by equation 13-4.
¼
h L
¼
f D
L
2
v
D 2g
(13-4)
Quite obviously, f D 4 f f . The student should be careful to note which friction factor he is using to properly calculate frictional head loss by either equation (13-3) or (13-4). The Fanning friction factor, f f , will be used exclusively in this text. The student may easily verify that the Fanning friction factor is the same as the skin friction coefficient C f . Our task now becomes that of determining suitable relations for f f from that theory and experimental data.
170
Chapter 13
Flow in Closed Conduits
13.2 FRICTION FACTORS FOR FULLY DEVELOPED LAMINAR, TURBULENT, AND TRANSITION FLOW IN CIRCULAR CONDUITS Laminar Flow Some analysis has been performed already for incompressible laminar flow. As fluid behavior can be described quite well in this regime according to Newton’s viscosity relation, we should expect no difficulty in obtaining a functional relationship for f f in the case of laminar flow. Recall that, for closed conduits, the flow may be considered laminar for values of the Reynolds number less than 2300. From Chapter 8, the Hagen–Poiseuille equation was derived for incompressible, laminar, conduit flow m avg dP ¼ 32 dx D2 v
(8-9)
Separating variables and integrating this expression along a length, L , of the passage, we get P
Z
dP
P0
¼ 32
mv avg
D2
L
Z
dx
0
and DP
¼ 32 m Davg2 L v
(13-5)
Recall that equation (8-9) held for the case of fully developed flow; thus v avg does not vary along the length of the passage. Forming an expression for frictional head loss from equation (13-5), we have L ¼ DrPg ¼ 32 mgravg D2 v
h L
(13-6)
Combining this equation with equation (13-3), the defining relation for f f L v 2
mv avg L
¼ 32 gr D2 ¼ 2 f D
h L
f
g
and solving for f f , we obtain 16 ¼ 16 D mavgr ¼ Re
f f
(13-7)
v
This very simple result indicates that f f is inversely proportional to Re in the laminar flow range; the friction factor is not a function of pipe roughnessfor values of Re < 2300, but varies only with the Reynolds number. This result has been experimentally verified and is the manifestation of the viscous effects in the fluid, damping out any irregularities in the flow caused by protrusions from a rough surface.
Turbulent Flow In the case of turbulent flow in closed conduits or pipes, the relation for f f is not so simply obtained or expressed as in the laminar case. No easily derived relation such as the Hagen– Poiseuille law applies; however, some use can be made of the velocity profiles expressed in Chapter 12 for turbulent flow. All development will be based on circular conduits; thus we
13.2
Friction Factors for Fully Developed Laminar, Turbulent
171
are primarily concerned with pipes or tubes. In turbulent flow a distinction must be made between smooth- and rough-surfaced tubes. Smooth Tubes.
The velocity profile in the turbulent core has been expressed as
þ ¼ 5:5 þ 2:5 ln yþ
v
where the variables
þ v
(12-63)
and yþ are defined according to the relations
þ p
v
v
ffiffi ffi ffi ffi ffiffi ffi ffi ffi p þ
and
t 0 / r
(12-58)
t 0 / r y v
y
(12-60)
The average velocity in the turbulent core for flow in a tube of radius R can be evaluated from equation (12-63) as follows: v avg
A 0 v dA
R ¼ p ffiffi ffi ffi ffi Z p ffiffi ffi ffi ffi ffi þ A
R
t 0 / r
¼
t 0 / r y
2:5 ln
v
0
5:5 2pr dr
p R2
Letting y
¼ R r , we obtain p t / rR 0 þ 1:75 t 0 / r (13-8) avg ¼ 2:5 t 0 / r ln v p The functions t / r and C are related according to equation (12-2). As C and f are
p ffiffi ffi ffi ffi ffiffi ffi ffi ffi p ffiffi ffi ffi ffi
v
ffiffi ffi ffi ffi 0
f
f
f
equivalent, we may write v
p t avg / r ¼
1
ffiffi ffi ffi ffi p ffiffi ffi ffi ffi p ffi ffi ffi ffi ffi p ffiffi ffi ffi ffi ¼ þ 0
f f /2
(13-9)
The substitution of equation (13-9) into equation (13-8) yields 1
f f /2
2:5 ln
R v
v avg
f f /2
1:75
(13-10)
Rearranging the argument of the logarithm into Reynolds number form, and changing to log10, we see that equation (13-10) reduces to 1
p ffiffi ffi ¼ f f
p ffiffi ffi
4:06 log10 Re
f f
0:60
(13-11)
This expression gives the relation for the friction factor as a function of Reynolds number for turbulent flow in smooth circular tubes. The preceding development was first performed by von Ka´ rma´n.1 Nikuradse,2 from experimental data, obtained the equation 1 4:0 log10 Re f f 0:40 (13-12)
p ffiffi ffi ¼ f f
which is very similar to equation (13-11).
1 2
T. von Ka´rma´n, NACA TM 611, 1931. J. Nikuradse, VDI-Forschungsheft , 356 , 1932.
p ffiffi ffi
172
Chapter 13
Flow in Closed Conduits
By an analysis similar to that used for smooth tubes, von Ka´rma´n developed equation (13-13) for turbulent flow in rough tubes 1 D 4:06 log10 2:16 (13-13)
Rough Tubes.
p ffiffi ffi ¼
e
f f
þ
which compares very well with the equation obtained by Nikuradse from experimental data 1 D 4:0 log10 2:28 (13-14)
p ffiffi ffi ¼
e
f f
þ
Nikuradse’s results for fully developed pipe flow indicated that the surface condition, that is, roughness, had nothing to do with the transition from laminar to turbulent flow. Once the Reynolds number becomes large enough so that flow is fully turbulent, then either equation (13-12) or (13-14) must be used to obtain the proper value for f f . These two equations are quite different in that equation (13-12) expresses f f as a function of Re only and equation (13-14) gives f f as a function only of the relative roughness. The difference is, of course, that the former equation is for smooth tubes and the latter for rough tubes. The question that naturally arises at this point is ‘‘what is ‘rough’?’’ It has been observed from experiment that equation (13-12) describes the variation in f f for a range in Re, even for rough tubes. Beyond some value of Re, this variation deviates from the smooth-tube equation and achieves a constant value dictated by the tube roughness as expressed by equation (13-14). The region wherein f f varies both with Re and e / D is called the transition region. An empirical equation describing the variation of f f in the transition region has been proposed by Colebrook. 3 1
p ffiffi ffi ¼ f f
4 log10
D e
þ 2:28 4log10
p ffiffi ffi þ
D / e
4:67
Re
f f
1
(13-15)
Equation (13-15) is applicable to the transition region above a value of ( D / e)/ (Re f f ) 0:01. Below this value, the friction factor is independent of the Reynolds number, and the flow is said to be fully turbulent . To summarize the development of this section, the following equations express the friction-factor variation for the surface and flow conditions specified:
p ffiffi ffi ¼
For laminar flow (Re < 2300)
16 ¼ Re
f f
(13-7)
For turbulent flow (smooth pipe, Re > 3000) 1
p ffiffi ffi ¼ f f
p ffiffi ffi p ffiffi ffi
4:0 log10 Re
0:40
f f
For turbulent flow (rough pipe, (Re > 3000; D / e)/(Re 1
p ffiffi ffi ¼ f f
And for transition flow 1
p ffiffi ffi ¼ f f
3
4 log10
D e
4:0 log10
(13-12)
f f ) < 0:01)
D e
þ 2:28
þ 2:28 4 log10
C. F. Colebrook, J. Inst. Civil Engr. (London) II, 133 (1938–39).
4:67
(13-14)
D / e
p ffiffi ffi þ
Re
f f
!
1
(13-15)
13.3
173
Friction Factor and Head-Loss Determination for Pipe Flow
13.3 FRICTION FACTOR AND HEAD-LOSS DETERMINATION FOR PIPE FLOW Friction Factor A single friction-factor plot based uponequations (13-7), (13-13), (13-14), and (13-15)has been presented by Moody.4 Figure 13.1 is a plot of the Fanning friction factor vs. the Reynolds number for a range of values of the roughness parameter e / D. 0.025 Laminar flow Complete turbulence rough pipe
0.02
0.05 0.04 0.015 0.03 0.025 0.02 0.015 0.01 f
f
, r o t c a f n o i t c a r f g n i n n a F
0.01 0.009
0.008
0.008
0.006
0.007
0.004
0.006
0.003
Uncertain region
0.002 0.0015
0.005
0.001 0.0008 0.0006
0.004
0.0004
, s s e n h g u o r e v i t a l e R
0.0002 0.0001
0.003
5 10–5
0.0025 5 10–6
3 10–5 2 10–5
Smooth pipe ( e = 0) 0.002 1 10–5 68 2 3 4 68 2 3 4 68 2 3 4 68 2 3 4 68 2 3 4 68 103 104 105 106 107 108 Reynolds number = D vavg / n
Figure 13.1 The Fanning friction factor as a function of Re and D / e.
4
L. F. Moody, Trans. ASME , 66 , 671 (1944).
D / e
174
Chapter 13
Flow in Closed Conduits
When using the friction-factor plot, Figure 13.1, it is necessary to know the value of the roughness parameter that applies to a pipe of given size and material. After a pipe or tube has been in service for some time, its roughness may change considerably, making the determination of e / D quite difficult. Moody has presented a chart, reproduced in Figure 13.2, by which a value of e / D can be determined for a given size tube or pipe constructed of a particular material.
0.05 0.04 0.03
0.07 0.06
0.02
0.05 Riveted steel
0.01 0.008 0.006 0.005 0.004 0.003 0.002 D / e , s s e n h g u o r e v i t a l e R
0.001 0.0008 0.0006 0.0005 0.0004 0.0003
0.035 Wood stave
e
2
s e
p 0.025 i
0.02 0.018
e
= 0 . 0 0 3
e
0.016
= 0 . 0 = 0 0 0 6 . 00 1 = 0 . 0 0 0 5 = 0 . 0 0 = 0 1 5 0 . 0 0 0 4
0.014
e
e
p h g u o r , e c n e l u b r u t e t e l p m o c r o f
0.012 f
4
e
0.01
e
= 0 . 0 0 0 , 0 0 5
1
= 0 . 0 3
e
= C 0 . 0 o m A s 0 p h 0 m 8 e r a 5 l c i t e a l d c a s t e e s t l o i r o r w n r o u g h t i r o n
e
0.000,01 0.000,008 0.000,006 0.000,005
e
= 0 . 0 1
D r a w n t u b i n g
0.000,02
0.03
C o n c r e t e
C a s G a l t i v a r o n n i z e d i r o n
0.0002
0.0001 0.000,08 0.000,06 0.000,05 0.000,04 0.000,03
0.04
0.009 0.008
3 4 5 6 8 10 20 30 40 60 80 100 Pipe diameter, D, in in.
200 300
Figure 13.2 Roughness parameters for pipes and tubes. (From L. F. Moody, Trans. ASME , (1944).) Values of e are given in feet.
The combination of these two plots enables the frictional head loss for a length, L , of pipe having diameter D to be evaluated, using the relation 2
¼ 2 f D L g
h L
v
f
(13-3)
13.3
Friction Factor and Head-Loss Determination for Pipe Flow
Recently Haaland5 has shown that over the range 10 8 friction factor may be expressed (within 1.5%) as
1
p ffiffi ffi ¼
3:6 log10
f f
Re 4 104, 0:05 e / D 0, the 10 / 9
þ 6: 9 Re
175
e
3:7 D
(13-15a)
This expression allows explicit calculation of the friction factor.
Head Losses Due to Fittings The frictional head loss calculated from equation (13-3) is only a part of the total head loss that must be overcome in pipe lines and other fluid-flow circuits. Other losses may occur due to the presence of valves, elbows, and any otherfittings that involve a change in the direction of flow or in the size of the flow passage. The head losses resulting from such fittings are functions of the geometry of the fitting, the Reynolds number, and the roughness. As the losses in fittings, to a first approximation, have been found to be independent of the Reynolds number, the head loss may be evaluated as
¼
h L
DP
r
2
v
¼ K 2g
(13-16)
where K is a coefficient depending upon the fitting. An equivalent method of determining the head loss in fittings is to introduce an equivalent length , L eq, such that
¼ 2 f
h L
f
L eq
2
v
D g
(13-17)
where L eq is the length of pipe that produces a head loss equivalent to the head loss in a particular fitting. Equation (13-17) is seen to be in the same form as equation (13-3), and thus the total head loss for a piping system may be determined by adding the equivalent lengths for the fittings to the pipe length to obtain the total effective length of the pipe. Friction loss factors for various pipe fittings
Table 13.1
Fitting
K
L eq / D
Gate valve, 34 open
7.5 3.8 0.15 0.85
350 170 7 40
Gate valve, 12 open
4.4
200
Globe valve, wide open Angle valve, wide open Gate valve, wide open
Gate valve, 14 open Standard 90 elbow Short-radius 90 elbow Long-radius 90 elbow Standard 45 elbow Tee, through side outlet Tee, straight through 180 Bend 8
8
8
8
8
5
20 0.7 0.9 0.4 0.35 1.5 0.4 1.6
S. E. Haaland, Trans. ASME, JFE , 105 , 89 (1983).
900 32 41 20 15 67 20 75
176
Chapter 13
Flow in Closed Conduits
Comparison of equations (13-16) and (13-17) shows that the constant K must be equal to 4 f f L eq / D. Although equation (13-17) appears to be dependent upon the Reynolds number because of the appearance of theFanning friction factor, it is not. Theassumption made in both equations (13-16) and (13-17) is that the Reynolds number is large enough so that the flow is fully turbulent. The friction coefficient for a given fitting, then, is dependent only upon the roughness of the fitting. Typical values for K and L eq / D are given in Table 13.1. Recall that the head loss due to a sudden expansion is calculated in Chapter 6, with the result given in equation (6-13).
Equivalent Diameter Equations (13-16) and (13-17) are based upon a circular flow passage. These equations may be usedtoestimatetheheadlossinaclosedconduitofanyconfigurationifan‘‘equivalentdiameter’’ for a noncircular flow passage is used. An equivalent diameter is calculated according to area of flow ¼ 4 cross-sectional wetted perimeter
Deq
(13-18)
The ratio of the cross-sectional area of flow to the wetted perimeter is called the hydraulic radius. The reader may verify that Deq corresponds to D for a circular flow passage. One type of noncircular flow passage often encountered in transfer processes is the annular area between two concentricpipes. The equivalent diameter for this configuration is determinedas follows:
¼ p4 D20 D2 Wetted perimeter ¼ pð D0 þ D Þ p /4 ð D20 D2 Þ ¼ D0 D Deq ¼ 4 p ð D0 þ D Þ
Cross-sectional area yielding
i
i
i
i
i
(13-19)
This value of D eq may now be used to evaluate the Reynolds number, the friction factor, and the frictional head loss, using the relations and methods developed previously for circular conduits.
13.4 PIPE-FLOW ANALYSIS Application of the equations and methods developed in the previous sections is common in engineering systems involving pipe networks. Such analyses are always straightforward but may vary as to the complexity of calculation. The following three example problems are typical, but by no means all-inclusive, of the types of problems found in engineering practice.
EXAMPLE 1
Water at 59 F flows through a straight section of a 6-in.-ID cast-iron pipe with an average velocity of 4 fps. The pipe is 120 ft long, and there is an increase in elevation of 2 ft from the inlet of the pipe to its exit. 8
Find the power required to produce this flow rate for the specified conditions. The control volume in this case is the pipe and the water it encloses. Applying the energy equation to this control volume, we obtain dQ
dt
dW s
dt
dW m
dt
Z Z ¼ þ ð : Þ r e
c:s:
P r
v n dA
þ
@ @ t
Z Z Z
c:v:
re dV
(6-10)
13.4
Pipe-Flow Analysis
177
An evaluation of each term yields dQ
dW s
¼0 dt
Z Z þ ð : Þ r e
c:s:
P r
v n dA
dt
¼ r A avg v
@ @ t
¼ W _
2 2
þ v
2
Z Z Z
gy2
2 1
P2
P1
v
þ r þ u2 2 gy1 r u1
¼ 0
re dV
c:v:
and dW m
dt
¼0
The applicable form of the energy equation written on a unit mass basis is now
_ / m _ W
2 2 ¼ 1 2 2 þ gð y1 y2 Þ þ P1 r P2 þ u1 u2 v
v
and with the internal energy change written as gh L , the frictional head loss, the expression for w becomes
_ / m W _
2 2 ¼ 1 2 2 þ gð y1 y2 Þ þ P1 r P2 gh v
v
L
Assuming the fluid at both ends of the control volume to be at atmospheric pressure, _ / m (P1 P2 )/ r 0, and for a pipe of constant cross section ( v 21 v 22 )/2 0, giving for W _
¼
_ / m W _
¼
¼ g( y1 y2) gh
L
Evaluating h L , we have 1 4 ð 2Þð Þ Re ¼ ¼ 164; 000 1:22 105 e ¼ 0:0017 (from Figure 14:2) D f ¼ 0:0059 (from equation (14-15a)) f
yielding 2
2
Þð120 ftÞð16ft /s Þ ¼ 1:401 ft ¼ 2ð0:0059 ð0:5 ftÞð32:2 ft/s2Þ
h L
The power required to produce the specified flow conditions thus becomes
"Þ
2 ftÞ 1:401ft ¼ gðð 550ftlbf /hp-s
_ W
¼ 0:300 hp
3
#
62:3 lbm /ft p 2 32:2 lbm ft/s lbf 4
1 ft 2
2
4
ft s
178
Chapter 13
EXAMPLE 2
Flow in Closed Conduits A heat exchanger is required, which will be able to handle 0.0567 m 3 /s of water through a smooth pipe with an equivalent length of 122 m. The total pressure drop is 103,000 Pa. What size pipe is required for this application? Once again, applying equation (6-10), we see that a term by term evaluation gives dQ
dW s
¼ 0 dt
Z Z þ ð : Þ Z Z Z r e
c:s:
@ @ t
P
v n dA
r
dt
¼ r A avg v
2 2
v
2
dW m
¼0
dt
¼0 2 1
P2
v
P1
þ gy2 þ r þ u2 2 gy1 r u1
¼ 0
re dV
c:v:
and the applicable equation for the present problem is 0
¼ P2 r P1 þ gh
L
The quantity desired, thediameter, is included in the head-lossterm but cannot be solvedfor directly, as the friction factor also depends on D. Inserting numerical values into the above equation and solving, we obtain 0
¼
103 000 Pa 1000 kg/m3
þ 2 f
f
2
0:0567 p D2 /4
m2 122 m g s2 D m g
or 0
¼ 103 þ 1:27 Df 5 f
The solution to this problem must now be obtained by trial and error. A possible procedure is the following: 1. Assume a value for f f . 2. Using this f f , solve the above equation for D. 3. Calculate Re with this D. 4. Using e / D and the calculated Re, check the assumed value of f f . 5. Repeat this procedure until the assumed and calculated friction factor values agree.
Carrying out these steps for the present problem, the required pipe diameter is 0.132 m (5.2 in.).
EXAMPLE 3
An existing heat exchanger has a cross section as shown in Figure 13.3 with nine 1-in.-OD tubes inside a 5-in.-ID pipe. For a 5-ft length of heat exchanger, what flow rate of water at 60 F can be achieved in the shell side of this unit for a pressure drop of 3 psi? 8
An energy-equation analysis using equation (6-10) will follow the same steps as in example 13.2, yielding, as the governing equation 0
¼ P2 r P1 þ gh
L
13.5
Friction Factors for Flow in the Entrance to a Circular Conduit
179
1 in.
5 in.
5 ft
Figure 13.3 Shell-and-tube head-exchanger configuration.
The equivalent diameter for the shell is evaluated as follows:
¼ p4 ð25 9Þ ¼ 4p in:2 Wetted perimeter ¼ pð5 þ 9Þ ¼ 14p in: Flow area
thus
¼ 4 144pp ¼ 1:142 in:
Deq
Substituting the proper numerical values into the energy equation for this problem reduces it to 2
0
2
2
: ð144in: /ft Þ ¼ 3 lbf /in þ 2 f 1:94 slugs/ft3
2
f v avg ft
2 2
/s
ð
5 ft g 1:142/12 ft g
Þ
or 0
¼ 223 þ 105 f
2
f v avg
As f f cannot be determined without a value of Re, which is a function of v avg, a simple trial-and-error procedure such as the following might be employed: 1. Assume a value for f f . 2. Calculate
v avg
from the above expression.
3. Evaluate Re from this value of v avg. 4. Check the assumed value of f f using equation (13-15a). 5. If the assumed and calculated values for f f do not agree, repeat this procedure until they do.
Employing this method, we find the velocity to be23.6fps, giving a flow rate for this problem of 2:06ft3 /min 0:058m3 /s : Notice that in each of the last two examples in which a trial-and-error approach was used, the assumption of f f was made initially. This was not, of course, the only way to approach these problems; however, in both cases a value for f f could be assumed within a much closer range than either D or v avg.
ð
Þ
13.5 FRICTION FACTORS FOR FLOW IN THE ENTRANCE TO A CIRCULAR CONDUIT The development and problems in the preceding section have involved flow conditions that did not change along the axis of flow. This condition is often met, and the methods just described will be adequate to evaluate and predict the significant flow parameters.
180
Chapter 13
Flow in Closed Conduits
In many real flow systems this condition is never realized. A boundary layer forms on the surface of a pipe, and its thickness increases in a similar manner to that of the boundary layer on a flat plate as described in Chapter 12. The buildup of the boundary layer in pipe flow is depicted in Figure 13.4.
v
x
Figure 13.4 Boundary-layer buildup in a pipe.
A boundary layer forms on the inside surface and occupies a larger amount of the flow area for increasing values of x , the distance downstream from the pipe entrance. At some value of x , the boundary layer fills the flow area. The velocity profile will not change the downstream from this point, and the flow is said to be fully developed. The distance downstream from the pipe entrance to where flow becomes fully developed is called the entrance length, symbolized as L e. Observe that the fluid velocity outside the boundary layer increases with x , as is required to satisfy continuity. The velocity at the center of the pipe finally reaches a value of 2 v 1 for fully developed laminar flow. The entrance length required for a fully developed velocity profile to form in laminar flow has been expressed by Langhaar 6 according to L e D
¼ 0:0575 Re
(13-20)
where D represents the inside diameter of the pipe. This relation, derived analytically, has been found to agree well with experiment. There is no relation available to predict the entrance length for a fully developed turbulent velocity profile. An additional factor that affects the entrance length in turbulent flow is the nature of the entrance itself. The reader is referred to the work of Deissler 7 for experimentally obtained turbulent velocity profiles in the entrance region of the circular pipes. A general conclusion of the results of Deissler and others is that the turbulent velocity profile becomes fully developed after a minimum distance of 50 diameters downstream from the entrance. The reader should realize that the entrance length for the velocity profile differs considerably from the entrance length for the velocity gradient at the wall. As the friction factor is a function of d v / dy at the pipe surface, we are also interested in this starting length. Two conditions exist in the entrance region, which cause the friction factor to be greater than in fully developed flow. The first of these is the extremely large wall velocity gradient right at the entrance. The gradient decreases in the downstream direction, becoming constant before the velocity profile becomes fully developed. The other factor is the existence of a ‘‘core’’ of fluid outside the viscous layer whose velocity must increase as 6 7
H. L. Langhaar, Trans. ASME , 64 , A-55 (1942). R. G. Deissler, NACA TN 2138 (1950).
13.5
Friction Factors for Flow in the Entrance to a Circular Conduit
o i t a r r o t c a f n o i t 1 c i r F
0
Fully developed laminar flow
/ D x
181
Figure 13.5 Velocity profile and friction-factor variation for laminar flow in the region near a pipe entrance.
dictated by continuity. The fluid in the core is thus being accelerated, thereby producing an additional drag force whose effect is incorporated in the friction factor. The friction factor for laminar flow in the entrance to a pipe has been studied by Langhaar.8 His results indicated the friction factor to be highest in the vicinity of the entrance, then to decrease smoothly to the fully developed flow value. Figure 13.5 is a qualitative representation of this variation. Table 13.2 gives the results of Langhaar for the average friction factor between the entrance and a location, a distance x from the entrance. For turbulent flow in the entrance region, the friction factor as well as the velocity profile is difficult to express. Deissler9 has analyzed this situation and presented his results graphically. Even for very high free-stream velocities, there will be some portion of the entrance over which the boundary layer is laminar. The entrance configuration, as well as the Reynolds number, affects the length of the pipe over which the laminar boundary layer exists before becoming turbulent. A plot similar to Figure 13.5 is presented in Figure 13.6 for turbulent-flow friction factors in the entrance region. Table 13.2 Average friction factor for laminar flow in the entrance to a circular pipe x / D
Re 0.000205 0.000830 0.001805 0.003575 0.00535 0.00838 0.01373 0.01788 0.02368 0.0341 0.0449 0.0620 0.0760
8 9
Op cit. R. G. Deissler, NACA TN 3016 (1953).
f f
x
D
0.0530 0.0965 0.1413 0.2075 0.2605 0.340 0.461 0.547 0.659 0.845 1.028 1.308 1.538
182
Chapter 13
Flow in Closed Conduits
o i t a r r o t c a f n o i t 1 c i r F
Laminar boundary layer 0 0
Fully developed flow (turbulent)
Turbulent boundary layer
Figure 13.6 Velocity profile and friction-factor variation in turbulent flow in the region near a pipe entrance.
x / D
The foregoing description of the entrance region has been qualitative. For an accurate analytical consideration of a system involving entrance-length phenomena, Deissler’s results portrayed in Figure 13.7 may be utilized. 0.07 0.06 0.05 ) ) x 2 0.04 P / –
2
v
1 r
P ( ( x D 4 0.03
0.02
Reynolds number = Dvavg / n 104
0.01
3 104
6 104 105
0 0
2
4
6
8
10
12
14
16
18
20
x / D
Figure 13.7 Static pressure drop due to friction and momentum change in the entrance to a smooth, horizontal, circular tube (Deissler).
It is important to realize that in many situations flow is never fully developed; thus the friction factor will be higher than that predicted from the equations for fully developed flow or the friction-factor plot.
13.6 CLOSURE The information and techniques presented in this chapter have included applications of the theory developed in earlier chapters supported by correlations of experimental data. The chapters to follow will be devoted to heat and mass transfer. One specific type of transfer, momentum transfer, has been considered up to this point. The student will find that he is able to apply much of the information learned in momentum transfer to counterparts in the areas of heat and mass transfer.
Problems
183
PROBLEMS 13.1 An oil with kinematic viscosity of 0 :08 103 ft2 /s and a density of 57 lbm /ft3 flows through a horizontal tube 0.24 in. in diameter at the rate of 10 gal/h. Determine the pressure drop in 50 ft of tube.
commercial steel.The flow rate through thepump is 500gal/min. Use the (incorrect) assumption that the flow is fully developed.
13.2 A lubricating line has an inside diameter of 0.1 in. and is 30 in. long. If the pressure drop is 15 psi, determine the flow rate of the oil. Use the properties given in Problem 13.1.
13.11 The siphon of Problem 6.31 is made of smooth rubber hose and is 23 ft long. Determine the flow rate and the pressure at point B.
13.3 The pressure drop in a section of a pipe is determined from tests with water. A pressure drop of 13 psi is obtained at a flow rate of 28:3 lbm /s. If the flow is fully turbulent, what will be the pressure drop when liquid oxygen ( r 70lbm /ft3 ) flows through the pipe at the rate of 35 lb m /s?
13.12 A galvanized rectangular duct 8 in. square is 25 ft long and carries 600 ft 3 /min of standard air. Determine the pressure drop in inches of water.
¼
13.4 A 280-km-long pipeline connects two pumping stations. If 0:56 m3 /s are to be pumped through a 0.62-m-diameter line, the discharge station is 250m lower in elevation than the upstream station, and the discharge pressure is to be maintained at 300,000 Pa, determine thepower required to pump the oil. The oil has a kinematic viscosity of 4 :5 106 m2 /s and a density of 810kg/m3 . The pipe is constructed of commercial steel. The inlet pressure may be taken as atmospheric.
13.10 The pipe in Problem 6.33 is 35 m long and made of commercial steel. Determine the flow rate.
13.13 A cast-iron pipeline 2 m long is required to carry 3 million gal of water per day. The outlet is 175 ft higher than the inlet. The costs of three sizes of pipe when in place are as follows:
10-in. diameter 12-in. diameter 14-in. diameter
$11.40 per ft $14.70 per ft $16.80 per ft
13.5 In the previous problem, a 10-km-long section of the pipeline is replaced during a repair process with a pipe with internal diameter of 0.42 m. Determine the total pumping power required when using the modified pipeline. The total pipeline length remains 280 km.
Power costs are estimated at $0.07 per kilowatt hour over the 20-year life of the pipeline. If the line can be bonded with 6.0% annual interest, what is the most economical pipe diameter? The pump efficiency is 80%, and the water inlet temperature is expected to be constant at 42 F.
13.6 Oil having a kinematic viscosity of 6 :7 106 m2 /s and density of 801kg/m3 is pumped through a pipe of 0.71m diameter at an average velocity of 1.1 m/s. The roughness of the pipe is equivalent to that of a commercial steel pipe. If pumping stations are 320 km apart, find the head loss (in meters of oil) between the pumping stations and the power required.
13.14 Estimate the flow rate of water through 50 ft of garden hose from a 40-psig source for
13.7 The cold-water faucet in a house is fed from a water main through the following simplified piping system: a. A 160 ft length of 3/4-in.-ID copper pipe leading from the main line to the base of the faucet. b. Six 90 standard elbows. 8
c. One wide-open angle valve (with no obstruction). d. The faucet. Consider the faucet to be made up of two parts: (1) a conventional globe valve and (2) a nozzle having a cross-sectional area of 0.10 in. 2.
8
a. A 1/2-in.-ID hose; b. A 3/4-in.-ID hose.
60 m and h 2 13.15 Two water reservoirs of height h 1 30 m are connected by a pipe that is 0.35 m in diameter. The exit of the pipe is submerged at distance h3 8 m from the reservoir surface.
¼
¼
a. Determine the flow rate through the pipe if the pipe is 80 m long and the friction factor f f 0:004. The pipe inlet is set flush with the wall.
¼
b. If the relative roughness e / D factor and flow rate.
The pressure in the main line is 60 psig (virtually independent of flow),and thevelocity there is negligible. Find themaximum rate of discharge from the faucet. As a first try, assume for the pipe 0:007. Neglect changes in elevation throughout the system. f f
¼
13.8 Water at the rate of 118 ft3 /min flows through a smooth horizontal tube 250 ft long. The pressure drop is 4.55 psi. Determine the tube diameter. 13.9 Calculate the inlet pressure to a pump 3 ft above the level of a sump. The pipe is 6 in. in diameter, 6 ft long, and made of
¼
h1
L =
80 m
¼ 0:004, determine the friction
h3
h2
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Chapter 13
Flow in Closed Conduits
13.16 An 8-km-long, 5-m-diameter headrace tunnel at the Paute river hydroelectric project in Ecuador supplies a powerstation 668m below the entrance of the tunnel. If the tunnel surface is concrete, find the pressure at the end of the tunnel if the flow rate is 90 m 3 /s. 13.17 Determine the flow rate through a 0.2-m gate valve with upstream pressure of 236 kPa when the valve is a. open;
13.24 Water flows at a volumetric flow rate of 0.25 m 3 /s from reservoir 1 to reservoir 2 through three concrete pipes connected in series. Pipe 1 is 900 m long and has a diameter of 0.16 m. Pipe 2 has a length of 1500 m and a diameter of 0.18 m. Pipe 3 is 800 m long and the diameter is 0.20 m. Neglecting minor losses, determine the difference in surface elevations. 13.25 A system consists of three pipes in series. The total pressure drop is 180 kPa, and the decrease in elevation is 5 m. Data for the three pipes are as follows:
b. 1/4 closed; c. 1/2 closed; d. 3/4 closed. 13.18 Water at 20 C flows through a cast-iron pipe at a velocity of 34 m/s. The pipe is 400 m long and has a diameter of 0.18 m. Determine the head loss due to friction.
Pipe
13.19 A 2.20-m diameter pipe carries water at 15 C. The head loss due to friction is 0.500 m per 300 m of pipe. Determine the volumetric flow rate of the water leaving the pipe.
1 2 3
8
8
13.20 Water at 20 C is being drained from an open tank through a cast-iron pipe 0.6 m diameter and 30 m long. The surface of the water in the pipe is at atmospheric pressure and at an elevation of 46.9 m, andthe pipe discharges to the atmosphere at an elevation 30 m. Neglecting minor losses due to configuration, bends, valves, and fittings, determine the volumetric flow rate of the water leaving the pipe.
Length, m
Diameter, cm
Roughness, mm
125 150 100
8 6 4
0.240 0.120 0.200
8
13.21 A 15-cm diameter wrought-iron pipe is to carry water at 20 C. Assuming a level pipe, determine the volumetric flow rate at the discharge if the pressure loss is not permitted to exceed 30.0 kPa per 100 m. 8
13.22 A level 10-m-long water pipe has a manometer at both the inlet and the outlet. The manometers indicate pressure head of 1.5 and 0.2 m, respectively. The pipe diameter is 0.2 m and the pipe roughness is 0.0004 m. Determine the mass flow rate in the pipe in kg/s. 13.23 Determine the depth of water behind the dam in the figure that will provide a flow rate of 5 :675 104 m3 /s through a 20-m-long, 1.30 cm commercial steel pipe.
13.26 Two concrete pipes areconnected in series. Theflow rate of water at 20 C through the pipes is 0.18 m 3 /s, with a total head loss of 18 m for both pipes. Each pipehas a length of 312.5 m and a relative roughness of 0.0035 m. Neglecting minor losses, if one pipe has a diameter of 0.30 m, determine the diameter of the other. 8
13.27 A 0.2-m-diameter cast-iron pipe and a 67-mmdiameter commercial steel pipe are parallel, and both run from the same pump to a reservoir. The pressure drop is 210 kPa and the lines are 150 m long. Determine the flow rate of water in each line. 13.28 A system consists of three pipes in parallel with a total head loss of 24 m. Data for the three pipes are as follows:
Pipe 1 2 3
h
Length, m 100 150 80
Diameter, cm 8 6 4
Roughness, mm 0.240 0.120 0.200
1.30 cm
For water at 20 C, neglect minor losses and determine the volumetric flow rate in the system. 8
L
