2nd – Year Fluid Mechanics, Faculty of Engineering and Computing, Curtin University
FLUID MECHANICS 230
For Second-Year Chemical, Civil and Mechanical Engineering
FLUID MECHANICS LECTURE NOTES
CHAPTER 6 VISCOUS FLOW IN PIPES
6.1 Introduction
In this chapter, we will consider viscous incompressible flow in pipes where the fluid is confined and bounded. Pipe systems are widely used in practice. Typical examples include drinking water distribution pipe systems, oil pipe lines etc. Figure 6-1 presents the schematic diagram of a typical pipe system.
Figure 6-1 A typical pipe system [1]
As shown in Figure 6-1, a pipe system may include 1. Individual straight pipes
2. Pipe connectors (such as Tee-union, elbow connector etc.), for connecting pipes
3. Flow rate control devices (such as valves) for adjusting the flow rate
4. Inlet and outlet
5. Pumps which add energy into the fluid
where items 2, 3 and 4 are often called pipe components .
6.2 Real pipe flow
For inviscid, incompressible, steady and irrotational flows, Bernoulli’s Equation applies
streamline a along const V gz P ,22
=++ρρwhich can also be written as 0)2
(2
=++ΔV gz P ρρ. (E6-1)
Equation E6-1 indicates that for ideal fluid (0=μ), if there is no change in the sum of the fluid elevation and dynamic pressure, i.e.0)2()(2=Δ+ΔV gz ρρ, the overall pressure drop across a pipe system should be zero.
However, for real fluid, energy loss hence pressure drop L P Δ across a pipe system is inevitable because of the shear force (friction) between the pipe and the fluid. This friction is a result of the viscous nature of real fluid (0≠μ). To deliver the fluid through a pipe system, pumps are often introduced into the pipe system in order to provide pressure rise work P Δ. Therefore, E6-1 is no longer applicable to real pipe flows and we need to consider both L P Δ and work P Δ, i.e.
0)2
(2
=Δ?Δ++
+ΔL work P P V gz P ρρ. (E6-2)
The pressure drop L P Δ can be due to pressure drop in straight pipes (Item 1 in Figure 6-1), called major loss , and pressure drop in pipe components (Items 2, 3 and 4 in Figure 6-1), called minor loss . Therefore, we have or L major L L P P P min ,,Δ+Δ=Δ. (E6-3)
Equation E6-2 shows that to deliver fluid through a pipe system, a pump needs to be properly selected to provide enough pressure rise to overcome the pressure drop due to friction in the pipe system, the changes in elevation and the dynamic pressure. It indicates that to design and operate a real pipe system, we need to have sufficient knowledge on:
1. the relationship among pressure drop across a straight pipe (i.e. major loss ), the pipe properties and flow properties (This is discussed in Chapter 6)
2. the relationship among pressure drop across pipe components (i.e. minor loss ), the properties of the pipe component and flow properties (This is discussed in Chapter 7)
3. the energy gain by devices such as pumps (This is the topic of Chapter 10)
6.3 Pioneer work on measuring pressure drop across a pipe 6.3.1 Pressure-drop test [2]
Figure 6-2 illustrates is a simple experimental setup for measuring pressure drop across a pipe. Liquid flew from the tank (by elevation energy) to the pipe. There would be a long section where the flow was not uniform, before the fluid entering the test section to produce a
uniform flow. The test section has a length of x Δ
and the pressures at both ends were measured as P 1 and P 2.
A flow-regulating valve was introduced to control the flow rate so that tests could be conducted to find the correlation between the pressure gradient across the test section and the fluid flow rate in pipe. Figure 6-3 presents the original experimental results of Osborne Reynolds in 1883 of one specific fluid and one specific pipe [3]. Extensive experiments showed that these observations are generic for pipe flow, regardless of the type of liquid and kind of pipe used in such experiments.
Figure 6-2 Experimental setup for pressure-drop test [2].
Figure 6-3 Measured pressure gradient
x P
ΔΔ (i.e.x
P P Δ?12) of a specific pipe as a function of volumetric flow rate Q of a specific fluid [2], originally from reference [3].
Depending on the flow rate Q , Figure 6-3 shows that the measured pressure gradient of the test section may fall into one of the three regions:
Region I : At a low flow rate, the pressure gradient is proportional to flow rate;
Region II : At an intermediate flow rate, the flow behaviour seemed to be unpredictable so that the experimental data are scattered within the regions of the two curves which are extrapolated from those in Region I and III.
Region III : At a high flow rate, the pressure gradient is proportional to the flow rate to the 1.8 (for smooth pipe) or 2.0 power (for rough pipe);
The questions are: why are there three regions in Figure 6-3? How should we interpret Figure 6-3 and obtain enough knowledge on pipe flows to guide the design of pipe systems?
6.3.2 Laminar, transitional and turbulent flow in pipe [1]
In 1883, Osborne Reynolds [3] did the pioneer work to understand Figure 6-3. Flow in the three regions were visualised using a transparent pipe with a dye streak as a tracer. The flow patterns of the three regions in Figure 6-3 are shown in Figure 6-4 while the time dependence of fluid velocity at point A is shown in Figure 6-5.
1. Laminar flow: the dye streak remains a steady line as it flows through the pipe. All the flow motion is in axial direction, there is no mixing perpendicular to the axis of the pipe. This is observed at low flow rates, i.e. Region I in Figure 6-3.
2. Transitional flow: the dye streak fluctuates in time and space, and intermittent bursts of irregular behaviour appear along the streak. The flow motion is not solely in axial direction anymore. Some mixing perpendicular to the axis of the pipe may occur. This is observed at higher flow rates, i.e. Region II in Figure 6-
3.
3. Turbulent flow: the dye streak spreads across the entire pipe in a random fashion. The flow motion is chaotic in all directions, causing rapid, crosswise mixing. This is observed at high enough flow rates, i.e., Region III in Figure 6-3.
It has been found that the Reynolds number, μ
ρVD
=Re , is a key dimensionless number that
determines the flow in pipe as laminar, transitional or turbulent flow. Reynolds number indicates the ratio between inertial force and viscous force exert on the fluid in pipe. At low Reynolds numbers, viscous force dominates and the flow motion is well-defined. The flow is laminar flow. At high Reynolds numbers, inertial force dominates so that flow becomes turbulent and chaotic. The flow becomes turbulent flow. With the intermittent Reynolds numbers, the flow is in transition from laminar to turbulent, called transitional flow.
Figure 6-4 Laminar, transitional and turbulent flows in pipe. Figure is from Reference [1], the
work was originally done by Osborne Reynolds in 1883.
Figure 6-5 Time dependence of fluid velocity at point A for laminar, transitional and turbulent
flows in a pipe; Figure is from reference [1].
It should be pointed out that besides the flow rate (corresponding to the average velocity V ), the flow character is also determined by fluid density ρ, viscosity μ and the pipe diameter D (i.e., the fluid’s Reynolds number). For flow in a round pipe, we have Re < 2100 Laminar flow 2100 < Re < 4000 Transitional flow Re > 4000 Turbulent flow
Therefore, the existence of the three flow regimes leads to the experimental observations of the three regions in Figure 6-3. The first region corresponds to laminar pipe flow (Re < 2100). The second region, where pressure drop is unpredictable, is due to transitional pipe flow (2100
6.3.3 Entrance region flow and fully developed pipe flow
When fluid enters a pipe, there exists a section, where the pipe flow is not fully developed, i.e., a uniform flow can only be produced after this section. The flow in this section is called entrance region flow . This is why during the pressure drop test in Figure 6-2, the test was done across the section where a uniform flow was produced. Such a uniform flow is called fully developed pipe flow .
The length of the entrance flow region of a pipe flow, L e , is generally short and a function of Reynolds number.
Re 06.0=D L e
for laminar flow (E6-4a)
6/1(Re)4.4=D
L e
for turbulent flow (E6-4b)
The pressure drop in the entrance region is higher than the fully developed region due to extra energy loss when fluid flows into the entrance. It is generally measured by experiments and will be discussed in Chapter 7. This chapter focuses on fully developed pipe flows.
6.4 Fully developed laminar pipe flow
At very low flow rates (fluid velocities), viscous effects dominate, the flow is laminar flow. In the fully develop region, the viscous forces are in equilibrium with pressure forces so that the velocity profile and pressure gradient remain constant along the pipe.
6.4.1 Force balance
It is simple to do force analysis as conditions in the fully developed laminar pipe flow are constant. Figure 6-6a illustrates the force balance system. For a fluid element at any given time, both pressure forces and shear force in action. Since there is no acceleration, these forces in the x-direction must sum to zero. At a radius r , we have Pressure force: F P = P(πr 2) – (P+δP)( πr 2)
Shear force:
Fs = (2πr δx)(shear stress at r)
Because there is no acceleration, according to Newton’s 2nd motion law F = ma , F = 0, i.e., the two forces are balanced. We can equate the two forces and take the limit 0→x δ, yielding
Shear stress at r, τ(r): dx
dP
r r 2)(?=τ (E6-5)
It should be noted that since the force balance is generic, E6-5 is applicable to all types of fluids, both Newtonian and non-Newtonian flows. Equation 6-5 indicates that 1. The pressure gradient across a pipe is a constant.
2. As shown in Figure 6-6b, the shear stress at the pipe centreline (r = 0) is zero and the shear stress reaches maximum at the pipe wall (r = R ). The shear stress is linear between the centreline and the wall;
3. The wall shear stress is dx
dP
R w 2?
=τ. The wall stress is proportional to pressure gradient across the pipe.
(a) (b)
Figure 6-6 (a) Force balance system in pipe flow; (b) Distribution of shear stress in pipe
6.4.2 Velocity profile
For Newtonian flow, assume at a radius r , the flow velocity is V(r), Newton’s law of viscosity applies. We can write the shear stress as
Newton’s law of viscosity:
dr
r dV r )
()(μτ?=. Substituting into E6-5, we have
)(2)(dx
dP
r dr r dV ??=μ (r)
where
dx
dP
is a constant. Therefore, we can integrate the equation and apply the boundary conditions V(R) = 0, producing the following velocity profile:
???
?
????????????=2
21)(4)(R r dx dp R r V μ (E6-6)
Equation E6-6 indicates that
1. The velocity at the pipe wall is zero, i.e., the fluid is non-slip at the wall;
2. The fluid velocity V c along the pipe centreline (where r = 0), V c , reaches maximum.
We have: (42max dx
dp R V V c ?=
=μ; 3. The velocity profile is parabolic, as it can be expressed in terms of V c ,
???
?
???????????=21)(R r V r V c , as shown in Figure 6-7.
Figure 6-7 Velocity profile of laminar pipe flow Figure 6-8 Flow through a pipe
6.4.3 Flow rate
The flow is axisymmetric about the pipe centreline. As shown in Figure 6-8, through a ring at radius r when its thickness dr is thin enough at the cross-section (with an area of rdr dA π2=) of a pipe, the velocity can be considered as constant. Therefore, the flow rate Q through the pipe can be calculated as
rdr
R r V rdr R r dx dp R dA
r V Q R
c R
pipe ∫∫∫???
????????????=???
?????????????==0202
21221)(4)(ππμ
i.e. (D is the diameter of the pipe, D = 2R )
??
?
????=???????==dx dp D dx dp R V R Q c μπμππ12882442
The average velocity V can be calculated as
???
????=???????=?
??
????===dx dp D dx dp R R
dx dp R V A Q V c μμπμπ32882222
4
If the pressure drop across a pipe with a length of L is ΔP , i.e., L
P
dx dp Δ=
?, we have L
P
D Q Δ=μπ1284 (E6-7)
L
P
D V Δ=
μ322 (E6-8) Note, E6-7 and E6-8 are ONLY suitable for laminar flow!
6.5 Fully developed turbulent pipe flow 6.5.1 Randomness of turbulent pipe flow
In turbulent flow (Re > 4000), the fluid experiences random, chaotic motion, including strong eddy transport on a macro scale, compared with the molecular motion in laminar flow. As a result, there is much more dissipation of energy as fluid molecules experience velocity changes in all directions. We can consider the two velocity components, i.e. j i v u V +=. Figure 6-9 shows an example of the axial component, u(t),
measured at a given location.
Figure 6-9 Axial component of velocity at a given location in turbulent pipe flow
Although the flow is chaotic, the velocity can be described in terms of a mean value (denoted with an overbar) on which the fluctuations (denoted with a prime) are superimposed.
'u u u += (E6-9a) where
∫+?=T
t t dt u T u 00
1
0'100
'
=?=∫+T t t dt u T u .
Similarly, in the radial direction of the pipe flow, we have 'v v v += (E6-9b) where
0=v 0'100
'=?=∫+T
t t dt v T v .
6.5.2 Turbulent shear stress
In turbulent flow, in addition to the motion of fluid particles discussed in laminar flow, there is also motion across the flow direction. The parcels of fluid experience relatively larger shear stress, resulting in momentum transfer. Figure 6-10 illustrates the momentum transport in a control volume in turbulent pipe flow.
Figure 6-10 Momentum transport in turbulent pipe flow
In the x-direction we have:
0int =???
?
????+????????+CV the o transfered momentum of rate Net CV the in increase momentum of Rate forces applid
of
Sum
1. We need to consider: Pressure force: F P = P(πr 2) – (P+δP)( πr 2) Viscous shear force: Fs = – ( 2πr δx τ) Sum of applied forces: – τ 2πr δx – πr 2δP
2. Rate of momentum increase in the CV: 0
3. Net rate of momentum transferred into the CV: x r u u v δπρ2)(''+
Therefore, we get
02)(2''2=++??x r u u v P r x r δπρδπδπτ Collecting the terms, rearranging then taking the limit 0→x δ,
2'''r
dx dP v u u v ?????
??=??ρρτ
Applying time-averaging gives
2'''r
dx P d v u u v ?????????=??ρρτ As 0'=v , it reduces to
2)(''r
dx P d v u ???
??????=?+ρτ (E6-10)
where dr
u
d μ
τ=. From E6-5, if we consider a total shear stress at r , total τ, which include fluid viscous shear stress and the shear stress due to eddy transport, we have
2r
dx P d total ???
??????=τ . (E6-11) A comparison between E6-10 and E6-11 leads to
()
turb lam total u v dr
u
d ττρμτ+=′′?+=. (E6-12)
Therefore, in turbulent flow, the turbulent shear stress consists of two components:
? dr
u
d lam μττ==, as a result of fluid viscous effect;
? )(''v u turb ρτ?
=, termed as Reynolds stress, as result of eddy transport.
Figure 6-11 Distribution of shear stress in turbulent pipe flow
For turbulent pipe flow, the distribution of the two components of turbulent shear stress is shown in Figure 6-11. Governed by E6-11, the total shear stress is linear between the centreline and the wall. However, the contribution of viscous shear stress and Reynolds stress are significantly different from the pipe wall to the pipe centreline. As shown in Figure 6-11, near the pipe wall, viscous shear stress dominates while at the pipe centreline, Reynolds stress dominates and viscous shear stress is negligible. Therefore, the structure of turbulent pipe flow may be divided into three parts, as shown in Figure 6-12. 1. Viscous sublayer , where turb lam ττ>>.
In the viscous sublayer, the viscous effects are dominant, and the flow is laminar. The mean velocity, u , is zero at the wall and increases rapidly with r . 2. Buffer layer or Overlap layer , where turb lam ττ≈
Since both turbulent and laminar stresses are acting, this layer demonstrates properties of both laminar and viscous flow.
3. Turbulent core or Outer layer , where turb lam ττ<<
Viscous stresses are negligible, resulting in almost constant velocity and little shear. This zone occupies 80-90% of the cross-sectional area of the pipe.
Figure 6-12 Structure of turbulent pipe flow
6.5.3 Velocity profile of turbulent pipe flow
For laminar flow, the velocity profile can be analytically derived as E6-6. However, it is impossible to do so for a turbulent flow so that we can only obtain semi-empirical results. As the importance of viscous shear stress and Reynolds stress varies in different layers (see Figures 6-11 and 6-12), it is expected that the velocity profile varies in various layers.
Viscous sublayer
We define a friction velocity, ρτw u =*. Please note that *u is not a real physical quantity
but *u has the same unit of velocity. In the viscous sublayer, experimental data show that
*
*
u y u
u ν= (E6-13) where ρμν/= and r R y ?=. Equation E6-13 is only valid within the viscous sublayer, which is taken as
5*
≤ν
yu . (E6-14)
Therefore, Equation E6-14 is often used to estimate the thickness of viscous sublayer.
In the buffer layer and turbulent core
Experimental data in these two layers can be described by the following empirical equation.
4.5ln
5.2**+??
????=u y u u ν (E6-15) For the buffer layer: 505*
≤<νyu (E6-16)
For the turbulent core:
50*
>ν
yu (E6-17)
The turbulent flow profile is shown in Figure 6-13. Experimental data are plotted along with the predictions using E6-13 and E6-15.
Figure 6-13 Turbulent flow profile, modified from [1]
6.5.4 Velocity profile of turbulent pipe flow – the power law
For turbulent pipe flow, the viscous sublayer is generally thin and the flow structure is dominated by the turbulent core. In 1932, Nikuradse proposed the following empirical power law velocity profile :
n
R r u u
1max 1??
????
?=,
(E6-18) where n is a function of Reynolds number, and u max is the velocity on the pipe centreline. For many practical fluid, n = 7 is a reasonable approximation. Certainly, E6-18 CANNOT be used near the wall (i.e., viscous sublayer).
Figure 6-14 Correlation between n and Reynolds number [1]
*
u u
Figure 6-15 Correlation between n and Reynolds number, modified from [1]
Figure 6-14 shows that the value of n increases with Reynolds number, indicating the stronger the turbulence, the higher the n value. Figure 6-15 shows the velocity profiles at various n values. It can be seen that at strong turbulence (with a high n value), the pipe flow has nearly uniform velocity profile, for example the profile at n = 10.
6.6 Friction factor 6.6.1 Definition
Let’s introduce a dimensionless number, friction factor f . For a pipe with a length of L and a diameter of D , friction factor (Darcy friction factor) is defined as
2
/)/(2V L D P f ρΔ= (E6-19)
To understand the meaning of the friction factor in E6-19, we recall that E6-5 applies for all flows therefore the wall stress can be written as
)/(4
1
2L D P L P R w Δ=Δ=τ (E6-20)
Substituting into E6-19, we have the definition of friction factor f as
2
/42
V f w
ρτ= (E6-21) Fundamentally, the friction factor f is the ratio of wall shear stress and inertial force of the flow. E6-19 and E6-21 is generic for laminar or turbulent flows. With the introduction of friction factor f , rewriting E6-19, the pressure drop can be calculated as
2
2
V D L f P ρ=Δ (E6-22)
or in head form
g
V D L f h major L 22
,= (E6-22a)
E6-22 (or E6-22a) is very important for the engineering design and operation of pipe systems. It shows that the pressure drop of a fully-developed pipe flow is a function of three parameters, i.e., the friction factor f , pipe geometry (L/D) and dynamic pressure 2/2V ρ (or velocity head g V 2/2). Therefore, pipe flow problems are converted to find f for the flows under different conditions.
6.6.2 Friction factor for laminar pipe flow
For laminar pipe flow, according to E6-7, we know 2/32D VL P μ=Δ This is for laminar flow only! If we divided both sides by the dynamic pressure (2
2
V ρ) and
D
L
, we have
Re 64642//322/)/(222=???
?????==ΔVD L D
V D VL V L D P ρμρμρ. Therefore, for laminar flow, the friction factor is proportional to Re
1
, i.e. Re
64
=
f . (E6-23)
6.6.3 Friction factor for turbulent flow
Equations E6-19, E6-21 and E6-22 are applicable to turbulent pipe flow. However, for turbulent flow, it is impossible to analytically derive the friction factor f , which can ONLY be obtained from experimental data. In addition, most pipes, except glass tubing, have rough surfaces. The pipe surface roughness is quantified by a dimensionless number, relative pipe roughness (D /ε), where ε is pipe roughness and D is pipe diameter. For laminar pipe flow, the flow is dominated by viscous effects hence surface roughness is not a consideration.
However, for turbulent flow, the surface roughness may protrude beyond the laminar sub-layer and affect the flow to a certain degree. Therefore, the friction factor f can be generally written as a function of Reynolds number Re and pipe relative roughness D
ε
:
)(Re,
D
f ε
Φ= (E6-24)
For smooth pipes
For turbulent flow in smooth pipes with Re < 105, Blasius [4] calculated the friction factor for a large variety of pipe-flow experiments and found that the friction factor f is indeed only a function of Reynolds number,
4/1Re
316
.0=f (E6-25)
For rough pipes
Depending on the depth of surface protrusions, one would expect that it is possible to determine the following three regimes,
1. If the surface protrusions are within the viscous sublayer, the pipe can be treated as hydraulically smooth so that the flow is dependent only on the Reynolds number, i.e. E6-24 can be simplified to (Re)Φ=f , i.e., E6-25.
2. If the surface protrusions extend into just the buffer layer, the flow is dependent on both the Reynolds number and the relative roughness. Details of E6-24 needs to be obtained from experimental investigations.
3. For larger protrusions into the turbulent core, the effect of surface roughness is so
strong that the flow is dependent only on pipe roughness, i.e. )(D
f ε
Φ=.
6.6.4 Moody chart
As discussed in the previous two subsections, the friction factor for turbulent flow is a function of Reynolds number and pipe roughness, described in the generic equation E6-24. In order to find the detailed equation of E6-24, large quantities of experiments have been carried out using pipes of various relative surface roughness under different flow conditions.
Figure 6-16 shows the functional dependence of f on Re and D ε and is called the Moody chart [5]. The Moody Chart covers all three flow regions: laminar, transitional and completely turbulent. Table 6-1 lists the typical roughness from various new, clean pipe surfaces. To use the Moody chart, we need to
1. Calculate the relative roughness (D ε) and the Reynolds number (μρVD =Re ) for the flow under consideration.
2. Follow the line for the particular roughness value until we locate the Reynolds number, and read the friction factor from the left axis.
3. Take extra care when reading the Reynolds number axis, which goes up to 108. In particular, note the pattern of values along the axis, e.g., 105 2 4 6 8 106, etc. If a value 8.5 falls into the region, the exponent needs to be read TO THE LEFT (i.e. 105), not to the right. This is a significant source of error when determining friction factors, and hence leading to possible faulty pipe designs.
Table 6-1 Equivalent roughness for new pipes [5]
Equivalent roughness, ε
Pipe
Feet Millimetres
Glass, plastic Smooth Smooth Drawn tubing (e.g. copper) 0.000005 0.0015 Commercial steel or wrought iron 0.00015 0.045 Galvanised iron 0.0005 0.15 Cast iron 0.00085 0.26 Concrete 0.001 – 0.01 0.3 – 3 Riveted steel 0.003 – 0.03 0.9 – 9.0
It should be noted that the Moody chart is to help us understand the fundamentals and practice hand solutions. Working with the Moody chart directly can sometimes be tedious and error-prone. Most modern engineers prefer to use computer programs to solve flow problems. In this case, besides the analytical solutions from laminar flow, empirical correlations are developed to simulate partially regions of the Moody chart with good accuracy, including
For laminar flow: Re 64
=f (E6-26a) For turbulent flow in smooth pipes, Re < 105 4/1Re 316
.0=f (E6-26b)
For the entire turbulent region:
???
?????+?=f D f Re 51.27.3/log 0.21ε (E6-26c)
6.7 Revisit of Figure 6-3 Region I: Laminar pipe flow
The test section length is Δx . For laminar pipe flow, according to E6-22, we have the pressure gradient as
2
12
V
D f x P ρ?
?????=ΔΔ (E6-27) The fluid velocity V can be calculated from flow rate Q and pipe cross-section area A (which is determined by pipe diameter D ) ,
2
4D
Q
A Q V π== (E6-28) According to E6-23, for laminar pipe flow (with the substitution of V using E6-28),
Q D VD f ρπμρμ1664Re 64=== (E6-29) Substituting E6-28 and E6-29 into E6-27, yielding
Q D D Q D Q D V
D f x P 2
222
128)4(2111621πμπρρπμρ==???
????=ΔΔ In experiments leading to Figure 6-3, 2
128D
πμ
is a constant (assumed as c 1), therefore, we have Q c x
P
1=ΔΔ (E6-30)
In Region I, E6-30 therefore holds. The pressured gradient should be proportional to the flow rate. This was indeed validated by the experimental data in Figure 6-3.
Figure 6-16 The Moody Chart, adapted from [1] which the original data were from [5]
Prepared by Dr Hongwei Wu, since 2006 Chapter 6 – Page 17 of 19
Region III: Turbulent flow
E6-27 and E6-28 are still applicable. We need to use the friction factor in turbulent flow.
1. For rough pipes
From the Moody chart, one can see that when Reynolds number is high enough, the flow is wholly turbulent flow. The friction factor f is only a function of relative pipe roughness, independent of Reynolds number. The same pipe was used in experiment therefore the relative pipe roughness should be a constant. Consequently, the friction factor should be constant (denoted as c ),
c D
f =Φ=)(ε
(E6-31)
Substituting E2-28 and E2-31 into E6-27, we have
252222
8)4(21121Q D f D Q D f V
D f x P πρπρρ==???????=ΔΔ. In the experiments leading to Figure 6-3, 5
28D
f πρ
is a constant (denoted as c 2), we have
22Q c x
P
=ΔΔ (E6-30) Therefore the pressure gradient is proportional to the square of the flow rate.
2. For smooth pipes
Blasius equation E6-25 (also E6-26b) can be used to calculate the friction factor f for
turbulent flow in smooth pipes. Substituting μ
ρVD
=Re and E6-28 into E6-25, we have
()()()25.04
/14/14/14
/14/14316.0316.0Re 316.0????
?????===Q D V D f ρμπρμ (E6-32)
Substituting E2-28 and E2-32 into E6-27, we have
75.1224
/12
)4
(214316.021Q D
D D V
D f x P πρρμπρ???
?????=???????=ΔΔ. In the experiments leading to Figure 6-3, 224
/1)4(21
4316.0D D D πρρμπ???
????? is a constant (denoted
as c 3), we have
75.13Q c x
P
=ΔΔ (E6-33)
Hence the pressure gradient is proportional to the flow rate to ~1.8 power.
6.8 Noncircular Conduits
Many practical engineering situations use conduits of non-circular cross section. Certainly, the details of the flows in noncircular conduits depend on the exact cross-sectional shape. However, the discussion on flow in round pipe can be applied in noncircular conduits with slight modifications.
We introduce an easy-to-use parameter, called hydraulic diameter, D H , defined as
P
A
D H 4=
(E6-34) where A is the cross-sectional area of the noncircular pipe and P is the wetted perimeter. Therefore, for noncircular pipe, we used D H in
friction factor 2/)
/(2
V L D P f H ρΔ=, Reynolds number
μ
ρH
VD =Re , relative roughness
H
D ε
.
For round pipe of diameter D, we can calculate D D D D H ==ππ2. Some typical value of D H
is listed in Table 6-2.
Rectangular
Concentric Annulus
References
[1]. Munson BR, Young DF and Okiishi TH, Fundamentals of Fluid Mechanics, 4th Edition, John Wiley & Sons, Brisbane, 2002.
[2]. Noel de Nevers, Fluid Mechanics for Chemical Engineers, 3rd Edition, McGraw-Hill’s Chemical Engineering Series, Sydney 2005.
[3]. Reynolds O. An experimental investigation of the circumstances which determine whether the motion of water shall be direct or sinuous and of the law of resistance in parallel channels, Philosophy Transactions of the Royal Society, 174, 1883. [4]. White FM, Viscous Fluid Flow, McGraw-Hill, New York, 1979.
[5]. Moody LF, Friction factors for pipe flow, Transactions of the ASME, 1944; 66: 671.
第一章小结 1、流体的特征 与固体的区别：静止状态下，只能承受压力，一般不能承受拉力与抵抗拉伸变形。 在任意剪切力作用下，流体将发生连续的剪切变形（流动），剪切力大小正比于剪切变形速率。固体所受剪切力大小则正比于剪切变形量。 液体与气体的区别：难于压缩；有一定的体积，存在一个自由液面； 2、连续介质 连续介质模型：把流体视为没有间隙地充满它所占据的整个空间的一种连续介质，且其所有的物理量都是空间坐标和时间的连续函数的一种假设模型。 流体质点：几何尺寸同流动空间相比是极小量,又含有大量分子的微元体。 3、粘性 流体在运动（流动）的状态下，产生内摩擦力以抵抗流体变形的性质。粘性是流体的固有属性。 牛顿内摩擦定律（粘性定律）：液体运动时，相邻液层间所产生的切应力与剪切变形的速率成正比。 动力粘性系数μ：反映流体粘滞性大小的系数。 国际单位：牛·秒/米2， N.s/m2 或：帕·秒 运动粘性系数ν：ν=μ/ρ国际单位：米2/秒， m2/s 粘度的影响因素：温度是影响粘度的主要因素。当温度升高时，液体的粘度减小，气体的粘度增加。 粘滞性是流体的主要物理性质，它是流动流体抵抗剪切变形的一种性质，不同的流体粘滞性大小用动力粘度μ或运动粘度v来反映。其中温度是粘度的影响因素：随温度升高，气体粘度上升、液体粘度下降。 复习题 1.连续介质假设意味着。 (A)流体分子互相紧连 (B) 流体的物理量是连续函数 (C) 流体分子间有空隙 (D) 流体不可压缩 2.流体的体积压缩系数k 是在条件下单位压强变化引起的体积变化率。 (A) 等压 (B) 等温 (C) 等密度 3.水的体积弹性模数空气的弹性模数。
第三章 流体静力学 【3-2】 图3-35所示为一直煤气管，为求管中静止煤气的密度，在高度差 H =20m 的两个截面装U 形管测压计，内装水。已知管外空气的密度ρa =1.28kg/m3，测压计读数h 1=100mm ，h 2=115mm 。与水相比，U 形管中气柱的影响可以忽略。求管内煤气的密度。 图3-35 习题3-2示意图 【解】 1air 1O H 1gas 2 p gh p +=ρ 2air 2O H 2gas 2 p gh p +=ρ 2gas gas 1gas p gH p +=ρ 2air air 1air p gH p +=ρ 2gas gas 1air 1O H 2 p gH p gh +=+ρρ gH gh p p air 2O H 1air 2gas 2ρρ-=- gH gh gH gh air 2O H gas 1O H 2 2 ρρρρ-+= H H h h gas air 2O H 1O H 2 2 ρρρρ=+- () 3air 21O H gas kg/m 53.028.120 115 .01.010002 =+-?=+-=ρρρH h h 【3-10】 试按复式水银测压计（图3-43）的读数算出锅炉中水面上蒸汽的绝对压强p 。已知：H =3m ，h 1=1.4m ，h 2=2.5m ，h 3=1.2m ，h 4=2.3m ，水银的密度ρHg =13600kg/m 3。
图3-43 习题3-10示意图 【解】 ()p h H g p +-=1O H 12 ρ ()212Hg 1p h h g p +-=ρ ()232O H 32p h h g p +-=ρ ()a 34Hg 3p h h g p +-=ρ ()()212Hg 1O H 2 p h h g p h H g +-=+-ρρ ()()a 34Hg 232O H 2 p h h g p h h g +-=+-ρρ ()()a 3412Hg 321O H 2 p h h h h g p h h h H g +-+-=+-+-ρρ ()()()()() Pa 14.3663101013252.15.24.13807.910004.15.22.13.2807.913600a 321O H 1234Hg 2=+-+-??--+-??=+-+---+-=p h h h H g h h h h g p ρρ ()()()()()Pa 366300.683 1013252.15.24.1380665.910004.15.22.13.280665.913600a 321O H 1234Hg 2=+-+-??--+-??=+-+---+-=p h h h H g h h h h g p ρρ 【3-15】 图3-48所示为一等加速向下运动的盛水容器，水深h =2m ，加速度a =4.9m/s 2。试确定：（1）容器底部的流体绝对静压强；（2）加速度为何值时容器底部所受压强为大气压强？（3）加速度为何值时容器底部的绝对静压强等于零？ 图3-48 习题3-15示意图 【解】 0=x f ，0=y f ，g a f z -=
6-5 某蒸汽冷凝器内有250根平行的黄铜管，通过的冷却水流量Q =8 l /s ，水温为10o C ，为了使黄铜管内冷却水保持为紊流（此时黄铜管的热交换性能比层流时好），问黄铜管的直径不得超过多少? 解：查表1.3有10℃的水621.310*10/m s ν-= 由2 14 Q n d v π= ① 及临界雷诺数R e 2300vd ν = = ② 联立有 14d m m = 即为直径最大值 6.7 某管道的半径0r 15cm =，层流时的水力坡度J 0.15=，紊流时的水力坡度J 0.20=，试求管壁处的切应力0τ和离管轴r 10cm =轴处的切应力。 解：层流时： 2 f 3 000h r r 1510 g g J 1.0109.80.15110.25Pa 2l 2 2 τρρ-?===??? ?= 2 3 r 1010 g J 1.0109.80.1573.5Pa 2 2 τρ-?==??? ?= 紊流时： 2 f 3 000h r r 1510 g g J 1.0109.80.20147Pa 2l 2 2 τρρ-?===??? ?= 2 '3 r 1010 g J 1.0109.80.2098Pa 2 2 τρ-?==??? ?= 6.9为了确定圆管内径，在管内通过ν为0.013 cm 2/s 的水，实测流量为35cm 3 /s ，长15m ，管段上的水头损失为2㎝水柱，试求此圆管的内径。 解: 设管内为层流 4 2 2 12832264gd lQ gd l g d l d h f πνυνυ υν= = = 1 1 4 4 1281280.013150035 1.949802f lQ d cm gh νππ???????=== ? ? ????? ? ? 校核 1768013 .094.135 44Re =???= = = πν πν υd Q d 层流 6-18 利用圆管层流Re 64=λ，紊流光滑区25 .0Re 3164.0=λ和紊流粗糙区25 .011.0? ? ? ??=d k s λ这三
【最新整理，下载后即可编辑】 第一章 1-1 906 10500453.06 =?= =-V m ρkg/m 3 906.01000 906==d 1-2 544.0140027327334.11013252732730=?+?=+=p t ρρkg/m 3 1-3 1 1 21211V V V t t V dV dt V --==α 98.616060)2080(10550)(611122=+?-??=+-=-V V t t V V αm 3/h 1-4 9 3 3 6661121210510 11011099510102111----?=??-?-?-=---=-=V V V p p V dV dp κ1/P a 1-5 47109.26781028.4--?=??==νρμ Pa·s 1-6 63 103.14 .999103.1--?=?= =ρμνm 2/s 1-7 (1) 17.26605000 1.014.360=??==dn u π m/s 521023.510 005.017.260?=?=-=-δu dy du 1/s (2) 222d dy du dL d dy du A d F M μπμ=== 3 5 221033.51023.5108.01.014.35.322-?=?????==du dy L d M πμ Pa·s (3) 3531079.21023.51033.5?=???==-dy du μτPa 1-8 (1)y dy du μμτ2== (2)μμμμτ2122=?===y dy du 1-9 (1) h u bL dy du A F 022μμ== (2) 当2 h y =时，h u dy du 0μμτ== (3)当h y 2 3 =时，0u u = 所以0==dy du μτ 1-10 2903 .03 .0133)(112121=? ?==+=+=μμμμdy du A dy du A F F F N
五、流体静力学基本方程的意义 流体静力学基本方程的另一种形式： (一)几何意义 流体静力学基本方程的意义 z为位置水头 为压强水头 为测压管水头 静止流体中测压管水头为常数。 (一)能量意义 质量为M 的流体质点距离某一基准面O-O的高度为z，该质点具有位势能：mgz ；每单位重量的流体所具有的位势能为： 任意点的具有长度量纲， 称为测压管高度，也称为压强水头， 为单位重量流体具有的压能， 也是潜在的位势能。 五流体静力学基本方程的意义 (一)能量意义 流体静力学基本方程的能量意义： Z为单位重量流体具有的位能
为单位重量流体具有的压能 为单位重量流体具有的总势能 静止流体中单位重量流体具有的总势能守恒 静止流体中测压管水头为常 数 测压管水头 六、静水压强分布图 由于静水压强是一种分布力，故可以用压强分布图表示： 静水压强分布图绘制原则： 压强大小：根据计算，用成比例线段表示； 压强方向：根据内法线方向确定，用箭头表示。 六、静水压强分布图
绘制原则： 压强大小：根据计算(只计相对压强），用成比例的线段表示； 压强方向：根据内法线方向确定，用箭头表示。 如两侧受力：分别绘制即可。 曲面壁上静水压强分布图 前述两个原则同样适用 七测压计 等压面：与质量力正交。两互不相混的流体平衡时，交界面是等压面，而且是水平面。静止流体中，水平面是等压面的条件： 静止、同种、连通 同时成立，缺一不可。 2、3 点压强相等 6、7 点压强相等 p2>p1 ; p3=p4 ; p5>p4 ; 此处 p0 大于大气压 题6-9，题6-11
正确判断等压面----静止、同种、连通的水平面， 用公式计算。 题6-9 （2008）盛水容器A和B的上方密封，测压管水面位置如图所示，其底部压强分别为 p A和 p B。若两容器内水深相等，则 p A和 p B的关系为（） 解：在仅受重力作用的静止液体中，等压面为水平面，据此分别绘容器A和B的等压面 M-M 及N-N，液面均为大气压。 由于可知： 表示大气压 答案：A 例6-11（2010年） 如图所示，在上部为气体下部为水的封闭容器上装有U形水银测压计，其中1、2、3点位于同一平面上，其压强关系为（） (A) p1< p2< p3 (B) p1> p2> p3 (C) p2< p1< p3 (D) p2= p1 = p3 解：水平面是等压面的条件---静止、同种、连通。故点1、 2、3压强不相等。绘等压面M-M 及N-N ,再比较三点压强. 空气的重度远远小于水的重度，可忽略不计
闻建龙主编的《工程流体力学》习题参考答案 第一章 绪论 1-1 物质是按什么原则分为固体和液体两大类的？ 解：从物质受力和运动的特性将物质分成两大类：不能抵抗切向力，在切向力作用下可以无限的变形（流动），这类物质称为流体。如空气、水等。而在同等条件下，固体则产生有限的变形。 因此，可以说：流体不管是液体还是气体，在无论多么小的剪应力（切向）作用下都能发生连续不断的变形。与此相反，固体的变形与作用的应力成比例，经一段时间变形后将达到平衡，而不会无限增加。 1-2 何谓连续介质假设？引入连续介质模型的目的是什么？在解决流动问题时，应用连续介质模型的条件是什么？ 解：1753年，欧拉首次采用连续介质作为流体宏观流动模型，即不考虑流体分子的存在，把真实的流体看成是由无限多流体质点组成的稠密而无间隙的连续介质，甚至在流体与固体边壁距离接近零的极限情况也认为如此，这个假设叫流体连续介质假设或稠密性假设。 流体连续性假设是流体力学中第一个根本性假设，将真实流体看成为连续介质，意味着流体的一切宏观物理量，如密度、压力、速度等，都可看成时间和空间位置的连续函数，使我们有可能用数学分析来讨论和解决流体力学问题。 在一些特定情况下，连续介质假设是不成立的，例如：航天器在高空稀薄气体中飞行，超声速气流中激波前后，血液在微血管（1μm ）内的流动。 1-3 底面积为2 5.1m 的薄板在液面上水平移动（图1-3），其移动速度为s m 16，液层 厚度为mm 4，当液体分别为C 020的水和C 0 20时密度为3 856m kg 的原油时，移动平板 所需的力各为多大？ 题1-3图 解：20℃ 水：s Pa ??=-3 10 1μ 20℃，3 /856m kg =ρ， 原油：s Pa ??='-3 102.7μ 水： 23 3 /410 416 101m N u =??=? =--δμτ N A F 65.14=?=?=τ
工程流体力学禹华谦习题答案第6章
第六章 理想流体动力学 6-1平面不可压缩流体速度分布为 Vx=4x+1；Vy=-4y. (1) 该流动满足连续性方程否? (2) 势函数φ、流函数ψ存在否?（3） 求φ、ψ 解：（1）由于 044=-=??+??y Vy x Vx ，故该流动满足连续性方程 （2）由ωz = 21(y Vx x Vy ??-??)=)44(21 +-＝0, 故流动有势，势函数φ存在，由于该流动满足连续性方程， 流函数ψ存在，. （3）因 Vx y x ??=??= ψ?=4x+1 Vy= y ??φ=-x ??ψ=-4y d φ= x ??φdx+y ??φdy=Vxdx+Vydy=(4x+1)dx+(-4y)dy φ= ?d φ=?x ??φ dx+y ??φdy=?Vxdx+Vydy=? (4x+1)dx+(-4y)dy =2x 2-2y 2+x d ψ= x ??ψ dx+y ??ψdy=-Vydx+Vxdy=4ydx+(4x+1)dy ψ= ?d ψ=?x ??ψ dx+y ??ψdy=?-Vydx+Vxdy=? 4ydx+(4x+1)dy =4xy+y 6-2 平面不可压缩流体速度分布: Vx=x 2-y 2+x; Vy=-(2xy+y). (1) 流动满足连续性方程否? (2) 势函数φ、流函数ψ存在否? (3)求φ、ψ .
解：（1）由于x Vx ??+x Vy ??=2x ＋1－（2x ＋1）＝0，故该流动满足连续性方程，流动存在. (2）由ωz = 21(y Vx x Vy ??-??)=))2(2(21 y y ---＝0, 故流动有势，势函数φ存在，由于该流动满足连续性方程，流函数ψ也存在. （3）因 Vx= x ??φ =y ??ψ= x 2-y 2+x, Vy=y ??φ=-x ??ψ=-(2xy+y). d φ= x ??φ dx+y ??φdy=Vxdx+Vydy=(x 2-y 2+x )dx+(-(2xy+y).)dy φ= ?d φ=? x ??φ dx+y ??φdy=?Vxdx+Vydy =? (x 2-y 2+x )dx+(- (2xy+y))dy =3 3 x -xy 2+(x 2-y 2)/2 d ψ= x ??ψdx+y ??ψdy=-Vydx+Vxdy ψ= ?d ψ=? x ??ψdx+y ??ψdy=?-Vydx+Vxdy =?(2xy+y)dx+ (x 2-y 2+x)dy =x 2y+xy-y 3/3 6-3平面不可压缩流体速度势函数 φ=x 2-y 2-x,求流场上A(-1,-1),及B(2,2)点处的速度值及流函数值 解: 因 Vx= x ??φ =y ??ψ=2x-1，V y ＝y x y 2-=??-=??ψ φ，由于x Vx ??+x Vy ??＝0，该 流动满足连续性方程，流函数ψ存在 d ψ= x ??ψdx+y ??ψdy=-Vydx+Vxdy ψ= ?d ψ=? x ??ψ dx+y ??ψdy=?-Vydx+Vxdy=?2ydx+(2x-1)dy=2xy-y
[陈书3-8] 已知流体运动的速度场为32x v yt at =+，2y v xt =，0z v =，式中a 为常数。试求：1t =时过(0,)b 点的流线方程。 解： 流线满足的微分方程为： x y z dx dy dz v v v == 将32x v yt at =+，2y v xt =，0z v =，代入上式，得： 3 22dx dy yt at xt = +（x-y 平面内的二维运动） 移向得：32(2)xtdx yt at dy =+ 两边同时积分：32(2)xtdx yt at dy =+??（其中t 为参数） 积分结果：223x t y t ayt C =++（此即流线方程，其中C 为积分常数） 将t=1, x=0, y=b 代入上式，得：20b ab C =++ ∴积分常数2C b ab =-- ∴t=1时刻，过(0,b)点的流线方程为：222()x y ay b ab =+-+ 整理得：222()0x y ay b ab --++= 陈书3-10 已知二元不可压缩流体流动的流线方程如下，问哪一个是无旋的？ （1）2Axy C =； （2）Ax By C +=； （3）()2ln A xy C =， 其中A ，B ，C 均为常数。
[解法一] （1）根据流线方程2Axy C =? 220Aydx Axdy += 当0 A ≠时，有 dx dy x y =- 令(),u xf x y =，(),v yf x y =- 根据流体的不可压缩性，从而 '''' 0x y x y u v f xf f yf xf yf x y ??+=+--=-=?? 再把流线方程2Axy C =对x 求导得到 ' ' 220y A y A xy y x +=?=- 所以 '''''' 20x y y y y u v xf yf xf y yf yf x y ??+ =-=-=-=?? y 是任意的，得到'0y f = 2 ' '' 0y x y u v y xf yf x f y x x ????-=+=-= ???? ? 无旋 （2）根据流线方程Ax By C +=? 0Adx Bdy += 令(),u Bf x y =，(),v Af x y =- 根据流体的不可压缩性，从而 ' ' 0x y u v Bf Af x y ??+=-=?? 再把流线方程Ax By C +=对x 求导得到 ' ' 0A A By y B +=?=- 所以' ' ' 20x y y u v Bf Af Af x y ??+ =-=-=?? 当0A =时，0 v =无旋 当0 A ≠时，'0y f = 2 ''' 0y x y u v A Bf Af B f y x B ????-=+=-= ????? 无旋 （3）根据流线方程()2ln A xy C = ?2 22 111220A y dx xydy A dx dy xy xy x y ????+=+= ? ?????
课后答案网 工程流体力学 第一章绪论 1-1. 20C 的水2.5m 3 ，当温度升至80C 时，其体积增加多少？ ［解］温度变化前后质量守恒，即 = 7V2 3 又20C 时，水的密度 d 二998.23kg / m 3 80C 时，水的密度 = 971.83kg/m 3 啦 3 V 2 =亠=2.5679m 「2 则增加的体积为 V 二V 2 -V^ 0.0679 m 3 1-2.当空气温度从 0C 增加至20C 时，运动粘度\增加15%，重度 减少10%，问此时动力粘度 」增加 多少(百分数)？ ［解］ 宀(1 0.15)、.原(1 -0.1)「原 = 1.035 原「原=1.035'I 原 ■' -「原1.035?L 原一」原 原 原——原二0.035 卩原 卩原 此时动力粘度 J 增加了 3.5% 2 1-3?有一矩形断面的宽渠道，其水流速度分布为 u =0.002 Jg(hy-0.5y )/」，式中'、」分别为水的 密度和动力粘度，h 为水深。试求h =0.5m 时渠底(y=0)处的切应力。 ［解］ 一 =0.002「g(h -y)/「 dy 当 h =0.5m , y=0 时 = 0.002 1000 9.807(0.5 —0) J du dy -0.002 'g(h -y)
= 9.807Pa 1-4.一底面积为45 x 50cm 2,高为1cm 的木块，质量为5kg ，沿涂有润滑油的斜面向下作等速运动，木块 运动速度u=1m/s ，油层厚1cm ,斜坡角22.620 （见图示），求油的粘度。 mg sin v I mg sin A U 0.4 0.45 — d 0.001 」-0.1047Pa s 1-5 .已知液体中流速沿 y 方向分布如图示三种情况，试根据牛顿内摩擦定律 沿y 方向的分布图。 ［解］木块重量沿斜坡分力 F 与切力T 平衡时，等速下滑 5 9.8 sin 22.62 -=一，定性绘出切应力 dy 1-6 ?为导线表面红绝缘，将导线从充满绝缘涂料的模具中拉过。已知导线直径 的粘度」=0.02Pa . s 。若导线以速率50m/s 拉过模具，试求所需牵拉力。 0.9mm ，长度20mm ，涂料 (1.O1N ) e y I
流体力学练习题 第一章 1-1解：设：柴油的密度为ρ，重度为γ；40C 水的密度为ρ0，重度为γ0。则在同一地点的相对密度和比重为： 0ρρ= d ，0 γγ=c 1-2解：336/1260101026.1m kg =??=-ρ 1-3解：269/106.191096.101.0m N E V V V V p p V V p p p ?=??=?- =?-=????-=ββ 1-4解：N m p V V p /105.210 4101000295 6 --?=?=??-=β 1-5解：1）求体积膨涨量和桶内压强 受温度增加的影响，200升汽油的体积膨涨量为： 由于容器封闭，体积不变，从而因体积膨涨量使容器内压强升高，体积压缩量等于体积膨涨量。故： 2）在保证液面压强增量0.18个大气压下，求桶内最大能装的汽油质量。设装的汽油体积为V ，那么：体积膨涨量为： 体积压缩量为：
因此，温度升高和压强升高联合作用的结果，应满足： 1-6解：石油的动力粘度：s pa .028.01.0100 28 =?= μ 石油的运动粘度：s m /1011.39 .01000028.025-?=?== ρμν 1-7解：石油的运动粘度：s m St /1044.0100 40 25-?=== ν 石油的动力粘度：s pa .0356.010*******.05=???==-ρνμ 1-8解：2/1147001 .01 147.1m N u =? ==δ μ τ 1-9解：()()2/5.1621196.012.02 1 5.0065.02 1 m N d D u u =-? =-==μ δ μ τ 第二章 2-4解：设：测压管中空气的压强为p 2，水银的密度为1ρ，水的密度为2ρ。在水银面建立等压面1-1，在测压管与容器连接处建立等压面2-2。根据等压面理论，有 21p gh p a +=ρ（1） gz p z H g p 2221)(ρρ+=++（2） 由式（1）解出p 2后代入（2），整理得： 2-5解：设：水银的密度为1ρ，水的密度为2ρ，油的密度为3ρ；4.0=h ，6.11=h ， 3.02=h ，5.03=h 。根据等压面理论，在等压面1-1上有： 在等压面2-2上有：
第六章 管内流动和水力计算 液体出流 【6-11】 加热炉消耗q m =300kg/h 的重油，重油的密度ρ=880kg/m 3，运动黏度ν=0.000025m 2/s 。如图6-54所示，压力油箱位于喷油器轴线以上h =8m 处，而输油管的直径d =25mm ，长度l =30m 。求在喷油器前重油的计示压强？[62504Pa]。 图6-54 习题6-11示意图 【解】 流速 ()s m 0.192915 025 .08803600/300444 1222=???====ππρπρ d q d q A q v m m V 2300192.915000025 .0025.00.192915<=?==νvd Re 输油管内流动是层流 沿程损失 ()m 3.91557 807 .9025.0915.1920.192915 30323226422222f =????=??===g d Re lv g v d l Re g v d l h λ 以油箱液面为1-1，喷油器前为2-2断面，列写伯努利方程： w 222 2 2112 1 122h g p z g v g p z g v a a +++=++ραρα 由于是层流，221==αα；f w h h =；01=a v ；v v a =1；h z z =-21；01=p 。 w 2 222h g p g v h ++=ρ ()[]()[] ()Pa 62489.5 0.1929153.915578807.98802 2w 2=--??=--=v h h g p ρ 【6-16】用新铸铁管输送25℃的水，流量q v =300L/s ，在l =1000m 长的管道上沿程损失为 h f =2m （水柱），试求必须的管道直径。 【解】 v d q V 24 1π=2 4d q v V π= 5 22 2 22f 8242gd lq g d q d l g v d l h V V πλπλλ=?? ? ??==λπf 22 5 8gh lq d V = d q d d q vd R e V V 1442 πννπν=== 新铸铁管 ε=0.25～0.42，25℃水的运动黏度ν=0.897×10-6m 2/s 。 取ε=0.3， λ=0.0175 得 d = 0.5790 m
工程流体力学三级项目流体对曲面壁作用力分析 指导老师： 班级： 小组成员： 制作时间：2012/10/12
目录 1.项目目的 (3) 2.项目要求 (3) 3.公式推导计算过程 (4) 4.Matlab软件编程求解过程 (6) 5.分析比较 (9) 6.小组成员感想 (9) 7.参考文献 (11)
一、项目目的 （1）学会流体作用在曲面壁上的力的分析方法 （2）掌握流体作用在曲面壁上的力的推导过程 （3）了解Matlab 软件的简单使用 二、项目要求 一个闸门的横截面如图所示，垂直于直面的深度是6m,外形x=2y+0.5y2 ，此闸门可以绕0点旋转，试以闸门浅的水深度为自变量，推导一下参量的表达式；水平分力；垂直分力；作用在闸门上对原点0的顺时针方 向转矩。
三、公式推导计算过程 流体对曲面壁受力分布图 此曲面在XOZ平面上的投影如图所示，在此截面上取微面积dA，设其形心在水面以下的深度为h，则此微分面积上所承受的压力为： dP=γhdA 此压力垂直于微元面积dA，并指向右下方，与水平面成ɑ角，可将其分解为
水平分力：dPx=dPcosα =γhdAcosɑ 垂直分力：dPz= dPsinɑ =γhdAsinɑ 由图三可知dPcosɑ为dA在垂面yoz上的投影面dAx ，dAsinɑ为dA在水平面xoy上的投影面积dAz因此上式可改写为 dPx=γhdAx dPz=γhdAz 将上式沿曲面ABCD相应的投影面积积分，可得此曲面所受液体的总压力P为 水平分力：Px=∫AxγhdAx 竖直分力：Pz=∫AxγhdAz 式中∫Ax hdAx 为曲面ABCD的垂直投影面积Ax绕y 轴的静力矩可表示为 ∫AxhdAx=h o Ax 式中，ho为投影面积在Ax的形心在水面下的深度，所以总压力的 水平分力： Px=γh0A x =bh2/2 =29400h2
第六章 理想流体动力学 6-1平面不可压缩流体速度分布为 Vx=4x+1；Vy=-4y. (1) 该流动满足连续性方程否? (2) 势函数φ、流函数ψ存在否?（3）求φ、ψ 解：（1）由于044=-=??+??y Vy x Vx ，故该流动满足连续性方程 （2）由ωz = 21(y Vx x Vy ??-??)=)44(21+-＝0, 故流动有势，势函数φ存在，由于该流动满足连续性方程， 流函数ψ存在，. （3）因 Vx y x ??=??=ψ?=4x+1 Vy=y ??φ=-x ??ψ=-4y d φ=x ??φdx+y ??φdy=Vxdx+Vydy=(4x+1)dx+(-4y)dy φ= ?d φ=?x ??φdx+y ??φdy=?Vxdx+Vydy=? (4x+1)dx+(-4y)dy =2x 2-2y 2+x d ψ=x ??ψdx+y ??ψdy=-Vydx+Vxdy=4ydx+(4x+1)dy ψ= ?d ψ=?x ??ψdx+y ??ψdy=?-Vydx+Vxdy=? 4ydx+(4x+1)dy =4xy+y 6-2 平面不可压缩流体速度分布: Vx=x 2-y 2+x; Vy=-(2xy+y). (1) 流动满足连续性方程否? (2) 势函数φ、流函数ψ存在否? (3)求φ、ψ . 解：（1）由于x Vx ??+x Vy ??=2x ＋1－（2x ＋1）＝0，故该流动满足连续性方程，流动存在. (2）由ωz = 21(y Vx x Vy ??-??)=))2(2(21y y ---＝0, 故流动有势，势函数φ存在，由于该流动满足连续性方程，流函数ψ也存在.
[陈书1-15] 图轴在滑动轴承中转动，已知轴的直径cm D 20=，轴承宽度cm b 30=，间隙cm 08.0=δ。间隙中充满动力学粘性系数s Pa 245.0?=μ的润滑油。若已知轴旋转时润滑油阻力的损耗功率W P 7.50=，试求轴承的转速?=n 当转速min 1000r n =时，消耗功率为多少？（轴承运动时维持恒定转速） 【解】轴表面承受的摩擦阻力矩为：2D M A τ= 其中剪切应力：dr du ρντ= 表面积：Db A π= 因为间隙内的流速可近似看作线性分布，而且对粘性流体，外表面上应取流速为零的条件，故径向流速梯度： δ ω2D dr du = 其中转动角速度：n πω2= 所以：2322nD D D nb M Db πμπμπδδ == 维持匀速转动时所消耗的功率为：3322D n b P M M n μπωπδ === 所以：Db P D n μπδπ1= 将： s Pa 245.0?=μ m cm D 2.020== m cm b 3.030== m cm 410808.0-?==δ W P 7.50= 14.3=π 代入上式，得：min r 56.89s r 493.1==n 当r 3 50min r 1000==n 时所消耗的功率为： W b n D P 83.6320233==δ μπ [陈书1-16]两无限大平板相距mm 25=b 平行（水平）放置，其间充满动力学粘性系数s Pa 5.1?=μ的甘油，在两平板间以m 15.0=V 的恒定速度水平拖动一面积为
2m 5.0=A 的极薄平板。如果薄平板保持在中间位置需要用多大的力？如果置于距一板10mm 的位置，需多大的力？ 【解】平板匀速运动，受力平衡。 题中给出平板“极薄”，故无需考虑平板的体积、重量及边缘效应等。 本题应求解的水平方向的拖力。 水平方向，薄板所受的拖力与流体作用在薄板上下表面上摩擦力平衡。 作用于薄板上表面的摩擦力为： A dz du A F u u u μτ== 题中未给出流场的速度分布，且上下两无限大平板的间距不大，不妨设为线性分布。 设薄板到上面平板的距离为h ，则有： h V dz du u = 所以：A h V F u μ= 同理，作用于薄板下表面的摩擦力为： A h b V F d -=μ 维持薄板匀速运动所需的拖力： ?? ? ??-+=+=h b h AV F F F d u 11μ 当薄板在中间位置时，m 105.12mm 5.123 -?==h 将m 1025mm 253-?==b 、s m 15.0=V 、2m 5.0=A 和s Pa 5.1?=μ代入，得： N 18=F 如果薄板置于距一板（不妨设为上平板）10mm 的位置，则： m 1010mm 103-?==h 代入上式得：N 75.18=F [陈书1-17]一很大的薄板放在m 06.0=b 宽水平缝隙的中间位置，板上下分别放有不同粘度的油，一种油的粘度是另一种的2倍。当以s m 3.0=V 的恒定速度水平拖动平板时，每平方米受的总摩擦力为N 29=F 。求两种油的粘度。 【解】平板匀速运动，受力平衡。 题中给出薄板”，故无需考虑平板的体积、重量及边缘效应等。 本题应求解的水平方向的拖力。
第三章 流体静力学 【3-2】 图3-35所示为一直煤气管，为求管中静止煤气的密度，在高度差H =20m 的两个截面装U 形管测压计，内装水。已知管外空气的密度ρa =1.28kg/m3，测压计读数h 1=100mm ，h 2=115mm 。与水相比，U 形管中气柱的影响可以忽略。求管内煤气的密度。 图3-35 习题3-2示意图 【解】 1air 1O H 1gas 2p gh p +=ρ 2air 2O H 2gas 2p gh p +=ρ 2gas gas 1gas p gH p +=ρ 2air air 1air p gH p +=ρ 2gas gas 1air 1O H 2 p gH p gh +=+ρρ gH gh p p air 2O H 1air 2gas 2ρρ-=- gH gh gH gh air 2O H gas 1O H 2 2 ρρρρ-+= H H h h gas air 2O H 1O H 2 2 ρρρρ=+- () 3air 21O H gas kg/m 53.028.120 115 .01.010002 =+-?=+-=ρρρH h h 【3-10】 试按复式水银测压计（图3-43）的读数算出锅炉中水面上蒸汽的绝对压强p 。已知：H =3m ， h 1=1.4m ，h 2=2.5m ，h 3=1.2m ，h 4=2.3m ，水银的密度ρHg =13600kg/m 3。 图3-43 习题3-10示意图 【解】 ()p h H g p +-=1O H 12ρ ()212Hg 1p h h g p +-=ρ
()232O H 32p h h g p +-=ρ ()a 34Hg 3p h h g p +-=ρ ()()212Hg 1O H 2 p h h g p h H g +-=+-ρρ ()()a 34Hg 232O H 2 p h h g p h h g +-=+-ρρ ()()a 3412Hg 321O H 2 p h h h h g p h h h H g +-+-=+-+-ρρ ()()()()() Pa 14.3663101013252.15.24.13807.910004.15.22.13.2807.913600a 321O H 1234Hg 2=+-+-??--+-??=+-+---+-=p h h h H g h h h h g p ρρ ()()()()()Pa 366300.683 1013252.15.24.1380665.910004.15.22.13.280665.913600a 321O H 1234Hg 2=+-+-??--+-??=+-+---+-=p h h h H g h h h h g p ρρ 【3-15】 图3-48所示为一等加速向下运动的盛水容器，水深h =2m ，加速度a =4.9m/s 2。试确定：（1） 容器底部的流体绝对静压强；（2）加速度为何值时容器底部所受压强为大气压强？（3）加速度为何值时容器底部的绝对静压强等于零？ 图3-48 习题3-15示意图 【解】 0=x f ，0=y f ，g a f z -= 压强差公式 () z f y f x f p z y x d d d d ++=ρ ()()z g a z f y f x f p z y x d d d d d -=++=ρρ ()?? --=h p p z g a p a d d ρ ()()()()??? ? ??-=-=----=-g a gh a g h g a h g a p p a 10ρρρρ ??? ? ??-+=g a gh p p a 1ρ () a g h p p a -=-ρh p p g a a ρ-- = （1） ()()()Pa 111138.39.480665.921000101325=-??+=-+=a g h p p a ρ
第一章 第二章 第三章 1.1 试谈牛顿内摩擦定律？产生摩擦力的根本原因是什么？（参考分数：8分） 答：流体内只要存在相对运动，流体内就会产生内摩擦力来抵抗此相对运动，牛顿经过大量牛 顿平板试验得出单位面积上的内摩擦力：τ=F/A=μ·du/dy 即为牛顿内摩擦定律。产生摩擦力的 根本原因是流体内存在着相对运动。 1.2 液体和气体的粘性随温度的升高或降低发生变化，变化趋势是否相同？为什么？（参考分 数：8分） 答：不相同，液体的粘度随温度升高而减小，气体的粘度却随温度升高而增大。其原因是，液 体分子间距小，内聚力强，粘性作用主要来源于分子内聚力，当液体温度升高时，其分子间距加大， 内聚力减小，粘度随温度上升而减小；而气体的内聚力极小，可以忽略，其粘性作用可以说完全是 分子热运动中动量交换的结果，当气体温度升高时，热运动加剧，其粘度随温度升高而增加。 1.3 何谓流体的连续介质模型？为了研究流体机械运动规律，说明引入连续介质模型的必要性。 答：流体的连续介质模型：假定流体是由连续分布的流体质点所组成，即认为流体所占据的空 间完全由没有任何空隙的流体质点所充满，流体质点在时间过程中作连续运动。根据流体的连续介 质假设，表征流体性质和运动特性的物理量和力学量一般为空间坐标和时间变量的连续函数，这样 就可以用数学分析方法来研究流体运动，解决流体力学问题。 1.4 什么是表面张力？试对表面张力现象作物理解释。 答：液体的表面张力是液体自由表面上相邻部分之间的拉力，其方向与液面相切，并与两相邻 部分的分界线垂直。表面张力是分子引力在液体表面上的一种宏观表现。例如，在液体和气体相接 触的自由表面上，液面上的分子受到液体内部分子的吸引力与其上部气体分子的吸引力不平衡，其 合力的方向与液面垂直并指向液体内部。在合力的作用下，表层中的液体分子都力图向液体内部收 缩，使液体具有尽量缩小其表面的趋势，这样沿液体的表面便产生了拉力，即表面张力。 1.5 动力粘度μ＝0.172Pa·s 的润滑油充满在两个同轴圆柱体的间隙中，外筒固定，内径D ＝ 12cm ，间隙h ＝0.02cm ，试求：（1）当内筒以速度U ＝1m/s 沿轴线方向运动时，内筒表面的切应力 τ1，如图1-3（a ）；（2）当内筒以转速n ＝180r/min 旋转时，内筒表面的切应力τ2，如图1-3（b ）。 (b) 解：内筒外径 96cm .1102.02122h D d =?-=-= （1）当内筒以速度U ＝1m/s 沿轴线方向运动时，内筒表面的切应力 N 8601002.01172.0h U dy du 2 -1=??===μμτ （2）当内筒以转速n ＝180r/min 旋转时，内筒的旋转角速度60 2n πω= ，内筒表面的切应力
第一章 1-1 90610500453.06 =?= = -V m ρkg/m 3 906.01000 906 == d 1-2 544 .01400 273273 34.11013252732730 =?+?=+=p t ρρkg/m 3 1-3 1 1 21211V V V t t V dV dt V --== α 98 .616060)2080(10550)(611122=+?-??=+-=-V V t t V V αm 3 /h 1-4 9 3 36661121210 51011011099510102111----?=??-?-?-=---=-=V V V p p V dV dp κ1/ Pa 1-5 4 7 109.26781028.4--?=??==νρμ Pa·s 1-6 6 3 103.14 .99910 3.1--?=?==ρμνm 2/s 1-7 (1) 17 .2660 5000 1.014.360 =??= = dn u π m/s 5 2 1023.510005.017.260?=?=-=-δu dy du 1/s (2) 2 22d dy du dL d dy du A d F M μπμ=== 3 5221033.510 23.5108.01.014.35.322-?=?????== du dy L d M πμ Pa·s (3) 3531079.21023.51033.5?=???==-dy du μ τPa 1-8 (1)y dy du μμτ2==
(2)μμμμτ2122=?===y dy du 1-9 (1) h u bL dy du A F 022μμ== (2) 当2 h y =时，h u dy du μμτ== (3)当h y 2 3=时，0 u u = 所以0==dy du μτ 1-10 2903 .03 .0133)(112121=??==+=+=μμμμdy du A dy du A F F F N 967 .01=μ Pa·s 933 .1212==μμ Pa·s 1-11 dr r r dr r r r dA dy du r dF dM α δπωμαπδωμμsin 2sin 203 =-=?=?= α δαπωμααδπωμαδπωμαδπωμαα α cos 24)(sin 2sin 2sin 234403030tg H Htg dr r dr r dM M Htg Htg Htg == ===???1-12 62 .260 200 25.014.360 =??= = dn u πm/s 3925 .050.025.014.3=??==dL A πm 2 3 3 1022.410 2.00 62.23925.082.0?=?-??==-dy du A F μN 05 .1162.21022.43=??==Fu P kW 1-13 0841 .0100092.0109144.04=???==-νρμ Pa·s 1459 .03048.01524.014.3=??==dL A πm 2 2 .736102 4.1526.152061459.00841.03 =?--??==-dy du A F μN 42 .462.736=?==Fv P kW