First year ﬂuid mechanics - University College Londonuceseug/Fluids1/2/Notes.pdf · First year...

First year fluid mechanics

Flows in pipes and pipelines

The steady flow energy equation

Bernoulli’s equation is an energy equation derived for frictionless (inviscid)conditions with no energy input or extraction. It is a special form of moregeneral steady flow energy equation, which includes viscose losses andwork transfer to the fluid. These effects are accounted for by introducingadditional terms into Bernoulli’s equation.

Let us consider the balance of energy inside volume abcd (figure 1), whichis bounded by the walls of a stream tube and two cross sections 1 and 2 atheights z1,2 from an arbitrary datum level. In this derivation we assume thatthe fluid velocity does not change across the stream tube. The fluid velocitiesat the inlet 1 and outlet 2 are V1,2 and the areas of the cross sections are A1,2.External work is applied to the fluid inside the control volume with powerP (e.g. pumps or turbines operate there), and Pf is the power of frictionalforces inside the control volume and on its boundaries. During the smalltime period ∆t the fluid inside the volume a′add′ enters the control volumeand the fluid inside the volume bcc′b′ leaves the control volume. Accordingthe the continuity principles for a steady flow these volumes are the same:

A1V1∆t = A2V2∆t = Q ∆t,

where Q is the volumetric flow rate of fluid. The energy entering the controlvolume with the fluid through the cross section 1 is

∆E1 = (V 2

1

2+ g z1) ρ Q ∆t,

and includes the kinetic energy of moving fluid and the potential energy offluid in the gravitational field. Analogously, the energy leaving the controlvolume through the cross section 2 is

∆E2 = (V 2

1

2+ g z2) ρ Q ∆t.

Pressure P1 acting on the moving boundary a′d′ produces work P1A1V1∆t =P1 Q ∆t, which should be added to the incoming energy E1, and similarly

1

2

��

��

��

��

t

V2 ∆ t

c

V1 ∆

a’

d’

c’

b’

a

d

b

P1

∆

∆P2

A1

V1

V2

A2

∆P

z1

z2

∆P

E

t

t

f

2

1E1

2

Fig. 1:

the pressure work P2 Q ∆t should be added to the outgoing energy E2. Theenergy of the fluid in the control volume will also be increased by the workof external forces P ∆t and dissipated due to the work of friction Pf ∆t. Fora stationary flow the energy of the fluid inside the control volume does notchange, and the total gain of energy is equal to the total energy loss:

(ρ V 2

1

2+ρ g z1) Q ∆t+P1 Q ∆t+P∆t = (

ρ V 2

2

2+ρ g z2) Q ∆t+P2 Q ∆t+Pf∆t.

This gives the steady flow energy equation in the following form:

P1 +ρ V 2

1

2+ ρ g z1 +

P

Q= P2 +

ρ V 2

2

2+ ρ g z2 +

Pf

Q, (1)

which represents conservation of energy per unit of fluid volume. It shouldbe noted that the sign of the external power P is positive if the work doneon the fluid (pumps, compressors) and negative when work done by the fluid(turbines). In engineering it is conventional to use the energy equation perunit of fluid weight, which can be obtained dividing (1) by ρg. The energyof fluid per unit of weight has dimensions of meters (Joules/Newtons) and iscalled head. The value

H =P

ρ g+

V 2

2 g+ z

is called total head and represents the total energy of a unit weight of flowingfluid. It consists of pressure head, velocity head and potential head. Thesteady flow energy equation can now be formulated in terms of change oftotal head:

H2 − H1 =P

ρ g Q− hf , (2)

3

τ

P1 P2

d

UV(r)

x

R

rz 1

L

z 2

Fig. 2:

where hf = Pf/(ρ g Q) is head loss. Therefore, the total head of a steadilyflowing fluid is increased by external work transferred to the unit of fluidweight, and decreased by the head loss due to viscous dissipation.

Application to flow in a straight horizontal pipe

The steady flow energy equations is widely applied in engineering for speci-fying flows of viscous fluid through pipes and pipe systems. In this sectionwe consider a steady flow through a straight circular pipe of the internalradius R, diameter d (figure ). The flow in the pipe is parallel (this meansthat pressure is constant across the pipe), and the velocity profile does notchange along the pipe. Such flow is called fully developed, and occurs farenough from the pipe inlet. The pipe wall is the boundary of a stream tube,and using one-dimensional approximation we take the mean velocity of thefluid through the pipe as the flow velocity. From the last term you should re-member that the mean velocity U is a uniform velocity, which would providethe same flow rate Q through the pipe as the actual velocity profile V (r):

Q = π R2U = 2 π

R∫

0

r V (r) dr.

4

Taking an average velocity as the flow velocity we make a mistake in deter-mining the exact value of the kinetic energy of the flow because the mean ofthe square of velocity in the kinetic head does not equal to the square of themean velocity. The factor by which the term U2/2g must be multiplied toget the exact value of the kinetic head is known as kinetic energy correction

factor. Fortunately, for many practical flows this factor is close to 1. See therecommended literature for more details.

For a pipe with constant cross-section area U is constant along the pipe.Then, for a horizontal pipe the steady flow energy equation (2) takes theform:

P1 − P2 = ρ g hf . (3)

For viscous fluid hf > 0, which means that steady flow of such fluid in a pipeis possible only if the pressure gradient is applied along the pipe. The headloss hf along the pipe can be conveniently measured by tube manometers.Referring to figure we can see that Pa = P1 − ρ g z1 = P2 − ρ g z2, and

hf = z1 − z2.

The head loss along the length L of the pipe is due to the friction on thepipe walls. We chose a cylindrical fluid particle as shown on figure . Theparticle is moving with a constant velocity, that is the total force acting onit is zero. This means that the pressure force on the vertical surfaces of theparticle is balanced by the friction on the pipe wall. We can write:

π d2

4(P1 − P2) = π d L τw , (4)

where τ is the frictional shear stress on the pipe wall, which can be expressedin terms of the non-dimensional friction coefficient or friction factor of thepipe:

f =τw

ρ U2/2.

Substituting back into (3) and (4) we obtain the Darcy equation for headloss in a pipe:

hf = 4 fL

d

U2

2 g. (5)

Classical investigations of flows in pipes was performed by Osborn Reynolds

who published his classical experimental results in 18831. In one of his ex-periments Reynolds studied the dependence of pressure gradient along a pipe

1 O.Reynolds (1883) An experimental investigation of the circumstances which deter-mine whether the motion of water shall be direct or sinuous, and the law of resistancein parallel channels, Phil. Trans. Roy. Soc. 174, 935–82. Available online via JSTOR:http://www.jstor.org/view/03701662/ap000029/00a00170/0

5

Fig. 3: The experimental apparatus and results of experiments on flow inpipes from the original Reynolds paper.

6

from the pipe flow rate. The sketch of experimental apparatus and an ex-ample of the obtained results are illustrated on figure 3. Reynolds found thelinear dependence for low flow rates, when the flow in the pipe is laminar. Inlogarithmic coordinates used on the figure this dependence is represented bythe straight line with the slope 1. After the flow rate reaches some criticalvalue rapid changes of flow characteristics occurs in the narrow region of flowrates (critical region), and the line inclination decreases. This new behaviourroughly corresponds to a power low with an exponent n < 1. For these largervalues of flow rates flow in the pipe becomes turbulent leading to significantchanges of flow characteristics. Reynolds found that the divergence from thelaminar behaviour for a pipe flow always starts when the non-dimensionalparameter known now as Reynolds number

Re =U d

ν, (6)

reaches a certain critical value Rec. This value was found to be about 2300and does not depends on pipe diameter and could be slightly lower for pipeswith a rough surface.

Extensive experiments for determination of friction factor f for circularpipes of different diameters and wall roughness carrying flows of various meanvelocities had been carried by L.F.Moody in 1944. The results of theseexperiments are represented in the form of Moody diagram (figure 4),which is widely used for calculating flows trough pipes. It turns out thatfriction factor depends on the Reynolds number and on the relative roughnessof the pipe wall k/d.

Regions with different behaviour of f can be observed on the Moodydiagram. For small values of the Reynolds number (laminar flows) the frictionfactor does not depends on the wall roughness, and is specified by a simpleformula

f =16

Re. (7)

In the narrow critical zone flow becomes turbulent and for larger Reynoldsnumbers f depends on both Re and k/d (transition zone). For very largeReynolds numbers (complete turbulence) f depends on k/d only. Moodydiagram can be used to find the flow rate through a given pipe, when a givenpressure difference is applied to pipe ends. For a laminar flow, using (3), (5),(6) and (7) we obtain the Poiseuille equation:

Q =π R4

8 µ

∆P

L,

where µ = ρ ν is dynamic viscosity. In the case of complete turbulence,keeping in mind that f does not depend on Re and therefore is constant for

7

Fig. 4: Moody diagram

8

a given pipe, we get:

Q = π

√

R5

f ρ

∆P

L.

Note different dependence of flow rate on pressure gradient in the pipe forlaminar flow and complete turbulent flow. In the former case, the flow rateis proportional to pressure gradient, and in the latter case the flow rateproportional to the square root of pressure gradient.

Head losses in pipe systems

The steady flow energy equation and the theory of pipe flows with frictionallosses considered on the previous lecture provide the basis for calculation offlows through systems of pipes and for designing of pipelines. The Darcyequation (5) can be used to specify the value of the head loss in a pipe whichshould be compensated by a certain amount of energy transferred to the fluid(pumping, height or pressure difference between the pipe ends) to providethe required mean velocity U and therefore the required flow rate throughthat pipe. In the most of practical cases the flow in pipelines is turbulent,when the friction coefficient f depends only on the relative wall roughnessas can be seen from the Moody diagram (figure 4) Then the equation (5) foreach pipe in a pipeline can be rewritten as:

hf = Kf

U2

2 g(8)

where the loss coefficient Kf depends only on the properties of a particularpipe in the pipeline and does not depend on the flow rate.

Pipes are not the only elements of pipelines where head loss is possible.The head loss can also occur in pipe fittings, bends, contractions, etc. Flowrate through a pipeline can be regulated by changing a head loss in valves.Equation (8) provides the general form of equations used to calculate headloss in various elements of a pipeline, with a specific empirical (that is foundfrom an experiment) loss coefficient Kf for each such an element. Examplesof pipe line elements with loss coefficients can be found in the literature.

Pipes in series

If pipes or other elements are connected in series, that is from end to end, thetotal head loss is the sum of losses in all individual elements. It is convenientto express equations (5) and (8) for each element by using the flow rateQ = A U , which is the same for all elements connected in series. Then the

9

z 2

z 11

2

.

.

h f

Fig. 5:

head loss for an individual element i can be written as hi = κi Q2, with a

suitable coefficient κi for each element. Then the head loss of the entiresystem is

hf = Q2∑

i

κi .

Example:

Water flows between tanks with water levels z1 and z2 trough two identi-cal pipes of length l and diameter d (figure 5). The pipes are connectedby an elbow with the loss coefficient K1 = 0.1. Sharp edged inlet andoutlet have the loss coefficients K2 = 0.5 and K3 = 1 respectively. Thefriction coefficient f of the pipes is constant for given flow conditions.Find the flow rate of fluid in the pipes.

For a steady flow the head loss along the path 1–2 should be balanced bythe difference of the total head at points 1 and 2. Pressure at points 1 and2 is the same, and velocities there are negligibly small. Therefore, only thepotential head contributes to the total head difference. That is

hf = z1 − z2.

The head loss along the pipes is the sum of individual losses

hf =

(

4 fl

d+ 4 f

l

d+ K1 + K2 + K3

)

U2

2 g,

and the corresponding flow rate is

Q =πd2

4

√

2 g ( z1 − z2 )

4 f(l/d) + K1 + K2 + K3

.

10

21

B

A

C

Fig. 6:

Parallel pipes

The flow divides between two or more pipes and then comes together again.For such pipe systems the sum of flow rates through individual componentsis the entire flow rate through the system

Q =∑

i

Qi.

For each element of the parallel pipe system the difference of the total headbetween its ends is the same, which means that all elements have the samehead loss

κi Q2

i = hh .

For an N -element system (i = 1, 2, 3, . . .N) we usually have an unknownhead loss hf and N unknown flow rates Qi. To find these N + 1 values wecan use N + 1 equations above.

Example:

Two identical pipes A and B have length L and cross section area A.The pipes are connected in parallel to a pipeline with a constant flowrate Q (figure 6). A valve C is used to regulate flows through the pipes.When the valve is fully opened its loss coefficient is K = 0.3. Frictioncoefficient f of the pipes is constant under the given flow conditions,and the losses in fittings are negligible. Find the minimal and themaximal flow rate trough pipe A.

Head loss in each pipe between 1 and 2 is equal to the difference in the totalheads between these points:

H1 − H2 = hA = hB .

11

Head losses in A and B are:

hA,B = 4 fL

d

Q2

A

2 A2g= (4 f

L

d+ K)

Q2

B

2 A2g

and the sum of flow rates in A and B is the total flow rate of the system:

QA + QB = Q.

For the flow rate QA we obtain the following quadratic equation

Q2

A − 2 Q(1 +4 f L

d K) QA + Q2(1 +

4 f L

d K) = 0

with solutions

QA = Q

(

1 +4 f L

d K±√

4 f L

d K( 1 +

4 f L

d K)

)

. (9)

The flow rate QA should be less then Q, therefore we should take the solutionwith the minus sign. Maximal QA corresponds to the closed valve (K = ∞),when all fluid flows through the pipe A: QA = Q. The minimal QA corre-sponds to the fully opened valve, and its value can be found by substitutingthe value of K into the formula (9). Note, that if K = 0 (no extra valveresistance) both parallel branches are identical, and have the same flow rateQA = QB = Q/2. Equation (9) gives value QA = Q/2 in the limit K → 0.

Pipe branches

A classical problem with a branching pipeline is the three reservoir prob-

lem illustrated on figure 7. Three reservoirs with different water levels areconnected by three pipes with a junction point 0 and unknown flow rates.One of the difficulties of the problem is that we usually do not know theflow direction in one of the branches (branch 3) before solving the problem.General principles applied for solving the three reservoir problem and otherproblems with branching pipes are:

1. The uniqueness of the total head. This means that at each point of apipeline the total head have only one value. The important consequenceof this property is that the value of the total head at a junction pointis the same for all pipes.

2. The continuity principle is applied to a junction point. That is the flowinto the junction is the same as the flow from the junction.

12

z 1

z 2

z 3

z 0Q 1

Q 3

Q 2

0

.

.

.

3

1

2

Fig. 7:

3. Darcy’s equation is applied to each pipe with additional losses in fit-tings. Note, that for long pipelines the frictional losses in pipes providethe major contribution to the total head loss and minor losses in fittingcan often be neglected.

As before, we represent the head loss in each branch in the form

hi = κi Q2

i .

Assuming originally the direction of the flow in the pipe 0–3 as shown on thefigure 7 and using the principles stated above we can write:

z1 − H0 = κ1 Q2

1

H0 − z2 = κ1 Q2

2

H0 − z3 = κ1 Q2

3

Q1 = Q2 + Q3.

Thus, we have 4 equations to specify three unknown flow rates Q1, Q2, Q3,and the unknown total head H0 at the junction. The algebraic solution ofthese equations is tedious and not possible for more than 3 pipes. However,we can use the trial and error method, taking some value of H0 as an initialguess, calculating the flow rates and then checking the continuity condition.

13

Q

H 0

.

.

H∆Ps

Fig. 8:

I is useful to use trial points to plot the graph of Q1 − Q2 − Q3 as functionof H0. The required solution will be an intersection of this graph with thehorizontal axis. If the original choice of the flow direction in the pipe 0–3is wrong, the solution will not be possible. For an opposite flow in 0–3 theequations become:

z1 − H0 = κ1 Q2

1

H0 − z2 = κ1 Q2

2

z3 − H0 = κ1 Q2

3

Q1 + Q3 = Q2.

The two systems of equations become identical if H0 = z3 in which caseQ3 = 0. It is convenient to chose H0 = z3 as an initial guess to specifythe flow direction in 0–3. If Q1 > Q2, then the flow is from 0 to 3, and ifQ2 > Q1, the flow is from 3 to 0.

Pumping

A pump can can deliver water to a higher level by transmitting energy to theflow. Pumps are characterised by the total head ∆H applied to the fluid,discharge Q and the shaft horsepower Ps. A particular pump can provide ahead increase ∆H for a discharge Q and requires power Ps on the shaft. Thecurve expressing the relationship between pump discharge and head is calledthe characteristic curve or head curve, and the curve specifying the requiredpower is the power curve. A typical example of pump characteristics is given

15

on figure 9, where B.H.P. stands for brake horsepower, that is the power ofan engine driving the pump shaft. The power transmitted by a pump towater flow or water horsepower is specified as

Pw = ρ g Q ∆H,

and the pump efficiency isη = Pw/Ps,

that is the ratio of utilised power to the total power required by a pump,and the value of η is always less then 1. To specify the working regime of apump in a pipeline we have to equate the head provided by the pump at aspecific flow rate with the pumping head H0 (figure 8) and head losses in thepipeline for that flow rate:

∆H = H0 + κ Q2,

where the value of κ can be regulated by the valve, which will change theflow in the pipeline. The problem can be solved graphically, by plotting theparabola of the required head H0 + κ Q2 on the pump diagram and findingits intersection with the characteristic curve of the pump.

Quasi-steady flows

The energy equation have been derived for the case of a steady flow. For aflow in a pipeline this means that flow rates and heads at different points ofthe pipeline do not depend on time. However, if flow changes very slowlyand unsteady effects can be neglected, we can apply the steady flow energyequation with sufficient accuracy at any instant during the process. Unsteadyflows with slowly changing parameters which can be assumed steady at eachtime moment are called quasi-steady flows. For example, if a tank with alarge area of water surface A (figure 10) is drained through a pipe of a muchsmaller cross section a the water level in the tank will change very slowlyand we can calculate the flow rate Q through the pipe at each time instant tby taking the current value of the water level Z(t) and applying the steadyflow energy equation as if Z was constant:

H1 − H3 = KU2

2 g,

where the total heads of the water surface in the tank of the jet at the pipeoutlet are

H1 = Z +Pa

ρ gand H3 =

U2

2 g+

Pa

ρ g

16

Z

Q ( t ) = − A dZ / dtZ Z (

t )

1A

2

3a

.1

2

Fig. 10:

respectively, and K is the total loss coefficient of the pipeline including fric-tion losses in the pipe, fittings losses, entry losses, etc. This gives

Z(t) = (1 + K)Q2

2 a2 g,

and using the relation between the flow rate and the water level Q = −A dZ/dtwe obtain the following differential equation describing the evolution of thewater level in the tank:

dZ

dt= − a

A

√

2 g

1 + K

√

Z(t) .

We can see that when the value of the area ratio a/A is small and the waterlevel Z(t) is not too big, then the non-stationary term dZ/dt is small andthe application of the steady flow energy equation is justified. This equationcan be used to calculate the time required to change the level of water in thetank from the initial level Z(0) = z1 to any given level Z(t) = z:

t = −A

a

√

1 + K

2 g

z∫

z1

dZ√Z

=A

a

√

1 + K

2 g(√

z1 −√

z ) .

This equation can also be used to determine the level z left after a given timet has elapsed.

First year ﬂuid mechanics - University College Londonuceseug/Fluids1/2/Notes.pdf · First year...

Documents

Transcript of First year ﬂuid mechanics - University College Londonuceseug/Fluids1/2/Notes.pdf · First year...