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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 more
general steady flow energy equation, which includes viscose losses and
work transfer to the fluid. These effects are accounted for by introducing
additional terms into Bernoulli’s equation.
Let us consider the balance of energy inside volume abcd (figure 1), which
is bounded by the walls of a stream tube and two cross sections 1 and 2 at
heights z1,2 from an arbitrary datum level. In this derivation we assume that
the fluid velocity does not change across the stream tube. The fluid velocities
at 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 power
P (e.g. pumps or turbines operate there), and Pf is the power of frictional
forces inside the control volume and on its boundaries. During the small
time period ∆t the fluid inside the volume a′ add′ enters the control volume
and the fluid inside the volume bcc′ b′ leaves the control volume. According
the the continuity principles for a steady flow these volumes are the same:
A1 V1 ∆t = A2 V2 ∆t = Q ∆t,
where Q is the volumetric flow rate of fluid. The energy entering the control
volume with the fluid through the cross section 1 is
V12
+ g z1 ) ρ Q ∆t,
2
and includes the kinetic energy of moving fluid and the potential energy of
fluid in the gravitational field. Analogously, the energy leaving the control
volume through the cross section 2 is
∆E1 = (
V12
+ g z2 ) ρ Q ∆t.
2
Pressure P1 acting on the moving boundary a′ d′ produces work P1 A1 V1 ∆t =
P1 Q ∆t, which should be added to the incoming energy E1 , and similarly
∆E2 = (
1
2
b
P∆ t
a
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E
P1
z1
d’
V1
V2
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∆ E2
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c’
1
V1 ∆ t
A1
V2 ∆ t
z2
Pf ∆ t
d
Fig. 1:
the pressure work P2 Q ∆t should be added to the outgoing energy E2 . The
energy of the fluid in the control volume will also be increased by the work
of external forces P ∆t and dissipated due to the work of friction Pf ∆t. For
a stationary flow the energy of the fluid inside the control volume does not
change, and the total gain of energy is equal to the total energy loss:
ρ V12
ρ V22
(
+ρ g z1 ) Q ∆t+P1 Q ∆t+P∆t = (
+ρ g z2 ) Q ∆t+P2 Q ∆t+Pf ∆t.
2
2
This gives the steady flow energy equation in the following form:
P1 +
P
ρ V22
Pf
ρ V12
+ ρ g z1 + = P2 +
+ ρ g z2 +
,
2
Q
2
Q
(1)
which represents conservation of energy per unit of fluid volume. It should
be noted that the sign of the external power P is positive if the work done
on the fluid (pumps, compressors) and negative when work done by the fluid
(turbines). In engineering it is conventional to use the energy equation per
unit of fluid weight, which can be obtained dividing (1) by ρg. The energy
of fluid per unit of weight has dimensions of meters (Joules/Newtons) and is
called head. The value
P
V2
H=
+
+z
ρg
2g
is called total head and represents the total energy of a unit weight of flowing
fluid. It consists of pressure head, velocity head and potential head. The
steady flow energy equation can now be formulated in terms of change of
total head:
P
H2 − H1 =
− hf ,
(2)
ρgQ
3
z2
z1
r
R
U
P1
d
V(r)
P2
x
τ
L
Fig. 2:
where hf = Pf /(ρ g Q) is head loss. Therefore, the total head of a steadily
flowing fluid is increased by external work transferred to the unit of fluid
weight, 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 specifying flows of viscous fluid through pipes and pipe systems. In this section
we consider a steady flow through a straight circular pipe of the internal
radius R, diameter d (figure ). The flow in the pipe is parallel (this means
that pressure is constant across the pipe), and the velocity profile does not
change along the pipe. Such flow is called fully developed, and occurs far
enough 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 the
fluid through the pipe as the flow velocity. From the last term you should remember that the mean velocity U is a uniform velocity, which would provide
the same flow rate Q through the pipe as the actual velocity profile V (r):
Q = π R2 U = 2 π
ZR
0
r V (r) dr.
4
Taking an average velocity as the flow velocity we make a mistake in determining the exact value of the kinetic energy of the flow because the mean of
the square of velocity in the kinetic head does not equal to the square of the
mean velocity. The factor by which the term U 2 /2g must be multiplied to
get 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 the
recommended 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 the
form:
P1 − P2 = ρ g hf .
(3)
For viscous fluid hf > 0, which means that steady flow of such fluid in a pipe
is possible only if the pressure gradient is applied along the pipe. The head
loss 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 the
pipe walls. We chose a cylindrical fluid particle as shown on figure . The
particle is moving with a constant velocity, that is the total force acting on
it is zero. This means that the pressure force on the vertical surfaces of the
particle is balanced by the friction on the pipe wall. We can write:
π d2
(P1 − P2 ) = π d L τw ,
(4)
4
where τ is the frictional shear stress on the pipe wall, which can be expressed
in terms of the non-dimensional friction coefficient or friction factor of the
pipe:
τw
f=
.
ρ U 2 /2
Substituting back into (3) and (4) we obtain the Darcy equation for head
loss in a pipe:
L U2
hf = 4 f
.
(5)
d 2g
Classical investigations of flows in pipes was performed by Osborn Reynolds
who published his classical experimental results in 18831. In one of his experiments Reynolds studied the dependence of pressure gradient along a pipe
1
O.Reynolds (1883) An experimental investigation of the circumstances which determine whether the motion of water shall be direct or sinuous, and the law of resistance
in 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 in
pipes from the original Reynolds paper.
6
from the pipe flow rate. The sketch of experimental apparatus and an example of the obtained results are illustrated on figure 3. Reynolds found the
linear dependence for low flow rates, when the flow in the pipe is laminar. In
logarithmic coordinates used on the figure this dependence is represented by
the straight line with the slope 1. After the flow rate reaches some critical
value rapid changes of flow characteristics occurs in the narrow region of flow
rates (critical region), and the line inclination decreases. This new behaviour
roughly corresponds to a power low with an exponent n < 1. For these larger
values of flow rates flow in the pipe becomes turbulent leading to significant
changes of flow characteristics. Reynolds found that the divergence from the
laminar behaviour for a pipe flow always starts when the non-dimensional
parameter known now as Reynolds number
Ud
,
(6)
ν
reaches a certain critical value Rec . This value was found to be about 2300
and does not depends on pipe diameter and could be slightly lower for pipes
with a rough surface.
Extensive experiments for determination of friction factor f for circular
pipes of different diameters and wall roughness carrying flows of various mean
velocities had been carried by L.F.Moody in 1944. The results of these
experiments are represented in the form of Moody diagram (figure 4),
which is widely used for calculating flows trough pipes. It turns out that
friction factor depends on the Reynolds number and on the relative roughness
of the pipe wall k/d.
Regions with different behaviour of f can be observed on the Moody
diagram. For small values of the Reynolds number (laminar flows) the friction
factor does not depends on the wall roughness, and is specified by a simple
formula
16
f=
.
(7)
Re
In the narrow critical zone flow becomes turbulent and for larger Reynolds
numbers f depends on both Re and k/d (transition zone). For very large
Reynolds numbers (complete turbulence) f depends on k/d only. Moody
diagram can be used to find the flow rate through a given pipe, when a given
pressure difference is applied to pipe ends. For a laminar flow, using (3), (5),
(6) and (7) we obtain the Poiseuille equation:
Re =
Q=
π R4 ∆P
,
8µ 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
s
R5 ∆P
.
fρ L
Note different dependence of flow rate on pressure gradient in the pipe for
laminar flow and complete turbulent flow. In the former case, the flow rate
is proportional to pressure gradient, and in the latter case the flow rate
proportional to the square root of pressure gradient.
Head losses in pipe systems
The steady flow energy equation and the theory of pipe flows with frictional
losses considered on the previous lecture provide the basis for calculation of
flows through systems of pipes and for designing of pipelines. The Darcy
equation (5) can be used to specify the value of the head loss in a pipe which
should be compensated by a certain amount of energy transferred to the fluid
(pumping, height or pressure difference between the pipe ends) to provide
the required mean velocity U and therefore the required flow rate through
that pipe. In the most of practical cases the flow in pipelines is turbulent,
when the friction coefficient f depends only on the relative wall roughness
as can be seen from the Moody diagram (figure 4) Then the equation (5) for
each pipe in a pipeline can be rewritten as:
hf = Kf
U2
2g
(8)
where the loss coefficient Kf depends only on the properties of a particular
pipe 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. Flow
rate through a pipeline can be regulated by changing a head loss in valves.
Equation (8) provides the general form of equations used to calculate head
loss in various elements of a pipeline, with a specific empirical (that is found
from an experiment) loss coefficient Kf for each such an element. Examples
of 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, the
total head loss is the sum of losses in all individual elements. It is convenient
to express equations (5) and (8) for each element by using the flow rate
Q = A U, which is the same for all elements connected in series. Then the
9
1
hf
.
2
z2
z1
.
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 entire
system is
X
hf = Q2
κi .
i
Example:
Water flows between tanks with water levels z1 and z2 trough two identical pipes of length l and diameter d (figure 5). The pipes are connected
by an elbow with the loss coefficient K1 = 0.1. Sharp edged inlet and
outlet have the loss coefficients K2 = 0.5 and K3 = 1 respectively. The
friction 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 by
the difference of the total head at points 1 and 2. Pressure at points 1 and
2 is the same, and velocities there are negligibly small. Therefore, only the
potential head contributes to the total head difference. That is
hf = z1 − z2 .
The head loss along the pipes is the sum of individual losses
2
l
l
U
hf = 4 f + 4 f + K1 + K2 + K3
,
d
d
2g
and the corresponding flow rate is
s
2
πd
2 g ( z1 − z2 )
Q=
.
4
4 f (l/d) + K1 + K2 + K3
10
B
C
1
2
A
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 components
is the entire flow rate through the system
X
Q=
Qi .
i
For each element of the parallel pipe system the difference of the total head
between its ends is the same, which means that all elements have the same
head loss
κi Q2i = hh .
For an N-element system (i = 1, 2, 3, . . . N) we usually have an unknown
head loss hf and N unknown flow rates Qi . To find these N + 1 values we
can 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 flow
rate 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. Friction
coefficient f of the pipes is constant under the given flow conditions,
and the losses in fittings are negligible. Find the minimal and the
maximal flow rate trough pipe A.
Head loss in each pipe between 1 and 2 is equal to the difference in the total
heads between these points:
H1 − H2 = hA = hB .
11
Head losses in A and B are:
hA,B = 4 f
L Q2A
L
Q2B
=
(4
f
+
K)
d 2 A2 g
d
2 A2 g
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
Q2A − 2 Q(1 +
4f L
4f L
) QA + Q2 (1 +
)=0
dK
dK
with solutions
QA = Q
4f L
±
1+
dK
r
4f L
4f L
(1 +
)
dK
dK
!
.
(9)
The flow rate QA should be less then Q, therefore we should take the solution
with the minus sign. Maximal QA corresponds to the closed valve (K = ∞),
when all fluid flows through the pipe A: QA = Q. The minimal QA corresponds to the fully opened valve, and its value can be found by substituting
the value of K into the formula (9). Note, that if K = 0 (no extra valve
resistance) both parallel branches are identical, and have the same flow rate
QA = 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 problem illustrated on figure 7. Three reservoirs with different water levels are
connected 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 the
flow direction in one of the branches (branch 3) before solving the problem.
General principles applied for solving the three reservoir problem and other
problems with branching pipes are:
1. The uniqueness of the total head. This means that at each point of a
pipeline the total head have only one value. The important consequence
of this property is that the value of the total head at a junction point
is the same for all pipes.
2. The continuity principle is applied to a junction point. That is the flow
into the junction is the same as the flow from the junction.
12
1
.
3
.
Q1
Q3
z1
0
z3
Q2
2
z2
z0
.
Fig. 7:
3. Darcy’s equation is applied to each pipe with additional losses in fittings. Note, that for long pipelines the frictional losses in pipes provide
the major contribution to the total head loss and minor losses in fitting
can often be neglected.
As before, we represent the head loss in each branch in the form
hi = κi Q2i .
Assuming originally the direction of the flow in the pipe 0–3 as shown on the
figure 7 and using the principles stated above we can write:
z1 − H0 = κ1 Q21
H0 − z2 = κ1 Q22
H0 − z3 = κ1 Q23
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 of
these 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 initial
guess, calculating the flow rates and then checking the continuity condition.
13
.
H0
Q
∆H
Ps
.
Fig. 8:
I is useful to use trial points to plot the graph of Q1 − Q2 − Q3 as function
of H0 . The required solution will be an intersection of this graph with the
horizontal axis. If the original choice of the flow direction in the pipe 0–3
is wrong, the solution will not be possible. For an opposite flow in 0–3 the
equations become:
z1 − H0 = κ1 Q21
H0 − z2 = κ1 Q22
z3 − H0 = κ1 Q23
Q1 + Q3 = Q2 .
The two systems of equations become identical if H0 = z3 in which case
Q3 = 0. It is convenient to chose H0 = z3 as an initial guess to specify
the flow direction in 0–3. If Q1 > Q2 , then the flow is from 0 to 3, and if
Q2 > Q1 , the flow is from 3 to 0.
Pumping
A pump can can deliver water to a higher level by transmitting energy to the
flow. Pumps are characterised by the total head ∆H applied to the fluid,
discharge Q and the shaft horsepower Ps . A particular pump can provide a
head increase ∆H for a discharge Q and requires power Ps on the shaft. The
curve expressing the relationship between pump discharge and head is called
the characteristic curve or head curve, and the curve specifying the required
power is the power curve. A typical example of pump characteristics is given
κ1
κ2
H0
14
c BJM Pumps http://www.bjmpumps.com
Fig. 9: Typical pump characteristics. 15
on figure 9, where B.H.P. stands for brake horsepower, that is the power of
an engine driving the pump shaft. The power transmitted by a pump to
water flow or water horsepower is specified as
Pw = ρ g Q ∆H,
and the pump efficiency is
η = 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 a
pump in a pipeline we have to equate the head provided by the pump at a
specific flow rate with the pumping head H0 (figure 8) and head losses in the
pipeline for that flow rate:
∆H = H0 + κ Q2 ,
where the value of κ can be regulated by the valve, which will change the
flow in the pipeline. The problem can be solved graphically, by plotting the
parabola of the required head H0 + κ Q2 on the pump diagram and finding
its 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 a
flow in a pipeline this means that flow rates and heads at different points of
the pipeline do not depend on time. However, if flow changes very slowly
and unsteady effects can be neglected, we can apply the steady flow energy
equation with sufficient accuracy at any instant during the process. Unsteady
flows with slowly changing parameters which can be assumed steady at each
time moment are called quasi-steady flows. For example, if a tank with a
large area of water surface A (figure 10) is drained through a pipe of a much
smaller cross section a the water level in the tank will change very slowly
and we can calculate the flow rate Q through the pipe at each time instant t
by taking the current value of the water level Z(t) and applying the steady
flow energy equation as if Z was constant:
H1 − H3 = K
U2
,
2g
where the total heads of the water surface in the tank of the jet at the pipe
outlet are
Pa
U2
Pa
H1 = Z +
and H3 =
+
ρg
2g ρg
16
A
1
.
Z(t)
Z1
2
Z2
Q ( t ) = − A dZ / dt
a
3
Fig. 10:
respectively, and K is the total loss coefficient of the pipeline including friction 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/dt
we obtain the following differential equation describing the evolution of the
water level in the tank:
r
dZ
a
2g p
=−
Z(t) .
dt
A 1+K
We can see that when the value of the area ratio a/A is small and the water
level Z(t) is not too big, then the non-stationary term dZ/dt is small and
the application of the steady flow energy equation is justified. This equation
can be used to calculate the time required to change the level of water in the
tank from the initial level Z(0) = z1 to any given level Z(t) = z:
s
s
Zz
√
A 1+K
dZ
A 1+K √
√ =
t=−
( z1 − z ) .
a
2g
a
2g
Z
z1
This equation can also be used to determine the level z left after a given time
t has elapsed.