ENERGY LOSSES IN FLOW
Friction
in Pipes
Energy
Losses in Bends and Fittings
Pressure
Drop through Equipment
Equivalent
Lengths of Pipe
Compressibility
Effects for Gases
Calculation
of Pressure Drops in Flow Systems
Energy losses can occur through friction in pipes, bends and fittings,
and in equipment.
Friction in Pipes
In Bernouilli's equation
the symbol Eƒ
was used to denote the energy loss due to friction in the pipe. This loss
of energy due to friction was shown, both theoretically and experimentally,
to be related to the Reynolds number for the flow. It has also been found
to be proportional to the velocity pressure of the fluid and to a factor
related to the smoothness of the surface over which the fluid is flowing.
If
we define the wall friction in terms of velocity pressure of the fluid flowing,
we can write:
F/A = f rv2/2
(3.16)
where
F is the friction force, A is the area over which the friction
force acts, r
is the density of the fluid, v is the velocity of the fluid, and
f
is a coefficient called the friction factor.
Consider
an energy balance over a differential length, dL, of a straight
horizontal pipe of diameter D, as in Fig. 3.7.
Figure 3.7. Energy balance over a length of pipe.
Consider
the equilibrium of the element of fluid in the length dL. The
total force required to overcome friction drag must be supplied by a pressure
force giving rise to a pressure drop dP along the length dL.
The pressure drop
force is:
dP x Area of pipe = dP x pD2/4
The friction force is (force/unit area) x wall area of pipe
=
F/A x pD
x dL
so from eqn. (3.16),
= (frv2/2)
x pD
x dL
Therefore equating
prressure drop and friction force
(pD2/4)
dP = (f rv2/2)
pD
x dL,
therefore
dP = 4(f
rv2/2)
x dL/D
Integrating between
L1 and L2, in which interval P
goes from P1 to P2 we have:
dP
= 4(frv2/2)
x dL/D
P1 - P2
= (4frv2/2)(L1
- L2)/D
i.e.
DPf
= (4frv2/2)
x (L/D)
(3.17)
or
Eƒ = DPf/r
= (2fv2)(L/D)
(3.18)
where
L = L1 - L2 = length of pipe in which the
pressure drop, DPf
= P1 - P2 is the frictional pressure
drop, and Eƒ is the frictional loss of energy.
Equation
(3.17) is an important equation; it is known as the Fanning equation,
or sometimes the D'Arcy or the Fanning-D'Arcy equation. It is used to
calculate the pressure drop that occurs when liquids flow in pipes.
The
factor f in eqn.(3.17) depends upon the Reynolds number for the
flow, and upon the roughness of the pipe. In Fig.
3.8 experimental results are plotted, showing the relationship of
these factors. If the Reynolds number and the roughness factor are known,
then f can be read off from the graph.
Figure 3.8 Friction factors in pipe
(After Moody,1944)
It
has not been found possible to find a simple expression that gives analytical
equations for the curve of Fig. 3.8, although the curve can be approximated
by straight lines covering portions of the range. Equations can be written
for these lines. Some writers use values for fwhich differ from
that defined in eqn. (3.16) by numerical factors of 2 or 4. The same symbol,
f, is used so that when reading off values for f, its definition
in the particular context should always be checked. For example, a new
f = 4f removes one numerical factor from eqn. (3.17).
Inspection
of Fig. 3.8 shows that for low values of (Re), there appears to be a simple
relationship between ƒ
and (Re) independent of the roughness of the pipe. This is perhaps not
surprising, as in streamline flow there is assumed to be a stationary
boundary layer at the wall and if this is stationary there would be no
liquid movement over any roughness that might appear at the wall. Actually,
the friction factor f in streamline flow can be predicted theoretically
from the Hagen-Poiseuille equation, which gives:
f =
16/(Re)
(3.19)
and this applies
in the region 0 < (Re) < 2100.
In
a similar way, theoretical work has led to equations which fit other regions
of the experimental curve, for example the Blasius equation which
applies to smooth pipes in the range 3000 < (Re) < 100,000 and in
which:
f
|
ƒ
= |
0.316
|
(
Re)-0.25 |
(3.19) |
|
|
|
|
4
|
|
|
In
the turbulent region, a number of curves are shown in Fig. 3.8. It would
be expected that in this region, the smooth pipes would give rise to lower
friction factors than rough ones. The roughness can be expressed in terms
of a roughness ratio that is defined as the ratio of average height of
the projections, which make up the "roughness" on the wall of
the pipe, to the pipe diameter. Tabulated values are given showing the
roughness factors for the various types of pipe, based on the results
of Moody (1944). These factors e are
then divided by the pipe diameter D to give the roughness ratio
to be used with the Moody graph. The question of relative
roughness of the pipe is under some circumstances a difficult one to resolve.
In most cases, reasonable accuracy can be obtained by applying Table
3.1 and Fig. 3.8.
TABLE
3.1
RELATIVE ROUGHNESS FACTORS FOR PIPES
| Material |
Roughness
factor (e)
|
|
Material |
Roughness
factor (e)
|
| Riveted
steel |
0.001-
0.01 |
|
Galvanized
iron |
0.0002 |
| Concrete |
0.0003
- 0.003 |
|
Asphalted
cast iron |
0.001 |
| Wood
staves |
0.0002
- 0.003 |
|
Commercial
steel |
0.00005 |
| Cast
iron |
0.0003 |
|
Drawn
tubing |
Smooth |
EXAMPLE
3.10. Pressure drop in a pipe
Calculate the pressure drop along 170 m of 5 cm diameter horizontal steel
pipe through which olive oil at 20°C is flowing at the rate of 0.1
m3 min-1.
Diameter of pipe
= 0.05 m,
Area of cross-section A
= (p/4)D2
= p/4
x (0.05)2
= 1.96 x 10-3 m2
From Appendix 4,
Viscosity of olive
oil at 20°C = 84 x 10-3 Ns m-2 and density
= 910 kg m-3,
and velocity = (0.1 x 1/60)/(1.96 x 10-3) = 0.85 m
s-1,
Now
(Re) = (Dvr/m)
= [(0.05 x 0.85 x 910)/(84 x 10-3)]
= 460
so that the flow
is streamline, and from Fig. 3.8, for (Re) = 460
f =
0.03.
Alternatively for
streamline flow from (3.18), f = 16/(Re) = 16/460 =
0.03 as before.
And so the pressure
drop in 170 m, from eqn. (3.17)
DPf
= (4frv2/2)
x (L/D)
= [4 x 0.03
x 910 x (0.85)2 x 1/2] x [170 x 1/0.05]
= 1.34 x 105 Pa
= 134 kPa.
Energy Losses in Bends and Fittings
When
the direction of flow is altered or distorted, as when the fluid is flowing
round bends in the pipe or through fittings of varying cross-section,
energy losses occur which are not recovered. This energy is dissipated
in eddies and additional turbulence and finally lost in the form of heat.
However, this energy must be supplied if the fluid is to be maintained
in motion, in the same way, as energy must be provided to overcome friction.
Losses in fittings have been found, as might be expected, to be proportional
to the velocity head of the fluid flowing. In some cases the magnitude
of the losses can be calculated but more often they are best found from
tabulated values based largely on experimental results. The energy loss
is expressed in the general form,
Eƒ
= kv2/2
(3.20)
where k has
to be found for the particular fitting. Values of this constant k
for some fittings are given in Table 3.2.
TABLE
3.2
FRICTION LOSS
FACTORS IN FITTINGS
|
|
k
|
| Valves,
fully open: |
|
|
gate |
0.13
|
|
globe |
6.0
|
|
angle |
3.0
|
| Elbows: |
|
|
90° standard |
0.74
|
|
medium sweep |
0.5
|
|
long radius |
0.25
|
|
square |
1.5
|
| Tee,
used as elbow |
1.5
|
| Tee,
straight through |
0.5
|
| Entrance,
large tank to pipe: |
|
|
sharp |
0.5
|
|
rounded |
0.05
|
Energy is also lost
at sudden changes in pipe cross-section. At a sudden enlargement the loss
has been shown to be equal to:
Ef
= (v1 - v2)2/2
(3.21)
For a sudden contraction
Ef = kv22/2
(3.22)
where
v1 is the velocity upstream of the change in section and v2
is the velocity downstream of the change in pipe diameter from D1
to D2.
The coefficient k
in eqn. (3.22) depends upon the ratio of the pipe diameters (D2/D1)
as given in Table 3.3.
TABLE
3.3
LOSS FACTORS
IN CONTRACTIONS
| D2/D1 |
0.1
|
0.3
|
0.5
|
0.7
|
0.9
|
| k |
0.36
|
0.31
|
0.22
|
0.11
|
0.02
|
Pressure Drop through Equipment
Fluids
sometimes have to be passed through beds of packed solids; for example
in the air drying of granular materials, hot air may be passed upward
through a bed of the material. The pressure drop resulting is not easy
to calculate, even if the properties of the solids in the bed are well
known. It is generally necessary, for accurate pressure-drop information,
to make experimental measurements.
A
similar difficulty arises in the calculation of pressure drops through
equipment such as banks of tubes in heat exchangers. An equation of the
general form of eqn. (3.20) will hold in most cases, but values for k
will have to be obtained from experimental results. Useful correlations
for particular cases may be found in books on fluid flow and from works
such as Perry (1997) and McAdams
(1954).
Equivalent Lengths of Pipe
In
some applications it is convenient to clculate pressure drops in fittings
from added equivalent lengths of straight pipe, rather than directly in
terms of velocity heads or velocity pressures when making pipe-flow calculations.
This means that a fictitious length of straight pipe is added to the actual
length, such that friction due to the fictitious pipe gives rise to the
same loss as that which would arise from the fitting under consideration.
In this way various fittings, for example bends and elbows, are simply
equated to equivalent lengths of pipe and the total friction losses computed
from the total pipe length, actual plus fictitious. As Eƒ
in eqn. (3.20) is equal to Eƒ in eqn. (3.17), k
can therefore be replaced by 4ƒL/D where L is the length
of pipe (of diameter D) equivalent to the fitting.
Compressibility Effects for Gases
The
equations so far have all been applied on the assumption that the fluid
flowing was incompressible, that is its density remained unchanged through
the flow process. This is true for liquids under normal circumstances
and it is also frequently true for gases. Where gases are passed through
equipment such as dryers, ducting, etc., the pressures and the pressure
drops are generally only of the order of a few centimetres of water and
under these conditions compressibility effects can normally be ignored.
Calculation of Pressure Drops in Flow Systems
From
the previous discussion, it can be seen that in many practical cases of
flow through equipment, the calculation of pressure drops and of power
requirements is not simple, nor is it amenable to analytical solutions.
Estimates can, however, be made and useful generalizations are:
(1) Pressure drops through equipment are in general proportional to velocity
heads, or pressures; in other words, they are proportional to the square
of the velocity.
(2) Power requirements are proportional to the product of the pressure
drop and the mass rate of flow, which is to the cube of the velocity,
v2
x rAv
= rAv3.
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