ENERGY LOSSES IN FLOW
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Friction in Pipes
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Energy Losses in Bends and Fittings
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Pressure Drop through Equipment
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Equivalent Lengths of Pipe
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Compressibility Effects for Gases
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Calculation of Pressure Drops in Flow Systems
Energy losses can occur through friction in pipes, bends and fittings, and in equipment.
Friction in Pipes
If we define the wall friction in terms of velocity pressure of the fluid flowing, we can write:
![FIG. 3.7. Energy balance over a length of pipe]()
Figure 3.7. Energy balance over a length of pipe.
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
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Figure 3.8 Friction factors in pipe
(After Moody,1944)
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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.
And so the pressure drop in 170 m, from eqn. (3.17)
DPf = (4frv2/2) x (L/D)
Friction in PipesEnergy Losses in Bends and FittingsPressure Drop through EquipmentEquivalent Lengths of PipeCompressibility Effects for GasesCalculation of Pressure Drops in Flow SystemsEnergy 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 = (frv2/2) pD x dL,Integrating between L1 and L2, in which interval P goes from P1 to P2 we have:
therefore
dP = 4(frv2/2) x dL/D
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)
dP = 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 eare 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
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 pipeCalculate 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,From Appendix 4,
Area of cross-section A
= (p/4)D2
= p/4 x (0.05)2
= 1.96 x 10-3 m2
Viscosity of olive oil at 20°C = 84 x 10-3 Ns m-2 and density = 910 kg m-3,Now (Re) = (Dvr/m)
and velocity = (0.1 x 1/60)/(1.96 x 10-3) = 0.85 m s-1,
= [(0.05 x 0.85 x 910)/(84 x 10-3)]so that the flow is streamline, and from Fig. 3.8, for (Re) = 460
= 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.
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
Energy is also lost at sudden changes in pipe cross-section. At a sudden enlargement the loss has been shown to be equal to: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 | |
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
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 |
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).
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.
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.
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,
(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.