energy loses in pipes

8/12/2019 Energy Loses in Pipes

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FLUID-FLOW THEORY(cont'd)ENERGY LOSSES IN FLOW

Friction in PipesEnergy Losses in Bends and Fittings

Pressure Drop through EquipmentEquivalent Lengths of PipeCompressibility Effects for GasesCalculation 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 Ewas used to denote the energy loss due to friction in the pipe. This lossof 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 afactor 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 v2/2 (3.16)where Fis the friction force, Ais the area over which the friction force acts, is the density of the fluid, v isthe velocity of the fluid, and fis a coefficient called the friction factor.

Consider an energy balance over a differential length, dL, of a straight horizontal pipe of diameter D, as inFig. 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 overcomefriction drag must be supplied by a pressure force giving rise to a pressure drop dPalong the length dL.The pressure drop force is:

dPx Area of pipe = dPx D2/4The friction force is (force/unit area) x wall area of pipe

= F/Ax Dx dL

so from eqn. (3.16), = (fv2/2) x Dx dLTherefore equating prressure drop and friction force


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(D2/4) dP= (fv2/2) Dx dL,therefore

dP= 4(fv2/2) x dL/DIntegrating between L1and L2, in which interval Pgoes from P1to P2we have:

dP= 4(fv2/2) x dL/DP1- P2= (4fv

2/2)(L1- L2)/D

i.e.

Pf= (4fv2/2) x (L/D) (3.17)

or

E= Pf/= (2fv2)(L/D) (3.18)

where L = L1- L2= length of pipe in which the pressure drop, Pf= P1- P2is the frictional pressure drop,and Eis the frictional loss of energy.Equation (3.17) is an important equation; it is known as the Fanning equation, or sometimes the D'Arcy othe Fanning-D'Arcy equation. It is used to calculate the pressure drop that occurs when liquids flow inpipes.The factor fin eqn.(3.17) depends upon the Reynolds number for the flow, and upon the roughnessof thepipe. In Fig. 3.8experimental results are plotted, showing the relationship of these factors. If the Reynoldsnumber 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


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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, itsdefinition in the particular context should always be checked. For example, a new f= 4f removes onenumerical 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 and (Re) independent of the roughness of the pipe. This is perhaps not surprising, as in streamline flowthere is assumed to be a stationary boundary layer at the wall and if this is stationary there would be noliquid movement over any roughness that might appear at the wall. Actually, the friction factor f instreamline 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, forexample the Blasius equationwhich applies to smooth pipes in the range 3000 < (Re) < 100,000 and inwhich:

0.316 ( Re)-0.25 (3.19)f =

4In 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 beexpressed 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 givenshowing the roughness factors for the various types of pipe, based on the results of Moody (1944). These

factors are then divided by the pipe diameter Dto give the roughness ratio to be used with the Moodygraph. The question of relative roughness of the pipe is under some circumstances a difficult one toresolve. In most cases, reasonable accuracy can be obtained by applying Table 3.1and Fig. 3.8.

TABLE 3.1RELATIVE ROUGHNESS FACTORS FOR PIPES

Material Roughness factor () Material Roughness factor ()Riveted steel 0.001- 0.01 Galvanized iron 0.0002Concrete 0.0003 - 0.003 Asphalted cast iron 0.001Wood staves 0.0002 - 0.003 Commercial steel 0.00005Cast 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 at20C is flowing at the rate of 0.1 m

3min

-1.

Diameter of pipe = 0.05 m,Area of cross-section A


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= (/4)D2

= /4 x (0.05)2= 1.96 x 10

-3m

2From Appendix 4,Viscosity of olive oil at 20C = 84 x 10 -3Ns 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) = (Dv/)= [(0.05 x 0.85 x 910)/(84 x 10

-3)]

= 460so 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)

Pf= (4fv2/2) x (L/D)

= [4 x 0.03 x 910 x (0.85)2x 1/2] x [170 x 1/0.05]

= 1.34 x 105Pa

= 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 orthrough fittings of varying cross-section, energy losses occur which are not recovered. This energy isdissipated in eddies and additional turbulence and finally lost in the form of heat. However, this energymust be supplied if the fluid is to be maintained in motion, in the same way, as energy must be provided toovercome friction. Losses in fittings have been found, as might be expected, to be proportional to thevelocity head of the fluid flowing. In some cases the magnitude of the losses can be calculated but moreoften they are best found from tabulated values based largely on experimental results. The energy loss isexpressed in the general form,E= kv

2/2 (3.20)

where khas to be found for the particular fitting. Values of this constant kfor some fittings are given inTable 3.2.TABLE 3.2

FRICTION LOSS FACTORS IN FITTINGSk

Valves, fully open:


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gate 0.13globe 6.0angle 3.0

Elbows:90standard 0.74medium sweep 0.5long radius 0.25square 1.5

Tee, used as elbow 1.5Tee, straight through 0.5Entrance, large tank to pipe:

sharp 0.5rounded 0.05

Energy is also lost at sudden changes in pipe cross-section. At a sudden enlargement the loss has beenshown to be equal to:

Ef= (v1- v2)2/2 (3.21)

For a sudden contractionEf= kv2

2/2 (3.22)

where v1is the velocity upstream of the change in section and v 2is the velocity downstream of the changein pipe diameter from D1to 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 3.3LOSS FACTORS IN CONTRACTIONS

D2/D1 0.1 0.3 0.5 0.7 0.9k 0.36 0.31 0.22 0.11 0.02

Pressure Drop through EquipmentFluids sometimes have to be passed through beds of packed solids; for example in the air drying ofgranular materials, hot air may be passed upward through a bed of the material. The pressure dropresulting is not easy to calculate, even if the properties of the solids in the bed are well known. It isgenerally 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 inheat exchangers. An equation of the general form of eqn. (3.20) will hold in most cases, but values for kwill have to be obtained from experimental results. Useful correlations for particular cases may be found inbooks on fluid flow and from works such as Perry (1997)and McAdams (1954).


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Equivalent Lengths of Pipe

In some applications it is convenient to clculate pressure drops in fittings from added equivalent lengths ofstraight pipe, rather than directly in terms of velocity heads or velocity pressures when making pipe-flowcalculations. 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 underconsideration. In this way various fittings, for example bends and elbows, are simply equated to equivalentlengths of pipe and the total friction losses computed from the total pipe length, actual plus fictitious. As Ein eqn. (3.20) is equal to Ein eqn. (3.17), kcan therefore be replaced by 4L/Dwhere Lis the length ofpipe (of diameter D) equivalent to the fitting.

Compressibility Effects for GasesThe 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 normalcircumstances and it is also frequently true for gases. Where gases are passed through equipment suchas 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, thecalculation of pressure drops and of power requirements is not simple, nor is it amenable to analyticalsolutions. Estimates can, however, be made and useful generalizations are:

(1) Pressure drops through equipment are in general proportional to velocity heads, or pressures; in otherwords, 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,v

2x Av =Av3.

energy loses in pipes

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