2.10. DESIGN EQUATIONS FOR LAMINAR AND TURBULENT FLOW IN PIPES

2.10A. Velocity Profiles in Pipes

One of the most important applications of fluid flow is flow inside circular conduits, pipes, and tubes. Appendix A.5 gives sizes of commercial standard steel pipe. Schedule 40 pipe in the different sizes is the standard usually used. Schedule 80 has a thicker wall and will withstand about twice the pressure of schedule 40 pipe. Both have the same outside diameter so that they will fit the same fittings. Pipes of other metals have the same outside diameters as steel pipe to permit interchanging parts of a piping system. Sizes of tubing are generally given by the outside diameter and wall thickness. Perry and Green (P1) give detailed tables of various types of tubing and pipes.

When fluid is flowing in a circular pipe and the velocities are measured at different distances from the pipe wall to the center of the pipe, it has been shown that in both laminar and turbulent flow, the fluid in the center of the pipe is moving faster than the fluid near the walls. These measurements are made at a reasonable distance from the entrance to the pipe. Figure 2.10-1 is a plot of the relative distance from the center of the pipe versus the fraction of maximum velocity ν′/ν_max, where ν′ is local velocity at the given position and ν_max the maximum velocity at the center of the pipe. For viscous or laminar flow, the velocity profile is a true parabola, as derived in Eq. (2.9-9). The velocity at the wall is zero.

In many engineering applications the relation between the average velocity ν_av in a pipe and the maximum velocity ν_max is useful, since in some cases only the ν_max at the center point of the tube is measured. Hence, from only one point measurement this relationship between ν_max and ν_av can be used to determine ν_av. In Fig. 2.10-2 experimentally measured values of ν_av/ν_max are plotted as a function of the Reynolds numbers Dν_av ρ/μ and Dν_max ρ/μ.

The average velocity over the whole cross section of the pipe is precisely 0.5 times the maximum velocity at the center as given by the shell momentum balance in Eq. (2.9-13) for laminar flow. On the other hand, for turbulent flow, the curve is somewhat flattened in the center (see Fig. 2.10-1) and the average velocity is about 0.8 times the maximum. This value of 0.8 varies slightly, depending upon the Reynolds number, as shown in the correlation in Fig. 2.10-2. (Note: See Problem 2.6-3, where a value of 0.817 is derived using the -power law.)

2.10B. Pressure Drop and Friction Loss in Laminar Flow

1. Pressure drop and loss due to friction

When the fluid is in steady-state laminar flow in a pipe, then for a Newtonian fluid the shear stress is given by Eq. (2.4-2), which is rewritten for change in radius dr rather than distance dy, as follows:

Equation 2.10-1

Using this relationship and making a shell momentum balance on the fluid over a cylindrical shell, the Hagen–Poiseuille equation (2.9-11) for laminar flow of a liquid in circular tubes is obtained. This can be written as

Equation 2.10-2

where p₁ is upstream pressure at point 1, N/m²; p₂ is pressure at point 2; ν is average velocity in tube, m/s; D is inside diameter, m; and (L₂ − L₁) or ΔL is length of straight tube, m. For English units, the right-hand side of Eq. (2.10-2) is divided by g_c.

The quantity (p₁ − p₂)_f or Δp_f is the pressure loss due to skin friction. Then, for constant ρ, the friction loss F_f is

Equation 2.10-3

This is the mechanical-energy loss due to skin friction for the pipe in N · m/kg of fluid and is part of the Σ F term for frictional losses in the mechanical-energy-balance equation (2.7-28). This term (p₁ − p₂)_f for skin-friction loss is different from the (p₁ − p₂) term, owing to velocity head or potential head changes in Eq. (2.7-28). That part of Σ F which arises from friction within the channel itself by laminar or turbulent flow is discussed in Sections 2.10B and 2.10C. The part of friction loss due to fittings (valves, elbows, etc.), bends, and the like, which sometimes constitute a large part of the friction, is discussed in Section 2.10F. Note that if Eq. (2.7-28) is applied to steady flow in a straight, horizontal tube, we obtain (p₁ − p₂)/ρ = Σ F.

One of the uses of Eq. (2.10-2) is in the experimental measurement of the viscosity of a fluid by measuring the pressure drop and volumetric flow rate through a tube of known length and diameter. Slight corrections for kinetic energy and entrance effects are usually necessary in practice. Also, Eq. (2.10-2) is often used in the metering of small liquid flows.

EXAMPLE 2.10-1. Metering of Small Liquid Flows A small capillary with an inside diameter of 2.22 × 10−3 m and a length 0.317 m is being used to continuously measure the flow rate of a liquid having a density of 875 kg/m3 and μ = 1.13 × 10−3 Pa · s. The pressure-drop reading across the capillary during flow is 0.0655 m water (density 996 kg/m3). What is the flow rate in m3/s if end-effect corrections are neglected? Solution: Assuming that the flow is laminar, Eq. (2.10-2) will be used. First, to convert the height h of 0.0655 m water to a pressure drop using Eq. (2.2-4), Substituting into Eq. (2.10-2) the values μ = 1.13 × 10−3 Pa · s, L2 − L1 = 0.317 m, D = 2.22 × 10−3 m, and Δpf = 640 N/m2, and solving for ν, Equation 2.10-2 The volumetric rate is then Since it was assumed that laminar flow is occurring, the Reynolds number will be calculated to check this: Hence, the flow is laminar as assumed.

2. Use of friction factor for friction loss in laminar flow

A common parameter used in laminar and especially in turbulent flow is the Fanning friction factor, f, which is defined as the drag force per wetted surface unit area (shear stress τ_s at the surface) divided by the product of density times velocity head, or . The force is Δp_f times the cross-sectional area πR² and the wetted surface area is 2πR ΔL. Hence, the relation between the pressure drop due to friction and f is as follows for laminar and turbulent flow:

Equation 2.10-4

Rearranging, this becomes

Equation 2.10-5

Equation 2.10-6

For laminar flow only, combining Eqs. (2.10-2) and (2.10-5),

Equation 2.10-7

Equations (2.10-2), (2.10-5), (2.10-6), and (2.10-7) for laminar flow hold up to a Reynolds number of 2100. Beyond that, at a N_Re value above 2100, Eqs. (2.10-2) and (2.10-7) do not hold for turbulent flow. For turbulent flow, Eqs. (2.10-5) and (2.10-6), however, are used extensively along with empirical methods for predicting the friction factor f, as discussed in the next section.

EXAMPLE 2.10-2. Use of Friction Factor in Laminar Flow Assume the same known conditions as in Example 2.10-1 except that the velocity of 0.275 m/s is known and the pressure drop Δpf is to be predicted. Use the Fanning friction factor method. Solution: The Reynolds number is, as before, From Eq. (2.10-7) the friction factor f is Using Eq. (2.10-5) with ΔL = 0.317 m, ν = 0.275 m/s, D = 2.22 × 10−3 m, and ρ = 875 kg/m3, This, of course, agrees with the value in Example 2.10-1.

When the fluid is a gas and not a liquid, the Hagen–Poiseuille equation (2.10-2) can be written as follows for laminar flow:

Equation 2.10-8

where m = kg/s, M = molecular weight in kg/kg mol, T = absolute temperature in K, and R = 8314.3 N · m/kg mol · K. In English units, R = 1545.3 ft · lb_f/lb mol · °R.

2.10C. Pressure Drop and Friction Factor in Turbulent Flow

In turbulent flow, as in laminar flow, the friction factor also depends on the Reynolds number. However, it is not possible to predict theoretically the Fanning friction factor f for turbulent flow as was done for laminar flow. The friction factor must be determined empirically (experimentally), and it not only depends upon the Reynolds number but also on surface roughness of the pipe. In laminar flow the roughness has essentially no effect.

Dimensional analysis also shows the dependence of the friction factor on these factors. In Sections 3.11 and 4.14, methods of obtaining the dimensionless numbers and their importance are discussed.

A large number of experimental data on friction factors for smooth pipe and pipes of varying degrees of equivalent roughness have been obtained and the data correlated. For design purposes, to predict the friction factor f and, hence, the frictional pressure drop for round pipe, the friction-factor chart in Fig. 2.10-3 can be used. It is a log–log plot of f versus N_Re. This friction factor f is then used in Eqs. (2.10-5) and (2.10-6) to predict the friction loss Δp_f or F_f:

Equation 2.10-5

Equation 2.10-6

For the region with a Reynolds number below 2100, the line is the same as Eq. (2.10-7). For a Reynolds number above 4000 for turbulent flow, the lowest line in Fig. 2.10-3 represents the friction-factor line for smooth pipes and tubes, such as glass tubes and drawn copper and brass tubes. The other lines, for higher friction factors, represent lines for different relative roughness factors, ε/D, where D is the inside pipe diameter in m and ε is a roughness parameter, which represents the average height in m of roughness projections from the wall (M1). In Fig. 2.10-3, values for the equivalent roughness of new pipes are given (M1). The most common pipe, commercial steel, has a roughness of ε = 4.6 × 10⁻⁵ m (1.5 × 10⁻⁴ ft).

The reader should be cautioned about using friction factors f from other sources. The Fanning friction factor f in Eq. (2.10-6) is the one used here. Others use a friction factor that may be four times larger.

EXAMPLE 2.10-3. Use of Friction Factor in Turbulent Flow A liquid is flowing through a horizontal straight pipe at 4.57 m/s. The pipe used is commercial steel, schedule 40, 2-in. nominal diameter. The viscosity of the liquid is 4.46 cp and the density 801 kg/m3. Calculate the mechanical-energy friction loss Ff in J/kg for a 36.6-m section of pipe. Solution: The following data are given: From Appendix A.5, D = 0.0525 m, ν = 4.57 m/s, ρ = 801 kg/m3, ΔL = 36.6 m, and The Reynolds number is calculated as Hence, the flow is turbulent. For commercial steel pipe from the table in Fig. 2.10-3, the equivalent roughness is 4.6 × 10−5 m: For a NRe of 4.310 × 104, the friction factor from Fig. 2.10-3 is f = 0.0060. Substituting into Eq. (2.10-6), the friction loss is

In problems involving the friction loss F_f in pipes, F_f is usually the unknown, with the diameter D, velocity ν, and pipe length ΔL known. Then a direct solution is possible, as in Example 2.10-3. However, in some cases, the friction loss F_f is already set by the available head of liquid. Then if the volumetric flow rate and pipe length are set, the unknown to be calculated is the diameter. This solution must be by trial and error, since the velocity y appears in both N_Re and f, which are unknown. In another case, with F_f again being already set, the diameter and pipe length are specified. This solution is also by trial and error, to calculate the velocity. Example 2.10-4 indicates the method to be used to calculate the pipe diameter with F_f set. Others (M2) give a convenient chart to aid in these types of calculations.

EXAMPLE 2.10-4. Trial-and-Error Solution to Calculate Pipe Diameter Water at 4.4°C is to flow through a horizontal commercial steel pipe having a length of 305 m at the rate of 150 gal/min. A head of water of 6.1 m is available to overcome the friction loss Ff. Calculate the pipe diameter. Solution: From Appendix A.2, the density ρ = 1000 kg/m3 and the viscosity μ is The solution is by trial and error since v appears in NRe and f. Assume that D = 0.089 m for the first trial. For commercial steel pipe and using Fig. 2.10-3, ε = 4.6 × 10−5 m. Then, From Fig. 2.10-3 for NRe = 8.730 × 104 and ε/D = 0.00052, f = 0.0051. Substituting into Eq. (2.10-6), Solving for D, D = 0.0945 m. This does not agree with the assumed value of 0.089 m. For the second trial, D will be assumed to be 0.0945 m. From Fig. 2.10-3, f = 0.0052. It can be seen that f does not change much with NRe in the turbulent region: Solving, D = 0.0954 m or 3.75 in. Hence, the solution agrees closely with the assumed value of D.

2.10D. Pressure Drop and Friction Factor in Flow of Gases

The equations and methods discussed in this section for turbulent flow in pipes hold for incompressible liquids. They also hold for a gas if the density (or the pressure) changes by less than 10%. Then an average density, ρ_av in kg/m³, should be used and the errors involved will be less than the uncertainty limits in the friction factor f. For gases, Eq. (2.10-5) can be rewritten as follows for laminar and turbulent flow:

Equation 2.10-9

where ρ_av is the density at p_av = (p₁ + p₂)/2. Also, the N_Re used is DG/μ, where G is kg/m² · s and is a constant independent of the density and velocity variations for the gas. Equation (2.10-5) can also be written for gases as

Equation 2.10-10

where R is 8314.3 J/kg mol · K or 1545.3 ft · lb_f/lb mol · °R and M is molecular weight.

The derivation of Eqs. (2.10-9) and (2.10-10) applies only to cases with gases where the relative pressure change is small enough that large changes in velocity do not occur. If the exit velocity becomes large, the kinetic-energy term, which has been omitted, becomes important. For pressure changes above about 10%, compressible flow is occurring, and the reader should refer to Section 2.11. In adiabatic flow in a uniform pipe, the velocity in the pipe cannot exceed the velocity of sound.

EXAMPLE 2.10-5. Flow of Gas in Line and Pressure Drop Nitrogen gas at 25°C is flowing in a smooth tube having an inside diameter of 0.010 m at the rate of 9.0 kg/s · m2. The tube is 200 m long and the flow can be assumed to be isothermal. The pressure at the entrance to the tube is 2.0265 × 105 Pa. Calculate the outlet pressure. Solution: The viscosity of the gas from Appendix A.3 is μ = 1.77 × 10−5 Pa · s at T = 298.15 K. Inlet gas pressure p1 = 2.0265 × 105 Pa, G = 9.0 kg/s · m2, D = 0.010 m, M = 28.02 kg/kg mol, ΔL = 200 m, and R = 8314.3 J/kg mol · K. Assuming that Eq. (2.10-10) holds for this case and that the pressure drop is less than 10%, the Reynolds number is Hence, the flow is turbulent. Using Fig. 2.10-3, f = 0.0090 for a smooth tube. Substituting into Eq. (2.10-10), Solving, p2 = 1.895 × 105 Pa. Hence, Eq. (2.10-10) can be used, since the pressure drop is less than 10%.

2.10E. Effect of Heat Transfer on Friction Factor

The friction factor f in Fig. 2.10-3 is given for isothermal flow, that is, no heat transfer. When a fluid is being heated or cooled, the temperature gradient will cause a change in physical properties of the fluid, especially the viscosity. For engineering practice the following method of Sieder and Tate (P1, S3) can be used to predict the friction factor for nonisothermal flow for liquids and gases:

Hence, when the liquid is being heated, ψ is greater than 1.0 and the final f decreases. The reverse occurs on cooling the liquid.

2.10F. Friction Losses in Expansion, Contraction, and Pipe Fittings

Skin-friction losses in flow through straight pipe are calculated by using the Fanning friction factor. However, if the velocity of the fluid is changed in direction or magnitude, additional friction losses occur. This results from additional turbulence which develops because of vortices and other factors. Methods for estimating these losses are discussed below.

1. Sudden enlargement losses

If the cross section of a pipe enlarges very gradually, very little or no extra losses are incurred. If the change is sudden, it results in additional losses due to eddies formed by the jet expanding in the enlarged section. This friction loss can be calculated by the following for turbulent flow in both sections. The following equation was derived in Example 2.8-4 as Eq. (2.8-3.6):

Equation 2.10-15

where h_ex is the friction loss in J/kg, K_ex is the expansion-loss coefficent and equals (1 − A₁/A₂)², ν₁ is the upstream velocity in the smaller area in m/s, ν₂ is the downstream velocity, and α = 1.0. If the flow is laminar in both sections, the factor α in the equation becomes . For English units the right-hand side of Eq. (2.10-15) is divided by g_c. Also, h = ft · lb_f/lb_m.

2. Sudden contraction losses

When the cross section of the pipe is suddenly reduced, the stream cannot follow around the sharp corner, and additional frictional losses due to eddies occur. For turbulent flow, this is given by

Equation 2.10-16

where h_c is the friction loss, α = 1.0 for turbulent flow, ν₂ is the average velocity in the smaller or downstream section, and K_c is the contraction-loss coefficient (P1) and approximately equals 0.55 (1 − A₂/A₁). For laminar flow, the same equation can be used with α = (S2). For English units the right side is divided by g_c.

3. Losses in fittings and valves

Pipe fittings and valves also disturb the normal flow lines in a pipe and cause additional friction losses. In a short pipe with many fittings, the friction loss from these fittings could be greater than in the straight pipe. The friction loss for fittings and valves is given by the following equation:

Equation 2.10-17

where K_f is the loss factor for the fitting or valve and ν₁ is the average velocity in the pipe leading to the fitting. Experimental values for K_f are given in Table 2.10-1 for turbulent flow (P1) and in Table 2.10-2 for laminar flow.

Table 2.10-1. Friction Loss for Turbulent Flow Through Valves and Fittings

Type of Fitting or Valve		Frictional Loss, Numberof Velocity Heads, Kf	Frictional Loss, Equivalent Length of Straight Pipe in Pipe Diameters, Le/D
Elbow, 45°		0.35	17
Elbow, 90°		0.75	35
Tee		1	50
Return bend		1.5	75
Coupling		0.04	2
Union		0.04	2
Gate valve
	Wide open	0.17	9
	Half open	4.5	225
Globe valve
	Wide open	6.0	300
	Half open	9.5	475
Angle valve, wide open		2.0	100
Check valve
	Ball	70.0	3500
	Swing	2.0	100
Water meter, disk		7.0	350
Source: R. H. Perry and C. H. Chilton, Chemical Engineers' Handbook, 5th ed. New York: McGraw-Hill Book Company, 1973. With permission.

Table 2.10-2. Friction Loss for Laminar Flow Through Valves and Fittings (K1)

	Frictional Loss, Number of Velocity Heads, Kf Reynolds Number
Type of Fitting or Valve	50	100	200	400	1000	Turbulent
Elbow, 90°	17	7	2.5	1.2	0.85	0.75
Tee	9	4.8	3.0	2.0	1.4	1.0
Globe valve	28	22	17	14	10	6.0
Check valve, swing	55	17	9	5.8	3.2	2.0

As an alternative method, some texts and references (B1) give data for losses in fittings as an equivalent pipe length in pipe diameters. These data, also given in Table 2.10-1, are presented as L_e/D, where L_e is the equivalent length of straight pipe in m having the same frictional loss as the fitting, and D is the inside pipe diameter in m. The K values in Eqs. (2.10-15) and (2.10-16) can be converted to L_e/D values by multiplying the K by 50 (P1). The L_e values for the fittings are simply added to the length of the straight pipe to get the total length of equivalent straight pipe to use in Eq. (2.10-6).

4. Frictional losses in mechanical-energy-balance equation

The frictional losses from the friction in the straight pipe (Fanning friction), enlargement losses, contraction losses, and losses in fittings and valves are all incorporated in the ΣF term of Eq. (2.7-28) for the mechanical-energy balance, so that

Equation 2.10-18

If all the velocities, ν, ν₁, and ν₂, are the same, then by factoring, Eq. (2.10-18) becomes, for this special case,

Equation 2.10-19

The use of the mechanical-energy-balance equation (2.7-28) along with Eq. (2.10-18) will be shown in the following examples.

EXAMPLE 2.10-6. Friction Losses and Mechanical-Energy Balance An elevated storage tank contains water at 82.2°C, as shown in Fig. 2.10-4. It is desired to have a discharge rate at point 2 of 0.223 ft3/s. What must be the height H in ft of the surface of the water in the tank relative to the discharge point? The pipe used is commercial steel pipe, schedule 40, and the lengths of the straight portions of pipe are shown. Figure 2.10-4. Process flow diagram for Example 2.10-6. Solution: The mechanical-energy-balance equation (2.7-28) is written between points 1 and 2. Equation 2.10-20 From Appendix A.2, for water, ρ = 0.970(62.43) = 60.52 lbm/ft3 and μ = 0.347 cp = 0.347(6.7197 × 10−4) = 2.33 × 10−4 lbm/ft · s. The diameters of the pipes are The velocities in the 4-in. and 2-in. pipe are The Σ F term for frictional losses in the system includes the following: (1) contraction loss at tank exit, (2) friction in the 4-in. straight pipe, (3) friction in 4-in. elbow, (4) contraction loss from 4-in. to 2-in. pipe, (5) friction in the 2-in. straight pipe, and (6) friction in the two 2-in. elbows. Calculations for the six items are as follows: Contraction loss at tank exit. From Eq. (2.10-16), for contraction from A1 to A3 cross-sectional area, since A1 of the tank is very large compared to A3, Friction in the 4-in. pipe. The Reynolds number is Hence, the flow is turbulent. From Fig. 2.10-3, ε = 4.6 × 10−5 m (1.5 × 10−4 ft). Then, for NRe = 219 300, the Fanning friction factor f = 0.0047. Substituting into Eq. (2.10-6) for ΔL = 20.0 ft of 4-in. pipe, Friction in 4-in. elbow. From Table 2.10-1, Kf = 0.75. Then, substituting into Eq. (2.10-17), Contraction loss from 4- to 2-in. pipe. Using Eq. (2.10-16) again for contraction from A3 to A4 cross-sectional area, Friction in the 2-in pipe. The Reynolds number is The Fanning friction factor from Fig. 2.10-3 is f = 0.0048. The total length ΔL = 125 + 10 + 50 = 185 ft. Substituting into Eq. (2.10-6), Friction in the two 2-in. elbows. For a Kf = 0.75 and two elbows, The total frictional loss Σ F is the sum of items (1) through (6): Using as a datum level z2, z1 = H ft, z2 = 0. Since turbulent flow exists, α = 1.0. Also, ν1 = 0 and ν2 = ν4 = 9.57 ft/s. Since p1 and p2 are both at 1 atm abs pressure and ρ1 = ρ2, Also, since no pump is used, WS = 0. Substituting these values into Eq. (2.10-20), Solving, H(g/gc) = 33.77 ft · lbf/lbm (100.9 J/kg) and H is 33.77 ft (10.3 m) height of water level above the discharge outlet.

EXAMPLE 2.10-7. Friction Losses with Pump in Mechanical-Energy Balance Water at 20°C is being pumped from a tank to an elevated tank at the rate of 5.0 × 10−3 m3/s. All of the piping in Fig. 2.10-5 is 4-in. schedule 40 pipe. The pump has an efficiency of 65%. Calculate the kW power needed for the pump. Figure 2.10-5. Process flow diagram for Example 2.10-7. Solution: The mechanical-energy-balance equation (2.7-28) is written between points 1 and 2, with point 1 being the reference plane: Equation 2.7-28 From Appendix A.2, for water, ρ = 998.2 kg/m3 and μ = 1.005 × 10−3 Pa · s. For 4-in. pipe, from Appendix A.5, D = 0.1023 m and A = 8.219 × 10−3 m2. The velocity in the pipe is ν = 5.0 × 10−3/(8.219 × 10−3) = 0.6083 m/s. The Reynolds number is Hence, the flow is turbulent. The Σ F term for frictional losses includes the following: (1) contraction loss at tank exit, (2) friction in the straight pipe, (3) friction in the two elbows, and (4) expansion loss at the tank entrance. Contraction loss at tank exit. From Eq. (2.10-16), for contraction from a large A1 to a small A2, Friction in the straight pipe. From Fig. 2.10-3, ε = 4.6 × 10−5 m and ε/D = 4.6 × 10−5/0.1023 = 0.00045. Then for NRe = 6.181 × 104, f = 0.0051. Substituting into Eq. (2.10-6) for ΔL = 5 + 50 + 15 + 100 = 170 m, Friction in the two elbows. From Table 2.10-1, Kf = 0.75. Then, substituting into Eq. (2.10-7) for two elbows, Expansion loss at the tank entrance. Using Eq. (2.10-15), The total frictional loss is Σ F: Substituting into Eq. (2.7-28), where and (p2 − p1) = 0, Solving, WS = −153.93 J/kg. The mass flow rate is m = 5.0 × 10−3(998.2) = 4.991 kg/s. Using Eq. (2.7-30),

2.10G. Friction Loss in Noncircular Conduits

The friction loss in long, straight channels or conduits of noncircular cross section can be estimated by using the same equations employed for circular pipes if the diameter in the Reynolds number and in the friction-factor equation (2.10-6) is taken as the equivalent diameter. The equivalent diameter D is defined as four times the hydraulic radius r_H. The hydraulic radius is defined as the ratio of the cross-sectional area of the channel to the wetted perimeter of the channel for turbulent flow only. Hence,

Equation 2.10-21

For example, for a circular tube,

For an annular space with outside diameter D₁ and inside D₂,

Equation 2.10-22

For a rectangular duct of sides a and b ft,

Equation 2.10-23

For open channels and partly filled ducts in turbulent flow, the equivalent diameter and Eq. (2.10-6) are also used (P1). For a rectangle with depth of liquid y and width b,

Equation 2.10-24

For a wide, shallow stream of depth y,

Equation 2.10-25

For laminar flow in ducts running full and in open channels with various cross-sectional shapes other than circular, equations are given elsewhere (P1).

2.10H. Entrance Section of a Pipe

If the velocity profile at the entrance region of a tube is flat, a certain length of tube is necessary for the velocity profile to be fully established. This length for the establishment of fully developed flow is called the transition length or entry length. This is shown in Fig. 2.10-6 for laminar flow. At the entrance the velocity profile is flat; that is, the velocity is the same at all positions. As the fluid progresses down the tube, the thickness of the boundary layers increases until finally they meet at the center of the pipe and the parabolic velocity profile is fully established.

The approximate entry length L_e of a pipe of diameter D for a fully developed velocity profile to be formed in laminar flow is (L2)

Equation 2.10-26

For turbulent flow, no relation is available to predict the entry length for a fully developed turbulent velocity profile to form. As an approximation, the entry length is nearly independent of the Reynolds number and is fully developed after 50 diameters downstream.

EXAMPLE 2.10-8. Entry Length for a Fluid in a Pipe Water at 20°C is flowing through a tube of diameter 0.010 m at a velocity of 0.10 m/s. Calculate the entry length. Calculate the entry length for turbulent flow. Solution: For part (a), from Appendix A.2, ρ = 998.2 kg/m3, μ = 1.005 × 10−3 Pa · s. The Reynolds number is Using Eq. (2.10-26) for laminar flow, Hence, Le = 0.571 m. For turbulent flow in part (b), Le = 50(0.01) = 0.50 m.

The pressure drop or friction factor in the entry length is greater than in fully developed flow. For laminar flow the friction factor is highest at the entrance (L2) and then decreases smoothly to the fully developed flow value. For turbulent flow there will be some portion of the entrance over which the boundary layer is laminar and the friction-factor profile is difficult to express. As an approximation, the friction factor for the entry length can be taken as two to three times the value of the friction factor in fully developed flow.

2.10I. Selection of Pipe Sizes

In large or complex process piping systems, the optimum size of pipe to use for a specific situation depends upon the relative costs of capital investment, power, maintenance, and so on. Charts are available for determining these optimum sizes (P1). However, for small installations, approximations are usually sufficiently accurate. Representative values for ranges of velocity in pipes are shown in Table 2.10-3. For stainless-steel pipes, recent data (D1) show that the velocities in Table 2.10-3 for process lines or pump discharge should be increased by 70%.

Table 2.10-3. Representative Ranges of Velocities in Steel Pipes

		Velocity
Type of Fluid	Type of Flow	ft/s	m/s
Nonviscous liquid	Inlet to pump	2–3	0.6–0.9
	Process line or pump discharge	5.7	1.7
Viscous liquid	Inlet to pump	0.2–0.8	0.06–0.25
	Process line or pump discharge	3	0.9
Gas air	Process line	53	16
Steam 100 psig	Process line	38	11.6