I. GARCIA, Contributing Author, Houston, Texas; and A. GARCIA, Kuraray America Pasadena, Pasadena, Texas
Calculating pipe diameters through long, straight pipes is executed in classic fluid flow problems using traditional trial-and-error methods or by complex equations. Most fluid mechanics textbooks facilitate the solution assuming that the pipe length is long enough to disregard fittings losses, and losses are considered as due to pipe friction only. Conversely, most real piping systems contain valves and fittings that make the task more difficult, as the energy loss of the system is then the sum of the losses of the pipes and fittings.
This article presents numerous equations applicable in turbulent regime of flow (Re > 4,000) that estimate an equivalent length of pipe to account for valves and fittings. Two other equations obtain the precise value of pipe diameter that, by using a pipe data table, allows the selection of the optimum pipe size that satisfies the maximum allowable pressure drop requirement for a given flowrate in any piping system.
In this type of problem, the following information is known:
BASIC EQUATIONS
Equivalent length of pipe. Based in the Crane (L / D)e method and representing values of loss coefficients (K), the equivalent length of straight pipe (Le)—accounting for valves and fittings—is calculated by the equations below.
For smooth pipes (with any fluid) and for clean commercial steel pipe (handling liquids with a dynamic viscosity of μ > 1cP or gases with a kinematic viscosity of ν > 5 × 10–5 ft2 ⁄ sec), use Eqs. 1 or 2 or both, according to the case:
Le = 0.144 × ∑ (L / D)e × (μ / ρ)–0.17 × q0.41 × (Ls / hf )0.12 (1)
Le = 7.85 × ∑K × (μ / ρ)–0.16 × q0.5 × (Ls / hf )0.17 (2)
Pipe diameter. For the above condition, the diameter of the pipe required can be obtained using Eq. 3:
D = 0.24 × (μ / ρ)0.04 × q0.376 × (La / hf )0.208 (3)
For rough pipe (any fluid) and for clean commercial steel pipe (handling liquids with a dynamic viscosity of μ > 1cP or gases with kinematic viscosity of ν ≤ 5 × 10–5 ft2⁄ sec ) use Eqs. 4 or 5 or both, according to the case:
Le = 0.032 × ∑ (L / D)e × ε–0.211 × q0.396 × (Ls / hf )0.2 (4)
Le = 1.86 × ∑K × ε–0.2 × q0.48 × (Ls / hf )0.24 (5)
For the above condition, the required pipe diameter can be determined using Eq. 6:
D = 0.341 × ε0.05 × q0.38 × (La / hf )0.19 (6)
In any of the above cases, the adjusted length of pipe must be computed according to Eq. 7:
La = Ls + ∑(le ) (7)
CHECKING THE RESULTS
The mean velocity of flow, the Reynolds number, the relative roughness of pipe (if any), the friction factor, the equivalent length of pipe and the adjusted length of pipe are calculated by traditional fluid flow equations and the pipe diameter is checked by the rearranged Darcy equation (Eq. 8):
D = 0.4789 × q0.4 × [f × (La / hF )]0.2 (8)
An important step to be confirmed is that the requirement of the maximum pressure drop given in the problem is satisfied.
Two sample problems detailed below demonstrate the usefulness of this pipe diameter calculation.
Example 1. A line is being run from a pump discharge to a pressure vessel. The flow will be 275 gallons per minute (gpm) of a chemical solution (μ = 0.6 cP; ρ = 64.8 lb/ft3). A maximum pressure drop of 9 psi is available for the run. The path includes one swing check valve, five gate valves, eighteen 90° welded elbows, r/d = 1, one globe valve and 156 ft of new carbon-steel pipe. A loss coefficient of K0 = 4.5 will account for an orifice plate installed in the system and K0 = 1 for the pipe exit. With all valves wide open, determine the size of pipe required for this run and the system pressure drop.
The flowrate is calculated using Eq. 9:
q = 275 × 0.002228 = 0.6127 (9)
The maximum allowable head loss is determined using Eq. 10. TABLE 1 provides a summation of (L / D)e and K0.
hf = 144 × (9/64.8) = 20 (10)
Since μ < 1 cP Eqs. 4 and 5 are in use, the equivalent length of pipe is calculated using Eqs. 11 and 12:
Le = 0.032 × 790 × 0.000151–0.211 × 0.61270.396 × (156 / 20)0.2 = 201 (11)
Le = 1.86 × 5.5 × 0.000151–0.2 × 0.61270.48 × (156 / 20)0.24 = 77 (12)
The adjusted length of pipe is calculated using Eq. 13:
La = 156 + 201 + 77 = 434 (13)
The pipe diameter is calculated using Eq. 14:
D = 0.341 × 0.0001510.05 × 0.61270.38 × (434 / 20)0.19 = 0.32718 (14)
The selected pipe was a 4-in. Sch. 40, where D = 0.3355, d = 4.026, ft = 0.016 and ε = 0.000151. The results can be confirmed using Eqs. 15–23:
Mean velocity (Eq. 15):
v = (4 × 0.6127) / (3.1416 × 0.33552 ) = 6.9306 (15)
Reynolds number (Eq. 16):
Re = 124 × [(4.026 × 6.9306 × 64.87) / 0.6] = 374,075 (16)
Friction factor (Eq. 17):
f = 0.25 / [log((5.74/374,0750.9) + 0.27 × (0.000151 / 0.3355))]2 = 0.018 (17)
Summation of loss coefficients (Eq. 18):
∑K = 790 × 0.016 + 5.5 = 18.14 (18)
Equivalent length of pipe (Eq. 19):
Le = (18.14 / 0.018) 0.3355 = 338 (19)
Length of pipe adjusted (Eq. 20):
La = 156 + 338 = 494 (20)
Pipe diameter (Eq. 21):
D = 0.4789 × 0.61270.4 × [0.018 × (494 / 20)]0.2 = 0.33476 (21)
Pressure drop (Eqs. 22 and 23):
∆P = 2.161 × 10–4 × [(f × La × ρ × Q2 ) / d5 ] (22)
∆P = 2.161 × 10–4 × [(0.018 × 494 × 64.8 × 2752 ) / 4.0265 ] = 8.9 (23)
To satisfy the maximum drop requirement of 9 psi, a 4-in. Sch. 40 line is required.
Example 2. Low-pressure gas (μ = 0.02Cp; ρ = 0.11 lb/ft3 ) at a rate of 720 ft3will be passed through 400 ft of a stainless-steel smooth pipe. The system includes eight 90° mitre bends (α = 75°) and two open gate valves. A loss coefficient of Ko = 7 will account for an orifice plate installed in the straight length of the run. What pipe size is required if 7 in. of water pressure drop is allowable?
The flowrate is calculated using Eq. 24:
q = 720 / 60 = 12 (24)
The maximum head loss is calculated using Eq. 25:
hf = (62.36 / 0.11) × (7 / 12) = 330.7 (or ~331) (25)
The summation of (L / D)e and Ko from TABLE 1 is calculated in Eqs. 26 and 27:
(L / D)e = 8 × 40 + 2 × 8 = 336 (26)
∑Ko = 7 (27)
The equivalent length of pipe is calculated using Eqs. 28 and 29. For smooth pipe, Eqs. 1 and 2 are used.
Le = 0.144 × 336 × (0.02 / 0.11)–0.17 × 120.41 × (400 / 331)0.12 = 183 (28)
Le = 7.85 × 7 × (0.02 / 0.11)–0.16 × 120.5 × (400 / 331)0.17 = 258 (29)
The adjusted length of pipe is calculated using Eq. 30:
La = 400 + 183 + 258 = 841 (30)
The pipe diameter is calculated using Eq. 31:
D = 0.24 × (0.02 / 0.11)0.04 × 120.376 × (841 / 331)0.208 = 0.6928 (31)
The pipe selected is dnom = 8-in. Sch. 5S, where the d = 8.407 and D = 0.7 ft = 0.014D.
The results can be confirmed using Eqs. 32–38. The mean gas velocity is calculated using Eq. 32:
v = (4 × 12) / (3.1416 × 0.72 ) = 31.18 (32)
The Reynolds number is calculated using Eq. 33:
Re = 124 × [(8.407 × 31.18 × 0.11) / 0.02] = 178,773 (33)
The friction factor is calculated using Eq. 34:
f = 0.0056 + 0.5 × 178,773–0.32 = 0.016 (34)
The summation of loss coefficients is calculated using Eq 35:
∑ K = 336 × 0.014 + 7 = 11.704 (35)
The equivalent length of pipe is calculated using Eq. 36:
Le = (11.704 / 0.016) × 0.7 = 512 (36)
The adjusted length of pipe is calculated using Eq. 37:
La = 400 + 512 = 912 (37)
The pipe diameter is calculated using Eq. 38:
D = 0.4789 × 120.4 × [0.016 × (912 / 331)]0.2 = 0.693 (38)
An 8-in. Sch. 5S line is correct. Pressure drop is calculated using Eqs. 39 and 40:
WC = 2.238 × [(f × La × ρ × v2) / (ρw × d)] (39)
WC = 2.238 × [(0.016 × 912 × 0.11 × 31.182 ) /(62.34 × 8.407)] = 6.66 (40)
Takeaways. These proposed calculation methods determine the correct diameter of pipe in any real piping system under turbulent regime of flow for most fluids, satisfying the maximum pressure drop requirement where fittings cannot be ignored.
The pipe diameter calculated provides the engineer with the proper pipe size to be selected from standard pipe data. Of course, the engineer must make a realistic choice, considering the maximum temperature and pressure that the system must endure. There is obviously a lot of capital to be saved by using the lowest schedule pipe that will safely handle process conditions.
The values of constants (L / D)e, ft and Ko for valves and fittings can be found in the extensive tabulation of the Crane Technical Manual. HP
NOMENCLATURE
d = internal diameter of pipe, in.
dnom = nominal pipe size, in.
D = internal diameter of pipe, ft
f = friction factor
ft = friction factor in zone of complete turbulence
hf = head loss, ft.
Ko = loss coefficient for a specific type of fitting
Le = equivalent length of pipe, ft
Ls = straight length of pipe, ft
La = adjusted length of pipe, ft
(L / D)e = equivalent length of pipe in number of pipe diameters
P = gauge pressure, psi
q = volumetric flowrate, ft3/sec
Q = volumetric flowrate, gpm
WC = pressure drop, inches of water at 60°F
Re = Reynolds number
v = mean velocity of flow, ft/sec
∆ = differential between two points
Ɛ = absolute roughness of pipe, ft
μ = absolute (dynamic) viscosity, cP
ν = kinematic viscosity, ft2/sec
ρ = density of fluid, lb/ft3
ρw = density of water at 60°F, lb/ft3
∑ = summation
LITERATURE CITED
Israel Garcia graduated from the University of Cienfuegos, Cuba, with an MS degree in mechanical engineering. Rodriguez has been attached to the mechanical engineering faculty of that university since 1985 as a Professor in fluid mechanics, heat transfer and science materials. Garcia has more than 30 yr of industrial experience in chemical plants and power stations, and has presented several papers on the design of heat exchangers, pressure vessels and piping systems. He now works as a Consulting Engineer in Houston, Texas. The author can be reached at isgaro47@gmail.com.
Alejandro Garcia is a mechanical engineer who graduated from the University of Cienfuegos, Cuba. He received his MS degree in mechanical engineering with a specialty in materials from the Autonomous University of Nuevo Leon, Mexico. He gained several years of experience in chemicals and power plants as a static and dynamic equipment specialist. He now works in the Kuraray America Pasadena, Texas Plant as Project Engineer II. The author can be reached alessandromilan88@gmail.com.
