For a pipe with an inside diameter of .75 inch, plot the Reynolds number of all
ID: 2213538 • Letter: F
Question
For a pipe with an inside diameter of .75 inch, plot the Reynolds number of all flows from 0 to 10 gal/min. This is approximately equivalent to a garden hose. on your plot, identify the transition from laminar to turbulent flow. the properties of automotive lube oil are given below. SAE Multigrade Viscosity Density Centistokes 10-6 reyns (lb s/in2) kg/m3 lb/in3 40 oC 100 oC 104 oF 212 oF 5W-30 64 11 8.2 1 860 0.0311 10W-30 69 11 8.8 1.1 865 0.0312 10W-40 94 14.3 11.9 1.5 865 0.0312 20W-50 166 18.7 21.3 2.7 872 0.0315Explanation / Answer
A complete detail of Reynold No Fluid flow within a system is either steady-state or transient. Under steady-state conditions, flow properties remain constant with respect to time; consequently, the accumulation term in Equation 1-2 equals zero. Transient conditions indicate that one or more flow properties change with time, as observed during refinery plant start-up operations or in the simpler case of fluid draining from a vessel by gravity. The primary method of fluid transfer is a piping or conduit network of circular cross-sectional area. Flow through a pipe is assumed to occupy the entire cross section at any specific point and can be characterized as laminar, transitional, or turbulent. Under laminar flow conditions, all fluid elements compose a smooth, unidirectional stream. Turbulent flow involves bulk fluid movement mixed with an irregular motion of eddies except for a laminar film at the pipe or conduit wall called a boundary layer. The laminar and turbulent flow regimes are separated by a transition region where fluid behavior does not have a distinct pattern. The nature of fluid flow can be determined through calculation of a dimensionless parameter known as the Reynolds number, NRe (Osborne Reynolds, 1842–1912). NRe is defined as a ratio of inertial to viscous forces: **Where ? is fluid density (lbm/ft3); ? is fluid velocity through the pipe or conduit (ft/s); D is the pipe inner diameter (ft); and µ is fluid viscosity (lbm/ft•s), converted from cP. In laminar flow viscous forces dominate, whereas in turbulent flow inertial forces prevail. In an engineering context, flow is considered laminar if NRe is less than 2000 and turbulent if it is greater than 4000. The range 2000 = NRe = 4000 corresponds to transitional values. Fluid transport phenomena are largely empirical. Only a handful of problems, such as those involving laminar flow, have exact mathematical solutions; this points out the importance of laminar flow to the process industry as well as to designers of process equipment and instrumentation. As an example involving compressible flow, mass flow controllers (MFCs) typically utilize flow-bypass thermal technology for gases, in which an appropriately sized restriction element within the device induces laminar flow so that an accurate measurement can be obtained and reproduced. The majority of fluid flow problems are challenging as they depend on approximations derived from correction factors determined from extensive experimentation. The following sections present topics in modeling incompressible fluid flow through piping, orifices, valves, and pumps. Relevant problems that involve both steady-state and transient flow conditions are discussed, with an emphasis on refinery plant applications. Crane Technical Paper No. 410 (see Bibliography) serves as a primary resource for relations, empirical pipe flow data, and engineering data. 2. The DARCY-WEISHBACH EQUATION, Bernoulli Theorem, and Flow Coefficient Cv. The Darcy-Weisbach equation (Henry Darcy, 1803–1858, and Julius Ludwig Weisbach, 1806–1871) is the general relation for pressure drop and is valid for laminar or turbulent flow barring cavitation. The Crane paper presents several variations of the relation in terms of relevant process variables, demonstrating its versatility as a flow modeling tool. The Darcy-Weisbach equation will be derived here with respect to pressure drop ?P and volumetric flow rate q, with the development of each expression facilitated through the use of a mechanical energy balance commonly known as the Bernoulli theorem (Daniel Bernoulli, 1700–1782). The resulting derivations will be applied to valve characterization, including the flow coefficient Cv. Fluid flow is assumed to be at steady state. The Bernoulli theorem accounts for two points in adiabatic flow when friction forces dominate energy transfer: **where P is pressure (psi or lbf/in2), with the applicable conversion factor; gc is a conversion factor (32.174 lbm •ft/lbf •s2); g is the gravitational constant (32.174 ft/s2); Z is elevation (ft); hL represents frictional loss in the line and valves/fittings (ft) Each term in Equation 2-1 is referred to as a head, with units of ft. The sum of the pressure head, velocity head, and elevation head for either point (1) or point (2) defines the corresponding total head. Friction loss head hL designates the sum of all frictional contributions in the line between points (1) and (2) and typically is represented by individual source terms. The conversion factor gc provides unit detail relating lbf to lbm; note that some technical resources, including Crane, omit gc, resulting in minor dimensional inconsistencies for English units. For incompressible flow through a horizontal pipe, Equation 2-1 reduces to: The Darcy-Weisbach equation in its simplest form presents friction loss head as the product of velocity head and a resistance coefficient: **where f is the Darcy friction factor, and L is pipe length (ft). To determine f, the Reynolds number NRe from Equation 1-3 must be calculated to ascertain the nature of flow. For laminar flow, f is independent of piping or conduits and can be calculated from the quotient (64/NRe). For turbulent flow, f is a function of NRe and the relative roughness e of the pipe walls as a ratio to inner diameter (e/D) and becomes relatively constant with higher values of NRe. Moody plots (Lewis Ferry Moody, 1880–1953) provided in Crane, pp. A-23 and A-24, relate f to NRe and e/D for any commercial pipe, and Crane p. A-25 relates f to NRe and D specifically for clean steel pipe. Pipe length L equals the sum of straight pipe length and equivalent length of bends, fittings, and valves, where the equivalent length of a bend, fitting, or valve is the length of straight pipe that produces the same frictional loss. Representative equivalent length data, specified as L/D, are provided in Crane, p. A-30. The Darcy-Weisbach equation can be written in terms of pressure drop using Equations 2-3 and 2-2: where K represents Given sufficient turbulence, K is constant and pressure drop increases with the square of velocity. From the continuity relation in Equation 1-2, velocity relates to volumetric flow rate at a specific cross-sectional area: where q is volumetric flow rate (ft3/s), and d is the pipe inner diameter (in) with the applicable conversion factor. Substituting Equation 2-5 into Equation 2-4 yields the Darcy-Weisbach equation in terms of flow rate: Flow rate therefore varies with the square of the pipe inner diameter, along with the square root of the pressure drop, provided K is constant. A convenient means of expressing valve and pipe fitting capacity and flow characteristics, particularly with control valves, is through the use of a parameter called the flow coefficient Cv. By definition, Cv is the volumetric flow rate of water, in gal/min or gpm, at 60°F that effects a pressure drop of 1 psid across the valve or fitting. Note that Cv has units of gpm/v psid and can be calculated with the following relation using any liquid with a viscosity close to that of water: where Q is volumetric flow rate (gpm), and s.g. is specific gravity Combining Equations 2-6 and 2-7 with appropriate unit conversions (shown) and applicable substitution of conditions relates Cv to the pipe inner diameter: The resulting equation shows that the Cv of a pipe fitting or valve changes with the square of the pipe inner diameter, assuming constant K. Commercially, valves are provided with pressure drop and Cv test data for each available diameter. Model flow curves, including corresponding flow coefficients, are shown in Figure 2-1. Figure 2-1 Pressure drop and flow coefficient data, commercial valve The values for Cv indicate flow curve readings at a pressure drop of 1 psid. Each curve presents a parabolic relationship between flow rate and pressure drop, which follows from Equation 2-6. Flow rate and Cv likewise exhibit a nonlinear relationship with valve diameter given a fixed pressure drop; both were shown as second-order in Equations 2-6 and 2-8, respectively. 3. NOZZLES and ORIFICES Nozzles and orifices are piping ancillaries that are used primarily as metering devices in pipelines and perforated plates. The volumetric flow rate through a nozzle or orifice, neglecting velocity of approach, presents a comparative model to the Darcy-Weisbach equation in Equation 2-6: where C is the flow coefficient (dimensionless; not to be confused with Cv for valves), and A is the cross-sectional area of the nozzle or orifice (ft2). Rewriting Equation 3-1 in a slightly condensed, more convenient form: where d0 is the nozzle or orifice diameter (in) with the applicable conversion factor. Converting Equation 3-2 to units of gpm: Calculation of the flow rate q or Q requires an iterative sequence as the flow coefficient C is found by trial and error. The differential pressure and fluid properties are known. Calculate ß, the diameter ratio of the orifice to the inner pipe d0/d. Estimate C based on ß and the perceived nature of flow by using the applicable plot in Crane, p. A-19. Determine q from Equation 3-2, velocity v based on d from Equation 2-5 or Crane, p. B-14 (if applicable), and the Reynolds number NRe from Equation 1-3. Find C based on NRe and ß by using the selected plot and compare with step 2. If C from step 4 is not within 5% of the initial guess, repeat the sequence with an appropriately considered new guess. A square-edged orifice can also be sized to place in a pipe. The differential head has to be specified; the flow rate and fluid properties are known. Determine v from Equation 2-5 or Crane, p. B-14 (if applicable), and NRe from Equation 1-3. Assume a ratio for ß (0.5 is suitable) and calculate d0. Find C based on ß and NRe by using the applicable plot in Crane, p. A-19. Calculate d0 from Equation 3-2 or 3-3 and compare with step 2. If d0 from step 4 is less than the initial guess by at least 5%, repeat the sequence with a smaller guess for ß. The following example illustrates the calculation of an orifice. Figure 3-1 Strategy: In both considerations, the new orifice diameter can be solved from Equation 3-3 as a ratio of new process variables to the original. A constant flow coefficient C thus is eliminated, simplifying the problem substantially. For an unknown C, the five-step approach to size a square-edged orifice will be useful. Original and new variables will be designated with the respective subscripts orig and new. Given: Piping: 6” Schedule 40 welded steel d = 6.065 in (Crane, p. B-16) d0, orig = 3 in Flow rates through pipe: Qorig = 250 gpm Qnew = 2Qorig = 500 gpm Applicable properties of water: ?60°F = 62.37 lbm/ft3, µ60°F = 1.129 cP Calculations: New orifice characteristics assuming constant flow coefficient C Express volumetric flow rate Q in terms of original and new flow rates by using Equation 3-3. Combine the expressions and solve for d0, new: Quantify new orifice size: New orifice characteristics with unknown flow coefficient C (five-step approach) 1) Determine velocity vnew and Reynolds number NRe, new. - Look up vnew in Crane, p. B-14 (500 gpm, 6” Schedule 40). - Calculate NRe, new by using Equation 1-3. 2) Provide initial guess for ßnew. - Calculate ratio using d0, new obtained assuming constant C: 3) Estimate flow coefficient Cnew: Read from provided chart based on results from steps 1 and 2. 4) Calculate d0, new and compare with step 2. Rearrange Equation 3-3 accordingly and combine both process representations. Determine flow coefficient Corig: - Calculate Reynolds number NRe, orig: - Calculate diameter ratio ßorig: - Read from chart. 5) Calculate the percent error e between new orifice diameters. Use d0, new from step 2 as the base or “actual” value. Since e = 5%, repeat from step 2 with a lesser guess. Comparison of d0, new between the two approaches Perform similar error calculation. -Depending on process sensitivity, assuming constant C may be acceptable. 4. DRAINING CYLINDRICAL TANK The rate of depth change as liquid drains from a small circular aperture in an open cylinder-shaped tank is derived in the following illustration. The effects of friction will be ignored here and subsequently explored by using separate transient models. Figure 4.1 Frictionless Model: The Torricelli Equation From the continuity equation with respect to mass m, Equation 1-2 reduces to: Assuming incompressible flow, Equation 4-1 becomes: The volume of liquid in the tank at any time t can be expressed as: Differentiating both sides of Equation 4-3 with respect to t: Equating Equations 4-2 and 4-4: where ß represents the diameter ratio D0/D. To solve Equation 4-5, the exit velocity must be expressed in terms of an arbitrary, fixed height of liquid in the tank, which can be considered separately as a steady-state condition and modeled by using the Bernoulli theorem. Considering point (1) located at the surface of the liquid in the tank and point (0) at the exit, Equation 2-1 reduces to The velocities at points (1) and (0) are related by continuity: Substituting Equation 4-7 into Equation 4-6 and rearranging yields the Torricelli equation (Evangelista Torricelli, 1608–1647): The term (1 – ß4) typically makes a minimal contribution and can be ignored but will be included in the remainder of the derivation for completeness. Substituting Equation 4-8 into Equation 4-5 yields the following simple differential equation: The solution presents the time theoretical minimum required for the tank to drain: Friction Model: Square-Edged Orifice The tank now is modeled as a vertical pipe containing a square-edged orifice. Volumetric flow rate as presented in Equation 3-1 represents draining liquid. Combining Equation 3-1 with Equation 4-2 yields Equating Equation 4-11 with Equation 4-4 and performing applicable substitutions: Unlike the frictionless model, the pressure drop must be considered at the orifice, which reduces to a function of liquid height under static conditions. Considering points (1) and (0) as before: This relation can be used to calculate the pressure exerted by any vertical column of liquid or the pressure drop corresponding to a friction loss head. Substituting Equation 4-13 into Equation 4-12 and separating variables yields the following differential equation: Since the flow coefficient C is found by trial and error, a property-dependent representation is beyond the scope of the model. The solution therefore assumes that C remains constant over the drainage period, which is reasonable for varying levels of turbulence ( NRe = 10000 ) on the diameter ratio. For instance, if ß is assumed to have a maximum value of 0.3, C is constant for Reynolds numbers greater than about 20000. As a technical reminder, velocity is based on tank diameter D in determining NRe. Integrating Equation 4-14 presents the time-depth relation for the friction model: The solution for the frictionless model in Equation 4-10 is therefore proportional to Equation 4-15 by a factor of C, neglecting the term (1 – ß4) and considering sufficiently turbulent flow. For example, if C is determined as 0.6, then the time theoretical minimum for drainage will be approximately 60% of the actual.
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