Question

4. Consider upwards, laminar flow in a vertical pipe of length L. (a) Starting with the NS equations given in 6.128 c. simplify the eq'ns, state the BCs in the coordinate system shown, solve for vír), wall shear stress tw, and the average flow speed V=Q/rR; (b) rearrange your result into the form , like EB CV relation, (AP/pg +1)=f* (LD)V/(29) for arbitrary length L, and show that the friction factor f= C2/Rc and evaluate C2 as a number. Follow the analysis of section 6.9.3 for the horizontal tube, (see also 8.2.1, 8.2.2). (c) If the fluid is water, D = 1cm, and L = 5m, evaluate the magnitude & sign required for AP to get an average flow speed of V = 2 m/s. (Recall that AP/L = @P/@z). Show steps and algebra clcarly for full credit Approx. Ans: (a) v- (r-RP); 1-(R2); Vavg -- D /u; (b) C2-70; (c) AP -45 kPa

Answer 1

In this solution some basic concepts and formulas of Mathematics are used. For more information, refer to any standard textbook or drop a comment below. Please give a Thumbs Up, if solution is helpful.

Solution :

​​​​​I have given thorough derivations below. Just put in case of formulas for inclined pipe. For any furthet help feel free to contact.

poll = 206 = @

8 4. LAMINAR FLOW IN PIPES We mentioned in Section 8-2 that flow in pipes is laminar for Res 2300, and that the flow is fully developed if the pipe is sufficiently long (relative to the entry length) so that the entrance effects are negligible. In this section we consider the steady laminar flow of an incompressible fluid with con stant properties in the fully developed region of a straight circular pipe. We obtain the momentum equation by applying a momentum balance to a dif ferential volume element, and obtain the velocity profile by solving it. Then we use it to obtain a relation for the friction factor. An important aspect of the analysis here is that it is one of the few available for viscous flow in fully developed laminar flow, each fluid particle moves at a constant axial velocity along a streamline and the velocity profile ur) remains unchanged in the flow direction. There is no motion in the radial direction and thus the velocity component in the direction normal to flow is every where zero. There is no acceleration since the flow is steady and fully developed Now consider a ring-shaped differential volume element of radius r thick ness dr, and length dx oriented coaxially with the pipe, as shown in Fig. 8-11. The volume element involves only pressure and viscous effects and Thus the pressure and shear forces must balance each other. The pressure force acting on a submerged plane surface is the product of the pressure at the controid of the surface and the surface area. A force balance on the volume element in the flow direction gives 12 dr P).- (2 dr P) + 2 x ), - ( 2 x ). -0 (3-9) FIGURE 8.11 Froe body diagram of a ring-shaped differential fluid clomant of radius thickness dr. and length dx oriented coaxially with a horizontal pipe in fully developed laminarflow FLUID MECHANICS 2R which indicates that in fully developed flow in a horizontal pipe, the viscous and pressure forces balance each other. Dividing by 2 wdrdx and rearranging Puth - P. (1) -(I), (8-10) + P+P) de or Taking the limit as dr, dx Ogives dP007) - 0 Substituting -- (du/dr) and taking - constant gives the desired equation, #d du - dP 120 Forang P RIP Sging - R -O The quantity duldr is negative in pipe flow, and the negative sign is included to obtain positive values for (Ordu/dr -Ouldy since y=R-r) The left side of Eq. B-12 is a function of r, and the right side is a function of The equality must hold for any value ofr and x, and an equality of the form = 9) can be satisfied only if both ) and g(x) are equal to the same constant. Thus we conclude that didx constant. This can be verified by writing a force balance on a volume element of radius R and thickness de (a slice of the pipe), which gives (Fig. 8-12) dp 21 R FIGURE 8-12 Free-body diagram of a fluid disk element of radius R and length x in fully developed laminarflow in a horizontal pipe Here T, is constant since the viscosity and the velocity profile are constants in the fully developed region. Therefore, dldx - constant. Equation 8-12 can be solved by rearranging and integrating it twice to give free )+C Inr+C The velocity profile u(r) is obtained by applying the boundary conditions aular - atr - (because of symmetry about the centerline) and u-0 at T= R (the no-slip condition at the pipe surface) We get 05-15 ox Therefore, the velocity profile in fully developed laminar flow in a pipe is parabolic with a maximum at the centerline and minimum ero) at the pipe wall. Also, the axial velocity u is positive for any r, and thus the axial pres sure gradient dP/dx must be negative e pressure must decrease in the flow direction because of viscous effects) The average velocity is determined from its definition by substituting Eq. 8-15 into Eq. 8-2, and performing the integration. It gives Vonglurrar - - - ) -- Combining the last two equations, the velocity profile is rewritten as u(t) = 21,01- 8-17)

CHAPTER 8 This is a convenient form for the velocity profile since V. can be deter- mined easily from the flow rate information The maximum velocity occurs at the centerline and is determined from Eq 8-17 by substituting r = 0, -26 Therefore, the average velocity in fully developed laminar pipe flow is one- half of the maximum velocity Pressure Drop and Head Loss A quantity of interest in the analysis of pipe Now is the pressure drop AP since it is directly related to the power requirements of the fan or pump to maintain flow We note that did = constant and integrating from x = x where the pressure is P, to X X + L where the pressure is gives dPP-P expression in Eq. 8-16, the pressure Substituting Eq. 8-19 into the drop can be expressed as Laminar SLV_32LV BP-P-P-B D 8 201 The symbol A is typically used to indicate the difference between the final and initial values, like ay - y - y But in fluid flow, AP is used to desig nale pressure drop, and thus it is P -P A pressure drop due lo viscous effects represents an irreversible pressure loss, and it is called pressure loss AP to emphasize that it is a loss (ust like the head loss h, which is pro- portional lo it). Note from Eq. 8-20 that the pressure drap is proportional to the viscosity u of the nuld, and AP would be zero if there were no friction. Therefore, the drop of pressure from P, 10 P, in this case is due entirely to viscous effects, and Eq. 8-20 represents the pressure loss AP when a lluid of vis- cosity flows through a pipe of constant diameter and length L at aver age velocity V In practice, it is found convenient to express the pressure loss for all types of fully developed internal flows (laminar or turbulent flows, circular or noncircular pipes, smooth or rough surfaces, horizontal or inclined pipes) as (Fig. 8-13) Pressure oss: AP- Head los no Pressure oss: AP. - Bing 18-211 where pV2 is the dynamic pressure and f is the Darcy friction factor 18-22) it is also called the Darcy-Weisbach friction factor, named after the Frenchman Herry Darcy (1803-1858) and the German Julius Weisbach (1806-1871), the two engineers who provided the greatest contribution in its development. It should not be confused with the friction coefficient C FIGURE 8-13 The relation for pressure loss (and head loss) is one of the most general relations in fluid mechanics, and it is valid for laminar or turbulent flows circular or noncircular pipes, and pipes with smooth or rough surfaces FLUID MECHANICS (also called the Fanning friction factor named after the American engineer John Fanning (1837-1911), which is defined as C = 21 Nov?) =114 Setting Eq. 8-20 and 8-21 equal to each other and solving for f gives the friction factor for fully developed laminar Now in a circular pipe. Circola r e This equation shows that in laminar flow, the friction factor is a function of the Reynolds number only and is independent of the roughness of the pipe surface In the analysis of piping systems, pressure losses are commonly expressed in terms of the equivalent fluid column height, called the head loss h. Nol- ing from fluid statics that AP = pgh and thus a pressure difference of AP corresponds to a fluid height of h APlpg, the pipe head loss is obtained by dividing AP by pg to give AP. LV Head loss OD 2 The head loss h represents the additional height that the fluid needs to be raised by a pump in order to overcome the frictional losses in the pipe. The head loss is caused by viscosity, and it is directly related to the wall shear stress. Equations 8-21 and 8-24 are valid for both laminar and turbulent flows in both circular and noncircular pipes, but Eq. 8-23 is valid only for fully developed laminar flow in circular pipes Once the pressure loss for head loss) is known, the required pumping power to overcome the pressure loss is determined from Wpvg. - VAP - Veghị - ng.

lp -VA-BUL DK Weara - V3 - Vogh - ghi (8251 where V is the volume flow rate and is the mass flow rate The average velocity for laminar flow in a horizontal pipe is from Eq. 8-20, P.-PR (P. -PJD APD Horizontal pipe SL32L 32 L Then the volume flow rate for laminar flow through a horizontal pipe of diameter D and length L becomes P.-PRP - P SP D 128 128 This equation is known as Poiseuille's law, and this flow is called Hagen Poiseuille flow in honor of the works of G. Hagen (1797-1884) and J. Poiseuille (1799-1869) on the subject. Note from Eq B-27 that for a speci fied flow rate, the pressure drop and thus the required pumping power is pro- portional to the length of the pipe and the viscosity of the fluid, but it is inversely proportional to the fourth power of the radius (or diameter of the pipe. Therefore, the pumping power requirement for a piping system can be reduced by a factor of 16 by doubling the pipe diameter (Fig. 8-14). Or course the benefits of the reduction in the eng costs must be weighed against the increased cost of construction due to using a larger diameter pipe. the pressure drop Pequals the pressure loss Up in the case of a hor frontal pipe, but this is not the case for inclined pipes or pipes with vari able CTOSS-sectional area. This can be demonstrated by writing the energy FIGURE 8-14 The pumping power requirement for aminan piping system can be reduced by a factor of 16 by doubling the pipe di equation for steady, incompressible one-dimensional flow in terms of heads as (see Chap 5) P. Vi PV 19 20 21 +2+huth -28) where h e is the useful pump head delivered to the fluid, h . is the turbine head extracted from the fluid, he is the irreversible head loss between sections 1 and 2 V. and V, are the average velocities at sections 1 and 2, respectively, anda, and or are the kinetic energy correction factors at sections 1 and 2 (it can be shown that ar -2 for fully developed laminar flow and about 1.05 for fully developed turbulent flow). Equation B-28 can be rearranged as P.-P, - pleV3 - V2 + pal; -2.) + h -hepth) (8-29) Therefore, the pressure drop AP = P, - Pand pressure loss APL = pgh for a given flow section are equivalent if (i) the flow section is horizontal so that there are no hydrostatic or gravity effects (2, 2), (2) the flow sec- tion does not involve any work devices such as a pump or a turbine since they change the fluid pressure (hum = hsie = 0), (3) the cross-sectional area of the flow section is constant and thus the average flow velocity is constant (V = V), and (4) the velocity profiles at sections 1 and 2 are the same shape ( 2) FIGURE 8-1 Free-body diagram of a ring-shape differential fluid element of radius thickness dr. and length d oriente coaxially with an inclined pipe in full developed laminarflow Inclined Pipes Relations for inclined pipes can be obtained in a similar manner from a force balance in the direction of flow. The only additional force in this case is the component of the fluid weight in the flow direction, whose magnitude is W -W sin -p resin -pg(2-r dr da) sine 8-301 where is the angle between the horizontal and the flow direction (Fig. 8-15) The force balance in EQ 8-9 now becomes (2e dr P). -2mdr . +2 -). - 2 x) -92 rda)sins- which results in the differential equation Horizontal pipe: V. Inclined . Downwandco Following the same solution procedure, the velocity profile can be shown to be - tag sine)(- It can also be shown that the average velocity and the volume flow rate rela Lions for laminar flow through inclined pipes are, respectively CAP-i ndu- (AP - polsin D' - 32 12BL which are identical to the corresponding relations for horizontal pipes, except that AP is replaced by AP - pgl sine. Therefore, the results already obtained for horizontal pipes can also be used for inclined pipes provided that AP is replaced by AP-POL sin 8 (Fig. 8-16). Note that 8 >0 and thus Sin for uphill flow and and thus sine for downhill flow FIGURE B1 The relations developed for full developed laminar flow throug horizontal pipes can also be use for inclined pipes by replacin AP with APPL sin

FLUID MECHANICS In inclined pipes, the combined effect of pressure difference and gravity drives the flow. Gravity helps downhill flow but opposes uphill flow. There- fore, much greater pressure differences need to be applied to maintain a specified flow rate in uphill flow although this becomes important only for liquids, because the density of gases is generally low. In the special case of no flow (V-0), we have AP - pgl sin 8, which is what we would obtain from Nuid stalics (Chap. 3) Laminar Flow in Noncircular Pipes The friction factor relations are given in Table 8-1 for fully developed lam inar flow in pipes of various cross sections. The Reynolds number for flow in these pipes is based on the hydraulic diameter D = 4A./p, where A is the cross-sectional area of the pipe and p is its wetted perimeter TABLE 8-1 Friction factor for sections - AA developed and Re-V a bow in pipes of various CrOSS D Friction Factor Tube Geometry Circle 6400 Re Rectangle 56.92/Re 62.20 Re 61.36/Re 72.92/Re 78.BO/Re 82.32/Re 96.00 Re Ellipse 64.00/Re 67.28 Re 72.96/Re 76.60Re 78.16 Re 50.80 Re 52.28 Re 53.32 Re 52.6 Re 50.96 Re EXAMPLE 8-1 Flow Rates in Horizontal and Inclined Pipes Oil at 20°C (-888 m and -0.800 km - 5l is flowing steadily through a 5 cm diameter 40 m long pipe (Fig. 8-17). The pressure at the pipe niet and outlet are measure to be 745 and 97 kPa, respectively Determine the toate of oil through the pipe assuming the pipe is alhor rental inclined 15" pard, dinclined 15 downward. Also verify that the flow through the pipelaminar SOLUTION The pressure readings at the inlet and outlet of a pipe are given The flow rates are to be determined for three different orientations, and the How is to be shown to be laminar Assumptions 1 Tha Fions steady and incompressible. 2 The entrance ellects are neglig ble, and thus the flow is fully developed. 3 The pipe oves no components such as bends, valves, and connectors. 4 The piping section involves to work devices such as a pump or a turbine Properties The density and dynamic Viscosity of oil are given to be p - kim ard -OHOO kein. 5, respectively Analysis The pressure drop across the pipe and the pipe cross sectional arra are FIGURE 8-17 Schematic for Example 9-1 AP-P-P-745-97-648 kPa A - DM - (0.05 m) 44 -0.001963 m? (a) The flow rate for all three cases can be determined from Eq. 8-34, ü (AP - polsin) 128 L where is the angle the pipe makes with the horizontal. For the horizontal casa, 8 - 0 and thus sin 8-0. Therefore,

where is the angle the pipe makes with the horizontal. For the horizontal Case, - O and thus sing-0. Therefore, . AP D (648 kPa)(0.05 m)' (1000 Nm/1 kg mis 124 1280 800 kg/ ms)(40 m) 1 KP I N ) =0.00311 m/s Col For uphill flow with an inclination of 15", we have - +15", and ü (AP - POL sin D 120 [648,000 Pa - (888 kg/m (9.81 m/s) (40 m) sin 15°]*(0.06 m) /1 kg. m/s 128(0.800 kg/m 5) (40 m) 1 P . ) = 0.00267 m/s (c) For downhill flow with an inclination of 15', we have 8 = -15', and User (AP - pgl sin 6)D 12BuL 1648,000 Pa - (858 kg/m (9.81 m/s)(40 m) sin(-15)) (0.05 m) /1 kg. mis 129(0.800 kg/m - 5)(40 m) -0.00354 ms T Pam 360/2036 FLUID MECHANICS The flow rate is the highest for the downhill flow case, as expected. The average fluid velocity and the Reynolds number in this case are U 0.00354 m/s Mọi 0001963 m = 1.80 mis PVD (888 kg/m)(1.80 m/s)(0.05 m) - 100 0.800 kg/m - S which is much less than 2300. Therefore the flow is laminer for all three Cases and the analysis is valid Discussion Note that the flow is driven by the combined effect of pressure difference and gravity. As can be seen from the flow rates we calculated, eravity opposes uphill flow, but enhance downhill flow Gravity has no effect on the flow rate in the horizontal case. Downhill flow can occur even in the absence of an applied pressure difference for the case of P - R - 97 kPa (ie, no applied pressure difference), the pressure throughout the entire pipe would remain constant at 97 Pa, and the fluid would flow through the pipe at a rate of 0.00043 mrs under the influence of gravity. The flow rate increases as the tilt angle of the pipe from the horizontal is increased in the negative direction and would reach its maximum value when the pipe is vertical. EXAMPLE 8 2 Pressure Drop and Head Loss in a Pipe Water at 40°F (- 62.42 lbmitt and - 1.038 x 10-3 ihmitt.s) is flowing through a 0.12-in- (- 0.010 Ft) diameter 30-fl-long horizontal pipe steadily at an average velocity of 3.0 ft/s (Fig. 8-181. Determine (al the head loss, (b) the pressure drop, and (c) the pumping power reruirement to over come this pressure drop. FIGURE B-1B Schematic for Example 8-2 SOLUTION The average flow velocity in a pipe is given. The head loss, the pressure drop, and the pumping power are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are neglie ble, and thus the flow is fully developed. 3 The pipe involves components such as bends, valves, and connectors Properties The dersity and dynamic viscosity of water are given to be pa 62.42 Ibmit and -1.038 X 10 bmvits, respectively Analysis (a) First we need to determine the flow regime. The Reynolds num baris PVD (62.42 Ibm ) (3 001) = 1803 1038 x 10ibmlft. which is less than 2300. Therefore, the flow is laminar. Then the friction factor and the head loss become 64 -0.0355 14.9 Re 1803 LV- 5301 (3 tus)? -roza 0.BIBS 0.01 ft 2(322 ) (bl Noting that the pipe is horizontal and its diameter is constant, the pres- sure drop in the pipe is due entirely to the frictional losses and is equivalent to the pressure loss,

335 CHAPTER 8 SP = APL = P = 0.0355 30 ft (62.42 Ibm/ft)(3 ft/s)? ( 1 lbf - D 2 .000.01 ft 2 (32.2 lbm · ft/s) = 929 lbf/ft2 = 6.45 psi (c) The volume flow rate and the pumping power requirements are j = Vavg A. = Vavg(T-D214) = (3 ft/s)[+(0.01 ft)214] = 0.000236 ft?/s 1W W pump = j AP = (0.000236 ft?ls)(929 1bf/ft?) G alicia) = 0.30 W Therefore, power input in the amount of 0.30 W is needed to overcome the frictional losses in the flow due to viscosity. Discussion The pressure rise provided by a pump is often listed by a pump manufacturer in units of head (Chap. 14). Thus, the pump in this flow needs to provide 14.9 ft of water head in order to overcome the irreversible head loss.