fluid_mechanics

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394 Chapter 8 ■ Viscous Flow in Pipes ¢z is the change in elevation2 is a hydrostatic type pressure term. If there is no flow, V 0 and ¢p g/ sin u g¢z, as expected for fluid statics. E XAMPLE 8.2 Laminar Pipe Flow GIVEN An oil with a viscosity of m 0.40 N # sm 2 and density r 900 kgm 3 flows in a pipe of diameter D 0.020 m. through the pipe at the same rate as in part 1a2, but with p 1 p 2 ? (b) How steep a hill, u, must the pipe be on if the oil is to flow (c) For the conditions of part 1b2, if p 1 200 kPa, what is the FIND (a) What pressure drop, p 1 p 2 , is needed to produce pressure at section x 3 5 m, where x is measured along the pipe? a flowrate of Q 2.0 10 5 m 3 s if the pipe is horizontal with x 1 0 and x 2 10 m? SOLUTION (a) If the Reynolds number is less than 2100 the flow is laminar and the equations derived in this section are valid. Since the average velocity is V QA 12.0 10 5 m 3 s2 3p10.0202 2 m 2 44 0.0637 ms, the Reynolds number is Re rVDm 2.87 6 2100. Hence, the flow is laminar and from Eq. 8.9 with / x 2 x 1 10 m, the pressure drop is or (Ans) (b) If the pipe is on a hill of angle u such that ¢p p 1 p 2 0, Eq. 8.12 gives or ¢p p 1 p 2 128m/Q pD 4 12810.40 N # sm 2 2110.0 m212.0 10 5 m 3 s2 p10.020 m2 4 sin u 12810.40 N # sm 2 212.0 10 5 m 3 s2 p1900 kgm 3 219.81 ms 2 210.020 m2 4 Thus, u 13.34°. ¢p 20,400 Nm 2 20.4 kPa sin u 128mQ prgD 4 (1) (Ans) COMMENT This checks with the previous horizontal result as is seen from the fact that a change in elevation of ¢z / sin u 110 m2 sin113.34°2 2.31 m is equivalent to a pressure change of ¢p rg ¢z 1900 kgm 3 219.81 ms 2 2 12.31 m2 20,400 Nm 2 , which is equivalent to that needed for the horizontal pipe. For the horizontal pipe it is the work done by the pressure forces that overcomes the viscous dissipation. For the zero-pressure-drop pipe on the hill, it is the change in potential energy of the fluid “falling” down the hill that is converted to the energy lost by viscous dissipation. Note that if it is desired to increase the flowrate to Q 1.0 10 4 m 3 s with p 1 p 2 , the value of u given by Eq. 1 is sin u 1.15. Since the sine of an angle cannot be greater than 1, this flow would not be possible. The weight of the fluid would not be large enough to offset the viscous force generated for the flowrate desired. A larger diameter pipe would be needed. (c) With p 1 p 2 the length of the pipe, /, does not appear in the flowrate equation 1Eq. 12. This is a statement of the fact that for such cases the pressure is constant all along the pipe 1provided the pipe lies on a hill of constant slope2. This can be seen by substituting the values of Q and u from case 1b2 into Eq. 8.12 and noting that ¢p 0 for any /. For example, ¢p p 1 p 3 0 if / x 3 x 1 5 m. Thus, p 1 p 2 p 3 so that p 3 200 kPa (Ans) COMMENT Note that if the fluid were gasoline 1m 3.1 10 4 N # sm 2 and r 680 kgm 3 2, the Reynolds number would be Re 2790, the flow would probably not be laminar, and use of Eqs. 8.9 and 8.12 would give incorrect results. Also note from Eq. 1 that the kinematic viscosity, n mr, is the important viscous parameter. This is a statement of the fact that with constant pressure along the pipe, it is the ratio of the viscous force 1m2 to the weight force 1g rg2 that determines the value of u. Poiseuille’s law can be obtained from the Navier–Stokes equations. 8.2.2 From the Navier–Stokes Equations In the previous section we obtained results for fully developed laminar pipe flow by applying Newton’s second law and the assumption of a Newtonian fluid to a specific portion of the fluid— a cylinder of fluid centered on the axis of a long, round pipe. When this governing law and assumptions are applied to a general fluid flow 1not restricted to pipe flow2, the result is the Navier –Stokes equations as discussed in Chapter 6. In Section 6.9.3 these equations were solved for the specific geometry of fully developed laminar flow in a round pipe. The results are the same as those given in Eq. 8.7.

8.2 Fully Developed Laminar Flow 395 The governing differential equations can be simplified by appropriate assumptions. We will not repeat the detailed steps used to obtain the laminar pipe flow from the Navier – Stokes equations 1see Section 6.9.32 but will indicate how the various assumptions used and steps applied in the derivation correlate with the analysis used in the previous section. General motion of an incompressible Newtonian fluid is governed by the continuity equation 1conservation of mass, Eq. 6.312 and the momentum equation 1Eq. 6.1272, which are rewritten here for convenience: 0V 0t V 0 V V p r g n2 V (8.13) (8.14) For steady, fully developed flow in a pipe, the velocity contains only an axial component, which is a function of only the radial coordinate 3V u1r2î 4. For such conditions, the left-hand side of the Eq. 8.14 is zero. This is equivalent to saying that the fluid experiences no acceleration as it flows along. The same constraint was used in the previous section when considering F ma for the fluid cylinder. Thus, with g gkˆ the Navier –Stokes equations become V 0 p rgkˆ m 2 V (8.15) The flow is governed by a balance of pressure, weight, and viscous forces in the flow direction, similar to that shown in Fig. 8.10 and Eq. 8.10. If the flow were not fully developed 1as in an entrance region, for example2, it would not be possible to simplify the Navier –Stokes equations to that form given in Eq. 8.15 1the nonlinear term V V would not be zero2, and the solution would be very difficult to obtain. Because of the assumption that V u1r2î, the continuity equation, Eq. 8.13, is automatically satisfied. This conservation of mass condition was also automatically satisfied by the incompressible flow assumption in the derivation in the previous section. The fluid flows across one section of the pipe at the same rate that it flows across any other section 1see Fig. 8.82. When it is written in terms of polar coordinates 1as was done in Section 6.9.32, the component of Eq. 8.15 along the pipe becomes 0p 0x rg sin u m 1 r 0 0u ar 0r 0r b (8.16) Since the flow is fully developed, u u1r2 and the right-hand side is a function of, at most, only r. The left-hand side is a function of, at most, only x. It was shown that this leads to the condition that the pressure gradient in the x direction is a constant— 0p0x ¢p/. The same condition was used in the derivation of the previous section 1Eq. 8.32. It is seen from Eq. 8.16 that the effect of a nonhorizontal pipe enters into the Navier–Stokes equations in the same manner as was discussed in the previous section. The pressure gradient in the flow direction is coupled with the effect of the weight in that direction to produce an effective pressure gradient of ¢p/ rg sin u. The velocity profile is obtained by integration of Eq. 8.16. Since it is a second-order equation, two boundary conditions are needed—112 the fluid sticks to the pipe wall 1as was also done in Eq. 8.72 and 122 either of the equivalent forms that the velocity remains finite throughout the flow 1in particular u 6 q at r 02 or, because of symmetry, that 0u0r 0 at r 0. In the derivation of the previous section, only one boundary condition 1the no-slip condition at the wall2 was needed because the equation integrated was a first-order equation. The other condition 10u0r 0 at r 02 was automatically built into the analysis because of the fact that t m dudr and t 2t w rD 0 at r 0. The results obtained by either applying F ma to a fluid cylinder 1Section 8.2.12 or solving the Navier –Stokes equations 1Section 6.9.32 are exactly the same. Similarly, the basic assumptions regarding the flow structure are the same. This should not be surprising because the two methods are based on the same principle—Newton’s second law. One is restricted to fully developed laminar pipe flow from the beginning 1the drawing of the free-body diagram2, and the other starts with the general governing equations 1the Navier –Stokes equations2 with the appropriate restrictions concerning fully developed laminar flow applied as the solution process progresses.

8.2 Fully Developed Laminar Flow 395<br />

The governing<br />

differential equations<br />

can be simplified<br />

by<br />

appropriate assumptions.<br />

We will not repeat the detailed steps used to obtain the laminar pipe flow from the Navier –<br />

Stokes equations 1see Section 6.9.32 but will indicate how the various assumptions used and steps applied<br />

in the derivation correlate with the analysis used in the previous section.<br />

General motion of an incompressible Newtonian <strong>fluid</strong> is governed by the continuity equation<br />

1conservation of mass, Eq. 6.312 and the momentum equation 1Eq. 6.1272, which are rewritten<br />

here for convenience:<br />

0V<br />

0t<br />

V 0<br />

V V p<br />

r g n2 V<br />

(8.13)<br />

(8.14)<br />

For steady, fully developed flow in a pipe, the velocity contains only an axial component, which<br />

is a function of only the radial coordinate 3V u1r2î 4. For such conditions, the left-hand side of<br />

the Eq. 8.14 is zero. This is equivalent to saying that the <strong>fluid</strong> experiences no acceleration as it<br />

flows along. The same constraint was used in the previous section when considering F ma for<br />

the <strong>fluid</strong> cylinder. Thus, with g gkˆ the Navier –Stokes equations become<br />

V 0<br />

p rgkˆ m 2 V<br />

(8.15)<br />

The flow is governed by a balance of pressure, weight, and viscous forces in the flow direction,<br />

similar to that shown in Fig. 8.10 and Eq. 8.10. If the flow were not fully developed 1as in an entrance<br />

region, for example2, it would not be possible to simplify the Navier –Stokes equations to that<br />

form given in Eq. 8.15 1the nonlinear term V V would not be zero2, and the solution would be<br />

very difficult to obtain.<br />

Because of the assumption that V u1r2î, the continuity equation, Eq. 8.13, is automatically<br />

satisfied. This conservation of mass condition was also automatically satisfied by the<br />

incompressible flow assumption in the derivation in the previous section. The <strong>fluid</strong> flows across<br />

one section of the pipe at the same rate that it flows across any other section 1see Fig. 8.82.<br />

When it is written in terms of polar coordinates 1as was done in Section 6.9.32, the component<br />

of Eq. 8.15 along the pipe becomes<br />

0p<br />

0x rg sin u m 1 r<br />

0 0u<br />

ar<br />

0r 0r b<br />

(8.16)<br />

Since the flow is fully developed, u u1r2 and the right-hand side is a function of, at most, only<br />

r. The left-hand side is a function of, at most, only x. It was shown that this leads to the condition<br />

that the pressure gradient in the x direction is a constant— 0p0x ¢p/. The same condition<br />

was used in the derivation of the previous section 1Eq. 8.32.<br />

It is seen from Eq. 8.16 that the effect of a nonhorizontal pipe enters into the Navier–Stokes<br />

equations in the same manner as was discussed in the previous section. The pressure gradient in<br />

the flow direction is coupled with the effect of the weight in that direction to produce an effective<br />

pressure gradient of ¢p/ rg sin u.<br />

The velocity profile is obtained by integration of Eq. 8.16. Since it is a second-order equation,<br />

two boundary conditions are needed—112 the <strong>fluid</strong> sticks to the pipe wall 1as was also done<br />

in Eq. 8.72 and 122 either of the equivalent forms that the velocity remains finite throughout the<br />

flow 1in particular u 6 q at r 02 or, because of symmetry, that 0u0r 0 at r 0. In the derivation<br />

of the previous section, only one boundary condition 1the no-slip condition at the wall2 was<br />

needed because the equation integrated was a first-order equation. The other condition<br />

10u0r 0 at r 02 was automatically built into the analysis because of the fact that t m dudr<br />

and t 2t w rD 0 at r 0.<br />

The results obtained by either applying F ma to a <strong>fluid</strong> cylinder 1Section 8.2.12 or solving<br />

the Navier –Stokes equations 1Section 6.9.32 are exactly the same. Similarly, the basic assumptions<br />

regarding the flow structure are the same. This should not be surprising because the two methods<br />

are based on the same principle—Newton’s second law. One is restricted to fully developed laminar<br />

pipe flow from the beginning 1the drawing of the free-body diagram2, and the other starts with<br />

the general governing equations 1the Navier –Stokes equations2 with the appropriate restrictions<br />

concerning fully developed laminar flow applied as the solution process progresses.

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