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Unformatted text preview: 7.28. CHAPTER 7, PROBLEM 28 983 7.28 Chapter 7, Problem 28 Problem: The radius of the pipe section shown below and the inlet velocity are constant and equal to R and U , respectively. At the outlet, the velocity is given by u = uo (r), where r is distance from the pipe centerline. The difference between inlet pressure, pi , and outlet pressure, po , is pi − po = 1.05ρ U 2 , where ρ is the density of the (incompressible) fluid in the pipe. The pipe lies in a horizontal plane so that potential energy is constant. Determine Umax as a function of U , compute α1 and α2 , and compute the head loss, hL , as a function of U and the acceleration of gravity, g . Up ......................................................................................................................................................................................................... . ................................................................................................................................................................................................... ............................................................................................................................................................................................................................. . ........................................................ .... .... .... .... .... .... .... .... . . .......... ..... ..... ..... ............................................................................................................................................................................................................ . . ......................................... o max ...................................... .................................. . .. . . ................................ ........ .................................. ............................... .............................. . .. .. . ............................... .............................. ................... ................ .. .......................... ............................... ................................... ............................... ...................................... ............................................ ..................................... ... ..... . ........................................................................................................................................................................................................................... ............................................................................................................................................................................................ .. . . . ............................................................................................................................................................................................. .................................................................................................................................................................................................................. . .. .................................................................................................................................................................................................. .............................................................................................................................................................................................. . . .......................................................................................................................................................................................... .................................................................................................................................................................................. . ..................................................................................................................................................................... .......................................................................................................................................... u (r) = U 1− r R uo po Solution: Select the stationary control volume shown in the figure Mass Principle: In general, the mass principle tells us that d ρ dV + ρ ure · n dS = 0 dt V S Because the flow is steady and incompressible, the instantaneous rate of change of mass vanishes. Also, ure = u. Hence, the mass principle simplifies to S ρ u · n dS = 0 Thus, noting the axial symmetry of the pipe, we have ρ −π U R2 + 2πρUmax In et R 0 1− r rdr = 0 R Out et Introducing a dimensionless integration variable η = r/R, we have 2Umax R2 1 0 (1 − η )η dη = UR2 Then performing the indicated integration 2Umax η2 η3 − 2 3 η =1 =U η =0 =⇒ Umax = 3U 984 CHAPTER 7. ENERGY PRINCIPLE Kinetic-Energy Correction Factors: By definition, the quantity α is α≡ 1 π R2 2π 0 u u R 0 3 rdrdθ = 2 R2 u u R 0 3 rdr Now, at the inlet, u = u = U , wherefore α1 = 2 R2 r R R rdr = 0 2 r =R =1 r =0 At the outlet, it is obvious from the mass principle that u = U . Thus, α2 = 2 R2 Umax U R 0 3 1− r R 3 rdr = 54 R2 R 1− 0 r R 3 rdr Finally, define η ≡ r/R, so that 1 α2 = 54 0 (1 − η )3 η dη = 54 η2 3 η5 = 54 − η3 + η4 − 2 4 5 1 0 η − 3η 2 + 3η 3 − η 4 dη η =1 = η =0 54 = 2.7 20 Energy Principle: So, applying the approximate energy equation, we have p1 u2 ¯ p2 u2 ¯ + α1 1 + z1 = + α2 2 + z2 + hL ρg 2g ρg 2g Now, we are given u1 = u2 = U , p1 = pi and p2 = pa . Also, because the pipe lies in a horizontal plane, z1 = z2 . Thus, using the computed values of α1 and α2 , pi U2 pa U2 + = + 2.7 + hL ρg 2g ρg 2g =⇒ hL = U2 pi − pa − 1.7 ρg 2g Finally, noting that pi − pa = 1.05ρ U 2 , hL = 1.05 ρU 2 U2 U2 − 0.85 = 0.20 ρg g g ...
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