MC 2312 Mecánica De Fluidos – Potter (Solucionario Capítulo 4) pdf

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(1)CHAPTER 4. The Integral Forms of the Fundamental Laws 4.1. 4.2. 4.3 4.4. v a) No net force may act on the system: ΣF = 0. b) The energy transferred to or from the system must be zero: Q - W = 0. v c) If V3n = V3 ⋅ nˆ 2 = 10iˆ ⋅ (− jˆ ) = 0 is the same for all volume elements then v D v v D v ΣF = V ∫ dm , or ΣF = (mV ). Since mass is constant for a system Dt v v Dt v v DV DV v v ΣF = m . Since = a, Σ F = ma. Dt Dt. v 1 Extensive properties: Mass, m; Momentum, mV ; kinetic energy, mV 2 ; 2 potential energy, mgh; enthalpy, H. v Associated intensive properties (divide by the mass): unity, 1; velocity, V; V2 /2; gh; H/m = h (specific enthalpy). Intensive properties: Temperature, T; time, t; pressure, p; density, ρ; viscosity, µ. (B). System (t ) = V 1 c.v.(t ) = V 1 System (t + ∆t ) = V 1 + V c.v.(t + ∆t ) = V. 4.5. System ( t ) = V 1 + V c.v.( t ) = V 1 + V System (t + ∆t ) = V. 1. 2. 2. 1. 2. 2 2. +V. c.v.( t + ∆t ) = V 1 + V. 1. 3. 2 pump. 2. 50. 3.

(2) 4.6. a) The energy equation (the 1st law of Thermo). b) The conservation of mass. c) Newton’s 2nd law. d) The energy equation. e) The energy equation.. 4.7. n̂. n̂. v. v. n̂. v. ω. v. n̂. 4.8. n̂. n̂ n̂ v. v. v. n̂. v. v. 4.9. n̂. n̂. v. v v. n̂. n̂. v. v. n̂. n̂ 4.10. 4.11. v. 1 $ 1 $ n$ 1 = − i− j = −0.707(i$ + $j ) . n$ 2 = 0.866 $i − 0.5 $j . 2 2 v V1n = V1 ⋅ nˆ1 = 10iˆ ⋅ [ −0.707( iˆ + ˆj )] = −7.07 fps v V2 n = V2 ⋅ n$ 2 = 10i$ ⋅ ( 0.866i$ − 0.5 $j ) = 8.66 fps v V3n = V3 ⋅ nˆ 2 = 10iˆ ⋅ (− jˆ ) = 0. v flux = ηρn$ ⋅ VA flux1 = ηρ[−0.707 (i$ + $j )] ⋅ 10i$A / 0.707 = −10ηρA flux2 = ηρ( 0.866i$ − 0.5 $j ) ⋅ 10i$A / 0.866 = 10ηρA flux3 = ηρ( − $j ) ⋅ 10$iA = 0 3. 51. n$ 3 = − $j .. v. n̂.

(3) 4.12. v (B ⋅ n$ ) A = 15(0.5i$ + 0.866 $j ) ⋅ $j (10 × 12) = 15 × 0.866 × 120 = 1559 cm 3 Volume = 15 sin 60 o × 10 × 12 = 1559 cm 3. 4.13. The control volume must be independent of time. Since all space coordinates are integrated out on the left, only time remains; thus, we use an ordinary derivative to differentiate a function of time. But, on the right, we note that ρ and η may be functions of (x, y, z, t); hence, the partial derivative is used.. 4.14 2 c.v. (0) = c.v. (∆t) = volume 1. 1. 4.15. system (∆t) is in volumes 1 and 2. 2 3. 1. system (∆t) = V1 + V 2 + V 3 c.v. (∆ t) = V1 + V2. 1. 4.16 system boundary at (t + ∆t). 4.17. If fluid crosses the control surface only on areas A1 and A2 , v v v $ $ $ ρ n ⋅ VdA = ρ n ⋅ VdA + ρ n ⋅ V dA = 0 ∫ ∫ ∫ c .s .. A1. A2. For uniform flow all quantities are constant over each area: v v ρ 1n$ 1 ⋅ V 1 ∫ dA + ρ 2 n$ 2 ⋅ V 2 ∫ dA = 0 A1. A. 2 v v Let A 1 be the inlet so n$ 1 ⋅ V1 = −V1 and A 2 be the outlet so n$ 2 ⋅ V2 = V 2 . Then. −ρ 1V1 A 1 + ρ 2 V2 A 2 = 0 or. ρ 2 A 2V 2 = ρ 1A 1V 1. 52.

(4) 4.18. Use Eq. 4.4.2 with m V representing the mass in the volume: v dm V dm V 0= + ∫ ρn$ ⋅ V dA = + ρA 2 V 2 − ρA 1V 1 dt dt c. s . dm V = + ρQ − m& . dt Finally, dmV = m& − ρQ. dt. 4.19. Use Eq. 4.4.2 with m S representing the mass in the sponge: v dm S dm S 0= + ∫ ρn$ ⋅ VdA = + ρA 2 V2 + ρA 3V 3 − ρA 1V1 dt dt =. dm S + m& 2 + ρA 3V 3 − ρQ 1 . dt. Finally, dm S = ρQ1 − m& 2 − ρA 3V 3 . dt m& = ρ AV =. 4.20. (D). 4.21. A1V1 = A2V2.. p 200 AV = π × 0.042 × 70 = 0.837 kg/s . RT 0.287 × 293. π×. 1.25 2 2.5 2 × 60 = π × V2. 144 144. ∴V2 = 15 ft/sec.. 1.25 2 1.25 2 m& = ρAV = 1.94π × 60 = 3.968 slug/sec. Q = AV = π × 60 = 2.045 ft 3 / sec. 144 144 4.22. A1 V1 = A2 V2 . π × .0252 × 10 = (2π × .6 × .003)V2 . ∴V2 = 1.736 m/s. 2 m & = ρAV = 1000π ×.025 × 10 = 19.63 kg/s. Q = AV=π × .0252 × 10 = 0.01963 m 3 / s.. 4.23. & in = ρA1 V1 + ρA2 V2 . 200 = 1000 π × .0252 × 25 + 1000 Q2 . ∴Q2 = 0.1509 m 3 / s. m. 4.24. ρ1 =. p1 40 × 144 7 × 144 = = .006455 slug/ft3 . ρ 2 = = .000963 slug/ft3 . RT1 1716 × 520 1716 × 610 & m .2 & = ρAV . m ∴V1 = = . ∴V1 = 355 fps. 2 ρ 1 A 1 (π × 2 / 144 ).006455 m & 2 = 0.2 =.000963 × (2 × 3 / 144)V2 . ∴V2 = 4984 fps.. 53.

(5) 4.25. p1 500 kg 1246 kg = = 4.433 3 . ρ 2 = = 8.317 3 RT .287 × 393 m .287 × 522 m 2 2 4.433 π × .05 × 600 = 8.317 π × .05 V2 . ∴V2 = 319.8 m/s. m & = ρ 1 A1V1 = 20.89 kg/s. Q 1 = A1V1 = 4.712 m 3 / s . Q2 = 2.512 m 3 / s .. 4.26. ρ 1 A1V1 = ρ 2 A2V2 p1 p A 1V1 = 2 A 2V 2 RT1 RT2 200 120 π × 0.05 2 × 40 = π × 0.03 2 × 120. 293 T2. ρ1 A1V1 = ρ 2 A2V2 .. ∴ T 2 = 189.9 K. 4.27. ρ1 =. or. −83 o C.. d 22 a) A 1V 1 = A 2 V 2 . (2 × 1.5 + 1.5 × 1.5) 3 = π × 2. 4 d2 2 b) (2 × 1.5 + 1.5 × 1.5) 3 = π 2 × . ∴d2 = 4.478 m 4 2 1 R c) (2 × 1.5 + 1.5 × 1.5) 3 =  πR 2 − ×.866 R × 2. 3  2 ∴R = 3.581 m. ∴d2 = 7.162 m. 4.28. (A). 4.29.  r a) v = 10 1 −  .  r0 . ∴d2 = 3.167 m cosθ = 1/2 θ = 60o θ. R. Refer to the circle of Problem 4.27: 75.7 × 2 Q = AV = (π × 0.42 × − 0.10 × 0.40 × sin75.5o ) × 3 = 0.516 m 3 /s. 360 r0   r r2  πr V = ∫ vdA = ∫ 10 1 −  2πrdr = 20π ∫  r − dr . r0   r0  0 0 0 r0. r0. 2 0. 20  r02 r02  10 ∴V = 2  −  = = 3.333 m/s. r0  2 3 3 m& = ρA V = 1000 × π ×.04 2 × 3.33 = 16.75 kg / s.. Q = AV = 0.01675 m 3 / s.. r0  r2    r2 r2  r2  b) v = 10 1 − 2  . πr02V = ∫ 10 1 − 2  2πrdr = 20π  0 − 0  . ∴V = 5 m/s r0  4  r0    2 0 2 m& = ρA V = 1000 × π ×.04 × 5 = 25.13 kg / s. Q = AV = 0.02513 m 3 / s..  r c) v = 20 1 −  . r0  . πr02V =.  r 2 20 1 −  ∫r / 2  r0  2πrdr + 10πr0 / 4. 0 r0. m & = ρAV = 1000 × π ×.04 2 × 5.833 = 29.32 kg / s.. 54. ∴V = 5.833 m/s. Q = 0.02932 m 3 / s..

(6) 4.30. a) Since the area is rectangular, V = 5 m/s. m& = ρA V = 1000 × .08 × .8 × 5 = 320 kg / s.. Q=. & m = 0.32 m 3 / s . ρ.  y y2 b) v = 40 − 2  with y = 0 at the lower wall. h h   y y2 h ∴ Vhw = ∫ 40 − 2  wdy = 40 × w. ∴V = 6.667 m/s. 6 h h  0 h. m& = ρA V = 1000 × .08 × .8 × 6.667 = 426.7 kg / s.. c) V × .08 = 10 × .04 + 5 × .02 + 5 × .02.. Q = 0.4267 m 3 / s.. ∴V = 7.5 m/s. m& Q& = = 0.48 m 3 / s . ρ. m& = ρA V = 1000 × .08 × .8 × 7.5 = 480 kg / s.. 4.31. a) A 1V 1 = ∫ v 2 dA . With r0 =. 1 , 24. b) A 1V 1 = ∫ v 2 dA . With h =. 1 , 24. c) A1V1 = ∫ v2 dA.. 2 r0  r2  r02  1 π ×   × 6 = ∫ v max  1 − 2  2πrdr = 2πv m a x .  24  r0  4  0. 1 ×w × 6= 12. ∴ v ( y ) = 9(1 − 576y 2 ) fps..  r2  r02 π × 0.01 × 2 = ∫ vmax  1 − 2  2π rdr = 2π vmax .  r  4 0  0  2. r0. ∴ v ( r ) = 4(1 − 10000r 2 ) m/s.. vmax = 4 m/s..  y2  4h 0.02 × w × 2 = ∫ vmax  1 − 2 wdy = vmax w .   3 h  −h  h. With h = 0.01 m, 4.32.  y2 4h ∫− hv m a x  1 − h 2  wdy = v m a x w 3 . h. v max = 9 fps.. With r0 = 0.01 m, d) n̂. ∴ v (r ) = 12( 1 − 576r 2 ) fps.. v max = 12 fps.. vmax = 3 m/s.. ∴ v( y ) = 3(1 − 10000 y 2 ) m/s.. If dm / dt = 0 , then ρ 1 A 1V1 = ρ 2 A 2 V 2 + ρ 3 A 3 V3 . In terms of m& 2 and Q 3 this becomes, letting ρ 1 = ρ 2 = ρ 3 , 1000 × π × 0.02 2 × 12 = m& 2 + 1000 × 0.01. r1. 4.33. ∫ v dA = A V . 1. 0. ∴ 2πv max. 2. 2. ∴ m& 2 = 5.08 kg / s ..  r2  2 ∫0 v m a x  1 − r12  2πrdr = π ×.0025 × 2. r1. .005 2 = π ×.0025 2 × 2. 4. ∴ v max = 1 m/s.. 55.  r2  ∴ v (r ) =  1 −  m / s. .005 2  .

(7) 4.34.  .1  2 ρ ×.2 × 2 × 10 =  ρ ∫ 10(20 y − 100y )2 dy + ρ ×.1 × 2 × 10 + m& .  0  Note: We see that at y = 0.1 m the velocity u(.1) = 10 m/s. Thus we integrate to y = 0.1, and between y = 0.1 and 0.2 the velocity u = 10. 4  ∴ m& = 0.6667ρ = 0.82 kg/s. 4ρ =  ρ + 2ρ  + m& . 3  m& in = m& out + m& .. h. 4.35. h. V1 h1 = ∫ u( y )dy .. 10×.05 = ∫ 10( 20y − 100y 2 )dy. 0. 0. 100 3   2 = 10 10 h − h .   3 ∴666.7 h3 − 200 h2 = −1. This can be solved by trial-and-error: h = .06: −.576 ? −1. h = .07: −.751 ? −1. h = .08: −.939 ? −1. h = .083: −.997 ? −1. h = .084: −1.016 ? −1. ∴h = 0.0832: or 8.32 cm.. Note: Fluid does not cross a streamline so all the flow that enters on the left leaves on the right. The streamline simply moves further from the wall. 4.36. m& = ∫ ρVdA =. 1/ 3. ∫ 2.2(1−.3545y )(6 y − 9 y. 2. )2 × 5dy. 0. 1/ 3. (. ). = 22 ∫ 6 y − 2.127 y 2 − 9 y 2 + 3.19 y 3 dy = 4.528 slug/sec. 0. 2 2 4 u m a x = × 2 = fps. (See Prob. 4.31b). 3 3 3 2.2 + 1.94 4  1 ρ= = 2.07 slug/ft3 . ∴ ρV A = 2.07 × ×  5 ×  = 4.6 slug/sec. 2 3  3. V =. Thus, ρV A ≠ m& since ρ = ρ(y) and V = V(y) so that ρV ≠ ρV . 4.37. A 1V 1 = A 2 V 2 .. 4.38. 3 4 m 3 of air 3 m of H 2 O 2000 × π ×.0015 × 9000 × 5 = 1.5 × (1.5h). 3 m 3 of air s. 4.39. Use Eq. 4.3.3:. π ×.012 × 8 = ( 2π ×.2×.04 )V 2 cos 30 o .. ∂ρ ∴ ρ 1 A 1V1 = − tire . V ∂t. 0=. ∂ρ. ∫ ∂t d −V. v + ρ 1V1 ⋅ n$ 1 A 1 .. ∴V2 = 0.05774 m/s.. v V1 ⋅ n$ 1 = −V1 .. ( 37 + 14.7)144 1 ∂ρ × π ×   × 180 = × 17.  96  1716 × 520 ∂t 2. 56. ∴h = 0.565 m..

(8) ∴. 4.40. slug ∂ρ = 3.01 × 10 −5 3 . ∂t ft − sec. m & in = m &2+m & 3.. V1 = 20 m/s (see Prob. 4.31c). 20 × 1000π ×.02 = 10 + 1000π ×.02 2 V3 . ∴ V3 = 12.04 m/s. 2. 4.41. 0= ∴. 4.42. 4.43. 4.44. d d & net = m c. v. + m &2 +m &3 −m &1 m c .v. + m dt dt. d & 1 −m &2 −m & 3 = 1000 × π ×.02 2 × 20 − 10 − 1000π ×.02 2 × 10 m c . v. = m dt = 2.57 kg/s.. The control surface is close to the Ve interface at the instant shown. n̂ ∴Vi = interface velocity. ρ e A eV e = ρ i A i Vi . 8000 1.5 × π ×.15 2 × 300 = π × 12 V i . .287 × 673 ∴Vi = 0.244 m/s. Assume an incompressible flow: 4Q1 = A 2 V 2 . 4 × 1500 / 60 = (2 × 4)V 2 .. ∴ V 2 = 12.5 fps.. For an incompressible flow (low speed air flow). ∫ udA = A V 2. 0. 2. 2. .. ∫ 20 y. 1/ 5. × 0.8dy = π × 0.15 2 V 2 .. 0. A1. 5 20 × 0.8 0.2 6/ 5 = π × 0.15 2 V 2 . 6. 4.45. n̂. Vi. ∴ V 2 = 27.3 m / s.. A 1V1 + ∫ v 2 dA = A e Ve.  r2  2 200 1 −  ∫0  0.025 2  2πrdr = π × 0.1 V e 0.1178 + 0.1963 = 0.0314V e . ∴ V e = 10.0 m / s. 0 .025. π ( 0.1 2 − 0.025 2 ) × 4 +. 4.46. Draw a control volume around the entire set-up: dm tissue 0= + ρV 2 A 2 − ρV1 A 1 dt  d2 − d2  & 2 = m& tissue + ρπ  2  h2 − ρπ ( h1 tan φ ) h&1  4 . 57.

(9) or  d 2 − d 22 &  & tissue = ρπ  m h2 + h 21 h& 1 tan 2 φ .  4 . 4.47. The width w of the channel is constant throughout the flow. Then dm d 0= + ρA 2 V 2 − ρA 1V 1 . 0 = ( ρwhL ) + ρA 2 V 2 − ρA 1V 1 dt dt dh 0=ρ w × 100 + ρ 0.2 w × 8 − ρ 4 w × 0.2. ∴ h& = 0.008 m / s . dt. 4.48. 0=. 4.49. dm + ρA 2 V 2 − ρA 1V 1 dt & + 1000( π × 0.003 2 × 0.02 − 10 × 10 −6 / 60). =m. ρ 1 A 1 V1 = ρ 2 A 2 V 2 .. m& 1 = ρ 2 A 2V 2 .. 400e −10/ 100 × 10 −6 × 900 = 0.2 × π × 0.05 2 Ve .. 4.50. ∴ m& = 3.99 × 10 −4 kg / s.. ∴ V e = 207 m / s.. dm + ρ 3Q 3 − ρ 1 A 1V1 − m& 2 where m = ρAh. dt a) 0 = 1000π × 0.6 2 h& + 1000 × 0.6 / 60 − 1000π × 0.02 2 × 10 − 10. ∴ h& = 0.0111 m / s or 11.1 mm / s 0=. b) 0 = 1000π × 0.6 2 h& + 1000 × 0.01 − 0 − 20. ∴ h& = 0.00884 m / s or 8.84 mm / s . c) 0 = 1000π × 0.62 h& + 1000 × 1.0/60 − 1000π × 0.022 × 5 − 10. ∴ h& = 0.000339 m/s or 0.339 mm/s. 4.51. A 1V1 = A 2V 2 where A 2 is an area just under the top surface. dh a) π × 0.02 2 × 10e − t/ 10 = π × ( h tan 60 o ) 2 dt 2 − t / 10 ∴ h dh = 0.001333 e dt . ∴ h 3 = −0.04 e− t/ 10 + 0.04. Finally, h( t ) = 0.342(1 − e− t/ 10 ) 1/ 3 . b) 0.04 × 10 × 10e − t/ 10 = ( h tan 60 o ) × 10 h& ∴ hdh = 0.2309e − t/ 10 dt. Finally,. ∴ h 2 = −4.62 e− t/ 10 + 4.62.. h (t ) = 2.15(1 − e − t /10 )1 / 2 .. 58.

(10) 4.52. du W& = T ω + pAV + µ A belt dy. = 20 × 500 × 2π / 60 + 400 × 0.4 × 0.5 × 10 + 1.81 × 10 −5 × 100 × 0.5 × 0.8 = 1047 + 800 + 0.000724 = 1847 W 4.53. If the temperature is essentially constant, the internal energy of the c.v. does not change and the flux of internal energy into the pipe is the same as that leaving the pipe. Hence, the two integral terms are zero. The losses are equal to the heat transfer exiting the pipe.. 4.54. 80% of the power is used to increase the pressure while 20% increases the internal energy (Q& = 0 because of the insulation). Hence, m& ∆~ u = 0.2W&. 1000 × 0.02 × 4.18 ∆T = 0.2 × 500.. 4.55. W& P V22 − V12 p − p1 = + 2 . γQ 2g γ. (D). ∴W& P = 40 kW. 4.56 4.57. ∴ ∆T = 0.836 oC .. W& P 1200 − 200 = . γ × 0.040 γ. and energy req'd =. 40 = 47.1 kW. 0.85. Q × 9800 × 20 & P = Qγ H P . W 5 × 746 = . 0.87 ηp & W − T = −40 × 0.89. & mg a) W& = 40 × 0.89 × 200 × 9.81 = 69 850 W. ∴ Q = 0.01656 m 3 / s .. T. b) W& T = 40 × 0.89 × ( 90 000 / 60 ) × 9.81 = 523 900 W & T = 40 × 0.89 × (8 × 10 6 / 3600) × 9.81 = 776 100 W c) W 4.58 4.59. W&T 10000000 = ηT ∆z . = 0.89 × 50. ∴V = 1.273 m/s ρ AVg 100 × 3 × 60 × V × 9.8 3 ft V12 p 1 V2 p + + z1 = 2 + 2 + z2 . V1 2g γ 2g γ. −. 2. h2 V2. 2. 12 36 +6 = + h2 . 2 × 32.2 64.4 h22 20.1 8.236 = 2 + h2 . h2. Continuity: 3 × 12 = h2 V2 .. 59.

(11) 4.60. This can be solved by trial-and-error. h2 = 8': 8.24 ? 8.31 h2 = 7.9': 8.24 ? 8.22. Q h2 = 7.93' .. h2 = 1.8': 8.24 ? 8.00. h2 = 1.75': 8.24 ? 8.31. Q h2 = 1.76'.. V12 V 22 + z1 = + z 2 + hL . 2g 2g. 42 16 ∴ +2= + h 2 + 0.2. 2 × 9.81 2 × 9.81h22. ∴ 2.615 = 0.815 / h 22 + h 2 . Trial-and-error provides the following:. 4.61. h2 = 2.5: 2.615 ?= 2.63. h2 = 2.45: 2.615 =? 2.59.. ∴ h2 = 2.47 m. h2 = 0.65: 2.615 =? 2.58. h2 = 0.64: 2.615 =? 2.63.. ∴ h2 = 0.646 m. Manometer: Position the datum at the top of the right mercury level. V2 9810 ×. 4 + 9810 z 2 + p 2 + 2 × 1000 = ( 9810 × 13.6)×.4 + 9810 × 2 + p1 2 p2 V 22 p Divide by γ = 9810: .4 + z 2 + + = 13.6×.4 + 2 + 1 . (1) γ 2g γ Energy:. V12 p 1 V 22 p 2 + + z1 = + + z2 . 2g γ 2g γ V12 = 12.6 ×.4. 2g. Subtract (1) from (2): With z1 = 2 m, 4.62. (2). The manometer equation (see Prob. 4.61) is p V2 p 0.4 + z 2 + 2 + 2 = 13.6×.4 + 2 + 1 . γ 2g γ. ∴V1 = 9.94 m/s. (1). V12 p 1 V2 p V2 + + z 1 = 2 + 2 + z 2 + 0.05 2 . (2) 2g γ 2g γ 2g Subtract (1) from (2): With z1 = 2 m, and with V2 = 4V1 (continuity) 1.8V12 = 12.6 × 0.4. ∴V1 = 7.41 m/s. 2g Energy:. 0=. V22 − V12. +. p 2 − p1. −1202 p . 0= + 2 . ∴ p2 = 7 200000 Pa. 2 × 9.8 9810. 4.63. (A). 4.64. 1 Q = 120 × 0.002228 = π ×   V 1.  12 . 2g. γ. 2. ∴V1 = 12.25 fps.. 60.

(12) 2. 2. Continuity:. 1 1.5 π ×   V1 = π ×   V 2 .  12   12 . Energy:. V12 p 1 V 22 p 2 V12 + = + + 0.37 . 2g γ 2g γ 2g. ∴V2 = 5.44 fps..  12.25 2 5.44 2  ∴ p 2 = 60 × 144 + 62.4 0.63 − = 8702.9 psf or 60.44 psi 64.4 64.4  . 4.65. Q = 600 × 10-3 /60 = π × .022 V1 .. α=. 1. ∴V1 = 7.958 m/s.. ∫ V dA = AV 3 0.02 × w × 6.673 A 1V1 .04 2 × 7.958 V2 = = = 3.537 m/s. A2 .06 2. Energy:. 3. y2  10 1 − wdy  ∫ 2   0.02 0  . 0.02. 1. 3. 3. V12 p1 V 22 p 2 + = + + hL . 2g γ 2g γ. 7 .958 2 − 3.537 2 690 000 − 700 000 ∴ hL = + = 1.571 m 2 × 9.81 9810 4.66. V1 = Q / A1 = Energy:. 4.67. 0.08 = 28.29 m/s. ∴V2 = 9V1 = 254.6 m/s. π ×.03 2 V12 p1 V22 p2 V2 + = + +.2 1 . 2g γ 2g γ 2g  254.6 2 28.29 2  6 ∴ p1 = 9810 − 0.8  = 32.1 × 10 Pa. 2 × 9 . 81 2 × 9 . 81  . π × .07 2V1 = π × .0252 V2 .. a) Across the nozzle:. ∴V2 = 7.84 V1 .. V p V p 7.84 2 − 1 2 + 1 = + 2. ∴ p1 = 9810 V1 . 2g γ 2g γ 2 × 9.81 For the contraction: π ×.07 2 V1 = π ×.052 V3 . ∴V3 = 1.96 V1 . Energy:. Energy:. 2 1. 2 2. V12 p1 V32 p3 + = + . 2g γ 2g γ p1 p = 12.6×.15 + 3 . γ γ 2 2 2 V V1 V + 12.6 ×.15 = 3 = 1.96 2 1 . 2g 2g 2g ∴V1 = 3.612 m/s. ∴p1 = 394 400 Pa.. Manometer: γ ×.15 + p1 = 13.6γ ×.15 + p3 . Subtract the above 2 eqns: ∴ (1.96 2 − 1) V12 = 12.6 ×.15 × 2g.. 61. ∴.

(13) From the reservoir surface to section 1: V02 p 0 V12 p 1 + + z0 = + + z1 2g γ 2g γ 3.612 2 394 400 H= + = 40.0 m. 19.62 9810 p p b) Manometer: γ ×.2 + p1 = 13.6γ ×.2 + p3 . ∴ 1 = 12.6 ×.2 + 3 . γ γ 2 2 V p V1 p Energy: + 1 = 3 + 3. Also, V3 = 1.96 V1 . 2g γ 2g γ V12 1.96 2 V12 ∴ + 12.6 ×.2 = . ∴V1 = 4.171 m/s. 2g 2g The nozzle is the same as in part (a): ∴p1 = 534 700 Pa. From the reservoir surface to the nozzle exit: V02 p 0 V2 p V2 32.7 2 + + z0 = 2 + 2 + z2 . ∴H = 2 = = 54.5 m. 2g γ 2g γ 2 g 2 × 9.81 4.68 a) Energy:. V 02 p 0 V2 p + + z 0 = 2 + 2 + z 2 . ∴ V 2 = 2 gz 0 = 2 × 9.81 × 2.4 = 6.862 m/s. 2g γ 2g γ. Q = AV = .8 × 1 × 6.862 = 5.49 m 3 / s . For the second geometry the pressure on the surface is zero but it increases with depth. The elevation of the surface is 0.8 m. V2 ∴ z 0 = 2 + h. ∴ V2 = 2g( z 0 − h) = 2 × 9.81 × 2 = 6.264 m/s. 2g ∴Q = .8 × 6.264 = 5.01 m 3 / s . Note: z0 is measured from the channel bottom in the 2nd geometry. ∴z0 = H + h.. V02 p 0 V22 p2  b) + + z0 = + + z2 . ∴ V 2 = 2 gz 0 = 2 × 32.2 ×  6 +  2g γ 2g γ ∴Q = AV = (2 × 1) × 21.23 = 42.5 cfs. For the second geometry, the bottom is used as the datum: V2 V2 ∴ z 0 = 2 + 0 + h. ∴ 2 = ( H + h) − h. 2g 2g ∴ V2 = 2gH = 2 × 32.2 × 6 = 19.66 fps. ∴Q = 39.3 cfs.. 62. 2  = 21.23 fps. 2.

(14) 4.69. 4.70. From the reservoir surface to the exit: Continuity: 2 2 2 V0 p0 V2 p2 V1 .03 2 + + z0 = + + z2 + K . V1 = V2 = .1406 V2 . .08 2 2g γ 2g γ 2g V2 .1406 2 V22 10 = 2 + 5 × 2g 2g ∴V2 = 13.36 m/s. ∴Q = 13.36 × π × .0152 = 0.00944 m 3 / s. The velocity in the pipe is V1 = 1.878 m/s. p 1.8782 1.8782 Energy 0 → A: 10 = + A +.8 + 3. ∴pA = 65 500 Pa. 2 × 9.81 9810 2 × 9.81 p 1.8782 1.8782 Energy 0 → B: 10 = + B + 2.0 + 10. ∴pB = −5290 Pa. 2 × 9.81 9810 2 × 9.81 p 1.878 2 1878 . 2 Energy 0 → C: 10 = + C + 12 + 2.8 . ∴pC = −26 300 Pa. 2 × 9.81 9810 2 × 9.81 p 1.8782 1.878 2 Energy 0 → D: 10 = + D +0+ 5 . ∴pD = 87 500 Pa. 2 × 9.81 9810 2 × 9.81 V 02 p 0 V2 p + + z 0 = 2 + 2 + z 2. 2g γ 2g γ. V 22 80 000 +4= . 9810 2 × 9.81. ∴V2 = 19.04 m/s.. a) Q = A 2 V2 = π ×.025 2 × 19.04 = 0.0374 m 3 / s. b) Q = A 2 V2 = π ×.09 2 × 19.04 = 0.485 m 3 / s . c) Q = A 2 V2 = π ×.05 2 × 19.04 = 0.1495 m 3 / s. 4.71. a). p0 V2 V2 + z 0 = 2 + 1.54 1 . γ 2g 2g. 16V12 V2 80 000 +4= + 1.54 1 . 9810 2g 2g. ∴V1 = 3.687 m/s.. Q = A 1V1 = π ×.05 2 × 3.687 = 0.0290 m 3 / s.. .09 2 V = 3.24V 2 . .05 2 2 V2 3.24 2 V 22 80 000 + 4 = 2 + 2.3 . ∴V2 = 3.08 m/s. ∴ Q = A 2V 2 = 0.0784 m 3 / s. 9810 2g 2g. b) A 1V 1 = A 2 V 2 .. c). 4.72. V1 =. V2 V2 80 000 + 4 = 2 + 1.5 2 . 9810 2g 2g. (C). ∴V2 = 9.77 m/s.. Manometer: γ H + p1 = ρ g Energy: K. V22 + p2 2g. 7.962 100000 = . 2 × 9.81 9810. or 9810 × 0.02 + p1 = ρ g. ∴ K = 3.15.. 63. ∴ Q = A 2V 2 = 0.0767 m 3 / s . V22 . 2g.

(15) Combine the equations: 9810 × 0.02 = 1.2 ×. 4.73. Manometer: γH + γz + p1 = 13.6γH + γz + p 2 . Energy:. p1 p = 12.6 H + 2 . γ γ. V 22 − V 12 12.6 H = . 2g. V2 =. d12 d 22.  d14  ∴ V = 12.6 H × 2 g /  4 − 1 .  d2  2 1. V1 .. d 2 π  12.6 H × 2 g  ∴ Q = V1π 1 =  4  4 4  d1 / d 24 − 1 . 4.74. ∴. ∴V1 = 18.1 m/s.. p1 V 12 p 2 V22 + = + . γ 2g γ 2g. Combine energy and manometer: Continuity:. V12 . 2. 1/ 2.  H  d = 12.35 d d  4   d1 − d24  2 1. 2 1. 1/ 2. 2 2. Use the result of Problem 4.73: 1/ 2 .2   a) Q = 12.35 × .16 2 ×.08 2  4  = 0.0365 m 3 / s .  .16 −.084  1/2 .4   b) Q = 12.35 × .24 2 ×.082  4  = 0.0503 m 3 / s .  .24 −.08 4 .  H  Q = 22.37 d d  4 4  d1 − d 2 . c) Using English units with g = 32.2: 2. 2. 1 1 10 / 12  Q = 22.37 ×      4   2   4   .5 −.25 4  2.  1  15 / 12  d) Q = 22.37 ×1 ×    4   3  1 −.3333 4 . 2 1. 4.75. 4.76. (B). 1/ 2. .. 1/ 2. = 1.318 cfs. 1/ 2. = 2.796 cfs.. 2. V 2 ∆p hL = K = . 2g γ. 2 2. Q 0.040 = = 7.96 m/s. A π × 0.042 7.962 100000 K = . ∴ K = 3.15. 2 × 9.81 9810 V=. V 22 = H. 2g. a) Energy from surface to outlet: Energy from constriction to outlet:. 64. ∴ V 22 = 2 gH .. p1 V 12 p 2 V22 + = + . γ 2g γ 2g.

(16) Continuity: V1 = 4V 2 . With p1 = pv = 2450 Pa and p2 = 100 000 Pa, 2450 16 100 000 1 + × 2 gH = + × 2 gH . ∴H = 0.663 m. 9810 2 × 9.81 9810 2 × 9.81 b) With p1 = 0.34 psia, p2 = 14.7 psia, .34 × 144 16 14.7 × 144 1 + 2 gH = + 2 gH . 62.4 2g 62.4 2g 4.77. Continuity:. ∴H = 2.21 ft.. V 22 V1 = 4V 2 . Energy  surface to exit: = H. 2g. Energy constriction to exit:. p v V 12 p 2 V 22 + = + . γ 2g γ 2g. V 22 − 16V 22 ∴ pv = p2 + γ = p 2 − 15 H γ = 100 000 − 15×.65 × 9810 = 4350 Pa. 2g From Table B.1, T = 30°C. 4.78. Energy  surface to surface:. z 0 = z 2 + hL .. ∴ 30 = 20 + 2. V 22 . 2g. ∴V 12 = 160 g. ∴V 22 = 10 g. 160 g (−94 000) Energy  surface to constriction: 30 = + + z1 2g 9810. Continuity: V1 = 4V2 .. ∴z1 = −40.4 m. 4.79. Continuity: Energy:. ∴H = 40.4 + 20 = 60.4 m.. 10 2 V = 2.778 V1 . 62 1 V12 p 1 V 22 p 2 + = + . 2g γ 2g γ. V2 =. ∴V 1 = 7.67 m/s. 4.80. V12 200 000 2.778 2 V12 2450 + = + . 2g 9810 2g 9810. ∴Q = π × .052 × 7.67 = 0.0602 m 3 / s.. Velocity at exit = Ve . Velocity in constriction = V1 . Velocity in pipe = V2 . Energy — surface to exit:. V e2 = H. 2g. ∴ V e2 = 2 gH .. D2 V . Also, V1 = 4V 2 . d2 e V12 pv Energy — surface to constriction: H = + . 2g γ Continuity across nozzle: V2 =. 65.

(17) a) 5 =. 1  D4  −97 550 .  16 × 4 × 2 g × 5 + 2g  .2 9810 . b) 15 =. 4.81. ∴ D = 0.131 m. 1  D4  (.34 − 14.7 )144 × 2 g × 15 + .  16 4 2 g  ( 8 / 12) 62.4 . V22 V 22 Energy — surface to exit: 3 = +4 . 2g 2g. ∴ D = 0. 446 ′ or 5.35 ′′. ∴ V 22 = 11.77.. 11.77 1176 − 100 000 11.77 2 Energy — surface to “A”: 3 = + + ( H + 3) + 1.5 . 2 × 9.81 9810 2 × 9.81 ∴ H = 8.57 m . 2. 4.82. 1 m& = ρA V = 1.94 × π ×   × 120 = 5.079 slug / sec.  12  2 2 & = 5.079 × 32.2  30 − 120 + 120 × 144  / 0.85 = 12 ,950 ft - lb or 23.5 Hp . W P  2 × 32.2 62. 4  sec . 4.83. m& = ρA V = 1000 × π ×.02 2 × 40 = 50.27 kg / s.. ∆p   10 2 − 40 2 20 000 = 50.27 × 9.81 + / 0.82. 9810   2 × 9.81. 4.84. (C). ∴ ∆p = 1.088 × 10 6 Pa .. W& P V22 − V12 ∆p = + . γQ 2g γ. W&P 16 W& P = Q∆ p = 0.040 × 400 = 16 kW. = = 18.0 kW. η 0.89 4.85. 4.86.  0 − 10.2 2 −600 000  & −W T = 2 × 1000 × 9.81 + × 0.87. 9810   2 × 9.81 2 We used V2 = Q / A 2 = = 10.2 m / s. π ×.25 2. ∴ W& T = 1.304 × 10 6 W .. 450 450 = 15.9 fps. V2 = = 10.19 fps . 2 π ×3 π × 3.75 2  1019 . 2 − 15.9 2 (18 − 140)144   1  − 10,000  × 550 = 450 × 1.94 × 32.2  + ηT .  .746 62.4  2 × 32.2  ∴ ηT = 0.924. V1 =. 66.

(18) 4.87. V 2 − V12 p2 p1 c  &  2 a) Q& − W& S = mg + − + z2 − z1 + v ( T2 − T1 ) . γ2 γ1 g  2g  The above is Eq. 4.5.17 with Eq. 4.5.18 and Eq. 1.7.13. p g 85 × 9.81 600 × 9.81 20 500 γ1 = 1 = = 9.92 N / m 3 . γ 2 = = . RT1 .287 × 293 .287 T2 T2  200 2 600 000T2 85 000 716.5  ∴ −( −1 500 000) = 5 × 9.81 + − + (T2 − 293) . 20 500 9.92 9.81  2 × 9.81  o ∴ T 2 = 572 K or 299 C .. Be careful of units! p 2 = 600 000 Pa, cv = 716.5 b) −60 000 + 1 500 000 = same as above. 4.88. γ1 =. J K ⋅ kg. ∴ T2 = 560 K. or. 287 o C.. p 1 g 14.7 × 144 × 32.2 lb 60 × 144 × 32.2 lb = = 0.0764 3 . γ 2 = = 0.213 3 . RT1 1716 × 520 ft 1716 × 760 ft. ft - lb c v = 4296 . slug - oR. 2. 1 & = ρAVg = γAV =.213 × π ×   × 600 =.697 lb /sec. mg  24 . Use Eq. 4.5.17 with Eqs. 4.5.18 and 1.7.13:  V 2 − V 12 p 2 p1 c v  &  2 Q& + W& c = mg + − + (T2 − T1 ) + z 2 − z 1 . γ2 γ1 g  2g   600 2 60 × 144 14.7 × 144 4296  −10 × 778 ×.697 + W& c =.697  + − + ( 300 − 60) .213 .0764 32.2  2 × 32.2  ft - lb ∴ W& c = 40 600 or 73.8 Hp. sec. 4.89. 4.90. V22 V22  & η = mg & Energy — surface to exit: −W − 20 + 4 . 5  . T T 2 g 2 g   15 V2 = mg & = Qγ = 15 × 9810 = 147 150 N / s. 2 = 13.26 m / s. π ×.6  13.26 2 13.26 2  & − WT × 0.8 = 147 150 − 20 + 4.5 ∴ W& T = 5390 kW. . 2 × 9 . 81 2 × 9 . 81   (D). 4.582 pB 7.16 2 36.0 + 15 = + + 3.2 . 2 × 9.81 9810 2 × 9.81 In the above energy equation we used. hL = K. ∴ pB = 416 000 Pa. V2 Q 0.2 with V = = = 4.42 m/s. 2g A π × 0.2 2. 67.

(19) 4.91. Energy — surface to “C”:  10 2 200 000 10 2  W& P ×.8 + mg & × 10 =  + + 7.7  770.5. 9810 2 × 9.81  2 × 9.81 & = ρA V g = 1000 × π ×.05 2 × 10 × 9.81 = 770.5 N / s.) ∴ W& P = 52 700 W . (mg p 10 2 10 2 + A + 1.5 . ∴ p A = 169 300 Pa . 2 × 9.81 9810 2 × 9.81  V 2 − V O2 p B − pO V2  &  B Energy — surface to “B”: W& PηP = mg + + z B − zO + K B  γ 2 g   2 g Energy — surface to “A”: 30 =.  102 p 10 2  52 700×.8 = 770.5 + B − 30 + 15 . . 2 × 9.81   2 × 9.81 9810 4.92. V2 20 20 + γz 1 + p1 = 13.6γ × + γz 2 + p 2 + 2 ρ. 12 12 2 2 p p V 20 20 ∴ + z 1 + 1 = 13.6 × + z2 + 2 + 2 . 12 γ 12 γ 2g. Manometer: γ ×. V12 p p V2 + z1 + 1 = H T + z 2 + 2 + 2 . 2g γ γ 2g. Energy:. ∴. 20 20 V 12 = 13.6 × + − HT. 12 12 2 g. ∴ H T = 12.6 ×. 20 51.6 2 + = 62.3' . 12 2 × 32.2. V1 =. 18 = 51.6 fps. 2 1  π ×   3. W& T = γQ ηT H T = 62.4 × 18×.9 × 62.3 = 62,980. 4.93. ∴ p B = 706 100 Pa.. Energy—across the nozzle:. ∴. p1 V12 p2 V22 + = + . γ 2g γ 2g. V12 6.252 V12 400 000 + = . 9810 2 × 9.81 2 × 9.81. ft - lb sec. V2 =. or. 52 22. 115 Hp .. V1 = 6.25V1.. ∴ V1 = 4.58 m/s , VA = 7.16 m/s , V2 = 28.6 m/s.. Energy—surface to exit: 28.62 4.58 2 7.16 2 H P + 15 = + 1.5 + 3.2 . ∴ H P = 36.8 m. 2 × 9.81 2 × 9.81 2 × 9.81 ∴W& P = γ QH P / ηP = 9810 × (π × .012 )× 28.6 × 36.8/.85 = 3820 W.. 68.

(20) Energy —surface to “A”: 7.16 2 p 7.162 15 = + A + 3.2 . ∴ p A = 39 400 Pa 2 × 9.81 9810 2 × 9.81 Energy —surface to “B”: 4.582 p 7.16 2 36.0 + 15 = + B + 3.2 . ∴ pB = 416 000 Pa 2 × 9.81 9810 2 × 9.81 4.94. (A). V=. Q 0.1 = = 19.89 m / s. A π ×.04 2. V 22 p 2 V 22 HP = + + z2 + K . 2g γ 2g. Energy —surface to entrance:. 19.89 2 180 000 19.89 2 + + 50 + 5.6 = 201. 4 m . 2 × 9.81 9810 2 × 9.81 ∴ W& P = γQH P / η P = 9810 × 0.1 × 201.4 / 0.75 = 263 000 W . ∴ HP =. 4.95. Energy —surface to exit: ∴ V 2 = 7.83 m / s.. 4.96. V 22 p 2 V 22 10 = + + z 2 + 2.2 . 2g γ 2g. Q = 0.02 = 7.83 × πd 22 / 4.. ∴ d 2 = 0.0570 m .. Depth on raised section = y 2 . Continuity: 3 × 3 = V2 y 2 . V22 32 +3 = + (0.4 + y 2 ). 2g 2g. Energy (see Eq. 4.5.21): 92 ∴ 3.059 = + y2, 2 g y 22 Trial-and-error:. or. y2 = 2.0: y2 = 1.8: y 2 = 2.1: y2 = 2.3:. y 32 − 3.059 y 22 + 4.128 = 0. − .11 ? 0. ∴ y2 = 1.85 m.  + .05 ? 0. − .1 ? 0.  ∴ y 2 = 2.22 m. + .1 ? 0.. The depth that actually occurs depends on the downstream conditions. We cannot select a “correct” answer between the two. . m 3. 4.97. Mass flux occurs as shown. The velocity of all fluid elements leaving the top and bottom is approximately 32 m/s. The distance where u = 32 m /s is y = ± 2 m.. 69. . m 2. . m 1. . m 3.

(21) To find m& 3 use continuity: 2. m& 1 = m& 2 + 2 m& 3 .. ρ 4 × 10 × 32 = ρ 2 ∫ ( 28 + y 2 )10dy + 2 m& 3 . 0. 8  ∴ m& 3 = 640 ρ − 10 ρ 28 × 2 +  = 53.3ρ. 3 . Rate of K.E. loss = m& 1. 2 V12 V2 u3 − 2m& 3 1 − ρ 2 ∫ 10 dy 2 2 2 0. 2 32 2 2 = 1280 ρ − 53.3 ρ 32 − 10ρ ∫ ( 28 + y 2 ) 3 dy 2 0. = ρ[655360 − 54579 − 507320] = 115000 W . 4.98. The average velocity at section 2 is also 8 m/s. The kinetic-energycorrection factor for a parabola is 2 (see Example 4.9). The energy equation is: V12 p 1 V2 p + = α 2 2 + 2 + hL . 2g γ 2g γ 82 150 000 82 110 000 + =2 + + hL . 2 × 9.81 9810 2 × 9.81 9810. ∴ hL = 0.815 m . 4.99. 1 12 1 23  2 VdA = ( 28 + y ) dy = 28 × 2 + = 29.33 m / s A∫ 2 ∫0 2  3  2 1 1 3 α= V dA = ( 28 + y 2 ) 3 dy 3 ∫ 3 ∫ AV 2 × 29.33 0 1 = 28 3 × 2 + 3 × 28 2 × 2 3 / 3 + 3 × 28 × 2 5 / 5 + 2 7 / 7 = 1.005 3 2 × 29.33 V =. [. 4.100 a) V =. ]. 0.01  1 1 r2 10 1 −  ∫ VdA = ∫  0.012 A π × 0.012 0 . 1 α= AV. 1 ∫ V dA = π × 0.012 × 5 3 3. 3. =. 0 . 01. ∫ 0.  20  0.012 0.014  −  2π rdr =   = 5 m/s 0.012  2 4 × 0.012   3.  r2  10  1 −  2πrdr  0.01 2  3. 2000  0.01 2 3 × 0.01 4 3 × 0.016 0.018  − + −   = 2.00 0.012 × 5 3  2 4 × 0.012 6 × 0.014 8 × 0.016 . 70.

(22) b) V =. 1 1 0.02  y2 VdA = 10 1 −  ∫ ∫ 2 A 0.02w 0  0.02.  10  0.023 wdy = 0.02 −   0.02  3 × 0.022  3. y2  α= V dA = 10 1 − wdy  ∫ ∫ 2   AV 3 0.02 × w × 6.673 0 0.02   3 1000  3 × 0.02 3 × 0.02 5 0.02 7  = 0 . 02 − + −   = 1.541 0.02 × 6.67 3  3 × 0.02 2 5 × 0.02 4 7 × 0.02 6  1. 3. 0.02. 1. 1 1 R r  4.101 V = ∫ VdA = u 1−  2 ∫ max   A πR 0 R. 1/ n. 3.   = 6.67 m/s . n   n 2πrdr = −2u max  −   2n + 1 n + 1. V 2  ρR 3  r K. E. = ρ ∫ V  dA = ∫ u max  1 −    2  20 R. 3 /n. n    n 2πrdr = ρπu 3max − R 2  −   3 + 2n 3 + n   . 5 a) V = −2u m a x  −  11. 5  = 0.758 u m a x 6 5 5 2 3 2 3 K. E. = ρπR u max  −  = 0.24 ρπR u max  8 13  0.24 ρπR 2u 3m a x K . E. α= = = 1.102 1 1 3 2 3 3 ρAV ρπR × 0.758 u m a x 2 2. 7 7 b) V = −2u max  −  = 0.817 u max  15 8  7 3 2 7 2 3 K. E. = ρπu max R  −  = 0.288 ρπR u max  10 17  3 0.288 ρπR 2 u max K. E. α= = = 1.056 V 2  0.817 2 u 2max  2 ρAV   ρπR × 0.817 u m a x    2  2   9 9 c) V = −2u m a x  −  = 0.853 u max  19 10  9  9 2 3 2 3 K. E. = ρπR u max  −  = 0.321 ρπR u m a x  12 21 0.321ρπR 2 u m3 a x K . E. α= = = 1.034 1 1 3 2 3 3 ρAV ρπR × 0.853 u m a x 2 2. 71.

(23)  V 2 − V 12 ~ ~  4.102 Engine power = FD × V∞ + m&  2 + u2 − u1  2    V22 − V12  & & m f g f = FD V ∞ + m  + cv (T2 − T1 ) 2  . W& η = FD × V. 4.103. 10− 3  m 3   kJ  100  km  1340 100 000  kg  . = ×   × 930 3  × q f   ×   × 015 m  5  km   kg  3600  s  1000 3600 ∴ q f = 48 030 kJ / kg. 4.104 0 = α 2. V 22 p 2 V2 p νLV + + z 2 − 1 − 1 − z 1 + 32 2g γ 2g γ gD 2. V2 10 −6 × 180V − 0.35 + 32 × . 2 × 9.81 9.81 × 0.02 2 V 2 + 14.4V − 3.434 = 0. ∴ V = 0.235 m / s 0= 2. 4.105 Energy from surface to surface:. and. Q = 7.37 × 10 −5 m 3 / s. V22 p 2 V12 p 1 V2 HP = + + z2 − − − z1 + K . 2g γ 2g γ 2g. Q2 a) H P = 40 + 5 = 40 + 50.7 Q 2 2 π × 0.04 × 2 × 9.81 Try Q = 0.25: H P = 43.2 (energy). H P = 58 (curve) Try Q = 0.30: H P = 44.6 (energy). H P = 48 (curve) Solution: Q = 0.32 m 3 / s . 20 Q 2 = 40 + 203 Q 2 π × 0.04 2 × 2 × 9.81 Try Q = 0.25: H P = 52.7 (energy). H P = 58 (curve). b) H P = 40 +. Solution: Q = 0.27 m 3 / s Note: The curve does not allow for significant accuracy.. 4.106 Continuity:. A 1 V1 = A 2 V2 + A 3 V3. π × 0.06 2 × 5 = π × 0.02 2 × 20 + π × 0.03 2 V 3 . ∴ V 3 = 1111 . m/ s Energy: energy in + pump energy = energy out V 2 p  V 2 p  V 2 p  m& 1  1 + 1  + W& P × η P = m& 2  2 + 2  + m& 3  3 + 3  ρ ρ ρ  2  2  2. 72.

(24)  5 2 120 000   20 2 300 000  2 1000π × 0.06 2 × 5 + +  + 0.85W& P = 1000π × 0.02 × 20  1000  1000   2  2  11.112 500 000  +1000π × 0.03 2 × 11.11 +  1000   2. ∴ W& P = 26 700 W 4.107 (A). After the pressure is found, that pressure is multiplied by the area of the window. The pressure is relatively constant over the area.. V12 p 1 V 22 p 2 4.108 + = + . 2g γ 2g γ. V2 =. V12 16 V 12 200 000 a) + = . 2 × 9.81 9810 2 × 9.81 p1 A 1 − F = m& (V 2 − V 1 ).. d2 ( d /2) 2. V1 = 4 V1 .. ∴ V 1 = 5.164 m / s.. 200 000π ×.03 2 − F = 1000π ×.03 2 × 5.164( 4 × 5.164 − 5.164 ). ∴ F = 339 N . b). c). d). e). f). V12 16 V 12 400 000 + = . ∴ V 1 = 7.303 m / s. 2 × 9.81 9810 2 × 9.81 400 000π ×.03 2 − F = 1000π ×.03 2 × 7.303( 4 × 7.303 − 7.303). ∴ F = 679 N .. V12 200 000 16 V12 + = . ∴V1 = 5.164 m/s. 2 × 9.81 9810 2 × 9.81 200 000π × .06 2 − F = 1000π × .062 × 5.164(4 × 5.164 − 5.164). ∴ F = 1356 N. V12 30 × 144 16 V12 + = . ∴ V1 = 17.24 fps. 2 × 32.2 62.4 2 × 32.2 30 × π × 1.52 − F = 1.94 × π × (1.5/12) 2 × 17.242 (4 − 1).. ∴ F = 127 lb.. V12 60 × 144 16 V12 + = . ∴ V1 = 24.38 fps. 2 × 32.2 62.4 2 × 32.2 60 × π ×1.52 − F = 1.94 × π × (1.5/12) 2 × 24.382 (4 − 1).. ∴ F = 254 lb.. 30 × 144 16 = . ∴ V1 = 17.24 fps. 2 × 32.2 62.4 2 × 32.2 30 × π × 32 − F = 1.94 × π × (3/12) 2 × 17.242 (4 − 1). V12. +. V12. V12 p 1 V 22 p 2 4.109 + = + . 2g γ 2g γ. 92 V2 = 2 V1 = 9V1 . 3. V12 81 V12 2 000 000 + = . 2 × 9.81 9810 2 × 9.81. ∴V12 = 50.. 73. ∴ F = 509 lb..

(25) p1 A 1 − F = m& ( V2 − V1 ) = m& 8V 1 2 000 000π ×.045 2 − F = 1000π ×.045 2 × 8 × 50 ∴ F = 10 180 N .. 4.110. V12 p1 V22 p2 + = + . 2g γ 2g γ ΣFx = m& (V 2 x − V 1x ).. V0π × .012 = Ve × .006 × .15.. 10 2 V = 1.562 V 1. 82 1 ∴ p 1A 1 − F = m& (V 2 − V 1 ). a) V2 =. ∴Ve = 11.1 m/s.. V12 400 000 2.441 V12 + = . ∴ V 1 = 23.56 m / s . 2 × 9.81 9810 2 × 9.81. 400 000π ×.05 2 − F = 1000π ×.05 2 × 23.56(.562 × 23.56). ∴ F = 692 N . V12 400 000 7.716 V12 10 2 b) V2 = 2 V 1 = 2.778 V 1. + = . ∴ V 1 = 10.91 m / s. 6 2g 9810 2g 400 000π ×.05 2 − F = 1000π ×.05 2 × 10.91(1.778 × 10.91). ∴ F = 1479 N . V12 400 000 39.06 V12 10 2 c) V2 = 2 V 1 = 6.25 V 1 . + = . ∴ V 1 = 4.585 m / s. 4 2g 9810 2g 400 000π ×.05 2 − F = 1000π ×.05 2 × 4.585( 5.25 × 4.585). ∴ F = 2275 N . V12 400 000 625 V12 10 2 d) V2 = 2 V 1 = 25 V1 . + = . ∴ V 1 = 1132 . m / s. 2 2g 9810 2g 400 000π ×.05 2 − F = 1000π ×.05 2 × 1.132( 24 × 1.132). 4.111 (C). V12 p1 V22 p + = + 2. 2g γ 2g γ. p1 = 9810 ×. ∴ F = 2900 N .. (6.252 − 1) × 12.732 = 3085000 Pa. 2 × 9.81. p1 A1 − F = ρQ(V2 − V1 ). 3085000 × π × 0.052 − F = 1000 × 0.1 × 12.73(6.25 − 1) ∴ F = 17500 N. 4.112 V2 = 4V1 = 120 fps.. V22 − V12   120 2 − 30 2  p1 = γ   = 62.4   = 13,080 psf.  2 × 32.2   2g  2. 2.  1.5   1.5  F = p1 A1 − m& (V2 x − V1x ) = 13,080 π   − 1.94 × π   × 30( −120 −30) = 1072 lb.  12   12  V 12 p1 V22 p 2 15 V12 p1 + = + . ∴ = . 2g γ 2g γ 2g γ 2 × 9.81 a) V12 = × 200 000 = 26.67. ∴ V 1 = 5.16 m / s, V2 = 20.7 m / s. 15 × 9810 p1 A 1 − Fx = m& (V 2x − V1x ). ∴ Fx = 200 000π ×.04 2 + 1000π ×.04 2 × 5.16 2 = 1139 N.. 4.113 V2 = 4 V1 .. 74.

(26) ∴ Fy = 1000π ×.04 2 × 5.16( 20.7 ) = 537 N .. Fy = m& (V 2 y − V1y ).. 2 × 9.81 × 400 000 = 53.33. ∴ V 1 = 7.30 m / s, V 2 = 29.2 m / s. 15 × 9810 p1 A 1 − Fx = m& (V 2x − V1x ). ∴ Fx = 400 000π ×.04 2 + 1000π ×.04 2 × 7.3 2 = 2280 N .. b) V12 =. Fy = m& (V 2 y − V1y ) = 1000π ×.04 2 × 7.3 × ( 29.2) = 1071 N . c) V12 =. 2 × 9.81 × 800 000 = 106.7. 15 × 9810. ∴ V 1 = 10.33 m / s, V 2 = 41.3 m / s.. Fx = p1 A1 + ρ A1V12 = 800 000π × .042 + 1000π × .042 × 10.332 = 4560 N. Fy = m& (V 2 y ) = 1000π ×.04 2 × 10.33( 41.3) = 2140 N . V12 p 1 V22 p 2 + = + 2g γ 2g γ. 40 2 4.114 V2 = 2 V 1 = 80 m / s. 10. F. p1A 1. V2.  80 5  ∴ p 1 = 9810  − = 3.19 × 10 6 Pa.   2 × 9.81 2 × 9.81  p1 A 1 − F = m& (V2 x − V1x ). ∴ F = 3.19 × 10 6 π ×.2 2 − 1000π ×.2 2 × 5(80 − 5) = 353 000 N. 2. 4.115 A 1V1 = A 2V 2 .. 2. π ×.025 2 × 4 = π (.025 2 −.02 2 )V 2 .. ∴ V 2 = 11.11 m / s.. p1 V 12 p 2 V22 + = + . γ 2g γ 2g.  11.112 − 42 p1 = 9810   2 × 9.81  p1 A1 − F = m& (V2 − V1 ).. p 1A 1. F.   = 53700 Pa. . ∴ F = 53 700π × .0252 − 1000π × .0252 × 4(11.11 − 4) = 49.6 N.. 4.116 Continuity: Energy:. .7 V 1 =.1 V2 .. ∴ V 2 = 7 V 1.. V12 p1 V2 p + + z1 = 2 + 2 + z 2 2g γ 2g γ. F1 Rx. V12 49V12 +.7 = +.1. ∴ V1 = 0.495 , V 2 = 3.467 m / s. 2 × 9.81 2 × 9.81 Momentum: F1 − F2 − R x = m& ( V2 − V1 ) 9810 ×.35(.7 × 1.5) − 9810×.05( 0.1 × 1.5) − R x = 1000 × (.1 × 1.5) × 3.467( 3. 467 −.495) ∴ R x = 1986 N. ∴ R x acts to the left on the water, and to the right on the obstruction.. 75. F2.

(27) 4.117 Continuity: 6 V1 =.2 V 2 . ∴ V 2 = 30 V1 . Energy (along bottom streamline): V12 p1 V22 p 2 + + z1 = + + z2 2g γ 2g γ. F F1. V22 /900 V22 +6= + 0.2. 2 × 9.81 2 × 9.81 ∴V2 = 10.67, V1 = .36 m/s. Momentum: F1 − F2 − F = m& (V 2 − V 1 ) 9810 × 3( 6 × 4) − 9810 ×.1(.2 × 4) − F = 1000 × (.2 × 4 ) × 10.67 (10.67 −.36) ∴ F = 618 000 N . (F acts to the right on the gate.) 4.118 a) 8 ×.6 = V 2 y 2 .. F1 − F2 = m& (V 2 − V 1 ).. y2  8 ×.6  y 2 w = ρ.6w × 8 − 8 . 2  y2  .6 − y 2 γ 4.8 × 8 × 2 (.36 − y 22 ) = 4.8 ρ × 8 . ∴ (.6 + y 2 ) y 2 = . 2 y2 9.81. γ ×.3×.6 w − γ. y 22 +.6 y 2 − 7 .829 = 0.. ∴ y 2 = 2.51 m .. (See Example 4.12.). b) y 2 =.  1 1 8 8 2 2 2 2 ×.4 × 12  = 3.23 m .  − y 1 + y 1 + y 1V 1  = −.4 + .4 + 2  g 9.81  2  . c) y 2 =.  1 1 8 8 2 2 2 2  × 2 × 20  = 6.12 ft . − y 1 + y 1 + y 1V1  = −2 + 2 + 2  g 32.2  2  . d) y 2 =.  1 1 8 8 2 2 2 2  × 3 × 30  = 11.54 ft . − y 1 + y 1 + y 1V1  =  −3 + 3 + 2  g 32.2  2  . 4.119 Continuity:. V2 y 2 = V1 y 1 = 4V 2 y 1 .. Use the result of Example 4.12: a) y 2 = 4×.8 = 3.2 m .. 3.2 =. ∴ y 2 = 4y 1 .   1 8 y 2 = − y 1 +  y 12 + y 1V 12  2 g   . 1 8    −.8 +  .82 + × .8 × V12  2  9.81  . 1/2 . . . b) y 2 = 4 × 2 = 8 ft. 8=. 1 8  2 2 × 2 × V1   −2 +  2 +   2 32.2. 1/ 2.  . . 76. 1/ 2.   . ∴ V 1 = 8.86 m / s.. ∴ V 1 = 25.4 fps.. F2.

(28) V12 12 +3= + y1 . 2 × 9.81 2 × 9.81. 9 4.120 V = = 1 m / s. 3×3. V1 y 1 = 1 × 3..  3.05 ? 2.93  = V 3  V1 = 7.19 m / s. ∴ 3.05 = + . Trial-and-error:  19.62 V 1 y 1 =. 417 m. V1 = 7.2: 3.05 =? 3.06  1/2 1 8    y2 =  −.417 +  .4172 + × .417 × 7.19 2   = 1.90 m. 2  9.81    V2 × 1.9 = 7 .19×.417. V2 = 1.58 m / s. V 1 = 7:. 2 1. 4.121 Refer to Example 4.12: γ. y1  60  y 1 w − γ × 3 × 6w = ρ × 6 w × 10 10 −  . (V1 y 1 = 6 ⋅ 10). 2 y1  .  y − 6 γ 1200 ∴ ( y 12 − 36) = 600 ρ 1  . ∴ ( y 1 + 6) y 1 = = 37.27. ∴ y 1 = 3.8 ft , V1 = 15.8 fps. 2 32.2  y1  4.122 Continuity: Momentum:. 20 × π ×.015 2 = V2 π ×.03 2 .. p2A2. p1A1. ∴ V 2 = 5 m / s. p1 A 1 − p 2 A 2 = m& (V 2 − V 1 ).. 60 000π ×.03 2 − p 2π ×.03 2 = 1000π ×.015 2 × 20( 5 − 20). ∴ p 2 = 135 kPa.. 4.123 V1 A1 = 2V2 A2 .. V2 =15. p1 V12 p 2 V 22 + = + . γ 2g γ 2g. ΣFx = m& (V2 x − V1x ) .. π × .052 2π × .0252. = 30 m/s.. 30 2 − 15 2 ∴ p 1 = 9810 = 337 500 Pa. 2 × 9.81. p1 A1 − F = m& ( −V1 ).. ∴ F = p1 A 1 + m& V1 = 337 500π ×.05 2 + 1000π ×.05 2 × 15 2 = 4420 N .. 4.124 m& 1 = 1000π ×.03 2 × 12 = 33.93 kg / s. m& 3 = 1000π ×.02 2 × 8 = 10.05 kg / s. ∴ m& 2 = m& 1 − m& 3 = 23.88 = 1000π ×.03 2 V2 . ∴ V 2 = 8.446 m / s.. R x Ry p1A1. p2A2 p3A 3. Energy from 1 → 2:. V12 p 1 V 22 p2 + = + . 2g γ 2g γ. 77. 12 2 − 8.446 2 × 9810 2 × 9.81 = 536 300 Pa.. ∴ p2 = 500 000 +.

(29) V12 p 1 V 32 p 3 Energy from 1 → 3: + = + . 2g γ 2g γ 12 2 − 8 2 9810 = 540 000 Pa. 2 × 9.81 p1 A 1 − p 2 A 2 − R x = m& 2V 2x + m& 3V 3 x − m& 1V 1x .. ∴ p 3 = 500 000 +. ∴ R x = 500 000π ×.03 2 − 536 300π ×.03 2 + 33.93 × 12 − 23.88 × 8.446 = 103 N . p3 A 3 − R y = m& 3 V3 y + m& 2V 2y − m& 1V 1y .. ∴ Ry = 540 000π ×.02 2 − 10.05 × ( −8) = 759 N. 4.125 a) ΣFx = m& (V 2x − V1x ).. & 1. − F = −mV =. ∴ F = 300 × 38.2 = 11 460 N .. V1 =. m& ρA 1. 300 = 38.2 m/s 1000π ×.05 2. V2. F V1. b) −F = m& r (V 1 − VB )(cosα − 1). 28.2 ∴ F = 300 × (38.2 − 10) = 6250 N . 38.2 c) −F = m& r (V 1 − VB )(cosα − 1). 48.2 ∴ F = 300 × ( 38.2 − ( −10)) = 18 250 N . 38.2 2 1.25  4.126 a) −F = m& (V 2 x − V1x ). 200 = 1.94π   × V 12 .  12 . ∴ V 1 = 55 fps.. 2. 1.25  b) −F = m& r (V 1 − VB )(cosα − 1). 200 = 1.94π   (V1 − 30) 2 . ∴ V 1 = 85 fps.  12  2. 1.25  c) −F = m& r (V 1 − VB )(cosα − 1). 200 = 1.94π   (V1 + 30) 2 . ∴ V 1 = 25 fps.  12  4.127 a) −F = m& (V 2 x − V1x ). −700 = 1000π ×.04 2 V 1 (V1 × cos 30 o − V1 ). ∴ V 1 = 32.24 m / s. ∴ m& = ρA 1V 1 = 1000π ×.04 2 × 32.24 = 162.1 kg / s. b) −F = m& r (V 1 − VB )(cosα − 1). −700 = 1000π ×.04 2 (V 1 − 8) 2 (.866 − 1). ∴ V 1 = 40.24 m / s. ∴ m& = ρA 1V 1 = 1000π ×.04 2 × 40.24 = 202 kg / s . c) −F = m& r (V 1 − VB )(cosα − 1). −700 = 1000π ×.04 2 (V 1 + 8 ) 2 (.866 − 1). ∴ V 1 = 24.24 m / s. ∴ m& = ρA 1V 1 = 1000π ×.04 2 × 24.24 = 121.8 kg / s . 4.128 (D). − Fx = m& (V2 x − V1x ) = 1000× 0.01 × 0.2 × 50(50cos60o − 50) = −2500 N.. 78.

(30) 2. 4.129 a) −R x = m& (V 2x. 1 − V 1x ) = 1.94π ×   × 120(120 cos 60 o − 120 ). ∴ R x = 305 lb.  12  2. R y = m& (V2 y. 1 − V1y ) = 1.94π   × 120 × (120 ×.866).  12 . ∴ R y = 528 lb.. 2. 1 b) −R x = m& r (V1 − V B )(cos α − 1) = 1.94π ×   × 60 × 60(.5 − 1). ∴ R x = 76.2 lb.  12  2. 1 R y = m& r (V 1 − VB ) sin α = 1.94π   × 60 × ( 60×.866 ).  12 . ∴ R y = 132 lb.. 2. 1 c) −R x = m& r (V1 − V B )(cos α − 1) = 1.94π ×   × 180 × 180(.5 − 1). ∴ R x = 686 lb.  12  2. 1 R y = m& r (V 1 − VB ) sin α = 1.94π   × 180 × (180×.866). ∴ R y = 1188 lb.  12  4.130 VB = R ω = 0.5 × 30 = 15 m / s. −R x = m& (V1 − V B )(cos α − 1) = 1000π ×.025 2 × 40 × 25(.5 − 1). ∴ R x = 982 N. ∴ W& = 10 R V = 10 × 982 × 15 = 147 300 W . x. B. 4.131 a) −R x = m& (V 2x − V1x ) = 4π .02 2 × 400( −400 cos 60 o − 400).. ∴ R x = 1206 N.. R y = m& (V2 y − V1y ) = 4π ×.02 × 400( 400 sin 60 ). ∴ R y = 696 N. o 2 2 & r (V1 − VB )(cos 120 − 1) = 4π .02 × 300 ( −.5 − 1). ∴ R x = 679 N . b) −R x = m o. 2. R y = m& r (V 1 − VB )sin α = 4π ×.02 2 × 300 2 ×.866. ∴ R y = 392 N. o 2 2 & r (V1 − VB )(cos 120 − 1) = 4π .02 × 500 ( −.5 − 1). ∴ R x = 1885 N. c) −R x = m R y = m& r (V 1 − VB )sin α = 4π ×.02 2 × 500 2 ×.866.. ∴ R y = 1088 N.. 4.132 −Fx = m& (V1 − V B )(cos 120 o − 1) = 4π ×.02 2 × ( 400 − 180) 2 (−.5 − 1). ∴ R x = 365 N. VB = 1.2 × 150 = 180 m / s. W& = 15 × 365 × 180 = 986 000 W. The y-component force does no work. 4.133 (A). − Fx = m& (Vr 2 x − Vr1x ) = 1000 × π × 0.022 × 60 × (40cos45o − 40) = 884 N.. Power = Fx ×VB = 884 × 20 = 17700 W..  V = 507 fps. ∴ r1  = Vr 2 750cos β1 − 300 = Vr1 cos45o  Note: V2 x − V1x = −V r 2 cos α 2 + V B − V r1 cos α 1 − VB = −Vr1 (cos α 2 + cos α 1 ).. 4.134 a) Refer to Fig. 4.16:. 750sin β1 = Vr1 sin45o. 2. .5 & r1 (cos α 2 + cos α 1 ) =.015π   × 750 × 507(cos 30 o + cos 45 o ) = 48.9 lb. ∴ R x = mV  12 . 79.

(31) & = 15 R V = 15 × 48.9 × 300 = 220,000 ft - lb or 400 Hp. ∴W x B sec o 750 sin β 1 = V r1 sin 60  b) V = 554 fps = V r 2 . o 750 cos β 1 − 300 = V r1 cos 60  r1 2 .5 & r1 (cos α 2 + cos α 1 ) =.015π ×   × 750 × 554(cos 30 o + cos 60 o ) = 46.4 lb. ∴ R x = mV  12  & = 15 R V = 15 × 46.4 × 300 = 209 ,000 ft - lb or 380 Hp. ∴W x B sec o 750 sin β 1 = V r1 sin 90  c) V = 687 fps = Vr 2 . o 750 cos β 1 − 300 = V r1 cos 90  r1 2 .5 & r1 (cos α 2 + cos α 1 ) =.015π ×   × 750 × 687(cos 30 o + 0) = 36.5 lb. ∴ R x = mV  12  & = 15 R V = 15 × 36.5 × 300 = 164 ,300 ft - lb or 299 Hp. ∴W x B sec. 100 sin 30 o = Vr 1 sin α 1  o  ∴ α 1 = 36.9 , Vr1 = 83.3 m / s. 100 cos 30 o − 20 = Vr 1 cos α 1 . 4.135 a) Refer to Fig. 4.16:.  o  V2 = 71.5, α 2 = 48 . o V2 cos60 = 83.3cos α 2 − 20  − V1x ) = 1000π ×.015 2 × 100( −71.5 cos 60 o − 100 cos 30 o ). ∴ R x = 8650 N . V2 sin60o = 83.3sinα 2. −R x = m& (V 2x ∴ W& = 12V B R x = 12 × 20 × 8650 = 2.08 × 10 6 W .. b). 100 sin 30 o = V r1 sin α 1  o  ∴ α 1 = 47 , V r1 = V r2 = 68.35 m / s. 100 cos 30 o − 40 = Vr 1 cos α 1 .  o  V2 = 38.9 m/s, α 2 = 29.5 . o V2 cos60 = 68.35cos α 2 − 40 −R x = m& (V 2x − V 1x ) = 1000π ×.015 2 × 100( −38.9 cos 60 o − 100 cos 30 o ). ∴ R x = 7500 N . ∴ W& = 12V R = 12 × 40 × 7500 = 3.60 × 10 6 W. V2 sin60o = 68.35sinα 2. B. c). x. 100 sin 30 o = V r1 sin α 1  o  ∴ α 1 = 53.8 , V r1 = V r2 = 61.96 m / s. 100 cos 30o − 50 = V r1 cos α 1   o  V2 = 19.32 m/s, α 2 = 15.66 . o V2 cos60 = 61.96cos α 2 − 50 V2 sin60o = 61.76sinα 2. 80.

(32) −R x = m& (V 2x − V 1x ) = 1000π ×.015 2 × 100( −19.32 cos 60 o − 100 cos 30 o ). ∴ R x = 6800 N . ∴ W& = 12 R V = 12 × 6800 × 50 = 4.08 × 10 6 W . x. B. 50 sin 30 o = V r1 sin α 1  2 2  ∴ V r1 = 2500 − 86.6V B + V B o 50 cos 30 − VB = V r1 cos α 1 . 4.136 a) Refer to Fig. 4.16:.  2 2 2  ∴ Vr 2 = Vr1 = 900 + 30VB + VB . o 30cos60 − Vr 2 cos α 2 = VB  Combine the above: VB = 13.72 m / s. Then, α 1 = 59.4 o , α 2 = 42.1o . −R x = m& (V 2x − V 1x ) = 1000π ×.01 2 × 50( −30 cos 60 o − 50 cos 30 o ). ∴ R x = 916 N . ∴ W& = 15V R = 15 × 13.72 × 916 = 188 500 W . 30sin60o = Vr 2 sin α 2. B. b). x. 50 sin 30 o = V r1 sin α 1  2 2  ∴ V r1 = 2500 − 86.6V B + V B 50 cos 30 o − VB = V r1 cos α 1 . ∴ V B = 14.94 m / s..  2 2 o o  ∴ Vr 2 = 900 + 20.52VB + VB . α 1 = 41.4 , α 2 = 48.2 o 30cos70 − Vr 2 cos α 2 = VB  − Rx = m& (V2 x − V1x ) = 1000π × .012 × 50( −30cos70o − 50cos30o ). ∴ Rx = 841 N. ∴ W& = 15V R = 15 × 14.94 × 841 = 188 500 W . 30sin70o = Vr 2 sin α 2. B. x. 50 sin 30 o = V r1 sin α 1  2 2 c)  ∴ Vr 1 = 2500 − 86.6VB + V B ∴ V B = 16.49 m / s o 50 cos 30 − VB = V r1 cos α 1   2 2 o o  ∴Vr 2 = 900 + 10.42VB + VB . α 1 = 43 , α 2 = 53.7 o 30cos80 − Vr 2 cos α 2 = VB  −R x = m& (V 2x − V 1x ) = 1000π ×.012 × 50( −30 cos 80 o − 50 cos 30 o ). ∴ R x = 762 N . ∴ W& = 15V R = 15 × 16.49 × 762 = 188 500 W . 30sin80o = Vr 2 sin α 2. B. x. 4.137 To find F, sum forces normal to the plate: Σ Fn = m &  (Vout )n − V1n  .   o a) ∴ F = 1000×.02×.4 × 40 −( −40 sin 60 ) = 11 080 N . (We have neglected friction). [. ]. Σ Ft = 0 = m& 2V2 + m&3 (− V3 )− m& 1 × 40sin30o. Bernoulli: V1 = V 2 = V 3 . ∴ 0 = m& 2 − m& 3 −.5 m& 1  ∴ m& 2 =.75m& 1 =.75 × 320 = 240 kg / s.  m& 3 = 80 kg / s. Continuity: m& 1 = m& 2 + m& 3  1 20 × ×120( −120sin60o ) = 3360 lb. (We have neglected friction) 12 12 ΣFt = 0 = m& 2V 2 + m& 3 ( −V 3 ) − m& 1 × 120 sin 30 o . Bernoulli: V1 = V 2 = V 3 .. b)∴ F = −1.94 ×. 81.

(33) 20 × 120 ∴ 0 = m& 2 − m& 3 − 0.5m& 1  ∴ m& 2 = .75m& 1 = .75 × 1.94 × 144  Continuity: m& 1 = m&2 + m&3  = 22.6 slug/sec. and m& 3 = 9.7 slug/sec.. 4.138 F = m& r (V1r ) n = 1000 ×.02 ×. 4 × ( 40 + 20) 2 sin 60 o = 24 940 N. F = 24 940 cos 30 o = 21 600 N. ∴ W& = 21 600 × 20 = 432 000 W . x. 4.139 F = m& r (V1r ) n = 1000 ×.02×.4( 40 − VB ) 2 sin 60 o . Fx = 8( 40 − V B2 ) sin 2 60 o . W& = V F = 8V ( 40 − V ) 2 ×.75 = 6(1600V − 80V 2 + V 3 ). B x. B. B. B. dW& = 6(1600 − 160V B + 3VB2 ) = 0. dVB 4.140 (A). B. B. ∴ V B = 13.33 m / s.. Let the vehicle move to the right. The scoop then diverts the water to the right. Then F = m& (V2 x − V1x ) = 1000 × 0.05 × 2 × 60 × [60 − ( −60)] = 720000 N.. 4.141 F = m& r (V1 − VB )(cos α − 1) = 1000 × .1 × .6VB (−VB )( −2) = 120VB2 . 2.  120 × 1000  At t = 0 : F = 120 ×   = 133 300 N.  3600  133 300 ao = = 1.33 m/s2 100 000 16 .67 t − F dVB − 120VB2 dVB = = . ∴ ∫ − 2 =.0012 ∫ dt . m dt 100 000 VB 33.33 0 1   1 ∴ − =.0012 t.  16.67 33.33 . V2 F V1 = 0. ∴ t = 26.6 sec.. 4.142 F = m& r (V1 − VB )(cos α − 1) = 90 × .8 × 2.5 × 13.89 ×( −13.89)( −1) = 34700 N. 50 × 1000   = 13.89 m/s  ∴ W& = 34700 ×13.89 = 482 000 W or 647 Hp.  VB = 3600   4.143 See the figure in Problem 4.141. F =m & r (V1 − VB )(cosα − 1) = 1000×.06×.2 × VB ( −VB )( −2) = 24 VB2 . dV B dV B . ∴ −24V B2 = 5000V B . dx dx x 24 dx 27. 78 dV B 24 −∫ = ∫ . − x = ln 27.78 − ln 250. 5000 VB 5000 0 250 −F = mV B. 82. ∴ x = 458 m ..

(34) 2.  1.25 V2 4.144 − F = m& r (V1 − VB )(cos α − 1) = 1.94π ×   (V1 − VB ) 2 ( −2).  12  F dV ∴ F = 0.1323(V1 − V B ) 2 = 20 B . VB dt dV At t = 0, V B = 0. Then 20 B = 0.1323V12 . dt dVB With = 6, V1 = 30.1 fps . dt VB 2 dVB 1 1 For t > 0, ∫ = 0.006615 ∫ dt . 0.01323 = − . ∴VB = 8.57 fps. 2 ( 30.1 − VB ) 30.1 − VB 30.1 0 0 4.145 For this steady-state flow, we fix the boat and move the upstream air. This provides us with the steady-state flow of Fig. 4.17. This is the same as observing the flow while standing on the boat. & = FV . 20 000 = F 50 × 1000 . ∴ F = 1440 N. (V = 13.89 m / s) W 1 1 3600 V + 13.89 F = m& (V 2 − V 1 ). 1440 = 1.23π × 12 2 (V 2 − 13.89). ∴ V 2 = 30.6 m / s. 2 30.6 + 13.89 ∴ Q = A 3 V 3 = π × 12 = 69.9 m 3 / s. 2 V 13.89 ηp = 1 = = 0.625 or 62.5%. V 3 22.24 200 × 1000 = 55.56 m / s. 3600 320 × 1000 55.56 + 88.89 V2 = = 88.89 m / s. ∴ m& = 1.2 × π × 1.12 = 329.5 kg / s. 3600 2 F = 329.5( 88.89 − 55.56) = 10 980 N. = ∆pπ × 1.12 . ∴ ∆p = 2890 Pa. & = F × V = 10 980 × 55.56 = 610 000 W or 818 Hp. W. 4.146 Fix the reference frame to the aircraft so that V1 =. 1. 88 = 29.33 fps. 60 2 88 10 29.33 + 58.67 V2 = 40 × = 58.67 fps. ∴ F = m& (V 2 − V1 ) = 1.94π ×   ( 58.67 − 29.33)  12  60 2 = 5460 lb. ft lb & = F × V = 5460 × 29.33 = 160 ,000 W or 291 Hp. 1 sec. 4.147 Fix the reference frame to the boat so that V1 = 20 ×. 83.

(35) 10 29.33 + 58.67 m& = 1.94 × π ×   = 186.2 slug / sec.  12  2 2. 4.148 Fix the reference frame to the boat: V1 = 10 m / s, V2 = 20 m / s. ∴Thrust = m& (V 2 − V1 ) = 1000 × 0.2( 20 − 10) = 2000 N . & = F × V1 = 2000 × 10 = 20 000 W or 26.8 Hp. W 4.149 0.2 = V1 A 1 = V 1 ×.2 × 1.0. ∴ V1 = 1 m / s. ∴ V1 m a x = 2 m / s. ∴ V1 ( y ) = 20( 0.1 − y ). .1. .1. .13 flux in = 2 ∫ ρV dy = 2∫ 1000 × 20 (.1 − y ) dy = 800 000 = 267 N . 3 0 0 The slope at section 1 is −20. ∴ V 2 ( y ) = −20y + A . 2. 2. Continuity: A 1V1 = A 2 V 2 .. 2. ∴ V2 = 2V1 = 2 m / s.. 2 = A − 1 / 2. ∴ A = 2.5.. ∴ V 2 ( y ) = 2.5 − 20y .. V 2 (0) = A.   ∴ V2 = A − 1 / 2. V2 (.05) = A − 1 .05.  ( y −.125) 3  800 000 flux out = 2 ∫ 1000( 2.5 − 20y ) dy = 800 000  [0.00153]  = 3 3  0 0 = 408.3 N . ∴change = 408 − 267 = 141 N. .05. 2. .1. 2 ∫ V dA. 2∫ 20 2 (.1 − y ) 2 dy. .1 3 4 = . V 2A 12 ×.2 × 1.0 3 3 b) See Problem 4.149: V2 ( y ) = 20( 0.125 − y ), .05 ≥ y ≥ 0. V 2 = 2 m / s.. 4.150 a) β =. =. 0. = 4000. . 05. β=. 2 ∫ V dA 2. V A. =. 2 ∫ 20 2 ( y −.125 ) 2 dy 0. 2 ×.1 × 1.0 2. = 2000. ( y −.125) 3 3. 4.151 From the c.v. shown: ( p1 − p 2 )πr02 = τ w 2πro L. ∆p ro du du 0.03 × 144×.75 / 12 ∴τ w = =µ . ∴ = 2L dr w dr w 2 × 30 × 2.36 × 10 − 5 ft / sec = 191 . ft.  r2  4.152 Write the equation of the parabola: V (r ) = V m a x  1 − 2  . r0  . 84. .05 0. = 1.021. τw2πroL. p1A1. p2A2.

(36) .006. ∫. Continuity: π ×.006 2 × 8 =. 0.  r2  V max 1 −  2πrdr.  .006 2 . ∴ V max = 16 m / s.. & 1. Momentum: p1 A 1 − p 2 A 2 − FDrag = ∫ ρV 2 dA − mV .006. 40 000π ×.006 − FDrag = 2. ∫ 0. 2.  r2  2 1000 × 16  1 −  2πrdr − 1000 × π ×.006 × 8 × 8  .006 2  2. 4.524 − FDrag = 9.651 − 7.238.. ∴ FDrag = 2.11 N .. 2   4.153 m& top = ρA 1V 1 − ρ ∫ V2 ( y )dA = 1.23 2 × 10 × 32 − ∫ ( 28 + y 2 )10 dy  = 65.6 kg / s.  0  2 F − = ∫ ρV 2 dA + m& top V 1 − m& 1V1 = 1.23∫ ( 28 + y 2 ) 2 10dy + 65.6 × 32 − 1.23 × 20 × 32 2 . 2 0 ∴ F = 3780 N .. .1   4.154 a) m& top = m& 1 − m& 2 = ρA 1V1 − ∫ ρu( y )dA = 1.23 .1 × 2 × 8 − ∫ ( 20 y − 100 y 2 )8 × 2 dy   0  = 0.656 kg / s. (Note: y = 0.1 for u ( y ) = 8).. .1. Momentum: −FDrag = ρ ∫ 64( 20 y − 100 y 2 ) 2 2 dy +.656 × 8 − ρ ×.1 × 2 × 8 2 0. = 1.23 × 6.83 + 5.25 − 1.23 × 12.8.. ∴ FDrag = 2.1 N. .1. b) To find h:. 8 h = ∫ 8(20 y − 100 y 2 )dy . 0. ∴h =. 20 ×.12 100 ×.001 − = 0.0667 m . 2 3 .1. Momentum: −FDrag = 1.23∫ 64( 20 y − 100 y 2 ) 2 2 dy − 1.23 ×.0667 × 2 × 8 2 . 0. = 1.23 × 6.83 − 10.50. 4.155 a) Energy:. V12 V2 + z1 = 2 + z 2 + hL . 2g 2g. ∴ FDrag = 2.1 N .. See Problem 4.118(a).. 82 1.912 2 + 0.6 = + 2.51 + h L . ∴ hL = 1166 . m. 2 × 9.81 2 × 9.81 ∴ losses = γA 1V1 hL = 9810 × (.6 × 1) × 8 × 1.166 = 54 900 W / m of width. V12 V 22 b) See Problem 4.120: + z1 = + z 2 + hL . 2g 2g. 85.

(37) 7.19 2 1.58 2 +.417 = + 1.9 + h L . ∴ hL = 1.025 m. 2 × 9.81 2 × 9.81 ∴ losses = γA 1V1 hL = 9810×.417 × 3 × 7.19 × 1.025 = 90 300 W 5.17 2 32 + 1.16 = + 2 + hL . ∴ hL = 0.0636 m. 2 × 9.81 2 × 9.81 ∴ losses = γA 1V1 hL = 9810 × 116 . × 5.17 × 0.0636 = 3740 W / m of width.. c) See Problem 4.121:. 4.156 See Problem 4.122: V1 = 20 m / s, V 2 = 5 m / s, p1 = 60 kPa, p2 = 135 kPa. V12 p 1 V 22 p 2 Then, + = + + hL . 2g γ 2g γ. 20 2 60 000 52 135 000 + = + + hL . 2 × 9.81 9810 2 × 9.81 9810. V12 20 2 ∴ h L = 11.47 m = K =K . 2g 2 × 9.81 4.157 Continuity: Energy:. V1 D 2 = Vd 2 .. ∴V1 =. V12 V2 + H (t) = . 2g 2g. d Momentum: ΣFx − ( FI ) x = dt. ∴ K = 0.562.. d2 V. D2. ∴ V = 2 gH ( t ).. ∫ ρV. x. d −V + m& (V 2x. c.v .. v  d2 s  − V1x ).  2  = ax .  dt  x. t πd 2 πd 2 ∴ −a x m(t ) = ρ V (V ). m( t ) = mo − ρ ∫ V ( t )dt . 4 4 0. But, V1 = −. dH dH d 2 . ∴− = 2 dt dt D. 2 gH . ∴ −. 2  ρπd 2  2gd  ∴ ax = 2 g  t + H o 2 4  2D . 2. dH d2 = H 1/ 2 D 2.  πd 2 ρ  4. t. ∫ 0. 2 gdt . ∴ H 1/ 2 =. 2 gd 2. 2D 2   2 gd 2   2 g  t + H dt − m o  o 2  2D  . t + Ho .. 4.158 This is a very difficult design problem. There is an optimum initial mass of water for a maximum height attained by the rocket. It will take a team of students many hours to work this problem. It involves continuity, energy, and momentum. 4.159 Ve = v MI =. m& 4 = = 19.89 m / s. ρA e 1000 × 4 × π ×.004 2. Velocity in a rm = V .. .3 v v r × ( 2 Ω × V ) ρ d V − = 4 ∫ ∫ ri$ × ( −2 Ωk$ × Vi$)ρAdr. c. v .. 0. 86.

(38) .3. = −8ρAVΩ k$ ∫ rdr = −0.36 ρAVΩ k$. 0. v d v v ΣM = 0 and ( r × V )ρd − V = 0. dt c∫. v . The z-component of. v v v $ ∫ r × V (V ⋅ n )ρdA =.3i$ × (.707V e $j +.707V e k$)Ve ρA e .. c .s .. v v v $ ∫ r × V (V ⋅ n )ρdA =.3×.707V e2 A e ρ.. c .s .. Finally, −( M I ) z = 0.36 ρAV Ω = 4 ×.3×.707Ve2 Ae ρ. Using AV = A eVe , 0.36Ω = 4 ×.3×.707 × 19.89. ∴ Ω = 46.9 rad / s. y. v 4.160 A moment M resists the motion thereby producing power. One of the arms is shown.. r V. x. Ω Ve. 10 /12 . 25 v M I = ∫ 4 ri$ × ( −2Ωk$ × Vi$ ) ρAdr = −8 ρAVΩk$ ∫ rdr = −2.778ρAVΩk$ . 0. 0. v d v v ΣM = M k$ , ( r × V )d −V = 0, and dt c∫.v .. 10 v v v $ ∫c.s.r × V (V ⋅ n )ρdA = 12 × V e2 ρA e × 4 k$.. 2. 2. .75 200 10 1/ 4 Thus, M + 2.778 × 1.94π   × × 30 = 200 2 × × 1.94π   × 4.  12   12  9 12 ∴ M = 309 ft - lb. & W = MΩ = 309 × 30 = 9270 ft - lb / sec. 4.161 m& = 10 = ρA V = 1000π ×.012 V0 . Continuity:. ∴ V 0 = 31.8 m / s.. V0 π ×.01 = Vπ ×.01 + Ve ×.006( r −.05). 2. 2. V0π × .012 = Ve × .006 × .15. ∴Ve = 11.1 m/s. ∴V = V0 − 19.1( r −.05)Ve = 42.4 − 212r . . 05 .2 v $ $ $ M I = ∫ 2 ri × ( +2Ωk × V0 i ) ρAdr + ∫ 2 ri$ × [ +2Ω k$ × ( 42.4 − 212 r )$i ] ρAdr 0. . 05. .05. .2. 0. . 05. = 4 ΩV0 ρAk$ ∫ rdr + 4 ΩρAk$ ∫ ( 42.4 r − 212 r 2 )dr = 4 Ω × 31.8 × 1000π ×.01 2 ×. .05 2 $ k + 4 Ω × 1000π ×.01 2 2 212 3  42.4 2 2 3  (. 2 − . 05 ) − (.2 −.05 ) k$  2 3 . 87.

(39) = ( 0.05Ω + 0.3Ω ) k$ = 0.35 Ωk$. .2. .2. . 05. . 05. 2 ∫ ri$ × (−V e $j )V e ρ ×.006 dr = −11.1 × 1000 ×.006 ∫ rdr k$ = −13.86 k$.. ∴ −0.35Ω = −13.86.. ∴ Ω = 39.6 rad /s.. 1000 4.162 1000 = MΩ. ∴M = = 2 N ⋅ m. 500 v M I = ∫ ri$r × ( −2Ωk$ × V ( r )$i r ) ρ 2πr ×.02dr R. = 0.08πΩ ∫ r 2V( r )drk$. 0. Continuity: V ( r) 2πr ×.02 = Vr cos 30 o 2πR ×.02.. ∴ V ( r) = 0.866 RV r / r .. v v v $ r ∫ × V (V ⋅ n )ρdA = − R(RΩ + Vr sin 30 o )Vr cos 30 o ρ2πR×.02 k$ = −.00301Vr (35 +.5Vr )k$.. c .s .. .15. ∴ −2 − 16.32V r ∫ r dr = −.00301V r ( 35+.5V r ).. ∴ V r2 − 52.1Vr − 1333 = 0.. 0. 1 ∴ V r = ( 52.1 ± 52.12 + 4 × 1333 ) = 70.9 m / s. 2 The flow rate is Q = A e Vr cos 30 o = 2π ×.15 ×.02 × 70.9 ×.866 = 1.16 m 3 / s .. .008 2 4.163 See Problem 4.159. Ve = 19.89 m / s. V = × 19.89 = 3.18 m / s. .02 2 .3 v   dΩ $  $  M I = 4 ∫ ri$ × ( −2 Ωk$ × Vi$ ) +  − k  × ri ρAdr. A = π ×.012 , Ae = π ×.004 2 .   dt   0 .3 .3 dΩ $ 2 dΩ $ = −8 ρAVΩ k$ ∫ rdr − 4 ρA k ∫ r dr = −360 AVΩk$ − 36 A k. dt 0 dt 0 v. v. v. ∫ (r × V ) (V ⋅ n$)ρdA = 212V z. 2 e. A e k$.. c .s .. dΩ dΩ = 212Ve2 A e or + 31.8 Ω = 373. dt dt The solution is Ω = Ce −31. 8t + 11.73. The initial condition is Ω( 0) = 0. ∴ C = −11.73.. Thus, 360 AV Ω + 36 A. Finally,. Ω = 11.73( 1 − e −31. 8t ) rad / s.. 4.164 This design problem would be good for a team of students to do as a project. How large a horsepower blower could be handled by an average person?. 88.

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