ch08solutions - Problem 8.1 [1] Given: Find: Solution: Air...

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Unformatted text preview: Problem 8.1 [1] Given: Find: Solution: Air entering duct Flow rate for turbulence; Entrance length The governing equations are Re = The given data is V⋅ D ν Recrit = 2300 From Table A.9 or, for turbulent, Lturb = 25D to 40D Q= π2 ⋅D ⋅V 4 − 4 ft 2 D = 6⋅ in Llaminar = 0.06⋅ Recrit⋅ D Q π Recrit = 4 ⋅D ν π 4 2 ν = 1.62 × 10 ⋅ s ⋅D or Q= − 4 ft 2 Hence Recrit⋅ π⋅ ν ⋅ D 4 1 2 ⋅ ft Q = 0.146⋅ ft s 3 Q = 2300 × For laminar flow For turbulent flow × 1.62 × 10 ⋅ s × Llaminar = 0.06⋅ Recrit⋅ D Lmin = 25⋅ D Llaminar = 0.06 × 2300 × 6⋅ in Lmin = 12.5⋅ ft Lmax = 40⋅ D Llaminar= 69.0⋅ ft Lmax = 20⋅ ft Problem 8.2 [2] Problem 8.3 [3] Given: Find: Solution: From Table A.9 The given data is Air entering pipe system Flow rate for turbulence in each section; Which become fully developed 2 ν = 1.62 × 10 L = 5⋅ ft − 4 ft ⋅ s D1 = 1⋅ in Recrit = 2300 D2 = 1 ⋅ in 2 D3 = 1 ⋅ in 4 The critical Reynolds number is Writing the Reynolds number as a function of flow rate QD ⋅ π 2ν ⋅D 4 Then the flow rates for turbulence to begin in each section of pipe are Re = ν = Q1 = Recrit⋅ π⋅ ν⋅ D1 4 Recrit⋅ π⋅ ν⋅ D2 4 Q1 = 2300 × π 4 ft s 3 V⋅ D or Q= Re⋅ π⋅ ν⋅ D 4 × 1.62 × 10 − 4 ft 2 ⋅ s × 1 12 ⋅ ft Q1 = 0.0244 ft s 3 Q2 = Q2 = 0.0122 Q3 = Recrit⋅ π⋅ ν⋅ D3 4 Q3 = 0.00610 ft s 3 Hence, smallest pipe becomes turbulent first, then second, then the largest. For the smallest pipe transitioning to turbulence (Q3) For pipe 3 or, for turbulent, Re3 = 2300 Lmin = 25⋅ D3 Llaminar = 0.06⋅ Re3⋅ D3 Lmin = 0.521 ft Lmax = 40⋅ D3 Llaminar = 2.87 ft Lmax = 0.833 ft Llaminar < L: Not fully developed Lmax/min < L: Not fully developed For pipes 1 and 2 Llaminar = 0.06⋅ ⎜ ⎛ 4⋅ Q3 ⎞ ⎟ ⋅ D1 ⎝ π⋅ ν ⋅ D 1 ⎠ ⎛ 4⋅ Q3 ⎞ ⎟ ⋅ D2 ⎝ π⋅ ν ⋅ D 2 ⎠ Llaminar = 2.87 ft Llaminar < L: Not fully developed Llaminar = 0.06⋅ ⎜ Llaminar = 2.87 ft Llaminar < L: Not fully developed For the middle pipe transitioning to turbulence (Q2) For pipe 2 Re2 = 2300 Llaminar = 0.06⋅ Re2⋅ D2 Llaminar = 5.75 ft Llaminar > L: Fully developed or, for turbulent, Lmin = 25⋅ D2 Lmin = 1.04 ft Lmax = 40⋅ D2 Lmax = 1.67 ft Lmax/min < L: Not fully developed For pipes 1 and 3 L1 = 0.06⋅ ⎜ ⎛ 4⋅ Q2 ⎞ ⎟ ⋅ D1 ⎝ π⋅ ν ⋅ D 1 ⎠ L3min = 0.521 ft L1 = 5.75 ft L3max = 40⋅ D3 L3max = 0.833 ft Lmax/min < L: Not fully developed L3min = 25⋅ D3 For the large pipe transitioning to turbulence (Q1) For pipe 1 Re1 = 2300 Llaminar = 0.06⋅ Re1⋅ D1 Llaminar = 11.5 ft Llaminar > L: Fully developed or, for turbulent, Lmin = 25⋅ D1 Lmin = 2.08 ft Lmax = 40⋅ D1 Lmax = 3.33 ft Lmax/min < L: Not fully developed For pipes 2 and 3 L2min = 25⋅ D2 L2min = 1.04 ft L2max = 40⋅ D2 L2max = 1.67 ft Lmax/min < L: Not fully developed L3min = 25⋅ D3 L3min = 0.521 ft L3max = 40⋅ D3 L3max = 0.833 ft Lmax/min < L: Not fully developed Problem 8.4 [2] Given: Find: Solution: That transition to turbulence occurs at about Re = 2300 Plots of average velocity and volume and mass flow rates for turbulence for air and water From Tables A.8 and A.10 ρair = 1.23⋅ V⋅ D ν kg m 3 νair = 1.45 × 10 −5 m 2 ⋅ s ρw = 999⋅ kg m 3 νw = 1.14 × 10 −6 m 2 ⋅ s The governing equations are Re = Recrit = 2300 For the average velocity V= Recrit⋅ ν D 2300 × 1.45 × 10 −5 m 2 ⋅ Hence for air Vair = s D −6 m 2 m 0.0334⋅ s Vair = D 2 2300 × 1.14 × 10 For water Vw = D ⋅ s m 0.00262⋅ s Vw = D π⋅ Recrit⋅ ν 4 2 For the volume flow rates Q = A⋅ V = π 4 ⋅D ⋅V = 2 π 4 ⋅D ⋅ 2 Recrit⋅ ν D 2 = ⋅D Hence for air Qair = π −5 m × 2300 × 1.45⋅ 10 ⋅ ⋅D 4 s π −6 m × 2300 × 1.14⋅ 10 ⋅ ⋅D 4 s 2 Qair = 0.0262⋅ m ×D s m ×D s 2 2 For water Qw = Qw = 0.00206⋅ Finally, the mass flow rates are obtained from volume flow rates mair = ρair⋅ Qair mw = ρw⋅ Qw These results are plotted in the associated Excel workbook kg mair = 0.0322⋅ ×D m⋅ s kg mw = 2.06⋅ ×D m⋅ s The relations needed are From Tables A.8 and A.10 the data required is ρair = ρw = D (m) 1.23 999 0.0001 kg/m3 kg/m3 0.001 33.350 2.62 0.01 3.335 0.262 2 νair = 1.45E-05 m /s 2 νw = 1.14E-06 m /s 0.05 0.667 1.0 2.5 5.0 7.5 10.0 V air (m/s) 333.500 V w (m/s) 26.2 3.34E-02 1.33E-02 6.67E-03 4.45E-03 3.34E-03 5.24E-02 2.62E-03 1.05E-03 5.24E-04 3.50E-04 2.62E-04 Q air (m3/s) 2.62E-06 2.62E-05 2.62E-04 1.31E-03 2.62E-02 6.55E-02 1.31E-01 1.96E-01 2.62E-01 Q w (m3/s) 2.06E-07 2.06E-06 2.06E-05 1.03E-04 2.06E-03 5.15E-03 1.03E-02 1.54E-02 2.06E-02 m air (kg/s) 3.22E-06 3.22E-05 3.22E-04 1.61E-03 3.22E-02 8.05E-02 1.61E-01 2.42E-01 3.22E-01 m w (kg/s) 2.06E-04 2.06E-03 2.06E-02 1.03E-01 2.06E+00 5.14E+00 1.03E+01 1.54E+01 2.06E+01 Average Velocity for Turbulence in a Pipe 1.E+04 V (m/s) 1.E+02 Velocity (Air) 1.E+00 Velocity (Water) 1.E-02 1.E-04 1.E-04 1.E-03 1.E-02 D (m) 1.E-01 1.E+00 1.E+01 Flow Rate for Turbulence in a Pipe 1.E+01 Q (m3/s) 1.E-01 Flow Rate (Air) 1.E-03 Flow Rate (Water) 1.E-05 1.E-07 1.E-04 1.E-03 1.E-02 D (m) 1.E-01 1.E+00 1.E+01 Mass Flow Rate for Turbulence in a Pipe 1.E+02 m flow (kg/s) 1.E+00 Mass Flow Rate (Air) 1.E-02 Mass Flow Rate (Water) 1.E-04 1.E-06 1.E-04 1.E-03 1.E-02 D (m) 1.E-01 1.E+00 1.E+01 Problem 8.5 [4] Part 1/2 Problem 8.5 [4] Part 2/2 Problem 8.6 [2] Problem 8.7 [2] Problem 8.8 [3] Problem 8.9 [2] 2h y x Given: Find: Solution: Basic equation Laminar flow between flat plates Shear stress on upper plate; Volume flow rate per width τyx = μ⋅ du dy 2 u ( y) = − 2 2 h dp ⎡ ⎛ y⎞ ⎤ ⋅ ⋅ ⎢1 − ⎜ ⎟ ⎥ 2⋅ μ dx ⎣ ⎝h⎠ ⎦ (from Eq. 8.7) Then At the upper surface τyx = y=h −h dp ⎛ 2⋅ y ⎞ dp ⋅ ⋅⎜ − = −y⋅ 2⎟ 2 dx dx ⎝h⎠ τyx = −1.5⋅ mm × h 1⋅ m 1000⋅ mm × 1.25 × 10 ⋅ 3 N m ⋅m 2 τyx = −1.88 Pa 2⋅ h ⋅ b dp ⋅ 3⋅ μ dx 2 3 The volume flow rate is ⌠ h 2 ⌠ ⌠ h ⋅ b dp ⎮ ⎮ Q = ⎮ u dA = ⎮ u⋅ b dy = − ⋅ ⋅⎮ ⌡− h 2⋅ μ dx ⌡ ⌡ 3 −h ⎡ ⎛ y ⎞ 2⎤ ⎢1 − ⎜ ⎟ ⎥ dy ⎣ ⎝h⎠ ⎦ 2 Q=− Q 2 1⋅ m ⎞ m 3N = − × ⎛ 1.5⋅ mm × ⎜ ⎟ × 1.25 × 10 ⋅ 2 × b 3⎝ 0.5⋅ N ⋅ s 1000⋅ mm ⎠ m ⋅m Q −6m = −5.63 × 10 b s Problem 8.10 [2] Problem 8.11 [3] D p1 a F L Given: Find: Solution: Basic equation For the system Piston cylinder assembly Rate of oil leak Q a ⋅ Δp = l 12⋅ μ⋅ L 3 Q= π⋅ D⋅ a ⋅ Δp 12⋅ μ⋅ L 3 (from Eq. 8.6c; we assume laminar flow and verify this is correct after solving) F 4⋅ F Δp = p1 − patm = = 2 A π⋅ D Δp = 4 π × 4500⋅ lbf × ⎛ ⎜ 1 ⎝ 4⋅ in × 12⋅ in ⎞ ⎟ 1⋅ ft ⎠ 2 Δp = 358⋅ psi lbf⋅ s ft 2 At 120oF (about 50oC), from Fig. A.2 3 μ = 0.06 × 0.0209⋅ 2 μ = 1.25 × 10 − 3 lbf⋅ s ⋅ 2 ft Q= π 12 × 4⋅ in × ⎛ 0.001⋅ in × ⎜ ⎝ 1⋅ ft ⎞ lbf 144⋅ in ft 1 × × ⎟ × 358⋅ 2 × 2 −3 12⋅ in ⎠ 1⋅ ft 1.25 × 10 lbf⋅ s 2⋅ in in Q a ⋅ π⋅ D V= 1 π × 1.25 × 10 − 5 ft 3 2 Q = 1.25 × 10 2 − 5 ft 3 ⋅ s Q = 0.0216⋅ in s 3 Check Re: V= Q A = s ft 2 × 1 1 12⋅ in ⎞ × ×⎛ ⎜ ⎟ .001⋅ in 4⋅ in ⎝ 1⋅ ft ⎠ ν = 6.48 × 10 − 4 ft 2 V = 0.143⋅ ft s Re = V⋅ a ν ft s × 0.001⋅ in × ν = 6 × 10 1⋅ ft 12⋅ in −5 × 10.8 s ⋅ s (at 120oF, from Fig. A.3) Re = 0.143⋅ × s 6.48 × 10 −4 2 Re = 0.0184 so flow is very much laminar ft The speed of the piston is approximately Vp = Q Vp = 4 1 12⋅ in ⎞ − 5 ft × 1.25 × 10 ×⎛ × ⎜ ⎟ π s ⎝ 4⋅ in 1⋅ ft ⎠ 3 2 ⎛ π⋅ D 2 ⎞ ⎜ ⎟ ⎝4⎠ Vp = 1.432 × 10 − 4 ft ⋅ s The piston motion is negligible so our assumption of flow between parallel plates is reasonable Problem 8.12 [3] Problem 8.13 [3] Problem 8.14 [3] Given: Find: Solution: Hydrostatic bearing Required pad width; Pressure gradient; Gap height 2⋅ x ⎞ For a laminar flow (we will verify this assumption later), the pressure gradient is constant( x) = pi⋅ ⎛ 1 − p ⎜ ⎟ W⎠ ⎝ where pi = 700 kPa is the inlet pressure (gage) ⌠ ⎮ Hence the total force in the y direction due to pressure is F = b⋅ where b is the pad width into the paper ⎮ p dx ⌡ ⎮ 2⋅ x ⎞ F = b⋅⎮ pi ⋅ ⎛ 1 − ⎜ ⎟ dx W⎠ ⎝ ⎮ ⌡− W 2 W ⌠2 b⋅W F = pi ⋅ 2 This must be equal to the applied load F. Hence W= 2F ⋅ pi b W = 2× 3 m 2 3 × 50000⋅ N m W = 0.143 m 700 × 10 ⋅ N The pressure gradient is then dp Δp 2⋅ Δp 700 × 10 ⋅ N 1 MPa =− =− = −2 × × = −9.79⋅ W 2 dx W 0.143⋅ m m m Q h ⎛ dp ⎞ =− ⋅⎜ ⎟ l 12⋅ μ ⎝ dx ⎠ 2 3 The flow rate is given (Eq. 8.6c) 1 3 Hence, for h we have ⎛ 12⋅ μ⋅ Q ⎜ l h = ⎜− dp ⎜ dx ⎝ h = ⎢−12 × ⎜ − ⎞ ⎟ ⎟ ⎟ ⎠ At 35oC, from Fig. A.2 μ = 0.15⋅ 1 3 N ⋅s m 2 ⎡ ⎢ ⎣ −6 3 ⎞ ⎤ ⎟ × 0.15⋅ N ⋅ s × 1⋅ mL × 10 ⋅ m × 1⋅ min⎥ 2 min⋅ m ⎜ 9.79 × 106⋅ N ⎟ 60⋅ s ⎥ 1⋅ mL m ⎝ ⎠ ⎦ ⎛ m 3 h = 1.452 × 10 2 −5 m Check Re: Re = V⋅ D DQ hQ 1Q =⋅=⋅ =⋅ ν νA ν b⋅h νl s 1.6 × 10 −4 ν = 1.6 × 10 3 −4m s −4 (at 35oC, from Fig. A.3) Re = ⋅m 2 × 1⋅ mL 10 ⋅ m 1⋅ min × × min⋅ m 1⋅ mL 60⋅ s −6 Re = 1.04 × 10 so flow is very much laminar Problem 8.15 [4] Problem 8.16 [2] Given: Find: Solution: Navier-Stokes Equations Derivation of Eq. 8.5 The Navier-Stokes equations are ∂u ∂v ∂w + + =0 ∂x ∂y ∂z ⎛ ∂ 2u ∂ 2u ∂ 2u ⎞ ⎛ ∂u ∂u ∂u ∂u ⎞ ∂p ρ ⎜ + u + v + w ⎟ = ρg x − + μ ⎜ 2 + 2 + 2 ⎟ ⎜ ∂t ⎜ ∂x ∂x ∂y ∂z ⎟ ∂x ∂y ∂z ⎟ ⎝ ⎠ ⎠ ⎝ 1 4 5 3 6 4 3 4 3 (5.1c) (5.27a) ρ⎜ ⎜ ⎛∂ v ∂ v ∂ v⎞ ⎛ ∂v ∂v ∂v ∂v ⎞ ∂p + u + v + w ⎟ = ρg y − + μ⎜ 2 + 2 + 2 ⎟ ⎟ ⎜ ∂x ∂x ∂y ∂z ⎠ ∂y ∂y ∂z ⎟ ⎝ ∂t ⎠ ⎝ 2 2 2 1 4 5 3 4 5 3 (5.27b) 1 3 3 3 3 ρ⎜ ⎜ ⎛∂ w ∂ w ∂ w⎞ ⎛ ∂w ∂w ∂w ∂w ⎞ ∂p +u +v +w ⎟ = ρg z − + μ⎜ 2 + 2 + 2 ⎟ ⎟ ⎜ ∂x ∂x ∂y ∂z ⎠ ∂z ∂y ∂z ⎟ ⎝ ∂t ⎠ ⎝ 2 2 2 3 3 3 3 (5.27c) The following assumptions have been applied: (1) Steady flow (given). (2) Incompressible flow; ρ = constant. (3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0. (4) Fully developed flow, so no properties except pressure p vary in the x direction; ∂/∂x = 0. (5) See analysis below. (6) No body force in the x direction; gx = 0 Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption (3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4) also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant. Since v is zero at the solid surface, then v must be zero everywhere. The fact that v = 0 reduces the Navier–Stokes equations further, as indicated by (5). Hence for the y direction ∂p = ρg ∂y which indicates a hydrostatic variation of pressure. In the x direction, after assumption (6) we obtain μ Integrating twice ∂ 2u ∂p − =0 ∂y 2 ∂x u= 1 ∂p 2 c1 y + y + c2 2 μ ∂x μ To evaluate the constants, c1 and c2, we must apply the boundary conditions. At y = 0, u = 0. Consequently, c2 = 0. At y = a, u = 0. Hence 0= which gives 1 ∂p 2 c1 a+ a 2 μ ∂x μ 1 ∂p a 2 μ ∂x c1 = − and finally u= 2 a 2 ∂p ⎡⎛ y ⎞ ⎛ y ⎞⎤ ⎢⎜ ⎟ − ⎜ ⎟ ⎥ 2 μ ∂x ⎢⎝ a ⎠ ⎝ a ⎠⎥ ⎣ ⎦ Problem 8.17 [5] Problem 8.18 [5] Problem 8.19 [3] Given: Find: Solution: Laminar velocity profile of power-law fluid flow between parallel plates Expression for flow rate; from data determine the type of fluid The velocity profile is n+ 1⎤ ⎡ ⎢ n⎥ h Δp ⎞ n⋅ h y⎞ ⎥ u=⎛ ⋅ ⋅ ⎢1 − ⎛ ⎟ ⎜ ⎟⋅ ⎜ ⎝ k L ⎠ n + 1 ⎣ ⎝ h⎠ ⎦ 1 n The flow rate is then ⌠ Q = w⋅ ⎮ u dy ⌡− h h or, because the flow is symmetric ⌠ Q = 2⋅ w⋅ ⎮ u dy ⌡0 h The integral is computed as ⌠ n+ 1 2⋅ n+ 1⎤ ⎡ ⎮ ⎢ n n⎥ ⎮ n y⎞ ⎮ 1 − ⎛ y⎞ ⎥ dy = y⋅ ⎢1 − ⋅⎛ ⎟ ⎜⎟ ⎜ ⎮ ⎝ h⎠ ⎣ 2⋅ n + 1 ⎝ h ⎠ ⎦ ⌡ 1 n 1 n Using this with the limits 2⋅ n+ 1⎤ ⎡ ⎢ h Δp ⎞ n⋅ h n n⎥ Q = 2⋅ w ⋅ ⎛ ⋅ ⋅ h⋅ ⎢1 − ⋅ ( 1) ⎜ ⎟⋅ ⎥ ⎝ k L ⎠ n + 1 ⎣ 2⋅ n + 1 ⎦ h Δp ⎞ 2⋅ n⋅ w⋅ h Q=⎛ ⋅ ⎜ ⎟⋅ ⎝ k L ⎠ 2⋅ n + 1 2 The associated Excel spreadsheet shows computation of n. The data is Δp (kPa) Q (L/min) 10 0.451 20 0.759 30 1.01 40 1.15 50 1.41 60 1.57 70 1.66 80 1.85 90 2.05 100 2.25 We can fit a power curve to the data Flow Rate vs Applied Pressure for a Non-Newtonian Fluid 10.0 Q (L/min) Data Power Curve Fit 1.0 y = 0.0974x0.677 R2 = 0.997 0.1 10 Δp (kPa) 1/n = 0.677 n= 1.48 100 Hence Problem 8.20 [2] Problem 8.21 [2] Problem 8.22 [2] U1 d U2 y x Given: Find: Solution: Laminar flow between moving plates Expression for velocity; Volume flow rate per depth Using the analysis of Section 8-2, the sum of forces in the x direction is ⎡ ∂ dy ⎛ ⎛ dy ⎞⎤ dx dx ⎞ ∂ ∂ ∂ ⎢τ + τ ⋅ − ⎜ τ − τ ⋅ ⎟⎥ ⋅ b ⋅ dx + ⎜ p − p ⋅ − p + p ⋅ ⎟ ⋅ b ⋅ dy = 0 ∂y 2 ⎠⎦ ∂x 2 ∂x 2 ⎠ ⎣ ∂y 2 ⎝ ⎝ Simplifying Integrating twice Boundary conditions: Hence dτ dp = =0 dy dx u = c1⋅ y + c2 u ( 0) = −U1 c2 = −U1 or μ⋅ du dy 2 2 =0 u ( y = d ) = U2 u ( y) = 75⋅ y − 0.25 d c1 = U1 + U2 d y u ( y) = U1 + U2 ⋅ − U1 d ( ) (u in m/s, y in m) ⌠ ⌠ ⎮ ⎮ The volume flow rate is Q = u dA = b⋅ ⎮ u dy ⎮ ⌡ ⌡ ⌠ y ⎤ Q = b⋅ ⎮ ⎡ U1 + U2 ⋅ − U1⎥ dx ⎮⎢ d ⎦ ⌡0 ⎣ ( ) Q = b⋅ d⋅ ( U2 − U1 2 ) Q 1⋅ m 1 m = 10⋅ mm × × × ( 0.5 − 0.25) × b 1000⋅ mm 2 s Q = 0.00125 m s 3 m Problem 8.23 [3] Given: Find: Solution: Laminar flow of two fluids between plates Velocity at the interface Using the analysis of Section 8-2, the sum of forces in the x direction is ⎡ ∂ dy ⎛ ∂ dy ⎞⎤ ⎛ ∂ dx dx ⎞ ∂ ⎢τ + τ ⋅ − ⎜ τ − τ ⋅ ⎟⎥ ⋅ b⋅ dx + ⎜ p − p ⋅ − p + p ⋅ ⎟ ⋅ b⋅ dy = 0 ∂x 2 ⎠ ⎣ ∂y 2 ⎝ ∂y 2 ⎠⎦ ⎝ ∂x 2 Simplifying dτ dp = =0 dy dx or μ⋅ du dy 2 2 =0 u2 = c3⋅ y + c4 Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields We need four BCs. Three are obvious y=0 u1 = 0 y = h u1 = u2 u1 = c1⋅ y + c2 y = 2⋅ h u2 = U The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same y=h Using these four BCs Hence From the 2nd and 3rd equations Hence 0 = c2 c2 = 0 c1⋅ h − U = −c3⋅ h μ1⋅ du1 du2 = μ2⋅ dy dy U = c3⋅ 2⋅ h + c4 μ1⋅ c1 = μ2⋅ c3 c1⋅ h + c2 = c3⋅ h + c4 and μ1⋅ c1 = μ2⋅ c3 c1 = U h⋅ ⎜ 1 + μ1 c1⋅ h − U = −c3⋅ h = − ⋅ h⋅ c1 μ2 ⎛ ⎝ μ1 ⎞ μ2 ⎟ ⎠ ⋅y Hence for fluid 1 (we do not need to complete the analysis for fluid 2) u1 = U 20⋅ Evaluating this at y = h, where u1 = uinterface uinterface = ft s 1⎞ ⎟ 3⎠ ⎛ μ1 ⎞ h⋅ ⎜ 1 + ⎟ ⎝ μ2 ⎠ ⎛1 + ⎜ ⎝ uinterface = 15⋅ ft s Problem 8.24 [3] Given: Find: Solution: Given data Properties of two fluids flowing between parallel plates; applied pressure gradient Velocity at the interface; maximum velocity; plot velocity distribution k= dp dx = −1000⋅ N⋅ s m 2 Pa m h = 2.5⋅ mm μ2 = 2⋅ μ1 μ2 = 1⋅ N⋅ s m 2 μ1 = 0.5⋅ (Lower fluid is fluid 1; upper is fluid 2) Following the analysis of Section 8-2, analyse the forces on a differential CV of either fluid The net force is zero for steady flow, so ⎡τ + dτ ⋅ dy − ⎛ τ − dτ ⋅ dy ⎞⎤ ⋅ dx⋅ dz + ⎡p − dp ⋅ dx ⎢ ⎜ ⎟⎥ ⎢ dy 2 dy 2 ⎠⎦ dx 2 ⎣ ⎝ ⎣ dτ dp = =k dy dx − ⎛p + ⎜ ⎝ dp dx ⎞⎤ ⋅ ⎟⎥ ⋅ dy⋅ dz = 0 dx 2 ⎠⎦ d 2 2 Simplifying so for each fluid μ⋅ u=k dy Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields u1 = k 2 ⋅ y + c1⋅ y + c2 2⋅ μ1 u2 = k 2 ⋅ y + c3⋅ y + c4 2⋅ μ2 For convenience the origin of coordinates is placed at the centerline We need four BCs. Three are obvious y = −h y=0 y=h u1 = 0 u1 = u2 u2 = 0 (1) (2) (3) The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same y=0 μ1⋅ du1 du2 = μ2⋅ dy dy (4) Using these four BCs 0= k 2 ⋅ h − c1⋅ h + c2 2⋅ μ1 c2 = c4 0= k 2 ⋅ h + c3⋅ h + c4 2⋅ μ2 μ1⋅ c1 = μ2⋅ c3 Hence, after some algebra k⋅ h μ2 − μ1 c1 = ⋅ 2⋅ μ1 μ2 + μ1 The velocity distributions are then ( ( ) ) k⋅ h c2 = c4 = − μ2 + μ1 2 k⋅ h μ2 − μ1 c3 = ⋅ 2⋅ μ2 μ2 + μ1 ( ( ) ) u1 = k 2⋅ μ 1 ⋅ ⎢y + y⋅ h⋅ ⎡ ⎣ 2 (μ2 − μ1)⎤ − k⋅ h2 ⎥ (μ2 + μ1)⎦ μ2 + μ1 u2 = k 2⋅ μ2 ⋅ ⎢y + y⋅ h⋅ ⎡ ⎣ 2 (μ2 − μ1)⎤ − k⋅ h2 ⎥ (μ2 + μ1)⎦ μ2 + μ1 Evaluating either velocity at y = 0, gives the velocity at the interface k⋅ h μ2 + μ1 2 −3m uinterface = − uinterface = 4.17 × 10 s The plots of these velocity distributions are shown in the associated Excel workbook, as is the determination of the maximum velocity. −3 m From Excel umax = 4.34 × 10 ⋅ s The data is k= h= μ1 = μ2 = -1000 2.5 0.5 1.0 Pa/m mm N.s/m2 N.s/m2 The velocity distribution is y (mm) u 1 x 103 (m/s) u 2 x 103 (m/s) -2.50 -2.25 -2.00 -1.75 -1.50 -1.25 -1.00 -0.75 -0.50 -0.25 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 0.000 0.979 1.83 2.56 3.17 3.65 4.00 4.23 4.33 4.31 4.17 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA 4.17 4.03 3.83 3.57 3.25 2.86 2.42 1.91 1.33 0.698 0.000 The lower fluid has the highest velocity We can use Solver to find the maximum (Or we could differentiate to find the maximum) y (mm) -0.417 u max x 103 (m/s) 4.34 Velocity Distribution Between Parallel Plates 2.5 2.0 y (mm) 1.5 1.0 0.5 0.0 -0.5 0.0 -1.0 -1.5 -2.0 -2.5 1.0 2.0 3.0 4.0 Lower Velocity Upper Velocity u x 103 (m/s) 5.0 Problem 8.25 [2] Given: Find: Solution: Velocity profile between parallel plates Pressure gradients for zero stress at upper/lower plates; plot From Eq. 8.8, the velocity distribution is u= U⋅ y a + a ⎛ ∂ ⎞ ⎡⎛ y ⎞ y⎤ ⋅ ⎜ p ⎟ ⋅ ⎢⎜ ⎟ − ⎥ 2⋅ μ ⎝ ∂x ⎠ ⎣⎝ a ⎠ a⎦ 2 2 The shear stress is τyx = μ⋅ du dy = μ⋅ U a + a ⎛∂ ⎞ ⎛ y 1 ⎞ ⋅ ⎜ p ⎟ ⋅ 2⋅ − 2 ⎝ ∂x ⎠ ⎜ a2 a ⎟ 2 ⎝ ⎠ (a) For τyx = 0 at y = a 0 = μ⋅ U a∂ +⋅p a 2 ∂x 2 2 ∂ ∂x p=− 2⋅ U⋅ μ a 2 The velocity distribution is then u= U⋅ y a 2⋅ U⋅ μ ⎡⎛ y ⎞ y⎤ − ⋅ ⋅ ⎢⎜ ⎟ − ⎥ 2 ⎣⎝ a ⎠ a 2⋅ μ a⎦ a U a∂ −⋅p a 2 ∂x 2 2 y y⎞ u = 2⋅ − ⎛ ⎟ ⎜ a ⎝a⎠ U ∂ ∂x 2⋅ U⋅ μ a 2 2 (b) For τyx = 0 at y = 0 0 = μ⋅ p= The velocity distribution is then U⋅ y a 2⋅ U⋅ μ ⎡⎛ y ⎞ y⎤ u= + ⋅ ⋅ ⎢⎜ ⎟ − ⎥ a a⎦ 2⋅ μ a2 ⎣⎝ a ⎠ u y⎞ =⎛ ⎟ ⎜ U ⎝a⎠ 2 The velocity distributions are plotted in the associated Excel workbook y /a (a) u /U (b) u /U 0.0 0.000 0.000 0.1 0.190 0.010 0.2 0.360 0.040 0.3 0.510 0.090 0.4 0.640 0.160 0.5 0.750 0.250 0.6 0.840 0.360 0.7 0.910 0.490 0.8 0.960 0.640 0.9 0.990 0.810 1.0 1.00 1.000 Zero-Stress Velocity Distributions 1.00 0.75 y /a 0.50 0.25 0.00 0.00 0.25 0.50 u /U 0.75 1.00 Zero Stress Upper Plate Zero Stress Lower Plate Problem 8.26 [2] Given: Find: Solution: Computer disk drive Flow Reynolds number; Shear stress; Power required For a distance R from the center of a disk spinning at speed ω V = R⋅ ω ρ⋅ V⋅ a μ s V = 25⋅ mm × V⋅ a ν −6 1⋅ m 1000⋅ mm × 8500⋅ rpm × 2⋅ π⋅ rad rev −5 m 2 × 1⋅ min 60⋅ s V = 22.3⋅ m s The gap Reynolds number is Re = = ν = 1.45 × 10 ⋅m × s 1.45 × 10 −5 ⋅ s from Table A.10 at 15oC Re = 0.384 Re = 22.3⋅ The flow is definitely laminar The shear stress is then τ = μ⋅ du dy m × 0.25 × 10 ⋅m 2 = μ⋅ V a μ = 1.79 × 10 1 0.25 × 10 −6 − 5 N ⋅s ⋅ 2 from Table A.10 at 15oC τ = 1.60⋅ kPa 2 −5 2 m − 5 N ⋅s τ = 1.79 × 10 ⋅ × 22.3⋅ × 2 s m The power required is P = T⋅ ω N m 2 m ⋅m with A = ( 5⋅ mm) 2⋅ π⋅ rad rev 1⋅ min 60⋅ s A = 2.5 × 10 m where torque T is given by −5 2 T = τ⋅ A ⋅ R P = τ⋅ A ⋅ R ⋅ ω P = 1600⋅ × 2.5 × 10 ⋅ m × 25⋅ mm × 1⋅ m 1000⋅ mm × 8500⋅ rpm × × P = 0.890 W Problem 8.27 [2] Problem 8.28 [2] Problem 8.29 [2] Given: Find: Solution: Velocity distribution on incline Expression for shear stress; Maximum shear; volume flow rate/mm width; Reynolds number 2 ρ⋅ g⋅ sin( θ) ⎛ y⎞ ⋅⎜ h⋅y − ⎟ 2⎠ μ ⎝ From Example 5.9 For the shear stress τ is a maximum at y = 0 u ( y) = τ = μ⋅ du dy = ρ⋅ g⋅ sin( θ) ⋅ ( h − y) τmax = ρ⋅ g⋅ sin( θ) ⋅ h = SG⋅ ρH2O⋅ g⋅ sin( θ) ⋅ h τmax = 1.2 × 1000 kg m 3 × 9.81⋅ m s 2 × sin( 15⋅ deg) × 0.007⋅ m × N ⋅s kg⋅ m 2 τmax = 21.3 Pa This stress is in the x direction on the wall The flow rate is ⌠ h 2 ⌠ ⌠ y⎞ ⎮ ρ⋅ g⋅ sin( θ) ⎛ ⎮ Q = ⎮ u dA = w⋅ ⎮ u ( y) dy = w⋅ ⎮ ⋅ ⎜ h ⋅ y − ⎟ dy ⌡0 2⎠ μ ⎝ ⌡ ⌡0 2 2 m −4 s 3 h Q= ρ⋅ g⋅ sin( θ) ⋅ w⋅ h 3⋅ μ mm s 3 3 Q 1 kg m m N ⋅s 3 = × 1.2 × 1000 × 9.81⋅ × sin( 15⋅ deg) × ( 0.007⋅ m ) × ⋅ = 2.18 × 10 3 2 w 3 1.60⋅ N ⋅ s kg⋅ m m s mm s 3 m Q = 217 w mm The average velocity is V= Q Q = A w⋅ h ρ⋅ V ⋅ h μ kg m 3 V = 217⋅ mm × 1 7⋅ mm V = 31.0 mm s The gap Reynolds number is Re = Re = 1.2 × 1000 The flow is definitely laminar × 31⋅ mm m 1⋅ m ⎞ × 7⋅ mm × ×⎛ ⎜ ⎟ s 1.60⋅ N⋅ s ⎝ 1000⋅ mm ⎠ 2 2 Re = 0.163 Problem 8.30 [3] Given: Find: Solution: Given data Data on flow of liquids down an incline Velocity at interface; velocity at free surface; plot h = 2.5⋅ mm θ = 30⋅ deg ν1 = 2 × 10 −4 m 2 ⋅ s ν2 = 2⋅ ν1 (The lower fluid is designated fluid 1, the upper fluid 2) From Example 5.9 (or Exanple 8.3 with g replaced with gsinθ), a free body analysis leads to (for either fluid) d 2 2 u=− dy ρ⋅ g⋅ sin ( θ) μ Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields u1 = − ρ⋅ g⋅ sin( θ) 2 ⋅ y + c1⋅ y + c2 2⋅ μ1 y=0 y=h u2 = − u1 = 0 u1 = u2 ρ⋅ g⋅ sin( θ) 2 ⋅ y + c3⋅ y + c4 2⋅ μ2 (1) (2) We need four BCs. Two are obvious The third BC comes from the fact that there is no shear stress at the free surface y = 2⋅ h μ2⋅ du2 dy =0 (3) The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same y=h Using these four BCs c2 = 0 − μ1⋅ du1 du2 = μ2⋅ dy dy (4) ρ⋅ g⋅ sin ( θ) 2 ρ⋅ g⋅ sin ( θ) 2 ⋅ h + c1⋅ h + c2 = − ⋅ h + c3⋅ h + c4 2⋅ μ2 2⋅ μ1 −ρ⋅ g⋅ sin ( θ) ⋅ 2⋅ h + μ2⋅ c3 = 0 −ρ⋅ g⋅ sin ( θ) ⋅ h + μ1⋅ c1 = −ρ⋅ g⋅ sin ( θ) ⋅ h + μ2⋅ c3 Hence, after some algebra c1 = 2⋅ ρ⋅ g⋅ sin ( θ) ⋅ h μ1 2⋅ ρ⋅ g⋅ sin ( θ) ⋅ h μ2 ρ⋅ g⋅ sin ( θ) 2 ⋅ 4⋅ y⋅ h − y 2⋅ μ1 c2 = 0 2 μ2 − μ1 c4 = 3⋅ ρ⋅ g⋅ sin ( θ) ⋅ h ⋅ 2⋅ μ1⋅ μ2 c3 = ( ) ) The velocity distributions are then u1 = ( ) u2 = ρ⋅ g⋅ sin ( θ) ⎡ 2 μ2 − μ1 2⎤ ⋅ ⎢3⋅ h ⋅ + 4⋅ y⋅ h − y ⎥ 2⋅ μ2 μ1 ⎣ ⎦ ( Rewriting in terms of ν1 and ν2 (ρ is constant and equal for both fluids) u1 = g⋅ sin ( θ) 2 ⋅ 4⋅ y⋅ h − y 2⋅ ν1 ( ) u2 = g⋅ sin ( θ) ⎡ 2 ν2 − ν1 2⎤ ⋅ ⎢3⋅ h ⋅ + 4⋅ y⋅ h − y ⎥ 2⋅ ν2 ν1 ⎣ ⎦ ( ) (Note that these result in the same expression if ν1 = ν2, i.e., if we have one fluid) Evaluating either velocity at y = h, gives the velocity at the interface uinterface = 3⋅ g⋅ h ⋅ sin ( θ) 2⋅ ν1 2 uinterface = 0.23 m s Evaluating u2 at y = 2h gives the velocity at the free surface ufreesurface = g⋅ h ⋅ sin ( θ) ⋅ 2 (3⋅ ν2 + ν1) 2⋅ ν1⋅ ν2 ufreesurface = 0.268 m s The velocity distributions are plotted in the associated Excel workbook h= 2.5 mm θ= 30 deg 2 ν1 = 2.00E-04 m /s 2 ν2 = 4.00E-04 m /s y (mm) u 1 (m/s) u 2 (m/s) 0.000 0.250 0.500 0.750 1.000 1.250 1.500 1.750 2.000 2.250 2.500 2.750 3.000 3.250 3.500 3.750 4.000 4.250 4.500 4.750 5.000 0.000 0.0299 0.0582 0.0851 0.110 0.134 0.156 0.177 0.196 0.214 0.230 Velocity Distributions down an Incline 5.0 4.0 y (mm) 3.0 2.0 1.0 0.0 0.00 Lower Velocity Upper Velocity 0.230 0.237 0.244 0.249 0.254 0.259 0.262 0.265 0.267 0.268 0.268 0.05 0.10 0.15 u (m/s) 0.20 0.25 0.30 Problem 8.31 [3] Given: Find: Solution: Flow between parallel plates Shear stress on lower plate; Plot shear stress; Flow rate for pressure gradient; Pressure gradient for zero shear; Plot a dp ⎡⎛ y ⎞ y⎤ ⋅ ⋅ ⎢⎜ ⎟ − ⎥ 2⋅ μ dx ⎣⎝ a ⎠ a⎦ 2 2 From Section 8-2 u ( y) = U⋅ y a y a + For dp/dx = 0 u = U⋅ ⌠ Q =⎮ ⌡0 l τ = μ⋅ a ⌠ y U⋅ a u ( y) dy = w⋅ ⎮ U⋅ dy = ⎮ a 2 ⌡0 = μ⋅ U a when dp/dx = 0 a Q= 1 2 × 5⋅ ft s × 0.1 12 ⋅ ft Q = 0.0208 ft s 3 ft For the shear stress du dy μ = 3.79 × 10 − 7 lbf⋅ s ⋅ 2 (Table A.9) ft The shear stress is constant - no need to plot! τ = 3.79 × 10 ft 12 1⋅ ft ⎞ − 7 lbf⋅ s ⋅ × 5⋅ × ×⎛ ⎜ ⎟ 2 s 0.1⋅ ft ⎝ 12⋅ in ⎠ ft 2 τ = 1.58 × 10 −6 psi Q will decrease if dp/dx > 0; it will increase if dp/dx < 0. For non- zero dp/dx: At y = 0.25a, we get τ = μ⋅ du dy = μ⋅ U a + a⋅ U a dp ⎛ y 1 ⎞ ⋅⎜ − ⎟ dx ⎝ a 2 ⎠ + a⋅ dp ⎛ 1 1 ⎞ U a dp ⋅ ⎜ − ⎟ = μ⋅ − ⋅ dx ⎝ 4 2 ⎠ a 4 dx 2 τ ( y = 0.25⋅ a) = μ⋅ lbf Hence this stress is zero when 2 ft dp 4⋅ μ⋅ U ft 12 ⎞ − 7 lbf ⋅ s − 4 psi = = 4 × 3.79 × 10 ⋅ × 5⋅ × ⎛ = 7.58 × 10 ⎜ ⎟ = 0.109⋅ 2 2 dx s ⎝ 0.1⋅ ft ⎠ ft ft ft a 0.1 0.075 y (in) 0.05 0.025 − 1×10 −4 0 1×10 −4 2×10 −4 3×10 −4 Shear Stress (lbf/ft3) Problem 8.32 [3] Given: Find: Solution: Flow between parallel plates Location and magnitude of maximum velocity; Volume flow in 10 s; Plot velocity and shear stress U⋅ y b dp ⎡⎛ y ⎞ y⎤ + ⋅ ⋅ ⎢⎜ ⎟ − ⎥ b 2⋅ μ dx ⎣⎝ b ⎠ b⎦ 2 2 2 From Section 8-2 u ( y) = For umax set du/dx = 0 du U b dp ⎛ 2⋅ y 1 ⎞ U 1 dp =0= + ⋅ ⋅⎜ − ⎟= + ⋅ ⋅ ( 2⋅ y − b) 2 dy a b 2⋅ μ dx b b 2⋅ μ dx ⎝ ⎠ Hence u = umax at y= b 2 − μ⋅ U dp b⋅ dx From Table A.8 at 15oC μ = 1.14 × 10 − 3 N⋅ s ⋅ 2 m y= 0.0025⋅ m 2 − 1.14 × 10 ⎛ m3 ⎞ m 1 − 3 N⋅ s ⎟ ⋅ × 0.25⋅ × × ⎜− 2 s 0.0025⋅ m ⎝ 175⋅ N ⎠ m U⋅ y b + b dp ⎡⎛ y ⎞ y⎤ ⋅ ⋅ ⎢⎜ ⎟ − ⎥ 2⋅ μ dx ⎣⎝ b ⎠ b⎦ 2 2 y = 1.90 × 10 −3 ⋅m y = 1.90⋅ mm Hence m s ×⎛ ⎜ umax = 1.90 ⎞ b with y = 1.90 mm m s umax = 0.25⋅ 2 2 1 m 175⋅ N ⎞ ⎡⎛ 1.90 ⎞ 2 ⎛ 1.90 ⎞⎤ × ⎛− × ⎢⎜ ⎟ + × ( 0.0025⋅ m) × ⎟ −⎜ ⎟⎥ ⎜ −3 3⎟ ⎝ 2.5 ⎠ 2 ⎝ 2.5 ⎠⎦ 1.14 × 10 ⋅ N⋅ s ⎝ m ⎠ ⎣⎝ 2.5 ⎠ umax = 0.278 ⌠⎡ b 2 2 Q⌠ b dp ⎡⎛ y ⎞ ⎮ U⋅ y ⎮ u ( y) dy = w⋅ ⎢ ⎢⎜ ⎟ − = + ⋅⋅ ⎮ w ⌡0 b 2⋅ μ dx ⎣⎝ b ⎠ ⌡0 ⎣ m 1 Q 1 3 = × 0.25⋅ × 0.0025⋅ m − × ( 0.0025⋅ m) × s 12 w 2 Q −4 m ⋅ Δt = 5.12 × 10 × 10⋅ s w s 2 y⎤⎤ b dp U⋅ b ⎥⎥ dy = − ⋅ b⎦⎦ 2 12⋅ μ dx 3 m 2 −3 1.14 × 10 ⋅ N⋅ s × ⎛− ⎜ 175⋅ N ⎞ m 3 ⎝ ⎟ ⎠ Q −4 m = 5.12 × 10 s w 3 −3 m 2 Flow = Flow = 5.12 × 10 2 −3 m = 5.12 × 10 2 m The velocity profile is u y b dp ⎡ y ⎞ y⎤ =+ ⋅ ⋅ ⎢⎛ ⎟ − ⎥ ⎜ U b 2⋅ μ⋅ U dx ⎣⎝ b ⎠ b⎦ 2 For the shear stress τ = μ⋅ du U b dp y⎞ ⎤ = μ⋅ + ⋅ ⋅ ⎡2⋅ ⎛ ⎟ − 1⎥ ⎢⎜ dy b 2 dx ⎣ ⎝ b ⎠ ⎦ The graphs on the next page can be plotted in Excel 1 0.8 0.6 0.4 0.2 y/b 0 0.2 0.4 0.6 0.8 1 1.2 u/U 1 0.8 0.6 0.4 0.2 − 0.2 − 0.1 y/b 0 0.1 0.2 0.3 0.4 Shear Stress (Pa) Problem 8.33 [4] Part 1/2 Problem 8.33 [4] Part 2/2 Problem 8.34 [4] Problem 8.35 [3] Problem 8.36 [2] Given: Find: Solution: Navier-Stokes Equations Derivation of Eq. 8.5 The Navier-Stokes equations are (using the coordinates of Example 8.3, so that x is vertical, y is horizontal) ∂u ∂v ∂w + + =0 ∂x ∂y ∂z ⎛ ∂ 2u ∂ 2u ∂ 2u ⎞ ⎛ ∂u ∂u ∂u ∂u ⎞ ∂p ρ ⎜ + u + v + w ⎟ = ρg x − + μ ⎜ 2 + 2 + 2 ⎟ ⎜ ∂t ⎜ ∂x ∂x ∂y ∂z ⎟ ∂x ∂y ∂z ⎟ ⎝ ⎠ ⎠ ⎝ 1 4 5 3 4 3 4 3 (5.1c) (5.27a) ρ⎜ ⎜ ⎛∂ v ∂ v ∂ v⎞ ⎛ ∂v ∂v ∂v ∂v ⎞ ∂p + u + v + w ⎟ = ρg y − + μ⎜ 2 + 2 + 2 ⎟ ⎟ ⎜ ∂x ∂x ∂y ∂z ⎠ ∂y ∂y ∂z ⎟ ⎝ ∂t ⎠ ⎝ 2 2 2 1 4 5 3 6 4 5 3 (5.27b) 1 3 3 3 3 ρ⎜ ⎜ ⎛∂ w ∂ w ∂ w⎞ ⎛ ∂w ∂w ∂w ∂w ⎞ ∂p +u +v +w ⎟ = ρg z − + μ⎜ 2 + 2 + 2 ⎟ ⎟ ⎜ ∂x ∂x ∂y ∂z ⎠ ∂z ∂y ∂z ⎟ ⎝ ∂t ⎠ ⎝ 2 2 2 3 3 3 3 (5.27c) The following assumptions have been applied: (1) Steady flow (given). (2) Incompressible flow; ρ = constant. (3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0. (4) Fully developed flow, so no properties except possibly pressure p vary in the x direction; ∂/∂x = 0. (5) See analysis below. (6) No body force in the y direction; gy = 0 Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption (3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4) also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant. Since v is zero at the solid surface, then v must be zero everywhere. The fact that v = 0 reduces the Navier–Stokes equations further, as indicated by (5). Hence for the y direction ∂p =0 ∂y which indicates the pressure is a constant across the layer. However, at the free surface p = patm = constant. Hence we conclude that p = constant throughout the fluid, and so ∂p =0 ∂x In the x direction, we obtain μ Integrating twice ∂ 2u + ρg = 0 ∂y 2 u=− c 1 ρgy 2 + 1 y + c2 2μ μ To evaluate the constants, c1 and c2, we must apply the boundary conditions. At y = 0, u = 0. Consequently, c2 = 0. At y = a, du/dy = 0 (we assume air friction is negligible). Hence τ (y = δ ) = μ which gives and finally du dy =− y =δ 1 μ ρgδ + c1 μ =0 c1 = ρgδ u=− 2 1 ρg ρg 2 ⎡ ⎛ y ⎞ 1 ⎛ y ⎞ ⎤ ρgy 2 + δ ⎢⎜ ⎟ − ⎜ ⎟ ⎥ y= 2μ μ μ ⎢⎝ δ ⎠ 2 ⎝ δ ⎠ ⎥ ⎣ ⎦ Problem 8.37 [4] Problem 8.38 [4] Using the result for average velocity from Example 8.3 Problem 8.39 [5] Problem 8.40 (In Excel) Given: Expression for efficiency Find: Plot; find flow rate for maximum efficiency; explain curve Solution: q 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 η 0.0% 7.30% 14.1% 20.3% 25.7% 30.0% 32.7% 33.2% 30.0% 20.8% 0.0% Efficiency of a Viscous Pump 35% 30% 25% η 20% 15% 10% 5% 0% 0.00 0.10 0.20 q 0.30 0.40 0.50 For the maximum efficiency point we can use Solver (or alternatively differentiate) q 0.333 η 33.3% The efficiency is zero at zero flow rate because there is no output at all The efficiency is zero at maximum flow rate Δp = 0 so there is no output The efficiency must therefore peak somewhere between these extremes Problem 8.41 [5] Problem 8.42 [5] Part 1/2 Problem 8.42 [5] Part 2/2 Problem 8.43 [5] Problem 2.59 Problem 8.44 [3] Given: Find: Solution: The given data is Data on a journal bearing Time for the bearing to slow to 10 rpm D = 50⋅ mm μ = 0.1⋅ N ⋅s m 2 L = 1⋅ m ωi = 60⋅ rpm I⋅ α = Torque = −τ ⋅ A⋅ I = 0.055⋅ kg⋅ m ωf = 10⋅ rpm D 2 2 δ = 1⋅ mm The equation of motion for the slowing bearing is where α is the angular acceleration and τ is the viscous stress, and A = π⋅ D⋅ L is the surface area of the bearing As in Example 8.2 the stress is given by τ = μ⋅ U δ = μ⋅ D⋅ ω 2⋅ δ where U and ω are the instantaneous linear and angular velocities. Hence I⋅ α = I⋅ dω dt =− 3 μ⋅ D ⋅ ω 2⋅ δ ⋅ π⋅ D ⋅ L ⋅ D 2 =− μ ⋅ π⋅ D ⋅ L ⋅ω 4⋅ δ 3 Separating variables dω μ ⋅ π⋅ D ⋅ L =− ⋅ dt ω 4⋅ δ⋅ I − μ⋅ π⋅ D ⋅ L ⋅t 4⋅ δ⋅ I − μ⋅ π⋅ D ⋅ L ⋅t 4⋅ δ⋅ I 3 3 Integrating and using IC ω = ω0 The time to slow down to ωf ω ( t) = ωi⋅ e = 10 rpm is obtained from solving 4⋅ δ⋅ I μ ⋅ π⋅ D ⋅ L 3 ωf = ωi⋅ e so t=− ⋅ ln ⎜ ⎛ ωf ⎞ ⎟ ⎝ ωi ⎠ t = 10 s Problem 8.45 [2] Problem 8.46 [2] Problem 8.47 [2] d p1 D F L Given: Find: Solution: Basic equation For the system Hyperdermic needle Volume flow rate of saline Q= π⋅ Δp⋅ d 128⋅ μ⋅ L 4 (Eq. 8.13c; we assume laminar flow and verify this is correct after solving) F 4⋅ F Δp = p1 − patm = = 2 A π⋅ D Δp = 4 π × 7.5⋅ lbf × ⎛ ⎜ 1 × 12⋅ in ⎞ ⎟ 1⋅ ft ⎠ 2 ⎝ 0.375⋅ in ft Δp = 67.9⋅ psi μ = 5⋅ μH2O μ = 1.05 × 10 2 − 4 lbf ⋅ s ⋅ 2 At 68oF, from Table A.7 μH2O = 2.1 × 10 π 128 − 5 lbf ⋅ s ⋅ 2 ft Q= × 67.9⋅ lbf in 2 3 × 144⋅ in 1⋅ ft 2 2 × ⎛ 0.005⋅ in × ⎜ ⎝ 1⋅ ft ⎞ ft 1 12⋅ in × × ⎟× −4 12⋅ in ⎠ 1⋅ in 1⋅ ft 1.05 × 10 lbf ⋅ s Q = 1.43 × 10 3 − 3 in 4 Q = 8.27 × 10 Q A − 7 ft ⋅ s V= 4 π × 8.27 × 10 − 7 ft 3 ⋅ s 2 Q = 0.0857⋅ ft s in min 3 Check Re: V= = Q π⋅ d 4 2 s ×⎛ ⎜ 1⎞ ⎛ 12⋅ in ⎞ ⎟ ×⎜ ⎟ ⎝ .005⋅ in ⎠ ⎝ 1⋅ ft ⎠ 2 V = 6.07⋅ Re = ρ⋅ V ⋅ d μ ρ = 1.94⋅ slug ft 3 (assuming saline is close to water) ft 2 −4 Re = 1.94⋅ slug ft 3 × 6.07⋅ ft 1⋅ ft × 0.005⋅ in × × s 12⋅ in × slug⋅ ft s ⋅ lbf 2 Re = 46.7 Flow is laminar 1.05 × 10 ⋅ lbf ⋅ s Problem 8.48 [3] Given: Find: Solution: The given data is Data on a tube "Resistance" of tube; maximum flow rate and pressure difference for which electrical analogy holds for (a) kerosine and (b) castor oil L = 100⋅ mm D = 0.3⋅ mm From Fig. A.2 and Table A.2 Kerosene: μ = 1.1 × 10 − 3 N ⋅s ⋅ 2 ρ = 0.82 × 990⋅ kg m 3 = 812⋅ kg m 3 m Castor oil: μ = 0.25⋅ N ⋅s m 2 ρ = 2.11 × 990⋅ kg m 3 = 2090⋅ kg m 3 For an electrical resistor V = R⋅ I (1) The governing equation for the flow rate for laminar flow in a tube is Eq. 8.13c Q= π⋅ Δp⋅ D 128⋅ μ⋅ L 4 or Δp = 128⋅ μ⋅ L π⋅ D 4 ⋅Q (2) By analogy, current I is represented by flow rate Q, and voltage V by pressure drop Δp. Comparing Eqs. (1) and (2), the "resistance" of the tube is R= 128⋅ μ⋅ L π⋅ D 4 The "resistance" of a tube is directly proportional to fluid viscosity and pipe length, and strongly dependent on the inverse of diameter The analogy is only valid for Re < 2300 ρ⋅ Writing this constraint in terms of flow rate Q π2 ⋅D 4 μ ⋅D < 2300 or Qmax = 2300⋅ μ⋅ π⋅ D 4⋅ ρ or ρ⋅ V ⋅ D < 2300 μ The corresponding maximum pressure gradient is then obtained from Eq. (2) 128⋅ μ⋅ L π⋅ D 4 Δpmax = ⋅ Qmax = 32⋅ 2300⋅ μ ⋅ L ρ⋅ D 3 2 (a) For kerosine Qmax = 7.34 × 10 3 −7m s Δpmax = 406 kPa (b) For castor oil Qmax = 6.49 × 10 3 −5m s Δpmax = 8156 MPa The analogy fails when Re > 2300 because the flow becomes turbulent, and "resistance" to flow is then no longer linear with flow rate Problem 8.49 [4] Problem 8.50 [4] Problem 8.51 [4] Part 1/2 Problem 8.51 [4] Part 2/2 Problem 8.52 [4] Part 1/2 Problem 8.52 [4] Part 2/2 Problem 8.53 [4] Problem 8.52 Problem 8.54 [3] Given: Find: Solution: Given data Two-fluid flow in tube Velocity distribution; Plot D = 0.2⋅ in L = 50⋅ ft Δp = −1⋅ psi μ1 = 0.02⋅ lbf ⋅ s ft 2 μ2 = 0.03⋅ lbf ⋅ s ft 2 From Section 8-3 for flow in a pipe, Eq. 8.11 can be applied to either fluid u= 2 r ⎛ ∂ ⎞ c1 ⋅ ⎜ p ⎟ + ⋅ ln ( r) + c2 4⋅ μ ⎝ ∂x ⎠ μ 2 r Δp c1 ⋅ + ⋅ ln ( r) + c2 4⋅ μ1 L μ1 2 r Δp c3 ⋅ + ⋅ ln ( r) + c4 4⋅ μ2 L μ2 Applying this to fluid 1 (inner fluid) and fluid 2 (outer fluid) u1 = u2 = We need four BCs. Two are obvious r= D 2 u2 = 0 (1) r= D 4 u1 = u2 (2) The third BC comes from the fact that the axis is a line of symmetry r=0 du1 dr =0 (3) The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same r= 2 D 4 μ1⋅ du1 2 du2 = μ2⋅ dr dr (4) 2 Using these four BCs ⎛D⎞ ⎜⎟ ⎝ 2 ⎠ ⋅ Δp + c3 ⋅ ln⎛ D ⎞ + c = 0 ⎜⎟ 4 4⋅ μ2 L μ2 ⎝ 2 ⎠ c1 =0 r → 0 μ1⋅ r lim ⎛D⎞ ⎛D⎞ ⎜⎟ ⎜⎟ 4 ⎠ Δp c1 ⎛ D ⎞ ⎝ ⎝ 4 ⎠ ⋅ Δp + c3 ⋅ ln⎛ D ⎞ + c ⋅ + ⋅ ln ⎜ ⎟ + c2 = ⎜⎟ 4 4⋅ μ1 L 4⋅ μ2 L μ1 ⎝ 4 ⎠ μ2 ⎝ 4 ⎠ D Δp 4⋅ c3 D Δp 4⋅ c1 ⋅ + =⋅ + D 8L D 8L Hence, after some algebra c1 = 0 (To avoid singularity) 2 D ⋅ Δp μ2 + 3⋅ μ1 c2 = − μ1⋅ μ2 64⋅ L ( ) ( c3 = 0 D ⋅ Δp c4 = − 16⋅ L⋅ μ2 2 Δp ⎡ 2 ⎛ D ⎞ ⎤ ⋅ ⎢r − ⎜ ⎟ ⎥ 4⋅ μ2⋅ L ⎣ ⎝2⎠ ⎦ 2 The velocity distributions are then 2 Δp ⎡ 2 ⎛ D ⎞ μ2 + 3⋅ μ1 ⎢r − u1 ( r) = ⋅ ⎜ ⎟⋅ 4⋅ μ1⋅ L ⎢ 4⋅ μ2 ⎣ ⎝2⎠ )⎤ ⎥ ⎥ ⎦ u2 ( r) = (Note that these result in the same expression if μ1 = μ2, i.e., if we have one fluid) Evaluating either velocity at r = D/4 gives the velocity at the interface 2 2 2 2 uinterface = − 3⋅ D ⋅ Δp 64⋅ μ2⋅ L uinterface = − 3 ⎛ 0.2 ⎞ ft 1 ⎛ lbf ⎞ × 144⋅ in × ×⎜ ⋅ ft⎟ × ⎜ −1⋅ × 2⎟ 2 64 ⎝ 12 ⎠ 0.03⋅ lbf ⋅ s 50⋅ ft ⎝ in ⎠ 1⋅ ft uinterface = 1.25 × 10 − 3 ft s Evaluating u1 at r = 0 gives the maximum velocity D ⋅ Δp⋅ μ2 + 3⋅ μ1 umax = − 64⋅ μ1⋅ μ2⋅ L 2 ( ) umax = − 1 ⎛ 0.2 ⎞ ⎛ lbf ⎞ × 0.03 + 3 × 0.02 ⋅ ft × 1 ×⎜ ⋅ ft⎟ × ⎜ −1⋅ 2⎟ 64 ⎝ 12 ⎠ 0.02 × 0.03 lbf ⋅ s 50⋅ ft ⎝ in ⎠ 2 2 umax = 1.88 × 10 − 3 ft s 0.1 0.075 Inner fluid Outer fluid r (in) 0.05 0.025 −4 −3 −3 −3 0 5×10 1×10 1.5×10 2×10 Velocity (ft/s) The velocity distributions can be plotted in Excel Problem 8.55 [2] Given: Find: Solution: Basic equation Turbulent pipe flow Wall shear stress (Eq. 4.18a) Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow With these assumptions the x momentum equation becomes p1 ⋅ π⋅ D π⋅ D + τw⋅ π⋅ D⋅ L − p2 ⋅ =0 4 4 2 2 or τw = (p2 − p1)⋅ D 4⋅ L =− Δp⋅ D 4⋅ L 1 1⋅ m 1 3N τw = − × 35 × 10 ⋅ × 150⋅ mm × × 2 4 1000⋅ mm 10⋅ m m Since τw is negative it acts to the left on the fluid, to the right on the pipe wall τw = −131 Pa Problem 8.56 [3] Given: Find: Solution: Basic equation Pipe glued to tank Force glue must hold when cap is on and off (Eq. 4.18a) First solve when the cap is on. In this static case π⋅ D ⋅ p1 4 Second, solve for when flow is occuring: Fglue = 2 where p1 is the tank pressure Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow With these assumptions the x momentum equation becomes p1 ⋅ π⋅ D π⋅ D + τw⋅ π⋅ D⋅ L − p2 ⋅ =0 4 4 2 2 Here p1 is again the tank pressure and p2 is the pressure at the pipe exit; the pipe exit pressure is patm = 0 kPa gage. Hence Fpipe = Fglue = −τw⋅ π⋅ D⋅ L = π⋅ D ⋅ p1 4 2 We conclude that in each case the force on the glue is the same! When the cap is on the glue has to withstand the tank pressure; when the cap is off, the glue has to hold the pipe in place against the friction of the fluid on the pipe, which is equal in magnitude to the pressure drop. Fglue = π 4 × ⎛ 2.5⋅ cm × ⎜ 1⋅ m ⎞ 3N ⎟ × 250 × 10 ⋅ 2 100⋅ cm ⎠ m 2 ⎝ Fglue = 123 N Problem 8.57 [2] Given: Find: Solution: Basic equation Flow through channel Average wall stress (Eq. 4.18a) Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow With these assumptions the x momentum equation becomes p1⋅ W⋅ H + τw⋅ 2⋅ L⋅ ( W + H) − p2⋅ W⋅ H = 0 1⋅ ft 12⋅ in ⎛ ×⎜ 30⋅ ft or W⋅ H τw = p2 − p1 ⋅ 2⋅ ( W + H) ⋅ L H L ( ) τw = −Δp⋅ 2⋅ ⎛ 1 + ⎜ H⎞ W⎠ psi ⎝ ⎟ 1 lbf 144⋅ in τw = − × 1⋅ × × 2 2 2 ft in 2 1⋅ in × ⎞ 1⋅ ft ⎟ 9.5⋅ in × ⎜ ⎟ 12⋅ in ⎜1 + ⎟ 30⋅ ft ⎝ ⎠ 1 τw = −0.195 lbf ft 2 τw = −1.35 × 10 −3 Since τw < 0, it acts to the left on the fluid, to the right on the channel wall Problem 8.58 [2] Given: Find: Solution: Given data Data on pressure drops in flow in a tube Which pressure drop is laminar flow, which turbulent ∂ ∂x p1 = −4.5⋅ kPa m ∂ ∂x p2 = −11⋅ kPa m D = 30⋅ mm From Section 8-4, a force balance on a section of fluid leads to R∂ D∂ τw = − ⋅ p = − ⋅ p 4 ∂x 2 ∂x Hence for the two cases τw1 = − D∂ ⋅p 4 ∂x 1 D∂ ⋅p 4 ∂x 2 τw1 = 33.8 Pa τw2 = − τw2 = 82.5 Pa Because both flows are at the same nominal flow rate, the higher pressure drop must correspond to the turbulent flow, because, as indicated in Section 8-4, turbulent flows experience additional stresses. Also indicated in Section 8-4 is that for both flows the shear stress varies from zero at the centerline to the maximums computed above at the walls. The stress distributions are linear in both cases: Maximum at the walls and zero at the centerline. Problem 8.59 [3] Problem 8.60 [3] Problem 8.61 [3] Given: Data on mean velocity in fully developed turbulent flow Find: Trendlines for each set; values of n for each set; plot Solution: y/R 0.898 0.794 0.691 0.588 0.486 0.383 0.280 0.216 0.154 0.093 0.062 0.041 0.024 u/U 0.996 0.981 0.963 0.937 0.907 0.866 0.831 0.792 0.742 0.700 0.650 0.619 0.551 y/R 0.898 0.794 0.691 0.588 0.486 0.383 0.280 0.216 0.154 0.093 0.062 0.037 u/U 0.997 0.998 0.975 0.959 0.934 0.908 0.874 0.847 0.818 0.771 0.736 0.690 Equation 8.22 is Mean Velocity Distributions in a Pipe 1.0 u/U 0.1 0.01 y/R Re = 50,000 Re = 500,000 Power (Re = 500,000) Power (Re = 50,000) 0.10 1.00 Applying the Trendline analysis to each set of data: At Re = 50,000 u/U = 1.017(y/R )0.161 2 with R = 0.998 (high confidence) Hence 1/n = 0.161 n = 6.21 At Re = 500,000 u/U = 1.017(y/R )0.117 2 with R = 0.999 (high confidence) Hence 1/n = 0.117 n = 8.55 Both sets of data tend to confirm the validity of Eq. 8.22 Problem 8.62 [3] Problem 8.63 [3] Part 1/2 Problem 8.63 [3] Part 2/2 Problem 8.64 [3] Given: Find: Solution: Laminar flow between parallel plates Kinetic energy coefficient, α Basic Equation: The kinetic energy coefficient, α is given by ∫ α= From Section 8-2, for flow between parallel plates A ρ V 3dA & mV 2 (8.26b) 2 2 ⎡⎛ ⎞⎤ 3 ⎡ ⎛ y⎞⎤ y⎟⎥ ⎟⎥ u = umax ⎢1 − ⎜ = V ⎢1 − ⎜ ⎜a ⎟ ⎥ 2 ⎢ ⎜a ⎟ ⎥ ⎢ ⎢ ⎝ 2⎠ ⎥ ⎢ ⎝ 2⎠ ⎥ ⎣ ⎣ ⎦ ⎦ since 3 umax = V . 2 Substituting ∫ α= Then A ρV 3dA & mV 2 ∫ = ρV A V 2 A ρu 3dA 1 ⎛u⎞ 1 = ∫ ⎜ ⎟ dA = A A⎝V ⎠ wa 3 − 2 ⎛u⎞ ⎛u⎞ ∫a ⎜ V ⎟ wdy = a ∫ ⎜ V ⎟ dy ⎝⎠ ⎠ 0⎝ 2 a 2 3 a 2 3 3 31 3 1 2 a ⎛ u ⎞ ⎛ umax ⎞ ⎛ y ⎞ ⎛ 3 ⎞ 23 ⎜ ⎟= α= ∫ ⎜ umax ⎟ ⎜ V ⎟ d ⎜ a ⎟ ⎜ 2 ⎟ ∫ (1 − η ) dη ⎜ ⎟ a 2 0⎝ ⎝ ⎠0 ⎠ ⎠⎝ ⎝ 2⎠ where η = Evaluating, y a 2 (1 − η ) The integral is then 23 = 1 − 3η 2 + 3η 4 − η 6 3 ⎛ 3⎞ α =⎜ ⎟ ⎝2⎠ 31 ⎛ 3⎞ ∫ (1 − 3η + 3η − η )dη = ⎜ 2 ⎟ ⎝⎠ 0 2 4 6 3 5 1 7 ⎤ 27 16 ⎡ 3 ⎢η − η + 5 η − 7 η ⎥ = 8 35 = 1.54 ⎣ ⎦0 1 Problem 8.65 [3] Problem 8.66 [3] Given: Find: Solution: Equation 8.26b is Definition of kinetic energy correction coefficient α α for the power-law velocity profile; plot ⌠ 3 ⎮ ⎮ ρ⋅ V dA ⌡ mrate⋅ Vav 2 α= where V is the velocity, mrate is the mass flow rate and Vav is the average velocity V = U⋅ ⎛ 1 − ⎜ r⎞ 1 n For the power-law profile (Eq. 8.22) ⎝ R⎠ 2. ⎟ For the mass flow rate mrate = ρ⋅ π⋅ R ⋅ Vav 2 2 3 Hence the denominator of Eq. 8.26b is mrate⋅ Vav = ρ⋅ π⋅ R ⋅ Vav We next must evaluate the numerator of Eq. 8.26b ⌠ ⎮ ⎮ ⌡ R ⌠ 3 ⎮ n ⎮ r⎞ 3 3 ρ⋅ V dA = ⎮ ρ⋅ 2⋅ π⋅ r⋅ U ⋅ ⎛ 1 − ⎟ dr ⎜ ⎮ ⎝ R⎠ ⌡ ⌠ 3 ⎮ ⎮ 22 3 n ⎮ ρ⋅ 2⋅ π⋅ r⋅ U3⋅ ⎛ 1 − r ⎞ dr = 2⋅ π⋅ ρ⋅ R ⋅ n ⋅ U ⎜ ⎟ ⎮ ( 3 + n) ⋅ ( 3 + 2⋅ n) ⎝ R⎠ ⌡0 To integrate substitute m = 1− r R dm = − dr R Then r = R⋅ ( 1 − m) R dr = −R⋅ dm ⌠ 0 3 ⎮ ⌠ 3 ⎮ ⎮ n r⎞ ⎮ ρ⋅ 2⋅ π⋅ r⋅ U3⋅ ⎛ 1 − ⎟ dr = −⎮ ρ⋅ 2⋅ π⋅ R⋅ ( 1 − m) ⋅ m n ⋅ R dm ⎜ ⌡1 ⎮ ⎝ R⎠ ⌡0 Hence ⌠ ⎮ ⎮ ⌡ ⌠ 3⎞ ⎛3 ⎮ +1 ⎜n 3 n⎟ ρ⋅ V dA = ⎮ ρ⋅ 2⋅ π⋅ R⋅ ⎝ m − m ⎠ ⋅ R dm ⌡0 1 22 3 ⌠ 2⋅ R ⋅ n ⋅ ρ⋅ π⋅ U 3 ⎮ ρ⋅ V dA = ⎮ ( 3 + n) ⋅ ( 3 + 2⋅ n) ⌡ Putting all these results together α= ⌠ 3 ⎮ ⎮ ρ⋅ V dA ⌡ mrate⋅ Vav 3 2⋅ R ⋅ n ⋅ ρ⋅ π⋅ U ( 3+ n) ⋅ ( 3+ 2⋅ n) = 2 2 3 22 3 ρ⋅ π⋅ R ⋅ Vav 2 2⋅ n ⎛U⎞⋅ α=⎜ Vav ⎟ ( 3 + n) ⋅ ( 3 + 2⋅ n) ⎝ ⎠ To plot α versus ReVav we use the following parametric relations n = −1.7 + 1.8⋅ log Reu Vav U 2 () (Eq. 8.23) = 2⋅ n ( n + 1) ⋅ ( 2⋅ n + 1) Vav U 3 (Eq. 8.24) ReVav = ⋅ ReU 2 U⎞ 2⋅ n α=⎛ ⎜ V ⎟ ⋅ ( 3 + n) ⋅ ( 3 + 2⋅ n) ⎝ av ⎠ (Eq. 8.27) A value of ReU leads to a value for n; this leads to a value for Vav/U; these lead to a value for ReVav and α The plots of α, and the error in assuming α = 1, versus ReVav are shown in the associated Excel workbook A value of Re U leads to a value for n ; this leads to a value for V av/U ; these lead to a value for Re Vav and α Re U 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 2.50E+06 5.00E+06 7.50E+06 1.00E+07 n 5.50 6.22 6.76 7.08 7.30 8.02 8.56 8.88 9.10 9.82 10.4 10.7 10.9 V av/U 0.776 0.797 0.811 0.818 0.823 0.837 0.846 0.851 0.854 0.864 0.870 0.873 0.876 Re Vav 7.76E+03 1.99E+04 4.06E+04 6.14E+04 8.23E+04 2.09E+05 4.23E+05 6.38E+05 8.54E+05 2.16E+06 4.35E+06 6.55E+06 8.76E+06 α 1.09 1.07 1.06 1.06 1.05 1.05 1.04 1.04 1.04 1.03 1.03 1.03 1.03 α Error 8.2% 6.7% 5.9% 5.4% 5.1% 4.4% 3.9% 3.7% 3.5% 3.1% 2.8% 2.6% 2.5% Kinetic Energy Coefficient vs Reynolds Number 1.10 1.08 α 1.05 1.03 1.00 1E+03 1E+04 1E+05 Re Vav 1E+06 1E+07 Error in assuming α = 1 vs Reynolds Number 10.0% 7.5% 5.0% 2.5% 0.0% 1E+03 Error 1E+04 1E+05 Re Vav 1E+06 1E+07 Problem 8.67 [2] Given: Find: Solution: Data on flow through elbow Head loss 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ hlT Basic equation ⎜ + α⋅ + z1⎟ − ⎜ + α⋅ + z2⎟ = = HlT 2⋅ g 2⋅ g g ⎝ ρ⋅ g ⎠ ⎝ ρ⋅ g ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 p1 − p2 ρ⋅ g V1 − V2 2⋅ g 2 2 Then HlT = + + z1 − z2 HlT = ( 70 − 45) × 10 ⋅ 3 2 2 2 m kg⋅ m s 1 s 3N 2 2 m⎞ × × × + × 1.75 − 3.5 ⋅ ⎛ ⎟ × + ( 2.25 − 3) ⋅ m ⎜ 2 1000⋅ kg 2 9.81⋅ m 2 ⎝ s ⎠ 9.81⋅ m ( ) m s ⋅N HlT = 1.33 m N⋅ m kg In terms of energy/mass hlT = g⋅ HlT hlT = 9.81⋅ m s 2 × 1.33⋅ m × N⋅ s kg⋅ m 2 hlT = 13.0⋅ Problem 8.68 [2] Given: Find: Solution: Data on flow in a pipe Head loss for horizontal pipe; inlet pressure for different alignments; slope for gravity feed Given or available data D = 50⋅ mm ρ = 1000⋅ kg m 3 The governing equation between inlet (1) and exit (2) is 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ ⎜ ρ + α1⋅ 2 + g⋅ z1⎟ − ⎜ ρ + α2⋅ 2 + g⋅ z2⎟ = hlT ⎝ ⎠⎝ ⎠ (8.29) Horizontal pipe data p1 = 588⋅ kPa z1 = z2 p2 = 0⋅ kPa V1 = V2 hlT = 588⋅ J kg (Gage pressures) Equation 8.29 becomes hlT = p1 − p2 ρ For an inclined pipe with the same flow rate, the head loss will be the same as above; in addition we have the following new data z1 = 0⋅ m Equation 8.29 becomes p1 = p2 + ρ⋅ g⋅ z2 − z1 + ρ⋅ hlT z2 = 25⋅ m ( ) p1 = 833⋅ kPa For a declining pipe with the same flow rate, the head loss will be the same as above; in addition we have the following new data z1 = 0⋅ m Equation 8.29 becomes p1 = p2 + ρ⋅ g⋅ z2 − z1 + ρ⋅ hlT z2 = −25⋅ m ( ) p1 = 343⋅ kPa For a gravity feed with the same flow rate, the head loss will be the same as above; in addition we have the following new data p1 = 0⋅ kPa Equation 8.29 becomes hlT z2 = z1 − g (Gage) z2 = −60 m Problem 8.69 [2] Given: Find: Solution: Basic equation Data on flow through elbow Inlet velocity 2 2 ⎛p ⎞ ⎛p ⎞h V1 V2 ⎜1 ⎟ ⎜2 ⎟ lT + α⋅ + z1⎟ − ⎜ + α⋅ + z2⎟ = = HlT ⎜ ρ⋅ g 2⋅ g ρ⋅ g 2⋅ g g ⎝ ⎠⎝ ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 2 2 Then V2 − V1 = 2⋅ V1 V1 = ( ) 2 − V1 = 3⋅ V1 = 2 2 2⋅ p1 − p2 + 2⋅ g⋅ z1 − z2 − 2⋅ g⋅ HlT ρ ( ) ( ) ⎤ 2 ⎡ p1 − p2 ⋅⎢ + g⋅ z1 − z2 − g⋅ HlT ⎥ ρ 3⎣ ⎦ ( ) ( ) V1 = 2 3 × ⎢50 × 10 ⋅ ⎡ ⎢ ⎣ 3 ⎤ m kg⋅ m 9.81⋅ m m 3N × × + × ( −2) ⋅ m − 9.81⋅ × 1⋅ m⎥ 2 1000⋅ kg 2 2 2 ⎥ m s ⋅N s s ⎦ V1 = 3.70 m s Problem 8.70 [2] Given: Find: Solution: Increased friction factor for water tower flow How much flow is decreased Basic equation from Example 8.7 V2 = 2⋅ g⋅ z1 − z2 L ⎞ f ⋅ ⎛ + 8⎟ + 1 ⎜ D ⎝ ⎠ D = 4⋅ in ft s ft s 2 ( ) where With f = 0.0308, we obtain We need to recompute with f = 0.04 L = 680⋅ ft V2 = 8.97⋅ V2 = z1 − z2 = 80⋅ ft and Q = 351 gpm × 80⋅ ft × 1 0.04⋅ ⎛ ⎜ 680 4 12 2 × 32.2⋅ ⎜ ⎝ + 8⎟ + 1 ⎞ V2 = 7.88 ft s ⎟ ⎠ Hence Q = V2⋅ A = V2⋅ Q = 7.88⋅ ft s × π 4 π⋅ D 4 ×⎛ ⎜ 2 2 ⎞ 12 ⎠ ⎝ 4 ⋅ ft ⎟ × 7.48⋅ gal 1⋅ ft 3 × 60⋅ s 1⋅ min Q = 309 gpm (From Table G.2 1 ft3 = 7.48 gal) Hence the flow is decreased by ( 351 − 309) ⋅ gpm = 42 gpm Problem 8.71 [2] Given: Find: Solution: Increased friction factor for water tower flow, and reduced length How much flow is decreased Basic equation from Example 8.7 V2 = 2⋅ g⋅ z1 − z2 L ⎞ f ⋅ ⎛ + 8⎟ + 1 ⎜ D ⎝ ⎠ D = 4⋅ in ft s 2 ( ) where now we have We need to recompute with f = 0.04 L = 380⋅ ft V2 = 2 × 32.2⋅ × 80⋅ ft × z1 − z2 = 80⋅ ft 1 0.04⋅ ⎛ ⎜ 380 4 12 ⎜ ⎝ + 8⎟ + 1 ⎞ V2 = 10.5 ft s ⎟ ⎠ Hence Q = V2⋅ A = V2⋅ Q = 10.5⋅ ft s × π 4 π⋅ D 4 ×⎛ ⎜ 2 2 ⎞ 12 ⎠ ⎝ 4 ⋅ ft ⎟ × 7.48⋅ gal 1⋅ ft 3 × 60⋅ s 1⋅ min Q = 411 gpm (From Table G.2 1 ft3 = 7.48 gal) Problem 8.72 [2] Given: Find: Solution: Basic equation Data on flow through Alaskan pipeline Head loss 2 2 ⎞ ⎛p ⎞h ⎛p V1 V2 lT ⎜1 ⎟⎜2 ⎟ + α⋅ + z1 − + α⋅ + z2 = = HlT ⎜ ρ ⋅g ⎟ ⎜ ρ ⋅g ⎟ 2⋅ g 2⋅ g g ⎝ oil ⎠ ⎝ oil ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) SG = 0.9 (Table A.2) Then HlT = p1 − p2 SGoil⋅ ρH2O⋅ g + z1 − z2 HlT = ( 8250 − 350) × 10 ⋅ 3 2 1 m kg⋅ m s 3N × × × × + ( 45 − 115) ⋅ m 2 0.9 1000⋅ kg 2 9.81⋅ m m s ⋅N HlT = 825 m N⋅ s kg⋅ m 2 In terms of energy/mass hlT = g⋅ HlT hlT = 9.81⋅ m s 2 × 825⋅ m × hlT = 8.09⋅ kN⋅ m kg Problem 8.73 [2] Problem 8.74 [2] Problem 8.75 [2] Problem 8.76 [3] Given: Find: Solution: Basic equations Data on flow from reservoir Head from pump; head loss 2 2 ⎛p ⎞ ⎛p ⎞h V3 V4 ⎜3 ⎟ ⎜4 ⎟ lT ⎜ ρ⋅ g + α⋅ 2⋅ g + z3⎟ − ⎜ ρ⋅ g + α⋅ 2⋅ g + z4⎟ = g = HlT ⎝ ⎠⎝ ⎠ for flow from 3 to 4 2 2 ⎛p ⎞ ⎛p ⎞ Δh V3 V2 ⎜3 ⎟ ⎜2 ⎟ pump = Hpump for flow from 2 to 3 ⎜ ρ⋅ g + α⋅ 2⋅ g + z3⎟ − ⎜ ρ⋅ g + α⋅ 2⋅ g + z2⎟ = g ⎝ ⎠⎝ ⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V2 = V3 = V4 (constant area pipe) Then for the pump Hpump = p3 − p2 ρ⋅ g 3 2 m kg⋅ m s 3N × × × 2 1000⋅ kg 2 9.81⋅ m Hpump = ( 450 − 150) × 10 ⋅ In terms of energy/mass hpump = g⋅ Hpump HlT = p3 − p4 ρ⋅ g + z3 − z4 m s ⋅N Hpump = 30.6 m N⋅ s kg⋅ m 2 hpump = 9.81⋅ m s 2 × 30.6⋅ m × hpump = 300⋅ N⋅ m kg For the head loss from 3 to 4 HlT = ( 450 − 0) × 10 ⋅ In terms of energy/mass hlT = g⋅ HlT 3 2 m kg⋅ m s 3N × × × + ( 0 − 35) ⋅ m 2 1000⋅ kg 2 9.81⋅ m m s ⋅N HlT = 10.9 m hlT = 107⋅ N⋅ m kg hlT = 9.81⋅ m s 2 × 10.9⋅ m × N⋅ s 2 kg⋅ m Problem 8.77 [2] Problem 8.78 [2] Problem 8.79 [2] Given: Find: Solution: Given data From Appendix A Data on flow in a pipe Friction factor; Reynolds number; if flow is laminar or turbulent D = 75⋅ mm ρ = 1000⋅ kg m 3 Δp Pa = 0.075⋅ L m μ = 4⋅ 10 − 4 N ⋅s ⋅ 2 kg mrate = 0.075⋅ s m The governing equations between inlet (1) and exit (2) are 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α1⋅ + g⋅ z1⎟ − ⎜ + α2⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ (8.29) hl = f ⋅ For a constant area pipe LV ⋅ D2 2 (8.34) V1 = V2 = V f= 2⋅ D L⋅ V V= 2 Hence Eqs. 8.29 and 8.34 become ⋅ (p1 − p2) ρ = 2⋅ D Δp ⋅ 2L ρ⋅ V V = 0.017 m s For the velocity mrate ρ⋅ π2 ⋅D 4 Hence f= 2⋅ D Δp ⋅ 2 ρ⋅ V L ρ⋅ V ⋅ D μ f = 0.0390 The Reynolds number is Re = Re = 3183 This Reynolds number indicates the flow is turbulent. (From Eq. 8.37, at this Reynolds number the friction factor for a smooth pipe is f = 0.043; the friction factor computed above thus indicates that, within experimental error, the flow corresponds to turbulent flow in a smooth pipe) Problem 8.80 [3] Solution: Using the add-in function Friction factor from the web site e/D = Re 500 1.00E+03 1.50E+03 2.30E+03 1.00E+04 1.50E+04 1.00E+05 1.50E+05 1.00E+06 1.50E+06 1.00E+07 1.50E+07 1.00E+08 0 0.0001 0.0002 0.0005 0.001 f 0.1280 0.0640 0.0427 0.0473 0.0309 0.0278 0.0180 0.0166 0.0116 0.0109 0.0081 0.0076 0.0059 0.1280 0.0640 0.0427 0.0474 0.0310 0.0280 0.0185 0.0172 0.0134 0.0130 0.0122 0.0121 0.0120 0.1280 0.0640 0.0427 0.0474 0.0312 0.0282 0.0190 0.0178 0.0147 0.0144 0.0138 0.0138 0.0137 0.1280 0.0640 0.0427 0.0477 0.0316 0.0287 0.0203 0.0194 0.0172 0.0170 0.0168 0.0167 0.0167 0.1280 0.0640 0.0427 0.0481 0.0324 0.0296 0.0222 0.0214 0.0199 0.0198 0.0197 0.0197 0.0196 0.1280 0.0640 0.0427 0.0489 0.0338 0.0313 0.0251 0.0246 0.0236 0.0235 0.0234 0.0234 0.0234 0.1280 0.0640 0.0427 0.0512 0.0376 0.0356 0.0313 0.0310 0.0305 0.0304 0.0304 0.0304 0.0304 0.1280 0.0640 0.0427 0.0549 0.0431 0.0415 0.0385 0.0383 0.0380 0.0379 0.0379 0.0379 0.0379 0.1280 0.0640 0.0427 0.0619 0.0523 0.0511 0.0490 0.0489 0.0487 0.0487 0.0486 0.0486 0.0486 0.1280 0.0640 0.0427 0.0747 0.0672 0.0664 0.0649 0.0648 0.0647 0.0647 0.0647 0.0647 0.0647 0.002 0.005 0.01 0.02 0.04 Friction Factor vs Reynolds Number 1.000 0.100 f 0.010 e/D = 0 0.0002 0.001 0.005 0.02 0.0001 0.0005 0.002 0.01 0.04 0.001 1.0E+02 Re 1.0E+03 1.0E+04 1.0E+05 1.0E+06 1.0E+07 1.0E+08 Problem 8.81 Using the above formula for f 0, and Eq. 8.37 for f 1 e/D = Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0.0310 0.0244 0.0208 0.0190 0.0179 0.0149 0.0131 0.0122 0.0116 0.0090 0.0081 0.0066 0.0060 0.0311 0.0247 0.0212 0.0195 0.0185 0.0158 0.0145 0.0139 0.0135 0.0124 0.0122 0.0120 0.0120 0.0313 0.0250 0.0216 0.0200 0.0190 0.0167 0.0155 0.0150 0.0148 0.0140 0.0139 0.0138 0.0137 0.0318 0.0258 0.0226 0.0212 0.0204 0.0186 0.0178 0.0175 0.0173 0.0168 0.0168 0.0167 0.0167 0.0327 0.0270 0.0242 0.0230 0.0223 0.0209 0.0204 0.0201 0.0200 0.0197 0.0197 0.0196 0.0196 0 0.0001 0.0002 0.0005 0.001 f0 0.0342 0.0291 0.0268 0.0258 0.0253 0.0243 0.0239 0.0238 0.0237 0.0235 0.0235 0.0234 0.0234 0.0383 0.0342 0.0325 0.0319 0.0316 0.0309 0.0307 0.0306 0.0305 0.0304 0.0304 0.0304 0.0304 0.0440 0.0407 0.0395 0.0390 0.0388 0.0383 0.0381 0.0380 0.0380 0.0379 0.0379 0.0379 0.0379 0.0534 0.0508 0.0498 0.0494 0.0493 0.0489 0.0488 0.0487 0.0487 0.0487 0.0486 0.0486 0.0486 0.0750 0.0731 0.0724 0.0721 0.0720 0.0717 0.0717 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.002 0.005 0.01 0.02 0.05 Using the add-in function Friction factor from the Web e/D = Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0 0.0001 0.0002 0.0005 0.001 f 0.0309 0.0245 0.0209 0.0191 0.0180 0.0150 0.0132 0.0122 0.0116 0.0090 0.0081 0.0065 0.0059 0.0310 0.0248 0.0212 0.0196 0.0185 0.0158 0.0144 0.0138 0.0134 0.0123 0.0122 0.0120 0.0120 0.0312 0.0250 0.0216 0.0200 0.0190 0.0166 0.0154 0.0150 0.0147 0.0139 0.0138 0.0138 0.0137 0.0316 0.0257 0.0226 0.0212 0.0203 0.0185 0.0177 0.0174 0.0172 0.0168 0.0168 0.0167 0.0167 0.0324 0.0268 0.0240 0.0228 0.0222 0.0208 0.0202 0.0200 0.0199 0.0197 0.0197 0.0196 0.0196 0.0338 0.0288 0.0265 0.0256 0.0251 0.0241 0.0238 0.0237 0.0236 0.0235 0.0234 0.0234 0.0234 0.0376 0.0337 0.0322 0.0316 0.0313 0.0308 0.0306 0.0305 0.0305 0.0304 0.0304 0.0304 0.0304 0.0431 0.0402 0.0391 0.0387 0.0385 0.0381 0.0380 0.0380 0.0380 0.0379 0.0379 0.0379 0.0379 0.0523 0.0502 0.0494 0.0492 0.0490 0.0488 0.0487 0.0487 0.0487 0.0486 0.0486 0.0486 0.0486 0.0738 0.0725 0.0720 0.0719 0.0718 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.002 0.005 0.01 0.02 0.05 The error can now be computed e/D = Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0 0.0001 0.0002 0.0005 0.001 0.002 0.005 0.01 0.02 0.05 0.29% 0.39% 0.63% 0.69% 0.71% 0.65% 0.52% 0.41% 0.33% 0.22% 0.49% 1.15% 1.44% 0.36% 0.24% 0.39% 0.38% 0.33% 0.04% 0.26% 0.41% 0.49% 0.51% 0.39% 0.15% 0.09% 0.43% 0.11% 0.19% 0.13% 0.06% 0.28% 0.51% 0.58% 0.60% 0.39% 0.27% 0.09% 0.06% 0.61% 0.21% 0.25% 0.35% 0.43% 0.64% 0.64% 0.59% 0.54% 0.24% 0.15% 0.05% 0.03% Error (%) 0.88% 1.27% 0.60% 1.04% 0.67% 1.00% 0.73% 0.95% 0.76% 0.90% 0.72% 0.66% 0.59% 0.47% 0.50% 0.37% 0.43% 0.31% 0.16% 0.10% 0.10% 0.06% 0.03% 0.02% 0.02% 0.01% 1.86% 1.42% 1.11% 0.93% 0.81% 0.48% 0.31% 0.23% 0.19% 0.06% 0.03% 0.01% 0.00% 2.12% 1.41% 0.98% 0.77% 0.64% 0.35% 0.21% 0.15% 0.12% 0.03% 0.02% 0.01% 0.00% 2.08% 1.21% 0.77% 0.58% 0.47% 0.24% 0.14% 0.10% 0.08% 0.02% 0.01% 0.00% 0.00% 1.68% 0.87% 0.52% 0.38% 0.30% 0.14% 0.08% 0.06% 0.05% 0.01% 0.01% 0.00% 0.00% The maximum discrepancy is 2.12% at Re = 10,000 and e/D = 0.01 0.100 f0 0.010 e/D = 0 e/D = 0.0001 e/D = 0.0002 e/D = 0.0005 e/D = 0.001 e/D = 0.002 e/D = 0.005 e/D = 0.01 e/D = 0.02 e/D = 0.05 0.001 1E+04 1E+05 1E+06 1E+07 1E+08 Re Problem 8.82 Using the above formula for f 0, and Eq. 8.37 for f 1 e/D = Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0.0309 0.0244 0.0207 0.0189 0.0178 0.0148 0.0131 0.0122 0.0116 0.0090 0.0081 0.0066 0.0060 0.0310 0.0245 0.0210 0.0193 0.0183 0.0156 0.0143 0.0137 0.0133 0.0123 0.0122 0.0120 0.0120 0.0311 0.0248 0.0213 0.0197 0.0187 0.0164 0.0153 0.0148 0.0146 0.0139 0.0139 0.0138 0.0138 0.0315 0.0254 0.0223 0.0209 0.0201 0.0183 0.0176 0.0173 0.0172 0.0168 0.0168 0.0167 0.0167 0.0322 0.0265 0.0237 0.0226 0.0220 0.0207 0.0202 0.0200 0.0199 0.0197 0.0197 0.0197 0.0197 0 0.0001 0.0002 0.0005 0.001 f0 0.0335 0.0285 0.0263 0.0254 0.0250 0.0241 0.0238 0.0237 0.0236 0.0235 0.0235 0.0235 0.0235 0.0374 0.0336 0.0321 0.0316 0.0313 0.0308 0.0306 0.0305 0.0305 0.0304 0.0304 0.0304 0.0304 0.0430 0.0401 0.0391 0.0387 0.0385 0.0382 0.0381 0.0381 0.0380 0.0380 0.0380 0.0380 0.0380 0.0524 0.0502 0.0495 0.0492 0.0491 0.0489 0.0488 0.0488 0.0488 0.0487 0.0487 0.0487 0.0487 0.0741 0.0727 0.0722 0.0720 0.0719 0.0718 0.0717 0.0717 0.0717 0.0717 0.0717 0.0717 0.0717 0.002 0.005 0.01 0.02 0.05 Using the add-in function Friction factor from the Web e/D = Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0 0.0001 0.0002 0.0005 0.001 f 0.0309 0.0245 0.0209 0.0191 0.0180 0.0150 0.0132 0.0122 0.0116 0.0090 0.0081 0.0065 0.0059 0.0310 0.0248 0.0212 0.0196 0.0185 0.0158 0.0144 0.0138 0.0134 0.0123 0.0122 0.0120 0.0120 0.0312 0.0250 0.0216 0.0200 0.0190 0.0166 0.0154 0.0150 0.0147 0.0139 0.0138 0.0138 0.0137 0.0316 0.0257 0.0226 0.0212 0.0203 0.0185 0.0177 0.0174 0.0172 0.0168 0.0168 0.0167 0.0167 0.0324 0.0268 0.0240 0.0228 0.0222 0.0208 0.0202 0.0200 0.0199 0.0197 0.0197 0.0196 0.0196 0.0338 0.0288 0.0265 0.0256 0.0251 0.0241 0.0238 0.0237 0.0236 0.0235 0.0234 0.0234 0.0234 0.0376 0.0337 0.0322 0.0316 0.0313 0.0308 0.0306 0.0305 0.0305 0.0304 0.0304 0.0304 0.0304 0.0431 0.0402 0.0391 0.0387 0.0385 0.0381 0.0380 0.0380 0.0380 0.0379 0.0379 0.0379 0.0379 0.0523 0.0502 0.0494 0.0492 0.0490 0.0488 0.0487 0.0487 0.0487 0.0486 0.0486 0.0486 0.0486 0.0738 0.0725 0.0720 0.0719 0.0718 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.0716 0.002 0.005 0.01 0.02 0.05 The error can now be computed e/D = Re 1.00E+04 2.50E+04 5.00E+04 7.50E+04 1.00E+05 2.50E+05 5.00E+05 7.50E+05 1.00E+06 5.00E+06 1.00E+07 5.00E+07 1.00E+08 0 0.0001 0.0002 0.0005 0.001 0.002 0.005 0.01 0.02 0.05 0.01% 0.63% 0.85% 0.90% 0.92% 0.84% 0.70% 0.59% 0.50% 0.07% 0.35% 1.02% 1.31% 0.15% 0.88% 1.19% 1.30% 1.34% 1.33% 1.16% 0.99% 0.86% 0.17% 0.00% 0.16% 0.18% 0.26% 1.02% 1.32% 1.40% 1.42% 1.25% 0.93% 0.72% 0.57% 0.01% 0.09% 0.18% 0.19% 0.46% 1.20% 1.38% 1.35% 1.28% 0.85% 0.48% 0.30% 0.20% 0.11% 0.15% 0.19% 0.20% Error (%) 0.64% 0.73% 1.22% 1.03% 1.21% 0.84% 1.07% 0.65% 0.94% 0.52% 0.47% 0.16% 0.19% 0.00% 0.07% 0.07% 0.01% 0.10% 0.15% 0.18% 0.18% 0.19% 0.20% 0.20% 0.20% 0.20% 0.55% 0.51% 0.28% 0.16% 0.09% 0.07% 0.13% 0.16% 0.17% 0.19% 0.20% 0.20% 0.20% 0.19% 0.11% 0.00% 0.06% 0.09% 0.15% 0.18% 0.18% 0.19% 0.20% 0.20% 0.20% 0.20% 0.17% 0.14% 0.16% 0.17% 0.18% 0.19% 0.20% 0.20% 0.20% 0.20% 0.20% 0.20% 0.20% 0.43% 0.29% 0.24% 0.23% 0.22% 0.21% 0.20% 0.20% 0.20% 0.20% 0.20% 0.20% 0.20% The maximum discrepancy is 1.42% at Re = 100,000 and e/D = 0.0002 0.100 f 0.010 e/D = 0 e/D = 0.0001 e/D = 0.0002 e/D = 0.0005 e/D = 0.001 e/D = 0.002 e/D = 0.005 e/D = 0.01 e/D = 0.02 e/D = 0.05 0.001 1E+04 1E+05 1E+06 1E+07 1E+08 Re Problem 8.83 [2] Problem 8.84 [2] Problem 8.85 [2] Given: Find: Solution: Basic equations Flow through gradual contraction Pressure after contraction; compare to sudden contraction 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlm ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 hlm = K⋅ V2 2 Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal 2 ⎛ D2 ⎞ ⎛ 2.5 ⎞ = 0.25 we find from Table 8.3 = 0.27 =⎜ For an included angle of 120 and an area ratio K ⎟ =⎜ ⎟ A1 ⎝5⎠ ⎝ D1 ⎠ 2 2 2 ⎛p V2 ⎜ 1 V1 ⎞ ⎛ p2 V2 ⎞ ⎟⎜ ⎟ 4⋅ Q Hence the energy equation becomes ⎜ with + − + = K⋅ V1 = 2 ρ 2 ⎟ ⎜ρ 2⎟ 2 ⎝ ⎠⎝ ⎠ π⋅ D o A2 2 V2 = 4⋅ Q π⋅ D 2 2 1 ρ 8⋅ ρ⋅ Q ⎡ ( 1 + K) 1⎤ 2 2 p2 = p1 − ⋅ ⎡( 1 + K) ⋅ V2 − V1 ⎤ = p2 − ⋅⎢ − ⎥ ⎣ ⎦ 2 4 4 2 ⎢ D2 π D1 ⎥ ⎣ ⎦ 2 3 3 8 kg ⎡ 0.003⋅ mm ⎛ 1⋅ m ⎞ ⎤ 3N ⎡ p2 = 200 × 10 ⋅ − × 1000⋅ ×⎢ ⋅⎜ ⎟ ⎥ × ⎢( 1 + 0.27) × 2 2 3⎣ s ⎝ 1000⋅ mm ⎠ ⎦ 2 1 ( 0.025⋅ m) 4 − m π m ⎣ ⎤ × N⋅ s 4⎥ kg⋅ m ( 0.05⋅ m) ⎦ 1 2 p2 = 200⋅ kPa No change because the flow rate is miniscule! Repeating the above analysis for an included angle of 180o (sudden contraction) K = 0.41 2 3 3 8 kg ⎡ 0.003⋅ mm ⎛ 1⋅ m ⎞ ⎤ 3N ⎡ p2 = 200 × 10 ⋅ − × 1000⋅ ×⎢ ⋅⎜ ⎟ ⎥ × ⎢( 1 + 0.41) × 2 2 3⎣ s ⎝ 1000⋅ mm ⎠ ⎦ m π m ⎣ ⎤ × N⋅ s − 4 4⎥ kg m ( 0.025⋅ m) ( 0.05⋅ m) ⎦ 1 1 3 2 p2 = 200⋅ kPa No change because the flow rate is miniscule! m s 3 The flow rate has a typo: it is much too small, and should be Q = 0.003⋅ 2 3 8 kg ⎛ 0.003⋅ m ⎞ 3N ⎟ × ⎡( 1 + 0.27) × p2 = 200 × 10 ⋅ − × 1000⋅ ×⎜ 2 2 3⎝ s ⎠⎢ not 1 1 Q = 0.003⋅ 2 mm s m π m ⎣ ( 0.025⋅ m) 4 − ⎤ N⋅ s ⎥ × kg⋅ m 4 ( 0.05⋅ m) ⎦ K = 0.41 p2 = 177⋅ kPa Repeating the above analysis for an included angle of 180o (sudden contraction) 2 3 8 kg ⎛ 0.003⋅ m ⎞ 3N ⎟ × ⎡( 1 + 0.41) × p2 = 200 × 10 ⋅ − × 1000⋅ ×⎜ 2 2 3⎝ s ⎠⎢ 1 ( 0.025⋅ m) 4 − m π m ⎣ ⎤ N⋅ s ⎥ × kg⋅ m 4 ( 0.05⋅ m) ⎦ 1 2 p2 = 175⋅ kPa There is slightly more loss in the sudden contraction Problem 8.86 [2] Given: Find: Solution: Flow through sudden expansion Inlet speed; Volume flow rate 2 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ Basic equations ⎜ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlm 2 2 ⎝ρ ⎠ ⎝ρ ⎠ hlm = K⋅ V1 2 Q = V⋅ A Δp = ρH2O⋅ g⋅ Δh Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal Hence the energy equation becomes 2 2 2 ⎛p V1 ⎜ 1 V1 ⎞ ⎛ p2 V2 ⎞ ⎟⎜ ⎟ ⎜ ρ + 2 ⎟ − ⎜ ρ + 2 ⎟ = K⋅ 2 ⎝ ⎠⎝ ⎠ A1 From continuity V2 = V1⋅ = V1⋅ AR A2 Hence 2 2 2 2 ⎛p V1 ⎜ 1 V1 ⎞ ⎛ p2 V1 ⋅ AR ⎞ ⎟⎜ ⎟ ⎜ ρ + 2 ⎟ −⎜ ρ + ⎟ = K⋅ 2 2 ⎝ ⎠⎝ ⎠ Solving for V1 V1 = ρ⋅ 1 − AR − K ( 2⋅ p2 − p1 2 ( ) ) 2 ⎛ D1 ⎞ 75 ⎞ AR = ⎜ =⎛ ⎟ ⎜ ⎟ = 0.111 ⎝ 225 ⎠ ⎝ D2 ⎠ 2 2 so from Fig. 8.14 K = 0.8 Also kg m 5 N⋅ s p2 − p1 = ρH2O⋅ g⋅ Δh = 1000⋅ × 9.81⋅ × ⋅m × = 49.1⋅ Pa 3 2 1000 kg⋅ m m s V1 = 2 × 49.1⋅ N m 2 Hence × 2 m × 1.23⋅ kg 3 (1 − 0.1112 − 0.8) 1 × 2 kg⋅ m N⋅ s 2 V1 = 20.6 m s m s 3 Q = V1⋅ A1 = π⋅ D 1 4 ⋅ V1 Q= π ⎛ 75 m ⎞ ×⎜ ⋅ m⎟ × 20.6⋅ 4 ⎝ 1000 ⎠ s Q = 0.0910⋅ Q = 5.46⋅ m min 3 Problem 8.87 [2] Given: Find: Solution: Flow through sudden contraction Volume flow rate 2 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ Basic equations ⎜ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlm 2 2 ⎝ρ ⎠ ⎝ρ ⎠ hlm = K⋅ V2 2 Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal Hence the energy equation becomes 2 2 2 ⎛p V2 ⎜ 1 V1 ⎞ ⎛ p2 V2 ⎞ ⎟⎜ ⎟ ⎜ ρ + 2 ⎟ − ⎜ ρ + 2 ⎟ = K⋅ 2 ⎝ ⎠⎝ ⎠ A2 From continuity V1 = V2⋅ = V2⋅ AR A1 Hence 2 2⎞ 2 2 ⎛p V2 ⎜ 1 V2 ⋅ AR ⎟ ⎛ p2 V2 ⎞ ⎜ ⎟ ⎜ρ + ⎟ − ⎜ ρ + 2 ⎟ = K⋅ 2 2 ⎝ ⎠⎝ ⎠ Solving for V2 V2 = ρ⋅ 1 − AR + K 2 × 0.5⋅ lbf in 2 ( 2⋅ p1 − p2 2 ( ) ) 2 2 ⎛ D2 ⎞ ⎛ 1⎞ AR = ⎜ ⎟ = ⎜ ⎟ = 0.25 ⎝ 2⎠ ⎝ D1 ⎠ 2 so from Fig. 8.14 K = 0.4 Hence V2 = ×⎛ ⎜ 2 12⋅ in ⎞ ⎝ 1⋅ ft ⎠ ⋅ V2 ⎟× ft 3 1.94⋅ slug × (1 − 0.252 + 0.4) 2 1 × slug⋅ ft lbf ⋅ s 2 3 V2 = 7.45⋅ ft s 3 Q = V2⋅ A2 = π⋅ D 2 4 π ⎛1 ⎞ ft Q= × ⎜ ⋅ ft⎟ × 7.45⋅ 4 ⎝ 12 ⎠ s ft Q = 0.0406⋅ s ft Q = 2.44⋅ min Q = 18.2 gpm Problem 8.88 [2] Given: Find: Solution: Given data Data on a pipe sudden contraction Theoretical calibration constant; plot D1 = 400⋅ mm D2 = 200⋅ mm The governing equations between inlet (1) and exit (2) are 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α1⋅ + g⋅ z1⎟ − ⎜ + α2⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ (8.29) where hl = K⋅ V2 2 2 (8.40a) Hence the pressure drop is (assuming α = 1) 2 ⎛V 2 V 2 V2 ⎞ ⎜2 ⎟ 1 Δp = p1 − p2 = ρ⋅ ⎜ − + K⋅ 2 2⎟ ⎝2 ⎠ For the sudden contraction V1⋅ π π 2 2 ⋅ D1 = V2⋅ ⋅ D2 = Q 4 4 2 or ⎛ D1 ⎞ V2 = V1⋅ ⎜ ⎟ ⎝ D2 ⎠ so 2 4 ⎤ ρ⋅ V1 ⎡⎛ D1 ⎞ ⎢ ⎥ Δp = ⋅⎜ ( 1 + K) − 1 ⎢D⎟ ⎥ 2 ⎣⎝ 2 ⎠ ⎦ For the pressure drop we can use the manometer equation Δp = ρ⋅ g⋅ Δh 2 4 ⎤ ρ⋅ V1 ⎡⎛ D1 ⎞ ⎢ ⎥ ρ⋅ g⋅ Δh = ⋅⎜ ( 1 + K) − 1 ⎟ ⎥ 2 ⎢ D2 Hence ⎣⎝ ⎠ ⎦ In terms of flow rate Q ρ ρ⋅ g⋅ Δh = ⋅ 2 ⎤ ⎡⎛ D ⎞ 4 ⎢1 ⎥ ⋅⎜ ⎟ ( 1 + K) − 1⎥ 2⎢ D ⎦ ⎛ π ⋅ D 2⎞ ⎣⎝ 2 ⎠ ⎜ ⎟ 4 1⎠ ⎝ Q 2 2 or ⎤ ⎡⎛ D ⎞ 4 ⎢1 ⎥ g⋅ Δh = ⋅⎜ ⎟ ( 1 + K) − 1⎥ 2 4⎢ D π ⋅ D1 ⎣⎝ 2 ⎠ ⎦ 8⋅ Q Q = k⋅ Δh 2 4 Hence for flow rate Q we find where k= g⋅ π ⋅ D1 ⎤ ⎡⎛ D ⎞ 4 ⎢1 ⎥ 8⋅ ⎜ ( 1 + K) − 1 ⎢D⎟ ⎥ ⎣⎝ 2 ⎠ ⎦ 2 For K, we need the aspect ratio AR ⎛ D2 ⎞ AR = ⎜ ⎟ ⎝ D1 ⎠ K = 0.4 AR = 0.25 From Fig. 8.15 Using this in the expression for k, with the other given values g⋅ π ⋅ D1 4 2 4 5 2 k= 8⋅ ⎜ ⎢ ⎤ ⎡⎛ D ⎞ ⎢1 ⎥ ⎟ ( 1 + K) − 1⎥ ⎣⎝ D2 ⎠ ⎦ = 0.12⋅ m s For Δh in mm and Q in L/min L min k = 228 1 2 mm The plot of theoretical Q versus flow rate Δh is shown in the associated Excel workbook D1 = D1 = K= k= 400 200 0.4 228 mm mm The values for Δh are quite low; this would not be a good meter L/min/mm1/2 it is not sensitive enough. In addition, it is non-linear. Δh (mm) Q (L/min) 0.010 23 0.020 32 0.030 40 0.040 46 0.050 51 0.075 63 0.100 72 0.150 88 0.200 102 0.250 114 0.300 125 0.400 144 0.500 161 0.600 177 0.700 191 0.767 200 Calibration Curve for a Sudden Contraction Flow Meter 1000 Q (L/mm) 100 10 0.01 0.10 Δh (mm) 1.00 Problem 8.89 [3] Given: Find: Solution: Contraction coefficient for sudden contraction Expression for minor head loss; compare with Fig. 8.15; plot We analyse the loss at the "sudden expansion" at the vena contracta The governing CV equations (mass, momentum, and energy) are Assume: 1) Steady flow 2) Incompressible flow 3) Uniform flow at each section 4) Horizontal: no body force 5) No shaft work 6) Neglect viscous friction 7) Neglect gravity Vc⋅ Ac = V2⋅ A2 pc⋅ A2 − p2⋅ A2 = Vc⋅ −ρ⋅ Vc⋅ Ac + V2⋅ ρ⋅ V2⋅ A2 pc − p2 = ρ⋅ Vc⋅ Ac A2 ⋅ V2 − Vc (1) The mass equation becomes The momentum equation becomes ( ) ( ) (2) or (using Eq. 1) ( ) ) The energy equation becomes pc p2 ⎛ ⎛ 2⎞ 2⎞ Qrate = ⎜ uc + + Vc ⎟ ⋅ −ρ⋅ Vc⋅ Ac + ⎜ u2 + + V2 ⎟ ⋅ ρ⋅ V2⋅ A2 ρ ρ ⎝ ⎠ ⎝ ⎠ ( ( ) or (using Eq. 1) hlm = u2 − uc − = mrate Qrate Vc − V2 2 2 2 + pc − p2 ρ (3) Combining Eqs. 2 and 3 hlm = Vc − V2 2 2 2 + Vc⋅ Ac A2 ⋅ V2 − Vc ( ) 2 2 Vc ⎡ ⎛ V2 ⎞ ⎤ ⎤ ⎢ ⎥ 2 Ac ⎡⎛ V2 ⎞ hlm = ⋅ 1−⎜ ⎟ ⎥ + Vc ⋅ ⋅ ⎢⎜ ⎟ − 1⎥ ⎢ 2 Vc A2 Vc ⎣ ⎝ ⎠⎦ ⎣⎝ ⎠ ⎦ From Eq. 1 Cc = Ac A2 Vc = 2 V2 Vc Hence hlm = 2 2 ⋅ ⎛ 1 − Cc ⎞ + Vc ⋅ Cc⋅ Cc − 1 ⎝ ⎠ 2 ( ) hlm = Vc 2 Vc 2 2 ⋅ ⎛ 1 − Cc + 2⋅ Cc − 2⋅ Cc⎞ ⎝ 2 2 ⎠ (4) 2 hlm = ⋅ 1 − Cc 2 ( ) 2 But we have hlm = K⋅ V2 2 Vc ⎛ V2 ⎞ Vc 2 = K⋅ ⋅⎜ ⋅ Cc ⎟ = K⋅ 2 2 Vc 2 2 2 (5) ⎝ ⎠ Hence, comparing Eqs. 4 and 5 (1 − Cc) K= Cc 2 2 So, finally 1 ⎞ K=⎛ ⎜ C − 1⎟ ⎝c ⎠ 2 where ⎛ A2 ⎞ Cc = 0.62 + 0.38⋅ ⎜ ⎟ ⎝ A1 ⎠ 3 This result, and the curve of Fig. 8.15, are shown in the associated Excel workbook. The agreement is reasonable. The CV analysis leads to A 2/A 1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 K CV 0.376 0.374 0.366 0.344 0.305 0.248 0.180 0.111 0.052 0.013 0.000 K Fig. 8.15 0.50 0.40 0.30 0.20 Loss Coefficient for a Sudden Contraction 1.0 0.8 K 0.5 0.3 0.0 0.00 Theoretical Curve Fig. 8.15 0.10 0.01 0.00 (Data from Fig. 8.15 is "eyeballed") Agreement is reasonable 0.25 0.50 Area Ratio AR 0.75 1.00 Problem 8.90 [2] Given: Find: Solution: Flow through short pipe Volume flow rate; How to improve flow rate 2 2 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ Basic equations ⎜ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlT 2 2 ⎝ρ ⎠ ⎝ρ ⎠ V2 L V2 hlT = hl + hlm = f ⋅ ⋅ + K⋅ D2 2 Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) L << so ignore hl 5) Reentrant Hence between the free surface (Point 1) and the exit (2) the energy equation becomes V1 2 2 2 2 + g⋅ z1 − V2 2 = K⋅ V2 2 A2 From continuity V1 = V2⋅ A1 Hence V2 ⎛ A2 ⎞ V2 V2 ⋅⎜ = K⋅ ⎟ + g⋅ h − 2 A1 2 2 2 2 2 2 ⎝ ⎠ Solving for V2 V2 = 2⋅ g⋅ h 2 ⎡ ⎛ A2 ⎞ ⎤ ⎢ ⎥ ⎢1 + K − ⎜ A ⎟ ⎥ ⎣ ⎝ 1⎠ ⎦ and from Table 8.2 K = 0.78 Hence V2 = 2 × 9.81⋅ m s 2 × 1⋅ m × 1 ⎡ 350 ⎞ ⎢1 + 0.78 − ⎛ ⎜ ⎟⎥ 3500 ⎠ ⎦ ⎣ ⎝ 2⎤ V2 = 3.33 m s Q = V2⋅ A2 Q = 3.33⋅ m 1⋅ m ⎞ 2 × 350⋅ mm × ⎛ ⎜ ⎟ s ⎝ 1000⋅ mm ⎠ 2 Q = 1.17 × 10 3 −3m s Q = 0.070 m min 3 The flow rate could be increased by (1) rounding the entrance and/or (2) adding a diffuser (both somewhat expensive) Problem 8.91 [3] Problem 8.92 [2] Problem 8.93 [3] Given: Find: Solution: Basic equations Flow out of water tank Volume flow rate using hole; Using short pipe section; Using rounded edge 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlT ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 2 V2 L V2 hlT = hl + hlm = f ⋅ ⋅ + K⋅ D2 2 Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl << 5) L << so hl = 0 Hence for all three cases, between the free surface (Point 1) and the exit (2) the energy equation becomes g⋅ z1 − Solving for V2 V2 = V2 2 2 2 = K⋅ V2 2 2⋅ g⋅ h ( 1 + K) Kpipe = 0.78 Kround = 0.28 V2 = 20.7⋅ 0.5 ⋅ ft ⎟ × ft s 60⋅ s 1⋅ min ft s Q = 11.6⋅ gpm Q = 12.7⋅ gpm for a short pipe (rentrant) From Table 8.2 Khole = 0.5 for a hole (assumed to be square-edged) r 0.01⋅ in Also, for a rounded edge so from Table 8.2 = = 0.02 D 0.5⋅ in Hence for the hole V2 = 2 × 32.2⋅ ft s 2 × 10⋅ ft × 1 ( 1 + 0.5) ft s × π 4 1 ( 1 + 0.78) 2 2 Q = V2⋅ A2 Q = 20.7⋅ ×⎛ ⎜ ⎝ 12 ⎞ ⎠ 7.48⋅ gal 1⋅ ft 3 × Hence for the pipe V2 = 2 × 32.2⋅ ft s 2 × 10⋅ ft × V2 = 19.0⋅ Q = V2⋅ A2 Q = 19.0⋅ ft π ⎛ 0.5 ⎞ 7.48⋅ gal 60⋅ s × ×⎜ ⋅ ft ⎟ × × 3 s 4 ⎝ 12 ⎠ 1⋅ min 1⋅ ft Hence the change in flow rate is 11.6 − 12.7 = −1.1⋅ gpm The pipe leads to a LOWER flow rate ft s 2 Hence for the rounded V2 = 2 × 32.2⋅ × 10⋅ ft × 1 ( 1 + 0.28) 2 V2 = 22.4⋅ ft s Q = 13.7⋅ gpm Q = V2⋅ A2 Q = 22.4⋅ ft π ⎛ 0.5 ⎞ 7.48⋅ gal 60⋅ s × ×⎜ ⋅ ft⎟ × × 3 s 4 ⎝ 12 ⎠ 1⋅ min 1⋅ ft Hence the change in flow rate is 13.7 − 12.7 = 1.0⋅ gpm The rounded edge leads to a HIGHER flow rate Problem 8.94 [2] Given: Find: Solution: Given data Data on inlet and exit diameters of diffuser Minimum lengths to satisfy requirements D1 = 100⋅ mm D2 = 150⋅ mm The governing equations for the diffuser are V1 2 1 AR 2 2 hlm = K⋅ and Cpi = 1 − = Cpi − Cp ⋅ 2 ( ) V1 2 (8.44) (8.42) Combining these we obtain an expression for the loss coefficient K K = 1− 1 AR 2 2 − Cp (1) The area ratio AR is ⎛ D2 ⎞ AR = ⎜ ⎟ ⎝ D1 ⎠ AR = 2.25 The pressure recovery coefficient Cp is obtained from Eq. 1 above once we select K; then, with Cp and AR specified, the minimum value of N/R1 (where N is the length and R1 is the inlet radius) can be read from Fig. 8.15 (a) K = 0.2 Cp = 1 − N = 5.5 R1 N = 5.5⋅ R1 (b) K = 0.35 Cp = 1 − N =3 R1 N = 3⋅ R 1 N = 150⋅ mm 1 AR From Fig. 8.15 2 1 AR 2 −K D1 2 Cp = 0.602 From Fig. 8.15 R1 = R1 = 50⋅ mm N = 275⋅ mm −K Cp = 0.452 Problem 8.95 [3] Given: Find: Solution: Basic equations Data on geometry of conical diffuser; flow rate Static pressure rise; loss coefficient p2 − p1 1 2 ⋅ ρ⋅ V 1 2 Cp = (8.41) hlm = K⋅ V1 2 2 = Cpi − Cp ⋅ 2 ( ) V1 2 (8.44) Cpi = 1 − 1 AR 2 (8.42) Given data D1 = 2⋅ in D2 = 3.5⋅ in N = 6⋅ in (N = length) Q = 750⋅ gpm From Eq. 8.41 1 2 Δp = p2 − p1 = ⋅ ρ⋅ V1 ⋅ Cp 2 (1) K = 1− 1 AR 2 Combining Eqs. 8.44 and 8.42 we obtain an expression for the loss coefficient K − Cp (2) The pressure recovery coefficient Cp for use in Eqs. 1 and 2 above is obtained from Fig. 8.15 once compute AR and the dimensionless length N/R1 (where R1 is the inlet radius) ⎛ D2 ⎞ The aspect ratio AR is AR = ⎜ ⎟ ⎝ D1 ⎠ R1 = D1 2 2 AR = ⎛ ⎜ ⎟ ⎝2⎠ Hence 3.5 ⎞ 2 AR = 3.06 N =6 R1 R1 = 1⋅ in From Fig. 8.15, with AR = 3.06 and the dimensionless length N/R1 = 6, we find Cp = 0.6 To complete the calculations we need V1 1⋅ ft 1⋅ min ⎛ 1 ⎞ × × π π min 7.48⋅ gal 60⋅ s ⎜ 2 ⎟ 2 ⋅D ⎜ ⋅ ft ⎟ 41 ⎝ 12 ⎠ 1 2 We can now compute the pressure rise and loss coefficient from Eqs. 1 and 2 Δp = ⋅ ρ⋅ V1 ⋅ Cp 2 V1 = Q V1 = 4 × 750⋅ gal × 1 slug ⎛ ft ⎞ lbf ⋅ s 1⋅ ft ⎞ × 1.94⋅ × ⎜ 76.6⋅ ⎟ × 0.6 × ×⎛ ⎜ ⎟ 3 2 s⎠ slug⋅ ft ⎝ 12⋅ in ⎠ ⎝ ft 1 1 K = 1− − Cp K = 1− − 0.6 2 2 AR 3.06 Δp = 2 2 2 3 2 V1 = 76.6⋅ ft s Δp = 23.7⋅ psi K = 0.293 Problem 8.96 [4] Problem 8.97 [3] Given: Find: Solution: Sudden expansion Expression for minor head loss; compare with Fig. 8.15; plot The governing CV equations (mass, momentum, and energy) are Assume: 1) Steady flow 2) Incompressible flow 3) Uniform flow at each section 4) Horizontal: no body force 5) No shaft work 6) Neglect viscous friction 7) Neglect gravity V1⋅ A1 = V2⋅ A2 p1⋅ A2 − p2⋅ A2 = V1⋅ −ρ⋅ V1⋅ A1 + V2⋅ ρ⋅ V2⋅ A2 p1 − p2 = ρ⋅ V1⋅ A1 A2 ⋅ V2 − V1 (1) The mass equation becomes The momentum equation becomes ( ) ( ) (2) or (using Eq. 1) ( ) ) The energy equation becomes p1 p2 ⎛ ⎛ 2⎞ 2⎞ Qrate = ⎜ u1 + + V1 ⎟ ⋅ −ρ⋅ V1⋅ A1 + ⎜ u2 + + V2 ⎟ ⋅ ρ⋅ V2⋅ A2 ρ ρ ⎝ ⎠ ⎝ ⎠ ( ( ) or (using Eq. 1) hlm = u2 − u1 − = mrate V1 − V2 2 2 2 Qrate V1 − V2 2 2 2 + p1 − p2 ρ (3) Combining Eqs. 2 and 3 hlm = + V1⋅ A1 A2 ⋅ V2 − V1 ( ) 2 2 V1 ⎡ ⎛ V2 ⎞ ⎤ ⎤ ⎢ ⎥ 2 A1 ⎡⎛ V2 ⎞ hlm = ⋅ 1−⎜ ⎟ ⎥ + V1 ⋅ ⋅ ⎢⎜ ⎟ − 1⎥ ⎢ V1 2 A2 V1 ⎣ ⎝ ⎠⎦ ⎣⎝ ⎠ ⎦ From Eq. 1 AR = A1 A2 V1 2 V1 2 = 2 V2 V1 Hence hlm = ⋅ 1 − AR 2 ( ( 2 ) + V12⋅ AR⋅ (AR − 1) 2 hlm = ⋅ 1 − AR + 2⋅ AR − 2⋅ AR 2 2 V1 2 2 ) hlm = K⋅ Finally V1 2 = ( 1 − AR) ⋅ 2 2 K = ( 1 − AR) This result, and the curve of Fig. 8.15, are shown in the associated Excel workbook. The agreement is excellent From the CV analysis AR 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 K CV 1.00 0.81 0.64 0.49 0.36 0.25 0.16 0.09 0.04 0.01 0.00 K Fig. 8.15 1.00 0.60 Loss Coefficient for a Sudden Expansion 1.0 0.8 Theoretical Curve Fig. 8.15 0.38 0.25 0.10 K 0.5 0.3 0.01 0.00 (Data from Fig. 8.15 is "eyeballed") Agreement is excellent 0.0 0.00 0.25 0.50 Area Ratio AR 0.75 1.00 Problem 8.98 [3] Problem 8.99 [2] Given: Find: Solution: Sudden expansion Expression for upstream average velocity The governing equation is 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ + α1⋅ + g⋅ z1⎟ − ⎜ + α2⋅ + g⋅ z2⎟ = hlT ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ (8.29) V hlT = hl + K⋅ 2 Assume: The mass equation is 2 1) Steady flow 2) Incompressible flow 3) hl = 0 4) α1 = α2 = 1 5) Neglect gravity V1⋅ A1 = V2⋅ A2 V2 = AR⋅ V1 p1 ρ V1 2 2 so A1 V2 = V1⋅ A2 (1) Equation 8.29 becomes + = p1 ρ + V1 2 2 + K⋅ V1 2 2 or (using Eq. 1) V1 p2 − p1 Δp 2 = = ⋅ 1 − AR − K ρ ρ 2 V1 = 2⋅ Δp 2 2 ( ) Solving for V1 ρ⋅ 1 − AR − K 2⋅ Δp ( ) 2 If the flow were frictionless, K = 0, so Vinviscid = ρ⋅ 1 − AR ( ) < V1 Hence the flow rate indicated by a given Δp would be lower V1 2 2 If the flow were frictionless, K = 0, so Δpinvscid = V1 2 2 ⋅ 1 − AR ( 2 ) compared to Δp = ⋅ 1 − AR − K ( 2 ) Hence a given flow rate would generate a larger Δp for inviscid flow Problem 8.100 [4] Flow Nozzle Short pipe Given: Find: Solution: Basic equations Flow out of water tank through a nozzle Change in flow rate when short pipe section is added; Minimum pressure; Effect of frictionless flow 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlT ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 2 V2 L V2 hlT = hl + hlm = f ⋅ ⋅ + K⋅ 2 D2 Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl << 5) L << so hl = 0 Hence for the nozzle case, between the free surface (Point 1) and the exit (2) the energy equation becomes g⋅ z1 − Solving for V2 V2 = V2 2 2 2 = Knozzle⋅ 2 2⋅ g⋅ z1 V2 (1 + Knozzle) Knozzle = 0.28 V2 = 15.9⋅ 2 For a rounded edge, we choose the first value from Table 8.2 Hence V2 = 2 × 32.2⋅ ft s 2 × 5⋅ ft × 1 ( 1 + 0.28) ft π ft s Q = 9.73⋅ gpm Q = 0.0217 ft 3 Q = V2⋅ A2 0.5 ⎞ 7.48⋅ gal 60⋅ s Q = 15.9⋅ × × ⎛ × ⎜ ⋅ ft⎟ × 3 s 4 ⎝ 12 ⎠ 1⋅ min 1⋅ ft 2 2 s When a small piece of pipe is added the energy equation between the free surface (Point 1) and the exit (3) becomes g⋅ z1 − From continuity V3 2 2 = Knozzle⋅ 2 A2 V2 + Ke⋅ V2 2 V3 = V2⋅ = V2⋅ AR A3 V2 = 2⋅ g⋅ z1 Solving for V2 ⎛ AR2 + K ⎞ nozzle + Ke⎠ ⎝ 2 ⎛ D2 ⎞ 1⎞ AR = =⎜ = ⎛ ⎟ = 0.25 ⎟ ⎜ A3 ⎝ 2⎠ ⎝ D3 ⎠ We need the AR for the sudden expansion A2 2 From Fig. 8.14 for AR = 0.25 Ke = 0.6 Hence V2 = 2 × 32.2⋅ ft s 2 × 5⋅ ft × (0.25 1 2 + 0.28 + 0.6 ) 2 V2 = 18.5⋅ ft s Q = 0.0252 ft s 3 Q = V2⋅ A2 Q = 18.5⋅ ft π ⎛ 0.5 ⎞ 7.48⋅ gal 60⋅ s × ×⎜ ⋅ ft⎟ × × 3 s 4 ⎝ 12 ⎠ 1⋅ min 1⋅ ft Q = 11.32⋅ gpm Comparing results we see the flow increases from 0.0217 ft3/s to 0.0252 ft3/s ΔQ 0.0252 − 0.0217 = = 16.1⋅ % Q 0.0217 The flow increases because the effect of the pipe is to allow an exit pressure at the nozzle LESS than atmospheric! The minimum pressure point will now be at Point 2 (it was atmospheric before adding the small pipe). The energy equation between 1 and 2 is 2 2 ⎛p V2 ⎜ 2 V2 ⎞ ⎟ g⋅ z1 − ⎜ + = Knozzle⋅ 2⎟ 2 ⎝ρ ⎠ Solving for p2 2 ⎡ ⎤ V2 ⎢ ⎥ p2 = ρ⋅ ⎢g⋅ z1 − ⋅K + 1)⎥ 2 ( nozzle ⎣ ⎦ Hence p2 = 1.94⋅ slug ft 3 × ⎢32.2⋅ ⎡ ⎢ ⎣ ft s 2 × 5⋅ ft − ⎤ lbf ⋅ s ft ⎞ × ⎛ 18.5⋅ ⎟ × ( 0.28 + 1)⎥ × ⎜ ⎥ slug⋅ ft 2⎝ s⎠ 1 2 2 ⎦ p2 = −113 2⋅ g⋅ z1 lbf ft 2 p2 = −0.782 psi If the flow were frictionless the the two loss coeffcients would be zero. Instead of V2 = ⎛ AR2 + K ⎞ nozzle + Ke⎠ ⎝ 2⋅ g⋅ z1 AR 2 We'd have If V2 is larger, then p2, through Bernoulli, would be lower (more negative) V2 = which is larger Problem 8.101 [2] Problem 8.102 [4] Given data: L = 15.3 m D = 3.18 mm K ent = 1.4 α= 2 Computed results: Tabulated or graphical data: ν = 1.00E-06 m2/s 3 ρ= 998 kg/m (Appendix A) Re = 2300 (Transition Re ) V = 0.723 m/s α= 1 (Turbulent) f = 0.0473 (Turbulent) d = 6.13 m Energy equation: (Using Solver ) (Vary d to minimize error in energy equation) Error 0.00% Left (m2/s) Right (m2/s) 59.9 59.9 Note that we used α = 1 (turbulent); using α = 2 (laminar) gives d = 6.16 m Problem 8.104 [2] Problem 8.105 [3] Given: Find: Solution: The given data is Data on a tube "Resistance" of tube for flow of kerosine; plot L = 100⋅ mm μ = 1.1 × 10 V = R⋅ I − 3 N⋅ s ⋅ 2 D = 0.3⋅ mm ρ = 0.82 × 990⋅ kg m 3 From Fig. A.2 and Table A.2 For an electrical resistor = 812⋅ kg m 3 (Kerosene) (1) m The governing equations for turbulent flow are 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ ⎜ ρ + α1⋅ 2 + g⋅ z1⎟ − ⎜ ρ + α2⋅ 2 + g⋅ z2⎟ = hl ⎝ ⎠⎝ ⎠ ⎞ ⎛e 2 ⎜D LV 1 2.51 ⎟ (8.34) hl = f ⋅ ⋅ = −2.0⋅ log⎜ + ⎟ D2 f ⎝ 3.7 Re⋅ f ⎠ (8.29) (8.37) Simplifying Eqs. 8.29 and 8.34 for a horizontal, constant-area pipe ⎛Q⎞ ⎜ π 2⎟ ⎜ ⋅D ⎟ 2 p1 − p2 LV L4 ⎠ = f⋅ ⋅ = f⋅ ⋅ ⎝ ρ D2 D 2 2 or Δp = 8⋅ ρ⋅ f ⋅ L π ⋅D 2 5 ⋅Q 2 (2) By analogy, current I is represented by flow rate Q, and voltage V by pressure drop Δp. Comparing Eqs. (1) and (2), the "resistance" of the tube is R= Δp 8⋅ ρ⋅ f ⋅ L⋅ Q = 25 Q π ⋅D The "resistance" of a tube is not constant, but is proportional to the "current" Q! Actually, the dependence is not quite linear, because f decreases slightly (and nonlinearly) with Q. The analogy fails! The analogy is hence invalid for Re > 2300 ρ⋅ Writing this constraint in terms of flow rate Flow rate above which analogy fails The plot of "resistance" versus flow rate is shown in the associated Excel workbook Q ⋅D π2 ⋅D 4 > 2300 μ or ρ⋅ V ⋅ D > 2300 μ or Q> 2300⋅ μ⋅ π⋅ D 4⋅ ρ Q = 7.34 × 10 3 −7m s Given data: L= D= 100 0.3 mm mm Tabulated or graphical data: μ= SG ker = ρw = 1.01E-03 0.82 990 N.s/m2 kg/m3 ρ= kg/m3 812 (Appendix A) Computed results: "R" (109 Pa/m3/s) 1133 1855 3085 4182 5202 6171 10568 18279 25292 31900 Q (m3/s) V (m/s) 1.0E-06 14.1 2.0E-06 28.3 4.0E-06 56.6 6.0E-06 84.9 8.0E-06 113.2 1.0E-05 141.5 2.0E-05 282.9 4.0E-05 565.9 6.0E-05 848.8 8.0E-05 1131.8 Re 3.4E+03 6.8E+03 1.4E+04 2.0E+04 2.7E+04 3.4E+04 6.8E+04 1.4E+05 2.0E+05 2.7E+05 f 0.0419 0.0343 0.0285 0.0257 0.0240 0.0228 0.0195 0.0169 0.0156 0.0147 The "resistance" is not constant; the analogy is invalid for turbulent flow "Resistance" of a Tube versus Flow Rate 1.E+06 1.E+04 "R" (109 Pa/m3/s) 1.E+02 1.E+00 1.0E-06 1.0E-05 3 Q (m /s) 1.0E-04 Given data: L= D= α= 100 10 1 Tabulated or graphical data: μ = 1.01E-03 m ρ= mm 998 (All flows turbulent) (Table A.8) K ent = 0.5 (Table 8.2) N.s/m2 kg/m3 (Square-edged) Computed results: Q (L/min) V (m/s) 1 2 3 4 5 6 7 8 9 10 0.2 0.4 0.6 0.8 1.1 1.3 1.5 1.7 1.9 2.1 Re 2.1E+03 4.2E+03 6.3E+03 8.4E+03 1.0E+04 1.3E+04 1.5E+04 1.7E+04 1.9E+04 2.1E+04 f 0.0305 0.0394 0.0350 0.0324 0.0305 0.0291 0.0280 0.0270 0.0263 0.0256 d (m ) 0.704 3.63 7.27 11.9 17.6 24.2 31.6 39.9 49.1 59.1 Required Reservoir Head versus Flow Rate 75 50 d (m) 25 0 0 2 4 6 Q (L/min) 8 10 12 Problem 8.107 [3] Given: Find: Solution: Basic equations Flow of oil in a pipe Percentage change in loss if diameter is reduced 2 LV hl = f ⋅ ⋅ D2 V= 64 f= Re V= Laminar 3 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2 Turbulent Here Q 4⋅ Q = 2 A π⋅ D V⋅ D ν 4 m ⎛1⎞ × 0.003⋅ ×⎜ ⎟ π s ⎝ 0.04⋅ m ⎠ m s × 0.04⋅ m × s V = 2.39 m s Then Re = Re = 2.39⋅ 0.00005⋅ m 2 2 2 Re = 1912 The flow is LAMINAR hl = f ⋅ LV ⋅ D2 2 hl = 64 L V ⋅⋅ Re D 2 ⎛ 2.39⋅ m ⎞ ⎜ ⎟ 64 25⋅ m s⎠ hl = × ×⎝ 2 1912 0.04⋅ m hl = 643⋅ ft s 2 2 When the diameter is reduced V= Q A = 4⋅ Q π⋅ D 2 V= 4 π × 0.003⋅ m 1⎞ ×⎛ ⎜ ⎟ s ⎝ 0.01⋅ m ⎠ s 3 2 V = 38.2 m s Re = V⋅ D ν Re = 38.2⋅ m s × 0.01⋅ m × 0.00005⋅ m e = 0.046⋅ mm 2 Re = 7640 The flow is TURBULENT For a steel pipe, from table 8.1 1 Given ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ LV ⋅ D2 2 f = 0.0389 hl = f ⋅ ⎛ 38.2⋅ m ⎞ ⎜ ⎟ 25⋅ m s⎠ hl = 0.0389 × ×⎝ 0.01⋅ m 2 2 hl = 7.64 × 10 ⋅ 2 5 ft 2 s 7.64 × 10 The increase in loss is 643 2 5 ft 2 s ft s 2 = 1188 2 This is a HUGH increase! As a percentage increase of 118800%. Hence choice of diameter is very important! The increase is because the diameter reduces by a factor of four and the velocity therefore increases by a factor of 16, and is squared! Problem 8.108 [2] Problem 8.109 [2] Problem 8.109 [3] Given: Data on reservoir/pipe system Find: Plot elevation as a function of flow rate; fraction due to minor losses Solution: L= D= e/D = K ent = K exit = 250 50 0.003 0.5 1.0 m mm 200 Required Head versus Flow Rate 150 Δz (m) 100 ν = 1.01E-06 m2/s Q (m /s) V (m/s) 0.0000 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030 0.0035 0.0040 0.0045 0.0050 0.0055 0.0060 0.0065 0.0070 0.0075 0.0080 0.0085 0.0090 0.0095 0.0100 0.000 0.255 0.509 0.764 1.02 1.27 1.53 1.78 2.04 2.29 2.55 2.80 3.06 3.31 3.57 3.82 4.07 4.33 4.58 4.84 5.09 3 Re 0.00E+00 1.26E+04 2.52E+04 3.78E+04 5.04E+04 6.30E+04 7.56E+04 8.82E+04 1.01E+05 1.13E+05 1.26E+05 1.39E+05 1.51E+05 1.64E+05 1.76E+05 1.89E+05 2.02E+05 2.14E+05 2.27E+05 2.40E+05 2.52E+05 f Δz (m) h lm /h lT 50 0.000 0.0337 0.562 0.882% 0.0306 2.04 0.972% 0.0293 4.40 1.01% 0.0286 7.64 1.04% 0.0282 11.8 1.05% 0.0279 16.7 1.07% 0.0276 22.6 1.07% 0.0275 29.4 1.08% 0.0273 37.0 1.09% 0.0272 45.5 1.09% 0.0271 54.8 1.09% 0.0270 65.1 1.10% 0.0270 76.2 1.10% 0.0269 88.2 1.10% 0.0269 101 1.10% 0.0268 115 1.11% 0.0268 129 1.11% 0.0268 145 1.11% 0.0267 161 1.11% 0.0267 179 1.11% 0 0.0000 0.0025 0.0050 Q (m3/s) 0.0075 0.0100 Minor Loss Percentage versus Flow Rate 1.2% 1.1% h lm /h lT 1.0% 0.9% 0.8% 0.0000 0.0025 0.0050 3 Q (m /s) 0.0075 0.0100 Problem 8.110 [2] Given: Find: Solution: Basic equations Flow from pump to reservoir Pressure at pump discharge 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlT ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 2 V1 L V1 hlT = hl + hlm = f ⋅ ⋅ + Kexit⋅ D2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V2 << Hence the energy equation between Point 1 and the free surface (Point 2) becomes 2 2 ⎛ p1 V2 ⎞ LV V ⎜ + ⎟ − ( g⋅ z2) = f ⋅ ⋅ + Kexit⋅ 2⎠ 2 D2 ⎝ρ 2 2 2 ⎛ V LV V⎞ p1 = ρ⋅ ⎜ g⋅ z2 − + f⋅ ⋅ + Kexit⋅ ⎟ 2 2⎠ D2 ⎝ Solving for p1 From Table A.7 (68oF) ρ = 1.94⋅ Re = slug ft 3 ν = 1.08 × 10 Re = 10⋅ ft s × − 5 ft 2 ⋅ s s 1.08 × 10 −5 2 V⋅ D ν 9 12 ⋅ ft × Re = 6.94 × 10 e = 0.0002 D 5 Turbulent ⋅ ft For commercial steel pipe e = 0.00015⋅ ft (Table 8.1) 1 so Flow is turbulent: Given For the exit Kexit = 1.0 slug ft 3 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + f = 0.0150 ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2 ⎛ LV⎞ so we find p1 = ρ⋅ ⎜ g⋅ z2 + f ⋅ ⋅ ⎟ D 2⎠ ⎝ 4⋅ mile 5280⋅ ft 1 ⎛ ft ⎞ × × × ⎜ 10⋅ ⎟ 0.75⋅ ft 1mile 2⎝ s⎠ 2⎤ p1 = 1.94⋅ × ⎢32.2⋅ ⎡ ⎢ ⎣ ft s 2 × 50⋅ ft + .0150 × ⎥ × lbf ⋅ s ⎥ slug⋅ ft ⎦ 2 p1 = 4.41 × 10 ⋅ 4 lbf 2 ft p1 = 306⋅ psi Problem 8.111 [3] Given: Find: Solution: Basic equations Flow through three different layouts Which has minimum loss 2 2 ⎛p ⎞⎜ ⎞ 2 V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ LV + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlT hlT = hl + hlm = f ⋅ ⋅ + ⎜ρ 2 2 D2 ⎝ ⎠ ⎝ρ ⎠ Minor ∑ ⎛ Le V 2 ⎞ ⎜f ⋅ ⋅ ⎟ ⎝ D 2⎠ 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore additional length of elbows For a flow rate of Q = 350⋅ L min V= 2 −6 m Q A = 4⋅ Q π⋅ D 2 V= 4 π × 350⋅ m s L min × 0.001⋅ m 1⋅ min ⎛ 1 ⎞ × ×⎜ ⎟ 1⋅ L 60⋅ s ⎝ 0.05⋅ m ⎠ s 1.01 × 10 −6 2 3 V = 2.97 m s 5 For water at 20oC ν = 1.01 × 10 ⋅ s Re = V⋅ D ν Re = 2.97⋅ × 0.05⋅ m × Re = 1.47 × 10 ⋅m Flow is turbulent. From Table 8.1 e = 0.15⋅ mm e −4 = 6.56 × 10 D f = 0.0201 Le D Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 5.25 + 2.5 ⋅ m p1 ρ p2 ρ 2 2 2 For Case (a) L= L = 5.81 m Le V LV ⋅ + 2⋅ f ⋅ ⋅ D2 D2 2 Two 45o miter bends (Fig. 8.16), for each = 13 Hence the energy equation is − = f⋅ Solving for Δp 2 Le ⎞ V ⎛L Δp = p1 − p2 = ρ⋅ f ⋅ ⋅ ⎜ + 2⋅ ⎟ 2 ⎝D D⎠ Δp = 1000⋅ kg m 3 × .0201 × ⎛ 2.97⋅ ⎜ ⎝ m⎞ ⎛ 5.81 + 2⋅ 13⎞ × N⋅ s ⎟ ×⎜ ⎟ s⎠ ⎝ 0.05 ⎠ kg⋅ m L = 7.75 m One standard 90o elbow (Table 8.4) 2 2 Δp = 25.2 kPa Le D For Case (b) L = ( 5.25 + 2.5) ⋅ m p1 p2 2 = 30 Hence the energy equation is Le V LV − = f⋅ ⋅ + f⋅ ⋅ D2 D2 ρ ρ 2 Solving for Δp 2 V ⎛ L Le ⎞ Δp = p1 − p2 = ρ⋅ f ⋅ ⋅⎜ + ⎟ 2 ⎝D D ⎠ Δp = 1000⋅ kg m 3 × .0201 × ⎜ 2.97⋅ ⎛ ⎝ m⎞ ⎛ 7.75 + 30⎞ × N⋅ s ⎟ ×⎜ ⎟ s⎠ ⎝ 0.05 ⎠ kg⋅ m Three standard 90o elbows, for each 2 2 Δp = 32.8 kPa Le D For Case (c) L = ( 5.25 + 2.5) ⋅ m p1 ρ p2 ρ 2 L = 7.75 m 2 = 30 Hence the energy equation is − = f⋅ Le V LV ⋅ + 3⋅ f ⋅ ⋅ D2 D2 Solving for Δp 2 Le ⎞ V ⎛L Δp = p1 − p2 = ρ⋅ f ⋅ ⋅ ⎜ + 3⋅ ⎟ 2 ⎝D D⎠ Δp = 1000⋅ kg m 3 × .0201 × ⎜ 2.97⋅ ⎛ ⎝ m⎞ ⎛ 7.75 + 3 × 30⎞ × N⋅ s ⎟ ×⎜ ⎟ s⎠ ⎝ 0.05 ⎠ kg⋅ m 2 2 Δp = 43.4 kPa Hence we conclude Case (a) is the best and Case (c) is the worst Problem 8.112 [2] Problem 8.113 [3] h LA LB Given: Find: Solution: Basic equations Pipe friction experiment Required average speed; Estimate feasibility of constant head tank; Pressure drop over 5 m 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ ⎜ ρ + α⋅ 2 + g⋅ z1⎟ − ⎜ ρ + α⋅ 2 + g⋅ z2⎟ = hlT ⎝ ⎠⎝ ⎠ 2 2 LA V A LB V B hlT = hA + hB = fA⋅ ⋅ + fB⋅ ⋅ DA 2 DB 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor losses We wish to have Hence, from ReB = 10 ReB = 5 V B⋅ D B ν 5 VB = 2 −6 m ReB⋅ ν DB 1 0.025⋅ m m s ×⎛ ⎜ 2.5 ⎞ 2 and for water at 20oC ν = 1.01 × 10 m s m s 2 −6 m ⋅ s VB = 10 × 1.01 × 10 We will also need ⋅ s × VB = 4.04 ⎛ DB ⎞ V A = V B⋅ ⎜ ⎟ ⎝ DA ⎠ VA⋅ DA ReA = ν 2 VA = 4.04⋅ ⎝5⎠ s 1.01 × 10 −6 ⎟ VA = 1.01 ReA = 1.01⋅ m s × 0.05⋅ m × ⋅m 2 ReA = 5 × 10 4 Both tubes have turbulent flow For PVC pipe (from Googling!) e = 0.0015⋅ mm For tube A Given ⎛e ⎞ ⎜D ⎟ A 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎜ 3.7 ReA⋅ fA ⎟ fA ⎝ ⎠ ⎛e ⎞ ⎜D ⎟ 1 B 2.51 ⎟ = −2.0⋅ log⎜ + ⎜ 3.7 ReB⋅ fB ⎟ fB ⎝ ⎠ fA = 0.0210 For tube B Given fB = 0.0183 Applying the energy equation between Points 1 and 3 g⋅ LA + h − ( ) VB 2 2 LA V A LB V B = fA⋅ ⋅ + fB⋅ ⋅ DA 2 DB 2 2 2 Solving for LA LA = VB ⎛ LB ⎞ ⋅ ⎜ 1 + fB⋅ ⎟ − g⋅ h 2 DB 2 ⎝ ⎠ ⎛ ⎜ ⎜g − ⎝ 2 fA VA ⎞ ⎟ ⋅ DA 2 ⎟ ⎠ LA = m⎞ 20 ⎞ m 1⎛ × ⎜ 4.04⋅ ⎟ × ⎛ 1 + 0.0183 × ⎜ ⎟ − 9.81⋅ 2 × 0.5⋅ m s⎠ 0.025 ⎠ 2⎝ ⎝ s 0.0210 m⎞ 1 9.81⋅ − × × ⎛ 1.01⋅ ⎟ ⎜ 2 2 s⎠ 0.05⋅ m ⎝ s m 2 2 LA = 12.8 m Most ceilings are about 3.5 m or 4 m, so this height is IMPRACTICAL Applying the energy equation between Points 2 and 3 2 2 2 ⎛p ⎜ 2 VB ⎞ ⎛ p3 VB ⎞ ⎟⎜ ⎟ L VB ⎜ ρ + 2 ⎟ − ⎜ ρ + 2 ⎟ = fB⋅ D ⋅ 2 ⎝ ⎠⎝ ⎠ B or 2 2 L VB Δp = ρ⋅ fB⋅ ⋅ DB 2 Δp = 29.9⋅ kPa 2 Δp = 1000⋅ kg m 3 × 0.0183 2 × 5⋅ m 0.025⋅ m × ⎛ 4.04⋅ ⎜ m⎞ ⎝ s⎠ ⎟× N⋅ s kg⋅ m Given data: L= D= 20 75 m mm Tabulated or graphical data: 0.26 mm (Table 8.1) μ = 1.00E-03 N.s/m2 ρ= kg/m3 999 (Appendix A) Gate valve L e/D = 8 Elbow L e/D = 30 (Table 8.4) e= Computed results: Q (m /s) V (m/s) 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.060 2.26 3.40 4.53 5.66 6.79 7.92 9.05 10.2 11.3 12.4 13.6 3 Re 1.70E+05 2.54E+05 3.39E+05 4.24E+05 5.09E+05 5.94E+05 6.78E+05 7.63E+05 8.48E+05 9.33E+05 1.02E+06 Δp (kPa) f 0.0280 28.3 0.0277 63.1 0.0276 112 0.0276 174 0.0275 250 0.0275 340 0.0274 444 0.0274 561 0.0274 692 0.0274 837 0.0274 996 Required Pressure Head for a Circuit 1200 1000 Δp (kPa) 800 600 400 200 0 0.00 0.01 0.02 0.03 Q (m /s) 3 0.04 0.05 0.06 0.07 Problem 8.115 [3] Given: Find: Solution: Basic equations Same flow rate in various ducts Pressure drops of each compared to round duct 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ Dh = 4⋅ A Pw e=0 (Smooth) Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor losses The energy equation simplifies to LV Δp = p1 − p2 = ρ⋅ f ⋅ ⋅ Dh 2 But we have V= Q A V = 1250⋅ 2 2 or Δp fV = ρ⋅ ⋅ L Dh 2 V = 20.8 ft s 2 ft 1⋅ min 1 × × 2 min 60⋅ s 1⋅ ft ρ = 0.00234⋅ ft s s 1.62 × 10 4 π −4 2 3 From Table A.9 ν = 1.62 × 10 V⋅ Dh ν − 4 ft ⋅ slug ft 3 s at 68oF 5 Hence Re = Re = 20.8⋅ × ⋅ ft × Dh = 1.284 × 10 ⋅ Dh (Dh in ft) For a round duct Dh = D = Dh = h= 4⋅ A Pw = 4⋅ A π 4⋅ b⋅ h 2⋅ ( b + h) so 2 Dh = = 2⋅ h⋅ ar 1 + ar × 1⋅ ft 2 Dh = 1.13 ft ar = b h A ar and Dh = 2⋅ ar ⋅A 1 + ar For a rectangular duct where But The results are: Round b ar h= b⋅ h A = ar ar or h= Dh = 1.13⋅ ft 51 Re = 1.284 × 10 ⋅ ⋅ Dh ft Re = 1.45 × 10 5 Given ⎛e ⎞ ⎜ Dh 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ Dh = 2⋅ ar ⋅A 1 + ar f = 0.0167 fV Δp = ρ⋅ ⋅ Dh 2 L 2 Δp − 3 lbf = 7.51 × 10 ⋅ 3 L ft Re = 1.28 × 10 5 ar = 1 Dh = 1 ft 51 Re = 1.284 × 10 ⋅ ⋅ Dh ft 2 Given ⎛e ⎞ ⎜ Dh ⎟ 1 2.51 = −2.0⋅ log ⎜ + ⎟ 3.7 f Re⋅ f ⎠ ⎝ f = 0.0171 fV Δp = ρ⋅ ⋅ Dh 2 L Δp − 3 lbf = 8.68 × 10 ⋅ 3 L ft Hence the square duct experiences a percentage increase in pressure drop of 8.68 × 10 −3 − 7.51 × 10 −3 −3 = 15.6 % 7.51 × 10 ar = 2 Dh = 2⋅ ar ⋅A 1 + ar Dh = 0.943 ft 51 Re = 1.284 × 10 ⋅ ⋅ Dh ft Re = 1.21 × 10 5 Given ⎞ ⎛e ⎜ Dh 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0173 Δp fV = ρ⋅ ⋅ L Dh 2 2 Δp − 3 lbf = 9.32 × 10 ⋅ 3 L ft −3 Hence the 2 x 1 duct experiences a percentage increase in pressure drop of 9.32 × 10 − 7.51 × 10 −3 −3 = 24.1 % 7.51 × 10 2⋅ ar ⋅A 1 + ar ar = 3 Dh = Dh = 0.866 ft 51 Re = 1.284 × 10 ⋅ ⋅ Dh ft Re = 1.11 × 10 5 Given ⎛e ⎞ ⎜ Dh ⎟ 1 2.51 = −2.0⋅ log ⎜ + ⎟ 3.7 f Re⋅ f ⎠ ⎝ f = 0.0176 Δp fV = ρ⋅ ⋅ L Dh 2 2 Δp lbf = 0.01⋅ 3 L ft −3 Hence the 3 x 1 duct experiences a percentage increase in pressure drop of 0.01 − 7.51 × 10 7.51 × 10 −3 = 33.2 % Note that f varies only about 7%; the large change in Δp/L is primarily due to the 1/Dh factor Problem 8.116 [3] Part 1/2 Problem 8.116 [3] Part 2/2 Problem 8.117 [3] Given: Find: Solution: Basic equations Flow through fire hose and nozzle Supply pressure 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlT ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ LV hlT = hl + hlm = f ⋅ ⋅ + D2 2 Minor ∑ ⎛ V2 ⎞ ⎜ K⋅ ⎟ ⎝ 2⎠ Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) p2 = patm so p2 = 0 gage Hence the energy equation between Point 1 at the supply and the nozzle exit (Point n); let the velocity in the hose be V p1 ρ From continuity − Vn 2 2 Vn LV V = f⋅ ⋅ + Ke + 4⋅ Kc ⋅ + Kn⋅ 2 2 D2 2 ( ) 2 2 D⎞ Vn = ⎛ ⎜ D ⎟ ⋅V ⎝ 2⎠ ρ⋅ V 2 2 2 and V= Q A = 4⋅ Q π⋅ D 2 V= 4 π × 0.75⋅ ft × s 3 1 ⎛ 1 ⋅ ft ⎞ ⎜ ⎟ ⎝4 ⎠ 2 V = 15.3 ft s Solving for p1 From Table A.7 (68oF) p1 = ⋅ ⎢f ⋅ 4 ⎡L ⎤ D⎞ ⎥ + Ke + 4⋅ Kc + ⎛ ⎜ D ⎟ ⋅ (1 + Kn)⎥ ⎢D ⎣ ⎝ 2⎠ ⎦ 3 ρ = 1.94⋅ slug ft ν = 1.08 × 10 ft s − 5 ft 2 ⋅ s ⋅ ft × s 1.08 × 10 −5 2 Re = For the hose V⋅ D ν Re = 15.3⋅ × 3 12 Re = 3.54 × 10 5 Turbulent ⋅ ft e = 0.004 D Given Flow is turbulent: 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2 f = 0.0287 4 2 p1 = ⎡ ⎤ lbf ⋅ s 1 slug ⎛ ft ⎞ 250 3⎞ × 1.94⋅ × ⎜ 15.3⋅ ⎟ × ⎢0.0287 × + 0.5 + 4 × 0.5 + ⎛ ⎟ × ( 1 + 0.02)⎥ × ⎜ 3 1 2 s⎠ ⎝ ⎝ 1⎠ ⎢ ⎥ slug⋅ ft ft ⎣ 4 ⎦ p1 = 2.58 × 10 ⋅ 4 lbf 2 ft p1 = 179⋅ psi Problem 8.118 [3] Given: Find: Solution: Basic equations Flow down corroded iron pipe Pipe roughness; Power savings with new pipe 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ hl = f ⋅ LV ⋅ D2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses Hence the energy equation becomes 2 ⎛ p1 ⎞ ⎛ p2 ⎞ LV ⎜ + g⋅ z1⎟ − ⎜ + g⋅ z2⎟ = f ⋅ ⋅ D2 ⎝ρ ⎠ ⎝ρ ⎠ and V= Q A = 4⋅ Q π⋅ D 2 V= 4 π × 0.2⋅ m 1⋅ min 1 × × 2 min 60⋅ s ( 0.025⋅ m) 3 V = 6.79 m s In this problem we can compute directly f and Re, and hnece obtain e/D ⎞ 2⋅ D ⎛ p1 − p2 Solving for f f= ⋅⎜ + g z1 − z2 ⎟ 2 ρ ⎠ L⋅ V ⎝ ( ) f = 2× 0.025 6 ×⎛ ⎜ ⋅ From Table A.8 (20oF) ν = 1.01 × 10 2 −6 m 2 3 ⎤ kg⋅ m m ⎞ × ⎡( 700 − 525) × 103⋅ N × m ⎢ × + 9.81⋅ × 6⋅ m⎥ ⎟ 2 1000⋅ kg 2 2 ⎥ ⎝ 6.79⋅ m ⎠ ⎢ m s ⋅N s ⎣ ⎦ s f = 0.0423 5 s Re = V⋅ D ν Re = 6.79⋅ m s × 0.025⋅ m × s 1.01 × 10 −6 Flow is turbulent: New pipe (Table 8.1) Given e = 0.15⋅ mm ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ e = 0.006 D 1 ⋅m 2 Re = 1.68 × 10 e = 0.0134 D Given ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0326 In this problem 2 ⎡ L V⎤ Δp = p1 − p2 = ρ⋅ ⎢g⋅ z2 − z1 + f ⋅ ⋅ ⎥ D2 ⎣ ( ) ⎦ 2 2 0.0326 6 m ⎞ ⎤ N⋅ s × × ⎛ 6.79⋅ ⎟ ⎥ × ⎜ 2 0.025 ⎝ s ⎠ ⎥ kg⋅ m Hence Δpnew = 1000⋅ kg m 3 × ⎢9.81⋅ ⎡ ⎢ ⎣ m s 2 × ( −6⋅ m) + ⎦ Δpnew = 121⋅ kPa Δpold − Δpnew Δpold = 30.6⋅ % Compared to Δpold = 175⋅ kPa we find Choose data: L= D= e= α= = 1.0 3.0 0.0 2 1 m mm mm (Laminar) (Turbulent) Tabulated or graphical data: μ = 1.00E-03 N.s/m 3 ρ = 999 kg/m (Appendix A) K ent = 0.5 (Square-edged) (Table 8.2) 2 Computed results: Q (L/min) V (m/s) Re 0.200 0.472 1413 0.225 0.531 1590 0.250 0.589 1767 0.275 0.648 1943 0.300 0.707 2120 0.325 0.766 2297 0.350 0.825 2473 0.375 0.884 2650 0.400 0.943 2827 0.425 1.002 3003 0.450 1.061 3180 Regime Laminar Laminar Laminar Laminar Laminar Laminar Turbulent Turbulent Turbulent Turbulent Turbulent f 0.0453 0.0403 0.0362 0.0329 0.0302 0.0279 0.0462 0.0452 0.0443 0.0435 0.0428 H (m ) 0.199 0.228 0.258 0.289 0.320 0.353 0.587 0.660 0.738 0.819 0.904 Required Reservoir Head versus Reynolds Number 1.00 0.75 H (m) 0.50 0.25 0.00 1000 Laminar Turbulent 1500 2000 Re 2500 3000 3500 The flow rates are realistic, and could easily be measured using a tank/timer system The head required is also realistic for a small-scale laboratory experiment Around Re = 2300 the flow may oscillate between laminar and turbulent: Once turbulence is triggered (when H > 0.353 m), the resistance to flow increases requiring H >0.587 m to maintain; hence the flow reverts to laminar, only to trip over again to turbulent! This behavior will be visible: the exit flow will switch back and forth between smooth (laminar) and chaotic (turbulent) Problem 8.120 [4] Problem 8.121 [2] Problem 8.122 [3] Given: Find: Solution: Basic equations Flow in horizontal pipe Flow rate 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ hl = f ⋅ LV ⋅ D2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses Hence the energy equation becomes p1 ρ Solving for V − p2 ρ = Δp ρ = f⋅ LV ⋅ D2 2 V= 2⋅ D⋅ Δp L ⋅ ρ⋅ f 2⋅ D⋅ Δp L⋅ ρ V⋅ D ν 1 3 V= k f lbf in 2 (1) k= k= 2× 300 × 40⋅ ×⎛ ⎜ × ⎟× ⎝ 1⋅ ft ⎠ 1.94⋅ slug s2⋅ lbf (2) s 1.08 × 10 (3) −5 2 12⋅ in ⎞ 2 ft 3 slugft ⋅ k = 2.57⋅ D ν ft s We also have From Table A.7 (68oF) Re = or − 5 ft 2 Re = c⋅ V c= 1 3 ⋅ ft × where c= ν = 1.08 × 10 ⋅ s c = 3.09 × 10 ⋅ 4s In addition 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ k f k f k f ⋅ ft ft Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f Make a guess for f f = 0.1 then V= V = 8.12⋅ ft s ft s ft s Re = c⋅ V Re = 2.51 × 10 5 Given ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 1 f = 0.0573 V= V = 10.7⋅ Re = c⋅ V Re = 3.31 × 10 5 Given ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ Q = V⋅ π⋅ D 4 2 f = 0.0573 V= V = 10.7⋅ 2 Re = c⋅ V Re = 3.31 × 10 5 The flow rate is then Q = 10.7⋅ ft π ⎛ 1 ⎞ 7.48⋅ gal 60⋅ s × × ⎜ ⋅ ft⎟ × × 3 s 4 ⎝3 ⎠ 1⋅ min 1⋅ ft Q = 419⋅ gpm Note that we could use Excel's Solver for this problem Problem 8.123 [3] Given: Find: Solution: Basic equations Drinking of a beverage Fraction of effort of drinking of friction and gravity 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ hl = f ⋅ LV ⋅ D2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses Hence the energy equation becomes, between the bottom of the straw (Point 1) and top (Point 2) g⋅ z1 − ⎜ ⎛ p2 ⎝ρ + g⋅ z2⎟ = f ⋅ ⎞ ⎠ LV ⋅ D2 2 where p2 is the gage pressure in the mouth The negative gage pressure the mouth must create is therefore due to two parts pgrav = −ρ⋅ g⋅ z2 − z1 ( ) pfric = −ρ⋅ f ⋅ LV ⋅ D2 2 Assuming a person can drink 12 fluid ounces in 5 s Assuming a straw is 6 in long diameter 0.2 in, with roughness V= 4⋅ Q π⋅ D 2 − 5 ft 2 12 ⋅ gal 3 128 1⋅ ft Q= × 5⋅ s 7.48⋅ gal e = 5 × 10 V= 4 π −5 Q = 2.51 × 10 − 3 ft 3 s in (from Googling!) − 3 ft 3 × 2.51 × 10 s ×⎛ ⎜ 1 12⋅ in ⎞ × ⎟ ⎝ 0.2⋅ in 1⋅ ft ⎠ 2 V = 11.5 ft s From Table A.7 (68oF) ν = 1.08 × 10 Re = V⋅ D ν ⋅ s (for water, but close enough) Re = 11.5⋅ ft 0.2 s × ⋅ ft × −5 2 s 12 1.08 × 10 ft Re = 1.775 × 10 4 Given ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ pgrav = −1.94⋅ pfric = −1.94⋅ slug ft 3 f = 0.0272 1 lbf⋅ s ⋅ ft × 2 slugft ⋅ 2 2 2 Then × 32.2⋅ ft s 2 × pgrav = −31.2 pfric = −105 pgrav pfric + pgrav lbf ft 2 2 pgrav = −0.217 psi pfric = −0.727 psi and slug ft 3 × 0.0272 × pfric 6 1 ft ⎞ lbf ⋅ s × × ⎛ 11.5⋅ ⎟ × ⎜ 0.2 2 ⎝ s⎠ slug⋅ ft and gravity is lbf ft Hence the fraction due to friction is pfric + pgrav = 77 % = 23 % These results will vary depending on assumptions, but it seems friction is significant! Problem 8.124 [2] Problem 8.125 [2] Problem 8.126 [2] Problem 8.127 [2] Problem 8.128 [2] Problem 8.129 [3] Given: Find: Solution: Basic equations Galvanized drainpipe Maximum downpour it can handle 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 hl = f ⋅ LV ⋅ D2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses LV Hence the energy equation becomes g⋅ z1 − g⋅ z2 = g⋅ z1 − z2 = g⋅ h = f ⋅ ⋅ D2 ( ) h=L Solving for V V= 2⋅ D⋅ g⋅ h L⋅ f 2⋅ D ⋅ g V⋅ D ν = 2⋅ D⋅ g f V= k f m s 2 (1) m s k= k= 2 × 0.075⋅ m × 9.81⋅ k = 1.21 D ν We also have Re = or −6 m 2 Re = c⋅ V (2) s 1.01 × 10 (3) −6 where c= From Table A.7 (20oC) ν = 1.01 × 10 ⋅ s c = 0.075⋅ m × ⋅m 2 c = 7.43 × 10 ⋅ 4s m In addition 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ e = 0.15 mm (Table 8.1) Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f Make a guess for f f = 0.01 e D then V= k f k f k f k f V = 12.13 m s Re = c⋅ V Re = 9.01 × 10 5 Given 1 ⎞ ⎛ ⎜ 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0236 V= V = 7.90 m s m s m s Re = c⋅ V Re = 5.86 × 10 5 Given ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 1 f = 0.0237 V= V = 7.88 Re = c⋅ V Re = 5.85 × 10 5 Given ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0237 V= V = 7.88 Re = c⋅ V Re = 5.85 × 10 5 The flow rate is then Q = V⋅ π⋅ D 4 2 Q = 7.88⋅ 3 mπ 2 × × ( 0.075⋅ m) s4 Q = 0.0348⋅ m s 3 The downpour rate is then Q = Aroof m 0.0348⋅ s 500⋅ m 2 × 100⋅ cm 60⋅ s cm × = 0.418⋅ 1⋅ m 1⋅ min min The drain can handle 0.418 cm/min Note that we could use Excel's Solver for this problem Problem 8.130 [3] Part 1/2 Problem 8.130 [3] Part 2/2 Fluid is not specified: use water Given data: Δp = D= L= 100 25 100 kPa mm m Tabulated or graphical data: μ = 1.00E-03 N.s/m 3 ρ= 999 kg/m (Water - Appendix A) 2 Computed results: 3 4 e/D V (m/s) Q (m /s) x 10 0.000 1.50 7.35 0.005 1.23 6.03 0.010 1.12 5.49 0.015 1.05 5.15 0.020 0.999 4.90 0.025 0.959 4.71 0.030 0.925 4.54 0.035 0.897 4.40 0.040 0.872 4.28 0.045 0.850 4.17 0.050 0.830 4.07 Re 37408 30670 27953 26221 24947 23939 23105 22396 21774 21224 20730 Regime Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Δp (kPa) f 0.0223 100 0.0332 100 0.0400 100 0.0454 100 0.0502 100 0.0545 100 0.0585 100 0.0623 100 0.0659 100 0.0693 100 0.0727 100 Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% Flow Rate versus Tube Relative Roughness for fixed Δp 8 6 Q (m3/s) x 104 4 2 0 0.00 It is not possible to roughen the tube sufficiently to slow the flow down to a laminar flow for this Δp . Even a relative roughness of 0.5 (a physical impossibility!) would not work. 0.01 0.02 e/D 0.03 0.04 0.05 Fluid is not specified: use water Given data: Δp = D= 100 25 m mm Tabulated or graphical data: μ = 1.00E-03 N.s/m 3 ρ= 999 kg/m (Water - Appendix A) 2 Flow Rate vs Tube Length for Fixed Δp 10.0 Computed results: L (km) V (m/s) Q (m3/s) x 104 1.0 1.5 2.0 2.5 5.0 10 15 19 21 25 30 0.40 0.319 0.270 0.237 0.158 0.105 0.092 0.092 0.092 0.078 0.065 1.98 1.56 1.32 1.16 0.776 0.516 0.452 0.452 0.452 0.383 0.320 Re 10063 7962 6739 5919 3948 2623 2300 2300 2300 1951 1626 Regime Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Laminar Laminar Laminar Laminar Δp (kPa) f 0.0308 100 0.0328 100 0.0344 100 0.0356 100 0.0401 100 0.0454 100 0.0473 120 0.0278 90 0.0278 99 0.0328 100 0.0394 100 Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 20.2% 10.4% 1.0% 0.0% 0.0% Laminar Q (m /s) 4 x 10 1.0 3 Turbulent 0.1 0 10 20 L (km) 30 40 The "critical" length of tube is between 15 and 20 km. For this range, the fluid is making a transition between laminar and turbulent flow, and is quite unstable. In this range the flow oscillates between laminar and turbulent; no consistent solution is found (i.e., an Re corresponding to turbulent flow needs an f assuming laminar to produce the Δp required, and vice versa!) More realistic numbers (e.g., tube length) are obtained for a fluid such as SAE 10W oil (The graph will remain the same except for scale) Given data: Δp = D= L= 153 75 100 kPa mm m Tabulated or graphical data: μ = 1.00E-03 N.s/m2 3 ρ= 999 kg/m (Water - Appendix A) Computed results: e/D V (m/s) 0.000 3.98 0.005 2.73 0.010 2.45 0.015 2.29 0.020 2.168 0.025 2.076 0.030 2.001 0.035 1.937 0.040 1.882 0.045 1.833 0.050 1.790 Q (m3/s) 0.0176 0.0121 0.0108 0.0101 0.00958 0.00917 0.00884 0.00856 0.00832 0.00810 0.00791 Re 2.98E+05 2.05E+05 1.84E+05 1.71E+05 1.62E+05 1.56E+05 1.50E+05 1.45E+05 1.41E+05 1.37E+05 1.34E+05 Regime Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Δp (kPa) f 0.0145 153 0.0308 153 0.0382 153 0.0440 153 0.0489 153 0.0533 153 0.0574 153 0.0612 153 0.0649 153 0.0683 153 0.0717 153 Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% It is not possible to roughen the tube sufficiently to slow the flow down to a laminar flow for this Δp. Computed results: L (m) V (m/s) 100 200 300 400 500 600 700 800 900 1000 1.37 1.175 1.056 0.975 0.913 0.865 0.825 0.791 0.762 0.737 Q (m3/s) 0.00606 0.00519 0.00467 0.00431 0.004036 0.003821 0.003645 0.003496 0.003368 0.003257 Re 1.03E+05 8.80E+04 7.92E+04 7.30E+04 6.84E+04 6.48E+04 6.18E+04 5.93E+04 5.71E+04 5.52E+04 Regime Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Δp (kPa) f 0.1219 153 0.0833 153 0.0686 153 0.0604 153 0.0551 153 0.0512 153 0.0482 153 0.0459 153 0.0439 153 0.0423 153 Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% Flow Rate versus Tube Relative Roughness for fixed Δp 0.020 0.015 Q (m /s) 3 0.010 0.005 0.000 0.00 0.01 0.01 0.02 0.02 e/D 0.03 0.03 0.04 0.04 0.05 0.05 Flow Rate versus Tube Relative Roughness for fixed Δp 0.010 Q (m3/s) 0.005 0.000 0 200 400 600 L (m) 800 1000 1200 Given data: p2 = 600 (Closed) D = 150 L = 200 Q = 0.75 (Open) kPa mm m m3/min Tabulated or graphical data: e = 0.26 mm (Table 8.1) μ = 1.00E-03 N.s/m2 ρ = 999 kg/m3 (Water - Appendix A) Computed results: Closed: H= 61.2 (Eq. 1) m Fully open: V= Re = f= 5.91 m/s 8.85E+05 0.0228 Partially open: 3 Q= 0.75 m /min V= 0.71 m/s Re = 1.06E+05 f = 0.0243 p2 = 591 kPa Eq. 2, solved by varying V using Solver : Left (m2/s) Right (m2/s) 601 601 Q= 0.104 Error 0% m3/s (Eq. 3) Problem 8.135 [3] Given: Find: Solution: Basic equations Syphon system Flow rate; Minimum pressure 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlT ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ hlT = f ⋅ LV ⋅ + hlm D2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 Hence the energy equation applied between the tank free surface (Point 1) and the tube exit (Point 2, z = 0) becomes g⋅ z1 − From Table 8.2 for reentrant entrance For the bend Solving for V R =9 D so from Fig. 8.16 V= V2 2 2 Le V V LV V = g⋅ z1 − = f⋅ ⋅ + Kent⋅ + f⋅ ⋅ 2 2 D2 D2 Kent = 0.78 Le D = 28 2 2 2 2 for a 90o bend so for a 180o bend (1) and Le D = 56 2⋅ g⋅ h ⎡ ⎛ L Le ⎞⎤ ⎢1 + Kent + f ⋅ ⎜ + ⎟⎥ ⎣ ⎝ D D ⎠⎦ Le = 2.8 m or −6 m 2 h = 2.5⋅ m The two lengths are We also have From Table A.7 (15oC) Le = 56⋅ D Re = V⋅ D ν L = ( 0.6 + π⋅ 0.45 + 2.5) ⋅ m Re = c⋅ V (2) c = 0.05⋅ m × s 1.14 × 10 (3) −6 L = 4.51 m where c= D ν 4s ν = 1.14 × 10 ⋅ s e D In addition 1 ⎞ ⎛ ⎜ 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ ⋅m 2 c = 4.39 × 10 ⋅ m e = 0.0015 mm (Table 8.1) Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f Make a guess for f f = 0.01 V= then 2⋅ g⋅ h ⎡ ⎛ L Le ⎞⎤ ⎢1 + Kent + f ⋅ ⎜ + ⎟⎥ ⎣ ⎝ D D ⎠⎦ V = 3.89 m s Re = c⋅ V Re = 1.71 × 10 5 Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ h Le ⎞⎤ f = 0.0164 V= Given ⎡ ⎛L ⎢1 + Kent + f ⋅ ⎜ + ⎟⎥ ⎣ ⎝ D D ⎠⎦ e ⎞ ⎛ ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ h V = 3.43 m s Re = c⋅ V Re = 1.50 × 10 5 f = 0.0168 V= Given ⎡ ⎛ L Le ⎞⎤ ⎢1 + Kent + f ⋅ ⎜ + ⎟⎥ ⎣ ⎝ D D ⎠⎦ ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ h V = 3.40 m s Re = c⋅ V Re = 1.49 × 10 5 f = 0.0168 V= ⎡ ⎛ L Le ⎞⎤ ⎢1 + Kent + f ⋅ ⎜ + ⎟⎥ ⎣ ⎝ D D ⎠⎦ V = 3.40 m s Re = c⋅ V Re = 1.49 × 10 5 Note that we could use Excel's Solver for this problem The minimum pressure occurs at the top of the curve (Point 3). Applying the energy equation between Points 1 and 3 2 ⎛p ⎞ 2 2 Le V 2 ⎛ p3 V2 ⎞ ⎜ 3 V3 ⎟ LV V g⋅ z1 − ⎜ + + g⋅ z3⎟ = g⋅ z1 − ⎜ + + g⋅ z3⎟ = f ⋅ ⋅ + Kent⋅ + f⋅ ⋅ 2 2 2 D2 D2 ⎝ρ ⎠ ⎝ρ ⎠ where we have Le D = 28 for the first 90o of the bend, and π × 0.45 ⎞ L = ⎛ 0.6 + ⎜ ⎟⋅m 2 ⎝ ⎠ L = 1.31 m 2 ⎡ ⎛ L Le ⎞⎤⎤ V⎡ ⎢g⋅ (z1 − z3) − ⋅ ⎢1 + Kent + f ⋅ ⎜ + ⎟⎥⎥ p3 = ρ⋅ 2⎣ ⎣ ⎝ D D ⎠⎦⎦ p3 = 1000⋅ kg m 3 × ⎢9.81⋅ ⎡ ⎢ ⎣ m s 2 × ( −0.45⋅ m) − ⎛ 3.4⋅ ⎜ ⎝ m⎞ ⎟ s⎠ 2 ⋅ ⎡1 + 0.78 + 0.0168⋅ ⎛ ⎢ ⎜ ⎣ 1.31 ⎞⎤⎤ N⋅ s + 28⎟⎥⎥ × ⎝ 0.05 ⎠⎦⎥ kg⋅ m 2 ⎦ p3 = −35.5 kPa Problem 8.136 [4] Given: Find: Solution: Basic equations Tank with drainpipe Flow rate for rentrant, square-edged, and rounded entrances 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlT ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ hlT = f ⋅ LV V ⋅ + Kent⋅ D2 2 2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 Hence the energy equation applied between the tank free surface (Point 1) and the pipe exit (Point 2, z = 0) becomes g⋅ z1 − Solving for V V= V2 2 2 V LV V = g⋅ z1 − = f⋅ ⋅ + Kent⋅ 2 2 D2 2⋅ g⋅ h 2 2 2 ⎛1 + K + f⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ V⋅ D ν or −6 m 2 (1) and h = ( 1.5 + 3.5) ⋅ m h = 5m We also have From Table A.7 (20oC) Re = Re = c⋅ V (2) ⋅ c = 0.025⋅ m × s 1.01 × 10 (3) −6 where c= D ν 4s ν = 1.01 × 10 s e D In addition 1 ⎞ ⎛ ⎜ 2.51 ⎟ = −2.0⋅ log⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ ⋅m 2 c = 2.48 × 10 ⋅ m e = 0.26⋅ mm (Table 8.1) Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f For a reentrant entrance, from Table 8.2 Kent = 0.78 then 2⋅ g⋅ h Make a guess for f f = 0.01 V= ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ f = 0.0388 V = 6.42 m s Re = c⋅ V Re = 1.59 × 10 5 Given ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ V= 2⋅ g⋅ h ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ V = 4.89 m s Re = c⋅ V Re = 1.21 × 10 5 Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0389 V= 2⋅ g⋅ h ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ V = 4.88 m s Re = c⋅ V Re = 1.21 × 10 5 Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ h f = 0.0389 V= ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ 2 V = 4.88 m s Re = c⋅ V Re = 1.21 × 10 5 Note that we could use Excel's Solver for this problem The flow rate is then Q = V⋅ π⋅ D 4 Q = 4.88⋅ mπ 2 × × ( 0.025⋅ m) s4 Q = 2.4 × 10 3 −3 m ⋅ s Q = 8.62⋅ m hr 3 For a square-edged entrance, from Table 8.2 Kent = 0.5 then 2⋅ g⋅ h Make a guess for f f = 0.01 V= ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ f = 0.0388 V = 6.83 m s Re = c⋅ V Re = 1.69 × 10 5 Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ h V= ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ V = 5.06 m s Re = c⋅ V Re = 1.25 × 10 5 Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ h f = 0.0389 V= ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ Q = V⋅ π⋅ D 4 2 V = 5.06 m s Re = c⋅ V Re = 1.25 × 10 5 The flow rate is then Q = 5.06⋅ r 3.75 = = 0.15 D 25 then 2⋅ g⋅ h mπ 2 × × ( 0.025⋅ m) s4 Q = 2.48 × 10 3 −3 m ⋅ s Q = 8.94⋅ m hr 3 For a rounded entrance, from Table 8.2 Kent = 0.04 Make a guess for f f = 0.01 V= ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ V = 7.73 m s Re = c⋅ V Re = 1.91 × 10 5 Given 1 ⎛e ⎞ ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ h f = 0.0387 V= ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ V = 5.40 m s Re = c⋅ V Re = 1.34 × 10 5 Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ h f = 0.0389 V= ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ V = 5.39 m s Re = c⋅ V Re = 1.34 × 10 5 Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ h f = 0.0389 V= ⎛1 + K + f ⋅ L ⎞ ⎜ ⎟ ent D⎠ ⎝ 2 V = 5.39 m s Re = c⋅ V Re = 1.34 × 10 5 Note that we could use Excel's Solver for this problem The flow rate is then Q = V⋅ π⋅ D 4 m 3 Q = 5.39⋅ m s × π 4 × ( 0.025⋅ m) 2 Q = 2.65 × 10 3 −3 m ⋅ s Q = 9.52⋅ 3 m 3 hr In summary: Renentrant: Q = 8.62⋅ hr Square-edged: Q = 8.94⋅ m 3 hr Rounded: Q = 9.52⋅ m hr Problem 8.137 [4] Problem 8.138 [5] Given: Find: Solution: Basic equations Tank with drain hose Flow rate at different instants; Estimate of drain time 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ hl = f ⋅ LV ⋅ D2 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor loss at entrance (L >>; verify later) Hence the energy equation applied between the tank free surface (Point 1) and the hose exit (Point 2, z = 0) becomes g⋅ z1 − Solving for V V= V2 2 2 V LV = g⋅ z1 − = f⋅ ⋅ 2 D2 2 2 2⋅ g⋅ h ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ V⋅ D ν or 2 −6 m 2 (1) and h = 10⋅ ft initially We also have Re = Re = c⋅ V (2) ft 10.8⋅ s 1⋅ s m s 2 2 where c= D ν From Fig. A.2 (20oC) ν = 1.8 × 10 ⋅ s × ν = 1.94 × 10 − 5 ft s c= 1 12 ⋅ ft × 1.94 × 10 −5 2 c = 4.30 × 10 ⋅ 3s ⋅ ft ft In addition ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ (3) with e = 0.01⋅ in D = 1 in Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f Make a guess for f f = 0.01 V= then ft s ft s 4 2⋅ g⋅ h ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ f = 0.0393 V = 9.59⋅ Re = c⋅ V Re = 4.12 × 10 Given ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ V= 2⋅ g⋅ h ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ V = 5.12⋅ Re = c⋅ V Re = 2.20 × 10 4 Given ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 1 f = 0.0405 V= Given ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ g⋅ h ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ 2⋅ g⋅ h ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ 2 V = 5.04⋅ ft s Re = c⋅ V Re = 2.17 × 10 4 f = 0.0405 V= V = 5.04⋅ ft s Re = c⋅ V Re = 2.17 × 10 4 Note that we could use Excel's Solver for this problem The flow rate is then Q = V⋅ π⋅ D 4 2 Note: Q = 5.04⋅ ft π ⎛ 1 ⎞ × × ⎜ ⋅ ft⎟ s 4 ⎝ 12 ⎠ f⋅ L = 24.3 D ft s 3 Ke = 0.5 hlm < hl Q = 0.0275⋅ Q = 12.3⋅ gpm Next we recompute everything for h = 5⋅ ft Given ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ Q = V⋅ f = 0.0405 V= 2⋅ g⋅ h ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ 2⋅ g⋅ h V = 3.57⋅ ft s ft s ft s ft s Re = c⋅ V Re = 1.53 × 10 4 Given f = 0.0415 V= ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ 2⋅ g⋅ h V = 3.52⋅ Re = c⋅ V Re = 1.51 × 10 4 Given f = 0.0415 V= ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ π 4 ×⎛ ⎜ V = 3.52⋅ Re = c⋅ V Re = 1.51 × 10 4 The flow rate is then π⋅ D 4 2 Q = 3.52⋅ ft s × ⎞ 12 ⎠ ⎝ 1 ⋅ ft⎟ 2 3 Q = 0.0192⋅ Q = 8.62⋅ gpm Next we recompute everything for h = 1⋅ ft Given ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 1 f = 0.0415 V= 2⋅ g⋅ h Given ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ π⋅ D 4 2 ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ 2⋅ g⋅ h V = 1.58⋅ ft s Re = c⋅ V Re = 6.77 × 10 3 f = 0.0452 V= ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ 2⋅ g⋅ h ⎛1 + f ⋅ L ⎞ ⎜ ⎟ D⎠ ⎝ 2 V = 1.51⋅ ft s ft s Re = c⋅ V Re = 6.50 × 10 3 Given 1 f = 0.0454 V= V = 1.51⋅ Re = c⋅ V Re = 6.48 × 10 3 The flow rate is then Q = V⋅ Q = 1.51⋅ ft π ⎛ 1 ⎞ × × ⎜ ⋅ ft⎟ s 4 ⎝ 12 ⎠ Q = 0.00824⋅ ft s 3 Q = 3.70⋅ gpm Initially we have dQ/dt = -12.3 gpm, then -8.62 gpm, then -3.70 gpm. These occur at h = 10 ft, 5 ft and 1 ft. The corresponding volumes in the tank are then Q = 7500 gal, 3750 gal, and 750 gal. Using Excel we can fit a power trendline to the dQ/dt versus Q data to find, approximately 1 dQ 2 = −0.12⋅ Q dt t= 1 ⋅ ( 7500 − 0.06 where dQ/dt is in gpm and t is min. Solving this with initial condition Q = 7500 gpm when t = 0 gives Q) 1 ⋅ ( 7500 − 0.06 750)⋅ min Hence, when Q = 750 gal (h = 1 ft) t= t = 987 min t = 16.4 hr Problem 8.139 [4] Part 1/2 Problem 8.139 [4] Part 2/2 Problem 8.140 [5] Part 1/2 Problem 8.140 [5] Part 2/2 Problem 8.141 Applying the energy equation between inlet and exit: Δp ρ =f LV2 D2 or Δp ρf V 2 = L D2 Δ p ⎛ Δ p ⎞ ⎛ Q0 ⎞ =⎜ ⎟⎜ ⎟ L ⎝ L ⎠0 ⎜ Q ⎟ ⎝⎠ 2 "Old school": D= e= ν= ρ= 1 0.00015 1.08E-05 1.94 in ft ft2/s slug/ft3 Q (gpm) 1.25 1.50 1.75 2.00 2.25 2.50 Flow 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00 5.25 5.50 5.75 6.00 6.25 6.50 6.75 7.00 7.25 7.50 7.75 0.02 8.00 8.25 8.50 8.75 9.00 Q (ft3/s) V (ft/s) Re f Δp (old Δp (psi/ft) school) (psi) 20 18 16 14 Flow (gpm) 12 10 8 6 4 2 0 0.00 Your boss was wrong! 0.00279 0.511 3940 0.00334 0.613 4728 0.00390 0.715 5516 0.00446 0.817 6304 0.00501 0.919 7092 0.00557versus Pressure 1.021 7881 Rate 0.00613 1.123 8669 0.00668 1.226 9457 0.00724 1.328 10245 0.00780 1.430 11033 0.00836 1.532 11821 0.00891 1.634 12609 0.00947 1.736 13397 0.01003 1.838 14185 0.01058 1.940 14973 0.01114 2.043 15761 0.01170 2.145 16549 0.01225 2.247 17337 0.01281 2.349 18125 0.01337 2.451 18913 0.01393 2.553 19701 0.01448 2.655 20489 0.01504 2.758 21277 0.01560 2.860 22065 0.01615 2.962 22854 0.01671 3.064 23642 0.01727 3.166 24430 0.04 0.06 0.01783 3.268 25218 Pressure Drop (psi/ft) 0.01838 3.370 26006 0.01894 3.472 26794 0.01950 3.575 27582 0.02005 3.677 28370 0.0401 0.00085 0.00085 0.0380 0.00122 0.00115 0.0364 0.00166 0.00150 0.0350 0.00216 0.00189 0.0339 0.00274 0.00232 0.0329 0.00338 0.00278 Drop 0.0321 0.00409 0.00328 0.0314 0.00487 0.00381 0.0307 0.00571 0.00438 0.0301 0.00663 0.00498 0.0296 0.00761 0.00561 0.0291 0.00865 0.00628 0.0286 0.00977 0.00698 0.0282 0.01095 0.00771 0.0278 0.01220 0.00847 0.0275 0.01352 0.00927 0.0272 0.01491 0.01010 0.0268 0.01636 0.01095 0.0265 0.01788 0.01184 0.0263 0.01947 0.01276 0.0260 0.02113 0.01370 0.0258 0.02285 0.01468 0.0255"Old 0.02465 School" 0.01569 0.0253Exact 0.02651 0.01672 0.0251 0.02843 0.01779 0.0249 0.03043 0.01888 0.0247 0.03249 0.10 0.02000 0.08 0.0245 0.03462 0.02115 0.0243 0.03682 0.02233 0.0242 0.03908 0.02354 0.0240 0.04142 0.02477 0.0238 0.04382 0.02604 Problem 8.142 [3] Given: Find: Solution: Basic equations Hydraulic press system Minimum required diameter of tubing 2 2 ⎛p ⎞⎜ ⎞ V1 V2 ⎜1 ⎟ ⎛ p2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 L V2 hl = f ⋅ ⋅ D2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses The flow rate is low and it's oil, so try assuming laminar flow. Then, from Eq. 8.13c 1 4 Δp = 128⋅ μ⋅ Q⋅ L π⋅ D 4 or 0.0209⋅ lbf⋅ s 2 128⋅ μ⋅ Q⋅ L ⎞ D=⎛ ⎜ ⎟ ⎝ π⋅ Δp ⎠ For SAE 10W oil at 100oF (Fig. A.2, 38oC) μ = 3.5 × 10 − 2 N ⋅s ⋅ × 2 m 1⋅ ft N ⋅s m 2 μ = 7.32 × 10 − 4 lbf⋅ s 2 ft Hence ⎡ 128 ft in 1⋅ ft ⎞ − 4 lbf ⋅ s D=⎢ × 7.32 × 10 × 0.02⋅ × 165⋅ ft × ×⎛ ⎜ ⎟ 2 ⎢π s ( 3000 − 2750) ⋅ lbf ⎝ 12⋅ in ⎠ ft ⎣ 3 2 2⎤ 1 4 ⎥ ⎥ ⎦ D = 0.0407 ft 2 D = 0.488 in Check Re to assure flow is laminar V= Q A = 4⋅ Q π⋅ D 2 12 1 ⎞ V= × 0.02⋅ ×⎛ ⋅⎟ ⎜ π s ⎝ 0.488 ft ⎠ 4 ft Re = SGoil⋅ ρH2O⋅ V⋅ D μ ft 2 −4 3 V = 15.4 ft s From Table A.2 SGoil = 0.92 so Re = 0.92 × 1.94⋅ slug ft 3 × 15.4⋅ ft 0.488 × ⋅ ft × s 12 × lbf ⋅ s lbf ⋅ s 2 7.32 × 10 slug⋅ ft Re = 1527 Hence the flow is laminar, Re < 2300. The minimum diameter is 0.488 in, so 0.5 in ID tube should be chosen Problem 8.143 [4] Given: Find: Solution: Basic equations Flow out of reservoir by pump Smallest pipe needed 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hlT ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 2 2 V2 Le V 2 L V2 hlT = hl + hlm = f ⋅ ⋅ + Kent⋅ + f⋅ ⋅ 2 D2 D2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl << Hence for flow between the free surface (Point 1) and the pump inlet (2) the energy equation becomes − p2 ρ − g⋅ z2 − V2 2 2 =− p2 ρ − g⋅ z2 − Le V V LV V = f⋅ ⋅ + Kent⋅ + f⋅ ⋅ 2 2 D2 D2 (1) 2 2 2 2 and p = ρ⋅ g⋅ h Solving for h2 = p2/ρg From Table 8.2 We also have and we are given 2 ⎤ V ⎡ ⎛ L Le ⎞ h2 = −z2 − ⋅ ⎢f ⋅ ⎜ + ⎟ + Kent⎥ 2⋅ g ⎣ ⎝ D D ⎠ ⎦ Kent = 0.78 for rentrant, and from Table 8.4 two standard elbows lead to −6 m 2 Le D = 2 × 30 = 60 e = 0.046⋅ mm (Table 8.1) ν = 1.51 × 10 Q = 6⋅ L s Q = 6 × 10 ⋅ 3 −3m s (Table A.8) z2 = 3.5⋅ m L = ( 3.5 + 4.5) ⋅ m L = 8m h2 = −6⋅ m s Equation 1 is tricky because D is unknown, so V is unknown (even though Q is known), L/D and Le/D are unknown, and Re and hence f are unknown! We COULD set up Excel to solve Eq 1, the Reynolds number, and f, simultaneously by varying D, but here we try guesses: D = 2.5⋅ cm V= 4⋅ Q π⋅ D 1 2 V = 12.2 m s Re = V⋅ D ν Re = 2.02 × 10 5 Given ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0238 2 ⎤ V ⎡ ⎛ L Le ⎞ h2 = −z2 − ⋅ ⎢f ⋅ ⎜ + ⎟ + Kent⎥ 2⋅ g ⎣ ⎝ D D ⎠ ⎦ h2 = −78.45 m V = 3.06 m s but we need -6 m! V⋅ D ν 5 D = 5⋅ cm V = 4⋅ Q π⋅ D 2 Re = Re = 1.01 × 10 Given ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0219 2 ⎤ V ⎡ ⎛ L Le ⎞ h2 = −z2 − ⋅ ⎢f ⋅ ⎜ + ⎟ + Kent⎥ 2⋅ g ⎣ ⎝ D D ⎠ ⎦ h2 = −6.16 m V = 2.94 m s but we need -6 m! V⋅ D ν 4 D = 5.1⋅ cm V= 4⋅ Q π⋅ D 2 Re = Re = 9.92 × 10 Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0219 2 ⎤ V ⎡ ⎛ L Le ⎞ h2 = −z2 − ⋅ ⎢f ⋅ ⎜ + ⎟ + Kent⎥ 2⋅ g ⎣ ⎝ D D ⎠ ⎦ h2 = −5.93 m To within 1%, we can use 5-5.1 cm tubing; this corresponds to standard 2 in pipe. Problem 8.144 [4] Given: Find: Solution: Basic equations Flow of air in rectangular duct Minimum required size 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ hl = f ⋅ LV ⋅ Dh 2 2 Dh = 4⋅ A Pw Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses Hence for flow between the inlet (Point 1) and the exit (2) the energy equation becomes p1 ρ − p2 ρ = Δp ρ = f⋅ LV ⋅ Dh 2 2 2 and Δp = ρH2O⋅ g⋅ Δh 2 4⋅ b⋅ h 2⋅ h ⋅ ar 2⋅ h⋅ ar For a rectangular duct Dh = = = h⋅ ( 1 + ar) 2⋅ ( b + h) 1 + ar 2 2 and also 2 A = b⋅ h = h ⋅ 2b h = h ⋅ ar Hence Δp = ρ⋅ f ⋅ L⋅ V ( 1 + ar) Q ( 1 + ar) ρ⋅ f ⋅ L⋅ Q ( 1 + ar) 1 ⋅ = ρ⋅ f ⋅ L ⋅ ⋅ = ⋅ ⋅ 2 2⋅ h⋅ ar 3 5 2 2⋅ h⋅ ar 4 2⋅ A ar h 2 Solving for h ρ⋅ f ⋅ L⋅ Q ( 1 + ar)⎥ h=⎢ ⋅ 3 ⎢ 4⋅ Δp ar ⎥ ⎡ ⎣ ⎤ 1 5 (1) We are given and also Q = 2850⋅ ft min 3 ⎦ L = 100⋅ ft Δp = 1.94⋅ slug ft ν = 1.62 × 10 3 e = 0⋅ ft × 32.2 2 ar = 2 × 1.25 lbf ⋅ s ⋅ ft × 12 slug⋅ ft (Table A.9) 2 Δp = ρH2O⋅ g⋅ Δh slug ft 3 ft s 2 Δp = 6.51⋅ lbf ft 2 ρ = 0.00234⋅ − 4 ft ⋅ s Equation 1 is tricky because h is unknown, so Dh is unknown, hence V is unknown (even though Q is known), and Re and hence f are unknown! We COULD set up Excel to solve Eq 1, the Reynolds number, and f, simmultaneously by varying h, but here we try guesses: ρ⋅ f ⋅ L⋅ Q ( 1 + ar)⎥ h=⎢ ⋅ 3⎥ ⎢ 4⋅ Δp ar ⎦ ⎣ Dh = 0.796⋅ ft f = 0.01 2⋅ h⋅ ar 1 + ar ⎡ 2 ⎤ 1 5 h = 0.597⋅ ft V⋅ Dh ν V= Q h ⋅ ar 2 V = 66.6⋅ ft s Dh = Re = Re = 3.27 × 10 5 Given ⎛e ⎞ ⎜ Dh 1 2.51 ⎟ = −2.0⋅ log⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0142 ρ⋅ f ⋅ L⋅ Q ( 1 + ar)⎥ h=⎢ ⋅ 3⎥ ⎢ 4⋅ Δp ar ⎦ ⎣ Dh = 2⋅ h⋅ ar 1 + ar ⎡ 2 ⎤ 1 5 h = 0.641⋅ ft V= Q h ⋅ ar 2 V = 57.8⋅ ft s 5 Dh = 0.855⋅ ft Re = V⋅ Dh ν Re = 3.05 × 10 Given 1 5 ⎛e ⎞ ⎜ Dh 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2 f = 0.0144 ρ⋅ f ⋅ L⋅ Q ( 1 + ar)⎥ h=⎢ ⋅ 3⎥ ⎢ 4⋅ Δp ar ⎦ ⎣ Dh = 2⋅ h⋅ ar 1 + ar ⎡ ⎤ h = 0.643⋅ ft V= Q h ⋅ ar 2 V = 57.5⋅ ft s Dh = 0.857⋅ ft Re = V⋅ Dh ν Re = 3.04 × 10 5 Given 1 5 ⎛e ⎞ ⎜ Dh 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2 f = 0.0144 ρ⋅ f ⋅ L⋅ Q ( 1 + ar)⎥ h=⎢ ⋅ 3 ⎢ 4⋅ Δp ar ⎥ ⎡ ⎣ ⎤ h = 0.643⋅ ft V= Q h ⋅ ar 2 V = 57.5⋅ ft s 5 ⎦ Dh = 0.857⋅ ft Re = Dh = 2⋅ h⋅ ar 1 + ar V⋅ Dh ν Re = 3.04 × 10 In this process h and f have converged to a solution. The minimum dimensions are 0.642 ft by 1.28 ft, or 7.71 in by 15.4 in Problem 8.145 [3] Problem 8.146 [4] Given: Find: Solution: Basic equations Flow of air in square duct Minimum required size 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ hl = f ⋅ LV ⋅ Dh 2 2 Dh = 4⋅ A Pw Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses Hence for flow between the inlet (Point 1) and the exit (2) the energy equation becomes p1 ρ For a square duct − p2 ρ = Δp ρ = f⋅ LV ⋅ Dh 2 2 and Δp = ρH2O⋅ g⋅ Δh 2 Dh = 4⋅ h⋅ h 2⋅ ( h + h) 2 =h 2 2 and also A = h⋅ h = h Hence Δp = ρ⋅ f ⋅ L⋅ V Q ρ⋅ f ⋅ L ⋅ Q = ρ⋅ f ⋅ L ⋅ = 2 5 2⋅ h 2⋅ h⋅ A 2⋅ h 1 5 Solving for h h=⎜ ⎛ ρ⋅ f ⋅ L ⋅ Q 2 ⎞ ⎟ ⎝ 2⋅ Δp ⎠ m s 3 (1) We are given and also Q = 2⋅ L = 25⋅ m Δp = 1000⋅ kg m ν = 1.50 × 10 3 e = 0.046⋅ mm × 9.81 2 (Table 8.1) N⋅ s kg⋅ m 2 Δp = ρH2O⋅ g⋅ Δh kg m 3 m s 2 × 0.015⋅ m × Δp = 147 Pa ρ = 1.21⋅ −5 m ⋅ s (Table A.10) Equation 1 is tricky because h is unknown, so Dh is unknown, hence V is unknown (even though Q is known), and Re and hence f are unknown! We COULD set up Excel to solve Eq 1, the Reynolds number, and f, simmultaneously by varying h, but here we try guesses: f = 0.01 h=⎜ ⎛ ρ⋅ f ⋅ L ⋅ Q 2 ⎞ ⎟ ⎝ 2⋅ Δp ⎠ 1 5 h = 0.333 m V⋅ Dh ν V= Q h 2 5 V = 18.0⋅ m s Dh = h Dh = 0.333 m Re = Re = 4.00 × 10 Given ⎛e ⎞ ⎜ Dh 1 2.51 ⎟ = −2.0⋅ log⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0152 h=⎜ ⎛ ρ⋅ f ⋅ L ⋅ Q 2 ⎞ ⎟ ⎝ 2⋅ Δp ⎠ 1 5 h = 0.362 m V= Q h V⋅ Dh ν 2 V = 15.2 m s 5 Dh = h Dh = 0.362⋅ m Re = Re = 3.68 × 10 Given 1 5 ⎞ ⎛e ⎜ Dh ⎟ 1 2.51 = −2.0⋅ log ⎜ + ⎟ 3.7 f Re⋅ f ⎠ ⎝ f = 0.0153 h=⎜ ⎛ ρ⋅ f ⋅ L ⋅ Q 2 ⎞ ⎟ ⎝ 2⋅ Δp ⎠ h = 0.363 m V= Q h 2 V = 15.2 m s In this process h and f have converged to a solution. The minimum dimensions are 0.363 m by 0.363 m, or 36.3 cm by 36.3 cm Fluid is not specified: use water (basic trends in plot apply to any fluid) Given data: Δp = L= 100 100 kPa m Tabulated or graphical data: μ = 1.00E-03 N.s/m2 3 ρ= 999 kg/m (Water - Appendix A) Flow Rate versus Tube Diameter for Fixed Δp 0.8 0.6 Q (m3/s) 4 x 10 0.4 Laminar Computed results: D (mm) V (m/s) Q (m3/s) x 104 Re 0.5 0.00781 0.0000153 4 1.0 0.0312 0.000245 31 2.0 0.125 0.00393 250 3.0 0.281 0.0199 843 4.0 0.500 0.0628 1998 5.0 0.460 0.0904 2300 6.0 0.530 0.150 3177 7.0 0.596 0.229 4169 8.0 0.659 0.331 5270 9.0 0.720 0.458 6474 10.0 0.778 0.611 7776 Regime Laminar Laminar Laminar Laminar Laminar Turbulent Turbulent Turbulent Turbulent Turbulent Turbulent Δp (kPa) f 16.4 100 2.05 100 0.256 100 0.0759 100 0.0320 100 0.0473 100 0.0428 100 0.0394 100 0.0368 100 0.0348 100 0.0330 100 Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.2% 0.0% 0.0% 0.0% 0.0% 0.0% 0.2 0.0 0.0 Turbulent 2.5 5.0 D (mm) 7.5 10.0 Problem 8.148 [4] Given: Find: Solution: Basic equations Flow of water in circular pipe Minimum required diameter 2 2 ⎛p ⎞ ⎛p ⎞ V1 V2 ⎜1 ⎟ ⎜2 ⎟ + α⋅ + g⋅ z1⎟ − ⎜ + α⋅ + g⋅ z2⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ hl = f ⋅ LV ⋅ D2 2 and also A= π⋅ D 4 2 Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses Hence for flow between the inlet (Point 1) and the exit (2) the energy equation becomes p1 ρ Hence − p2 ρ = Δp ρ = f⋅ 2 LV ⋅ D2 2 Δp = ρ⋅ f ⋅ LV LQ 8⋅ ρ⋅ f ⋅ L⋅ Q ⋅ = ρ⋅ f ⋅ ⋅ = 2 25 D2 D 2 A π ⋅D 2⎞ 1 5 2 2 Solving for D We are given and also 8⋅ ρ⋅ f ⋅ L⋅ Q ⎟ D=⎜ ⎜ π2⋅ Δp ⎟ ⎝ ⎠ Q = 1200⋅ gpm ρ = 1.94⋅ slug ft 3 ⎛ (1) L = 500⋅ ft ν = 1.08 × 10 e = 0.01⋅ ft − 5 ft 2 Δp = 50⋅ psi (Table A.7) ⋅ s Equation 1 is tricky because D is unknown, hence V is unknown (even though Q is known), and Re and hence f are unknown! We COULD set up Excel to solve Eq 1, the Reynolds number, and f, simultaneously by varying D, but here we try guesses: 1 5 f = 0.01 8⋅ ρ⋅ f ⋅ L⋅ Q ⎟ D=⎜ ⎜ π2⋅ Δp ⎟ ⎝ ⎠ 1 ⎛ 2⎞ D = 0.379⋅ ft V= 4⋅ Q π⋅ D 2 V = 23.7⋅ ft s Re = V⋅ D ν Re = 8.32 × 10 5 Given 1 5 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + f ⎟ = 0.0543 f ⎝ 3.7 Re⋅ f ⎠ 8⋅ ρ⋅ f ⋅ L⋅ Q ⎟ D=⎜ ⎜ π2⋅ Δp ⎟ ⎝ ⎠ ⎛ 2⎞ D = 0.531⋅ ft V= 4⋅ Q π⋅ D 2 V = 12.1⋅ ft s Re = V⋅ D ν Re = 5.93 × 10 5 Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0476 8⋅ ρ⋅ f ⋅ L⋅ Q ⎟ D=⎜ ⎜ π2⋅ Δp ⎟ ⎝ ⎠ ⎛ 2⎞ 1 5 D = 0.518⋅ ft V= 4⋅ Q π⋅ D 2 V = 12.7⋅ ft s Re = V⋅ D ν Re = 6.09 × 10 5 Given 1 5 ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⎞ f = 0.0481 8⋅ ρ⋅ f ⋅ L⋅ Q ⎟ D=⎜ ⎜ π2⋅ Δp ⎟ ⎝ ⎠ ⎛ D = 0.519⋅ ft V= 4⋅ Q π⋅ D 2 V = 12.7⋅ ft s Re = V⋅ D ν Re = 6.08 × 10 5 Given 1 5 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0480 8⋅ ρ⋅ f ⋅ L⋅ Q ⎟ D=⎜ ⎜ π2⋅ Δp ⎟ ⎝ ⎠ ⎛ 2⎞ D = 0.519⋅ ft V= 4⋅ Q π⋅ D 2 V = 12.7⋅ ft s Re = V⋅ D ν Re = 6.08 × 10 5 In this process D and f have converged to a solution. The minimum diameter is 0.519 ft or 6.22 in Problem 8.149 [3] Part 1/2 Problem 8.149 [3] Part 2/2 Problem 8.150 [3] Problem *8.151 [4] Problem *8.152 Problem 8.151 [5] Part 1/2 Problem *8.152 [5] Part 2/2 Problem 8.153 [2] Given: Find: Solution: Basic equations Flow through water pump Power required 2 ⎛p ⎞ ⎛p V 2 ⎞ ⎜ d Vd ⎟ ⎜s ⎟ s hpump = ⎜ + + g⋅ zd⎟ − ⎜ + + g⋅ zs⎟ 2 2 ⎝ρ ⎠ ⎝ρ ⎠ V= Q 4⋅ Q = 2 A π⋅ D Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow Hence for the inlet Vs = 4 π 4 π × 25⋅ lbm s lbm s × 1⋅ slug 32.2⋅ lbm 1⋅ slug 32.2⋅ lbm 2 2 × ft 12 1 ⎞ ×⎛ ⋅ ⎟ ⎜ 1.94⋅ slug ⎝ 3 ft ⎠ 3 3 2 Vs = 8.15 2 ft s ft s ps = −2.5⋅ psi For the outlet Vd = × 25⋅ × × ft 12 1 ⎞ ×⎛ ⋅ ⎟ ⎜ 1.94⋅ slug ⎝ 2 ft ⎠ and Vd = 18.3 pd = 50⋅ psi Then hpump = pd − ps ρ + Vd − Vs 2 Wpump = mpump⋅ hpump ⎛ p − p V 2 − V 2⎞ ⎜d s d s⎟ Wpump = mpump⋅ ⎜ + ⎟ 2 ⎝ρ ⎠ Note that the software cannot render a dot, so the power is Wpump and mass flow rate is mpump! ⎡ lbm 1⋅ slug lbf ⎛ 12⋅ in ⎞ ft 1 2 2 Wpump = 25⋅ × × ⎢( 50 − −2.5) ⋅ ×⎜ + × 18.3 − 8.15 ⎟× 2 ⎝ 1⋅ ft ⎠ s 32.2⋅ lbm ⎢ 1.94⋅ slug 2 in 2 3 ( ⎣ )⋅ ⎛ ft ⎞ ⎜⎟ ⎝ s⎠ 2 × lbf ⋅ s ⎤ ⎥× slug⋅ ft⎥ 2 1⋅ hp 550⋅ ft⋅ lbf s ⎦ Wpump = 5.69 hp For an efficiency of η = 70⋅ % Wrequired = Wpump η Wrequired = 8.13 hp Problem 8.154 [1] Given: Find: Solution: Basic equations Flow through water pump Power required 2 ⎛p ⎞ ⎛p V 2 ⎞ ⎜ d Vd ⎟ ⎜s ⎟ s hpump = ⎜ + + g⋅ zd⎟ − ⎜ + + g⋅ zs⎟ 2 2 ⎝ρ ⎠ ⎝ρ ⎠ V= Q 4⋅ Q = 2 A π⋅ D Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow In this case we assume Ds = Dd Then hpump = so pd − ps ρ = Δp ρ Vs = Vd and Wpump = mpump⋅ hpump Δp Δp Wpump = mpump⋅ = ρ⋅ Q ⋅ = Q⋅ Δp ρ ρ Note that the software cannot render a dot, so the power is Wpump and mass flow rate is mpump! L 0.001⋅ m 3N Wpump = 25⋅ × × 75 × 10 ⋅ 2 1⋅ L s m For an efficiency of η = 80⋅ % 3 Wpump = 1.88 kW Wpump η Wrequired = Wrequired = 2.34⋅ kW Problem 8.155 [3] Problem 8.156 [4] Problem 8.157 [3] Problem 8.158 [4] Problem 8.159 [4] Given: Find: Solution: Basic equations Fire nozzle/pump system Design flow rate; nozzle exit velocity; pump power needed 2 2 ⎛p ⎞ ⎛p ⎞ V2 V3 ⎜2 ⎟ ⎜3 ⎟ + α⋅ + g⋅ z2⎟ − ⎜ + α⋅ + g⋅ z3⎟ = hl ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 L V2 hl = f ⋅ ⋅ D2 for the hose Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 2 and 3 is approximately 1 4) No minor loss p3 ρ + V3 2 2 + g⋅ z3 = p4 ρ + V4 2 2 + g⋅ z4 for the nozzle (assuming Bernoulli applies) 2 2 ⎛p ⎞ ⎛p ⎞ V2 V1 ⎜2 ⎟ ⎜1 ⎟ ⎜ ρ + α⋅ 2 + g⋅ z2⎟ − ⎜ ρ + α⋅ 2 + g⋅ z1⎟ = hpump ⎝ ⎠⎝ ⎠ for the pump Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) No minor loss Hence for the hose 2 p2 − p3 Δp LV = = f⋅ ⋅ ρ ρ D2 or V= 2⋅ Δp⋅ D ρ⋅ f ⋅ L 2 −6 m We need to iterate to solve this for V because f is unknown until Re is known. This can be done using Excel's Solver, but here: Δp = 750⋅ kPa L = 100⋅ m e=0 2⋅ Δp⋅ D ρ⋅ f ⋅ L D = 3.5⋅ cm m s ρ = 1000⋅ V⋅ D ν kg m Make a guess for f f = 0.01 e D 3 ν = 1.01 × 10 ⋅ s V= V = 7.25 Re = Re = 2.51 × 10 5 Given 1 ⎞ ⎛ ⎜ 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ Δp⋅ D ρ⋅ f ⋅ L V = 5.92 m s f = 0.0150 V⋅ D ν 5 V= Re = Re = 2.05 × 10 Given 1 ⎞ ⎛e ⎜D 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ 2⋅ Δp⋅ D ρ⋅ f ⋅ L V = 5.81 m s f = 0.0156 V⋅ D ν 5 V= Re = Re = 2.01 × 10 Given ⎞ ⎛e ⎜D 1 2.51 ⎟ = −2.0⋅ log ⎜ + ⎟ f ⎝ 3.7 Re⋅ f ⎠ f = 0.0156 V= 2⋅ Δp⋅ D ρ⋅ f ⋅ L V = 5.80 2 m s Re = V⋅ D ν Re = 2.01 × 10 Q = 5.58 × 10 5 Q = V⋅ A = p3 ρ V3 2 2 π⋅ D ⋅V 4 Q= 2 π m 2 × ( 0.035⋅ m) × 5.80⋅ 4 s so 3 −3m s Q = 0.335 m min 3 For the nozzle + + g⋅ z3 = p4 V4 + + g⋅ z4 ρ 2 3 2 m kg⋅ m ⎛ m⎞ 3N × × + ⎜ 5.80⋅ ⎟ 2 1000⋅ kg 2 s⎠ ⎝ V4 = 2⋅ p3 − p4 2 + V3 ρ m s ( ) V4 = For the pump 2 × 700 × 10 ⋅ 2 m s ⋅N V4 = 37.9 ⎛p ⎞ ⎛p ⎞ V2 V1 ⎜2 ⎟ ⎜1 ⎟ + α⋅ + g⋅ z2⎟ − ⎜ + α⋅ + g⋅ z1⎟ = hpump ⎜ρ 2 2 ⎝ ⎠ ⎝ρ ⎠ 2 so hpump = p2 − p1 ρ p1 = 350⋅ kPa The pump power is Ppump = mpump⋅ hpump p2 = 700⋅ kPa + 750⋅ kPa p2 = 1450 kPa where Ppump and mpump are the pump power and mass flow rate (software cannot render a dot!) Ppump = 5.58 × 10 3 −3 m Ppump = ρ⋅ Q⋅ (p2 − p1) ρ = Q⋅ p2 − p1 ( ) ⋅ s × ( 1450 − 350) × 10 ⋅ 3N 2 m Ppump = 6.14 kW Prequired = Ppump η Prequired = 6.14⋅ kW 70⋅ % Prequired = 8.77 kW Problem 8.160 [4] Part 1/2 Problem 8.160 [4] Part 2/2 Problem 8.161 [4] Problem 8.162 [5] Given data: L= D= ηpump = 20 75 70% m mm Tabulated or graphical data: e= 0.26 mm (Table 8.1) 2 μ = 1.00E-03 N.s/m 3 ρ = 999 kg/m (Appendix A) Gate valve L e/D = 8 Elbow L e/D = 30 (Table 8.4) Computed results: 3 Q (m /s) V (m/s) Re 1.70E+05 2.54E+05 3.39E+05 4.24E+05 5.09E+05 5.94E+05 6.78E+05 7.63E+05 8.48E+05 9.33E+05 1.02E+06 f 0.0280 0.0277 0.0276 0.0276 0.0275 0.0275 0.0274 0.0274 0.0274 0.0274 0.0274 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055 0.060 0.0419 Power = 2.26 3.40 4.53 5.66 6.79 7.92 9.05 10.2 11.3 12.4 13.6 9.48 29.1 Δp (kPa) (Eq 1) 28.3 63.1 112 174 250 340 444 561 692 837 996 487 Δp (kPa) (Eq 2) 735 716 690 656 615 566 510 446 375 296 210 487 Circuit and Pump Pressure Heads 1200 1000 Δp (kPa) 800 600 400 200 0 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 Q (m3/s) Circuit Pump 7.11E+05 0.0274 kW (Eq. 3) Error Using Solver ! 0 Tabulated or graphical data: μ= ρ= lbf·s/ft2 2.10E-05 3 slug/ft 1.94 (Appendix A) e= V (ft/s) 4.58 5.50 6.42 7.33 8.25 9.17 10.08 11.00 11.92 12.83 13.75 12.0 750 hp Re 7.06E+05 8.47E+05 9.88E+05 1.13E+06 1.27E+06 1.41E+06 1.55E+06 1.69E+06 1.83E+06 1.98E+06 2.12E+06 1.84E+06 0.5 f 0.0531 0.0531 0.0531 0.0531 0.0531 0.0531 0.0531 0.0531 0.0531 0.0531 0.0531 0.0531 (Eq. 3) in Given data: L= D= ηpump = 2500 20 70% ft in Computed results: Q (ft /s) 10 12 14 16 18 20 22 24 26 28 30 26.1 Power = 3 Δp (psi) (Eq 1) Δp 11.3 16.2 22.1 28.9 36.5 45.1 54.6 64.9 76.2 88.4 101.4 76.8 (psi) (Eq 2) 135.0 130.6 125.4 119.4 112.6 105.0 96.6 87.4 77.4 66.6 55.0 76.8 Error 0.00 Using Solver ! Repeating, with smoother pipe Computed results: Q (ft /s) 10 12 14 16 18 20 22 24 26 28 30 27.8 Power = 3 e= V (ft/s) 4.58 5.50 6.42 7.33 8.25 9.17 10.08 11.00 11.92 12.83 13.75 12.8 702 hp Re 7.06E+05 8.47E+05 9.88E+05 1.13E+06 1.27E+06 1.41E+06 1.55E+06 1.69E+06 1.83E+06 1.98E+06 2.12E+06 1.97E+06 0.25 f 0.0410 0.0410 0.0410 0.0410 0.0410 0.0410 0.0410 0.0410 0.0410 0.0410 0.0410 0.0410 (Eq. 3) in Δp (psi) (Eq 1) Δp 8.71 12.5 17.1 22.3 28.2 34.8 42.1 50.1 58.8 68.2 78.3 67.4 (psi) (Eq 2) 135.0 130.6 125.4 119.4 112.6 105.0 96.6 87.4 77.4 66.6 55.0 67.4 Error 0.00 Using Solver ! Pump and Pipe Pressure Heads 160 140 120 Δp (psi) 100 80 60 40 20 0 10 15 20 Q (ft3/s) 25 30 Pipe (e = 0.5 in) Pipe (e = 0.25 in) Pump Given data: L= Dh = K= f= ρ= 30 0.5 12 0.03 1.1 m m Computed results: Q (m /s) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 2.51 3 kg/m 3 V (m/s) 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 10.06 Δp (Pa) (Eq 1) Δp 0 8 30 68 121 190 273 372 486 615 759 918 1093 768 (Pa) (Eq 2) 1020 1012 1000 984 965 942 915 884 850 812 770 724 675 768 Error 0.00 Using Solver ! Fan and Duct Pressure Heads 1200 1000 Δp (Pa) 800 600 400 200 0 0.0 0.5 1.0 1.5 Q (m /s) 3 Duct Fan 2.0 2.5 3.0 Problem 8.166 [5] Part 1/2 Problem 8.166 [5] Part 2/2 The workbook for Example 8.11 is modified for use in this problem Pipe Data: Pipe 0 1 2 3 4 L (m ) 300 400 100 100 75 D (mm) 75 75 75 75 75 e (mm) 0.15 0.15 0.15 0.15 0.15 Fluid Properties: ρ= μ= Available Head: Δp = 250 kPa 999 0.001 kg/m3 N.s/m2 Flows: Q 0 (m3/s) 0.00928 V 0 (m/s) 2.10 Re 0 1.57E+05 f0 0.0245 Q 1 (m3/s) 0.00306 V 1 (m/s) 0.692 Re 1 5.18E+04 f1 0.0264 Q 2 (m3/s) 0.00311 V 2 (m/s) 0.705 Re 2 5.28E+04 f2 0.0264 Q 3 (m3/s) 0.00311 V 3 (m/s) 0.705 Re 3 5.28E+04 f3 0.0264 Q 4 (m3/s) 0.00623 V 4 (m/s) 1.41 Re 4 1.06E+05 f4 0.0250 Heads: Δp 0 (kPa) 216.4 (1) Q 0 = Q 1 + Q4 0.00% (3) Δp = Δp 0 + Δp 1 0.03% (5) Δp 2 = Δp 3 0.00% Error: Δp 1 (kPa) 33.7 Δp 2 (kPa) 8.7 Δp 3 (kPa) 8.7 (2) Q 4 = Q 2 + Q 3 0.01% Δp 4 (kPa) 24.8 Constraints: (4) Δp = Δp 0 + Δp 4 + Δp 2 0.01% 0.05% Vary Q 0, Q 1, Q 2, Q 3 and Q 4 using Solver to minimize total error The workbook for Example Problem 8.11 is modified for use in this problem Pipe Data: Pipe A B C D Fluid Properties: ρ= μ= Available Head: Δp = 50 psi 1.94 2.10E-05 slug/ft3 lbf·s/ft2 L (ft) 150 150 150 150 D (in) 1.5 1.5 1 1.5 e (ft) 0.00085 0.00085 0.00085 0.00085 Flows: Q A (ft3/s) 0.103 V A (ft/s) 8.41 Re A 9.71E+04 fA 0.0342 Q B (ft3/s) 0.077 V B (ft/s) 6.28 Re B 7.25E+04 fB 0.0345 Q C (ft3/s) 0.026 V C (ft/s) 4.78 Re C 3.68E+04 fC 0.0397 Q D (ft3/s) 0.103 V D (ft/s) 8.41 Re D 9.71E+04 fD 0.0342 Heads: Δp A (psi) 19.5 (5) Q A = Q D 0.00% Δp B (psi) 11.0 Δp C (psi) 11.0 Δp D (psi) 19.5 (6) Q A = Q B + Q C 0.05% (8) Δp B = Δp C 0.00% Constraints: (7) Δp = Δp A + Δp B + Δp D 0.00% Error: 0.05% Vary Q A, Q B, Q C, and Q D using Solver to minimize total error Problem *8.169 [4] Problem 8.170 [5] Part 1/2 Problem 8.170 [5] Part 2/2 Problem 8.171 [2] Given: Find: Solution: Basic equation Flow through an orifice Pressure drop mactual = K⋅ At ⋅ 2⋅ ρ⋅ p1 − p2 = K⋅ At ⋅ 2⋅ ρ⋅ Δp ( ) Note that mactual is mass flow rate (the software cannot render a dot!) For the flow coefficient K = K ⎜ ReD1 , ⎛ ⎝ Dt ⎞ D1 ⎟ ⎠ At 65oC,(Table A.8) ρ = 980⋅ Q A kg m 3 ν = 4.40 × 10 4 π −7 m 2 ⋅ s × 20⋅ L × 0.001⋅ m 1⋅ L 3 V= V= V⋅ D ν ReD1 = β= From Fig. 8.20 Then Dt D1 s ( 0.15⋅ m ) m s ReD1 = 1.13⋅ × 0.15⋅ m × −7 2 s 4.40 × 10 ⋅ m β= 75 150 × 1 2 V = 1.13 m s 5 ReD1 = 3.85 × 10 β = 0.5 K = 0.624 2 2 ⎛ mactual⎞ 1 ρ⋅ Q ⎞ 1 ρ Q⎞ Δp = ⎜ ⋅ =⎛ ⋅ = ⋅⎛ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ K⋅ At ⎠ 2⋅ ρ ⎝ K⋅ At ⎠ 2⋅ ρ 2 ⎝ K⋅ At ⎠ 2 ⎤ ⎡ L 0.001⋅ m3 1 4 1 ⎥ Δp = × 980⋅ × ⎢20⋅ × × ×× 3⎢ 1⋅ L 2 s 0.624 π ( 0.075⋅ m) 2⎥ m ⎣ ⎦ 1 kg 2 Δp = 25.8 kPa Problem 8.172 [2] Problem 8.173 [2] Given: Find: Solution: Basic equation Flow through a venturi meter (NOTE: Throat is obviously 3 in not 30 in!) Flow rate C⋅ A t C⋅ A t 1−β 4 mactual = ⋅ 2⋅ ρ⋅ p1 − p2 = 4 1−β ( ) ⋅ 2⋅ ρ⋅ Δp Note that mactual is mass flow rate (the software cannot render a dot!) For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re β= Also Dt D1 β= 3 6 β = 0.5 Δp = ρHg⋅ g⋅ Δh = SGHg⋅ ρ⋅ g⋅ Δh Q= mactual ρ π = C⋅ A t ρ⋅ 1 − β 4 2 Then ⋅ 2⋅ ρ⋅ Δp = π⋅ C ⋅ D t 2 4 4⋅ ρ⋅ 1 − β ⋅ 2⋅ ρ⋅ SGHg⋅ ρ⋅ g⋅ Δh = ft s 2 π⋅ C ⋅ D t 2 4 4⋅ 1 − β ⋅ 2⋅ SGHg⋅ g⋅ Δh Q= 4× Hence V= Q A = 1⎞ × 0.99 × ⎛ ⋅ ft⎟ × ⎜ 4 ⎝4 ⎠ 1 − 0.5 4⋅ Q π⋅ D 1 2 2 × 13.6 × 32.2⋅ × 1⋅ ft Q = 1.49⋅ ft s ft s 3 V= 4 π × 1 ⎛ 1 ⋅ ft⎞ ⎜⎟ ⎝2 ⎠ 2 × 1.49⋅ ft s 3 V = 7.59⋅ At 75oF,(Table A.7) ν = 9.96 × 10 ReD1 = − 6 ft 2 ⋅ s ReD1 = 7.59⋅ ft s 3 V⋅ D1 ν ft s × 1 2 ⋅ ft × s 9.96 × 10 −6 2 ⋅ ft ReD1 = 3.81 × 10 5 Thus ReD1 > 2 x 105. The volume flow rate is Q = 1.49⋅ Given data: ΔH = L= D= Dt = β= 30 200 100 40 0.40 m m mm mm Tabulated or graphical data: K ent = 0.50 K exit = 1.00 Loss at orifice = 80% μ= 0.001 (Fig. 8.14) (Fig. 8.14) (Fig. 8.23) 2 N.s/m ρ= kg/m3 999 (Water - Appendix A) Computed results: Orifice loss coefficient: K= 0.61 (Fig. 8.20 Assuming high Re ) Flow system: V Q Re f = = = = 2.25 m/s m3/s 0.0176 2.24E+05 0.0153 Orifice pressure drop Δp = 265 kPa Eq. 1, solved by varying V AND Δp , using Solver : Left (m /s) 294 2 Right (m /s) 293 2 Error 0.5% Eq. 2 and m rate = ρQ compared, varying V AND Δp (From Q ) (From Eq. 2) m rate (kg/s) = 17.6 17.6 Total Error Error 0.0% 0.5% Procedure using Solver : a) Guess at V and Δp b) Compute error in Eq. 1 c) Compute error in mass flow rate d) Minimize total error e) Minimize total error by varying V and Δp Problem 8.175 [2] Problem 8.176 [2] Given: Find: Solution: Basic equation Flow through an venturi meter Flow rate C⋅ A t C⋅ A t 1−β 4 mactual = ⋅ 2⋅ ρ⋅ p1 − p2 = 4 1−β ( ) ⋅ 2⋅ ρ⋅ Δp Note that mactual is mass flow rate (the software cannot render a dot!) For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re β= Dt D1 mactual ρ π 4× Hence V= Q A = 1 − 0.5 4⋅ Q π⋅ D 1 2 4 β= 25 50 β = 0.5 2 Then Q= = C⋅ A t ρ⋅ 1 − β 4 ⋅ 2⋅ ρ⋅ Δp = π⋅ C ⋅ D t ⋅ 4 4⋅ 1 − β 2 2⋅ Δp ρ 3 m kg⋅ m 3N × × 2 1000⋅ kg 2 Q= × 0.99 × ( 0.025⋅ m ) × 2 × 150 × 10 ⋅ Q = 8.69 × 10 −3m 3 m s ⋅N s V= 4 π × 1 ( 0.05⋅ m ) 2 × 8.69 × 10 −3m 3 s V = 4.43 m s At 20oC (Table A.8) ν = 1.01 × 10 ReD1 = V⋅ D ν 2 −6 m ⋅ s ReD1 = 4.43⋅ Q = 8.69 × 10 m s × 0.05⋅ m × 3 s 1.01 × 10 −6 ⋅m 2 ReD1 = 2.19 × 10 Q = 0.522⋅ m 3 5 Thus ReD1 > 2 x 105. The volume flow rate is −3m s min Problem 8.177 [3] Problem 8.178 [4] Given: Find: Solution: Basic equation Flow through a venturi meter (NOTE: Throat is obviously 3 in not 30 in!) Maximum flow rate for incompressible flow; Pressure reading C⋅ A t C⋅ A t 1−β 4 mactual = ⋅ 2⋅ ρ⋅ p1 − p2 = 4 1−β ( ) ⋅ 2⋅ ρ⋅ Δp Note that mactual is mass flow rate (the software cannot render a dot!) Assumptions: 1) Neglect density change 2) Use ideal gas equation for density Then ρ= p Rair⋅ T ρ = 60⋅ lbf in 2 ×⎛ ⎜ × ⋅ ⎟× ⎝ 1⋅ ft ⎠ 53.33⋅ ft⋅ lbf 32.2⋅ lbm ( 68 + 460) ⋅ R 2 12⋅ in ⎞ 2 lbm⋅ R 1⋅ slug 1 ρ = 9.53 × 10 − 3 slug ⋅ 3 ft For incompressible flow V must be less than about 100 m/s or 330 ft/s at the throat. Hence mactual = ρ⋅ V2⋅ A2 β= Also Dt D1 mactual = 9.53 × 10 β= 3 6 ft π 1⎞ − 3 slug × 330⋅ × × ⎛ ⋅ ft⎟ ⎜ 3 s 4 ⎝4 ⎠ ft β = 0.5 Δh = 2 slug mactual = 0.154⋅ s Δp = ρHg⋅ g⋅ Δh 1 ⎛ mactual ⎞ 4 Δp = ⋅⎜ ⎟ ⋅ 1−β C⋅ A t 2⋅ ρ Δp ρHg⋅ g and in addition ⎝ ⎠ ( ) so ⎛ mactual ⎞ Δh = ⋅⎜ ⎟ 2⋅ ρ⋅ ρHg⋅ g ⎝ C⋅ A t ⎠ (1 − β4) 2 For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re Δh = (1 − 0.54) × 2 3 2 2 ⎡ ft s slug 1 4 ⎛ 4 ⎞⎤ × × × ⎢0.154 × × ×⎜ ⎟⎥ −3 13.6⋅ 1.94⋅ slug 32.2⋅ ft ⎣ s 0.99 π ⎝ 1⋅ ft ⎠ ⎦ 9.53 × 10 slug ft 3 2 Δh = 0.581⋅ ft Δh = 6.98⋅ in Hence V= 4⋅ mactual Q = 2 A π ⋅ ρ⋅ D 1 − 5 ft 2 V= 4 × π ft 3 −3 × slug 1 9.53 × 10 ⎛ 1 ⋅ ft⎞ ⎜⎟ ⎝2 ⎠ 2 × 0.154 slug s V = 82.3⋅ ft s At 68oF,(Table A.7) ν = 1.08 × 10 ReD1 = ⋅ s ReD1 = 82.3⋅ ft 1 s × ⋅ ft × −5 2 s2 1.08 × 10 ⋅ ft and pressure ReD1 = 3.81 × 10 Δh = 6.98 in 6 V⋅ D1 ν Thus ReD1 > 2 x 105. The mass flow rate is slug mactual = 0.154 s Hg Problem 8.179 [4] Problem 8.180 [5] Part 1/2 Problem 8.180 [5] Part 2/2 Problem 8.181 [1] V 1, A 1 V 2, A 2 Given: Find: Solution: Basic equations Flow through a diffuser Derivation of Eq. 8.42 Cp = p2 − p1 1 2 ⋅ ρ⋅ V 1 2 p1 ρ + V1 2 2 + g⋅ z1 = p2 ρ + V2 2 2 + g⋅ z2 Q = V⋅ A Assumptions: 1) All the assumptions of the Bernoulli equation 2) Horizontal flow 3) No flow separation p2 − p1 V1 V2 V1 ⎛ A1 ⎞ V1 = − = − ⋅⎜ ⎟ ρ 2 2 2 2 ⎝ A2 ⎠ p2 − p1 1 2 2 2 2 2 From Bernoulli using continuity Hence 2 ⎡ V 2 V 2 ⎛ A ⎞ 2⎤ ⎛ A1 ⎞ 1 1⎥ ⎢1 Cp = = ⋅ − ⋅⎜ ⎟ ⎥ = 1−⎜ ⎟ 2 1 1 2 2⎢ 2 ⎝ A2 ⎠ ⎦ ⎝ A2 ⎠ ⋅ ρ⋅ V 1 ⋅ V1 ⎣ 2 2 Finally Cp = 1 − 1 AR 2 which is Eq. 8.42. This result is not realistic as a real diffuser is very likely to have flow separation Problem 8.182 [5] Part 1/2 Problem 8.182 [5] Part 2/2 Problem 8.183 [4] Problem 8.184 [4] Part 1/2 Problem 8.184 [4] Part 2/2 Problem 8.185 Problem 8.158 [5] Part 1/3 Problem 8.158 8.158 Problem 8.158 Problem 8.158 Problem 8.185 [5] Part 2/3 Problem 8.185 [5] Part 1/3 ...
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This note was uploaded on 03/09/2010 for the course ME 309 taught by Professor Merkle during the Spring '08 term at Purdue University-West Lafayette.

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