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Problems and Answers Problems and Answers Part I Problem I.1 Two parallel flat plates are immersed vertically in water. The clearance between the two plates is b. Calculate the elevation of the liquid due to capillary force. The surface tension, σ, and the density of water are 0.0725 N/m and 3 997 kg/m , respectively. The contact angle, θc,is5, and b ¼ 0.5 mm. Problem I.2 Calculate the absolute pressure on the bottom of the sea. The depth of 3 the sea, H, is 250 m, the density of seawater, ρsw, is 1,030 kg/m , and the atmospheric pressure, p0, is 1,013.3 h Pa. Problem I.3 Mercury is used as the working fluid for measuring the pressure of water by a U-shaped manometer. The water pressure at one entrance of the manometer, p1, is 300 k Pa, and pressure at the other entrance, p2, is 230 k Pa. Calculate the mercury column height, H. The densities of the mercury and water are 13,600 kg/m3 and 1,000 kg/m3, respectively. Answer to Problem I.1 Let the width of each plate be W. The following force balance holds: bWp0 þ ρgbWH ¼ bWp0 þ 2Wσ cos θc (PI.1) The height, H, is expressed by H ¼ 2σ cos θc=ðρgbÞ (PI.2) H ¼ 2  0:0725  cos 5=ð997  9:80  0:5  10À3Þ (PI.3) ¼ 0:0296 m ¼ 29:6mm M. Iguchi and O.J. Ilegbusi, Basic Transport Phenomena in Materials Engineering, 227 DOI 10.1007/978-4-431-54020-5, © Springer Japan 2014 228 Problems and Answers Answer to Problem I.2 p ¼ p0 þ ρ gH ¼ 1; 013:3 hPa þ 1; 030  9:8  250 Pa sw (PI.4) ¼ 101:33 kPa þ 2523:5 kPa ¼ 2:62 MPa Answer to Problem I.3 The mercury column height is calculated from Eqs. (1–12) as follows: ¼ð À Þ=½ðρ À ρ Þ H p1 p2 Hg w g (PI.5) ¼ð300 À 230Þ103=½ð13,600 À 1; 000Þ9:8¼0:567 m Problem I.4 Calculate the Reynolds number, Re, for oil flow in a circular pipe. The diameter of the pipe is 50 mm, the density of the oil is 920 kg/m3, the volumetric oil flow rate is 56 L/min, and the dynamic viscosity of the oil is 40 m Pa s. Problem I.5 Calculate the mean velocity of water flow in a pipe. The Reynolds number is 3,000, the pipe diameter is 10 mm, and the kinematic viscosity of water is 1.01  10À6 m2/s. Problem I.6 Consider the flow in a pipe. The pipe diameter is 50 mm, the density of fluid is 920 kg/m3, the flow rate is 0.150 m3/min, and the kinematic viscosity is 56 m Pa s. The critical Reynolds number is 2,320. Determine if the flow is laminar or turbulent. Problem I.7 Calculate the volumetric flow rate of oil flowing in a circular pipe of 50 mm diameter. The Reynolds number is 2,320, the density of oil is 920 kg/m3, and the dynamic viscosity of oil is 50 m Pa s. Problem I.8 Show that the following relationship is obtained from the parabolic distribution of laminar pipe flow velocity: vm=vcl ¼ 1=2 (PI.6) Problem I.9 Show that the following relationship is obtained from the 1/7th power law of turbulent pipe flow velocity: vm=vcl ¼ 49=60 (PI.7) Problem I.10 Show that the following relationship is obtained from the 1/nth power law of turbulent pipe flow velocity: 2 vm=vcl ¼ 2n =½ðÞn þ 1 ðÞ2n þ 1 (PI.8) Problems and Answers 229 Answer to Problem I.4 The cross-sectional mean velocity, vm, is given by 2 v ¼ 4Q=ðπD2Þ¼4  56  10À3  ðÞ1=60 =½3:14 ð50  10À3Þ m (PI.9) ¼ 0:476 m=s The Reynolds number is calculated to give À3 À3 Re ¼ ρvmD=μ ¼ 920  0:476  50  10 =ð40  10 Þ¼547 (PI.10) where ρ and μ are the density and dynamic viscosity, respectively. Answer to Problem I.5 The Reynolds number is expressed by v D Re ¼ m (PI.11) νf where νf is the kinematic viscosity. The cross-sectional mean velocity, vm,is calculated from Eq. (pI.11) to give À6 À3 vm ¼ νfRe=D ¼ 1:01  10  3,000=ð10  10 Þ¼0:303 m=s (PI.12) Answer to Problem I.6 The cross-sectional mean velocity, vm, is given by 2 v ¼ 4Q=ðπD2Þ¼4  0:150  ðÞ1=60 =½3:14 ð50  10À3Þ m (PI.13) ¼ 1:27 m=s The Reynolds number becomes À3 À3 Re ¼ ρvmD=μ ¼ 920  1:27  50  10 =ð56  10 Þ¼1; 043 (PI.14) This Reynolds number is lower than the critical Reynolds number of 2,320, and the flow is laminar. Answer to Problem I.7 The flow rate, Q, is expressed by 2 2 Q ¼ πD vm=4 ¼ðπD =4ÞνfRe=D ¼ πDνf Re=4 ¼ πDμRe=ð4ρÞ ¼ 3:14  40  10À3  50  10À3  2,320=ð4  920Þ (PI.15) ¼ 3:96  10À3 m3=s 230 Problems and Answers Answer to Problem I.8 The cross-sectional mean velocity, vm, is defined as ð ð R 2 vm ¼ vdA=A ¼ 2π rvdr=ðπR Þ (PI.16) A 0 where A(¼πR2) is the cross-sectional area of the pipe. Substituting Eq. (2.68) into Eq. (pI.16) gives ð R ÀÁ 2 2 2 vm ¼ 2rvcl 1 À r =R dr=ðR Þ (PI.17) 0 The following new variable is introduced: x ¼ r=R (PI.18) Differentiation of Eq. (pI.18) yields dx ¼ dr=R (PI.19) Equation (pI.17) is transformed into ð 1 ÀÁ 2 vm ¼ 2vclx 1 À x dx (PI.20) 0 Integration of Eq. (pI.20) gives ð 1 ÀÁ 3 1 1 vcl vm ¼ 2vcl x À x dx ¼2vcl À ¼ (PI.21) 0 2 4 2 Problem I.11 Calculate the Reynolds number, given D ¼ 10 mm, vm ¼ 15 m/s, À5 2 and kinematic viscosity of air νa ¼ 1.5  10 m /s. Problem I.12 Calculate the cross-sectional mean velocity of water flow in a pipe, the Reynolds number, the friction factor, and the pressure drop. The pipe diameter is 50 mm, the pipe length is 20 m, the water flow rate is 1.00  10À3 m3/s, the density of water is 997 kg/m3, and the kinematic viscosity is 1.00  10À6 m2/s. Problem I.13 Obtain the dynamic viscosity, μ, given that D ¼ 1.0 mm, L ¼ 1.00 m, mass flow rate of oil ρ QL ¼ 0.260 kg/h, pressure drop, Δp ¼ 640 k Pa, and ρ ¼ 920 kg/m3. Answer to Problem I.11 The Reynolds number, Re, is given by À3 À5 4 Re ¼ vmD=νa ¼ 15  10  10 =ð1:5  10 Þ¼1:0  10 (PI.22) Problems and Answers 231 Answer to Problem I.12 The cross-sectional mean velocity, vm, is given by 2 v ¼ 4Q =ðπD2Þ¼4  1:00  10À3=½3:14 ð50  10À3Þ m L (PI.23) ¼ 0:510 m=s The Reynolds number, Re, is given by À3 À6 4 Re ¼ vmD=ν ¼ 0:510  50  10 =ð1:00  10 Þ¼2:55  10 (PI.24) The friction factor, λ, is calculated from the Blasius equation as À1=4 λ ¼ 0:3164 ReÀ1=4 ¼ 0:3164 ð2:55  104Þ ¼ 0:0250 (PI.25) The pressure drop, Δp, becomes Δp ¼ λðÞL=D ρv 2=2 ¼ 0:0250 ½20=ð50  10À3Þ Â 997  ðÞ0:510 2=2 m (PI.26) ¼ 1:30 kPa where L is the pipe length. Answer to Problem I.13 The pressure loss is expressed by L ρv 2 Δp ¼ λ m (PI.27) D 2 First, the flow in the pipe is assumed to be laminar. The friction factor, λ,is expressed as λ ¼ 64=Re ¼ 64νf =ðÞvmD (PI.28) Substituting Eq. (pI.28) into Eq. (pI.27) yields 2 Δp ¼ 32μLvm=D (PI.29) Eq. (pI.29) is rewritten by 2 μ ¼ ΔpD =ð32LvmÞ (PI.30) The flow rate, QL, is given by À8 3 QL ¼ 0:26=ð3,600  920Þ¼7:86  10 m =s; (PI.31) 232 Problems and Answers The cross-sectional mean velocity, vm, is calculated as follows: 2 v ¼ 4Q =ðπD2Þ¼4  7:86  10À8=½3:14 ð1:0  10À3Þ m L (PI.32) ¼ 0:100 m=s Substituting Eq. (pI.32) into Eq. (pI.30) gives 2 μ ¼ ΔpD2=ð32Lv Þ¼640  103 ð1:0  10À3Þ =ð32  1:0  0:100Þ m (PI.33) ¼ 0:200 Pa s It is necessary to check whether the laminar flow assumption is adequate or not: À3 Re ¼ ρvmD=μ ¼ 920  0:100  1:0  10 =0:200 ¼ 0:46 (PI.34) Accordingly, the laminar flow assumption is adequate. Problem I.14 A sphere of diameter, DpR, of 1.00 m is immersed in a river. The velocity of the approaching flow, VR, is 2.0 m/s. A model experiment is carried out using air as the working fluid to predict the drag force acting on the sphere.
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