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 10 3Þ (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 Þ=½ðρ ρ Þ H p1 p2 Hg w g (PI.5) ¼ð300 230Þ 103=½ð13,600 1; 000Þ 9:8 ¼0:567 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 10 6 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 =½ ðÞn 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 10 3 ðÞ1=60 =½3:14 ð50 10 3Þ m (PI.9) ¼ 0:476 m=s
The Reynolds number is calculated to give