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 , 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 103Þ (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 230Þ103=½ð13,600 1; 000Þ9:8¼0:567 m

Problem I.4 Calculate the , 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 1.01 106 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 56 103 ðÞ1=60 =½3:14 ð50 103Þ m (PI.9) ¼ 0:476 m=s

The Reynolds number is calculated to give

3 3 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

6 3 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 103Þ m (PI.13) ¼ 1:27 m=s

The Reynolds number becomes

3 3 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 40 103 50 103 2,320=ð4 920Þ (PI.15) ¼ 3:96 103 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 2 2 2 vm ¼ 2rvcl 1 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 2 vm ¼ 2vclx 1 x dx (PI.20) 0

Integration of Eq. (pI.20) gives ð 1 3 1 1 vcl vm ¼ 2vcl x x dx ¼2vcl ¼ (PI.21) 0 2 4 2

Problem I.11 Calculate the Reynolds number, given D ¼ 10 mm, vm ¼ 15 m/s, 5 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 1.00 103 m3/s, the density of water is 997 kg/m3, and the kinematic viscosity is 1.00 106 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

3 5 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 1:00 103=½3:14 ð50 103Þ m L (PI.23) ¼ 0:510 m=s

The Reynolds number, Re, is given by

3 6 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

1=4 λ ¼ 0:3164 Re1=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 103Þ 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

8 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 7:86 108=½3:14 ð1:0 103Þ 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 103Þ =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:

3 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. The diameter of the model sphere, DpM, is 0.10 m. Determine the approaching velocity of air flow, VM, in terms of the Reynolds number similitude by assuming that the temperatures of air and water are 20 C. (Ans: 300 m/s)

Problem I.15 A sphere of diameter, Dp, of 5.0 cm is moving in still water at a -6 2 velocity, V, of 2.0 m/s. The kinematic viscosity of water, νw, is 1.01 10 m /s. The drag coefficient, CD, is 0.44. Determine the hydrodynamic drag acting on the sphere. (Ans: 1.72 N) Problem I.16 A sphere of density of 3,000 kg/m3 and diameter of 5.0 cm is falling in still water of density of 998 kg/m3. The drag coefficient is 0.44. Determine the following quantities:

(a) Volume of sphere, Vp (b) Buoyancy force acting on sphere, FB (c) Weight of sphere, Fw (d) Drag force acting on sphere, FD (e) Terminal velocity of sphere, v1 (Ans: 1.72 m/s) Problem I.17 A man weighing 620 N descends in the atmosphere by parachute. The diameter and weight of the parachute are 5.8 m and 16.8 N, respectively. The density of air is 1.205 kg/m3. The drag coefficient is assumed to be 1.0. Determine the terminal velocity, v1.

Problem I.18 Calculate the cross-sectional mean velocity, vm, Reynolds number, Re, friction coefficient, λ, and pressure drop, Δp, of water flow given 3 3 6 2 that D ¼ 40 mm, L ¼ 30 m, QL ¼ 1.00 10 m /s, and ν ¼ 1.00 10 m /s. The critical Reynolds number, Rec, is 2,320. Problems and Answers 233

Problem I.19 A sphere of diameter of 3.0 cm is moving in still water at a terminal velocity of 3.0 m/s. The density and kinematic viscosity of water are 997 kg/m3 and 1.01 106 m2/s, respectively. Determine the hydrodynamic drag acting on the sphere. Answer to Problem I.14 According to the Reynolds similitude, the following relationship holds:

VRDpR=νR ¼ VMDpM=νM (PI.35) 5 6 VM ¼ VR DpR=DpM ðνM=νRÞ¼2:0 ðÞ½1:00=0:10 1:50 10 =ð1:0 10 Þ ¼ 300 m=s (PI.36) where the subscripts R and M denote the real process and model, respectively. Answer to Problem I.15 The density of water, ρ, is 997 kg/m3. The hydrodynamic drag is expressed by

2 F ¼ C A ρV2=2 ¼ 0:44 ½π ð5:0 102Þ =4997 ðÞ2:0 2=2 D D p (PI.37) ¼ 1:72 N

Answer to Problem I.16

(a) The volume of the sphere, Vp,is

3 2 3 5 3 Vp ¼ πDp =6 ¼ 3:14 ð5:0 10 Þ =6 ¼ 6:54 10 m (PI.38)

(b) The buoyancy force acting on the sphere, FB,is

¼ ρ ¼ : 5 : ¼ : FB wVpg 998 6 54 10 9 80 0 640 N (PI.39)

(c) The weight of the sphere, Fw,is

¼ ρ ¼ ; : 5 : ¼ : Fw pVpg 3 000 6 54 10 9 80 1 923 N (PI.40)

(d) The drag force acting on the sphere, FD, is calculated as

FD ¼ Fw FB ¼ 1:923 0:640 ¼ 1:283 N (PI.41) 234 Problems and Answers

(e) The terminal velocity of the sphere, v1, can be calculated from the following relation:

¼½ =ð ρ Þ1=2 v1 2FD CDAp w (PI.42)

2 2 2 3 2 Ap ¼ πDp =4 ¼ 3:14 ð5:0 10 Þ =4 ¼ 1:96 10 m (PI.43)

Substituting Eq. (pI.43) and the other quantities into Eq. (pI.42) gives

3 1=2 v1 ¼½2 1:283=ð0:44 1:96 10 998 ¼ 1:72 m=sÞ (PI.44)

Answer to Problem I.17 The hydrodynamic drag is given by

FD ¼ 620 þ 16:8 ¼ 636:8 N (PI.45)

The terminal velocity can be calculated as follows:

2 2 2 Ap ¼ πD =4 ¼ 3:14 ðÞ5:8 =4 ¼ 26:4m (PI.46)

1=2 1=2 v1 ¼½2F =ðC A ρ Þ ¼½2 636:8=ð1:0 26:4 1:205Þ D D p w (PI.47) ¼ 6:33 m=s

Answer to Problem I.18

3 2 vm ¼ Q=A ¼ 1:00 10 =½π ð0:040Þ =4¼0:796 m=s (PI.48)

6 4 Re ¼ vmD=ν ¼ 0:796 0:040=ð1:00 10 Þ¼3:18 10 (PI.49)

1=4 λ ¼ 0:3164Re1=4 ¼ 0:3164 ð3:18 104Þ ¼ 0:0237 (PI.50)

2 2 Δp ¼ λðÞL=D ρvm =2 ¼ 0:0237 ðÞ30=0:040 997 ðÞ0:796 =2 (PI.51) ¼ 8:77 103 Pa

Answer to Problem I.19 The Reynolds number is

2 6 4 Re ¼ V1D=νw ¼ 3:0 3:0 10 =ð1:01 10 Þ¼8:91 10 (PI.52) Problems and Answers 235

The drag coefficient follows Newton’s resistance law, and hence,

CD ¼ 0:44 (PI.53)

The hydrodynamic drag is given by

¼ ρ 2= FD CDAp wV1 2 (PI.54) ¼ 0:44 ½3:14 ðÞ0:03 2=4997 ð3Þ2=2 ¼ 1:39 N

Part II

Problem II.1 Derive Eq. (5.17). Problem II.2 The thickness and thermal conductivity of a flat wall made of concrete are 10 mm and 0.1 W/m K, respectively. The difference between the temperatures of the inner and outer surfaces of the wall is 35 C. Calculate the heat transfer by conduction in 2 h assuming that the wall surface area is 20 m2. Problem II.3 The outer surface of a flat plate made of iron is covered with an insulator. The thickness of the plate is 2.0 cm and the insulator is 5.0 cm thick. The inner wall of the plate is exposed to a gas at 400 C and the outer surface of the insulator is exposed to the atmosphere at 20 C. The heat-transfer coefficients on the flat plate side and insulator side are 5.0 and 1.0 W/m2 K, respectively. The thermal conductivities of the iron plate and the insulator are 40.0 and 0.050 W/ m K. Calculate the temperatures on the inner wall of the flat plate and on the outer wall of the insulator and the heat flux. Answer to Problem II.2 The heat transfer by conduction, Q, is given by

Q ¼ λ½ðθ1 θ2Þ=δAt ¼ 0:1 ½35=0:01 20 3,600 2 ¼ 5:04 107 J (PII.1)

Answer to Problem II.3 The heat flux, q, is expressed by

q ¼ðθa θbÞ=ð1=α1 þ δ1=λ1 þ δ2=λ2 þ 1=α2Þ (PII.2) 236 Problems and Answers

Substituting the abovementioned values into Eq. (pII.2) yields

q ¼ð400 20Þ=ðÞ1=5 þ 0:02=40 þ 0:05=0:05 þ 1=1 ¼ 380=ðÞ¼0:2 þ 0:0005 þ 1 þ 1 380=2:2005 (PII.3) ¼ 172:7W=m2

The wall temperatures are given by

θ1 ¼ θa q=α1 ¼ 400 172:7=5 ¼ 365:5 C (PII.4)

θ2 ¼ θb þ q=α2 ¼ 20 þ 172:7=1 ¼ 192:7 C (PII.5)

Problem II.4 The surfaces of a flat plate 2.0 cm thick are kept at 300 C and 10 C. Calculate the heat transfer in 2 h and heat flux assuming that the thermal conduc- tivity is 40 W/m K. Problem II.5 The difference between the inner and outer surface temperatures of a 50 mm thick pressure vessel is 40 C. The total surface area, A, of the vessel is 50 m2. The vessel is constructed of a material having thermal conductivity, λ,of 50 W/m K. Calculate the heat flux and the heat loss in 1 s. Problem II.6 A flat iron plate 4.0 cm thick is covered with insulator 4.0 cm thick. The inner wall of the plate is exposed to a gas at 350 C, while the outer wall of the insulator is surrounded by air at a temperature of 20 C. The heat- transfer coefficients on the inner and outer sides are 5.0 and 2.0 W/m2 K, respec- tively. The thermal conductivities of the plate material and insulator are 40.0 and 0.050 W/m K, respectively. Calculate the surface temperatures of the plate and insulator and the heat flux. Problem II.7 The thickness and thermal conductivity of a flat wall made of concrete are 10 mm and 0.15 W/m K, respectively. The difference in temperature between the inner and outer surfaces of the wall is 40 C. Calculate the heat transferred by conduction in 3 h assuming that the surface area of the wall is 15 m2. Problem II.8 Two plates made of different materials, A and B, have thermal conductivities λ1 and λ2, respectively. The plates have equal width, m, but different thicknesses, δ1 and δ2. As shown in Fig. PII.1a, b, two types of multilayer plates are constructed using four plates made of A and three plates made of B. Calculate the overall thermal conductivities of the two multilayer plates assuming that the surface temperatures are θ1, θ2(θ1 > θ2). Answer to Problem II.4

Q ¼ λ½ðθ1 θ2Þ=δAt ¼ 40 ½ðÞ300 10 =0:02 1 2 3,600 (PII.6) ¼ 4:18 109 J Problems and Answers 237

Fig. PII.1 Multilayer plates

Answer to Problem II.5 The wall is regarded as a flat plate. The heat transferred by conduction is given by

Q ¼ λ½ðθ1 θ2Þ=δAt ¼ 50 ðÞ40=0:050 50 1 (PII.7) ¼ 2:00 106 J

Answer to Problem II.6 The heat flux, q, is expressed by

q ¼ðθa θbÞ=ð1=α1 þ δ1=λ1 þ δ2=λ2 þ 1=α2Þ ¼ð350 20Þ=ðÞ1=5 þ 0:04=40 þ 0:04=0:05 þ 1=2 (PII.8) ¼ 330=ðÞ¼0:2 þ 0:001 þ 0:8 þ 0:5 330=1:501 ¼ 219:9W=m2

The wall temperatures are given by

θ1 ¼ θa q=α1 ¼ 350 219:9=5 ¼ 306:0 C (PII.9)

θ2 ¼ θb þ q=α2 ¼ 20 þ 219:9=2 ¼ 130:0 C (PII.10) 238 Problems and Answers

Answer to Problem II.7

Q ¼ λ½ðθ1 θ2Þ=δAt ¼ 0:15 ½40=0:010 15 3 3; 600 (PII.11) ¼ 9:72 107 J

Answer to Problem II.8

Case 1 The heat flux, q, can be expressed by

q ¼ðθ1 θ2Þ=ð4δ1=λ1 þ 3δ2=λ2Þ (PII.12)

The overall thermal conductivity, λ, is defined as follows:

q ¼ λðθ1 θ2Þ=ð4δ1 þ 3δ2Þ (PII.13)

Combining Eqs. (pII.12) and (pII.13) yields

λ ¼ λ1λ2ð4δ1 þ 3δ2Þ=ð4δ1λ2 þ 3δ2λ1Þ (PII.14)

Case 2 The heat flux for plate A is expressed by

qA ¼ λ1ðθ1 θ2Þ=m (PII.15)

In the same manner the heat flux for plate B is given by

qB ¼ λ2ðθ1 θ2Þ=m (PII.16)

Accordingly, the total heat transfer, Q, becomes

Q ¼ qA 4ðδ1lÞþqB 3ðδ2lÞ (PII.17) ¼ð4λ1δ1 þ 3λ2δ2Þðθ1 θ2Þl=m

The overall thermal conductivity, λ, is defined as

Q ¼ λðθ1 θ2Þð4δ1 þ 3δ2Þl=m (PII.18)

Combination of Eqs. (pII.17) and (pII.18) yields

λ ¼ð4λ1δ1 þ 3λ2δ2Þ=ð4δ1 þ 3δ2Þ (PII.19)

Problem II.9 Two materials, 1 and 2, used for reducing heat loss have thermal conductivities λ1 and λ2 (λ2 > λ1), respectively. Two types of insulation patterns are Problems and Answers 239

Fig. PII.2 Insulation pattern effect on heat loss chosen as shown in Fig. PII.2a, b. The amount of the material 1 in (a) is the same as that in (b). The same is true for material 2. Which pattern is more effective for reducing heat loss? Problem II.10 Steam flows in a hollow cylinder of outer diameter 110 mm and length 30 m. The cylinder is placed in the atmosphere and covered with insulator to reduce heat loss from the surface. The thickness and the thermal conductivity of the insulator are 5 cm and 0.12 W/m K, respectively. The inner and outer surfaces of the insulator are maintained at 100 C and 30 C, respectively. Calculate the heat loss from the outer surface of the insulator per unit time (1 s). Problem II.11 A long hollow cylinder of inner diameter 200 mm and thickness 10 mm is placed vertically in the atmosphere. The temperature of gas flowing in the cylinder is 400 C and that of air is 20 C. The heat-transfer coefficients on the inside and outside of the cylinder are 50.0 and 5.0 W/m2 K, respectively. The thermal conductivity of the cylinder is 40.0 W/m K. (a) Draw the cylinder with symbols, if necessary. (b) Calculate the heat transferred per unit length (1 m), i.e., the heat-transfer rate. (c) Calculate the temperature on the inner and outer walls of the cylinder.

Problem II.12 A hollow sphere has inner radius r1 and outer radius r2. The temperatures on the inner and outer walls are kept constant at θ1 and θ2 (θ1 > θ2), respectively. Calculate the heat transfer in the radial direction per unit time using Fourier’s law. Problem II.13 A sphere of 6 cm diameter is placed in the atmosphere. The temperature of the sphere is 317 C and that of air is 17 C. Calculate the heat loss due to natural . The and are 240 Problems and Answers

Table PII.1 Physical properties of air Temperature, θ (K) 300 400 600 800 Density, ρ (kg/m3) 1.176 0.882 0.588 0.441 2 Kinematic viscosity, νf (mm /s) 15.83 26.39 52.36 84.5 Thermal conductivity, λ (mW/m K) 26.14 33.05 45.6 56.9 , Pr () 0.717 0.715 0.710 0.719 expressed by the following relations and the physical properties of air are given in Table PII.1:

Nu ¼ 2 þ 0:60Pr1=3Gr1=4 (PII.20)

3 2 Gr ¼ gβd ðθs θfÞ=νf (PII.21)

Problem II.14 A sphere of radius 6 cm is placed in air flowing at velocity 20 m/s. The temperatures of the sphere and air are 317 C and 17 C, respectively. Calculate the heat loss due to forced convection. The Nusselt number is expressed by the following relation and the physical properties of air are given in Table PII.1:

Nu ¼ 2 þ 0:60Pr1=3Re1=2 (PII.22)

Problem II.15 Explain the following: (a) Three basic types of heat transfer (b) Reynolds number (definition and physical meaning) (c) Prandtl number (definition and physical meaning) (d) Nusselt number (definition and physical meaning) (e) Fourier’s law of conduction (f) Newton’s law of cooling (g) Natural convection (h) Forced convection Answer to Problem II.9 The temperature on the inner wall of the inner layer and that on the outer wall of the outer layer in case (a) are denoted by θ1 and θ2, respectively. In the same manner, 0 0 those in case (b) are denoted by θ1 and θ2 , respectively. The heat transfer per unit time in the two cases (a) and (b) are expressed by

Qa ¼ 2πðθ1 θ2Þ=½ð1=λ1Þ lnðÞþð 18=10 1=λ2Þ lnðÞ 30=18 (PII.23)

0 0 Qb ¼ 2πðθ1 θ2 Þ=½ð1=λ2Þ lnðÞþð 26=10 1=λ1Þ lnðÞ 30=26 (PII.24) Problems and Answers 241

The difference between the two denominators is expressed by

ð1=λ1Þ½þðlnðÞ 18=10 lnðÞ 30=26 1=λ2Þ½lnðÞ 30=18 lnðÞ 26=10

¼ð1=λ1Þ ln½18 26=ð10 30Þ ð1=λ2Þ ln½26 18=ð30 10Þ (PII.25) ¼ð1=λ1Þ ln½ð 156=100 1=λ2Þ ln½ 156=100

¼ð1=λ1 1=λ2Þ lnðÞ 156=100

This value is positive because λ2 > λ1. If we consider the case that (θ1 θ2) ¼ 0 0 (θ1 θ2 ), Qa < Qb. Consequently, (a) is more effective for reducing the heat loss. Answer to Problem II.10

θ1 ¼ 100 C (PII.26)

θ2 ¼ 30 C (PII.27)

r1 ¼ 110=2 ¼ 55 mm (PII.28)

r2 ¼ 55 þ 50 ¼ 105 mm (PII.29)

θ ¼ 2π lλðθ1θ2Þ= lnðÞr2=r1 ¼ 2 3:14 30 0:12 ðÞ100 30 = lnðÞ 0:105=0:055 (PII.30) ¼ 1:58 103=0:647 ¼ 2:44 103 W

Answer to Problem II.11

θa ¼ 400 C (PII.31)

θb ¼ 20 C (PII.32)

2 α1 ¼ 50:0W=m K (PII.33)

2 α2 ¼ 5:0W=m K (PII.34)

λ ¼ 40:0W=m K (PII.35)

r1 ¼ 200=2 ¼ 100 mm (PII.36)

r2 ¼ r1 þ 10 ¼ 100 þ 10 ¼ 110 mm (PII.37) 242 Problems and Answers

The heat-transfer rate is given by

Q ¼ 2πlðθa θbÞ=½1=ðα1r1Þþð1=λÞ lnðr2=r1Þþ1=ðα2r2Þ ¼ 2 3:14 1 ð400 20Þ=½1=ð50:0 0:100Þþð1=40:0Þ lnð0:110=0:100Þ þ 1=ð5:0 0:110Þ ¼ 2386.3=2:02 ¼ 1:18 103 W (PII.38)

The temperatures on the inner and outer walls are expressed by

θ1 ¼ θa Q=ð2πr1α1Þ¼400 1181=ð2 3:14 0:100 50:0Þ (PII.39) ¼ 400 37:62 ¼ 362:38 C

θ2 ¼ θb þ Q=ð2πr2α2Þ¼20 þ 1,181=ð2 3:14 0:110 5:0Þ (PII.40) ¼ 20 þ 341:92 ¼ 361:92 C

Answer to Problem II.12 Fourier’s law is expressed by

Q ¼λðdθ=drÞA (PII.41)

A ¼ 4πr2 (PII.42)

Substituting Eq. (pII.42) into Eq. (pII.41) yields

Q ¼4πλr2ðdθ=drÞ (PII.43)

dθ ¼Q=ð4πλÞdr=r2 (PII.44)

The boundary conditions are

θ ¼ θ1 at r ¼ r1 (PII.45)

θ ¼ θ2 at r ¼ r2 (PII.46)

Integrating Eq. (pII.44) subject to the boundary conditions of Eqs. (pII.45) and (pII.46) yields

Q ¼ 4πλðθ1 θ2Þ=½1=r1 1=r2 (PII.47) Problems and Answers 243

Answer to Problem II.13 The mean (film) temperature is given by

θfm ¼ ðÞ317 þ 17 =2 ¼ 167 C (PII.48)

The physical properties of air at this temperature are calculated from Table PII.1 as follows: Kinematic viscosity, νf

νf ¼ 26:39 þ 40 ð52:36 26:39Þ=ð600 400Þ¼26:39 þ 5:19 (PII.49) ¼ 31:58 mm2=s ¼ 31:58 106 m2=s

Thermal conductivity of air, λf

λf ¼ 33:05 þ 40 ð45:60 33:05Þ=ð600 400Þ¼33:05 þ 2:51 ¼ 35:56 m W=mK¼ 35:56 103 W=mK (PII.50)

Prandtl number, Pr

Pr ¼ 0:715 þ 40 ðÞ0:710 0:715 =ðÞ¼600 400 0:715 0:001 (PII.51) ¼ 0:714

The volume coefficient of expansion, β, is given by

β ¼ 1=ðÞ¼273 þ 17 1=290 (PII.52)

The Grashof and Nusselt numbers are calculated as follows:

¼ β 3ðθ θ Þ=ν2 Gr g d s f f 2 ¼ 9:80 ðÞ1=290 ðÞ0:06 3 ðÞ317 17 =ð31:58 106Þ (PII.53) ¼ 2:196 106

Nu ¼ 2 þ 0:60Pr1=3Gr1=4 = 1=4 ¼ 2 þ 0:60 ðÞ0:714 1 3 ð2:196 106Þ (PII.54) ¼ 2 þ 0:60 0:8938 38:49 ¼ 2 þ 20:64 ¼ 22:64½

The heat-transfer coefficient, α, is calculated as

α ¼ Nuλf=d (PII.55) ¼ 22:64 35:56 103=0:06 ¼ 13:42 W=m2 K 244 Problems and Answers

Accordingly, the heat transferred by natural convection per unit time is given by

2 Q ¼ αðθs θfÞA ¼ αðθs θfÞπd (PII.56) ¼ 13:42 ð317 17Þπ ð0:06Þ2 ¼ 45:53 W

Answer to Problem II.14 The diameter of the sphere, d,is

d ¼ 0:06 2 ¼ 0:12 m (PII.57)

The kinematic viscosity, νf, thermal conductivity, λf, and Prandtl number, Pr, are

6 2 νf ¼ 31:58 10 m =s (PII.58)

3 λf ¼ 35:56 10 W=m K (PII.59)

Pr ¼ 0:714 (PII.60)

The Reynolds number, Re, Nusselt number, Nu, and heat-transfer coefficient, α, are given by

6 4 Re ¼ Vd=νf ¼ 20 0:12=ð31:58 10 Þ¼7:60 10 ½ (PII.61)

Nu ¼ 2 þ 0:60Pr1=3Re1=2 = 1=2 ¼ 2 þ 0:60 ðÞ0:714 1 3 ð7:60 104Þ (PII.62) ¼ 2 þ 0:60 0:8938 275:7 ¼ 2 þ 147:8 ¼ 149:8 ½

α ¼ Nu λf=d (PII.63) ¼ 149:8 35:56 103=0:12 ¼ 44:40 W=m2 K

The heat-transfer rate by natural convection is given by

2 Q ¼ αðθs θfÞA ¼ αðθs θfÞπd (PII.64) ¼ 44:40 ð317 17Þπ ð0:12Þ2 ¼ 602:6W

Problem II.16 The temperature in a room is 30 C and that in the atmosphere is 5 C. The thickness of a window made of glass is 4.0 mm. The thermal conductivity of glass, λ, is 0.76 W/m K, and the heat-transfer coefficients inside and outside the 2 room, α1 and α2, are 10 and 50 W/m K, respectively. Calculate the heat flux, q, defined as heat flow per unit time and unit area. (Ans: 199.5 J/m2 s) Problems and Answers 245

Problem II.17 Two parallel plates are placed vertically at a distance L apart. The temperatures of the plates are θ1 and θ2 (θ1 > θ2). Calculate heat-transfer rates, Qi and Qii, in the following two cases: (i) Two cylindrical rods of cross-sectional area, A, and length, L, connect the two plates separately. The thermal conductivity of one rod is λa and that of the other rod is λb. Consider one-dimensional heat conduction, that is, the heat losses on the surfaces of the rods are negligible. (ii) Two cylindrical rods of cross-sectional area, 2A, and length, L/2, are combined into one cylindrical rod and then used to connect the two plates. The thermal conductivity of one rod is λa and that of the other rod is λb. Consider one-dimensional heat conduction.

Which heat-transfer rate is higher, Qi or Qii? Answer to Problem II.16 The heat flux, q, can be calculated from Eq. (5.16) to give

q ¼ðθa θbÞ=ð1=α1 þ δ=λ þ 1=α2Þ ¼ð30 5Þ=ðÞ1=10 þ 0:004=0:76 þ 1=50 (PII.65) ¼ 25=ðÞ¼0:1 þ 0:005263 þ 0:02 25=0:1253 ¼ 199:5W=m2

Answer to Problem II.17 Two heat-transfer rates can be evaluated in the following manner. Case 1 The heat-transfer rates through the cylindrical rods 1 and 2 are expressed by

Qa ¼ λa½ðθ1 θ2Þ=LÞAt (PII.66)

Qb ¼ λb½ðθ1 θ2Þ=LÞAt (PII.67)

Accordingly, Qi is given by

Qi ¼ Qa þ Qb (PII.68) ¼ðλa þ λbÞ½ðθ1 θ2Þ=LÞAt

Case 2 The heat-transfer rate, Qii, is readily derived from Eq. (5.20) as follows:

Qii ¼ðθ1 θ2Þð2AtÞ½ðL=2Þ=λa þðL=2Þ=λb (PII.69) ¼½4λaλb=ðλa þ λbÞ ½ðθ1 θ2Þ=LÞAt

We need to judge which heat-transfer rate is higher, Qi or Qii. At a glance, it is sufficient to examine the following function:

y ¼ðλa þ λbÞ4λaλb=ðλa þ λbÞ (PII.70) 246 Problems and Answers

Equation (pII.70) can be transformed into

2 y ¼½ðλa þ λbÞ 4λaλb=ðλa þ λbÞ 2 2 ¼ðλa þ 2λaλb þ λb 4λaλbÞ=ðλa þ λbÞ (PII.71) 2 2 ¼ðλa 2λaλb þ λb Þ=ðλa þ λbÞ 2 ¼ðλa λbÞ =ðλa þ λbÞ

Equation (pII.71) indicates that y is always positive for λa 6¼ λb, and we can conclude that Qi is higher than Qii under this condition.

Problem II.18 A long hollow cylinder of inner diameter, d1, of 180 mm and thickness, δ, of 8.0 mm is placed vertically. The temperature of the fluid inside the cylinder, θa, is 300 C and the temperature of the fluid outside, θb,is20 C. The heat-transfer coefficients inside and outside the cylinder, α1 and α2, are 50.0 W/m2 K and 5.0 W/m2 K, respectively. The heat conductivity of the cylinder, λ, is 50.0 W/m2 K. Calculate the heat-transfer rate over 2 m length of cylinder. Determine the inner and outer wall temperatures, θ1 and θ2. (Ans: Q ¼ 1,554 W, θ1 ¼ 272.5 C, θ2 ¼ 272.3 C) Problem II.19 A sphere of radius 4.0 cm is placed in the atmosphere. The temperature of the sphere is 427 C and the surrounding air is at 27 C. Calculate the heat loss due to natural convection. The Nusselt number and Grashof number are expressed by the following relations, and the physical properties of air are given in Table PII.1. (Ans: 104.3 W)

Nu ¼ 2 þ 0:60Pr1=3Gr1=4 (PII.72)

3 2 Gr ¼ gβd ðθs θfÞ=νf (PII.73)

Answer to Problem II.18 The heat-transfer rate per unit length is given by

Q=l ¼ 2πλðθ1 θ2Þ=½λ=ðα1r1Þþlnðr2=r1Þþλ=ðα2r2Þ (PII.74) where

r1 ¼ 180=2 ¼ 90 mm (PII.75)

r2 ¼ 90 þ 8:0 ¼ 98 mm (PII.76) Problems and Answers 247

Substituting these values and related quantities into Eq. (pII.74) yields

Q=l ¼ 2π 50 ð300 20Þ=½50=ð50:0 0:090ÞþlnðÞ 0:098=0:090 (PII.77) þ 50=ð5:0 0:098Þ ¼ 87920=113:24 ¼ 776:4W=m

Accordingly,

Q ¼ 776:4 2 ¼ 1,553 W (PII.78)

θ1 ¼ θa Q=ðα1A1Þ¼300 1,553=ð50:0 2 3:14 0:090 2Þ (PII.79) ¼ 300 27:47 ¼ 272:5 C

θ2 ¼ θb þ Q=ðα2A2Þ¼20 1,553=ð5:0 2 3:14 0:098 2Þ (PII.80) ¼ 20 þ 252:31 ¼ 272:3 C

Answer to Problem II.19

The mean temperature, θfm, is given by

θfm ¼ð427 þ 27Þ=2 ¼ 227 C ¼ 500 K (PII.81)

The physical properties can be obtained from Table PII.1 as follows:

2 6 2 νf ¼ð26:39 þ 52:36Þ=2 ¼ 39:38 mm =s ¼ 39:38 10 m =s (PII.82)

Pr ¼ð0:715 þ 0:710Þ=2 ¼ 0:713 ½ (PII.83)

λf ¼ð33:05 þ 45:6Þ=2 ¼ 39:33 mW=mK (PII.84) ¼ 39:33 103 W=mK

Substituting the above and following values into the definitions of the Grashof number yields Eq. (pII.90):

θf ¼ 27 þ 273 ¼ 300 K (PII.85)

β ¼ 1=300 ð1=KÞ (PII.86)

d ¼ 0:08 m (PII.87)

θs θf ¼ 427 27 ¼ 400 K (PII.88)

g ¼ 9:8m=s2 (PII.89) 248 Problems and Answers

2 Gr ¼ 9:8 ðÞ1=300 ðÞ0:08 3 400=ð39:38 106Þ ¼ 4:314 106 (PII.90)

From Eqs. (pII.83) and (pII.90) we have

= 1=4 Pr1=3 Gr1=4 ¼ ðÞ0:713 1 3 ð4:314 106Þ ¼ 40:65 (PII.91)

As this value is included in the applicable range of Eq. (pII.72), the Nusselt number is calculated from Eq. (pII.72):

Nu ¼ 2 þ 0:60Pr1=3 Gr1=4 = 1=4 ¼ 2 þ 0:60ðÞ 0:715 1 3 ð4:314 106Þ (PII.92) ¼ 2 þ 0:60 40:65 ¼ 26:39

The heat-transfer coefficient, α, is determined from the definition of the Nusselt number:

3 α ¼ 26:39λf=d ¼ 26:39 ð39:33 10 Þ=0:08 (PII.93) ¼ 12:97 W=m2 K

Q ¼ αðθ θ ÞA ¼ 12:97 ð427 27Þ3:14 ð0:08Þ2 s f (PII.94) ¼ 104:3W

Problem II.20 The temperature in a room is 30 C and the atmosphere is at 5 C. The thickness of the window made of glass is 4.0 mm. The thermal conductivity of glass, λ, is 0.76 W/m K, and the heat-transfer coefficients inside and outside the 2 room, α1 and α2, are 10 and 50 W/m K, respectively. Calculate the heat flux, q, defined as heat flow per unit time and unit area. (Ans: 199.5 J/m2 s) Answer to Problem II.20 The heat flux, q, is expressed by

q ¼ðθa θbÞ=ð1=α1 þ δ=λ þ 1=α2Þ ¼ð30 5Þ=ð1=10 þ 0:004=0:76 þ 1=50Þ (PII.95) ¼ 25=ð0:1 þ 0:005263 þ 0:02Þ¼25=0:1253 ¼ 199:5W=m2

Problem II.21 A sphere of diameter 7.0 cm is placed in a flow of air. The temperatures of the sphere and air are 427 C and 27 C, respectively. The approaching velocity of air flow is 17 m/s. Calculate the heat loss due to forced convection. The physical properties of air are given in Table PII.1. (Ans: 328.5 W; note that derivation is not given.) Problems and Answers 249

Problem II.22 Derive the following expression for the buoyancy force in Eq. (6.9a):

3 FB ¼ ρgd βðθs θfÞ (PII.96)

Answer to Problem II.22

The buoyancy force, FB, is expressed by

FB ¼ Vdρg (PII.97) where Vd is the volume of fluid displaced, ρ is the density of fluid, and g is the acceleration due to gravity. The volume of fluid affected by a sphere of diameter, d, can be approximated by d3 as the volume of the sphere is πd3/6. This volume of fluid is expanded due to the heat transferred from the sphere as follows:

3 3 3 d !½d þ d βðθs θfÞ (PII.98)

Accordingly, the volume of the fluid displaced, Vd, is given by

3 3 3 3 Vd ¼½d þ d βðθs θfÞ d ¼ d βðθs θfÞ (PII.99)

Substituting Eq. (pII.99) into Eq. (pII.97) gives

3 FB ¼ ρgd βðθs θfÞ (PII.100)

This expression is the same as Eq. (pII.96). Problem II.23 Derive the following expression for the shearing force in Eq. (6.9):

2 Fτ ¼ τAs ¼ μðÞv=d d (PII.101)

Answer to Problem II.23 Newton’s law of viscosity is described by

τ ¼ μdv=dy (PII.102) where τ is the shear stress acting on the surface of the sphere, μ is the dynamic viscosity of fluid, v is the velocity of fluid motion induced by convection around the sphere, and y is the distance measured from the surface of the sphere. The viscous force, Fτ, is expressed by

Fτ ¼ τAs (PII.103) 250 Problems and Answers

where As is the surface area of the sphere given by

2 2 As ¼ πd ! As ¼ d (PII.104)

In addition, the velocity gradient can be assumed to be

dv=dy ¼ v=d (PII.105)

Combining Eqs. (pII.101) and (pII.105) gives

τ ¼ μdv=dy ¼ μv=d (PII.106)

Substituting Eqs. (pII.104) and (pII.106) into Eq. (pII.103) yields

2 Fτ ¼ τAs ¼ μðÞv=d d (PII.107)

The dynamic viscosity, μ, is expressed by

μ ¼ ρνf (PII.108) where νf is the kinematic viscosity. The viscous force therefore is given by

2 Fτ ¼ ρνfðÞv=d d (PII.109)

Part III

Problem III.1 Helium gas at 25 C and 4 bars is stored in a spherical Pyrex container of 100 mm inside radius and 10 mm thickness. (a) Calculate the rate of mass loss from the container. (b) If mass flux rate is 8 1015 kg/s, find the pressure at which helium 13 2 3 3 is stored. (DAB ¼ 0.4 10 m /s, S ¼ 0.45 10 kmol/m.bar). (Ans: P ¼ 8bars) Answer to Problem III.1

Part A

0 ¼ CA;S1 CA;S2 NAr (PIII.1) Rm;diff 1 1 1 12 s ; ¼ ¼ : Rm diff 1 81 10 2 (PIII.2) 4πDAB R1 R2 m Problems and Answers 251

2 2 3 CA;S1 ¼ S PA ¼ 0:45 10 4 ¼ 1:8 10 k mol=m (PIII.3)

CA;S2 ¼ 0 (PIII.4)

nA;r ¼ mA NA;r (PIII.5)

kg 1:8 102 k mol=m3 n ; ¼ 4 (PIII.6) A r mol 1:81 1012 s=m3

25 kg n ; ¼ 4 10 (PIII.7) A r s

Part B

25 kg n ; ¼ 8 10 (PIII.8) A r s

15 k mol N ; ¼ 2 10 (PIII.9) A r s

3 3 CA;S1 ¼ S PA ¼ 0:45 10 k mol=m (PIII.10)

S P ¼ (PIII.11) A 0:45 103

PA ¼ 8 bars (PIII.12)

Problem III.2 Hydrogen flows in a tube at the rate of 8 10 6 k mol/s per meter length. The tube is 1 mm thick and has inner diameter of 50 mm. The outer surface is exposed to a gas stream for which the hydrogen partial pressure is 0.1 atm. The mass diffusivity and solubility of hydrogen in the tube material are 1.8 1011 m2/s and 160 k mol/m3 atm, respectively. Assuming the system is at 500 K, estimate the pressure at which the hydrogen flows within the tube. (Ans: P ¼ 8.65 atm) Answer to Problem III.2

mA ¼ 2kg=k mol for H2 (PIII.13)

Then,

6 k mol N ; ¼ 8 10 m (PIII.14) A r s 252 Problems and Answers

π ð ð Þ ð ÞÞ 2 DAB CA r1 CA r2 NA;r ¼ (PIII.15) ln D0 D1

π : 11ð ð Þ ð ÞÞ 6 k mol ¼ 2 1 8 10 CA r1 CA r2 8 10 m 51 (PIII.16) s ln 50

3 3 CAðr1ÞCAðr2Þ¼1:4 10 k mol=m (PIII.17)

3 CAðr2Þ¼160 0:1 ¼ 16 k mol=m (PIII.18)

3 CAðr1Þ¼1; 384 k mol=m (PIII.19)

CAðr1Þ PA;1 ¼ (PIII.20) SAB

PA;1 ¼ 8:65 atm (PIII.21)

Problem III.3 A rubber plug that is 40 mm thick and has surface area of 500 mm2 is used to contain CO2 at 25 C in a 20 L vessel. If the pressure is dropping at the rate of 9 1010 bar/s, what was the original pressure inside the vessel? Assume

k mol D ¼ 0:11 109m2=s ¼ and S ¼ 40:15 103 bar (PIII.22) AB m3

(Ans: PA,1 ¼ 13.15bars) Answer to Problem III.3

dP N RT A ¼ A (PIII.23) dt V where V ¼ 2 102 m3 and R is universal gas constant

N 8:314 102 298 9 1010 ¼ A (PIII.24) 2 102

k mol N ¼ 7:26 1013 (PIII.25) A s

C ; C ; N ¼ AD A 1 A 2 (PIII.26) A AB L Problems and Answers 253

k mol C ; C ; ¼ 0:528 (PIII.27) A 1 A 2 m3

CA;2 ¼ 0 (PIII.28)

k mol C ; ¼ 0:528 (PIII.29) A 1 m3

CA;1 P ; ¼ (PIII.30) A 1 S

PA;1 ¼ 13:15 bars (PIII.31)

Part IV

Problem IV.1 Air is injected into a cylindrical water bath of diameter, D,of 400 mm and depth, HL, of 600 mm through a centered single-hole bottom nozzle of inner diameter, dni, of 2.0 mm. The nozzle is wetted by water and the air flow 3 3 rate, Qg, is 500 cm /s. The densities of air and water, ρg and ρL, are 1.2 kg/m and 997 kg/m3, respectively. Determine the following quantities:

(a) Frequency of bubble formation, fBF (b) Modified defined in Eq. (9.64), Frm (c) Centerline value of gas holdup, αcl,atz ¼ 300 mm (d) Half-value radius of gas holdup, bα,atz ¼ 300 mm (e) Mean bubble rising velocity, uB,atz ¼ 300 mm (f) Centerline value of water velocity, ucl,atz ¼ 300 mm (g) Half-value radius of water flow velocity, bu,atz ¼ 300 mm Answer to Problem IV.1

: : f ¼ 12:1Q 0:133d 0:434 ¼ 12:1 ðÞ500 0 133 ðÞ0:2 0 434 BF g ni (PIV.1) ¼ 12:1 2:29 2:01 ¼ 55:7Hz

¼ ρ 2=ðρ 5Þ¼ : ð 6Þ2=½ : ð : 3Þ5 Frm gQg Lgdni 1 2 500 10 997 9 80 2 0 10 ¼ 1:2 2:5 107=ð3:13 1010Þ¼958 (PIV.2)

Problem IV.2 Estimate the frequency of bubble formation, fBF (Hz), and bubble diameter, dB, for a poorly wetted single-hole nozzle of inner diameter, dni,of 2.0 mm and outer diameter of 5.0 mm on the basis of Eq. (9.50). The gas flow 3 rate, Qg,is30cm/s. 254 Problems and Answers

Qg0

HL

Fig. PIV.1 Deepwater pool subjected to bottom gas Q injection gB

Answer to Problem IV.2 The frequency of bubble formation can be calculated from Eq. (9.50) as follows:

: : f ¼ 12:1Q 0:133d 0:434 ¼ 12:1 ð30Þ0 133 ð0:50Þ 0 434 BF g no (PIV.3) ¼ 12:1 1:572 1:35 ¼ 25:8Hz

The bubble diameter therefore is given by

: 1=3 0:5 0 289 dB ¼½6Qg=ðπfBFÞ ¼ 0:54½Qgdno (PIV.4) : 0:289 ¼ 0:54½30 ð0:50Þ0 5 ¼ 1:31 cm

Problem IV.3 Air is injected into a water bath through a single-hole bottom nozzle 4 3 at a flow rate, Qg,of3.0 10 m /s (see Fig. PIV.1). The bath diameter, D, and the 1 3 inner diameter of the nozzle, dni, are 2.00 10 m and 3.0 10 m, respectively. Calculate the cross-sectional mean air velocity, vn, and the , M,in the nozzle. The speed of sound, c, is assumed to be 340 m/s. In addition, calculate the superficial velocity of gas in the bath, jg,wherejg is defined as follows:

2 jg ¼ Qg=ðπD =4Þ (PIV.5)

3 ðAns: vn ¼ 42:5m=s; M ¼ 0:125; jg ¼ 9:55 10 m=sÞ Problems and Answers 255

Problem IV.4 Air is injected into a water pool through a single-hole bottom nozzle 4 3 at a flow rate, QgB,of3.0 10 m /s (see Fig. 9.10). The depth of the pool, H,is 15 m. Calculate the air flow rate at the surface of the pool, QgS. The density of water, 3 ρw, is 998 kg/m and the atmospheric pressure, p0, is 101.3 k Pa. The temperature of water, θ, is assumed to be constant everywhere in the bath. The pressure and temperature of the air at the bottom of the pool are assumed to be equal to their 4 3 respective values for the liquid there. (Ans: QgS ¼ 7.34 10 m /s) Answer to Problem IV.3

The mean air velocity, vn, is calculated in the following manner:

2 v ¼ 4Q =ðπd 2Þ¼4 3:0 104=½3:14 ð3:0 103Þ n g ni (PIV.6) ¼ 42:5m=s

The Mach number, M, is given by

M ¼ vn=c ¼ 42:5=340 ¼ 0:125 (PIV.7)

The superficial velocity of gas, jg, is expressed by

2 j ¼ 4Q =ðπD2Þ¼4 3:0 104=½3:14 ð2:00 101Þ g g (PIV.8) ¼ 9:55 103 m=s

Answer to Problem IV.4

The absolute pressure, pB, at the bottom of the pool is given by

3 pB ¼ ρgHL þ p0 ¼ 998 9:8 15 þ 101:3 10 (PIV.9) ¼ 146:7 103 þ 101:3 103 ¼ 248 k Pa

We assume that the perfect-gas law is valid under this condition:

pV ¼ mRθ (PIV.10) where p is the absolute pressure of the gas, V is the volume of the gas, m is the mass of the gas, and R is the gas constant. As m, R, and θ are constant under the present condition, Eq. (pIV.10) reduces to

pV ¼ const: (PIV.11)

The volume of gas, V, in this equation can be replaced by gas flow rate, Qg,as follows:

pQg ¼ const: (PIV.12) 256 Problems and Answers

The following relationship is obtained because the pressure on the pool surface is p0:

pBQgB ¼ p0QgS (PIV.13)

Consequently,

3 4 3 QgS ¼ pBQgB=p0 ¼ 248 10 3:0 10 =ð101:3 10 Þ (PIV.14) ¼ 7:34 104 m3=s

Such a big pool can be observed in Kaiyukan (one of the biggest aquaria in Japan) in Osaka Bay Area. A ring vortex was observed, generated by injecting air into the pool through an annular bottom nozzle expanded in the course of rising due to a decrease in the static pressure around it. Finally, it broke up into many smaller bubbles due to hydrodynamic instability. Problem IV.5 There are five particles of 1, 1.5, 2.0, 2.5, and 3.0 cm in diameter. Determine the mean diameters: D10 through D32. Problem IV.6 Air is injected into a water bath through a single-hole bottom nozzle 4 3 1 at a flow rate, Qg, of 3.0 10 m /s. The bath diameter, D, is 2.00 10 m. The 3 inner and outer diameters of the nozzle, dni and dno, are 2.0 10 m and 3 4.0 10 m, respectively. Calculate the frequency of bubble formation, fBF, and bubble diameter, dB, in the following two cases on the basis of empirical equation proposed by Davidson and Amick: (a) The nozzle is wetted by water. (b) The nozzle is poorly wetted by water. 2 (Ans: (a) fBF ¼ 52.1 Hz, dB ¼ 2.22 10 m, (b) fBF ¼ 38.6 Hz, 2 dB ¼ 2.46 10 m) Problem IV.7 Air is injected into a water bath through a single-hole bottom nozzle 4 3 1 at a flow rate, Qg, of 3.0 10 m /s. The bath diameter, D, is 2.00 10 m. The 3 inner and outer diameters of the nozzle, dni and dno, are 2.0 10 m and 3 4.0 10 m, respectively. Calculate the frequency of bubble formation, fBF, and bubble diameter, dB, in the following two cases on the basis of empirical equation 3 proposed by Iguchi et al. The densities of air and water, ρg and ρL, are 1.22 kg/m and 997 kg/m3, respectively. The surface tension of water, σ, is 73 mN/m. (a) The nozzle is wetted by water. (b) The nozzle is poorly wetted by water. 2 (Ans: (c) fBF ¼58.4 Hz, dB ¼ 2.14 10 m, (d) fBF ¼ 41.3 Hz, 2 dB ¼ 2.40 10 m) Index

A Circular cylinder, 54, 59, 105–113, 121, Abrupt contraction, 21, 42, 220 190–192 Abrupt expansion, 42, 220–221 Cold model, 197 Absolute pressure, 5–7, 11, 227, 255 Composite processing, 3 Absorptivity, 123, 125 Compressibility, 11–12 Advancing contact angle, 220 Concentration, 135–137, 139, 140, 142–145, Apparent activation energy, 141 198–200, 206, 211 Apparent dynamic viscosity, 73–75 Conduction, 89–91, 95–104, 108, 136, Arrhenius law, 141 235–237, 240, 245 Contact angle, 15, 16, 66, 81, 120, 218–220, 227 B Continuum, 4, 79, 84 Bernoulli equation, 25–28, 48 Convection, 90, 92, 96, 99, 102, 104–121, Bernoulli theorem, 3 131, 132, 139, 187, 239, 240, 244, Binary diffusion coefficient, 136, 138 246, 248, 249 Bingham model, 73 Coulter counter, 206–207 Black body, 123–126 Critical Reynolds number, 29, 32, 37, 41, Blake–Kozeny equation, 76, 77 52–54, 117, 228, 229, 232 Boiling heat transfer, 117–121 Bottom blown bubbling jet, 167–171, 173 Boundary layer, 25, 30, 49–54, 93, 94, 108, 116 D Boundary layer thickness, 49–52, 108 Density, 5, 7, 8, 11, 16, 18, 20, 21, 25–27, 30, Bubble frequency, 155, 158, 173–175, 177, 31, 34, 35, 45, 55, 56, 61–64, 66, 68, 76, 182, 183 93, 120, 121, 131, 136, 139, 153, 158, Bubble Reynolds number, 159, 160, 191 159, 161, 169, 191, 215, 218, 227–230, Bubble rising velocity, 156–159, 173–177, 232, 233, 240, 249, 253, 255, 256 182–184, 191, 253 Deposition rate, 141 Buckingham II theorem, 59, 62–66 Developing region, 30, 31, 34 Bulk modulus of elasticity, 11–12 Diffusion, 84, 94, 135–147, 190 Bursting phenomenon, 29 Drag coefficient, 54–56, 59, 64, 232, 235 Dynamical model, 157–158 Dynamic contact angle, 220 C Dynamic pressure, 26 Capacitance probe, 183, 209–210 Dynamic viscosity, 12–14, 22, 61–63, 66, Capillary force, 15–16, 82 72–75, 82, 92, 121, 131, 159, 188, , 215 228–230, 250

M. Iguchi and O.J. Ilegbusi, Basic Transport Phenomena in Materials Engineering, 257 DOI 10.1007/978-4-431-54020-5, © Springer Japan 2014 258 Index

E Gauge pressure, 5–7 Einstein model, 75 Grashof number, 107–109, 121, 131, 132, Einstein–Roscoe model, 75 239, 246 Electrophoresis method, 82 Electroresistivity probe, 161, 166, 171, 173, 182–185, 209 H Emissivity, 125, 127, 128 Half-value radius, 171, 173, 179, 253 Entrance length, 31 Heat flux, 91, 92, 95–98, 116, 119–121, Eo¨tvo¨s number, 159, 191 127–130, 235–238, 244, 245, 248 Equation of continuity, 3, 17–22, 28, 136 Heat transfer, 87, 89–94, 96–100, 102–121, Equation of motion, 17 123–132, 139, 142–144, 187–189, 223, Equilibrium contact angle, 15, 81, 218–220 235–240, 242–246, 248, 249 Equimolar counter-diffusion, 137 Heat transfer coefficient, 96–98, 105, 108, 109, Equivalent surface diameter, 163, 164 111–113, 120, 121, 132, 235, 236, 239, Equivalent volume diameter, 164 243, 244, 246, 248 Eyring model, 73 High-speed video camera, 166, 182, 207, 211 Hydraulic diameter, 31, 37, 39–41, 218 Hydrodynamic drag, 4, 57, 62, 205, 232–235 F Hydrodynamic loss, 26–28, 41, 48 Fick’s law of diffusion, 136 Flat plate, 23, 25, 49–53, 81, 93, 114–117, 127–128, 189–190, 218, 219, 235–237 I Flow pattern, 28, 54, 181, 199–203, 216–217, Inertial force, 29, 66, 68, 131, 159, 162, 219–222 173, 191 Flow quality, 153–154 Inviscid fluid, 25 Flow rate, 21, 27, 30, 31, 41, 48, 58, 82, 153, Ironmaking, 197 154, 156, 162, 167, 169, 174, 176, 177, 180, 181, 185, 186, 198, 199, 207, 218, 220, 221, 228–231, 254–256 K Flow rate measurement, 185 Karman’s vortex streets, 54, 58–59 Fluid, 3, 17, 71, 90, 96, 105, 131, 139, 152, Kinematic viscosity, 28–31, 35, 37, 38, 50–52, 197, 215 56, 58, 66, 76, 78, 80, 93, 94, 107, 108, Fluidized bed, 201–202, 210, 211 132, 159, 161, 177, 188, 190, 215, Flux, 91, 92, 95–98, 100, 116, 119–121, 127, 228–230, 232, 233, 240, 243, 244, 250 130, 136, 137, 140, 142–147, 155, 156, Kirchhoff’s law, 125 158, 235–238, 244, 245, 248, 250 Knudsen diffusion coefficient, 138 Forced convection, 105, 106, 110–118, 132, , 4, 15, 68, 79 139, 240 Fourier’s law, 91, 95–96, 99, 239, 240, 242 Frequency of bubble formation, 167–171, 253, L 254, 256 Laminar flow, 22, 29, 30, 32, 36, 37, 49–51, 76, Friction coefficient, 35–39, 50, 51, 53, 232 81, 115–116, 139, 228, 232 Froude number, 68, 172, 176, 179, 180, 193, 253 Laser-based optical method, 206 Fuel cell, 222–223 Laser optical sensor, 183–184 Fully developed region, 30, 31, 34, 35 Latent heat of fusion, 117 Fusion, 117 Logarithmic velocity distribution, 34 Lognormal distribution, 164 Loss coefficient, 42, 44–49 G Gas holdup, 155, 158, 171, 172, 174–177, 182, 183, 185, 253 M Gas-lift effect, 83, 151, 203 Mach number, 12, 67, 159, 162, 192, 254, 255 Gas-liquid two-phase flow, 151–194, 216–218, Mass concentration, 198, 199 220, 221 Mass flux, 142, 145, 250 Index 259

Mass transfer, 135–147, 174, 187–191 Porous media, 4, 75–78, 136–139 Mean diameter, 165–67, 202, 256 1/7-th Power law, 32, 33, 228 Mean free path, 4, 68, 79, 138 1/n-th Power law, 32, 33 Mean Nusselt number, 113, 115, 187 Prandtl number (Pr), 94, 107–114, 116, 121, Melting temperature, 117 132, 187, 188, 240, 243, 244, 247 MEMS. See Microelectromechanical systems Pressure drop, 3, 34–35, 45, 76–78, 216, 218, (MEMS) 230–232 Micro channel, 15, 78, 82, 84, 215–224 Pressure measurement, 185–186 Microelectromechanical systems (MEMS), Projected equivalent diameter, 163 84, 223 Micro reactor, 78–82, 84, 215, 216, 222 Mixing, 82–84, 198, 200, 211 R Mixing ratio, 198–200, 211 Radiation, 90, 93, 119, 123–132 Mixture model, 158 Radiation energy, 123, 125 Mole fraction, 136, 137 , 132 Molten metal, 3, 13, 71, 83, 94, 169, 183, 194, Receding contact angle, 220 197, 204 Reflectivity, 123 Momentum transfer rate, 29 Reverse transition, 30 Morton number, 159 Reynolds number, 29–33, 37, 38, 41, 51–56, 58, 59, 63, 66, 79, 80, 108, 112, 113, 115, 116, 159–161, 184, 188, 190–192, N 215, 228–232, 234, 240, 244 Natural convection, 105–110, 113–114, 119, Rheology, 71–76 131, 139 Navier-Stoke’s equation, 3, 22–25, 67 Newtonian fluid, 3, 13–14, 17, 71 S Newton’s law of cooling, 92, 96 , 94, 190 Newton’s law of viscosity, 12–14, 73, Shape factor, 125–126, 128 91, 92 Shear stress, 3, 4, 13, 34, 35, 50, 66, 72, 74, 92, Non-metallic inclusion, 151, 194, 197, 207 184, 249 Non-Newtonian fluid, 4, 13–14, 71–73 Sheath flow, 84, 207, 208 Normal distribution, 164, 174, 177 , 190 No-slip boundary condition, 3 Slip ratio, 154 Nusselt number, 107, 108, 110, 112–116, 120, Slurry, 3, 14, 74, 203–204 132, 187, 239, 240, 243, 244, 246 Solidification, 117 Solid–liquid two-phase flow, 197–199, 203–204, 209–211 O Soret coefficient, 137 Orifice, 42, 44, 47, 185, 220–222 Soret effect, 137 Ostwald-de Waele model, 73 Speed of sound, 12, 61, 67, 157, 159, 162, Over-all heat-transfer coefficient, 98 192, 254 Sphere, 26, 53–59, 62–64, 74, 75, 77, 78, 105–113, 121, 163, 167, 182, 187, 188, P 190, 191, 209, 232, 233, 235, 239, 240, Packed bed, 75–78 244, 246, 248–250 Particle image velocimetry (PIV), 84, 209 Static mixer, 84 Particle laden jet, 202 Steelmaking, 15, 151, 152, 177, 183, 197, 206, Peclet number, 132, 139 212, 219 Phase transformation, 117–121 Stefan–Boltzmann law, 93, 123–124 Phase velocity, 153–155 Still camera, 166, 182, 207 Pipe element, 4, 42–49, 220 Stoichiometric coefficient, 141 PIV. See Particle image velocimetry (PIV) (St), 58, 59, 67–68 Plunging jet, 83 Sudden contraction, 42, 45 Porosity, 76, 78, 138 Sudden expansion, 42, 45, 48 260 Index

Superficial velocity, 155, 194, 199, 201, 218, U 222, 254, 255 Ultrasonic sensor, 207 Surface tension, 8, 15–16, 64, 68, 80–81, 120, Universal gas constant, 141, 252 159, 191, 216–218, 227, 256 U-shaped manometer, 7, 8, 227

T V Taylor model, 75 Viscoelastic fluid, 14 Temperature measurement, 186 Viscous force, 29, 66, 80, 93, 108, 131, 159, Thermal boundary layer, 93, 94, 108 191, 249, 250 Thermal conductivity, 91, 93, 97, 107, 121, 132, Void fraction, 152–153, 157, 158, 171, 216 188, 235, 236, 238–240, 243–245, 248 Thermal diffusion coefficient, 141 Tortuosity, 138 W Transition to turbulence, 49, 52, 66, 161 Wall roughness, 35 Transmissivity, 123 Water model, 152, 174, 194, 209, 210 Turbulence intensity, 55–58, 187, 188, 190 , 68, 80, 161, 192, 215, 216 Turbulent flow, 28–31, 36, 49–51, 53, 77, 117, Weibull distribution, 164, 165 139, 228 Weissenberg effect, 14 Two-fluid model, 158 Wettability, 15, 16, 81, 167, 168, 216, 218–220