Fundamentals of heat and mass transfer frank p incropera david p dewitt solution manual ch08

PROBLEM 8.1 KNOWN: Flowrate and temperature of water in fully developed flow through a tube of prescribed diameter FIND: Maximum velocity and pressure gradient SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Isothermal flow -6 PROPERTIES: Table A-6, Water (300K): ρ = 998 kg/m , = 855 ì 10 N⋅s/m ANALYSIS: From Eq 8.6, Re D = & 4m ì 0.01 kg/s = = 596 Dà π ( 0.025m ) 855 × 10−6 kg ⋅ m/s Hence the flow is laminar and the velocity profile is given by Eq 8.15, u ( r) = 1 − ( r/ro )2    um The maximum velocity is therefore at r = 0, the centerline, where u ( ) = u m From Eq 8.5 um = & m ρπ D2 / = × 0.01 kg/s 998 kg/m3 × π ( 0.025m )2 = 0.020 m/s, hence u ( ) = 0.041 m/s Combining Eqs 8.16 and 8.19, the pressure gradient is dp 64 ρ u 2m =− dx Re D 2D dp 64 998 kg/m ( 0.020 m/s) =− × = −0.86 kg/m ⋅ s dx 596 × 0.025 m dp = −0.86N/m ⋅ m = −0.86 ×10 -5 bar/m dx < PROBLEM 8.2 KNOWN: Temperature and mean velocity of water flow through a cast iron pipe of prescribed length and diameter FIND: Pressure drop SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed flow, (3) Constant properties -6 PROPERTIES: Table A-6, Water (300K): = 997 kg/m , = 855 ì 10 N⋅s/m ANALYSIS: From Eq 8.22, the pressure drop is ∆p = f ρ u 2m L 2D Re D = ρ u mD 997 kg/m3 × 0.2 m/s × 0.15 m = = 3.50 ×10 -6 855 ì 10 N s/m With -4 the flow is turbulent and with e = 2.6 ×10 -3 × 10 and m for cast iron (see Fig 8.3), it follows that e/D = 1.73 f ≈ 0.027 Hence, 997 kg/m ( 0.2 m/s) ∆p = 0.027 ( 600m ) × 0.15 m ∆p = 2154 kg/s ⋅ m = 2154 N/m2 ∆p = 0.0215 bar < COMMENTS: For the prescribed geometry, L/D = (600/0.15) = 4000 >> (xfd,h/D)turb ≈ 10, and the assumption of fully developed flow throughout the pipe is justified PROBLEM 8.3 KNOWN: Temperature and velocity of water flow in a pipe of prescribed dimensions FIND: Pressure drop and pump power requirement for (a) a smooth pipe, (b) a cast iron pipe with a clean surface, and (c) smooth pipe for a range of mean velocities 0.05 to 1.5 m/s SCHEMATIC: ASSUMPTIONS: (1) Steady, fully developed flow PROPERTIES: Table A.6, Water (300 K): ρ = 997 kg/m3, µ = 855 × 10-6 N⋅s/m2, ν = µ/ρ = 8.576 × 10-7 m2/s ANALYSIS: From Eq 8.22a and 8.22b, the pressure drop and pump power requirement are ρ u 2m  = ∆p π D2 u L P = ∆pV ∆p = f m ( 2D ) (1,2) The friction factor, f, may be determined from Figure 8.3 for different relative roughness, e/D, surfaces or from Eq 8.21 for the smooth condition, 3000 ≤ ReD ≤ × 106, −2 f = 0.790 ln ( ReD ) − 1.64 (3) ( ) where the Reynolds number is u D 1m s × 0.25 m ReD = m = = 2.915 × 105 − ν 8.576 × 10 m s (4) (a) Smooth surface: from Eqs (3), (1) and (2), ( ( ) = 0.01451 ∆p = 0.01451 (997 kg m × 1m s × 0.25 m )1000 m = 2.89 × 10 kg s ⋅ m = 0.289 bar P = 2.89 × 104 N m (π × 0.252 m )1m s = 1418 N ⋅ m s = 1.42 kW ) f = 0.790 ln 2.915 × 105 − 1.64 −2 < < (b) Cast iron clean surface: with e = 260 µm, the relative roughness is e/D = 260 × 10-6 m/0.25 m = 1.04 × 10-3 From Figure 8.3 with ReD = 2.92 × 105, find f = 0.021 Hence, ∆p = 0.419 bar P = 2.06 kW < (c) Smooth surface: Using IHT with the expressions of part (a), the pressure drop and pump power requirement as a function of mean velocity, um, for the range 0.05 ≤ um ≤ 1.5 m/s are computed and plotted below Continued PROBLEM 8.3 (Cont.) deltap (bar) or P (kW) 0.6 0.2 0.08 0.04 0.01 0.5 1.5 Mean velocity, um (m/s) Pressure drop, deltap (bar) Pump power, P (kW) The pressure drop is a strong function of the mean velocity So is the pump power since it is proportional to both ∆p and the mean velocity COMMENTS: (1) Note that L/D = 4000 >> (xfg,h/D) ≈ 10 for turbulent flow and the assumption of fully developed conditions is justified (2) Surface fouling results in increased surface roughness and increases operating costs through increasing pump power requirements (3) The IHT Workspace used to generate the graphical results follows // Pressure drop: deltap = f * rho * um^2 * L / ( * D ) deltap_bar = deltap / 1.00e5 Power = deltap * ( pi * D^2 / ) * um Power_kW = Power / 1000 // Eq (1); Eq 8.22a // Conversion, Pa to bar units // Eq (2); Eq 8.22b // Useful for scaling graphical result // Reynolds number and friction factor: ReD = um * D / nu f = (0.790 * ln (ReD) - 1.64 ) ^ (-2) // Eq (3) // Eq (4); Eq 8.21, smooth surface condition // Properties Tool - Water: // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); x=0 // Quality (0=sat liquid or 1=sat vapor) rho = rho_Tx("Water",Tm,x) // Density, kg/m^3 nu = nu_Tx("Water",Tm,x) // Kinematic viscosity, m^2/s // Assigned variables: um = Tm = 300 D = 0.25 L = 1000 // Mean velocity, m/s // Mean temperature, K // Tube diameter, m // Tube length, m PROBLEM 8.4 KNOWN: Temperature and mass flow rate of various liquids moving through a tube of prescribed diameter FIND: Mean velocity and hydrodynamic and thermal entry lengths SCHEMATIC: ASSUMPTIONS: Constant properties PROPERTIES: (T = 300K) Liquid Table ρ(kg/m3) µ(N⋅s/m2) ν(m2/s) Engine oil Mercury Water A-5 A-5 A-6 884 13,529 1000 0.486 550 × 10 6400 -2 -6 0.152 × 10 0.113 ×10 0.0248 -3 -6 0.855 × 10 0.855 × 10 5.83 Pr -6 ANALYSIS: The mean velocity is given by & m 0.03 kg/s 61.1 kg/s ⋅ m um = = = ρ Ac ρπ ( 0.025m )2 / ρ The hydrodynamic and thermal entry lengths depend on ReD , & 4m × 0.03 kg/s 1.53 kg/s ⋅ m Re D = = = π Dµ π ( 0.025m ) µ µ -3 Hence, even for water (à = 0.855 ì 10 8.3 and 8.23 it follows that xfd,h = 0.05 D Re D = N⋅s/m ), ReD < 2300 and the flow is laminar From Eqs 1.91ì 103 kg/s 1.91 ×10−3 kg/s ) Pr ( xfd,t = 0.05 D ReD Pr = µ Hence: Liquid um(m/s) x fd,h(m) x fd,t(m) Oil Mercury Water 0.069 0.0045 0.061 0.0039 1.257 2.234 25.2 0.031 13.02 COMMENTS: Note the effect of viscosity on the hydrodynamic entry length and the effect of Pr on the thermal entry length PROBLEM 8.5 KNOWN: Number, diameter and length of tubes and flow rate for an engine oil cooler FIND: Pressure drop and pump power (a) for flow rate of 24 kg/s and (b) as a function of flow rate for  ≤ 30 kg/s the range 10 ≤ m SCHEMATIC: ASSUMPTIONS: (1) Fully developed flow throughout the tubes PROPERTIES: Table A.5, Engine oil (300 K): ρ = 884 kg/m3, µ = 0.486 kg/s⋅m ANALYSIS: (a) Considering flow through a single tube, find  ( 24 kg s ) 4m = = 251.5 Re D = π Dµ 25π ( 0.010 m ) 0.486 kg s ⋅ m (1) Hence, the flow is laminar and from Equation 8.19, 64 64 f= = = 0.2545 Re D 251.5 With  m ( 25 / 25 ) kg s ( ) um = = = 13.8 m s 2 884 kg m π ( 0.010 m ) ρ πD ( ) ( (2) ) (3) Equation 8.22a yields ∆p = f ρum 2D (884 kg m )(13.8 m s ) L = 0.2545 2 ( 0.010 m ) 2.5 m = 5.38 × 10 N m = 53.8 bar (4) 70 250 60 200 Pumping power (kW) P res s u re dro p (b ar) The pump power requirement from Equation 8.23b,  m 24 kg s  = ∆p ⋅ (5) = 1.459 × 10 N⋅m/s = 146 W P = ∆p ⋅ V = 5.38 × 106 N m ρ 884 kg m3 (b) Using IHT with the expressions of part (a), the pressure drop and pump power requirement as a  , for the range 10 ≤ m  ≤ 30 kg/s are computed and plotted below function of flow rate, m 50 40 < < 150 100 50 30 20 10 20 Flo w te (kg /s ) 30 10 20 30 Flow rate (kg/s) Continued PROBLEM 8.5 (Cont.) In the plot above, note that the pressure drop is linear with the flow rate since, from Eqs (2), the friction factor is inversely dependent upon mean velocity The pump power, however, is quadratic with the flow rate COMMENTS: (1) If there is a hydrodynamic entry region, the average friction factor for the entire tube length would exceed the fully developed value, thereby increasing ∆p and P (2) The IHT Workspace used to generate the graphical results follows /* Results: base case, part (a) P_kW ReD deltap_bar mdot 145.9 251.5 53.75 24 */ f mu rho um D N 0.2545 0.486 884.1 13.83 0.01 25 // Reynolds number and friction factor ReD = * mdot1 / (pi * D * mu) // Reynolds number, Eq (1) f = 64 / ReD // Friction factor, laminar flow, Eq 8.19, Eq (2) // Average velocity and flow rate mdot1 = rho * Ac * um // Flow rate, kg/s; single tube mdot = mdot1 * N // Total flow rate, kg/s; N tubes Ac = pi * D^2 / // Tube cross-sectional area, m^2 // Pressure drop and power deltap = f * rho * um^2 * L / (2 * D) deltap_bar = deltap * 1e-5 P = deltap * mdot / rho P_kW = P / 1000 // Input variables D = 0.01 mdot = 24 L = 2.5 N = 25 Tm = 300 // Pressure drop, N/m^2 // Pressure drop, bar // Power, W // Power, kW // Diameter, m // Total flow rate, kg/s // Tube length, m // Number of tubes // Mean temperature of oil, K // Engine Oil property functions : From Table A.5 rho = rho_T("Engine Oil",Tm) // Density, kg/m^3 mu = mu_T("Engine Oil",Tm) // Viscosity, N·s/m^2 PROBLEM 8.6 KNOWN: The x-momentum equation for fully developed laminar flow in a parallel-plate channel dP d 2u = constant = µ dx dy FIND: Following the same approach as for the circular tube in Section 8.1: (a) Show that the velocity profile, u(y), is parabolic of the form  y2  u ( y ) = um 1 −   ( a / )2  where um is the mean velocity expressed as um = a  dP  − 12 µ  dx  and -dp/dx = ∆p/L where ∆p is the pressure drop across the channel of length L; (b) Write the expression defining the friction factor, f, using the hydraulic diameter as the characteristic length, Dh; What is the hydraulic diameter for the parallel-plate channel? (c) The friction factor is estimated from the expression f = C ReDh where C depends upon the flow cross-section as shown in Table 8.1; What is the coefficient C for the parallel-plate channel (b/a → ∞ ) ? (d) Calculate the mean air velocity and the Reynolds number for air at atmospheric pressure and 300 K in a parallel-plate channel with separation of mm and length of 100 mm subjected to a pressure drop of ∆P = 3.75 N/m ; Is the assumption of fully developed flow reasonable for this application? If not, what effect does this have on the estimate for um? SCHEMATIC: ASSUMPTIONS: (1) Fully developed laminar flow, (2) Parallel-plate channel, a