Fundamentals of heat and mass transfer frank p incropera david p dewitt solution manual ch07

PROBLEM 7.1 KNOWN: Temperature and velocity of fluids in parallel flow over a flat plate FIND: (a) Velocity and thermal boundary layer thicknesses at a prescribed distance from the leading edge, and (b) For each fluid plot the boundary layer thicknesses as a function of distance SCHEMATIC: ASSUMPTIONS: (1) Transition Reynolds number is × 105 PROPERTIES: Table A.4, Air (300 K, atm): ν = 15.89 × 10-6 m2/s, Pr = 0.707; Table A.6, Water (300 K): ν = µ/ρ = 855 × 10-6 N⋅s/m2/997 kg/m3 = 0.858 × 10-6 m2/s, Pr = 5.83; Table A.5, Engine Oil (300 K): ν = 550 × 10-6 m2/s, Pr = 6400; Table A.5, Mercury (300 K): ν = 0.113 × 10-6 m2/s, Pr = 0.0248 ANALYSIS: (a) If the flow is laminar, the following expressions may be used to compute δ and δt, respectively, δ= 5x δt = Re1/ x δ Fluid Pr1/ Air Water Oil Mercury where u x 1m s ( 0.04 m ) 0.04 m s Re x = ∞ = = ν ν ν Rex δ (mm) δt (mm) 2517 4.66 × 104 72.7 3.54 × 105 3.99 0.93 23.5 0.34 4.48 0.52 1.27 1.17 (b) Using IHT with the foregoing equations, the boundary layer thicknesses are plotted as a function of distance from the leading edge, x 10 BL thickness, deltat (mm) BL thickness, delta (mm) 4 0 10 20 30 Distance from leading edge, x (mm) Air Water Oil Mercury 40 10 20 30 40 Distance from leading edge, x (mm) Air Water Oil Mercury COMMENTS: (1) Note that δ ≈ δt for air, δ > δt for water, δ >> δt for oil, and δ < δt for mercury As expected, the boundary layer thicknesses increase with increasing distance from the leading edge (2) The value of δt for mercury should be viewed as a rough approximation since the expression for δ/δt was derived subject to the approximation that Pr > 0.6 < PROBLEM 7.2 KNOWN: Temperature and velocity of engine oil Temperature and length of flat plate FIND: (a) Velocity and thermal boundary layer thickness at trailing edge, (b) Heat flux and surface shear stress at trailing edge, (c) Total drag force and heat transfer per unit plate width, and (d) Plot the boundary layer thickness and local values of the shear stress, convection coefficient, and heat flux as a function of x for ≤ x ≤ m SCHEMATIC: ASSUMPTIONS: (1) Critical Reynolds number is × 105, (2) Flow over top and bottom surfaces PROPERTIES: Table A.5, Engine Oil (Tf = 333 K): ρ = 864 kg/m3, ν = 86.1 × 10-6 m2/s, k = 0.140 W/m⋅K, Pr = 1081 ANALYSIS: (a) Calculate the Reynolds number to determine nature of the flow, u L 0.1m s × 1m ReL = ∞ = = 1161 ν 86.1×10−6 m s Hence the flow is laminar at x = L, from Eqs 7.19 and 7.24, and −1/ = 1m 1161 −1/ = 0.147 m δ = 5L ReL ( )( ) −1/ δ t = δ Pr −1/ = 0.147 m (1081) = 0.0143m < < (b) The local convection coefficient, Eq 7.23, and heat flux at x = L are hL = k Pr1/ = 0.140 W m ⋅ K 0.332 1161 1/ 1081 1/ = 16.25 W m ⋅ K 0.332 Re1/ ( ) ( ) L L 1m q′′x = h L ( Ts − T∞ ) = 16.25 W m ⋅ K ( 20 − 100 ) C = −1300 W m $ < Also, the local shear stress is, from Eq 7.20, ρu2 864 kg m3 τ s,L = ∞ 0.664 Re−L1/ = (0.1m s )2 0.664 (1161)−1/ 2 < τ s,L = 0.0842 kg m ⋅ s = 0.0842 N m (c) With the drag force per unit width given by D′ = 2Lτ s,L where the factor of is included to account for both sides of the plate, it follows that ( ) −1/ 2 −1/ = (1m ) 864 kg m3 (0.1m s ) / 1.328 (1161) D′ = 2L ρ u ∞ 1.328 Re L = 0.337 N m For laminar flow, the average value h L over the distance to L is twice the local value, hL, h L = 2h L = 32.5 W m ⋅ K The total heat transfer rate per unit width of the plate is $ q′ = 2Lh L ( Ts − T∞ ) = (1m ) 32.5 W m ⋅ K ( 20 − 100 ) C = −5200 W m < < Continued PROBLEM 7.2 (Cont.) (c) Using IHT with the foregoing equations, the boundary layer thickness, and local values of the convection coefficient and heat flux were calculated and plotted as a function of x deltax*10, hx*100, -q''x 5000 4000 3000 2000 1000 0 0.2 0.4 0.6 0.8 Distance from leading edge, x (m) BL thickness, deltax * 10 (mm) Convection coefficient, hx * 100 (N/m^2) Heat flux, - q''x (W/m^2) COMMENTS: (1) Note that since Pr >> 1, δ >> δt That is, for the high Prandtl liquids, the velocity boundary layer will be much thicker than the thermal boundary layer (2) A copy of the IHT Workspace used to generate the above plot is shown below // Boundary layer thickness, delta delta = * x * Rex ^-0.5 delta_mm = delta * 1000 delta_plot = delta_mm * 10 // Scaling parameter for convenience in plotting // Convection coefficient and heat flux, q''x q''x = hx * (Ts - Tinf) Nux = 0.332 * Rex^0.5 * Pr^(1/3) Nux = hx * x / k hx_plot = 100 * hx // Scaling parameter for convenience in plotting q''x_plot = ( -1 ) * q''x // Scaling parameter for convenience in plotting // Reynolds number Rex = uinf * x / nu // Properties Tool: Engine oil // Engine Oil property functions : From Table A.5 // Units: T(K) rho = rho_T("Engine Oil",Tf) // Density, kg/m^3 cp = cp_T("Engine Oil",Tf) // Specific heat, J/kg·K nu = nu_T("Engine Oil",Tf) // Kinematic viscosity, m^2/s k = k_T("Engine Oil",Tf) // Thermal conductivity, W/m·K Pr = Pr_T("Engine Oil",Tf) // Prandtl number // Assigned variables Tf = (Ts + Tinf) / Tinf = 100 + 273 Ts = 20 + 273 uinf = 0.1 x=1 // Film temperature, K // Freestream temperature, K // Surface temperature, K // Freestream velocity, m/s // Plate length, m PROBLEM 7.3 KNOWN: Velocity and temperature of air in parallel flow over a flat plate FIND: (a) Velocity boundary layer thickness at selected stations Distance at which boundary layers merge for plates separated by H = mm (b) Surface shear stress and v(δ) at selected stations SCHEMATIC: ASSUMPTIONS: (1) Steady flow, (2) Boundary layer approximations are valid, (3) Flow is laminar -6 PROPERTIES: Table A-4, Air (300 K, atm): ρ = 1.161 kg/m , ν = 15.89 × 10 m /s ANALYSIS: (a) For laminar flow, 5x δ = = Re1/2 x (u ∞ / ν ) 1/2 x ( m) δ ( mm ) 0.001 0.126 x1/2 = 0.01 0.399 5x1/2 = 3.99 ×10 −3 x1/2 (25 m/s/15,89×10-6 m / s ) 1/2 0.1 1.262 Boundary layer merger occurs at x = xm when δ = 1.5 mm Hence 0.0015 m x1/2 = 0.376 m1/2 x m = 141 mm m = -3 1/2 3.99 ×10 m < (b) The shear stress is τ s,x = 0.664 x ( m) ( τ s,x N/m ) ρu 2∞ / Re1/2 x = 0.001 6.07 ρu 2∞ / ( u∞ /ν )1 / x1/2 0.01 1.92 0.1 0.664 ×1.161 kg/m3 ( 25 m/s ) / 2 = ( ) / 1/2 25 m/s/15.89 ×10-6 m / s x = 0.192 x1/2 (N/m2 ) 0.61 1/2 The velocity distribution in the boundary layer is v = (1/2) (νu∞/x) (ηdf/dη - f) At y = δ, η ≈ 5.0, f ≈ 3.24, df/dη ≈ 0.991 1/2 0.5 v= 15.89 × 10−6 m2 / s × 25 m/s ( 5.0 × 0.991 − 3.28) = 0.0167/x1/2 m/s 1/2 x x ( m) v ( m/s ) ( ) 0.001 0.528 0.01 0.167 ( ) 0.1 0.053 COMMENTS: (1) v