PROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal conductivity k and inner temperature, T1 FIND: The outer temperature of the wall, T2 SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law, q cond = q x = q ′′ ⋅ A = -k x T −T dT ⋅ A = kA dx L Solving for T2 gives T2 = T1 − q cond L kA Substituting numerical values, find T2 = 415 C - 3000W × 0.025m 0.2W / m ⋅ K × 10m2 T2 = 415 C - 37.5 C T2 = 378 C COMMENTS: Note direction of heat flow and fact that T2 must be less than T1 < PROBLEM 1.2 KNOWN: Inner surface temperature and thermal conductivity of a concrete wall FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from -15 to 38°C SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties, (4) Outside wall temperature is that of the ambient air ANALYSIS: From Fourier’s law, it is evident that the gradient, dT dx = − q′′ k , is a constant, and x hence the temperature distribution is linear, if q′′ and k are each constant The heat flux must be x constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends only weakly on temperature The heat flux and heat rate when the outside wall temperature is T2 = -15°C are ) ( 25 C − −15 C dT T1 − T2 q′′ = − k =k = 1W m ⋅ K = 133.3W m x dx L 0.30 m q x = q′′ × A = 133.3 W m × 20 m = 2667 W x (1) (2) < Combining Eqs (1) and (2), the heat rate qx can be determined for the range of ambient temperature, -15 ≤ T2 ≤ 38°C, with different wall thermal conductivities, k 3500 Heat loss, qx (W) 2500 1500 500 -500 -1500 -20 -10 10 20 30 40 Ambient air temperature, T2 (C) Wall thermal conductivity, k = 1.25 W/m.K k = W/m.K, concrete wall k = 0.75 W/m.K For the concrete wall, k = W/m⋅K, the heat loss varies linearily from +2667 W to -867 W and is zero when the inside and ambient temperatures are the same The magnitude of the heat rate increases with increasing thermal conductivity COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane wall would not be linear PROBLEM 1.3 KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab Efficiency of gas furnace and cost of natural gas FIND: Daily cost of heat loss SCHEMATIC: ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties ANALYSIS: The rate of heat loss by conduction through the slab is T −T 7°C q = k ( LW ) = 1.4 W / m ⋅ K (11m × m ) = 4312 W t 0.20 m < The daily cost of natural gas that must be combusted to compensate for the heat loss is Cd = q Cg ηf ( ∆t ) = 4312 W × $0.01/ MJ 0.9 ×106 J / MJ ( 24 h / d × 3600s / h ) = $4.14 / d < COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation between it and the concrete PROBLEM 1.4 KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed thickness FIND: Thermal conductivity, k, of the wood SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be determined from Fourier’s law, Eq 1.2 Rearranging, k=q′′ x L W = 40 T − T2 m2 k = 0.10 W / m ⋅ K 0.05m ( 40-20 ) C < COMMENTS: Note that the °C or K temperature units may be used interchangeably when evaluating a temperature difference PROBLEM 1.5 KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions FIND: Heat loss through window SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from Fourier’s law, Eq 1.2 T −T q′′ = k x L  W (15-5 ) C q′′ = 1.4 x m ⋅ K 0.005m q′′ = 2800 W/m x Since the heat flux is uniform over the surface, the heat loss (rate) is q = q ′′ × A x q = 2800 W / m2 × 3m2 q = 8400 W COMMENTS: A linear temperature distribution exists in the glass for the prescribed conditions < PROBLEM 1.6 KNOWN: Width, height, thickness and thermal conductivity of a single pane window and the air space of a double pane window Representative winter surface temperatures of single pane and air space FIND: Heat loss through single and double pane windows SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-state conditions, (3) Enclosed air of double pane window is stagnant (negligible buoyancy induced motion) ANALYSIS: From Fourier’s law, the heat losses are Single Pane: $ T1 − T2 35 C = 19, 600 W qg = k g A = 1.4 W/m ⋅ K 2m L 0.005m ( ) ( ) T −T 25 $C Double Pane: qa = k a A = 0.024 2m2 = 120 W L 0.010 m COMMENTS: Losses associated with a single pane are unacceptable and would remain excessive, even if the thickness of the glass were doubled to match that of the air space The principal advantage of the double pane construction resides with the low thermal conductivity of air (~ 60 times smaller than that of glass) For a fixed ambient outside air temperature, use of the double pane construction would also increase the surface temperature of the glass exposed to the room (inside) air PROBLEM 1.7 KNOWN: Dimensions of freezer compartment Inner and outer surface temperatures FIND: Thickness of styrofoam insulation needed to maintain heat load below prescribed value SCHEMATIC: ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimensional conduction through walls of area A = 4m , (3) Steady-state conditions, (4) Constant properties ANALYSIS: Using Fourier’s law, Eq 1.2, the heat rate is q = q ′′ ⋅ A = k ∆T A total L Solving for L and recognizing that Atotal = 5×W , find k ∆ T W2 L = q  L= ( ) × 0.03 W/m ⋅ K 35 - (-10 ) C 4m   500 W L = 0.054m = 54mm < COMMENTS: The corners will cause local departures from one-dimensional conduction and a slightly larger heat loss PROBLEM 1.8 KNOWN: Dimensions and thermal conductivity of food/beverage container Inner and outer surface temperatures FIND: Heat flux through container wall and total heat load SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through bottom wall, (3) Uniform surface temperatures and one-dimensional conduction through remaining walls ANALYSIS: From Fourier’s law, Eq 1.2, the heat flux is $ T − T 0.023 W/m ⋅ K ( 20 − ) C ′′ = k = q = 16.6 W/m L 0.025 m < Since the flux is uniform over each of the five walls through which heat is transferred, the heat load is q = q′′ × A total = q′′  H ( 2W1 + 2W2 ) + W1 × W2    q = 16.6 W/m2  0.6m (1.6m + 1.2m ) + ( 0.8m × 0.6m ) = 35.9 W   < COMMENTS: The corners and edges of the container create local departures from onedimensional conduction, which increase the heat load However, for H, W1, W2 >> L, the effect is negligible PROBLEM 1.9 KNOWN: Masonry wall of known thermal conductivity has a heat rate which is 80% of that through a composite wall of prescribed thermal conductivity and thickness FIND: Thickness of masonry wall SCHEMATIC: ASSUMPTIONS: (1) Both walls subjected to same surface temperatures, (2) Onedimensional conduction, (3) Steady-state conditions, (4) Constant properties ANALYSIS: For steady-state conditions, the conduction heat flux through a one-dimensional wall follows from Fourier’s law, Eq 1.2, q ′′ = k ∆T L where ∆T represents the difference in surface temperatures Since ∆T is the same for both walls, it follows that L1 = L2 k1 q ′′ ⋅ k2 q1 ′′ With the heat fluxes related as q1 = 0.8 q ′′ ′′ L1 = 100mm 0.75 W / m ⋅ K × = 375mm 0.25 W / m ⋅ K 0.8 < COMMENTS: Not knowing the temperature difference across the walls, we cannot find the value of the heat rate PROBLEM 1.10 KNOWN: Thickness, diameter and inner surface temperature of bottom of pan used to boil water Rate of heat transfer to the pan FIND: Outer surface temperature of pan for an aluminum and a copper bottom SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction through bottom of pan ANALYSIS: From Fourier’s law, the rate of heat transfer by conduction through the bottom of the pan is T −T q = kA L Hence, T1 = T2 + qL kA where A = π D2 / = π (0.2m ) / = 0.0314 m Aluminum: T1 = 110 $C + Copper: T1 = 110 $C + 600W ( 0.005 m ) ( 240 W/m ⋅ K 0.0314 m 600W (0.005 m ) ( 390 W/m ⋅ K 0.0314 m2 ) ) = 110.40 $C = 110.25 $C COMMENTS: Although the temperature drop across the bottom is slightly larger for aluminum (due to its smaller thermal conductivity), it is sufficiently small to be negligible for both materials To a good approximation, the bottom may be considered isothermal at T ≈ 110 °C, which is a desirable feature of pots and pans PROBLEM 14.41 KNOWN: Initial concentration of hydrogen in a sheet of prescribed thickness Surface concentrations for time t > FIND: Time required for density of hydrogen to reach prescribed value at midplane of sheet SCHEMATIC: CA(x,0) = kmol/m = CA,i CA(0,tf) = 1.2 kg/m /2 kg/kmol CA(0,tf) = 0.6 kmol/m = CA CA(20 mm,t) = = CA,s ASSUMPTIONS: (1) One-dimensional diffusion in x, (2) Constant DAB, (3) No internal chemical reactions, (4) Uniform total molar concentration ANALYSIS: Using Heisler chart with heat and mass transfer analogy γ∗ = CA − CA,s CA,i − CA,s = 0.6 − ∗ = 0.2 = γ o 3.0 − With Bim = ∞, Fig D.1 may be used with ∗ θ o = 0.2, Bi −1 = Fo ≈ 0.75 Hence Fo m = DAB t f L2 = 0.75 t f = 0.75 ( 0.02 m ) /9 ×10−7 m2 / s < t f = 333s COMMENTS: If the one-term approximation to the infinite series solution θ∗ = ∞ ∑ Cn exp ( −ς n Fo ) cos ( ς n x∗ ) n =1 is used, it follows that ( ) ∗ γ o ≈ C1 exp − Fom = 0.2 ς2 Using values of ς1 = 1.56 and C1 = 1.27, it follows that exp  − (1.56 )2 Fom  = 0.157     Fo m = 0.76 which is in excellent agreement with the result from the chart PROBLEM 14.42 KNOWN: Sheet material has high, uniform concentration of hydrogen at the end of a process, and is then subjected to an air stream with a specified, low concentration of hydrogen Mass transfer parameters specified include: convection mass transfer coefficient, hm, and the mass diffusivity and solubility of hydrogen (A) in the sheet material (B), DAB and SAB, respectively FIND: (a) The final mass density of hydrogen in the material if the sheet is exposed to the air stream for a very long time, ρA,f, (b) Identify and evaluate the parameter that can be used to determine whether the transient mass diffusion process in the sheet can be characterized by a uniform concentration at any time; Hint: this situation is analogous to the lumped capacitance method for a transient heat transfer process; (c) Determine the time required to reduce the hydrogen concentration to twice the limiting value calculated in part (a) SCHEMATIC: ASSUMPTIONS: (1) One-dimensional diffusion, (2) Material B is stationary medium, (3) Constant properties, (4) Uniform temperature in air stream and material, and (5) Ideal gas behavior ANALYSIS: (a) The final content of H2 in the material will depend upon the solubility of H2 (A) in the material (B) and its partial pressure in the free stream From Eq 14.44, CA,f = SAB p A,∞ = 160 kmol / m3 ⋅ atm × 0.1 atm = 16 kmol / m3 ρ f = M A CA,f = kg / kmol × 16 kmol / m3 = 32 kg / m3 < (b) The parameters associated with transient diffusion in the material follow from the analogous treatment of Section 5.2 (Fig 5.3) and are represented in the schematic In the material, from Fick’s law, the diffusive flux is N ′′ A,dif = D AB CA,1 − CA,2 / L (1) At the surface, x = L, the rate equation, Eq 6.8, convective flux of species A is N ′′ A,conv = h m CA,s − CA,∞ Continued … PROBLEM 14.42 (Cont.) and substituting the ideal gas law, Eq 14.9, and introducing the solubility relation, Eq 14.44, N ′′ A,conv = hm SAB p A,s − SAB p A,∞ SAB Ru T∞ N ′′ A,conv = hm C2,s − CA,∞ SAB Ru T∞ 8 (2) where CA,∞ = CA,f, the final concentration in the material after exposure to the air stream a long time Considering a surface species flux balance, as shown in the schematic above, with the rate equations (1) and (2), DAB CA,1 − CA,2 hm CA,s − CA,f = L SAB Ru T∞ CA,1 − CA,2 Rm,dif ′′ h /S R T = m AB u ∞ = = Bi m CA,s − CA,f DAB / L Rm,conv ′′ (3) and introducing resistances to species transfer by diffusion, Eq 14.51, and convection Recognize from the analogy to heat transfer, Eq 5.10 and Table 14.2, that when Bim < 0.1, the concentration can be characterized as uniform during the transient process That is, the diffusion resistance is negligible compared to the convection resistance, Bi m = Bi m = h mL < 01 SAB Ru T∞ D AB (4) 115 m / h × 3600 s / h6 × 0.003 m 160 kmol / m3 ⋅ atm × 8.205 × 10-2 m3 ⋅ atm / kmol ⋅ K × 555 K × 2.68 × 10-8 m2 / s Bi m = 6.60 × 10 −3 < 0.1 Hence, the mass transfer process can be treated as a nearly uniform concentration situation From conservation of species on the material with uniform concentration,  ′′ − N A,conv = N A,st ′′ − hm d CA CA − CA,f = L SAB Ru T∞ dt Integrating, with the initial condition CA (0) = CA,i, find CA − CA,f CA,i − CA,f   = exp − hm t L SAB Ru T∞   (5) < Continued … PROBLEM 14.42 (Cont.) which is similar to the analogous heat transfer relation for the lumped capacitance analysis, Eq 5.6 (c) The time, to, required for the material to reach a concentration twice that of the limiting value, CA (To) = CA,f, can be calculated from Eq (5)  1.5 m / h × t o 12 − 16 × 16 kmol / m3 = exp − 3 ⋅ atm × 8.205 × 10 −2 m3 ⋅ atm / kmol ⋅ K × 555 K   0.003 m × 160 kmol / m  1320 -166 kmol / m t o = 42.9 hour < PROBLEM 14.43 KNOWN: Hydrogen-removal process described in Problem 14.3 (S), but under conditions for which -11 the mass diffusivity of hydrogen gas (A) in the sheet (B) is DAB = 1.8 × 10 m /s (instead of -8 2.6 × 10 m /s) With a smaller DAB, a uniform concentration condition may no longer be assumed to exist in the material during the removal process FIND: (a) The final mass density of hydrogen in the material if the sheet is exposed to the air stream for a very long time, ρA,f, (b) Identify and evaluate the parameters that describe the transient mass transfer process in the sheet; Hint: this situation is analogous to that of transient heat conduction in a plane wall; (c) Assuming a uniform concentration in the sheet at any time during the removal process, determine the time required to reach twice the limiting mass density calculated in part (a); (d) Using the analogy developed in part (b), determine the time required to reduce the hydrogen concentration to twice the limiting value calculated in part (a); Compare the result with that from part (c) SCHEMATIC: ASSUMPTIONS: (1) One-dimensional diffusion, (2) Material B is a stationary medium, (3) Constant properties, (4) Uniform temperature in air stream and material, and (5) Ideal gas behavior ANALYSIS: (a) The final content of H2 in the material will depend upon the solubility of H2 (A) in the material (B) at its partial pressure in the free stream From Eq 14.44, CA,f = SAB p A,∞ = 160 kmol / m3 ⋅ atm × 0.1 atm = 16 kmol / m3 ρ f = M A CA,f = kg / kmol × 16 kmol / m3 = 32 kg / m3 < (b) For the plane wall shown in the schematic below, the heat and mass transfer conservation equations and their initial and boundary conditions are Heat transfer ∂T ∂ 2T ∂t =α Mass (Species A) transfer ∂ CA ∂ CA = D AB ∂t ∂ x2 ∂ CA 10, t6 = ∂x ∂T 10, t6 = ∂x T x,0 = Ti −k ∂x CA x,0 = CA,i ∂T L, t = h T L, t − T∞ ∂x − DAB 6 ∂ CA hm L, t = CA x, t − C f ∂x SAB Ru T Continued … PROBLEM 14.43 (Cont.) The derivation for the species transport surface boundary condition is developed in the solution for Problem 14.3 (S) The solution to the mass transfer problem is identical to the analogous heat transfer problem provided the transport coefficients are represented as h h /S R T m AB u k DAB (1) (c) The uniform concentration transient diffusion process is analogous to the heat transfer lumpedcapacitance process From the solution of Problem 14.3 (S), the time to reach twice the limiting concentration, CA (to) = CA,f, can be calculated as   CA t o − CA,f hm to = exp − CA,i − CA,f L SAB Ru T   (2) < t o = 42.9 hour For the present situation, the mass transfer Biot number is Bi m = Bi m = hm L SAB Ru T DAB 115 m / h / 3600 s / h6 × 0.003 m 160 kmol / m3 ⋅ atm × 8.205 × 10-2 m3 ⋅ atm / kmol ⋅ K × 555 K × 1.8 × 10-11 m2 / s Bi m = 9.5 >> 0.1 and hence the concentration of A within B is not uniform (d) Invoking the analogy with the heat transfer situation, we can use the one-term series solution, Eq 5.40, with Bi m Bi and Fo m Fo Fo m = D AB t (3) L2 Continued … PROBLEM 14.43 (Cont.) With Bim = 9.5, find ζ1 = 1.4219 rad and C1 = 1.2609 from Table 5.1, so that Eq 5.41 becomes CA t o − CA,f = C1 exp −ς Fo m CA,i − CA,f 12 − 16 × 16 kmol / m3 = 12609 exp4−1.42192 Fo m 1320 − 166kmol / m3 Fo m = 18 × 10−11 m2 / s × t o 10.003 m62 = 1571 t o = 218 hour < COMMENTS: (1) Since Bim = 9.5, the uniform concentration assumption is not valid, and we expect the analysis to provide a longer time estimate to reach CA(to) = CA,f (2) Note that the uniform concentration analysis model of part (c) does not include DAB Why is this so? PROBLEM 14.44 KNOWN: Radius and temperature of air bubble in water FIND: Time to reach 99% of saturated vapor concentration at center SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial diffusion of vapor in air, (2) Constant properties, (3) Air is initially dry -4 PROPERTIES: Table A-8, Water vapor-air (300 K): DAB = 0.26 × 10 m /s ANALYSIS: Use Heisler charts with heat and mass transfer analogy, γ∗ ≡ CA − CA,s CA,i − CA,s =1 − CA CA,s ∗ − For γ o = − 0.99 = 0.01and Bim1 = 0, from Fig D.7 find Fom ≈ 0.52 Hence with Fo m = DAB t ro = 0.52 ( ) t = 0.52 10−6 m /0.26× 10−4 m /s = 0.02s < COMMENTS: (1) This estimate is likely to be conservative, since shear driven motion of air within the bubble would enhance vapor transport from the surface to the center (2) If the one-term approximation to the infinite series solution, θ∗ = ∞ ( ∑ Cn exp −ςn Fo n =1 ) ς(n r∗ ) sin ς n r∗ is used, it follows that with sin 0/0 = 1, ( ) ∗ γ o ≈ C1 exp −ζ1 Fo m = 0.1 Using values of C1 = 2.0 and ς1 = 3.11for Bi m = 100, it follows that 0.01 = 2.0 exp  − (3.11) Fo m      or Fo m = 0.55 which is in reasonable agreement with the Heisler chart result PROBLEM 14.45 KNOWN: Initial carbon content and prescribed surface content for heated steel FIND: Time required for carbon mole fraction to reach 0.01 at a distance of mm from the surface SCHEMATIC: ASSUMPTIONS: (1) Steel may be approximated as a semi-infinite medium, (2) One-dimensional diffusion in x, (3) Isothermal conditions, (4) No internal chemical reactions, (5) Uniform total molar concentration ANALYSIS: Conditions within the steel are governed by the species diffusion equation of the form ∂2 C A ∂CA = DAB ∂t ∂x or, in molar form, ∂2 x A ∂x = ∂x A DAB ∂t The initial and boundary conditions are of the form x A ( x,0 ) = 0.001 x A ( 0,t ) = x A,s = 0.02 x A ( ∞ , t ) = 0.001 The problem is analogous to that of heat transfer in a semi-infinite medium with constant surface temperature, and by analogy to Eq 5.57, the solution is x A ( x,t ) − x A,s x A,i − x A,s where   x   = erf  ( D t )1/2  AB   DAB = ×10 −5 exp [−17,000/1273] = 3.17 ×10 −11 m /s Hence     0.01 − 0.02 0.001m = 0.526 = erf   0.001 − 0.02  3.17 × 10−11t 1/2      ( ) where erf w = 0.526 → w ≈ 0.51, ( 0.51 = 0.001/2 3.17 ×10 −11 t ) 1/2 or t = 30,321s = 8.42 h < PROBLEM 14.46 KNOWN: Thick plate of pure iron at 1000°C subjected to a carburizing process with sudden exposure to a carbon concentration CC,s at the surface FIND: (a) Consider the heat transfer analog to the carburization process; sketch the mass and heat transfer systems; explain correspondence between variables; provide analytical solutions to the mass and heat transfer situation; (b) Determine the carbon concentration ratio, CC (x, t)/CC,s, at a depth of mm after hour of carburization; and (c) From the analogy, show that the time dependence of the 1/ mass flux of carbon into the plate can be expressed as n ′′ = ρ C,s D C − Fe / π t ; also, obtain an C expression for the mass of carbon per unit area entering the iron plate over the time period t SCHEMATIC: ASSUMPTIONS: (1) One-dimensional transient diffusion, (2) Thick plate approximates a semiinfinite medium for the transient mass and heat transfer processes, and (3) Constant properties ANALYSIS: (a) The analogy between the carburizing mass transfer process in the plate and the heat transfer process is illustrated in the schematic above The basis for the mass - heat transfer analogy stems from the similarity of the conservation of species and energy equations, the general solution to the equations, and their initial and boundary conditions For both processes, the plate is a semiinfinite medium with initial distributions, CC (x, t ≤ 0) = CC,i = and T (x, t ≤ 0) = Ti, suddenly subjected to a surface potential, CC (0, t > 0) = CC,s and T (0, t > 0) = Ts The heat transfer situation corresponds to Case 1, Section 5.7, from which the following relations were obtained Mass transfer Rate equation jC = − DAB ′′ Heat transfer ∂ Cc ∂x q ′′ = − k x Diffusion equation ∂ ∂x  ∂ CC  =  ∂ x  DAB ∂ CC ∂t 14.84 Polential distribution CC x, t − CC,s = − CC,s   1 ∂ ∂x  ∂ T =  ∂ x α    ∂T ∂x ∂T ∂t 2.15    T x, t − Ts x = erf Ti − Ts αt 1/    5.58 CC x, t x = erfc 1/ CC,s D AB t Continued … PROBLEM 14.46 (Cont.) Flux 1T k παst −1/T2i 6 q ′′ t = s See Part (c) 5.58 (b) Using the concentration distribution expression above, with L = mm, t = h and -11 DAB = × 10 m /s, find the concentration ratio,   CC 11 mm, h6   = 0.0314 0.001 m = erfc CC,s  243 × 10-11 m2 / s × 3600 s91/  < (c) From the heat flux expression above, the mass flux of carbon can be written as n′′ = C,s = ρc,s 1DC− Fe / π t61/ DC − Fe ρ C,s − π D C − Fe t 1/ < The mass per unit area entering the plate over the time period follows from the integration of the rate expression t 1/ m′′ ( t ) = ∫ n′′ dt = ρC,s ( DAB / π ) C C,s t ∫0 t -1/2 dt = ρ C,s ( DC− Fe t/π )1/ PROBLEM 14.47 KNOWN: Thickness, initial condition and bottom surface condition of a water layer FIND: (a) Time to reach 25% of saturation at top, (b) Amount of salt transfer in that time, (c) Final concentration of salt solution at top and bottom SCHEMATIC: ASSUMPTIONS: (1) One-dimensional diffusion, (2) Uniform total mass density, (3) Constant DAB ANALYSIS: (a) With constant ρ and DAB and no homogeneous chemical reactions, Eq 14.37b reduces to ∂ ρA ∂ρ A = DAB ∂t ∂x with the origin of coordinates placed at the top of the layer, the dimensionless mass density is ( ) ( ) γ ∗ x∗ ,Fom = ρ − ρA,s γ ρ = A = 1− A γ i ρ A,i − ρ A,s ρ A,s ( ) Hence, γ ∗ 0, Fo m,1 = − 0.25 = 0.75 The initial condition is γ ∗ x∗ ,0 = 1, and the boundary conditions are ∂γ ∗ ∂x∗ x∗ = =0 γ ∗ (1, Fom ) = where the condition at x∗ = corresponds to Bim = ∞ Hence, the mass transfer problem is analogous to the heat transfer problem governed by Eq 5.34 to 5.37 Assuming applicability of a oneterm approximation (Fom > 0.2), the solution is analogous to Eq 5.40 ( ) ( ) γ ∗ = C1 exp −ς1 Fom cos ς1x ∗ With Bi m = ∞, ς1 = π / = 1.571 rad and, from Table 5.1, C1 ≈ 1.274 Hence, for x∗ = 0, 0.75 = 1.274exp − (1.571) Fo m,1      Fo m,1 = − ln ( 0.75/1.274 ) / (1.571) = 0.215 Hence, (1 m )2 L2 t1 = Fom,1 = 0.215 = 1.79 ×108 s = 2071days −9 m / s DAB 1.2 ×10 < Continued … PROBLEM 14.47 (Cont.) (b) The change in the salt mass within the water is ( ) L ∆M A = M A ( t1 ) − M A,i = ∫ ρA − ρA,i dV = A∫ ρ Adx Hence, L ( ) ∆M ′′ = ρ A,s ∫ ρ A / ρ A,s dx A ( ) ∆M ′′ = ρ A,s L∫ − γ ∗ dx ∗ A ( ) ( ) dx∗ ∆M ′′ = ρA,s L ∫ 1 −C exp −ς1 Fo m cos ς1 x∗ A 0  ( ) ∆M ′′ = ρA,s L 1 − C1 exp −ς1 Fom sin ς1 / ς1  A     Substituting numerical values,  1.274exp  − (1.571)2 0.215 1       ∆M ′′ = 380 k g / m (1 m ) 1 − A  1.571rad     ∆M ′′ = 198.7 k g / m A < (c) Steady-state conditions correspond to a uniform mass density in the water Hence, ρ A ( 0, ∞ ) = ρ A ( L, ∞ ) = ∆M′′ / L = 198.7 k g / m3 A < COMMENTS: (1) The assumption of constant ρ is weak, since the density of salt water depends strongly on the salt composition (2) The requirement of Fom > 0.2 for the one-term approximation to be valid is barely satisfied PROBLEM 14.48 KNOWN: Temperature distribution expression for a semi-infinite medium, initially at a uniform temperature, that is suddenly exposed to an instantaneous amount of energy, Q o J / m2 ′′ " ' Analogous situation of a silicon (Si) wafer with a 1-µm layer of phosphorous (P) that is placed in a furnace suddenly initiating diffusion of P into Si FIND: (a) Explain the correspondence between the variables in the analogous temperature and concentration distribution expressions, and (b) Determine the mole fraction of P at a depth of 0.1 mm in the Si after 30 s SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, transient diffusion, (2) Wafer approximates a semi-infinite medium, (3) Uniform properties, and (4) Diffusion process for Si and P is initiated when the wafer reaches the elevated temperature as a consequence of the large temperature dependence of the diffusion coefficient -17 PROPERTIES: Given in statement: DP-Si = 1.2 × 10 m /s; Mass densities of Si and P: 2000 and 2300 kg/m ; Molecular weights of Si and P: 30.97 and 28.09 kg/kmol ANALYSIS: (a) For the thermal process illustrated in the schematic, the temperature distribution is  $ T x, t − Ti = Qo ′′  $ ρc παt 1/ " exp − x / 4αt ' (HT) where Ti is the initial, uniform temperature of the medium For the mass transfer process, the P concentration has the form  $ C P x, t =  M ′′ P,o π D P −Si t $ 1/ " exp − x2 / D P −Si t ' (MT) where M ′′ is the molar area density (kmol/m ) of P represented by the film of concentration CP P,o and thickness The correspondence between mass and heat transfer variables in the equations HT and MT involves the following conditions The LHS represents the increase with time of the temperature or concentration above the initial uniform distribution The initial concentration is zero, so only the CP (x, t) appears On the RHS note the correspondence of the terms in the exponential parenthesis and in the denominator The thermal diffusivity and diffusion coefficient are directly analogous; this can be seen by comparing the MT and HT diffusion equations, Eq 2.15 and 14.84 The terms Q ′′ / ρc and o M ′′ for HT and MT represent the energy and mass instantaneously appearing at the surface The P,o product ρc is the thermal capacity per unit area and appears in the storage term of the HT diffusion equation For MT, the “capacity” term is the volume itself Continued … PROBLEM 14.48 (Cont.) (b) The molar area density (kmol/m ) of P associated with the film of thickness = µm and concentration CP,o is  $ M ′′ = C P,o ⋅ d o = ρ P / M P d o P,o " ' M ′′ = 2000 kg / m3 / 30.97 kmol / kg × × 10−6 m P,o M ′′ = 6.458 × 10−5 kmol / m2 P,o Substituting numerical values into the MT equation, find  $ C p 0.1 mm, 30 s = " 6.458 × 10−5 kmol / m2 ' π × 1.2 × 10-17 m2 / s × 30 s  ' $ " exp − 0.0001 m / × 12 × 10 −7 m2 / s × 30 s C p = 0.08188 kmol / m3 The mole fraction of P in the Si wafer is  x P = C P / CSi = C P / ρ Si / MSi " $ x P = 0.08188 kmol / m3 / 2300 kg / m3 / 28.09 kmol / kg x P = 2.435 × 10−5 ' < ... the heat rate PROBLEM 1.10 KNOWN: Thickness, diameter and inner surface temperature of bottom of pan used to boil water Rate of heat transfer to the pan FIND: Outer surface temperature of pan... Electrical heater is perfectly insulated from dryer wall, (3) Heater and switch are isothermal at Tset, (4) Negligible heat transfer from sides of heater or switch, (5) Switch surface, As, loses heat. .. between hand and surroundings in the case of air flow ANALYSIS: The hand will feel colder for the condition which results in the larger heat loss The heat loss can be determined from Newton’s law of