2-11 Chapter 2 Example 2.2-1 ------------------------------------------------------------------------------ A solid slab of 5.15 wt% agar gel at 278oK is 10.16 mm thick and contains a uniform concentration of urea of 0.1 kmol/m3. Diffusion is only in the direction through two parallel flat surfaces 10.16 mm apart. The slab is suddenly immersed in pure turbulent water so that the surface resistance can be assumed to be negligible; i.e, the convective coefficient hm is very large. The diffusivity of urea in the agar is 4.72×10-10 m2/s. 1) Calculate the concentration at the midpoint of the slab (5.08 mm from the surface) and 2.54 mm from the surface after 10 hr. 2) If the thickness of the slab is halved, what would be the midpoint concentration in 10 hr? (Ref: Transport Processes and Separation Process by C.J. Geankoplis, Prentice Hall, 4th Edition, 2003) Solution ---------------------------------------------------------------------------------------------- 1) Calculate the concentration at the midpoint of the slab (5.08 mm from the surface) and 2.54 mm from the surface after 10 hr. The concentration at any point is given by the approximate solution *θ= *0θcos(ζ1x*) where *0θ= C1exp(-21ζFo) and Fo = 2ABD tL 2L = 10.16 mm => L = 5.08 mm = 5.08×10-3 m Fo = 2ABD tL = 103 2(4.72 10 )(36000)(5.08 10 )−−×× = 0.6584 > 0.2 At the midpoint of the slab x = 0 => x* = Lx = 0 Bim = 'mABh LK D = ∞ => ζ1 = 1.5708 and C1 = 1.2732 *θ= *0θcos(ζ1x*) = C1exp(-21ζFo) = 1.2732 exp(−1.57082×0.6584) *θ= ,, ,''A AA i Ac K cc K c∞∞−− = 0.251 Since cA,∞ = 0 => ,AA icc = 0.251 => cA = (0.251)(0.1) = 0.0251 kmol/m3. At x = 2.54 mm => x* = Lx = 0.52-12 *θ= C1exp(-21ζFo)cos(ζ1x*) = 1.2732 exp(−1.57082×0.6584)cos(1.5708×0.5) *θ= ,AA icc = (0.251)cos(π/4) = 0.177 cA = (0.177)(0.1) = 0.0177 kmol/m3. 2) If the thickness of the slab is halved, what would be the midpoint concentration in 10 hr? Fo = 2ABD tL = 103 2(4.72 10 )(36000)(2.54 10 )−−×× = 2.6336 *θ= *0θcos(ζ1x*) = C1exp(-21ζFo) = 1.2732 exp(−1.57082×2.6336) *θ= ,AA icc = 0.00192 cA = (0.00192)(0.1) = 0.000192 kmol/m3. Example 2.2-2 ------------------------------------------------------------------------------ A slab of white pine, 5 cm thick, has a moisture content of 20 wt% at the start of the drying process. The equilibrium moisture content is 5 wt% for the humidity conditions in the drying air. The ends and edges are covered with moisture-resistance coating to prevent evaporation. The diffusivity of water through pine is 1.0×10-9 m2/s. 1) If the two large surfaces are exposed to the drying air and the resistance to mass transfer outside the pine is negligible, estimate the time required to reduce the moisture content of the slab at the center to 10 wt% and the time to reduce the (average) water content to 10 wt%. 2) Repeat part (1) if the external mass transfer resistance is equal to the internal mass transfer resistance. 3) Repeat part (1) if only one large surface is exposed to the drying air while the other surface is also covered with moisture resistance coating. (Ref. Fundamentals of Momentum, Heat, and Mass Transfer by Welty, Wicks, and Wilson, 4th Edition, 2001) Solution ---------------------------------------------------------------------------------------------- 1) If the two large surfaces are exposed to the drying air and the resistance to mass transfer outside the pine is negligible, estimate the time required to reduce the moisture content of the slab at the center to 10 wt%. The dimensionless concentration can be expressed in terms of mass fraction ,**A AA i Ac cc c−− = ,**A AA i Aρ ρρ ρ−− = ,/ *// */A B A BA i B A Bρ ρ ρ ρρ ρ ρ ρ−− ωA = AA Bρρ ρ+ => ABρρ = 1AAωω−2-13 ,A iBρρ = 0.21 0.2− = 0.25 *ABρρ = 0.051 0.05− = 0.0526 ABρρ = 0.11 0.1− = 0.1111 ,**A AA i Ac cc c−− = ,/ *// */A B A BA i B A Bρ ρ ρ ρρ ρ ρ ρ−− = 0.1111 .05260.2500 0.0526−− = 0.2964 Bim = 'mABh LK D = ∞ => ζ1 = 1.5708 and C1 = 1.2732 0.2964 = *0θcos(ζ1x*) = C1exp(-21ζFo) = 1.2732 exp(−1.57082 Fo) Solving for Fo gives Fo = 2ABD tL = 0.591 Since 2L = 5 cm => L = 0.025 m t = 29(0.591)(0.025)10− = 3.69×105 sec. Estimate the time to reduce the (average) water content to 10 wt%. tMM∞= 1 − 11)sin(ζζ*0θ = ()( ),,*A i AA i AVVρ ρρ ρ−− = 0.2500 0.11110.2500 0.0526−− = 0.7036 Solving for *0θ gives *0θ = 1tMM∞ −  11sin( )ζζ = (1 − 0.7036) 1.5708sin(1.5708) = 1.2732 exp(−1.57082 Fo) Solving for Fo gives Fo = 2ABD tL = 0.408 t = 29(0.408)(0.025)10− = 2.55×105 sec.2-14 2) Repeat part (1) if the external mass transfer resistance is equal to the internal mass transfer resistance. Bim = 'mABh LK D = 1 => ζ1 = 0.8603 and C1 = 1.1191 0.2964 = *0θcos(ζ1x*) = C1exp(-21ζFo) = 1.1191 exp(−0.86032 Fo) Solving for Fo gives Fo = 2ABD tL = 1.795 t = 29(1.795)(0.025)10− = 1.12×106 sec. Estimate the time to reduce the (average) water content to 10 wt%. tMM∞= 1 − 11)sin(ζζ*0θ = ()( ),,*A i AA i AVVρ ρρ ρ−− = 0.2500 0.11110.2500 0.0526−− = 0.7036 Solving for *0θ gives *0θ = 1tMM∞ −  11sin( )ζζ = (1 − 0.7036) 0.8603sin(0.8603) = 1.1191 exp(−0.86032 Fo) Solving for Fo gives Fo = 2ABD tL = 1.624 t = 29(1.624)(0.025)10− = 1.015×106 sec.2-15 3) Repeat part (1) if only one large surface is exposed to the drying air while the other surface is also covered with moisture resistance coating. Bim = 'mABh LK D = ∞ => ζ1 = 1.5708 and C1 = 1.2732 Since mass transfer occurs only on one side we have L = 0.050 m and x is zero at the coating surface. At the center of the slab x = 0.025 m and x* = 0.5 0.2964 = *0θcos(ζ1x*) = C1exp(-21ζFo) cos(ζ1x*) 0.2964 = 1.2732 exp(−1.57082 Fo) cos(1.5708×0.5) Solving for Fo gives Fo = 2ABD tL = 0.45 t = 29(0.45)(0.05)10− = 1.126×105 sec. Estimate the time to reduce the (average) water content to 10 wt%. tMM∞= 1 − 11)sin(ζζ*0θ = ()( ),,*A i AA i AVVρ ρρ ρ−− = 0.2500 0.11110.2500 0.0526−− = 0.7036 Solving for *0θ gives *0θ = 1tMM∞ −  11sin( )ζζ = (1 − 0.7036) 1.5708sin(1.5708) = 1.2732 exp(−1.57082 Fo) Solving for Fo gives Fo = 2ABD tL = 0.408 t = 29(0.408)(0.05)10− = 1.02×105 sec.2-16