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Unformatted text preview: , y = L, from the top of the film, cCO 2 y / cCO 2 i = 0.5 . If we use Eqs. (399) and (3100), we do not have to compute the average masstransfer
coefficient.
Compute the value of η from (399):
1 − 0.5 = 0.5 = 0.7857 exp ( −5.1213η ) + 0.09726 exp ( −39.661η ) +
0.036093exp ( 106.25η ) Solving this nonlinear equation for η with a spreadsheet, η = 0.0894
From a rearrangement of Eq. (3100), L= 3δ2 u y η
2 DAB 3 (1.15 × 10−2 ) (4.86)(0.0894)
2 = 2 (1.96 ×10 −5 ) = 4.4 cm Exercise 3.28
Subject: Evaporation of water (A) from a film on a flat plate into air (B) at 1 atm (101.3 kPa)
and 25oC (298 K) that is flowing past the plate.
Given: Freestream air velocity = uo = 2 m/s. Plate length = L = 2 inches = 0.0508 m.
Diffusivity of water vapor in air = DA,B = 0.25 cm2/s = 0.25 x 104 m2/s
Assumptions: Ideal gas law. Laminar boundarylayer flow of the air. Negligible absorption of
air by the water film. Equilibrium at the airwater interface.
Find: Evaporation flux at 50% of L = x = 0.0254 m from the leading edge of the plate.
Analysis: First check to see if the boundary layer is laminar. From Perry's Handbook, for air
at 25oC, µ = 0.018 cP = 1.8 x 105 kg/ms. From the ideal gas law,
ρ = PM/RT = (101.3)(29)/(8.314)(298) = 1.186 kg/m3
From Eq. (3127), at x = 0.0254 m,
xu ρ (0.0254)(2)(1186)
.
N Re x = o =
= 3,350
−5
µ
18 × 10
.
Therefore, the boundary layer is laminar, since the criterion is less than 5 x 105.
Since the liquid film is pure water, the only masstransfer resistance for evaporation of water into
air is in the air boundary layer. From Eq. (3105), the molar masstransfer flux of water is,
nH 2 O
= k cx cH 2 O − cH 2 O
(1)
i
o
A
For a laminar boundary layer, the local masstransfer coefficient can be obtained from Eq. (3132), which can be rearranged to give:
k cx = 0.332 DA,B
N Re x
x 1/ 2 N Sc 1/ 3 (2) µ
18 × 10 −5
.
=
= 0.607
ρDA,B (1186) 0.25 × 10 −4
. From Eq. (3101), N Sc = From Eq. (2), k cx = 0.332 0.25 × 10 −4
0.0254 3,350 1/ 2 0.607 1/ 3 = 0.16 m / s At 25oC, vapor pressure of water = 0.46 psia, Therefore, mole fraction of water at the interface =
yH 2 O = 0.46/14.7 = 0.0313. cH 2 O = yH 2 O ρ/Mair = 0.0313(1.186)/29 = 0.0128
i kmol/m3.From Eq. (1), i nH 2O
A i = (0.016)[0.00128 − 0] = 2.05 × 10−5 kmol/sm 2 Exercise 3.29
Subject: Sublimation of a thin, flat plate of naphthalene (A) into air (B) at 100oC (373 K) and 1
atm (101.3 kPa) that is flowing past the plate.
Given: Length of plate, L, is 1m. Reynolds number at the trailing edge of the plate =
N Re L = 5 × 105 = limit of laminar boundarylayer flow.
Find: (a) Average rate of sublimation.
(b) Local rate of sublimation at x = 0.5 m from leading edge of plate.
Analysis: This exercise is similar to Example 3.14, which also deals with the sublimation of
naphthalene into air at the same temperature and pressure. Therefore, the physical properties
used in Example 3.14 apply here. The only masstransfer resistance is in the gas phase.
(a)...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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