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Q60

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file: GATE CH 2020 (with Solutions).pdf

pages: 35–37

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Where y is the vertical positon in meters measured from the top of the wall. In addition, use the following data and assumptions • The flow is fully developed • The width of the film is much larger than the thickness of the film, and the dissolved gas concentration is invariant over this width • The solubility of the gas in water, Cₐᵢ, is constant • Pure water enters at y =0 • The evaporation of water is negligible

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Answer

0.03

Solution

y = 0 /text{Film} /text{Thickness} /text{Wall} /delta{} y = L

h (w)

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V (/text{Average} /text{velocity}) = /frac{0.01m}{s}

Width of film = ’w’, height (L) = 1m

C_a_y = C_{ai}(1-e^{-30}^y)

Which means at y=0,Cₐ_y=0

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Convective rate of mass transfer of pure gas into water film = mass transfer rate at y = L – Mass transfer rate at y = 0

KL(CAi-CAy)A = wAy+dy-wA_y

Where, Kₗ is mass transfer coefficient A is Area over the differential block and is given by (wdy). Clearly for the gas, the available area of Cross section (wdy) Now, rate of mass transfer = Flux×Area Talking about area of mass transfer then we can say over the entire thickness of film, mass transfer is taking place which means area of mass transfer is (wδ).

K_l(C_{ai}-C_a_y)(wdy) = (N_a_y_{+}_d_y-N_a_y)(w/delta{})

Now, flux can be defined as Nₐ=Cₐ(uₐ−0) [Basic definition]

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K_l(C_{ai}-C_a_y)

dy = (C_a_y_{+}_d_y-C_a_y)

Can be written as dCₐ

K_ldy = /frac{V/delta{}dC_a}{(CAi-CAy)}

Integrating both sides,]

[VISUAL / REVIEW — NO INVENTED LATEX] Source math clip

K_l = /frac{/delta{}/mathrm{Vl}}{/ln[/frac{A_i}{(C_{ai}-/frac{C}{C_a}(y_1))} = ]]}

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i

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C_{ai}-C_a(y = 1) = e^{-30}⋅C_{ai}

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K_l = /frac{/delta{}/mathrm{Vl}}{/mathrm{Lne}^{30}}

K_l = /frac{(10^{-4} /times 0.01)}{1} /times 30

K_l = /frac{3 /times 10^{-5}m}{sec}

⇒K_l = /frac{0.03mm}{sec}

Hence, the correct answer is 0.03.

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Kₗ⋅L=δV−ln(Cₐᵢ−Cₐ(y)]ᶜ₀ᵃ⁽^y⁼¹⁾

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Source PDF: GATE CH 2020 (with Solutions).pdf

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