A large deep bath contains molten steel, the surface of which is in contact with air. The oxygen concentration in the bulk of the molten steel is 0.03% by mass and the rate of transfer of oxygen from the air is sufficiently high to maintain the surface layers saturated at a concentration of 0.16%

Question

A large deep bath contains molten steel, the surface of which is in contact with air. The oxygen concentration in the bulk of the molten steel is 0.03% by mass and the rate of transfer of oxygen from the air is sufficiently high to maintain the surface layers saturated at a concentration of 0.16% by weight. The surface of the liquid is disrupted by gas bubbles rising to the surface at a frequency of 120 bubbles per m² of surface per second, each bubble disrupts and mixes about 15 cm² of the surface layer into the bulk.
On the assumption that the oxygen transfer can be represented by a surface renewal model, obtain the appropriate equation for mass transfer by starting with Fick’s second law of diffusion and calculate:

(a) The mass transfer coefficient
(b) The mean mass flux of oxygen at the surface
(c) The corresponding film thickness for a film model, giving the same mass transfer rate.

Diffusivity of oxygen in steel = 1.2 × [latex]10^{-8}[/latex] m²/s. Density of molten steel = 7100 kg/m³ .

Accepted Answer

If C′ is defined as the concentration above a uniform datum value:

[latex]\frac{∂C^{\prime}}{∂t} =D\frac{∂^{2}C^{\prime}}{∂y^{2}}[/latex] (equation 10.100)

The boundary conditions are:
when [latex]\begin{matrix} t = 0, & 0 0 & y = 0 & C^{\prime} = C_{i}^{\prime} \ t > 0 &y = \infty & C^{\prime}=0 \end{matrix} [/latex]

The equation is most conveniently solved using Laplace transforms. The Laplace transform [latex]\overline{C^{\prime}}[/latex] of C′ is:

[latex]\overline{C^{\prime}} =\int_{0}^{\infty }e^{-pt}C^{\prime} dt[/latex] (equation 10.101)

Then: [latex]\frac{∂\overline{C^{\prime}}}{∂t} =\int_{0}^{\infty }e^{-pt}\frac{∂C^{\prime}}{∂t} dt[/latex] (equation 10.102)

[latex]=[e^{-pt}C^{\prime}]^{\infty }_{0}+p\int_{0}^{\infty }e^{-pt}C^{\prime}dt=p\overline{C^{\prime}}[/latex] (equation 10.103)

Since the Laplace transform operation is independent of y,

[latex]\frac{\overline{∂^{2}C^{\prime}} }{∂y^{2}} =\frac{\overline{∂^{2}C^{\prime}}}{ay^{2} }[/latex] (equation 10.104)

Taking Laplace transforms of both sides of equation 10.100:

[latex]p \overline{C^{\prime}}=D\frac{∂^{2}\overline{C^{\prime}} }{∂y^{2}}[/latex]

[latex]\frac{∂^{2}\overline{C^{\prime}} }{∂y^{2}}-\frac{p}{D} \overline{C^{\prime}}=0[/latex]

From which: [latex]\overline{C^{\prime}}=Ae^{\sqrt{(p/D)}y }+Be^{-\sqrt{(p/D)}y}[/latex] (equation 10.105)

When y = ∞, [latex]\overline{C^{\prime}}[/latex] = 0 and A = 0
When y = 0, [latex]\overline{C^{\prime}}[/latex] = [latex]C^{\prime}_{i}[/latex]/p and B = [latex]C^{\prime}_{i}[/latex]/p

[latex]\overline{C^{\prime}}=\frac{C^{\prime}_{i}}{p} e^{-\sqrt{(p/D)}y}[/latex]

[latex]\frac{d\overline{C^{\prime}}}{dy} =-\frac{C^{\prime}_{i}}{\sqrt{pD} } e^{-\sqrt{(p/D)}y}[/latex]

Inverting: -[latex]\frac{∂C^{\prime}}{∂y} =\frac {C^{\prime}_{i}}{\sqrt{D} } imes \frac{1}{\sqrt{\pi t} } e^{-y^{2}/4Dt}[/latex] (See Volume 1, Appendix Table 12)

The mass transfer rate at the surface, [latex](N_{A})_{t}=-D\left(\frac{∂C^{\prime}}{∂y} ight) _{y=0}=C^{\prime}_{i}\sqrt{\frac{D}{\pi t} } [/latex] at time t
The average rate of mass transfer in time t:

[latex]\frac{1}{t} \int_{0}^{t}C^{\prime}_{i}\sqrt{\frac{D}{\pi t} } dt=2C^{\prime}_{i}\sqrt{\frac{D}{\pi t} }[/latex]

Taking 1 m² of surface, the area disrupted by the bubbles per second is:
120 × 15/10000 = 0.18/s
∴ Average surface age duration = (1/0.18) = 5.55 s

[latex]C^{\prime}_{i}[/latex] = (0.16 - 0.03)/100 = 0.0013 kg O₂/kg steel = (0.0013/32) × 7100 = 0.2885 kmol/m³

Then:
(a) The mass transfer coefficient = [latex]2\sqrt{\frac{D}{\pi t} }[/latex] = 2(1.2 × [latex]10^{-8} /π × 5.55)^{0.5}[/latex] = 5.25 × [latex]10^{-5}[/latex] m/s

(b) The mean rate of transfer, [latex]N_{A}= 2 C_{i}^{\prime}(D/πt)^{0.5}[/latex] = 2 × 0.2885(1.2 × [latex]10^{-8}/π × 5.55)^{0.5}[/latex] = 1.51 × [latex]10^{-5}[/latex] kmol/m² s

(c) The film thickness L is given by: [latex]N_{A}= (D/L)\overline{C^{\prime}_{i}}[/latex]

and: L = (1.2 × [latex]10^{-8}[/latex] × 0.2885)/(1.51 × [latex]10^{-5}[/latex]) = 2.29 × [latex]10^{-4}[/latex] m = 0.23 mm

Question 10.24: A large deep bath contains molten steel, the surface of whic...

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