Splash Distribution in Oxygen Steelmaking

Abstract

The study of splashing is important in understanding oxygen steelmaking. Splashing creates large interfacial area between reacting surfaces and thereby directly affects the kinetics of steelmaking reactions. In the present cold modeling work, a study of splashing has been carried out for various lance heights and gas flow rates. Sampling of droplets has been done in both radial positions and vertical positions across the bath to investigate the effect of sampling positions in estimation of the droplet generation rate. A novel approach has been developed to estimate the droplet generation rate. The results of the study have been compared with previous investigations. Results show that positioning of sampling is a critical issue and can affect the estimation of droplets present in the emulsion. This study also demonstrates quantitatively how much the droplet generation rate is reduced when the cavity mode changes from splashing to penetrating for different Blowing numbers.

Appendices

Appendix A

Blowing Number Sample Calculation

Flow rate, Q (L/min) = 80;

Lance height, L (m) = 0.170;

Density of air at 273 K (0 °C) and 101325 Pa, ρ _g (kg/m³) = 1.29;

Nozzle diameter, d (m) = 0.003;

Density of water at 273 K (0 °C) and 101325 Pa, ρ _l (kg/m³) = 999.84;

Radius of the nozzle, r (m) = (d/2) = (0.003/2) = 0.0015;

Velocity at nozzle exit, $U_{0} = \frac{Q}{{\frac{\pi }{4} d^{2} }} = \frac{{\frac{80}{1000 \times 60}}}{{\frac{\pi }{4} \times (.003)^{2} }} = 188.63$ m/s

Impact velocity at bath surface[12]

$$ U_{j} = U_{0} \frac{0.97}{{\frac{aL}{r} + 0.29}} $$

$$ {\text{or}}, U_{\text{j}} = 188.63 \times \frac{0.97}{{\frac{0.07 \times 0.170}{0.0015} + 0.29}} = 22.25\;{\text{m/s}} $$

(Subagyo et al.[18] used the value of a = 0.07 for cold model experiments, 0.0382 for hot model data with no gas–metal reaction and 0.0393 for plant data with gas–metal reaction)

Critical gas velocity, $U_{g} = \eta U_{j} = 0.44721 \times 22.25 = 9.95$ ($\eta = {\text{constant}} = 0.44721$)

Blowing Number, N _B

$$ N_{\text{B}} = \frac{{\rho_{g} U_{g}^{2} }}{{2\sqrt {\sigma \rho_{l} g} }} = \frac{{1.29 \times (9.95)^{2} }}{{2 \times \sqrt {0.07 \times 9.81 \times 999.84} }} = 2.44 $$

Droplet Generation Rate Sample Calculation

At lance height, L = 0.170 m, flow rate 80 L/min

Droplet generation rate calculation for cylinder 3,

Amount of droplet collected (M5 to M7), W _sp = 0.056 × 10⁻³ kg

Sampling time, t _s = 10 seconds

No of sample pots, N _sp = 3

Volume of the sample pots, V _sp = π × (0.0065) ^2 × 0.014 = 0.000001858 m³

Volume of the third cylinder, V _b = π × (0.09^2 − 0.06^2) × 0.303 = 0.00428436 m³

$$ {\text{Droplet generation rate}}, R_{\text{B}} = \frac{{(\mathop \sum \nolimits Wsp )V_{\text{b}} }}{{t_{\text{s}} V_{\text{sp}} N_{\text{sp}} }} = \frac{{0.056 \times 10^{ - 3} \times 0.00428436}}{10 \times 0.000001858 \times 3} = 4.30 \times 10^{ - 3} {\text{kg/s}} $$

Appendix B

Error in Measurement of Weight of Droplets

At a particular operating condition and sampling place, droplets were collected twice and an average was taken. In estimation of the error, the variation of the weight of droplets collected with the number of experiments was analyzed.

For flow rate 80 L/min, lance height 0.160 m

Standard deviation of droplets weight (for 20 experiments), S ≈ σ = 0.083 × 10⁻³ kg

Confidence level 95 pct (level of significance, α = .05)

z = 1.96 (at 1 − α/2 = 1 − 0.05/2 = 0.9750)

Margin of error, ${\text{ME}} = \sqrt {\frac{{z^{2} \sigma^{2} }}{n}} = \sqrt {\frac{{1.96^{2} \times 0.000083^{2} }}{20}} = 0.04 \times 10^{ - 3 } {\text{kg}}$

It means that we are 95 pct confident that error in estimating the mean of droplet weight is no more than ±0.04 × 10⁻³ kg.

Error Due to Sampling Time Variation

Sampling time has varied from 3 to 150 seconds in different operating conditions. In this error estimation, it has been investigated how weight of droplets was affected by the sampling time variation.

For sample pot 2, L = 0.160 m, flow rate 80 L/min

Weight of droplets (sampling time 10 seconds) = 0.3332 × 10⁻³ kg

Weight of droplets (sampling time 5 seconds) = 0.18248 × 10⁻³ kg

Weight of droplets for $10{\text{ s }} = \, \left( {0. 1 8 2 4 8 { } \times { 1}0^{ - 3} / 5} \right) \, \times 10 \, = \, 0. 3 6 4 9 6 { } \times { 1}0^{ - 3 } {\text{kg}}$

Error for sample time variation (between sampling time 5 and 10 seconds) $= \, \left[ {\left( {0. 3 6 4 9 6- 0. 3 3 3 2} \right)/0. 3 3 3 2} \right] \times { 1}00 {\text{pct }} = { 9}. 5 3 {\text{pct}}$

Weight of droplets (sampling time 20 seconds) = 0.73966 × 10⁻³ kg

Weight of droplets for $10{\text{ s }} = \, \left( {0. 7 3 9 6 6/ 20} \right) \, \times { 1}0 \, = \, 0. 3 6 9 8 3 { } \times { 1}0^{ - 3 } {\text{kg}}$

Error for sample time variation (between sampling time 20 and 10 seconds)

$$ = \left[ {\left( {0. 3 6 9 8 3- 0. 3 3 3 2} \right)/0. 3 3 3 2} \right] \, \times { 1}00 {\text{pct }} = { 1}0. 9 9 {\text{pct}} $$