Group Hwk #4

# Group Hwk #4 - r(cm 0.8 1.0 1.2 1.4 1.6 1.8 2.0 t(s 0.00171...

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Engineering 10 CEE Module B Group assignment #4 Elysia Liaw, Jin Woo Roh and Yang Xia 1. The light intensity of the lamp g G = ± ²³´ µ = G.· ²³( ¸ ¹ ×².º»× ¼ ½ ) ½ ×(¾×².º») = 0.01326 ¿ ÀÁ ½ = 13.26 Á¿ ÀÁ ½ . 2. Choose Poliovirus I, whose 4-log UV dose is about 60 Á¿ ÀÁ ½ . According to the equation ln(Â/Â0) = −ΛgÃ , the lethality coefficient Λ = 4/60 = 0.0667. 3. Assume the value of the UV absorption coefficient is 80%. 4. Since the radius of the lamp is º · × 2.54 × Ä ² = 0.79375ÅÆ , we assume the radius of the outer casing varies from 0.8cm to 2.0cm. According to g(Ç) = g G È ´ µ ´ Ée ÊË(´Ê´ µ ) , we plot I(r) vs. r and get the following plot. 5. From the plot, we estimate the value of g ÌÍÎ = 6 Á¿ ÀÁ ½ . 6. For different radius for the outer casing, we calculate the exposure time (t) and expected log inactivation. The results are shown in the following table.
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Unformatted text preview: r (cm) 0.8 1.0 1.2 1.4 1.6 1.8 2.0 t (s) 0.00171 0.0638 0.140 0.230 0.333 0.450 0.580 −ΛgÃ − 0.0007 − 0.0255 − 0.0559 − 0.0917 − 0.1331 − 0.1800 − 0.2325 7. From the table above, we can see that no matter what the radius of the outer casing is, the expected log inactivation is less than 1-log. This is because of the fast flow rate of water in the pipe, which makes the exposure time t very low, thus makes −ΛgÃ too small. To fix this problem, without changing our pipe properties and the lamp shape, we will need a UV lamp with a much greater power (about 60W). 2 4 6 8 10 12 14 0.8 1 1.2 1.4 1.6 1.8 2 Light intensity I(r) mW/cm^2 Outer casing radius (cm) I (r)vs. r I vs. r...
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