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

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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 ײ.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) = −Λgà , 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 g 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 g 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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