Individual Hwk #4 - . For test tubes 5 and 6, the results...

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Engineering 10 CEE Module B Yang Xia Individual Assignment #4 20851683 We name the seven test tubes to be 1 to 7 and the test results are as following. Test tube # 1 2 3 4 5 6 7 Volume (100mL) 0.1 0.1 0.1 0.1 0.1 0.01 0.001 Results + + + + - - + Since the concentration C varies from 10 cells / 100mL to 30 cells / 100mL, we pick 10, 12, 14, 16, 18, 20, 22, 24, 26, 28 and 30 cells / 100mL for the concentration in order to graph g G±G²³ vs. concentrations. We find λ = V×C for each test tube for each concentration and plug the values into the Poisson equation g´µ¶ = λ · ¸ ¹º ·! . For test tubes 1, 2, 3, 4 and 7, the results are positive; g´»¼½¾¿¾ÀÁ¶ = g´µ > 0¶ = 1 − Á
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Unformatted text preview: . For test tubes 5 and 6, the results are negative; g = g = 0 = . Therefore, g GG = g g g g g g g . We graph g GG vs. concentrations and get the following chart. From the chart, we can tell that MPN is 0.2 cells / mL, which is 20 cells / 100mL. Comparing this to the undisinfected water bacterial concentration 10 cells / L, which is 10 cells / 100mL, the disinfection efficiency is N/N0 = 0.0002 = 2 10 , which is 4-log removal. 0.0002 0.0004 0.0006 0.0008 0.001 0.0012 0.0014 0.1 0.15 0.2 0.25 0.3 P_total Concentrations (cells/mL) P_total vs. Concentrations P_total vs. Concentrations...
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This note was uploaded on 02/09/2010 for the course ENGIN 10 taught by Professor Johnson during the Fall '08 term at University of California, Berkeley.

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