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Individual Hwk #4

# Individual Hwk #4 - => = Ppositive Pk 0 1 e Î For test...

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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 Ptotal vs. concentrations. We find λ = V×C for each test tube for each concentration and plug the values into the Poisson equation = - ! Pk λ ke λ k . For test

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tubes 1, 2, 3, 4 and 7, the results are positive;
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Unformatted text preview: = > = - - Ppositive Pk 0 1 e Î» . For test tubes 5 and 6, the results are negative; = = = - Pnegative Pk 0 e Î» . Therefore, Ptotal = . P1P2P3P4P5P6P7 We graph Ptotal 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 106 cells / L, which is 105 cells / 100mL, the disinfection efficiency is N/N0 = 0.0002 = 2 Ã—-10 4 , which is 4-log removal....
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