Lecture 14 - Pathogens Disinfection and Plug Flow Reactors...

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1 Pathogens, Disinfection and Plug Flow Reactors ± Lecture#14 ± plug flow reactors (PFR) ± microorganisms and chlorine disinfection ± ICP#12 ± Reading V: Ch. 5 (pg. 96-98; 108-109); pg. 180- 184 on pathogens; pg. 232, 273 on disinfection What you should have learned today ± What are pathogens? ± How can we remove pathogens from water? ± What are good disinfectants used in water treatment? ± What are the water quality standards for pathogens? What law sets the regulations? ± What is a plug flow reactor? ± How to we set up and solve PFR problems? ICP #11 (CMFR) Q in C in Q out C out J Rn A Volume: 1000 m 3 Area of Floor: 200 m 2 k = 7.6x10 -3 hr -1 J Rn = 1 pCi/m 2 -s Q air = V/10-hr = Q out C air = 0 C out = ? ± Q = V/10-hr = 1000m 3 /10-hr = 100 m 3 /hr ± Vk=(1000m 3 )x(7.6x10 -3 hr -1 ) = 7.6 m 3 /hr ± J Rn A=(1 pCi/m 2 -s)x(200m 2 )x(3600s/hr)= 7.2x10 5 pCi/hr ICP #11 (cont.) ± 0= J Rn A–Q out C out –VkC out ± C out (Q out + Vk) = J Rn A kV Q A J C C out Rn out ss + = = ICP #11 (cont.) ± C ss = 6.7 pCi/L ± action level of 4 pCi/L so ventilation may help alleviate the problem part b L pCi 7 . 6 L m 0 x1 m pCi x10 7 . 6 hr m 6 . 7 hr m 100 hr pCi x10 2 . 7 kV Q A J C C 3 3 3 3 3 3 5 out Rn out ss = = + + = + = = ICP #11 ± (b) If the ventilation rate is doubled, how long to reach 99% of the steady state concentration ± double ventilation rate Q out = 200 m 3 /hr ± new C ss = 3.47 pCi/L (see calculation below) ± 99% of steady-state value: ± 3.47/0.99 = 3.5 pCi/L L pCi 47 . 3 L m 0 x1 m pCi 10 x 47 . 3 hr m 6 . 7 hr m 200 hr pCi x10 2 . 7 kV Q A J C C 3 3 3 3 3 3 5 out Rn out ss = = + + = + = =
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2 ICP #11 ± C out = C ss + (C 0 –C ss )(exp(-t/ τ )) ± C out = C ss /0.99 = 3.5 pCi/m 3 ± C ss = 3.47x10 3 pCi/m 3 ± C 0 = 6.7x10 3 pCi/m 3 ± τ = V/(Q out + kV) = 1000/(200 + 7.6) = 4.8 hr ± t = 22.5 hr or ~4.7 τ C out = ( kV
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Lecture 14 - Pathogens Disinfection and Plug Flow Reactors...

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