FT_PPSS05-01

FT_PPSS05-01 - Solutions to Practice Problems No. 1 2. Qr =...

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Solutions to Practice Problems No. 1 2. Qr = 10 m3/s Qw = 20 L/s Cw = 200 mg/L A xs = 15 m2 [but note that this is not needed until the next problem] At steady state, change in storage = 0, so m’ r,mix = m’ r,pure + m’ w (QC) r,mix = (QC) r,pure + (QC) w (10 + 0.02 m3/s) C r,mix = (10 m3/s)(0 mg/L) + (20 mg/L)(200 L/s) C r,mix = [ (0) + (4000 mg/s)] /(10.02 m3/s) = 400 mg/m3 = 0.4 mg/L = 400 ug/L This is 8x higher than the drinking water std. 3. Without decay the stream at steady state has a uniform concentration of 400 ug/L everywhere. With decay, the conc drops over time, hence over space downstream. First order decay occurs over time, so need to know how much time elapses from discharge to the target point 12 km downstream. t 12km = L / v = L/(Q/A) = LA/Q = (12,000 m)(15 m2) / (10 m3/s) = 18,000 s = 5 h C o = 400 ug (the conc without decay) C 12km = C o e -kt = 400 mg/L [ exp-(0.14 h-1)(5 h) ] = 400(0.5) = 200 ug/L So the conc is only half as much, although still well above the standard. 4.
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FT_PPSS05-01 - Solutions to Practice Problems No. 1 2. Qr =...

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