

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Material Type: Assignment; Class: Environmental Engineering; Subject: Civil Engineering; University: Lafayette College; Term: Fall 2008;
Typology: Assignments
1 / 3
This page cannot be seen from the preview
Don't miss anything!
Surface loading rate = Q/ASo A s A (^) s = L*W, so L = As^ = Q/SLR = 2000 ms/W = 62.5 m^3 /d / 32 m/d = 62.5 m^2 /4 m = 15.6 m long^2 HRT = Vol/Q = 62.5 m 2 *2.4 m/2000 m 3 /d = 0.075 d = 1.8 hrs
A (^) s = Q/SLR = 2,000,000 gal/d / 800 gal/d-ft 2 = 2500 ft 2 Area = 1/4πd 2 = 2500 ft 2 , so d = 56.4 ft HRT = Vol/Q = (2500 ftSolve for depth, and you find depth 2 *7.48 gal/ft 3 ≥ *depth) / 2,000,000 gal/d must be 8.9 ft. ≥ 0.0833 days
But for settling, we must have a depth of at least 11 ft. So we set the design depth at 11 ft, which will increase the HRT to:HRT = Vol/Q = (2500 ft 2 *7.48 gal/ft 3 *11 ft) / 2,000,000 gal/d = 0.103 d = 2.5 hr
Apply the mass balance equation to the solids: QSince effluent has no suspended solids, C (^) inC (^) in = QeffCeff + Qbot C (^) bot Q (^) inC (^) in = QbotCbot eff=0, and thus Q (^) bot =Q (^) inC (^) in/Cbot = 18,500125/8000 = 289.1 m^3 /day Now apply a mass balance to the flow: QQ (^) in = Qeff + Q (^) bot eff = Qin - Q^ bot = 18,500 - 289 = 18,211 m^3 /day Solids flux = QinC (^) in = QbotCbot = 289 m^3 /day8000 mg/L*1000 L/m^3 1 kg/1,000,000 mg1day/24 hrs = 96.3 kg/day
Batch reaction, non-steady state solution: C = C Plot C = 1200 mg/L e 0 e-kt -2.5t (^) gives a standard exponential decay curve (plot not shown)
Reduction of 75% means the pollutant is reduced to 25% of its original concentration, so C/CSolving the equation for t gives: 0 = 0.
t = -1/kln(C/Ct = -1/kln(C/C (^) 0) = -1/2.5*ln(0.25) = 0.55 days = 13.2 hr
Use design equation for a CSTR, based on steady-state conditions:HRT = 1/k(C in/C^ out - 1) = 1/kfr/(1 - fr) = 1/2.5 day^ -1^ (0.9/0.1) = 3.6 days Vol = HRTQ = 3.6 day0.05 m 3 /sec86,400 sec/day = 15,552 m 3
Need to find design equations for CSTR and PFR, based on steady-state conditions and 0-order rxn. For a PFR, which is equivalent to a batch reactor with t=HRT, we have: VdC/dt = -kV, so dC = -kdt Integrating the left side from CC in to C (^) out , and the right side from 0 to HRT, we get: Solving for HRT:out^ - C^ in^ = -kHRT HRT = (C (^) in - C (^) out )/k For a CSTR, we have: 0 = QC0 = (C in - QCout - kV = Q(C (^) in - C (^) out ) - kV Solving for HRT:in^ - C^ out^ ) - kV/Q = (C^ in^ - C^ out^ ) - kHRT HRT = (C (^) in - C (^) out )/k Since the HRTs are the same for PFR and CSTR, the required volumes will also be the same. This result is because the reaction rate is a constant, independent of concentration. For any reaction orderthe PFR volume will be less than CSTR volume. greater than 0 ,