Monod Equation Plug Flow 1. Chlorine contact basin for disinfection
Where: Q = 0.25 m3/s A = channel cross section between baffles = 18 m 2 Rd = rate of microorganism kill in presence of chlorine = -kdX X = concentration of microorganism at any point in contact reactor Xo = influent concentration of microorganisms = 10 6 E. coli / 100 mL Kd = 5/hr Rc = rate of chlorine decay = -kcX Kc = 10-5 (mg-chlorine/L)/(#/ 100mL)(hr) *2 rate expressions, 2 constituents, 2 coupled mass balances FIND: a. Reactor volume and flow path length, L such that X L < 103 cells/100mL b. Chlorine concentration which must be added to insure that there is detectable chlorine at PFR exit (detection level = CL = 0.05mg/L) SOLUTION: 1. Steady – state mass balance on cells XL = Xo exp (-kd) = (1/kd)ln(Xo/XL) = (1/5)(hr)ln(106/103) = 1.4 hr V = Qt = 0.25m3/s(3600s/hr)(1.4 hr) = 1,260 m3 L = V/A = 1260m/18 = 70m 2. Steady - state mass balance on chlorine
Monod Equation Plug Flow
Monod Equation Plug Flow 2. The growth rate of E. coli be expressed by Monod kinetics with μm = 0.935 hr-1 and KS = 0.71 g/L. Assume that YX/S is 0.6 g dry cells per g substrate. CX0 is 1 g/L and CS0 = 10 g/L when the cells start to grow exponentially (i.e., at t = 0). show how CX, CS, and dCX/dt change with respect to time. SOLUTION: *Using PFR CS is varied from 10 g/L to 0. CX is calculated using as CX = 1 + 0.6 (10 – CS) t is calculated using as follows:
0.935t =(
1+10∗0.6 0.71∗0.6 ¿ 10 ) Cs 0.71∗0.6 Cx ( ) +1 ¿ ln + ¿ ln ¿ 1+10(0.6) 1
( )
(50) dCS / dt = -(-rS) (49) YX/S = rx/-rs (46) CX = CX0 exp[μ(t-t0)]
Monod Equation Plug Flow
3. The enzyme catalyzes the fermentation E substrate A (reactive), obtaining R. Find size required mixed reactor for 95% conversion of a feed stream (25 L / min) reagent (2 mol / L) and enzyme. Fermentation
Monod Equation Plug Flow kinetics at this enzyme concentration is given by A → R CA-rA = 0.1 / (1 + 0.5 CA) Solution Constant density System
τm=
Cao−Caf Cao−Caf = −ra 0.1Caf 1+0.5 Caf
Caf =Cao (1−Xa )=2 ( 1−0.95 )=0.1 τm= V=
(2−0.1)(1+0.5 ( 01 ) ) =199.5 min 0.1(0.1) τmVo=4987.5 L=5 m2
Monod Equation Plug Flow 4. You are given a microorganism that has a maximum growth rate (u m) of 0.39 hr-1. Under ideal conditions (maximum growth rate is achieved), how long will it take to obtain 1 x 10 10 CFU/ml if you begin with an inoculum of 2 x 107 CFU/ml? SOLUTION: ln X = ut + ln X0 ln 1 x 1010 CFU/ml = (0.39 hr-1)(t) + ln 2 x 107 CFU/ml t = 15.9 hr 5. An activated-sludge system is to be used for secondary treatment of 10,000 m3/d of municipal wastewater. After primary clarification, the BOD is 150 mg/L and it is desired to have not more than 5 mg/L of soluble BOD in the effluent. A completely mixed reactor is to be used, and a pilot-plant analysis has established the following kinetic values: Y = 0.5 kg/kg kd = 0.05 d-1 Assuming an MLSS concentration of 3,000 mg/L and an underflow concentration of 10,000 mg/L from the secondary clarifier, determine the following: a) the volume of the reactor b) the mass and volume of solids that must be wasted every day c) the recycle ratio SOLUTION: a) Select
c
= 10 days
1/ c = [Y(S0-S)/( X)] - kd 1/10 = [0.5 kg/kg(0.15-0.005 kg/m3)/ ( x 3.0 kg/m3)] - 0.05 d-1 = V/Q = V/(10,000 m3/d) 0.1d-1 = (0.02417)(10,000 m3/d)/V - 0.05 d-1 0.1d-1 = (241.67 m3/d)/V - 0.05 d-1 b)
c
V = 1,611 m3 = VX/(QWXu)
QWXu = VX/
c
Monod Equation Plug Flow QW= VX/( cXu) QW= [(1,611 m3)(3 kg/m3)]/[(10 d)(10 kg/m3)] QW= 48.3 m3/d c) (Q+Qr)X = (Q+Qr-Qw)Xe + (Qr+Qw)Xu = recycle factor, Qr/Q Assuming solids in the effluent are negligible compared with the influent and underflow: Xe = 0 Qr(Xu-X) = QX - QwXu Qr = (QX - QwXu)/(Xu-X) Qr = [(10,000 m3/d)(3 kg/m3)-483.3 kg/d]/ (10-3kg/m3) = 4,217 m3/d = Qr/Q = 4,217/10,000 = 42% 6. Calculate the time it will take to increase the cell number from 10 4 CFU/ml to 108 CFU/ml assuming a generation time of 1.5 hr. SOLUTION: The generation time is 1.5 hr, therefore determine the specific growth rate, u, from equation 3.6: 2 = eut ln 2 = u(1.5 hr) u = 0.46 hr-1 Plug this into equation 3.4: ln X = ut + ln X0 ln 108 CFU/ml = (0.46 hr-1)(t) + ln 104 CFU/ml t = 20 hr
