Reactor Models

Mixed Batch Reactors

Mass Balance

Rate Accumulated = 0 Rate In Rate Out 0 + Rate Generated Rate Accumulated = Rate Generated Rate Generated =

V (dC/dt) = r V Units: V(dC/dt) L [(mg/L)/s] = mg/s

Rate Accumulated = Rate Generated Or: V dC/dt = r V dC/dt = r Substitute appropriate model for r: Zero order (r = k) First order (r = kC) Second order (r = kC 2 ) Etc.

dC/dt = r Zero Order dC/dt = k (positive k – production) Separate variables and integrate: C t C 0 dC = k dt 0 C - C 0 = k t Negative k (consumption) C - C 0 = - k t

First order r = k C (production) dC/dt = k C C 0 C t dC/C = k dt 0 or: ln (C/C 0 ) = kt C = C 0 e kt r = - k C (consumption) ln(C/C 0 ) = - k t or: ln(C 0 /C) = kt therefore: C = C 0 e -kt

Example

Toxic organic compounds can be removed from batches of wastewater using activated carbon adsorption. In this process activated carbon is added to a batch of waste and mixed until the appropriate solution concentration is reached. Suppose laboratory tests have shown that the removal of methyl-ethyl-ketone (MEK) from a wastewater in a batch process can be modeled using first order kinetics with a reaction rate constant of –0.08 hr -1 . If a waste has 10 mg/L of MEK, how long will it take to reduce the concentration to 0.01 mg/l?

ln(C 0 /C) = k t ln(10/0.01) = 0.08 hr -1 t t = [ln(10/.01)]/0.08 = 86.3 hr

Plug Flow Reactors

Remember in a plug flow reactor, a slug of material is flowing through the reactor without mixing. In fact this slug may be thought of as a small mixed batch reactor, and thus the model of a mixed batch reactor applies to a plug flow reactor where the t in the equation is the hydraulic residence time of the reactor.

Zero Order C = C 0 + k t Where: t = V/Q and: C + C 0 - k t (production)

First Order ln(C/C 0 ) = k t (production) So: and: ln(C/C 0 ) = k (V/Q) V = (Q/k) ln(C/C 0 ) by the same method: V = -(Q/k) ln(C/C 0 ) or: V = (Q/k) ln(C 0 /C)

Example

Wastewater from the coking process at Geneva Steel Mill has a lot of ammonia. A treatment process is used to remove the ammonia, but it is not completely removed. The treated waste is discharged to a ditch that runs through the plant property and through a pond before it flows to Utah Lake. Ammonia removal by microorganisms in this system may be modeled as a first order reaction with a reaction rate constant of 0.5 day -1 . Suppose the channel is 2,500 feet long, 10 feet wide and 2.5 feet deep, and the pond is 250 feet long, 100 feet wide and 4 feet deep (dikes are constructed in the pond so that flow through it approximates plug flow). What is the maximum flow rate which may be treated by the system if the ammonia concentration from the treatment process is 30 mg/L and the water flowing to Utah Lake can have no more than 2 mg ammonia/L?

2,500’ x 10’ x 2.5 ‘ 250’ x 100’ x 4’

V = (Q/k) ln(C 0 /C) V = (2,500 x 10 x 2.5) + (250 x 100 x 4) = 162,500 ft 3 162,500 ft 3 x 7.48 gal/ft 3 = 1,215,500 gallons 1,215,500 = (Q/0.5 day -1 ) (ln(30/2)) Q = 224,400 gallons/day

CSTR’s

Rate Accumulated = Rate In Rate Out + Rate Generated

V(dC/dt) = Q C 0 - Q C + r V To solve this we can substitute the appropriate reaction rate equation and then solve the differential equation.

Generally, the differential equation can not be solved analytically. The equation could be solved by numerical methods or we can make an assumption to allow the solution.

Traditionally the solution is found by making the steady-state assumption.

0 = Q C 0 - Q C + r V

Zero Order

0 = Q C 0 - Q C + k V C = C 0 + k (V/Q) C = C 0 + k t C = C 0 - k t (production) (consumption) This is the same as for plug flow

First Order

0 = Q C 0 - Q C + k C V C 0 /C = 1 – k (V/Q) C = (Q C 0 )/(Q – kV) (production) C 0 /C = 1 + k(V/Q) C = (Q C 0 )/(Q + kV) or: V = (Q/k)[(C 0 /C) – 1] (consumption)

C = (Q C 0 )/(Q + kV) C = C 0 /(1 + k t ) (divide by V) In environmental engineering our objective is usually to make C small. As you can see from this equation there are several ways to do this: Increase k k t = k 0 e

M

(T – To) Increase V (remember t = V/Q) Decrease Q (again remember t = V/Q)

Example

CTSR’s are frequently used in wastewater treatment processes for industrial wastes. Normally these processes are designed with the objective of removing biodegradable organics from a waste stream by microbial decomposition, but other reactions may occur in the reactors. For example, volatile organics may be stripped from the liquid as it is aerated. Suppose a waste water contains 10 mg/L tri-chloroethylene (a solvent that is used to degrease parts), and that it is stripped in a first order reaction that has a reaction rate constant of 0.25 hr -1 . The treatment process utilizes a 500,000 gallon aeration basin and treats a flow of 2 mgd. How much TCE is put into the atmosphere in one day by this process?

C = C 0 /(1 + k t ) t = V/Q = 500,000 gal/2x10 6 gal/day = 0.25 day C = 10/(1 + 0.25 (6)) = 4 mg/L

W A Q, C 0 Mass Balance: Q, C 0 = Q C 0 - Q C - W A W A = [(2x10 6 )3.7854] 10 – [(2x10 6 )3.7854] 4 = 45.42x10

6 mg/day = 45 Kg/day

CSTR’s in Series

Consider two CSTR’s in series, first order reaction: By extension: Or: From the previous discussion, a mass balance on the first reactor: C 0 /C = 1 + k(V 0 /Q) Second reactor: C 1 /C 2 = 1 + k(V 0 /Q) V 0 is the volume of a individual reactor (V = n V 0 ) So: (C 0 /C)*(C 1 /C 2 ) = C 0 /C 2 = [1 + k(V 0 /Q)] 2 C 0 /C n = [1+ k(V 0 /Q)] n (C 0 /C n ) 1/n = 1 + k[(V/(n Q)]

CSTR’s

Reactor Performance

Reactor Performance

Consider a first-order reaction in a system that requires 90% removal: C 0 /C = 1/.1 = 10 V CSTR = Q/k [(C 0 /C) – 1] = Q/k (10 –1) = Q/k (9) V PFR = Q/k [ln(C 0 /C)] = Q/k ln 10 = Q/k 2.303

V CSTR /V PFR = 9/2.303 = 3.908

By this method we can see that the volume required for a CSTR will always be greater than the volume required by a PFR as long as the reaction order is greater than 0.